Xref: bloom-picayune.mit.edu rec.puzzles:18136 news.answers:3068 Newsgroups: rec.puzzles,news.answers Path: bloom-picayune.mit.edu]snorkelwacker.mit.edu]usc]wupost]uunet]questrel]chris From: uunet]questrel]chris (Chris Cole) Subject: rec.puzzles FAQ, part 1 of 15 Message-ID: Followup-To: rec.puzzles Summary: This posting contains a list of Frequently Asked Questions (and their answers). It should be read by anyone who wishes to post to the rec.puzzles newsgroup. Sender: chris@questrel.com (Chris Cole) Reply-To: uunet]questrel]faql-comment Organization: Questrel, Inc. Date: Mon, 21 Sep 1992 00:08:26 GMT Approved: news-answers-request@MIT.Edu Expires: Sat, 3 Apr 1993 00:08:21 GMT Lines: 1557 Archive-name: puzzles-faq/part01 Last-modified: 1992/09/20 Version: 3 Instructions for Accessing rec.puzzles Frequently Asked Questions List INTRODUCTION Below is a list of puzzles, categorized by subject area. Each puzzle includes a solution, compiled from various sources, which is supposed to be definitive. EMAIL To request a puzzle, send a letter to uunet]questrel]faql-request containing one or more lines of the form: send For example, to request decision/allais.p, send the line: send decision/allais.p or just: send allais The puzzle will be mailed via return email to the address in your request's "From:" line. If you are unsure of this address, and cannot edit this line, then include in your message BEFORE the first "send" line the line: return_address FTP The FAQL has been posted to news.answers. News.answers is archived in the periodic posting archive on pit-manager.mit.edu Ý18.172.1.27¨. Postings are located in the anonymous ftp directory /pub/usenet/news.answers, and are archived by "Archive-name". Other subdirectories of /pub/usenet contain periodic postings that may not appear in news.answers. Other news.answers/FAQ archives (which carry some or all of the FAQs in the pit-manager archive) are: archive.cs.ruu.nl Ý131.211.80.5¨ in the anonymous ftp directory /pub/NEWS.ANSWERS (also accessible via mail server requests to mail-server@cs.ruu.nl) cnam.cnam.fr Ý192.33.159.6¨ in the anonymous ftp directory /pub/FAQ ftp.uu.net Ý137.39.1.9 or 192.48.96.9¨ in the anonymous ftp directory /usenet ftp.win.tue.nl Ý131.155.70.100¨ in the anonymous ftp directory /pub/usenet/news.answers grasp1.univ-lyon1.fr Ý134.214.100.25¨ in the anonymous ftp directory /pub/faq (also accessible via mail server requests to listserv@grasp1.univ-lyon1.fr), which is best used by EASInet sites and sites in France that do not have better connectivity to cnam.cnam.fr (e.g. Lyon, Grenoble) Note that the periodic posting archives on pit-manager.mit.edu are also accessible via Prospero and WAIS (the database name is "usenet" on port 210). CREDIT The FAQL is NOT the original work of the editor (just in case you were wondering :^). In keeping with the general net practice on FAQL's, I do not as a rule assign credit for FAQL solutions. There are many reasons for this: 1. The FAQL is about the answers to the questions, not about assigning credit. 2. Many people, in providing free answers to the net, do not have the time to cite their sources. 3. I cut and paste freely from several people's solutions in most cases to come up with as complete an answer as possible. 4. I use sources other than postings. 5. I am neither qualified nor motivated to assign credit. However, I do whenever possible put bibliographies in FAQL entries, and I see the inclusion of the net addresses of interested parties as a logical extension of this practice. In particular, if you wrote a program to solve a problem and posted the source code of the program, you are presumed to be interested in corresponding with others about the problem. So, please let me know the entries you would like to be listed in and I will be happy to oblige. Address corrections or comments to uunet]questrel]faql-comment. INDEX ==> analysis/bugs.p <== Four bugs are placed at the corners of a square. Each bug walks directly toward the next bug in the clockwise direction. The bugs walk with constant speed always directly toward their clockwise neighbor. Assuming the bugs make at least one full circuit around the center of the square ==> analysis/c.infinity.p <== What function is zero at zero, strictly positive elsewhere, infinitely differentiable at zero and has all zero derivitives at zero? ==> analysis/cache.p <== Cache and Ferry (How far can a truck go in a desert?) A pick-up truck is in the desert beside N 50-gallon gas drums, all full. The truck's gas tank holds 10 gallons and is empty. The truck can carry one drum, whether full or empty, in its bed. It gets 10 miles to the gallon. ==> analysis/cats.and.rats.p <== If 6 cats can kill 6 rats in 6 minutes, how many cats does it take to kill one rat in one minute? ==> analysis/e.and.pi.p <== Which is greater, e^(pi) or (pi)^e ? ==> analysis/functional/distributed.p <== Find all f: R -> R, f not identically zero, such that (*) f( (x+y)/(x-y) ) = ( f(x)+f(y) )/( f(x)-f(y) ). ==> analysis/functional/linear.p <== Suppose f is non-decreasing with f(x+y) = f(x) + f(y) + C for all real x, y. Prove: there is a constant A such that f(x) = Ax - C for all x. (Note: continuity of f is not assumed in advance.) ==> analysis/integral.p <== If f is integrable on (0,inf), and differentiable at 0, and a > 0, show: inf ( f(x) - f(ax) ) ==> analysis/period.p <== What is the least possible integral period of the sum of functions of periods 3 and 6? ==> analysis/rubberband.p <== A bug walks down a rubberband which is attached to a wall at one end and a car moving away from the wall at the other end. The car is moving at 1 m/sec while the bug is only moving at 1 cm/sec. Assuming the rubberband is uniformly and infinitely elastic, will the bug ever reach the car? ==> analysis/series.p <== Show that in the series: x, 2x, 3x, .... (n-1)x (x can be any real number) there is at least one number which is within 1/n of an integer. ==> analysis/snow.p <== Snow starts falling before noon on a cold December day. At noon a snowplow starts plowing a street. It travels 1 mile in the first hour, and 1/2 mile in the second hour. What time did the snow start falling?? ==> analysis/tower.p <== A number is raised to its own power. The same number is then raised to the power of this result. The same number is then raised to the power of this second result. This process is continued forever. What is the maximum number which will yield a finite result from this process? ==> arithmetic/7-11.p <== A customer at a 7-11 store selected four items to buy, and was told that the cost was $7.11. He was curious that the cost was the same as the store name, so he inquired as to how the figure was derived. The clerk said that he had simply multiplied the prices of the four ==> arithmetic/clock/day.of.week.p <== It's restful sitting in Tom's cosy den, talking quietly and sipping a glass of his Madeira. I was there one Sunday and we had the usual business of his clock. ==> arithmetic/clock/thirds.p <== Do the 3 hands on a clock ever divide the face of the clock into 3 equal segments, i.e. 120 degrees between each hand? ==> arithmetic/consecutive.product.p <== Prove that the product of three or more consecutive natural numbers cannot be a perfect square. ==> arithmetic/consecutive.sums.p <== Find all series of consecutive positive integers whose sum is exactly 10,000. ==> arithmetic/digits/all.ones.p <== Prove that some multiple of any integer ending in 3 contains all 1s. ==> arithmetic/digits/arabian.p <== What is the Arabian Nights factorial, the number x such that x] has 1001 digits? How about the prime x such that x] has exactly 1001 zeroes on the tail end. (Bonus question, what is the 'rightmost' non-zero digit in x]?) ==> arithmetic/digits/circular.p <== What 6 digit number, with 6 different digits, when multiplied by all integers up to 6, circulates its digits through all 6 possible positions, as follows: ABCDEF * 1 = ABCDEF ABCDEF * 3 = BCDEFA ==> arithmetic/digits/divisible.p <== Find the least number using 0-9 exactly once that is evenly divisible by each of these digits? ==> arithmetic/digits/equations/123456789.p <== In how many ways can "." be replaced with "+", "-", or "" (concatenate) in .1.2.3.4.5.6.7.8.9=1 to form a correct equation? ==> arithmetic/digits/equations/1992.p <== 1 = -1+9-9+2. Extend this list to 2 - 100 on the left side of the equals sign. ==> arithmetic/digits/equations/383.p <== Make 383 out of 1,2,25,50,75,100 using +,-,*,/. ==> arithmetic/digits/extreme.products.p <== What are the extremal products of three three-digit numbers using digits 1-9? ==> arithmetic/digits/googol.p <== What digits does googol] start with? ==> arithmetic/digits/labels.p <== You have an arbitrary number of model kits (which you assemble for fun and profit). Each kit comes with twenty (20) stickers, two of which are labeled "0", two are labeled "1", ..., two are labeled "9". You decide to stick a serial number on each model you assemble starting ==> arithmetic/digits/nine.digits.p <== Form a number using 0-9 once with its first n digits divisible by n. ==> arithmetic/digits/palindrome.p <== Does the series formed by adding a number to its reversal always end in a palindrome? ==> arithmetic/digits/palintiples.p <== Find all numbers that are multiples of their reversals. ==> arithmetic/digits/power.two.p <== Prove that for any 9-digit number (base 10) there is an integral power of 2 whose first 9 digits are that number. ==> arithmetic/digits/prime/101.p <== How many primes are in the sequence 101, 10101, 1010101, ...? ==> arithmetic/digits/prime/all.prefix.p <== What is the longest prime whose every proper prefix is a prime? ==> arithmetic/digits/prime/change.one.p <== What is the smallest number that cannot be made prime by changing a single digit? Are there infinitely many such numbers? ==> arithmetic/digits/prime/prefix.one.p <== 2 is prime, but 12, 22, ..., 92 are not. Similarly, 5 is prime whereas 15, 25, ..., 95 are not. What is the next prime number which is composite when any digit is prefixed? ==> arithmetic/digits/reverse.p <== Is there an integer that has its digits reversed after dividing it by 2? ==> arithmetic/digits/rotate.p <== Find integers where multiplying them by single digits rotates their digits. ==> arithmetic/digits/sesqui.p <== Find the least number where moving the first digit to the end multiplies by 1.5. ==> arithmetic/digits/squares/leading.7.to.8.p <== What is the smallest square with leading digit 7 which remains a square when leading 7 is replaced by an 8? ==> arithmetic/digits/squares/length.22.p <== Is it possible to form two numbers A and B from 22 digits such that A = B^2? Of course, leading digits must be non-zero. ==> arithmetic/digits/squares/length.9.p <== Is it possible to make a number and its square, using the digits from 1 through 9 exactly once? ==> arithmetic/digits/squares/three.digits.p <== What squares consist entirely of three digits (e.g., 1, 4, and 9)? ==> arithmetic/digits/squares/twin.p <== Let a twin be a number formed by writing the same number twice, for instance, 81708170 or 132132. What is the smallest square twin? ==> arithmetic/digits/sum.of.digits.p <== Find sod ( sod ( sod (4444 ^ 4444 ) ) ). ==> arithmetic/digits/zeros/factorial.p <== How many zeros are in the decimal expansion of n]? ==> arithmetic/digits/zeros/lsd.factorial.p <== What is the least significant non-zero digit in the decimal expansion of n]? ==> arithmetic/digits/zeros/million.p <== How many zeros occur in the numbers from 1 to 1,000,000? ==> arithmetic/magic.squares.p <== Are there large squares, containing only consecutive integers, all of whose rows, columns and diagonals have the same sum? How about cubes? ==> arithmetic/pell.p <== Find integer solutions to x^2 - 92y^2 = 1. ==> arithmetic/prime/arithmetic.progression.p <== Is there an arithmetic progression of 20 or more primes? ==> arithmetic/prime/consecutive.composites.p <== Are there 10,000 consecutive non-prime numbers? ==> arithmetic/sequence.p <== Prove that all sets of n integers contain a subset whose sum is divisible by n. ==> arithmetic/sum.of.cubes.p <== Find two fractions whose cubes total 6. ==> arithmetic/tests.for.divisibility/eleven.p <== What is the test to see if a number is divisible by eleven? ==> arithmetic/tests.for.divisibility/nine.p <== What is the test to see if a number is divisible by nine? ==> arithmetic/tests.for.divisibility/seven.p <== What is the test to see if a number is divisible by 7? ==> arithmetic/tests.for.divisibility/three.p <== Prove that if a number is divisible by 3, the sum of its digits is likewise. ==> combinatorics/coinage/combinations.p <== How many ways are there to make change for a dollar? Count combinations of coins, not permuations. ==> combinatorics/coinage/dimes.p <== "Dad wants one-cent, two-cent, three-cent, five-cent, and ten-cent stamps. He said to get four each of two sorts and three each of the others, but I've forgotten which. He gave me exactly enough to buy them; just these dimes." How many stamps of each type does Dad want? ==> combinatorics/coinage/impossible.p <== What is the smallest number of coins that you can't make a dollar with? I.e., for what N does there not exist a set of N coins adding up to a dollar? It is possible to make a dollar with 1 current U.S. coin (a Susan B. Anthony), 2 coins (2 fifty cent pieces), 3 coins (2 quarters and a fifty cent piece), ==> combinatorics/color.p <== An urn contains n balls of different colors. Randomly select a pair, repaint the first to match the second, and replace the pair in the urn. What is the expected time until the balls are all the same color? ==> combinatorics/full.p <== Consider a string that contains all substrings of length n. For example, for binary strings with n=2, a shortest string is 00110 -- it contains 00, 01, 10 and 11 as substrings. Find the shortest such strings for all n. ==> combinatorics/gossip.p <== n people each know a different piece of gossip. They can telephone each other and exchange all the information they know (so that after the call they both know anything that either of them knew before the call). What is the smallest number of calls needed so that everyone knows everything? ==> combinatorics/grid.dissection.p <== How many (possibly overlapping) squares are in an mxn grid? ==> combinatorics/subsets.p <== Out of the set of integers 1,...,100 you are given ten different integers. From this set, A, of ten integers you can always find two disjoint subsets, S & T, such that the sum of elements in S equals the sum of elements in T. Note: S union T need not be all ten elements of ==> cryptology/Beale.p <== What are the Beale ciphers? ==> cryptology/Feynman.p <== What are the Feynman ciphers? ==> cryptology/Voynich.p <== What are the Voynich ciphers? ==> cryptology/swiss.colony.p <== What are the 1987 Swiss Colony ciphers? ==> decision/allais.p <== The Allais Paradox involves the choice between two alternatives: A. 89% chance of an unknown amount 10% chance of $1 million ==> decision/division.p <== N-Person Fair Division If two people want to divide a pie but do not trust each other, they can still ensure that each gets a fair share by using the technique that one ==> decision/dowry.p <== Sultan's Dowry A sultan has granted a commoner a chance to marry one of his hundred daughters. The commoner will be presented the daughters one at a time. ==> decision/envelope.p <== Someone has prepared two envelopes containing money. One contains twice as much money as the other. You have decided to pick one envelope, but then the following argument occurs to you: Suppose my chosen envelope contains $X, then the other envelope either contains $X/2 or $2X. Both cases are ==> decision/exchange.p <== At one time, the Mexican and American dollars were devalued by 10 cents on each side of the border (i.e. a Mexican dollar was 90 cents in the US, and a US dollar was worth 90 cents in Mexico). A man walks into a bar on the American side of the border, orders 10 cents worth of beer, and tenders a Mexican dollar ==> decision/newcomb.p <== Newcomb's Problem A being put one thousand dollars in box A and either zero or one million dollars in box B and presents you with two choices: ==> decision/prisoners.p <== Three prisoners on death row are told that one of them has been chosen at random for execution the next day, but the other two are to be freed. One privately begs the warden to at least tell him the name of one other prisoner who will be freed. The warden relents: 'Susie will ==> decision/red.p <== I show you a shuffled deck of standard playing cards, one card at a time. At any point before I run out of cards, you must say "RED]". If the next card I show is red (i.e. diamonds or hearts), you win. We assume I the "dealer" don't have any control over what the order of ==> decision/rotating.table.p <== Four glasses are placed upside down in the four corners of a square rotating table. You wish to turn them all in the same direction, either all up or all down. You may do so by grasping any two glasses and, optionally, turning either over. There are two catches: you are ==> decision/stpetersburg.p <== What should you be willing to pay to play a game in which the payoff is calculated as follows: a coin is flipped until in comes up heads on the nth toss and the payoff is set at 2^n dollars? ==> decision/switch.p <== Switch? (The Monty Hall Problem) Two black marbles and a red marble are in a bag. You choose one marble from the bag without looking at it. Another person chooses a marble from the bag and it ==> decision/truel.p <== A, B, and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 0.3, C's is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer ==> english/acronym.p <== What acronyms have become common words? ==> english/ambiguous.p <== What word in the English language is the most ambiguous? What is the greatest number of parts of speech that a single word can be used for? ==> english/antonym.p <== What words, when a single letter is added, reverse their meanings? Exclude words that are obtained by adding an "a-" to the beginning. ==> english/behead.p <== Is there a sentence that remains a sentence when all its words are beheaded? ==> english/capital.p <== What words change pronunciation when capitalized (e.g., polish -> Polish)? ==> english/charades.p <== A ....... surgeon was ....... to operate because he had ....... ==> english/contradictory.proverbs.p <== What are some proverbs that contradict one another? ==> english/contranym.p <== What words are their own antonym? ==> english/element.p <== The name of what element ends in "h"? ==> english/equations.p <== Each equation below contains the initials of words that will make the phrase correct. Figure out the missing words. Lower case is used only to help the initials stand out better. ==> english/fossil.p <== What are some examples of idioms that include obsolete words? ==> english/frequency.p <== In the English language, what are the most frequently appearing: 1) letters overall? 2) letters BEGINNING words? 3) final letters? ==> english/gry.p <== Find three completely different words ending in "gry." ==> english/homographs.p <== List all homographs (words that are spelled the same but pronounced differently) ==> english/homophones.p <== What words have four or more spellings that sound alike? ==> english/j.ending.p <== What words and names end in j? ==> english/ladder.p <== Find the shortest word ladders stretching between the following pairs: hit - ace pig - sty four - five ==> english/less.ness.p <== Find a word that forms two other words, unrelated in meaning, when "less" and "ness" are added. ==> english/letter.rebus.p <== Define the letters of the alphabet using self-referential common phrases (e.g., "first of all" defines "a"). ==> english/lipograms.p <== What books have been written without specific letters, vowels, etc.? ==> english/multi.lingual.p <== What words in multiple languages are related in interesting ways? ==> english/near.palindrome.p <== What are some long near palindromes, i.e., words that except for one letter would be palindromes? ==> english/palindromes.p <== What are some long palindromes? ==> english/pangram.p <== A "pangram" is a sentence containing all 26 letters. What is the shortest pangram (measured by number of letters or words)? What is the shortest word list using all 26 letters in alphabetical order? In reverse alphabetical order? ==> english/phonetic.letters.p <== What does "FUNEX" mean? ==> english/piglatin.p <== What words in pig latin also are words? ==> english/pleonasm.p <== What are some redundant terms that occur frequently (like "ABM missile")? ==> english/plurals/collision.p <== Two words, spelled and pronounced differently, have plurals spelled the same but pronounced differently. ==> english/plurals/doubtful.number.p <== A little word of doubtful number, a foe to rest and peaceful slumber. If you add an "s" to this, great is the metamorphosis. ==> english/plurals/drop.s.p <== What plural is formed by DROPPING the terminal "s" in a word? ==> english/plurals/endings.p <== List a plural ending with each letter of the alphabet. ==> english/plurals/french.p <== What English word, when spelled backwards, is its French plural? ==> english/plurals/man.p <== Words ending with "man" make their plurals by adding "s". ==> english/plurals/switch.first.p <== What plural is formed by switching the first two letters? ==> english/portmanteau.p <== What are some words formed by combining together parts of other words? ==> english/potable.color.p <== Find words that are both beverages and colors. ==> english/rare.trigraphs.p <== What trigraphs (three-letter combinations) occur in only one word? ==> english/records/pronunciation/silent.p <== What words have an exceptional number of silent letters? ==> english/records/pronunciation/spelling.p <== What words have exceptional ways to spell sounds? ==> english/records/pronunciation/syllable.p <== What words have an exceptional number of letters per syllable? ==> english/records/spelling/longest.p <== What is the longest word in the English language? ==> english/records/spelling/most.p <== What word has the most variant spellings? ==> english/records/spelling/operations.on.words/deletion.p <== What exceptional words turn into other words by deletion of letters? ==> english/records/spelling/operations.on.words/insertion.and.deletion.p <== What exceptional words turn into other words by both insertion and deletion of letters? ==> english/records/spelling/operations.on.words/insertion.p <== What exceptional words turn into other words by insertion of letters? ==> english/records/spelling/operations.on.words/movement.p <== What exceptional words turn into other words by movement of letters? ==> english/records/spelling/operations.on.words/substitution.p <== What exceptional words turn into other words by substitution of letters? ==> english/records/spelling/operations.on.words/transposition.p <== What exceptional words turn into other words by transposition of letters? ==> english/records/spelling/operations.on.words/words.within.words.p <== What exceptional words contain other words? ==> english/records/spelling/sets.of.words/nots.and.crosses.p <== What is the most number of letters that can be fit into a three by three grid of words, such that no letter is repeated in any row, column or diagonal? ==> english/records/spelling/sets.of.words/squares.p <== What are some exceptional word squares (square crosswords with no blanks)? ==> english/records/spelling/single.words.p <== What words have exceptional lengths, patterns, etc.? ==> english/repeat.p <== What is a sentence containing the most repeated words, without: using quotation marks, using proper names, using a language other than English, ==> english/repeated.words.p <== What is a sentence with the same word several times repeated? ==> english/rhyme.p <== What English words are hard to rhyme? "Rhyme is the identity in sound of an accented vowel in a word...and of all consonantal and vowel sounds following it; with a difference in ==> english/self.ref.letters.p <== Construct a true sentence of the form: "This sentence contains _ a's, _ b's, _ c's, ...," where the numbers filling in the blanks are spelled out. ==> english/self.ref.numbers.p <== What true sentence has the form: "There are _ 0's, _ 1's, _ 2's, ..., in this sentence"? ==> english/self.ref.words.p <== What sentence describes its own word, syllable and letter count? ==> english/sentence.p <== Find a sentence with words beginning with the letters of the alphabet, in order. ==> english/snowball.p <== Construct the longest coherent sentence you can such that the nth word is n letters long. ==> english/spoonerisms.p <== List some exceptional spoonerisms. ==> english/states.p <== What long words have all bigrams either a postal state code or its reverse? ==> english/telegrams.p <== Since telegrams cost by the word, phonetically similar messages can be cheaper. See if you can decipher these extreme cases: UTICA CHANSON MIGRATE INVENTION ANNUAL KNOBBY SORRY IN FACTUAL BEEN CLOVER. ==> english/trivial.p <== Consider the free non-abelian group on the twenty-six letters of the alphabet with all relations of the form = , where and are homophones (i.e. they sound alike but are spelled differently). Show that every letter is trivial. ==> english/weird.p <== Make a sentence containing only words that violate the "i before e" rule. ==> english/word.boundaries.p <== List some sentences that can be radically altered by changing word boundaries and punctuation. ==> english/word.torture.p <== What is the longest word all of whose contiguous subsequences are words? ==> games/chess/knight.control.p <== How many knights does it take to attack or control the board? ==> games/chess/mutual.check.p <== What position is a stalemate for both sides and is reachable in a legal game (including the requirement to prevent check)? ==> games/chess/mutual.stalemate.p <== What's the minimal number of pieces in a legal mutual stalemate? ==> games/chess/queens.p <== How many ways can eight queens be placed so that they control the board? ==> games/chess/size.of.game.tree.p <== How many different positions are there in the game tree of chess? ==> games/cigarettes.p <== The game of cigarettes is played as follows: Two players take turns placing a cigarette on a circular table. The cigarettes can be placed upright (on end) or lying flat, but not so that it touches any other cigarette on the table. This continues until one person looses by not ==> games/connect.four.p <== Is there a winning strategy for Connect Four? ==> games/craps.p <== What are the odds in craps? ==> games/crosswords/cryptic/clues.p <== What are some clues (indicators) used in cryptics? ==> games/crosswords/cryptic/double.p <== Each clue has two solutions, one for each diagram; one of the answers to 1ac. determines which solutions are for which diagram. All solutions are in Chamber's and Webster's Third except for one solution ==> games/crosswords/cryptic/intro.p <== What are the rules for cluing cryptic crosswords? ==> games/go-moku.p <== For a game of k in a row on an n x n board, for what values of k and n is there a win? Is (the largest such) k eventually constant or does it increase with n? ==> games/hi-q.p <== What is the quickest solution of the game Hi-Q (also called Solitair)? For those of you who aren't sure what the game looks like: ==> games/jeopardy.p <== What are the highest, lowest, and most different scores contestants can achieve during a single game of Jeopardy? ==> games/knight.tour.p <== For what board sizes is a knight's tour possible? ==> games/nim.p <== Place 10 piles of 10 $1 bills in a row. A valid move is to reduce the last i>0 piles by the same amount j>0 for some i and j; a pile reduced to nothing is considered to have been removed. The loser is the player who picks up the last dollar, and they must forfeit ==> games/othello.p <== How good are computers at Othello? ==> games/risk.p <== What are the odds when tossing dice in Risk? ==> games/rubiks.clock.p <== How do you quickly solve Rubik's clock? ==> games/rubiks.cube.p <== What is known about bounds on solving Rubik's cube? ==> games/rubiks.magic.p <== How do you solve Rubik's Magic? ==> games/scrabble.p <== What are some exceptional scrabble games? ==> games/square-1.p <== Does anyone have any hints on how to solve the Square-1 puzzle? ==> games/think.and.jump.p <== THINK & JUMP: FIRST THINK, THEN JUMP UNTIL YOU ARE LEFT WITH ONE PEG] O - O O - O / \ / \ / \ / \ O---O---O---O---O ==> games/tictactoe.p <== In random tic-tac-toe, what is the probability that the first mover wins? ==> geometry/K3,3.p <== Can three houses be connected to three utilities without the pipes crossing? _______ _______ _______ ! oil ! !water! ! gas ! ==> geometry/bear.p <== If a hunter goes out his front door, goes 50 miles south, then goes 50 miles west, shoots a bear, goes 50 miles north and ends up in front of his house. What color was the bear? ==> geometry/bisector.p <== If two angle bisectors of a triangle are equal, then the triangle is isosceles (more specifically, the sides opposite to the two angles being bisected are equal). ==> geometry/calendar.p <== Build a calendar from two sets of cubes. On the first set, spell the months with a letter on each face of three cubes. Use lowercase three-letter abbreviations for the names of all twelve months (e.g., "jan", "feb", "mar"). On the second set, ==> geometry/circles.and.triangles.p <== Find the radius of the inscribed and circumscribed circles for a triangle. ==> geometry/coloring/cheese.cube.p <== A cube of cheese is divided into 27 subcubes. A mouse starts at one corner and eats through every subcube. Can it finish in the middle? ==> geometry/coloring/dominoes.p <== There is a chess board (of course with 64 squares). You are given 21 dominoes of size 3-by-1 (the size of an individual square on a chess board is 1-by-1). Which square on the chess board can you cut out so that the 21 dominoes exactly cover the remaining ==> geometry/construction/4.triangles.6.lines.p <== Can you construct 4 equilateral triangles with 6 toothpicks? ==> geometry/construction/5.lines.with.4.points.p <== Arrange 10 points so that they form 5 rows of 4 each. ==> geometry/construction/square.with.compass.p <== Construct a square with only a compass and a straight edge. ==> geometry/cover.earth.p <== A thin membrane covers the surface of the earth. One square meter is added to the area of this membrane. How much is added to the radius and volume of this membrane? ==> geometry/dissections/circle.p <== Can a circle be cut into similar pieces without point symmetry about the midpoint? Can it be done with a finite number of pieces? ==> geometry/dissections/hexagon.p <== Divide the hexagon into: 1) 3 indentical rhombuses. 2) 6 indentical kites(?). 3) 4 indentical trapezoids. ==> geometry/dissections/square.70.p <== Since 1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2, can a 70x70 sqaure be dissected into 24 squares of size 1x1, 2x2, 3x3, etc.? ==> geometry/dissections/square.five.p <== Can you dissect a square into 5 parts of equal area with just a straight edge? ==> geometry/duck.and.fox.p <== A duck is swimming about in a circular pond. A ravenous fox (who cannot swim) is roaming the edges of the pond, waiting for the duck to come close. The fox can run faster than the duck can swim. In order to escape, the duck must swim to the edge of the pond before flying away. Assume that ==> geometry/earth.band.p <== How much will a band around the equator rise above the surface if it is made one meter longer? ==> geometry/ham.sandwich.p <== Consider a ham sandwich, consisting of two pieces of bread and one of ham. Suppose the sandwich was dropped into a machine and spindled, torn and mutiliated. Is it still possible to divide the ham sandwich with a straight knife cut such that both the ham and the bread are ==> geometry/hike.p <== You are hiking in a half-planar woods, exactly 1 mile from the edge, when you suddenly trip and lose your sense of direction. What's the shortest path that's guaranteed to take you out of the woods? Assume that you can navigate perfectly relative to your current location and ==> geometry/hole.in.sphere.p <== Old Boniface he took his cheer, Then he bored a hole through a solid sphere, Clear through the center, straight and strong, And the hole was just six inches long. ==> geometry/ladders.p <== Two ladders form a rough X in an alley. The ladders are 11 and 13 meters long and they cross 4 meters off the ground. How wide is the alley? ==> geometry/lattice/area.p <== Prove that the area of a triangle formed by three lattice points is integer/2. ==> geometry/lattice/equilateral.p <== Can an equlateral triangle have vertices at integer lattice points? ==> geometry/rotation.p <== What is the smallest rotation that returns an object to its original state? ==> geometry/smuggler.p <== Somewhere on the high sees smuggler S is attempting, without much luck, to outspeed coast guard G, whose boat can go faster than S's. G is one mile east of S when a heavy fog descends. It's so heavy that nobody can see or hear anything further than a few feet. Immediately ==> geometry/table.in.corner.p <== Put a round table into a (perpendicular) corner so that the table top touches both walls and the feet are firmly on the ground. If there is a point on the perimeter of the table, in the quarter circle between the two points of contact, which is 10 cm from one wall and 5 cm from ==> geometry/tesseract.p <== If you suspend a cube by one corner and slice it in half with a horizontal plane through its centre of gravity, the section face is a hexagon. Now suspend a tesseract (a four dimensional hypercube) by one corner and slice it in half with a hyper-horizontal hyperplane through ==> geometry/tetrahedron.p <== Suppose you have a sphere of radius R and you have four planes that are all tangent to the sphere such that they form an arbitrary tetrahedron (it can be irregular). What is the ratio of the surface area of the tetrahedron to its volume? ==> geometry/tiling/rational.sides.p <== A rectangular region R is divided into rectangular areas. Show that if each of the rectangles in the region has at least one side with rational length then the same can be said of R. ==> geometry/tiling/rectangles.with.squares.p <== Given two sorts of squares, (axa) and (bxb), what rectangles can be tiled? ==> geometry/tiling/scaling.p <== A given rectangle can be entirely covered (i.e. concealed) by an appropriate arrangement of 25 disks of unit radius. Can the same rectangle be covered by 100 disks of 1/2 unit radius? ==> geometry/tiling/seven.cubes.p <== Consider 7 cubes of equal size arranged as follows. Place 5 cubes so that they form a Swiss cross or a + (plus). ( 4 cubes on the sides and 1 in the middle). Now place one cube on top of the middle cube and the seventh below the middle cube, to effectively form a 3-dimensional ==> group/group.01.p <== AEFHIKLMNTVWXYZ BCDGJOPQRSU ==> group/group.01a.p <== 147 0235689 ==> group/group.02.p <== ABEHIKMNOPTXZ CDFGJLQRSUVWY ==> group/group.03.p <== BEJQXYZ DFGHLPRU KSTV CO AIW MN ==> group/group.04.p <== BDO P ACGJLMNQRSUVWZ EFTY HIKX ==> group/group.05.p <== CEFGHIJKLMNSTUVWXYZ ADOPQR B ==> group/group.06.p <== BCEGKMQSW DFHIJLNOPRTUVXYZ ==> induction/hanoi.p <== Is there an algorithom for solving the hanoi tower puzzle for any number of towers? Is there an equation for determining the minimum number of moves required to solve it, given a variable number of disks and towers? ==> induction/n-sphere.p <== With what odds do three random points on an n-sphere form an acute triangle? ==> induction/paradox.p <== What simple property holds for the first 10,000 integers, then fails? ==> induction/party.p <== You're at a party. Any two (different) people at the party have exactly one friend in common (the friend is also at the party). Prove that there is at least one person at the party who is a friend of everyone else. Assume that the friendship relation is symmetric and not reflexive. ==> induction/roll.p <== An ordinary die is thrown until the running total of the throws first exceeds 12. What is the most likely final total that will be obtained? ==> induction/takeover.p <== After graduating from college, you have taken an important managing position in the prestigious financial firm of "Mary and Lee". You are responsable for all the decisions concerning take-over bids. Your immediate concern is whether to take over "Financial Data". ==> logic/29.p <== Three people check into a hotel. They pay $30 to the manager and go to their room. The manager finds out that the room rate is $25 and gives $5 to the bellboy to return. On the way to the room the bellboy reasons that $5 would be difficult to share among three people so ==> logic/ages.p <== 1) Ten years from now Tim will be twice as old as Jane was when Mary was nine times as old as Tim. 2) Eight years ago, Mary was half as old as Jane will be when Jane is one year ==> logic/bookworm.p <== A bookworm eats from the first page of an encyclopedia to the last page. The bookworm eats in a straight line. The encyclopedia consists of ten 1000-page volumes. Not counting covers, title pages, etc., how many pages does the bookworm eat through? ==> logic/boxes.p <== Which Box Contains the Gold? Two boxes are labeled "A" and "B". A sign on box A says "The sign on box B is true and the gold is in box A". A sign on box B says ==> logic/calibans.will.p <== ---------------------------------------------- ! Caliban's Will by M.H. Newman ! ---------------------------------------------- ==> logic/camel.p <== An Arab sheikh tells his two sons that are to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower will win. The brothers, after wandering aimlessly for days, ask a wiseman for advise. After hearing the advice they jump on the ==> logic/centrifuge.p <== You are a biochemist, working with a 12-slot centrifuge. This is a gadget that has 12 equally spaced slots around a central axis, in which you can place chemical samples you want centrifuged. When the machine is turned on, the samples whirl around the central axis and do their thing. ==> logic/children.p <== A man walks into a bar, orders a drink, and starts chatting with the bartender. After a while, he learns that the bartender has three children. "How old are your children?" he asks. "Well," replies the bartender, "the product of their ages is 72." The man thinks for a ==> logic/condoms.p <== How can you have mutually safe sex with three women with only two condoms? ==> logic/dell.p <== How can I solve logic puzzles (e.g., as published by Dell) automatically? ==> logic/elimination.p <== 97 baseball teams participate in an annual state tournament. The way the champion is chosen for this tournament is by the same old elimination schedule. That is, the 97 teams are to be divided into pairs, and the two teams of each pair play against each other. ==> logic/family.p <== Suppose that it is equally likely for a pregnancy to deliver a baby boy as it is to deliver a baby girl. Suppose that for a large society of people, every family continues to have children until they have a boy, then they stop having children. ==> logic/flip.p <== How can a toss be called over the phone (without requiring trust)? ==> logic/friends.p <== Any group of 6 or more contains either 3 mutual friends or 3 mutual strangers. Prove it. ==> logic/hundred.p <== A sheet of paper has statements numbered from 1 to 100. Statement n says "exactly n of the statements on this sheet are false." Which statements are true and which are false? What if we replace "exactly" by "at least"? ==> logic/inverter.p <== Can a digital logic circuit with two inverters invert N independent inputs? The circuit may contain any number of AND or OR gates. ==> logic/josephine.p <== The recent expedition to the lost city of Atlantis discovered scrolls attributted to the great poet, scholar, philosopher Josephine. They number eight in all, and here is the first. ==> logic/locks.and.boxes.p <== You want to send a valuable object to a friend. You have a box which is more than large enough to contain the object. You have several locks with keys. The box has a locking ring which is more than large enough to have a lock attached. But your friend does not have the key to any ==> logic/mixing.p <== Start with a half cup of tea and a half cup of coffee. Take one tablespoon of the tea and mix it in with the coffee. Take one tablespoon of this mixture and mix it back in with the tea. Which of the two cups contains more of its original contents? ==> logic/number.p <== Mr. S. and Mr. P. are both perfect logicians, being able to correctly deduce any truth from any set of axioms. Two integers (not necessarily unique) are somehow chosen such that each is within some specified range. Mr. S. is given the sum of these two integers; Mr. P. is given the product of these ==> logic/riddle.p <== Who makes it, has no need of it. Who buys it, has no use for it. Who uses it can neither see nor feel it. Tell me what a dozen rubber trees with thirty boughs on each might be? ==> logic/river.crossing.p <== Three humans, one big monkey and two small monkeys are to cross a river: a) Only humans and the big monkey can row the boat. b) At all times, the number of human on either side of the river must be GREATER OR EQUAL to the number of monkeys ==> logic/ropes.p <== Two fifty foot ropes are suspended from a forty foot ceiling, about twenty feet apart. Armed with only a knife, how much of the rope can you steal? ==> logic/same.street.p <== Sally and Sue have a strong desire to date Sam. They all live on the same street yet neither Sally or Sue know where Sam lives. The houses on this street are numbered 1 to 99. ==> logic/self.ref.p <== Find a number ABCDEFGHIJ such that A is the count of how many 0's are in the number, B is the number of 1's, and so on. ==> logic/situation.puzzles.outtakes.p <== The following puzzles have been removed from my situation puzzles list, or never made it onto the list in the first place. There are a wide variety of reasons for the non-inclusion: some I think are obvious, some don't have enough of a story, some involve gimmicks that annoy me, ==> logic/situation.puzzles.p <== Jed's List of Situation Puzzles History: original compilation 11/28/87 ==> logic/smullyan/black.hat.p <== Three logicians, A, B, and C, are wearing hats, which they know are either black or white but not all white. A can see the hats of B and C; B can see the hats of A and C; C is blind. Each is asked in turn if they know the color of their own hat. The answers are: ==> logic/smullyan/fork.three.men.p <== Three men stand at a fork in the road. One fork leads to Someplaceorother; the other fork leads to Nowheresville. One of these people always answers the truth to any yes/no question which is asked of him. The other always lies when asked any yes/no question. The third person randomly lies and ==> logic/smullyan/fork.two.men.p <== Two men stand at a fork in the road. One fork leads to Someplaceorother; the other fork leads to Nowheresville. One of these people always answers the truth to any yes/no question which is asked of him. The other always lies when asked any yes/no question. By asking one yes/no question, can you ==> logic/smullyan/integers.p <== Two logicians place cards on their foreheads so that what is written on the card is visible only to the other logician. Consecutive positive integers have been written on the cards. The following conversation ensues: A: "I don't know my number." ==> logic/smullyan/liars.et.al.p <== Of a group of n men, some always lie, some never lie, and the rest sometimes lie. They each know which is which. You must determine the identity of each man by asking the least number of yes-or-no questions. ==> logic/smullyan/painted.heads.p <== While three logicians were sleeping under a tree, a malicious child painted their heads red. Upon waking, each logician spies the child's handiwork as it applied to the heads of the other two. Naturally they start laughing. Suddenly one falls silent. Why? ==> logic/smullyan/priest.p <== A priest takes confession of all the inhabitants in a small town. He discovers that in N married pairs in the town, one of the pair has committed adultery. Assume that the spouse of each adulterer does not know about the infidelity of his or her spouse, but that, since it is ==> logic/smullyan/stamps.p <== The moderator takes a set of 8 stamps, 4 red and 4 green, known to the logicians, and loosely affixes two to the forehead of each logician so that each logician can see all the other stamps except those 2 in the moderator's pocket and the two on her own head. He asks them in turn ==> logic/timezone.p <== Two people are talking long distance on the phone; one is in an East- Coast state, the other is in a West-Coast state. The first asks the other "What time is it?", hears the answer, and says, "That's funny. It's the same time here]" ==> logic/unexpected.p <== Swedish civil defense authorities announced that a civil defense drill would be held one day the following week, but the actual day would be a surprise. However, we can prove by induction that the drill cannot be held. Clearly, they cannot wait until Friday, since everyone will know it will be held that ==> logic/verger.p <== A very bright and sunny Day The Priest didst to the Verger say: "Last Monday met I strangers three None of which were known to Thee. ==> logic/weighing/balance.p <== You are given N balls and a balance scale and told that one ball is slightly heavier or lighter than the other identical ones. The scale lets you put the same number of balls on each side and observe which side (if either) is heavier. ==> logic/weighing/box.p <== You have ten boxes; each contains nine balls. The balls in one box weigh 0.9 kg; the rest weigh 1.0 kg. You have one weighing on a scale to find the box containing the light balls. How do you do it? ==> logic/weighing/gummy.bears.p <== Real gummy drop bears have a mass of 10 grams, while imitation gummy drop bears have a mass of 9 grams. Spike has 7 cartons of gummy drop bears, 4 of which contain real gummy drop bears, the others imitation. Using a scale only once and the minimum number of gummy drop bears, how ==> logic/weighing/weighings.p <== Some of the supervisors of Scandalvania's n mints are producing bogus coins. It would be easy to determine which mints are producing bogus coins but, alas, the only scale in the known world is located in Nastyville, which isn't on very friendly terms with Scandalville. In fact, Nastyville's ==> logic/zoo.p <== I took some nephews and nieces to the Zoo, and we halted at a cage marked Tovus Slithius, male and female. Beregovus Mimsius, male and female. ==> physics/balloon.p <== A helium-filled balloon is tied to the floor of a car that makes a sharp right turn. Does the balloon tilt while the turn is made? If so, which way? The windows are closed so there is no connection with the outside air. ==> physics/bicycle.p <== A boy, a girl and a dog go for a 10 mile walk. The boy and girl can walk 2 mph and the dog can trot at 4 mph. They also have bicycle which only one of them can use at a time. When riding, the boy and girl can travel at 12 mph while the dog can peddle at 16 mph. ==> physics/boy.girl.dog.p <== A boy, a girl and a dog are standing together on a long, straight road. Simulataneously, they all start walking in the same direction: The boy at 4 mph, the girl at 3 mph, and the dog trots back and forth between them at 10 mph. Assume all reversals of direction instantaneous. ==> physics/brick.p <== What is the maximum overhang you can create with an infinite supply of bricks? ==> physics/cannonball.p <== A person in a boat drops a cannonball overboard; does the water level change? ==> physics/dog.p <== A body of soldiers form a 50m-by-50m square ABCD on the parade ground. In a unit of time, they march forward 50m in formation to take up the position DCEF. The army's mascot, a small dog, is standing next to its handler at location A. When the ==> physics/magnets.p <== You have two bars of iron. One is magnetic, the other is not. Without using any other instrument (thread, filings, other magnets, etc.), find out which is which. ==> physics/milk.and.coffee.p <== You are just served a hot cup of coffee and want it to be as hot as possible when you drink it some number of minutes later. Do you add milk when you get the cup or just before you drink it? ==> physics/mirror.p <== Why does a mirror appear to invert the left-right directions, but not up-down? ==> physics/monkey.p <== Hanging over a pulley, there is a rope, with a weight at one end. At the other end hangs a monkey of equal weight. The rope weighs 4 ounces per foot. The combined ages of the monkey and it's mother is 4 years. The weight of the monkey is as many pounds as the mother ==> physics/particle.p <== What is the longest time that a particle can take in travelling between two points if it never increases its acceleration along the way and reaches the second point with speed V? ==> physics/pole.in.barn.p <== Accelerate a pole of length l to a constant speed of 90% of the speed of light (.9c). Move this pole towards an open barn of length .9l (90% the length of the pole). Then, as soon as the pole is fully inside the barn, close the door. What do you see and what actually happens? ==> physics/resistors.p <== What are the resistances between lattices of resistors in the shape of a: 1. Cube ==> physics/sail.p <== A sailor is in a sailboat on a river. The water (current) is flowing downriver at a velocity of 3 knots with respect to the land. The wind (air velocity) is zero, with respect to the land. The sailor wants to proceed downriver as quickly as possible, maximizing his downstream ==> physics/skid.p <== What is the fastest way to make a 90 degree turn on a slippery road? ==> physics/spheres.p <== Two spheres are the same size and weight, but one is hollow. They are made of uniform material, though of course not the same material. Without a minimum of apparatus, how can I tell which is hollow? ==> physics/wind.p <== Is a round-trip by airplane longer or shorter if there is wind blowing? ==> probability/amoeba.p <== A jar begins with one amoeba. Every minute, every amoeba turns into 0, 1, 2, or 3 amoebae with probability 25% for each case ( dies, does nothing, splits into 2, or splits into 3). What is the probability that the amoeba population ==> probability/apriori.p <== An urn contains one hundred white and black balls. You sample one hundred balls with replacement and they are all white. What is the probability that all the balls are white? ==> probability/cab.p <== A cab was involved in a hit and run accident at night. Two cab companies, the Green and the Blue, operate in the city. Here is some data: a) Although the two companies are equal in size, 85% of cab ==> probability/coincidence.p <== Name some amazing coincidences. ==> probability/coupon.p <== There is a free gift in my breakfast cereal. The manufacturers say that the gift comes in four different colours, and encourage one to collect all four (& so eat lots of their cereal). Assuming there is an equal chance of getting any one of the colours, what is the ==> probability/darts.p <== Peter throws two darts at a dartboard, aiming for the center. The second dart lands farther from the center than the first. If Peter now throws another dart at the board, aiming for the center, what is the probability that this third throw is also worse (i.e., farther from ==> probability/flips.p <== Consider a run of coin tosses: HHTHTTHTTTHTTTTHHHTHHHHHTHTTHT Define a success as a run of one H or T (as in THT or HTH). Use two different methods of sampling. The first method would consist of ==> probability/flush.p <== Which set contains more flushes than the set of all possible hands? (1) Hands whose first card is an ace (2) Hands whose first card is the ace of spades (3) Hands with at least one ace ==> probability/hospital.p <== A town has two hospitals, one big and one small. Every day the big hospital delivers 1000 babies and the small hospital delivers 100 babies. There's a 50/50 chance of male or female on each birth. Which hospital has a better chance of having the same number of boys ==> probability/icos.p <== The "house" rolls two 20-sided dice and the "player" rolls one 20-sided die. If the player rolls a number on his die between the two numbers the house rolled, then the player wins. Otherwise, the house wins (including ties). What are the probabilities of the player ==> probability/intervals.p <== Given two random points x and y on the interval 0..1, what is the average size of the smallest of the three resulting intervals? ==> probability/lights.p <== Waldo and Basil are exactly m blocks west and n blocks north from Central Park, and always go with the green light until they run out of options. Assuming that the probability of the light being green is 1/2 in each direction and that if the light is green in one direction it is red in the other, find the ==> probability/lottery.p <== There n tickets in the lottery, k winners and m allowing you to pick another ticket. The problem is to determine the probability of winning the lottery when you start by picking 1 (one) ticket. ==> probability/particle.in.box.p <== A particle is bouncing randomly in a two-dimensional box. How far does it travel between bounces, on avergae? Suppose the particle is initially at some random position in the box and is ==> probability/pi.p <== Are the digits of pi random (i.e., can you make money betting on them)? ==> probability/random.walk.p <== Waldo has lost his car keys] He's not using a very efficient search; in fact, he's doing a random walk. He starts at 0, and moves 1 unit to the left or right, with equal probability. On the next step, he moves 2 units to the left or right, again with equal probability. For ==> probability/reactor.p <== There is a reactor in which a reaction is to take place. This reaction stops if an electron is present in the reactor. The reaction is started with 18 positrons; the idea being that one of these positrons would combine with any incoming electron (thus destroying both). Every second, ==> probability/roulette.p <== You are in a game of Russian roulette, but this time the gun (a 6 shooter revolver) has three bullets _in_a_row_ in three of the chambers. The barrel is spun only once. Each player then points the gun at his (her) head and pulls the trigger. If he (she) is still ==> probability/unfair.p <== Generate even odds from an unfair coin. For example, if you thought a coin was biased toward heads, how could you get the equivalent of a fair coin with several tosses of the unfair coin? ==> series/series.01.p <== M, N, B, D, P ? ==> series/series.02.p <== H, H, L, B, B, C, N, O, F ? ==> series/series.03.p <== W, A, J, M, M, A, J? ==> series/series.03a.p <== G, J, T, J, J, J, A, M, W, J, J, Z, M, F, J, ? ==> series/series.03b.p <== A, J, B, C, G, T, C, V, J, T, D, F, K, B, H, ? ==> series/series.03c.p <== M, A, M, D, E, L, R, H, ? ==> series/series.04.p <== A, E, H, I, K, L, ? ==> series/series.05.p <== A B C D E F G H? ==> series/series.06.p <== Z, O, T, T, F, F, S, S, E, N? ==> series/series.06a.p <== F, S, T, F, F, S, ? ==> series/series.07.p <== 1, 1 1, 2 1, 1 2 1 1, ... What is the pattern and asymptotics of this series? ==> series/series.08a.p <== G, L, M, B, C, L, M, C, F, S, ? ==> series/series.08b.p <== A, V, R, R, C, C, L, L, L, E, ? ==> series/series.09a.p <== S, M, S, S, S, C, P, P, P, ? ==> series/series.09b.p <== M, S, C, P, P, P, S, S, S, ? ==> series/series.10.p <== D, P, N, G, C, M, M, S, ? ==> series/series.11.p <== R O Y G B ? ==> series/series.12.p <== A, T, G, C, L, ? ==> series/series.13.p <== M, V, E, M, J, S, ? ==> series/series.14.p <== A, B, D, O, P, ? ==> series/series.14a.p <== A, B, D, E, G, O, P, ? ==> series/series.15.p <== A, E, F, H, I, ? ==> series/series.16.p <== A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, X, Y? ==> series/series.17.p <== T, P, O, F, O, F, N, T, S, F, T, F, E, N, S, N? ==> series/series.18.p <== 10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, ___ , 100, 121, 10000 ==> series/series.19.p <== 1 01 01011 0101101011011 0101101011011010110101101101011011 etc. Each string is formed from the previous string by substituting '01' for '1' and '011' for '0' simultaneously at each occurance. ==> series/series.20.p <== 1 2 5 16 64 312 1812 12288 ==> series/series.21.p <== 5, 6, 5, 6, 5, 5, 7, 5, ? ==> series/series.22.p <== 3 1 1 0 3 7 5 5 2 ? ==> series/series.23.p <== 22 22 30 13 13 16 16 28 28 11 ? ==> series/series.24.p <== What is the next letter in the sequence: W, I, T, N, L, I, T? ==> series/series.25.p <== 1 3 4 9 10 12 13 27 28 30 31 36 37 39 40 ? ==> series/series.26.p <== 1 3 2 6 7 5 4 12 13 15 14 10 11 9 8 24 25 27 26 ? ==> series/series.27.p <== 0 1 1 2 1 2 1 3 2 2 1 3 1 2 2 4 1 3 1 3 2 2 1 4 2 ? ==> series/series.28.p <== 0 2 3 4 5 5 7 6 6 7 11 7 13 9 8 8 17 8 19 9 10 13 23 9 10 ? ==> series/series.29.p <== 1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1 2 2 3 2 3 3 4 2 3 3 4 3 4 ? ==> series/series.30.p <== I I T Y W I M W Y B M A D ==> series/series.31.p <== 6 2 5 5 4 5 6 3 7 ==> series/series.32.p <== 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 ==> series/series.33.p <== 2 12 360 75600 ==> series/series.34.p <== 3 5 4 4 3 5 5 4 3 ==> series/series.35.p <== 1 2 3 2 1 2 3 4 2 1 2 3 4 2 2 3 ==> trivia/area.codes.p <== When looking at a map of the distribution of telephone area codes for North America, it appears that they are randomly distributed. I am doubtful that this is the case, however. Does anyone know how the area codes were/are chosen? ==> trivia/eskimo.snow.p <== How many words do the Eskimo have for snow? ==> trivia/federal.reserve.p <== What is the pattern to this list: Boston, MA New York, NY Philadelphia, PA ==> trivia/jokes.self-referential.p <== What are some self-referential jokes? ==> analysis/bugs.p <== Four bugs are placed at the corners of a square. Each bug walks directly toward the next bug in the clockwise direction. The bugs walk with constant speed always directly toward their clockwise neighbor. Assuming the bugs make at least one full circuit around the center of the square before meeting, how much closer to the center will a bug be at the end of its first full circuit? ==> analysis/bugs.s <== Amorous Bugs ANSWER: 1 - e^(-2*pi) Let O(t) be the angle at time t of bug 1 relative to its starting point and r(O(t)) be its distance from the center of the square. Bug 1's vector trajectory is (using a Cartesian coordinate system with the origin at the center of the square): (1) X1 = Ýr(O) * cos(O), r(O) * sin(O)¨ By symmetry, bug 2's trajectory is the same only rotated by pi/2, viz.: (2) X2 = Ý-r(O) * sin(O), r(O) * cos(O)¨ Since bug 1 walks directly toward bug 2, the velocity of bug 1 must be proportional to the vector from bug 1 to bug 2: (3) d(X1)/d(t) = k * (X2 - X1) Equating each component of the vector equation (3) yields: (4) (d(r)/d(O) * cos(O) - r * sin(O)) * d(O)/d(t) = k * (-r * cos(O) - r * sin(O)) (5) (d(r)/d(O) * sin(O) + r * cos(O)) * d(O)/d(t) = k * (-r * sin(O) + r * cos(O)) These equations are solved by: (6) k = d(O)/d(t) and: (7) d(r)/d(O) = -r(O) (7) is solved by: (8) r(O) = e^-O Constant speed gives: (9) v^2 = constant = ((d(r)/d(O))^2+r^2)*(d(O)/d(t))^2 Substituting (8) into (9) yields (let V = v/sqrt(2)): (10) d(O)/d(t) = V * e^O Which is solved (using the boundary condition O(0) = 0) by: (11) O(t) = -ln(1 - V * t) Substituting (11) into (8) yields: (12) r(t) = r(0) - V * t The bug has made a full circle when O(T) = 2*pi; using (11): (13) T = 1/V * (1 - e^(-2*pi)) Substituting T into (12) yields the answer: (14) r(T) - r(0) = 1 - e^(-2*pi) ==> analysis/c.infinity.p <== What function is zero at zero, strictly positive elsewhere, infinitely differentiable at zero and has all zero derivitives at zero? ==> analysis/c.infinity.s <== exp(-1/x^2) This tells us why Taylor Series are a more limited device than they might be. We form a Taylor series by looking at the derivatives of a function at a given point; but this example shows us that the derivatives at a point may tell us almost nothing about its behavior away from that point. ==> analysis/cache.p <== Cache and Ferry (How far can a truck go in a desert?) A pick-up truck is in the desert beside N 50-gallon gas drums, all full. The truck's gas tank holds 10 gallons and is empty. The truck can carry one drum, whether full or empty, in its bed. It gets 10 miles to the gallon. How far away from the starting point can you drive the truck? ==> analysis/cache.s <== If the truck can siphon gas out of its tank and leave it in the cache, the answer is: { 1/1 + 1/3 + ... + 1/(2 * N - 1) } x 500 miles. Otherwise, the "Cache and Ferry" problem is the same as the "Desert Fox" problem described, but not solved, by Dewdney, July '87 "Scientific American". Dewdney's Oct. '87 Sci. Am. article gives for N=2, the optimal distance of 733.33 miles. In the Nov. issue, Dewdney lists the optimal distance of 860 miles for N=3, and gives a better, but not optimal, general distance formula. Westbrook, in Vol 74, #467, pp 49-50, March '90 "Mathematical Gazette", gives an even better formula, for which he incorrectly claims optimality: For N = 2,3,4,5,6: Dist = (600/1 + 600/3 + ... + 600/(2N-3)) + (600-100N)/(2N-1) For N > 6: Dist = (600/1 + 600/3 + ... + 600/9) + (500/11 + ... + 500/(2N-3)) The following shows that Westbrook's formula is not optimal for N=8: Ferry 7 drums forward 33.3333 miles (356.6667 gallons remain) Ferry 6 drums forward 51.5151 miles (300.0000 gallons remain) Ferry 5 drums forward 66.6667 miles (240.0000 gallons remain) Ferry 4 drums forward 85.7143 miles (180.0000 gallons remain) Ferry 3 drums forward 120.0000 miles (120.0000 gallons remain) Ferry 2 drums forward 200.0000 miles ( 60.0000 gallons remain) Ferry 1 drums forward 600.0000 miles --------------- Total distance = 1157.2294 miles (Westbrook's formula = 1156.2970 miles) Ý"Ferrying n drums forward x miles" involves (2*n-1) trips, each of distance x.¨ Other attainable values I've found: N Distance --- --------- (Ferry distances for each N are omitted for brevity.) 5 1016.8254 7 1117.8355 11 1249.2749 13 1296.8939 17 1372.8577 19 1404.1136 (The N <= 19 distances could be optimal.) 31 1541.1550 (I doubt that this N = 31 distance is optimal.) 139 1955.5509 (I'm sure that this N = 139 distance is not optimal.) So...where's MY formula? I haven't found one, and believe me, I've looked. I would be most grateful if someone would end my misery by mailing me a formula, a literature reference, or even an efficient algorithm that computes the optimal distance. If you do come up with the solution, you might want to first check it against the attainable distances listed above, before sending it out. (Not because you might be wrong, but just as a mere formality to check your work.) ÝWarning: the Mathematician General has determined that this problem is as addicting as Twinkies.¨ Myron P. Souris ! "If you have anything to tell me of importance, McDonnell Douglas ! for God's sake begin at the end." souris@mdcbbs.com ! Sara Jeanette Duncan @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ The following output comes from some hack programs that I've used to empirically verify some proofs I've been working on. Initial barrels: 12 (600 gallons) Attainable distance= 1274.175211 Barrels Distance Gas Moved covered left >From depot 1: 10 63.1579 480.0000 >From depot 2: 8 50.0000 405.0000 >From depot 3: 7 37.5000 356.2500 >From depot 4: 6 51.1364 300.0000 >From depot 5: 5 66.6667 240.0000 >From depot 6: 4 85.7143 180.0000 >From depot 7: 3 120.0000 120.0000 >From depot 8: 2 200.0000 60.0000 >From depot 9: 1 600.0000 0.0000 Initial barrels: 40 (2000 gallons) Attainable distance= 1611.591484 Barrels Distance Gas Moved covered left >From depot 1: 40 2.5316 1980.0000 >From depot 2: 33 50.0000 1655.0000 >From depot 3: 28 50.0000 1380.0000 >From depot 4: 23 53.3333 1140.0000 >From depot 5: 19 50.0000 955.0000 >From depot 6: 16 56.4516 780.0000 >From depot 7: 13 50.0000 655.0000 >From depot 8: 11 54.7619 540.0000 >From depot 9: 9 50.0000 455.0000 >From depot 10: 8 32.1429 406.7857 >From depot 11: 7 38.9881 356.1012 >From depot 12: 6 51.0011 300.0000 >From depot 13: 5 66.6667 240.0000 >From depot 14: 4 85.7143 180.0000 >From depot 15: 3 120.0000 120.0000 >From depot 16: 2 200.0000 60.0000 >From depot 17: 1 600.0000 0.0000 ==> analysis/cats.and.rats.p <== If 6 cats can kill 6 rats in 6 minutes, how many cats does it take to kill one rat in one minute? ==> analysis/cats.and.rats.s <== The following piece by Lewis Carroll first appeared in ``The Monthly Packet'' of February 1880 and is reprinted in _The_Magic_of_Lewis_Carroll_, edited by John Fisher, Bramhall House, 1973. /Larry Denenberg larry@bbn.com larry@harvard.edu Cats and Rats If 6 cats kill 6 rats in 6 minutes, how many will be needed to kill 100 rats in 50 minutes? This is a good example of a phenomenon that often occurs in working problems in double proportion; the answer looks all right at first, but, when we come to test it, we find that, owing to peculiar circumstances in the case, the solution is either impossible or else indefinite, and needing further data. The 'peculiar circumstance' here is that fractional cats or rats are excluded from consideration, and in consequence of this the solution is, as we shall see, indefinite. The solution, by the ordinary rules of Double Proportion, is as follows: 6 rats : 100 rats \ > :: 6 cats : ans. 50 min. : 6 min. / . . . ans. = (100)(6)(6)/(50)(6) = 12 But when we come to trace the history of this sanguinary scene through all its horrid details, we find that at the end of 48 minutes 96 rats are dead, and that there remain 4 live rats and 2 minutes to kill them in: the question is, can this be done? Now there are at least *four* different ways in which the original feat, of 6 cats killing 6 rats in 6 minutes, may be achieved. For the sake of clearness let us tabulate them: A. All 6 cats are needed to kill a rat; and this they do in one minute, the other rats standing meekly by, waiting for their turn. B. 3 cats are needed to kill a rat, and they do it in 2 minutes. C. 2 cats are needed, and do it in 3 minutes. D. Each cat kills a rat all by itself, and take 6 minutes to do it. In cases A and B it is clear that the 12 cats (who are assumed to come quite fresh from their 48 minutes of slaughter) can finish the affair in the required time; but, in case C, it can only be done by supposing that 2 cats could kill two-thirds of a rat in 2 minutes; and in case D, by supposing that a cat could kill one-third of a rat in two minutes. Neither supposition is warranted by the data; nor could the fractional rats (even if endowed with equal vitality) be fairly assigned to the different cats. For my part, if I were a cat in case D, and did not find my claws in good working order, I should certainly prefer to have my one-third-rat cut off from the tail end. In cases C and D, then, it is clear that we must provide extra cat-power. In case C *less* than 2 extra cats would be of no use. If 2 were supplied, and if they began killing their 4 rats at the beginning of the time, they would finish them in 12 minutes, and have 36 minutes to spare, during which they might weep, like Alexander, because there were not 12 more rats to kill. In case D, one extra cat would suffice; it would kill its 4 rats in 24 minutes, and have 24 minutes to spare, during which it could have killed another 4. But in neither case could any use be made of the last 2 minutes, except to half-kill rats---a barbarity we need not take into consideration. To sum up our results. If the 6 cats kill the 6 rats by method A or B, the answer is 12; if by method C, 14; if by method D, 13. This, then, is an instance of a solution made `indefinite' by the circumstances of the case. If an instance of the `impossible' be desired, take the following: `If a cat can kill a rat in a minute, how many would be needed to kill it in the thousandth part of a second?' The *mathematical* answer, of course, is `60,000,' and no doubt less than this would *not* suffice; but would 60,000 suffice? I doubt it very much. I fancy that at least 50,000 of the cats would never even see the rat, or have any idea of what was going on. Or take this: `If a cat can kill a rat in a minute, how long would it be killing 60,000 rats?' Ah, how long, indeed] My private opinion is that the rats would kill the cat. ==> analysis/e.and.pi.p <== Which is greater, e^(pi) or (pi)^e ? ==> analysis/e.and.pi.s <== Put x = pi/e - 1 in the inequality e^x > 1+x (x>0). ==> analysis/functional/distributed.p <== Find all f: R -> R, f not identically zero, such that (*) f( (x+y)/(x-y) ) = ( f(x)+f(y) )/( f(x)-f(y) ). ==> analysis/functional/distributed.s <== 1) Assuming f finite everywhere, (*) ==> x<>y ==> f(x)<>f(y) 2) Exchanging x and y in (*) we see that f(-x) = -f(x). 3) a <> 0 ==> f((a-a)/(a+a)) = (f(a)-f(a))/(f(a)+f(a)) ==> f(0) = 0. 4) a <> 0 ==> f((a+0)/(a-0)) = f(a)/f(a) ==> f(1) = 1. 5) x<>y, y<>0 ==> f(x/y) = f( ((x+y)/(x-y) + (x-y)/(x-y)) / ((x+y)/(x-y) - (x-y)/(x-y)) = f(x)/f(y) ==> f(xy) = f(x)f(y) by replacing x with xy and by noting that f(x*1) = f(x)*1 and f(x*0) = f(x)*f(0). 6) f(x*x) = f(x)*f(x) ==> f(x) > 0 if x>0. 7) Let a=1+\/2, b=1-\/2; a,b satisfy (x+1)/(x-1) = x ==> f(x) = (f(x)+1)/(f(x)-1) ==> f(a)=a, f(b)=b. f(1/\/2) = f((a+b)/(a-b)) = (a+b)/(a-b) = 1/\/2 ==> f(2) = 2. 8) By induction and the relation f((n+1)/(n-1)) = (f(n)+1)/(f(n)-1) we get that f(n)=n for all integer n. #5 now implies that f fixes the rationals. 9) If x>y>0 (*) ==> f(x) - f(y) = f(x+y)/f((x+y)/(x-y)) > 0 by #6. Thus f is order-preserving. Since f fixes the rationals *and* f is order-preserving, f must be the identity function. This was E2176 in _The American Mathematical Monthly_ (the proposer was R. S. Luthar). ==> analysis/functional/linear.p <== Suppose f is non-decreasing with f(x+y) = f(x) + f(y) + C for all real x, y. Prove: there is a constant A such that f(x) = Ax - C for all x. (Note: continuity of f is not assumed in advance.) ==> analysis/functional/linear.s <== By induction f(mx) = m(f(x)+C)-C. Let x=1/n, m=n and find that f(1/n) = (1/n)(f(1)+C)-C. Now let x=1/n and find that f(m/n) = (m/n)(f(1)+C)-C. f(-x+x) = f(-x) + f(x) + C ==> f(-x) = -2C - f(x) (since f(0) = -C) ==> f(-m/n) = -(m/n)(f(1)+C)-C. Since f is monotonic ==> f(x) = x*(f(1)+C)-C for all real x (Squeeze Theorem). ==> analysis/integral.p <== If f is integrable on (0,inf), and differentiable at 0, and a > 0, show: inf ( f(x) - f(ax) ) Int ---------------- dx = f(0) ln(a) 0 x ==> analysis/integral.s <== First, note that if f(0) is 0, then by substituting u=ax in the integral of f(x)/x, our integral is the difference of two equal integrals and so is 0 (the integrals are finite because f is 0 at 0 and differentiable there. Note I make no requirement of continuity). Second, note that if f is the characteristic function of the interval Ý0, 1¨--- i.e. 1, 0<=x<=1 f (x) = 0 otherwise then a little arithmetic reduces our integral to that of 1/x from 1/a to 1 (assuming a>1; if a <= 1 the reasoning is similar), which is ln(a) = f(0)ln(a) as required. Call this function g. Finally, note that the operator which takes the function f to the value of our integral is linear, and that every function meeting the hypotheses (incidentally, I should have said `differentiable from the right', or else replaced the characteristic function of Ý0,1¨ above by that of (-infinity, 1¨; but it really doesn't matter) is a linear combination of one which is 0 at 0 and g, to wit f(x) = f(0)g(x) + (f(x) - g(x)f(0)). ==> analysis/period.p <== What is the least possible integral period of the sum of functions of periods 3 and 6? ==> analysis/period.s <== Period 2. Clearly, the sum of periodic functions of periods 2 and three is 6. So take the function which is the sum of that function of period six and the negative of the function of period three and you have a function of period 2. ==> analysis/rubberband.p <== A bug walks down a rubberband which is attached to a wall at one end and a car moving away from the wall at the other end. The car is moving at 1 m/sec while the bug is only moving at 1 cm/sec. Assuming the rubberband is uniformly and infinitely elastic, will the bug ever reach the car? ==> analysis/rubberband.s <== Let w = speed of bug and N = ratio of car speed/bug speed = 100. Paint N+1 equally spaced stripes on the rubberband. When the bug is standing on one stripe, the next stripe is moving away from him at a speed slightly < w (relative to him). Since he is walking at w, clearly the bug can reach the next stripe. But once he reaches that stripe, the next one is only receeding at < w. So he walks on down to the car, one stripe at a time. The bug starts gaining on the car when he is at the next to last stripe. ==> analysis/series.p <== Show that in the series: x, 2x, 3x, .... (n-1)x (x can be any real number) there is at least one number which is within 1/n of an integer. ==> analysis/series.s <== Throw 0 into the sequence; there are now n numbers, so some pair must have fractional parts within 1/n of each other; their difference is then within 1/n of an integer. ==> analysis/snow.p <== Snow starts falling before noon on a cold December day. At noon a snowplow starts plowing a street. It travels 1 mile in the first hour, and 1/2 mile in the second hour. What time did the snow start falling?? You may assume that the plow's rate of travel is inversely proportioned to the height of the snow, and that the snow falls at a uniform rate. ==> analysis/snow.s <== 11:22:55.077 am. Method: Let b = the depth of the snow at noon, a = the rate of increase in the depth. Then the depth at time t (where noon is t=0) is at+b, the snowfall started at t_0=-b/a, and the snowplow's rate of progress is ds/dt = k/(at+b). If the snowplow starts at s=0 then s(t) = (k/a) log(1+at/b). Note that s(2 hours) = 1.5 s(1 hour), or log(1+2A/b) = 1.5 log(1+A/b), where A = (1 hour)*a. Letting x = A/b we have (1+2x)^2 = (1+x)^3. Solve for x and t_0 = -(1 hour)/x. The exact answer is 11:(90-30 SqrtÝ5¨). _American Mathematics Monthly_, April 1937, page 245 E 275. Proposed by J. A. Benner, Lafayette College, Easton. Pa. The solution appears, appropriately, in the December 1937 issue, pp. 666-667. Also solved by William Douglas, C. E. Springer, E. P. Starke, W. J. Taylor, and the proposer. See R.P. Agnew, "Differential Equations," 2nd edition, p. 39 ff. ==> analysis/tower.p <== A number is raised to its own power. The same number is then raised to the power of this result. The same number is then raised to the power of this second result. This process is continued forever. What is the maximum number which will yield a finite result from this process? ==> analysis/tower.s <== Tower of Exponentials ANSWER: e^(1/e) Let N be the number in question and R the result of the process. Then R can be defined recursively by the equation: (1) R = N^R Taking the logarithm of both sides of (1): (2) ln(R) = R * ln(N) Dividing (2) by R and rearranging: (3) ln(N) = ln(R) / R Exponentiating (3): (4) N = R^(1/R) We wish to find the maximum value of N with respect to R. Find the derivative of N with respect to R and set it equal to zero: (5) d(N)/d(R) = (1 - ln(R)) / R^2 = 0 For finite values of R, (5) is satisfied by R = e. This is a maximum of N if the second derivative of N at R = e is less than zero. (6) d2(N)/d2(R) ! R=e = (2 * ln(R) - 3) / R^3 ! R=e = -1 / e^3 < 0 The solution therefore is (4) at R = e: (7) Nmax = e^(1/e) ==> arithmetic/7-11.p <== A customer at a 7-11 store selected four items to buy, and was told that the cost was $7.11. He was curious that the cost was the same as the store name, so he inquired as to how the figure was derived. The clerk said that he had simply multiplied the prices of the four individual items. The customer protested that the four prices should have been ADDED, not MULTIPLIED. The clerk said that that was OK with him, but, the result was still the same: exactly $7.11. What were the prices of the four items? ==> arithmetic/7-11.s <== The prices are: $1.20, $1.25, $1.50, and $3.16 $7.11 is not the only number which works. Here are the first 160 such numbers, preceded by a count of distinct solutions for that price. Note that $7.11 has a single, unique solution. 1 - $6.44 1 - $7.83 2 - $9.20 3 - $10.89 1 - $6.51 1 - $7.86 1 - $9.23 1 - $10.95 1 - $6.60 3 - $7.92 1 - $9.24 2 - $11.00 1 - $6.63 1 - $8.00 1 - $9.27 1 - $11.07 1 - $6.65 1 - $8.01 2 - $9.35 1 - $11.13 1 - $6.72 1 - $8.03 3 - $9.36 1 - $11.16 2 - $6.75 5 - $8.10 1 - $9.38 1 - $11.22 1 - $6.78 1 - $8.12 5 - $9.45 2 - $11.25 1 - $6.80 1 - $8.16 2 - $9.48 2 - $11.27 2 - $6.84 2 - $8.19 1 - $9.54 1 - $11.30 1 - $6.86 1 - $8.22 1 - $9.57 1 - $11.36 1 - $6.89 1 - $8.25 1 - $9.59 1 - $11.40 2 - $6.93 3 - $8.28 2 - $9.60 2 - $11.43 1 - $7.02 3 - $8.33 1 - $9.62 2 - $11.52 1 - $7.05 1 - $8.36 2 - $9.63 2 - $11.55 2 - $7.07 1 - $8.37 1 - $9.66 2 - $11.61 1 - $7.08 2 - $8.40 1 - $9.68 1 - $11.69 1 - $7.11 1 - $8.45 2 - $9.69 1 - $11.70 1 - $7.13 2 - $8.46 1 - $9.78 1 - $11.88 2 - $7.14 1 - $8.52 2 - $9.80 1 - $11.90 3 - $7.20 5 - $8.55 1 - $9.81 1 - $11.99 1 - $7.25 1 - $8.60 1 - $9.87 1 - $12.06 1 - $7.26 4 - $8.64 4 - $9.90 1 - $12.15 2 - $7.28 1 - $8.67 1 - $9.92 1 - $12.18 1 - $7.29 1 - $8.69 2 - $9.99 1 - $12.24 3 - $7.35 1 - $8.73 1 - $10.01 1 - $12.30 1 - $7.37 2 - $8.75 1 - $10.05 1 - $12.32 1 - $7.47 1 - $8.76 2 - $10.08 1 - $12.35 1 - $7.50 1 - $8.78 1 - $10.17 2 - $12.42 1 - $7.52 5 - $8.82 1 - $10.20 1 - $12.51 4 - $7.56 1 - $8.85 2 - $10.26 1 - $12.65 1 - $7.62 1 - $8.88 3 - $10.29 2 - $12.69 4 - $7.65 2 - $8.91 3 - $10.35 1 - $12.75 1 - $7.67 1 - $8.94 2 - $10.44 1 - $12.92 2 - $7.70 1 - $8.96 1 - $10.53 1 - $12.96 3 - $7.74 3 - $9.00 1 - $10.56 1 - $13.23 1 - $7.77 1 - $9.02 1 - $10.64 1 - $13.41 1 - $7.79 2 - $9.03 2 - $10.71 1 - $13.56 2 - $7.80 1 - $9.12 3 - $10.80 1 - $14.49 1 - $7.82 2 - $9.18 1 - $10.85 1 - $15.18 There are plenty of solutions for five summands. Here are a few: $8.28 -- at least two solutions $8.47 -- at least two solutions $8.82 -- at least two solutions -- Mark Johnson mark@microunity.com (408) 734-8100 There may be many approximate solutions, for example: $1.01, $1.15, $2.41, and $2.54. These sum to $7.11 but the product is 7.1100061. ==> arithmetic/clock/day.of.week.p <== It's restful sitting in Tom's cosy den, talking quietly and sipping a glass of his Madeira. I was there one Sunday and we had the usual business of his clock. When the radio gave the time at the hour, the Ormolu antique was exactly 3 minutes slow. "It loses 7 minutes every hour", my old friend told me, as he had done so many times before. "No more and no less, but I've gotten used to it that way." When I spent a second evening with him later that same month, I remarked on the fact that the clock was dead right by radio time at the hour. It was rather late in the evening, but Tom assured me that his treasure had not been adjusted nor fixed since my last visit. What day of the week was the second visit? From "Mathematical Diversions" by Hunter + Madachy ==> arithmetic/clock/day.of.week.s <== The answer is 17 days and 3 hours later, which would have been a Wednesday. This is the only other time in the same month when the two would agree at all. In 17 days the slow clock loses 17*24*7 minutes = 2856 minutes, or 47 hours and 36 minutes. In 3 hours more it loses 21 minutes, so it has lost a total of 47 hours and 57 minutes. Modulo 12 hours, it has *gained* 3 minutes so as to make up the 3 minutes it was slow on Sunday. It is now (fortnight plus 3 days) exactly accurate. ==> arithmetic/clock/thirds.p <== Do the 3 hands on a clock ever divide the face of the clock into 3 equal segments, i.e. 120 degrees between each hand? ==> arithmetic/clock/thirds.s <== First let us assume that our clock has 60 divisions. We will show that any time the hour hand and the minute hand are 20 divisions (120 degrees) apart, the second hand cannot be an integral number of divisions from the other hands, unless it is straight up (on the minute). Let us use h for hours, m for minutes, and s for seconds. We will use =n to mean congruent mod n, thus 12 =5 7. We know that m =60 12h, that is, the minute hand moves 12 times as fast as the hour hand, and wraps around at 60. We also have s =60 60m. This simplifies to s/60 =1 m, which goes to s/60 = frac(m) (assuming s is in the range 0 <= s < 60), which goes to s = 60 frac(m). Thus, if m is 5.5, s is 30. Now let us assume the minute hand is 20 divisions ahead of the hour hand. So m =60 h + 20, thus 12h =60 h + 20, 11h =60 20, and, finally, h =60/11 20/11 (read 'h is congruent mod 60/11 to 20/11'). So all values of m are k + n/11 for some integral k and integral n, 0 <= n < 11. s is therefore 60n/11. If s is to be an integral number of units from m and h, we must have 60n =11 n. But 60 and 11 are relatively prime, so this holds only for n = 0. But if n = 0, m is integral, so s is 0. Now assume, instead, that the minute hand is 20 divisions behind the hour hand. So m =60 h - 20, 12h =60 h - 20, 11h =60 -20, h =60/11 -20/11. So m is still k + n/11. Thus s must be 0. But if s is 0, h must be 20 or 40. But this translates to 4 o'clock or 8 o'clock, at both of which the minute hand is at 0, along with the second hand. Thus the 3 hands can never be 120 degrees apart, Q.E.D. ==> arithmetic/consecutive.product.p <== Prove that the product of three or more consecutive natural numbers cannot be a perfect square. ==> arithmetic/consecutive.product.s <== Three consecutive numbers: If a and b are relatively prime, and ab is a square, then a and b are squares. (This is left as an exercise.) Suppose (n - 1)n(n + 1) = k^2, where n > 1. Then n(n^2 - 1) = k^2. But n and (n^2 - 1) are relatively prime. Therefore n^2 - 1 is a perfect square, which is a contradiction. Four consecutive numbers: n(n + 1)(n + 2)(n + 3) = (n^2 + 3n + 1)^2 - 1 Five consecutive numbers: Assume the product is a integer square, call it m. The prime factorization of m must have even numbers of each prime factor. For each prime factor, p, of m, p >= 5, p^2k must divide one of the consecutive naturals in the product. (Otherwise, the difference between two of the naturals in the product would be a positive multiple of a prime >= 5. But in this problem, the greatest difference is 4.) So we need only consider the primes 2 and 3. Each of the consecutive naturals is one of: 1) a perfect square 2) 2 times a perfect square 3) 3 times a perfect square 4) 6 times a perfect square. By the shoe box principle, two of the five consecutive numbers must fall into the same category. If there are two perfect squares, then their difference being less than five limits their values to be 1 and 4. (0 is not a natural number, so 0 and 1 and 0 and 4 cannot be the perfect squares.) But 1*2*3*4*5=120]=x*x where x is an integer. If there are two numbers that are 2 times a perfect square, then their difference being less than five implies that the perfect squares (which are multiplied by 2) are less than 3 apart, and no two natural squares differ by only 1 or 2. A similar argument holds for two numbers which are 3 times a perfect square. We cannot have the case that two of the 5 consecutive numbers are multiples (much less square multiples) of 6, since their difference would be >= 6, and our span of five consecutive numbers is only 4. Therefore the assumption that m is a perfect square does not hold. QED. In general the equation: y^2 = x(x+1)(x+2)...(x+n), n > 3 has only the solution corresponding to y = 0. This is a theorem of Rigge ÝO. Rigge, ``Uber ein diophantisches Problem'', IX Skan. Math. Kong. Helsingfors (1938)¨ and Erdos ÝP. Erdos, ``Note on products of consecutive integers,'' J. London Math. Soc. #14 (1939), pages 194-198¨. A proof can be found on page 276 of ÝL. Mordell, ``Diophantine Equations'', Academic Press 1969¨. ==> arithmetic/consecutive.sums.p <== Find all series of consecutive positive integers whose sum is exactly 10,000. ==> arithmetic/consecutive.sums.s <== Generalize to find X (and I) such that (X + X+1 + X+2 + ... + X+I) = T for any integer T. You are asking for all (X,I) s.t. (2X+I)(I+1) = 2T. The problem is (very) slightly easier if we don't restrict X to being positive, so we'll solve this first. Note that 2X+I and I+1 must have different parities, so the answer to the relaxed question is N = 2*(o_1+1)*(o_2+1)*...*(o_n+1), where 2T = 2^o_0*3^o_1*...*p_n^o_n (the prime factorization); this is easily seen to be the number of ways we can break 2T up into two positive factors of differing parity (with order). In particular, 20000 = 2^5*5^4, hence there are 2*(4+1) = 10 solutions for T = 10000. These are (2X+I,I+1): (32*1,5^4) (32*5,5^3) (32*5^2,5^2) (32*5^3,5) (32*5^4,1) (5^4,32*1) (5^3,32*5) (5^2,32*5^2) (5,32*5^3) (1,32*5^4) And they give rise to the solutions (X,I): (-296,624) (28,124) (388,24) (1998,4) (10000,0) (297,31) (-27,179) (-387,799) (-1997,3999) (-9999,19999) If you require that X>0 note that this is true iff 2X+I > I+1 and hence the number of solutions to this problem is N/2 (due to the symmetry of the above ordered pairs). ==> arithmetic/digits/all.ones.p <== Prove that some multiple of any integer ending in 3 contains all 1s. ==> arithmetic/digits/all.ones.s <== Let n be our integer; one such desired multiple is then ( 10^(phi(n))-1 )/9. All we need is that (n,10) = 1, and if the last digit is 3 this must be the case. A different proof using the pigeonhole principle is to consider the sequence 1, 11, 111, ..., (10^n - 1)/9. By previous reasoning we must have at some point that either some member of our sequence = 0 (mod n) or else some value (mod n) is duplicated. Assume the latter, with x_a and x_b, x_b>x_a, possesing the duplicated remainders. We then have that x_b - x-a = 0 (mod n). Let m be the highest power of 10 dividing x_b - x_a. Now since (10,n) = 1, we can divide by 10^m and get that (x_b - x_a)/10^m = 0 (n). But (x_b - x_a)/10^m is a number containing only the digit 1. Q.E.D. ==> arithmetic/digits/arabian.p <== What is the Arabian Nights factorial, the number x such that x] has 1001 digits? How about the prime x such that x] has exactly 1001 zeroes on the tail end. (Bonus question, what is the 'rightmost' non-zero digit in x]?) ==> arithmetic/digits/arabian.s <== The first answer is 450]. Determining the number of zeroes at the end of x] is relatively easy once you realize that each such zero comes from a factor of 10 in the product 1 * 2 * 3 * ... * x Each factor of 10, in turn, comes from a factor of 5 and a factor of 2. Since there are many more factors of 2 than factors of 5, the number of 5s determines the number of zeroes at the end of the factorial. The number of 5s in the set of numbers 1 .. x (and therefore the number of zeroes at the end of x]) is: z(x) = int(x/5) + int(x/25) + int(x/125) + int(x/625) + ... This series terminates when the powers of 5 exceed x. I know of no simple way to invert the above formula (i.e., to find x for a given z(x)), but I can approximate it by noting that, except for the "int" function, 5*z(x) - x = z(x) which gives: x = 4*z(x) (approximately). The given problem asked, "For what prime x is z(x)=1001". By the above forumla, this is approximately 4*1001=4004. However, 4004] has only 800 + 160 + 32 + 6 + 1 = 999 zeroes at the end of it. The numbers 4005] through 4009] all have 1000 zeroes at their end and the numbers 4010] through 4014] all have 1001 zeroes at their end. Since the problem asked for a prime x, and 4011 = 3*7*191, the only solution is x=4013. The problem of determining the rightmost nonzero digit in x] is somewhat more difficult. If we took the numbers 1, 2, ... , x and removed all factors of 5 (and an equal number of factors of 2), the remaining numbers multiplied together modulo 10 would be the answer. Note that since there are still many factors of 2 left, the rightmost nonzero digit must be 2, 4, 6, or 8 (x > 1). Letting r(x) be the rightmost nonzero digit in x], an expression for r(x) is: r(x) = (r(int(x/5)) * w * r(x mod 10)) mod 10, x >= 10. where w is 4 if int(x/10) is odd and 6 if it is even. The values of r(x) for 0 <= x <= 9 are 1, 1, 2, 6, 4, 2, 2, 4, 2, and 8. The way to see this is true is to take the numbers 1, 2, ..., x in groups of 10. In each group, remove 2 factors of 10. For example, from the set 1, 2, ..., 10, choose a factor of 2 from 2 and 6 and a factor of 5 from 5 and 10. This leaves 1, 1, 3, 4, 1, 3, 7, 8, 9, 2. Next, separate all the factors that came from multiples of 5. The rightmost nonzero digit of x] can now (hopefully) be seen to be: r(x) = (r(int(x/5)) * w * r(x mod 10)) mod 10 where w is the rightmost digit in the number formed by multiplying the numbers 1, 2, 3, ..., 10*int(x/10) after the factors of 10 and the factors left over by the multiples of 5 have been removed. In the example with x = 10, this would be (1 * 1 * 3 * 4 * 3 * 7 * 8 * 9) mod 10 = 4. The "r(x mod 10)" term takes care of the numbers from 10*int(x/10)+1 up to x. The "w" term can be seen to be 4 or 6 depending on whether int(x/10) is odd or even since, after removing 10*n+5 and 10*n+10 and a factor of 2 each from 10*n+2 and 10*n+6 from the group of numbers 10*n+1 through 10*n+10, the remaining factors (mod 10) always equals 4 and 4^t mod 10 = 4 if t is odd and 6 when t is even (t ]= 0). So, finally, the rightmost nonzero digit in 4013] is found as follows: r(4013) = (r(802) * 4 * 6) mod 10 r(802) = (r(160) * 6 * 2) mod 10 r(160) = (r(32) * 6 * 1) mod 10 r(32) = (r(6) * 4 * 2) mod 10 Using a table of r(x) for 0 <= x <= 9, r(6) = 2. Then, r(32) = (2 * 4 * 2) mod 10 = 6 r(160) = (6 * 6 * 1) mod 10 = 6 r(802) = (6 * 6 * 2) mod 10 = 2 r(4013) = (2 * 4 * 6) mod 10 = 8 Thus, the rightmost nonzero digit in 4013] is 8. ==> arithmetic/digits/circular.p <== What 6 digit number, with 6 different digits, when multiplied by all integers up to 6, circulates its digits through all 6 possible positions, as follows: ABCDEF * 1 = ABCDEF ABCDEF * 3 = BCDEFA ABCDEF * 2 = CDEFAB ABCDEF * 6 = DEFABC ABCDEF * 4 = EFABCD ABCDEF * 5 = FABCDE ==> arithmetic/digits/circular.s <== ABCDEF=142857 (the digits of the expansion of 1/7). ==> arithmetic/digits/divisible.p <== Find the least number using 0-9 exactly once that is evenly divisible by each of these digits? ==> arithmetic/digits/divisible.s <== Since the sum of the digits is 45, any permutation of the digits gives a multiple of 9. To get a multiple of both 2 and 5, the last digit must be 0, and thus to get a multiple of 8 (and 4), the tens digit must be even, and the hundreds digit must be odd if the tens digit is 2 or 6, and even otherwise. The number will also be divisible by 6, since it is divisible by 2 and 3, so 7 is all we need to check. First, we will look for a number whose first five digits are 12345; now, 1234500000 has a remainder of 6 when divided by 7, so we have to arrange the remaining digits to get a remainder of 1. The possible arrangements, in increasing order, are 78960, remainder 0 79680, remainder 6 87960, remainder 5 89760, remainder 6 97680, remainder 2 98760, remainder 4 That didn't work, so try numbers starting with 12346; this is impossible because the tens digit must be 8, and the hundreds digit cannot be even. Now try 12347, and 1234700000 has remainder 2. The last five digits can be 58960, remainder 6 59680, remainder 5, so this works, and the number is 1234759680. ==> arithmetic/digits/equations/123456789.p <== In how many ways can "." be replaced with "+", "-", or "" (concatenate) in .1.2.3.4.5.6.7.8.9=1 to form a correct equation? ==> arithmetic/digits/equations/123456789.s <== 1-2 3+4 5+6 7-8 9 = 1 +1-2 3+4 5+6 7-8 9 = 1 1+2 3+4-5+6 7-8 9 = 1 +1+2 3+4-5+6 7-8 9 = 1 -1+2 3-4+5+6 7-8 9 = 1 1+2 3-4 5-6 7+8 9 = 1 +1+2 3-4 5-6 7+8 9 = 1 1-2 3-4+5-6 7+8 9 = 1 +1-2 3-4+5-6 7+8 9 = 1 1-2-3-4 5+6 7-8-9 = 1 +1-2-3-4 5+6 7-8-9 = 1 1+2-3 4+5 6-7-8-9 = 1 +1+2-3 4+5 6-7-8-9 = 1 -1+2 3+4+5-6-7-8-9 = 1 -1 2+3 4-5-6+7-8-9 = 1 1+2+3+4-5+6+7-8-9 = 1 +1+2+3+4-5+6+7-8-9 = 1 -1+2+3-4+5+6+7-8-9 = 1 1-2-3+4+5+6+7-8-9 = 1 +1-2-3+4+5+6+7-8-9 = 1 1+2 3+4 5-6 7+8-9 = 1 +1+2 3+4 5-6 7+8-9 = 1 1+2 3-4-5-6-7+8-9 = 1 +1+2 3-4-5-6-7+8-9 = 1 1+2+3+4+5-6-7+8-9 = 1 +1+2+3+4+5-6-7+8-9 = 1 -1+2+3+4-5+6-7+8-9 = 1 1-2+3-4+5+6-7+8-9 = 1 +1-2+3-4+5+6-7+8-9 = 1 -1-2-3+4+5+6-7+8-9 = 1 1-2+3+4-5-6+7+8-9 = 1 +1-2+3+4-5-6+7+8-9 = 1 1+2-3-4+5-6+7+8-9 = 1 +1+2-3-4+5-6+7+8-9 = 1 -1-2+3-4+5-6+7+8-9 = 1 -1+2-3-4-5+6+7+8-9 = 1 -1+2 3+4 5-6 7-8+9 = 1 1-2 3-4 5+6 7-8+9 = 1 +1-2 3-4 5+6 7-8+9 = 1 -1+2 3-4-5-6-7-8+9 = 1 -1+2+3+4+5-6-7-8+9 = 1 1-2+3+4-5+6-7-8+9 = 1 +1-2+3+4-5+6-7-8+9 = 1 1+2-3-4+5+6-7-8+9 = 1 +1+2-3-4+5+6-7-8+9 = 1 -1-2+3-4+5+6-7-8+9 = 1 1+2-3+4-5-6+7-8+9 = 1 +1+2-3+4-5-6+7-8+9 = 1 -1-2+3+4-5-6+7-8+9 = 1 -1+2-3-4+5-6+7-8+9 = 1 1-2-3-4-5+6+7-8+9 = 1 +1-2-3-4-5+6+7-8+9 = 1 1-2 3+4+5+6+7-8+9 = 1 +1-2 3+4+5+6+7-8+9 = 1 1+2+3+4 5-6 7+8+9 = 1 +1+2+3+4 5-6 7+8+9 = 1 1 2+3 4+5-6 7+8+9 = 1 +1 2+3 4+5-6 7+8+9 = 1 1+2+3-4-5-6-7+8+9 = 1 +1+2+3-4-5-6-7+8+9 = 1 -1+2-3+4-5-6-7+8+9 = 1 1-2-3-4+5-6-7+8+9 = 1 +1-2-3-4+5-6-7+8+9 = 1 -1-2-3-4-5+6-7+8+9 = 1 -1-2 3+4+5+6-7+8+9 = 1 1-2+3 4-5 6+7+8+9 = 1 +1-2+3 4-5 6+7+8+9 = 1 1 2-3 4+5-6+7+8+9 = 1 +1 2-3 4+5-6+7+8+9 = 1 Total solutions = 69 69/19683 = 0.35 % for those who care (it's not very elegant but it did the trick): #include #include main (argc,argv) int argc; char *argvݨ; { int sresult, result, operatorÝ10¨,thisop; char bufÝ256¨,opsÝ3¨; int i,j,tot=0,temp; opsÝ0¨ = ' '; opsÝ1¨ = '-'; opsÝ2¨ = '+'; for (i=1; i<10; i++) operatorÝi¨ = 0; for (j=0; j < 19683; j++) { result = 0; sresult = 0; thisop = 1; for (i=1; i<10; i++) { switch (operatorÝi¨) { case 0: sresult = sresult * 10 + i; break; case 1: result = result + sresult * thisop; sresult = i; thisop = -1; break; case 2: result = result + sresult * thisop; sresult = i; thisop = 1; break; } } result = result + sresult * thisop; if (result == 1) { tot++; for (i=1;i<10;i++) printf("%c%d",opsÝoperatorÝi¨¨,i); printf(" = %d\n",result); } temp = 0; operatorÝ1¨ += 1; for (i=1;i<10;i++) { operatorÝi¨ += temp; if (operatorÝi¨ > 2) { operatorÝi¨ = 0; temp = 1;} else temp = 0; } } printf("Total solutions = %d\n" , tot); } cwren@media.mit.edu (Christopher Wren) ==> arithmetic/digits/equations/1992.p <== 1 = -1+9-9+2. Extend this list to 2 - 100 on the left side of the equals sign. ==> arithmetic/digits/equations/1992.s <== 1 = -1+9-9+2 2 = 1*9-9+2 3 = 1+9-9+2 4 = 1+9/9+2 5 = 1+9-sqrt(9)-2 6 = 1^9+sqrt(9)+2 7 = -1+sqrt(9)+sqrt(9)+2 8 = 19-9-2 9 = (1/9)*9^2 10= 1+(9+9)/2 11= 1+9+sqrt(9)-2 12= 19-9+2 13= (1+sqrt(9))]-9-2 14= 1+9+sqrt(9)]-2 15= -1+9+9-2 16= -1+9+sqrt(9)]+2 17= 1+9+9-2 18= 1+9+sqrt(9)]+2 19= -1+9+9+2 20= (19-9)*2 21= 1+9+9+2 22= (-1+sqrt(9))*(9-2) 23= (1+sqrt(9))]-sqrt(9)+2 24= -1+9*sqrt(9)-2 25= 1*9*sqrt(9)-2 26= 19+9-2 27= 1*9+9*2 28= 1+9+9*2 29= 1*9*sqrt(9)+2 30= 19+9+2 31= (1+sqrt(9))]+9-2 32= -1+sqrt(9)*(9+2) 33= 1*sqrt(9)*(9+2) 34= (-1+9+9)*2 35= -1+(9+9)*2 36= 1^9*sqrt(9)]^2 37= 19+9*2 38= 1*sqrt(9)]*sqrt(9)]+2 39= 1+sqrt(9)]*sqrt(9)]+2 40= (1+sqrt(9)])*sqrt(9)]-2 41= -1+sqrt(9)]*(9-2) 42= (1+sqrt(9))]+9*2 43= 1+sqrt(9)]*(9-2) 44= -1+9*(sqrt(9)+2) 45= 1*9*(sqrt(9)+2) 46= 1+9*(sqrt(9)+2) 47= (-1+sqrt(9)])*9+2 48= 1*sqrt(9)]*(sqrt(9)]+2) 49= (1+sqrt(9)])*(9-2) 50= (-1+9)*sqrt(9)]+2 51= -1+9*sqrt(9)]-2 52= 1*9*sqrt(9)]-2 53= -1+9*sqrt(9)*2 54= 1*9*sqrt(9)*2 55= 1+9*sqrt(9)*2 56= 1*9*sqrt(9)]+2 57= 1+9*sqrt(9)]+2 58= (1+9)*sqrt(9)]-2 59= 19*sqrt(9)+2 60= (1+9)*sqrt(9)*2 61= (1+sqrt(9)])*9-2 62= -1+9*(9-2) 63= 1*9*(9-2) 64= 1+9*(9-2) 65= (1+sqrt(9)])*9+2 66= 1*sqrt(9)]*(9+2) 67= 1+sqrt(9)]*(9+2) 68= -(1+sqrt(9))]+92 69= (1+sqrt(9))]+(9/.2) 70= (1+9)*(9-2) 71= -1-9+9^2 72= (1+sqrt(9))*9*2 73= -19+92 74= (-1+9)*9+2 75= -1*sqrt(9)]+9^2 76= 1-sqrt(9)]+9^2 77= (1+sqrt(9)])*(9+2) 78= -1+9*9-2 79= 1*9*9-2 80= 1+9*9-2 81= 1*9*sqrt(9)^2 82= -1+9*9+2 83= 1*9*9+2 84= 1+9*9+2 85= -1-sqrt(9)]+92 86= -1*sqrt(9)]+92 87= 1-sqrt(9)]+92 88= (1+9)*9-2 89= -1*sqrt(9)+92 90= 1-sqrt(9)+92 91= -1^9+92 92= (1+9)*9+2 93= 1^9+92 94= -1+sqrt(9)+92 95= 19*(sqrt(9)+2) 96= -1+99-2 97= 1*99-2 98= 1+99-2 99= 1*9*(9+2) 100= -1+99+2 ==> arithmetic/digits/equations/383.p <== Make 383 out of 1,2,25,50,75,100 using +,-,*,/. ==> arithmetic/digits/equations/383.s <== You can get 383 with ((2+50)/25+1)*100+75. Of course, if you expect / as in C, the above expression is just 375. But then you can get 383 with (25*50-100)/(1+2). Pity there's no way to use the 75. If we had a language that rounded instead of truncating, we could use ((1+75+100)*50)/(25-2) or (2*75*(25+100))/(50-1). I imagine your problem lies in not dividing things that aren't divisible. Dan Hoey Hoey@AIC.NRL.Navy.Mil ==> arithmetic/digits/extreme.products.p <== What are the extremal products of three three-digit numbers using digits 1-9? ==> arithmetic/digits/extreme.products.s <== There is a simple procedure which applies to these types of problems (and which can be proven with help from the arithmetic-geometric inequality). For the first part we use the "first large then equal" procedure. This means that are three numbers will be 7**, 8**, and 9**. Now the digits 4,5,6 get distributed so as to make our three number as close to each other as possible, i.e. 76*, 85*, 94*. The same goes for the remaining three digits, and we get 763, 852, 941. For the second part we use the "first small then different" procedure. Our three numbers will be of the form 1**, 2**, 3**. We now place the three digits so as to make our three numbers as unequal as possible; this gives 14*, 25*, 36*. Finishing, we get 147, 258, 369. Now, *prove* that these procedures work for generalizations of this problem. ==> arithmetic/digits/googol.p <== What digits does googol] start with? ==> arithmetic/digits/googol.s <== I'm not sure how to calculate the first googol of digits of log10(e), but here's the first 150(approximately) of them... 0.43429448190325182765112891891660508229439700580366656611445378316586464920 8870774729224949338431748318706106744766303733641679287158963906569221064663 We need to deal with the digits immediately after the decimal point in googol*log10(e), which are .187061 fracÝlog(googol])¨ = fracÝhalflog2pi + 50 + googol(100-log10(e))¨ = frac{halflog2pi + fracÝgoogol(100-log10(e))¨} = fracÝ.399090 + (1- .187061)¨ = .212029 10 ** .212029 = 1.629405 Which means that googol] starts with 1629 ==> arithmetic/digits/labels.p <== You have an arbitrary number of model kits (which you assemble for fun and profit). Each kit comes with twenty (20) stickers, two of which are labeled "0", two are labeled "1", ..., two are labeled "9". You decide to stick a serial number on each model you assemble starting with one. What is the first number you cannot stick. You may stockpile unused numbers on already assembled models, but you may not crack open a new model to get at its stickers. You complete assembling the current model before starting the next. ==> arithmetic/digits/labels.s <== The method I used for this problem involved first coming up with a formula that says how many times a digit has been used in all n models. n = k*10^i + m for some k,m with 0 2*n. ==> arithmetic/digits/nine.digits.p <== Form a number using 0-9 once with its first n digits divisible by n. ==> arithmetic/digits/nine.digits.s <== First, reduce the sample set. For each digit of ABCDEFGHI, such that the last digit, (current digit), is the same as a multiple of N : A: Any number 1-9 B: Even numbers 2,4,6,8 (divisible by 2). C: Any number 1-9 (21,12,3,24,15,6,27,18,9). D: Even numbers 2,4,6,8 (divisible by 4, every other even). E: 5 (divisible by 5 and 0 not allowed). F: Even numbers (12,24,6,18) G: Any number 1-9 (21,42,63,14,35,56,7,28,49). H: Even numbers (32,24,16,8) I: Any number 1-9 (81,72,63,54,45,36,27,18,9) Since E must be 5, I can eliminate it everywhere else. Since I will use up all the even digits, (2,4,6,8) filling in those spots that must be even. Any number becomes all odds, except 5. A: 1,3,7,9 B: 2,4,6,8 C: 1,3,7,9 D: 2,4,6,8 E: 5 F: 2,4,6,8 G: 1,3,7,9 H: 2,4,6,8 I: 1,3,7,9 We have that 2C+D=0 (mod 4), and since C is odd, this implies that D is 2 or 6; similarly we find that H is 2 or 6 ==> {B,F} = {4,8}. D+5+F=0 (mod 3) ==> if D=2 then F=8, if D=6 then F=4. We have two cases. Assume our number is of the form A4C258G6I0. Now the case n=8 ==> G=1,9; case n=3 ==> A+1+C=0 (mod 3) ==> {A,C}={1,7} ==> G=9, I=3. The two numbers remaining fail for n=7. Assume our number is of the form A8C654G2I0. The case n=8 ==> G=3,7. If G=3, we need to check to see which of 1896543, 9816543, 7896543, and 9876543 are divisible by 7; none are. If G=7, we need to check to see which of 1896547, 9816547, 1836547, and 3816547 are divisible by 7; only the last one is, which yields the solution 3816547290. ==> arithmetic/digits/palindrome.p <== Does the series formed by adding a number to its reversal always end in a palindrome? ==> arithmetic/digits/palindrome.s <== This is not known. If you start with 196, after 9480000 iterations you get a 3924257-digit non-palindromic number. However, there is no known proof that you will never get a palindrome. The statement is provably false for binary numbers. Roland Sprague has shown that 10110 starts a series that never goes palindromic. ==> arithmetic/digits/palintiples.p <== Find all numbers that are multiples of their reversals. ==> arithmetic/digits/palintiples.s <== We are asked to find numbers that are integer multiples of their reversals, which I call palintiples. Of course, all the palindromic numbers are a trivial example, but if we disregard the unit multiples, the field is narrowed considerably. Rouse Ball (_Mathematical_recreations_and_essays_) originated the problem, and G. H. Hardy (_A_mathematician's_apology_) used the result that 9801 and 8712 are the only four-digit palintiples as an example of a theorem that is not ``serious''. Martin Beech (_The_mathema- tical_gazette_, Vol 74, #467, pp 50-51, March '90) observed that 989*01 and 879*12 are palintiples, an observation he ``confirmed'' on a hand calculator, and conjectured that these are all that exist. I confirm that Beech's numbers are palintiples, I will show that they are not all of the palintiples. I will show that the palintiples do not form a regular language. And then I will prove that I have found all the palintiples, by describing the them with a generalized form of regular expression. The results become more interesting in other bases. First, I have a more reasonable method of confirming that these numbers are palintiples: Proof: First, letting "9*" and "0*" refer an arbitrary string of nines and a string of zeroes of the same length, I note that 879*12 = 879*00 + 12 = (880*00 - 100) + 12 = 880*00 - 88 219*78 = 219*00 + 78 = (220*00 - 100) + 78 = 220*00 - 22 989*01 = 989*00 + 1 = (990*00 - 100) + 1 = 990*00 - 99 109*89 = 109*00 + 89 = (110*00 - 100) + 89 = 110*00 - 11 It is obvious that 4x(220*00 - 22) = 880*00 - 88 and that 9x(110*00 - 11) = 990*00 - 99. QED. Now, to show that these palintiples are not all that exist, let us take the (infinite) language LÝ4¨ = (879*12 + 0*), and let Pal(LÝ4¨) refer to the set of palindromes over the alphabet LÝ4¨. It is immediate that the numbers in Pal(LÝ4¨) are palintiples. For instance, 8712 000 87912 879999912 879999912 87912 000 8712 = 4 x 2178 000 21978 219999978 219999978 21978 000 2178 (where I have inserted spaces to enhance readability) is a palintiple. Similarly, taking LÝ9¨ = (989*01 + 0*), the numbers in Pal(LÝ9¨) are palintiples. We exclude numbers starting with zeroes. The reason these do not form a regular language is that the sub-palintiples on the left end of the number must be the same (in reverse order) as the sub-palintiples on the right end of the number: 8712 8712 87999912 = 4 x 2178 2178 21999978 is not a palintiple, because 8712 8712 87999912 is not the reverse of 2178 2178 21999978. The pumping lemma can be used to prove that Pal(LÝ4¨)+Pal(LÝ9¨) is not a regular language, just as in the familiar proof that the palindromes over a non-singleton alphabet do not form a regular language. Now to characterize all the palintiples, let N be a palintiple, N=CxR(N), where R(.) signifies reversal, and C>1 is an integer. (I use "x" for multiplication, to avoid confusion with the Kleene star "*", which signifies the concatenated closure.) If D is a digit of N, let D' refer to the corresponding digit of R(N). Since N=CxR(N), D+10T = CxD'+S, where S is the carry in to the position occupied by D' when R(N) is multiplied by C, and T is the carry out of that position. Similarly, D'+10T'=CxD+S', where S', T' are carries in and out of the position occupied by D when R(N) is multiplied by C. Since D and D' are so closely related, I will use the symbol D:D' to refer to a digit D on the left side of a string with a corresponding digit D' on the right side of the string. More formally, an expression "xÝ1¨:yÝ1¨ xÝ2¨:yÝ2¨ ... xÝn¨:yÝn¨ w" will refer to a string "xÝ1¨ xÝ2¨ ... xÝn¨ w yÝn¨ ... yÝ2¨ yÝ1¨", where the xÝi¨ and yÝi¨ are digits and w is a string of zero or one digits. So 989901 may be written as 9:1 8:0 9:9 and 87912 may be written as 8:2 7:1 9. Thus Pal(LÝ4¨)+Pal(LÝ9¨) (omitting numbers with leading zeroes) can be represented as (8:2 7:1 9:9* 1:7 2:8 0:0*)* (0:0* + 0 + 8:2 7:1 ( 9:9* + 9:9* 9)) + (9:1 8:0 9:9* 0:8 1:9 0:0*)* (0:0* + 0 + 9:1 8:0 ( 9:9* + 9:9* 9)). (1) For each pair of digits D:D', there are a very limited--and often empty--set of quadruples S,T,S',T' of digits that satisfy the equations D +10T =CxD'+S D'+10T'=CxD +S', (2) yet such a quadruple must exist for "D:D'" to appear in a palintiple with multiplier C. Furthermore, the S and T' of one D:D' must be T and S', respectively, of the next pair of digits that appear. This enables us to construct a finite state machine to recognize those palintiples. The states ÝX#Y¨ refer to a pair of carries in D and D', and we allow a transition from state ÝT#S'¨ to state ÝS#T'¨ on input symbol D:D' exactly when equations (2) are satisfied. Special transitions for a single-digit input symbol (the central digit of odd-length palintiples) and the criteria for the initial and the accepting states are left as exercises. The finite state machines thus formed are State Symbol New Symbol New Symbol New Accept? State State State --> Ý0#0¨ Y 8:2 Ý0#3¨ 0:0 Ý0#0¨ 0 ÝA¨ Ý0#3¨ N 7:1 Ý3#3¨ Ý3#3¨ Y 1:7 Ý3#0¨ 9:9 Ý3#3¨ 9 ÝA¨ Ý3#0¨ N 2:8 Ý0#0¨ ÝA¨ Y for constant C=4, and State Symbol New Symbol New Symbol New Accept? State State State --> Ý0#0¨ Y 1:9 Ý0#8¨ 0:0 Ý0#0¨ 0 ÝA¨ Ý0#8¨ N 8:0 Ý8#8¨ Ý8#8¨ Y 0:8 Ý8#0¨ 9:9 Ý8#8¨ 9 ÝA¨ Ý8#0¨ N 9:1 Ý0#0¨ ÝA¨ Y for constant C=9, and the finite state machines for other constants accept only strings of zeroes. It is not hard to verify that the proposed regular expression (1) represents the union of the languages accepted by these machines, omitting the empty string and strings beginning with zero. I have written a computer program that constructs finite state machines for recognizing palintiples for various bases and constants. I found that base 10 is actually an unusually boring base for this problem. For instance, the machine for base 8, constant C=5 is State Symbol New Symbol New Symbol New Accept? State State State --> Ý0#0¨ Y 0:0 Ý0#0¨ 5:1 Ý0#3¨ 0 ÝA¨ Ý0#3¨ N 1:0 Ý1#1¨ 6:1 Ý1#4¨ Ý1#1¨ Y 0:1 Ý3#0¨ 5:2 Ý3#3¨ Ý3#0¨ N 1:5 Ý0#0¨ 6:6 Ý0#3¨ 6 ÝA¨ Ý3#3¨ Y 2:5 Ý1#1¨ 7:6 Ý1#4¨ Ý1#4¨ N 1:1 Ý4#1¨ 6:2 Ý4#4¨ 1 ÝA¨ Ý4#4¨ Y 2:6 Ý4#1¨ 7:7 Ý4#4¨ 7 ÝA¨ Ý4#1¨ N 1:6 Ý3#0¨ 6:7 Ý3#3¨ ÝA¨ Y for which I invite masochists to write the regular expression. If anyone wants more, I should remark that the base 29 machine for constant C=18 has 71 states] By the way, I did not find any way of predicting the size or form of the machines for the various bases, except that the machines for C=B-1 all seem to be isomorphic to each other. If anyone investigates the general behavior, I would be most happy to hear about it. Dan Hoey Hoey@AIC.NRL.Navy.Mil May, 1992 Ý A preliminary version of this message appeared in April, 1991. ¨ ================================================================ Dan ==> arithmetic/digits/power.two.p <== Prove that for any 9-digit number (base 10) there is an integral power of 2 whose first 9 digits are that number. ==> arithmetic/digits/power.two.s <== Let v = log to base 10 of 2. Then v is irrational. Let w = log to base 10 of these 9 digits. Since v is irrational, given epsilon > 0, there exists some natural number n such that {w} < {nv} < {w} + epsilon ({x} is the fractional part of x.) Let us pick n for when epsilon = log 1.00000000000000000000001. Then 2^n does the job. ==> arithmetic/digits/prime/101.p <== How many primes are in the sequence 101, 10101, 1010101, ...? ==> arithmetic/digits/prime/101.s <== Note that the sequence 101 , 10101, 1010101, .... can be viewed as 100**1 +1, 100**2 + 100**1 + 1, 100**3 + 100**2 + 100**1 +1 .... that is, the k-th term in the sequence is 100**k + 100**(k-1) + 100**(k-2) + ...+ 100**(1) + 1 = (100)**(k+1) - 1 ---------------- 11 * 9 = (10)**(2k+2) - 1 ---------------- 11 * 9 = ((10)**(k+1) - 1)*((10)**(k+1) +1) --------------------------------- 11*9 thus either 11 and 9 divide the numerator. Either they both divide the same factor in the numerator or different factors in the numerator. In any case, after dividing, they leave the numerators as a product of two integers. Only in the case of k = 1, one of the integers is 1. Thus there is exactly one prime in the above sequence: 101. ==> arithmetic/digits/prime/all.prefix.p <== What is the longest prime whose every proper prefix is a prime? ==> arithmetic/digits/prime/all.prefix.s <== 23399339, 29399999, 37337999, 59393339, 73939133 ==> arithmetic/digits/prime/change.one.p <== What is the smallest number that cannot be made prime by changing a single digit? Are there infinitely many such numbers? ==> arithmetic/digits/prime/change.one.s <== 200. Obviously, you would have to change the last digit, but 201, 203, 207, and 209 are all composite. For any smaller number, you can change the last digit, and get 2,11,23,31,41,53,61,71,83,97,101,113,127,131,149,151,163,173,181, or 191. 200+2310n gives an infinite family, because changing the last digit to 1 or 7 gives a number divisible by 3; to 3, a number divisible by 7; to 9, a number divisible by 11. ==> arithmetic/digits/prime/prefix.one.p <== 2 is prime, but 12, 22, ..., 92 are not. Similarly, 5 is prime whereas 15, 25, ..., 95 are not. What is the next prime number which is composite when any digit is prefixed? ==> arithmetic/digits/prime/prefix.one.s <== 149 ==> arithmetic/digits/reverse.p <== Is there an integer that has its digits reversed after dividing it by 2? ==> arithmetic/digits/reverse.s <== Assume there's such a positive integer x such that x/2=y and y is the reverse of x. Then x=2y. Let x = a...b, then y = b...a, and: b...a (y) x 2 -------- a...b (x) From the last digit b of x, we have b = 2a (mod 10), the possible values for b are 2, 4, 6, 8 and hence possible values for (a, b) are (1,2), (6,2), (2,4), (7,4), (3,6), (8,6), (4,8), (9,8). From the first digit a of x, we have a = 2b or a = 2b+1. None of the above pairs satisfy this condition. A contradiction. Hence there's no such integer. ==> arithmetic/digits/rotate.p <== Find integers where multiplying them by single digits rotates their digits. ==> arithmetic/digits/rotate.s <== 2 105263157894736842 3 1034482758620689655172413793 4 102564 153846 179487 205128 230769 5 142857 102040816326530612244897959183673469387755 6 1016949152542372881355932203389830508474576271186440677966 1186440677966101694915254237288135593220338983050847457627 1355932203389830508474576271186440677966101694915254237288 1525423728813559322033898305084745762711864406779661016949 7 1014492753623188405797 1159420289855072463768 1304347826086956521739 8 1012658227848 1139240506329 9 10112359550561797752808988764044943820224719 In base B, suppose you have an N-digit answer A whose digits are rotated when multiplied by K. If D is the low-order digit of A, we have (A-D)/B + D B^(N-1) = K A . Solving this for A we have D (B^N - 1) A = ----------- . B K - 1 In order for A >= B^(N-1) we must have D >= K. Now we have to find N such that B^N-1 is divisible by R=(BK-1)/gcd(BK-1,D). This always has a minimal solution N0(R,B) arithmetic/digits/sesqui.p <== Find the least number where moving the first digit ==> arithmetic/digits/sesqui.s <== Let's represent this number as a*10^n+b, where 1<=a<=9 and b < 10^n. Then the condition to be satisfied is: 3/2(a*10^n+b) = 10b+a 3(a*10^n+b) = 20b+2a 3a*10^n+3b = 20b+2a (3*10^n-2)a = 17b b = a*(3*10^n-2)/17 So we must have 3*10^n-2 = 0 (mod 17) (since a is less than 10, it cannot contribute the needed prime 17 to the factorization of 17b). (Also, assuming large n, we must have a at most 5 so that b < 10^n will be satisfied, but note that we can choose a=1). Now, 3*10^n-2 = 0 (mod 17) 3*10^n = 2 (mod 17) 10^n = 12 (mod 17) A quick check shows that the smallest n which satisfies this is 15 (the fact that one exists was assured to us because 17 is prime). So, setting n=15 and a=1 (obviously) gives us b=176470588235294, so the number we are looking for is 1176470588235294 and, by the way, we can set a=2 to give us the second smallest such number, 2352941176470588 Other things we can infer about these numbers is that there are 5 of them less than 10^16, 5 more less than 10^33, etc. ==> arithmetic/digits/squares/leading.7.to.8.p <== What is the smallest square with leading digit 7 which remains a square when leading 7 is replaced by an 8? ==> arithmetic/digits/squares/leading.7.to.8.s <== x=2996282391593370361328125 y=2824483699753370361328125 x^2=8977708170172487211329625006796419620513916015625 y^2=7977708170172487211329625006796419620513916015625 ==> arithmetic/digits/squares/length.22.p <== Is it possible to form two numbers A and B from 22 digits such that A = B^2? Of course, leading digits must be non-zero. ==> arithmetic/digits/squares/length.22.s <== No, the number of digits of A^2 must be of the form 3n or 3n-1. ==> arithmetic/digits/squares/length.9.p <== Is it possible to make a number and its square, using the digits from 1 through 9 exactly once? ==> arithmetic/digits/squares/length.9.s <== 567 and 854. ==> arithmetic/digits/squares/three.digits.p <== What squares consist entirely of three digits (e.g., 1, 4, and 9)? ==> arithmetic/digits/squares/three.digits.s <== The full set of solutions up to 10**12 is 1 -> 1 2 -> 4 3 -> 9 7 -> 49 12 -> 144 21 -> 441 38 -> 1444 107 -> 11449 212 -> 44944 31488 -> 9914 94144 70107 -> 49149 91449 3 87288 -> 14 99919 94944 956 10729 -> 9 14141 14499 11441 4466 53271 -> 199 49914 44949 99441 31487 17107 -> 9914 41941 99144 49449 2 10810 79479 -> 4 44411 91199 99149 11441 If the algorithm is used in the form I presented it before, generating the whole set P_n before starting on P_{n+1}, the store requirements begin to become embarassing. For n>8 I switched to a depth-first strategy, generating all the elements in P_i (i=9..12) congruent to a particular x in P_8 for each x in turn. This means the solutions don't come out in any particular order, of course. CPU time was 16.2 seconds (IBM 3084). In article <1990Feb6.025205.28153@sun.soe.clarkson.edu>, Steven Stadnicki suggests alternate triples of digits, in particular {1,4,6} (with many solutions) and {2,4,8} (with few). I ran my program on these as well, up to 10**12 again: 1 -> 1 2 -> 4 4 -> 16 8 -> 64 12 -> 144 21 -> 441 38 -> 1444 108 -> 11664 119 -> 14161 121 -> 14641 129 -> 16641 204 -> 41616 408 -> 1 66464 804 -> 6 46416 2538 -> 64 41444 3408 -> 116 14464 6642 -> 441 16164 12908 -> 1666 16464 25771 -> 6641 44441 78196 -> 61146 14416 81619 -> 66616 61161 3 33858 -> 11 14611 64164 2040 00408 -> 41 61616 64641 66464 6681 64962 -> 446 44441 64444 61444 8131 18358 -> 661 16146 41166 16164 40182 85038 -> 16146 61464 66146 61444 (Steven's last soln.) 1 20068 50738 -> 1 44164 46464 46111 44644 1 26941 38988 -> 1 61141 16464 66616 64144 1 27069 43631 -> 1 61466 41644 14114 64161 4 01822 24262 -> 16 14611 14664 16614 44644 4 05784 63021 -> 16 46611 66114 66644 46441 78 51539 12392 -> 6164 66666 14446 44111 61664 and 2 -> 4 22 -> 484 168 -> 28224 478 -> 2 28484 2878 -> 82 82884 (Steven's last soln.) 2109 12978 -> 44 48428 42888 28484 (so the answer to Steven's "Are there any more at all?" is "Yes".) The CPU times were 42.9 seconds for {1,4,6}, 18.7 for {2,4,8}. This corresponds to an interesting point: the abundance of solutions for {1,4,6} is associated with abnormally large sets P_n (!P_8! = 16088 for {1,4,6} compared to !P_8! = 5904 for {1,4,9}) but the deficiency of solutions for {2,4,8} is *not* associated with small P_n's (!P_8! = 6816 for {2,4,8}). Can anyone wave a hand convincingly to explain why the solutions for {2,4,8} are so sparse? I suspect we are now getting to the point where an improved algorithm is called for. The time to determine all the n-digit solutions (i.e. 2n-digit squares) using this last-significant-digit-first is essentially constant * 3**n. Dean Hickerson in <90036.134503HUL@PSUVM.BITNET>, and Ilan Vardi in <1990Feb5.214249.22811@Neon.Stanford.EDU>, suggest using a most-significant-digit-first strategy, based on the fact that the first n digits of the square determine the (integral) square root; this also has a running time constant * 3**n. Can one attack both ends at once and do better? Chris Thompson JANET: cet1@uk.ac.cam.phx Internet: cet1%phx.cam.ac.uk@nsfnet-relay.ac.uk Hey guys, what about 648070211589107021 ^ 2 = 419994999149149944149149944191494441 This was found by David Applegate and myself (about 5 minutes on a DEC 3100, program in C). This is the largest square less than 10^42 with the 149-property; checking took a bit more than an hour of CPU time. As somebody suggested, we used a combined most-significant/least-significant digits attack. First we make a table of p-digit prefixes (most significant p digits) that could begin a root whose square has the 149 property in its first p digits. We organize this table into buckets by the least significant q digits of the prefixes. Then we enumerate the s digit suffixes whose squares have the 149 property in their last s digits. For each such suffix, we look in the table for those prefixes whose last q digits match the first q of the suffix. For each match, we consider the p + s - q digit number formed by overlapping the prefix and the suffix by q digits. The squares of these overlap numbers must contain all the squares with the 149 property. The time expended is O(3^p) to generate the prefix table, O(3^s) to enumerate the suffixes, and O(3^(p+s) / 10^q) to check the overlaps (being very rough and ignoring the polynomial factors) By judiciously chosing p, q, and s, we can fix things so that each bucket of the table has around O(1) entries: set q = p log10(3). Setting p = s, we end up looking for squares whose roots have n = 2 - log10(3) digits, with an algorithm that takes time O( 3 ^ Ýn / (2 - log10(3)¨) ), roughly time O(3^Ý.66n¨). Compared to the O(3^n) performance of either single-ended algorithm, this lets us check 50% more digits in the same amount of time (ignoring polynomial factors). Of course, the space cost of the combined-ends method is high. -- Guy and Dave -- Guy Jacobson School of Computer Science Carnegie Mellon arpanet : guy@cs.cmu.edu Pittsburgh, PA 15213 csnet : Guy.Jacobson%a.cs.cmu.edu@csnet-relay (412) 268-3056 uucp : ...]{seismo, ucbvax, harvard}]cs.cmu.edu]guy Here is an algorithm which takes O(sqrt(n)log(n)) steps to find all perfect squares < n whose only digits are 1, 4 and 9. This doesn't sound too great *but* it doesn't use a lot of memory and only requires addition and <. Also, the actual run time will depend on where the first non-{1,4,9} digit appears in each square. set n = 1 set odd = 1 while(n < MAXVAL) { if(all digits of n are in {1,4,9}) { print n } add 2 to odd add odd to n } This works because (X+1)^2 - x^2 = 2x+1. That is, if you start with 0 and add successive odd numbers to it you get 0+1=1, 1+3=4, 4+5=9, 9+7=16 etc. I've started the algorithm at 1 for convenience. The "O" value comes from looking at at most all digits (log(n)) of all perfect squares < n (sqrt(n) of them) at most a constant number of times. I didn't save the articles with algorithms claiming to be O(3^log(n)) so I don't know if their calculations needed to (or did) account for multiplication or sqrt() of large numbers. O(3^log(n)) sounds reasonable so I'm going to assume they did unless I hear otherwise. Any comments? Please email if you just want to refresh my memory on the other algorithms. Andrew Charles acgd@ihuxy.ATT.COMM ==> arithmetic/digits/squares/twin.p <== Let a twin be a number formed by writing the same number twice, for instance, 81708170 or 132132. What is the smallest square twin? ==> arithmetic/digits/squares/twin.s <== 1322314049613223140496 = 36363636364 ^ 2. The key to solving this puzzle is looking at the basic form of these "twin" numbers, which is some number k = 1 + 10^n multiplied by some number a < 10^n. If ak is a perfect square, k must have some repeated factor, since a arithmetic/digits/sum.of.digits.p <== Find sod ( sod ( sod (4444 ^ 4444 ) ) ). ==> arithmetic/digits/sum.of.digits.s <== let X = 4444^4444 sod(X) <= 9 * (# of digits) < 145900 sod(sod(X)) <= sod(99999) = 45 sod(sod(sod(X))) <= sod(39) = 12 but sod(sod(sod(X))) = 7 (mod 9) thus sod(sod(sod(X))) = 7 ==> arithmetic/digits/zeros/factorial.p <== How many zeros are in the decimal expansion of n]? ==> arithmetic/digits/zeros/factorial.s <== The general answer to the question "what power of p divides x]" where p is prime is (x-d)/(p-1) where d is the sum of the digits of (x written in base p). So where p=5, 10 is written as 20 and is divisible by 5^2 (2 = (10-2)/4); x to base 10: 100 1000 10000 100000 1000000 x to base 5: 400 13000 310000 11200000 224000000 d : 4 4 4 4 8 trailing 0s in x] 24 249 2499 24999 249998 ==> arithmetic/digits/zeros/lsd.factorial.p <== What is the least significant non-zero digit in the decimal expansion of n]? ==> arithmetic/digits/zeros/lsd.factorial.s <== Reduce mod 10 the numbers 2..n and then cancel out the required factors of 10. The final step then involves computing 2^i*3^j*7^k mod 10 for suitable i,j and k. A small program that performs this calculation is appended. Like the other solutions, it takes O(log n) arithmetic operations. -kym === #include #include int pÝ6¨Ý4¨={ /*2*/ 2, 4, 8, 6, /*3*/ 3, 9, 7, 1, /*4*/ 4, 6, 4, 6, /*5*/ 5, 5, 5, 5, /*6*/ 6, 6, 6, 6, /*7*/ 7, 9, 3, 1, }; main(){ int i; int n; for(n=2;n<1000;n++){ i=lsdfact(n); printf("%d\n",i); } exit(0); } lsdfact(n){ int aÝ10¨; int i; int n5; int tmp; for(i=0;i<=9;i++)aÝi¨=alpha(i,n); n5=0; /* NOTE: order is important in following */ l5:; while(tmp=aÝ5¨){ /* cancel factors of 5 */ n5+=tmp; aÝ1¨+=(tmp+4)/5; aÝ3¨+=(tmp+3)/5; aÝ5¨=(tmp+2)/5; aÝ7¨+=(tmp+1)/5; aÝ9¨+=(tmp+0)/5; } l10:; if(tmp=aÝ0¨){ aÝ0¨=0; /* cancel all factors of 10 */ for(i=0;i<=9;i++)aÝi¨+=alpha(i,tmp); } if(aÝ5¨) goto l5; if(aÝ0¨) goto l10; /* n5 == number of 5's cancelled; must now cancel same number of factors of 2 */ i=ipow(2,aÝ2¨+2*aÝ4¨+aÝ6¨+3*aÝ8¨-n5)* ipow(3,aÝ3¨+aÝ6¨+2*aÝ9¨)* ipow(7,aÝ7¨); assert(i%10); /* must not be zero */ return i%10; } alpha(d,n){ /* number of decimal numbers in Ý1,n¨ ending in digit d */ int tmp; tmp=(n+10-d)/10; if(d==0)tmp--; /* forget 0 */ return tmp; } ipow(x,y){ /* x^y mod 10 */ if(y==0) return 1; if(y==1) return x; return pÝx-2¨Ý(y-1)%4¨; } ==> arithmetic/digits/zeros/million.p <== How many zeros occur in the numbers from 1 to 1,000,000? ==> arithmetic/digits/zeros/million.s <== In the numbers from 10^(n-1) through 10^n - 1, there are 9 * 10^(n-1) numbers of n digits each, so 9(n-1)10^(n-1) non-leading digits, of which one tenth, or 9(n-1)10^(n-2), are zeroes. When we change the range to 10^(n-1) + 1 through 10^n, we remove 10^(n-1) and put in 10^n, gaining one zero, so p(n) = p(n-1) + 9(n-1)10^(n-2) + 1 with p(1)=1. Solving the recurrence yields the closed form p(n) = n(10^(n-1)+1) - (10^n-1)/9. For n=6, there are 488,895 zeroes, 600,001 ones, and 600,000 of all other digits. ==> arithmetic/magic.squares.p <== Are there large squares, containing only consecutive integers, all of whose rows, columns and diagonals have the same sum? How about cubes? ==> arithmetic/magic.squares.s <== Here is an 8x8 example: 01 56 48 25 33 24 16 57 63 10 18 39 31 42 50 07 62 11 19 38 30 43 51 06 04 53 45 28 36 21 13 60 05 52 44 29 37 20 12 61 59 14 22 35 27 46 54 03 58 15 23 34 26 47 55 02 08 49 41 32 40 17 09 64 References: "Magic Squares and Cubes" W.S. Andrews The Open Court Publishing Co. Chicago, 1908 "Mathematical Recreations" M. Kraitchik Dover New York, 1953 ==> arithmetic/pell.p <== Find integer solutions to x^2 - 92y^2 = 1. ==> arithmetic/pell.s <== x=1 y=0 x=1151 y=120 x=2649601 y=276240 etc. Each successive solution is about 2300 times the previous solution; they are every 8th partial fraction (x=numerator, y=denominator) of the continued fraction for sqrt(92) = Ý9, 1,1,2,4,2,1,1,18, 1,1,2,4,2,1,1,18, 1,1,2,4,2,1,1,18, ...¨ Once you have the smallest positive solution (x1,y1) you don't need to "search" for the rest. You can obtain the nth positive solution (xn,yn) by the formula (x1 + y1 sqrt(92))^n = xn + yn sqrt(92). See Niven & Zuckerman's An Introduction to the Theory of Numbers. Look in the index under Pell's equation. ==> arithmetic/prime/arithmetic.progression.p <== Is there an arithmetic progression of 20 or more primes? ==> arithmetic/prime/arithmetic.progression.s <== There is an arithmetic progression of 21 primes: 142072321123 + 1419763024680 i, 0 <= i < 21. It was discovered on 30 November 1990, by programs running in the background on a network of Sun 3 workstations in the Department of Computer Science, University of Queensland, Australia. See: Andrew Moran and Paul Pritchard, The design of a background job on a local area network, Procs. 14th Australian Computer Science Conference, 1991, to appear. ==> arithmetic/prime/consecutive.composites.p <== Are there 10,000 consecutive non-prime numbers? ==> arithmetic/prime/consecutive.composites.s <== 9973]+2 through 9973]+10006 are composite. ==> arithmetic/sequence.p <== Prove that all sets of n integers contain a subset whose sum is divisible by n. ==> arithmetic/sequence.s <== Consider the set of remainders of the partial sums a(1) + ... + a(i). Since there are n such sums, either one has remainder zero (and we're thru) or 2 coincide, say the i'th and j'th. In this case, a(i+1) + ... + a(j) is divisible by n. (note this is a stronger result since the subsequence constructed is of *adjacent* terms.) Consider a(1) (mod n), a(1)+a(2) (mod n), ..., a(1)+...+a(n) (mod n). Either at some point we have a(1)+...+a(m) = 0 (mod n) or else by the pigeonhole principle some value (mod n) will have been duplicated. We win either way. ==> arithmetic/sum.of.cubes.p <== Find two fractions whose cubes total 6. ==> arithmetic/sum.of.cubes.s <== Restated: Find X, Y, minimum Z (all positive integers) where (X/Z)^3 + (Y/Z)^3 = 6 Again, a generalized solution would be nice. You are asking for the smallest z s.t. x^3 + y^3 = 6*z^3 and x,y,z in Z+. In general, questions like these are extremely difficult; if you're interested take a look at books covering Diophantine equations (especially Baker's work on effective methods of computing solutions). Dudeney mentions this problem in connection with #20 in _The Canterbury Puzzles_; the smallest answer is (17/21)^3 + (37/21)^3 = 6. For the interest of the readers of this group I'll quote: "Given a known case for the expression of a number as the sum or difference of two cubes, we can, by formula, derive from it an infinite number of other cases alternately positive and negative. Thus Fermat, starting from the known case 1^3 + 2^3 = 9 (which we will call a fundamental case), first obtained a negative solution in bigger figures, and from this his positive solution in bigger figures still. But there is an infinite number of fundamentals, and I found by trial a negative fundamental solution in smaller figures than his derived negative solution, from which I obtained the result shown above. That is the simple explanation." In the above paragraph Dudeney is explaining how he derived (*by hand*) that (415280564497/348671682660)^3 + (676702467503/348671682660)^3 = 9. He continues: "We can say of any number up to 100 whether it is possible or not to express it as the sum of two cubes, except 66. Students should read the Introduction to Lucas's _Theorie des Nombres_, p. xxx." "Some years ago I published a solution for the case 6 = (17/21)^3 + (37/21)^3, of which Legendre gave at some length a 'proof' of impossibility; but I have since found that Lucas anticipated me in a communication to Sylvester." ==> arithmetic/tests.for.divisibility/eleven.p <== What is the test to see if a number is divisible by eleven? ==> arithmetic/tests.for.divisibility/eleven.s <== If the alternating sum of the digits is divisible by eleven, so is the number. For example, 1639 leads to 9 - 3 + 6 - 1 = 11, so 1639 is divisible by 11. Proof: Every integer n can be expressed as n = a1*(10^k) + a2*(10^k-1)+ .....+ a_k+1 where a1, a2, a3, ...a_k+1 are integers between 0 and 9. 10 is congruent to -1 mod(11). Thus if (-1^k)*a1 + (-1^k-1)*a2 + ...+ (a_k+1) is congruent to 0mod(11) then n is divisible by 11. ==> arithmetic/tests.for.divisibility/nine.p <== What is the test to see if a number is divisible by nine? ==> arithmetic/tests.for.divisibility/nine.s <== If the sum of the digits is divisible by nine, so is the number. Proof: Every integer n can be expressed as n = a1*(10^k) + a2*(10^k-1)+ .....+ a_k+1 where a1, a2, a3, ...a_k+1 are integers between 0 and 9. Note that 10 is congruent to 1 (mod 9). Thus 10^k is congruent to 1 (mod 9) for every k >= 0. Thus n is congruent to (a1+a2+a3+....+a_k+1) mod(9). Hence (a1+a2+...+a_k+1) is divisible by 9 iff n is divisible by 9. ==> arithmetic/tests.for.divisibility/seven.p <== What is the test to see if a number is divisible by 7? ==> arithmetic/tests.for.divisibility/seven.s <== Take the last digit (n mod 10) and double it. Take the rest of the digits (n div 10) and subtract the doubled last digit from it. The resulting number is divisible by 7 iff the original number is divisible by 7. Example: Take 2009. Subtract (2009 mod 10) * 2 from (2009 div 10) - 9 * 2 + 200 = 182 Subtract (182 mod 10) * 2 from (182 div 10) - 2 * 2 + 18 = 14 so 2009 is divisible by 7. ==> arithmetic/tests.for.divisibility/three.p <== Prove that if a number is divisible by 3, the sum of its digits is likewise. ==> arithmetic/tests.for.divisibility/three.s <== First, prove 10^N = 1 mod 3 for all integers N >= 0. 1 = 1 mod 3. 10 = 1 mod 3. 10^N = 10^(N-1) * 10 = 10^(N-1) mod 3. QED by induction. Now let DÝ0¨ be the units digit of N, DÝ1¨ the tens digit, etc. Now N = Summation From k=0 to k=inf of DÝk¨*10^k. Therefore N mod 3 = Summation from k=0 to k=inf of DÝk¨ mod 3. QED ==> combinatorics/coinage/combinations.p <== How many ways are there to make change for a dollar? Count combinations of coins, not permuations. ==> combinatorics/coinage/combinations.s <== Assuming that you had coins of one cent, five cents, ten cents, 25 cents, 50 cents, and 100 cents, there are 293 ways to make change for a dollar. This can be calculated by determining the number of ways to make change using only a penny and then a penny and nickel, then penny, nickel, and dime, etc. The table is shown below: Amount 00 05 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 Coins .01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 .05 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 .10 1 2 4 6 9 12 16 20 25 30 36 42 49 56 64 72 81 90 100 110 121 .25 1 2 4 6 9 13 18 24 31 39 49 60 73 87 103 121 141 163 187 214 242 .50 1 2 4 6 9 13 18 24 31 39 49 62 77 93 112 134 159 187 218 253 292 1.0 1 2 4 6 9 13 18 24 31 39 49 62 77 93 112 134 159 187 218 253 293 The meaning of each entry is as follows: If you wish to make change for 50 cents using only pennies, nickels and dimes, go to the .10 row and the 50 column to obtain 36 ways to do this. To calculate each entry, you start with the pennies. There is exactly one way to make change for every amount. Then calculate the .05 row by adding the number of ways to make change for the amount using pennies plus the number of ways to make change for five cents less using nickels and pennies. This continues on for all denominations of coins. An example, to get change for 75 cents using all coins up to a .50, add the number of ways to make change using only .25 and down (121) and the number of ways to make change for 25 cents using coins up to .50 (13). This yields the answer of 134. ==> combinatorics/coinage/dimes.p <== "Dad wants one-cent, two-cent, three-cent, five-cent, and ten-cent stamps. He said to get four each of two sorts and three each of the others, but I've forgotten which. He gave me exactly enough to buy them; just these dimes." How many stamps of each type does Dad want? ÝJ.A.H. Hunter¨ ==> combinatorics/coinage/dimes.s <== The easy way to solve this is to sell her three each, for 3x(1+2+3+5+10) = 63 cents. Two more stamps must be bought, and they must make seven cents (since 17 is too much), so the fourth stamps are a two and a five. ==> combinatorics/coinage/impossible.p <== What is the smallest number of coins that you can't make a dollar with? I.e., for what N does there not exist a set of N coins adding up to a dollar? It is possible to make a dollar with 1 current U.S. coin (a Susan B. Anthony), 2 coins (2 fifty cent pieces), 3 coins (2 quarters and a fifty cent piece), etc. It is not possible to make exactly a dollar with 101 coins. ==> combinatorics/coinage/impossible.s <== The answer is 77: a) 5c = 1 or 5; b) 10c = 1 or 2 a's (1,2,6,10) c) 25c = 1 or 2 b's + 1 a d) 50c = 1 or 2 c's e) $1 = 1 or 2 d's total penny nickle dime quarter half 5 1 2 1 1 6 3 1 1 1 7 5 1 1 8 4 3 1 9 6 2 1 10 8 1 1 11 10 1 12 7 4 1 13 9 3 1 14 11 2 1 15 13 1 1 16 15 1 17 14 3 18 16 2 19 18 1 20 20 21 5 13 3 22 5 15 2 23 5 17 1 24 5 19 25 10 12 3 26 10 14 2 27 10 16 1 28 10 18 29 15 11 3 30 15 13 2 31 15 15 1 32 15 17 33 20 10 3 34 20 12 2 35 20 14 1 36 20 16 37 25 9 3 38 25 11 2 39 25 13 1 40 25 15 41 30 8 3 42 30 10 2 43 30 12 1 44 30 14 45 35 7 3 46 35 9 2 47 35 11 1 48 35 13 49 40 6 3 50 40 8 2 51 40 10 1 52 40 12 53 45 5 3 54 45 7 2 55 45 9 1 56 45 11 57 50 4 3 58 50 6 2 59 50 8 1 60 50 10 61 55 3 3 62 55 5 2 63 55 7 1 64 55 9 65 60 2 3 66 60 4 2 67 60 6 1 68 60 8 69 65 1 3 70 65 3 2 71 65 5 1 72 65 7 73 70 3 74 70 2 2 75 70 4 1 76 70 6 77 can't be done 78 75 1 2 79 75 3 1 80 75 5 81 can't be done 82 80 2 83 80 2 1 84 80 4 85 can't be done 86 can't be done 87 85 1 1 88 85 3 89 can't be done 90 can't be done 91 90 1 92 90 2 93-95 can't be done 96 95 1 97-99 can't be done 100 100 ==> combinatorics/color.p <== An urn contains n balls of different colors. Randomly select a pair, repaint the first to match the second, and replace the pair in the urn. What is the expected time until the balls are all the same color? ==> combinatorics/color.s <== (n-1)^2. If the color classes have sizes k1, k2, ..., km, then the expected number of steps from here is (dropping the subscript on k): 2 k(k-1) (j-1) (k-j) (n-1) - SUM ( ------ + SUM --------------- ) classes, 2 1 combinatorics/full.p <== Consider a string that contains all substrings of length n. For example, for binary strings with n=2, a shortest string is 00110 -- it contains 00, 01, 10 and 11 as substrings. Find the shortest such strings for all n. ==> combinatorics/full.s <== Knuth, Volume 2 Seminumerical Algorithms, section 3.2.2 discusses this problem. He cites the following results: Shortest length: m^n + n - 1, where m = number of symbols in the language. Algorithms: ÝExercise 7, W. Mantel, 1897¨ The binary sequence is the LSB of X computed by the MIX program: LDA X JANZ *+2 LDA A ADD X JNOV *+3 JAZ *+2 XOR A STA X ÝExercise 10, M. H. Martin, 1934¨ Set xÝ1¨ = xÝ2¨ = ... = xÝn¨ = 0. Set xÝi+1¨ = largest value < n such that substring of n digits ending at xÝi+1¨ does not occur earlier in string. Terminate when this is not possible. If we instead consider the strings as circular, we have a well known problem whose solution is given by any hamiltonian cycle in the de Bruijn (or Good) graph of dimension K. (Or equivalently an eulerian circuit in the de Bruijn graph of dimension K-1) As a string of length 2^K is produced, it must be optimal, and any shortest sequence must be an eulerian circuit in a dB graph. The de Bruijn graph Tn has as its vertex set the binary n-strings. Directed edges join n-strings that may be derived by deleting the left most digit and appending a 0 or 1 to the right end. de Bruijn + van Ardenne-Ehrenfest (in 1951) counted the number of eulerian circuits in Tn. There are 2^(2^(n-1)-n) of them. Some examples: K=2 1100 K=3 11101000 K=4 1111001011010000 The solution to the above problem (non-circular strings) can be found by duplicating the first K-1 digits of the solution string at the end of the string. These are not the only solutions, but they are of minimum length: 2^K + K-1. We can obtain a lower bound for the optimal sequence for the general case as follows: Consider first the simpler case of breaking into an answer machine which accepts d+1 digits, values 0 to n-1. We wish to find the minimal universal code that will allow us access to any such answering machine. Let us construct a digraph G = (V,E), where the n^d vertices are labelled with a d sequence of digits. Notation: let Ýv_{i,1},v_{i,2},...,v_{i,d}¨ denote the labelling on node v_i. An edge e = (v_i, v_j) is in E iff for k in 1, ..., d-1: v_{i,k+1} = v_{j,k}, i.e., the last d-1 digits in the labelling of the initial vertex of e is identical with the first d-1 digits in the labelling of the terminal vertex of e. We associate with each edge a value, t(e) = v_{j,d}, the last digit in the labelling of the terminal vertex. The intuition goes as follows: we are going to perform a Euler circuit of the digraph, where the label on the current vertex gives the last d digits in the output sequence so far. If we make a transition on edge e, we output the tone/digit t(e) as the next output value, thus preserving the invariant on the labelling. How do we know that a Euler circuit exists? Simple: a connected digraph has an Euler circuit iff for all vertices v: indegree(v) = outdegree(v). This property is trivially true for this digraph. So, in order to generate a universal code for the AM, we simply output 0^d (to satisfy the precondition for being in vertex Ý0,...,0¨), and perform an Euler circuit starting at node Ý0,...,0¨. Now, the total length of the universal sequence is just the number of edges traversed in the Euler circuit plus the initial precondition sequence, or n^d * n + d (number of vertices times the out-degree) or n^{d+1} + d. That this is a minimal sequence is obvious. Next, let us consider the machine AM' where the security code is of the form Ý0...n-1¨^d Ý0...m-1¨, i.e., d digits ranging from 0 to n-1, followed by a terminal digit ranging from 0 to m-1, m < n. We build a digraph G = (V, E) similar to the construction above, except for the following: an edge e is in E iff t(e) in 0 to m-1. This digraph is clearly non-Eulerian. In particular, there are two classes of vertices: (1) v is of the form Ý0...n-1¨^{d-1} Ý0...m-1¨ (``fat'' vertices) and (2) v is of the form Ý0...n-1¨^{d-1} Ým...n-1¨ (``thin'' vertices) Observations: there are (n^{d-1} * m) fat vertices, and (n^{d-1} * (n-m)) thin vertices. All vertices have out-degree of m. Fat vertices have in-degrees of n, and thin vertices have in-degrees of 0. Color all the edges blue. The question now becomes: can we put a bound on how many new (red) edges must we add to G in order to make a blue edge covering path possible? (Instead of thinking of edges being traversed multiple times in the blue edge covering path, we allow multiple edges between vertices and allow each edge to be traversed once.) Note that, in this procedure, we add edges only if it is allowed (the vertex labelling constraint). We will first obtain a lower bound on the length of a blue covering circuit, and then transform it into a bound for arbitrary blue covering paths. Clearly, we must add at least (n-m)*(n^{d-1}*m) edges incident from the fat vertices. Ý We need (n-m) new out-going edges for each of (n^{d-1}*m) vertices to bring the out-degree up to the in-degree. ¨ Let us partition our vertices into sets. Denote the range Ý0..m-1¨ by S, the range Ým..n-1¨ by L, and the range Ý0..n-1¨ by X. Let S_0 = { v: v = ÝX^{d-1}S¨ }. S_0 is just the set of fat vertices. Define in(S_0) = number of edges from vertices not in S to vertices in S. Define out(S_0) in the corresponding fashion, and let excess(S_0) = in(S_0)-out(S_0). Clearly, excess(S_0) = n^{d-1}m(n-m) from the argument above. Generalizing the requirement for Eulerian digraphs, we see that we must add excess(S_0) edges from S_0 if the blue edges connected to/within S_0 are to be covered by some circuit (edges may not be travered multiple times -- we add parallel edges to handle that case). In particular, edges from S_0 will be incident on vertices of the form ÝX^{d-2}SX¨. Furthermore, they can not be ÝX^{d-2}SS¨ since that is a subset of S_0 and adding those edges will not help excess(S_0). ÝNow, these edges may be needed if we are to have a circuit, but we do not consider them since they do not help excess(S_0).¨ So, we are forced to add excess(S_0) edges from S_0 to S_1 = { v: v = ÝX^{d-2}SL¨ }. Color these newly added edges red. Let us define in(S_1), out(S_1) and excess(S_1) as above for the modified digraph, i.e., including the red excess(S_0) edges that we just added. Clearly, in(S_1) = out(S_0) = n^{d-1}m(n-m), and out(S_1) = m*!S_1! = m*n^{d-2}m(n-m), so excess(S_1) = n^{d-2}m(n-m)^2. Consider S_0 union S_1. We must add excess(S_1) edges to S_0 union S_1 to make it possible for the digraph to be covered by a circuit, and these edges must go from {S_0 union S_1} to S_2 = { v: v = ÝX^{d-3}SL^2¨ } by a similar argument as before. Repeating this partitioning process, eventually we get to S_{d-1} = { v: v = ÝSL^{d-1}¨ }, where union of S_0 to S_{d-1} will need edges to S_d = { v: v = ÝL^d¨ }, where this process terminates. Note that at this time, excess(union of S_0 to S_{d-1}) = m(n-m)^d, but in(S_d) = 0 and out(S_d) = m(n-m)^d, and the process terminates. What have we shown? Adding up blue edges and the red edges gives us a lower bound on the total number of edges in a blue-edges covering circuit (not necessarily Eulerian) in the complete digraph. This comes out to be n^{d+1}-(n-m)^{d+1} edges. Next, we note that if we had an optimal path covering all the blue edges, we can transform it into a circuit by adding d edges. So, a minimal path can be no more than d edges shorter than the minimal circuit covering all blue edges. ÝOtherwise, we add d extra edges to make it into a shorter circuit.¨ So the shortest blue covering path through the digraph is at least n^{d+1}-{n-m}^{d+1}-d. With an initial pre-condition sequence of length d (to establish the transition invariant), the shortest universal answering machine sequence is of length at least n^{d+1}-(n-m)^{d+1}. While this has not been that constructive, it is easy to see that we can achieve this bound. If we looked at the vertices in each of the S_i's, we just add exactly the edges to S_{i+1} and no more. The resultant digraph would be Eulerian, and to find the minimal path we need only start at the vertex labelled Ý{n-1}^d¨, find the Euler circuit, and omit the last d edges from the tour. ==> combinatorics/gossip.p <== n people each know a different piece of gossip. They can telephone each other and exchange all the information they know (so that after the call they both know anything that either of them knew before the call). What is the smallest number of calls needed so that everyone knows everything? ==> combinatorics/gossip.s <== 1 for n=2 3 for n=3 2n-4 for n>=4 This can be achieved as follows: choose four professors (A, B, C, and D) as the "core group". Each professor outside the core group phones a member of the core group (it doesn't matter which); this takes n-4 calls. Now the core group makes 4 calls: A-B, C-D, A-C, and B-D. At this point, each member of the core group knows everything. Now, each person outside the core group calls anybody who knows everything; this again requires n-4 calls, for a total of 2n-4. The solution to the "gossip problem" has been published several times: 1. R. Tidjeman, "On a telephone problem", Nieuw Arch. Wisk. 3 (1971), 188-192. 2. B. Baker and R. Shostak, "Gossips and telephones", Discrete Math. 2 (1972), 191-193. 3. A. Hajnal, E. C. Milner, and E. Szemeredi, "A cure for the telephone disease", Canad Math. Bull 15 (1976), 447-450. 4. Kleitman and Shearer, Disc. Math. 30 (1980), 151-156. 5. R. T. Bumby, "A problem with telephones", Siam J. Disc. Meth. 2 (1981), 13-18. ==> combinatorics/grid.dissection.p <== How many (possibly overlapping) squares are in an mxn grid? ==> combinatorics/grid.dissection.s <== Given an n*m grid with n > m. Orient the grid so n is its width. Divide the grid into two portions, an m*m square on the left and an (n-m)*m rectangle on the right. Count the squares that have their upper right-hand corners in the m*m square. There are m^2 of size 1*1, (m-1)^2 of size 2*2, ... up to 1^2 of size m*m. Now look at the n-m columns of lattice points in the rectangle on the right, in which we find upper right-hand corners of squares not yet counted. For each column we count m new 1*1 squares, m-1 new 2*2 squares, ... up to 1 new m*m square. Combining all these counts in summations: m m total = sum i^2 + (n - m) sum i i=1 i=1 (2m + 1)(m + 1)m (n - m)(m + 1)m = ---------------- + --------------- 6 2 = (3n - m + 1)(m + 1)m/6 -- David Karr ==> combinatorics/subsets.p <== Out of the set of integers 1,...,100 you are given ten different integers. From this set, A, of ten integers you can always find two disjoint subsets, S & T, such that the sum of elements in S equals the sum of elements in T. Note: S union T need not be all ten elements of A. Prove this. ==> combinatorics/subsets.s <== First, a couple of points: (1) All empty subsets of the 10 integers are disjoint and have the same sum. This doesn't make for a very interesting problem. Thus, we impose the additional restriction that S and T be non-empty. (2) The 10 integers must be pairwise distinct. Consider, e.g., the 10 integers 1, 1, 1, 1, 1, 1, 1, 1, 1, and 1. There are no non-empty disjoint subsets with equal sums. Proof of puzzle: There are 2^10 = 1,024 subsets of the 10 integers, but there can be only 901 possible sums, the number of integers between the minimum and maximum sums. With more subsets than possible sums, there must exist at least one sum that corresponds to at least two subsets. Call two subsets with equal sums A and B. Let C = A intersect B; define S = A - C, T = B - C. Then S is disjoint from T, and sum(S) = sum(A-C) = sum(A) - sub(C) = sum(B) - sum(C) = sum(B-C) = sum(T). QED ==> cryptology/Beale.p <== What are the Beale ciphers? ==> cryptology/Beale.s <== The Beale ciphers are one of the greatest unsolved puzzles of all time. About 100 years ago, a fellow by the name of Beale supposedly buried two wagons-full of silver-coin filled pots in Bedford County, near Roanoke. There are local rumors about the treasure being buried near Bedford Lake. He wrote three encoded letters telling what was buried, where it was buried, and who it belonged to. He entrusted these three letters to a friend and went west. He was never heard from again. Several years later, someone examined the letters and was able to break the code used in the second letter. The code used either the text from the Declaration of Independence. A number in the letter indicated which word in the document was to be used. The first letter of that word replaced the number. For example, if the first three words of the document were "We hold these truths", the number 3 in the letter would represent the letter t. One of the remaining letters supposedly contains directions on how to find the treasure. To date, no one has solved the code. It is believed that both of the remaining letters are encoded using either the same document in a different way, or another very public document. For those interested, write to: The Beale Cypher Association P.O. Box 975 Beaver Falls, PA 15010 Item #904 is the 1885 pamphlet version ($5.00). #152 is the Cryptologia article by Gillogly that argues the hoax side ($2.00). A year's membership is $25, and includes 4 newsletters. TEXT for part 1 The Locality of the Vault. 71,194,38,1701,89,76,11,83,1629,48,94,63,132,16,111,95,84,341 975,14,40,64,27,81,139,213,63,90,1120,8,15,3,126,2018,40,74 758,485,604,230,436,664,582,150,251,284,308,231,124,211,486,225 401,370,11,101,305,139,189,17,33,88,208,193,145,1,94,73,416 918,263,28,500,538,356,117,136,219,27,176,130,10,460,25,485,18 436,65,84,200,283,118,320,138,36,416,280,15,71,224,961,44,16,401 39,88,61,304,12,21,24,283,134,92,63,246,486,682,7,219,184,360,780 18,64,463,474,131,160,79,73,440,95,18,64,581,34,69,128,367,460,17 81,12,103,820,62,110,97,103,862,70,60,1317,471,540,208,121,890 346,36,150,59,568,614,13,120,63,219,812,2160,1780,99,35,18,21,136 872,15,28,170,88,4,30,44,112,18,147,436,195,320,37,122,113,6,140 8,120,305,42,58,461,44,106,301,13,408,680,93,86,116,530,82,568,9 102,38,416,89,71,216,728,965,818,2,38,121,195,14,326,148,234,18 55,131,234,361,824,5,81,623,48,961,19,26,33,10,1101,365,92,88,181 275,346,201,206,86,36,219,324,829,840,64,326,19,48,122,85,216,284 919,861,326,985,233,64,68,232,431,960,50,29,81,216,321,603,14,612 81,360,36,51,62,194,78,60,200,314,676,112,4,28,18,61,136,247,819 921,1060,464,895,10,6,66,119,38,41,49,602,423,962,302,294,875,78 14,23,111,109,62,31,501,823,216,280,34,24,150,1000,162,286,19,21 17,340,19,242,31,86,234,140,607,115,33,191,67,104,86,52,88,16,80 121,67,95,122,216,548,96,11,201,77,364,218,65,667,890,236,154,211 10,98,34,119,56,216,119,71,218,1164,1496,1817,51,39,210,36,3,19 540,232,22,141,617,84,290,80,46,207,411,150,29,38,46,172,85,194 39,261,543,897,624,18,212,416,127,931,19,4,63,96,12,101,418,16,140 230,460,538,19,27,88,612,1431,90,716,275,74,83,11,426,89,72,84 1300,1706,814,221,132,40,102,34,868,975,1101,84,16,79,23,16,81,122 324,403,912,227,936,447,55,86,34,43,212,107,96,314,264,1065,323 428,601,203,124,95,216,814,2906,654,820,2,301,112,176,213,71,87,96 202,35,10,2,41,17,84,221,736,820,214,11,60,760 TEXT for part 2 (no title exists for this part) 115,73,24,807,37,52,49,17,31,62,647,22,7,15,140,47,29,107,79,84 56,239,10,26,811,5,196,308,85,52,160,136,59,211,36,9,46,316,554 122,106,95,53,58,2,42,7,35,122,53,31,82,77,250,196,56,96,118,71 140,287,28,353,37,1005,65,147,807,24,3,8,12,47,43,59,807,45,316 101,41,78,154,1005,122,138,191,16,77,49,102,57,72,34,73,85,35,371 59,196,81,92,191,106,273,60,394,620,270,220,106,388,287,63,3,6 191,122,43,234,400,106,290,314,47,48,81,96,26,115,92,158,191,110 77,85,197,46,10,113,140,353,48,120,106,2,607,61,420,811,29,125,14 20,37,105,28,248,16,159,7,35,19,301,125,110,486,287,98,117,511,62 51,220,37,113,140,807,138,540,8,44,287,388,117,18,79,344,34,20,59 511,548,107,603,220,7,66,154,41,20,50,6,575,122,154,248,110,61,52,33 30,5,38,8,14,84,57,540,217,115,71,29,84,63,43,131,29,138,47,73,239 540,52,53,79,118,51,44,63,196,12,239,112,3,49,79,353,105,56,371,557 211,505,125,360,133,143,101,15,284,540,252,14,205,140,344,26,811,138 115,48,73,34,205,316,607,63,220,7,52,150,44,52,16,40,37,158,807,37 121,12,95,10,15,35,12,131,62,115,102,807,49,53,135,138,30,31,62,67,41 85,63,10,106,807,138,8,113,20,32,33,37,353,287,140,47,85,50,37,49,47 64,6,7,71,33,4,43,47,63,1,27,600,208,230,15,191,246,85,94,511,2,270 20,39,7,33,44,22,40,7,10,3,811,106,44,486,230,353,211,200,31,10,38 140,297,61,603,320,302,666,287,2,44,33,32,511,548,10,6,250,557,246 53,37,52,83,47,320,38,33,807,7,44,30,31,250,10,15,35,106,160,113,31 102,406,230,540,320,29,66,33,101,807,138,301,316,353,320,220,37,52 28,540,320,33,8,48,107,50,811,7,2,113,73,16,125,11,110,67,102,807,33 59,81,158,38,43,581,138,19,85,400,38,43,77,14,27,8,47,138,63,140,44 35,22,177,106,250,314,217,2,10,7,1005,4,20,25,44,48,7,26,46,110,230 807,191,34,112,147,44,110,121,125,96,41,51,50,140,56,47,152,540 63,807,28,42,250,138,582,98,643,32,107,140,112,26,85,138,540,53,20 125,371,38,36,10,52,118,136,102,420,150,112,71,14,20,7,24,18,12,807 37,67,110,62,33,21,95,220,511,102,811,30,83,84,305,620,15,2,108,220 106,353,105,106,60,275,72,8,50,205,185,112,125,540,65,106,807,188,96,110 16,73,32,807,150,409,400,50,154,285,96,106,316,270,205,101,811,400,8 44,37,52,40,241,34,205,38,16,46,47,85,24,44,15,64,73,138,807,85,78,110 33,420,505,53,37,38,22,31,10,110,106,101,140,15,38,3,5,44,7,98,287 135,150,96,33,84,125,807,191,96,511,118,440,370,643,466,106,41,107 603,220,275,30,150,105,49,53,287,250,208,134,7,53,12,47,85,63,138,110 21,112,140,485,486,505,14,73,84,575,1005,150,200,16,42,5,4,25,42 8,16,811,125,160,32,205,603,807,81,96,405,41,600,136,14,20,28,26 353,302,246,8,131,160,140,84,440,42,16,811,40,67,101,102,194,138 205,51,63,241,540,122,8,10,63,140,47,48,140,288 CLEAR for part 2, made human readable. I have deposited in the county of Bedford about four miles from Bufords in an excavation or vault six feet below the surface of the ground the following articles belonging jointly to the parties whose names are given in number three herewith. The first deposit consisted of ten hundred and fourteen pounds of gold and thirty eight hundred and twelve pounds of silver deposited Nov eighteen nineteen. The second was made Dec eighteen twenty one and consisted of nineteen hundred and seven pounds of gold and twelve hundred and eighty eight of silver, also jewels obtained in St. Louis in exchange to save transportation and valued at thirteen Ýt¨housand dollars. The above is securely packed iÝn¨ Ýi¨ron pots with iron covÝe¨rs. ThÝe¨ vault is roughly lined with stone and the vessels rest on solid stone and are covered Ýw¨ith others. Paper number one describes thÝe¨ exact locality of the vaÝu¨lt so that no difficulty will be had in finding it. CLEAR for part 2, using only the first 480 words of the Declaration of Independence, then blanks filled in by inspection. ALL mistakes shown were caused by sloppy encryption. 0----5----10---15---20---25---30---35---40---45--- 0 ihavedepositedinthecountyofbedfordaboutfourmilesfr 50 ombufordsinanexcavationorvaultsixfeetbelowthesurfa 100 ceofthegroundthefollowingarticlesbelongingjointlyt 150 othepartieswhosenamesaregiveninnumberthreeherewith 200 thefirstdepositconsistcdoftenhundredandfourteenpou 250 ndsofgoldandthirtyeighthundredandtwelvepoundsofsil 300 verdepositednoveighteennineteenthesecondwasmadedec 350 eighteentwentyoneandconsistedofnineteenhundredands 400 evenpoundsofgoldandtwelvehundredandeightyeightofsi 450 lveralsojewelsobtainedinstlouisinexchangetosavetra 500 nsportationandvaluedatthirteenrhousanddollarstheab 550 oveissecurelypackeditronpotswithironcovtrsthtvault 600 isroughlylinedwithstoneandthevesselsrestonsolidsto 650 neandarecovereduithotherspapernumberonedescribesth 700 cexactlocalityofthevarltsothatnodifficultywillbeha 750 dinfindingit TEXT for part 3 Names and Residences. 317,8,92,73,112,89,67,318,28,96,107,41,631,78,146,397,118,98 114,246,348,116,74,88,12,65,32,14,81,19,76,121,216,85,33,66,15 108,68,77,43,24,122,96,117,36,211,301,15,44,11,46,89,18,136,68 317,28,90,82,304,71,43,221,198,176,310,319,81,99,264,380,56,37 319,2,44,53,28,44,75,98,102,37,85,107,117,64,88,136,48,154,99,175 89,315,326,78,96,214,218,311,43,89,51,90,75,128,96,33,28,103,84 65,26,41,246,84,270,98,116,32,59,74,66,69,240,15,8,121,20,77,80 31,11,106,81,191,224,328,18,75,52,82,117,201,39,23,217,27,21,84 35,54,109,128,49,77,88,1,81,217,64,55,83,116,251,269,311,96,54,32 120,18,132,102,219,211,84,150,219,275,312,64,10,106,87,75,47,21 29,37,81,44,18,126,115,132,160,181,203,76,81,299,314,337,351,96,11 28,97,318,238,106,24,93,3,19,17,26,60,73,88,14,126,138,234,286 297,321,365,264,19,22,84,56,107,98,123,111,214,136,7,33,45,40,13 28,46,42,107,196,227,344,198,203,247,116,19,8,212,230,31,6,328 65,48,52,59,41,122,33,117,11,18,25,71,36,45,83,76,89,92,31,65,70 83,96,27,33,44,50,61,24,112,136,149,176,180,194,143,171,205,296 87,12,44,51,89,98,34,41,208,173,66,9,35,16,95,8,113,175,90,56 203,19,177,183,206,157,200,218,260,291,305,618,951,320,18,124,78 65,19,32,124,48,53,57,84,96,207,244,66,82,119,71,11,86,77,213,54 82,316,245,303,86,97,106,212,18,37,15,81,89,16,7,81,39,96,14,43 216,118,29,55,109,136,172,213,64,8,227,304,611,221,364,819,375 128,296,1,18,53,76,10,15,23,19,71,84,120,134,66,73,89,96,230,48 77,26,101,127,936,218,439,178,171,61,226,313,215,102,18,167,262 114,218,66,59,48,27,19,13,82,48,162,119,34,127,139,34,128,129,74 63,120,11,54,61,73,92,180,66,75,101,124,265,89,96,126,274,896,917 434,461,235,890,312,413,328,381,96,105,217,66,118,22,77,64,42,12 7,55,24,83,67,97,109,121,135,181,203,219,228,256,21,34,77,319,374 382,675,684,717,864,203,4,18,92,16,63,82,22,46,55,69,74,112,134 186,175,119,213,416,312,343,264,119,186,218,343,417,845,951,124 209,49,617,856,924,936,72,19,28,11,35,42,40,66,85,94,112,65,82 115,119,233,244,186,172,112,85,6,56,38,44,85,72,32,47,63,96,124 217,314,319,221,644,817,821,934,922,416,975,10,22,18,46,137,181 101,39,86,103,116,138,164,212,218,296,815,380,412,460,495,675,820 952 Evidence in favor of a hoax- . Too many players. . Inflated quantities of treasure. . Many discrepancies exist in all documents. . The Declaration of Independence is too hokey a key. . Part 3 (list of 30 names) considered too little text. . W.F. Friedman couldn't crack it. . Why even encrypt parts 1 & 3? . Why use multi-part text, and why different keys for each part? . Difficult to keep treasure in ground if 30 men know where it was buried. . Who'd leave it with other than your own family? . The Inn Keeper waited an extra 10 years before opening box with ciphers in it? Who would do this, curiousity runs too deep in humans? . Why did anybody waste time deciphering paper 2, which had no title? 1 & 3 had titles] These should have been deciphered first? . Why not just one single letter? . Statistical analysis show 1&3 similar in very obscure ways, that 2 differs. Did somebody else encipher it? And why? Check length of keytexts, and forward/backward next word displacement selections. . Who could cross the entire country with that much gold and only 10 men and survive back then? . Practically everybody who visited New Mexico before 1821, left by way of the Pearly Gates, as the Spanish got almost every tourist:-) References: "The Beale Treasure: A History of a Mystery", by Peter Viemeister, Bedord, VA: Hamilton's, 1987. ISBN: 0-9608598-3-7. 230 pages. "The Codebreakers", by David Kahn, pg 771, CCN 63-16109. 1967. "Gold in the Blue Ridge, The True Story of the Beale Treasure", by P.B. Innis & Walter Dean Innis, Devon Publ. Co., Wash, D.C. 1973. "Signature Simulation and Certain Cryptographic Codes", Hammer, Communications of the ACM, 14 (1), January 1971, pp. 3-14. "How did TJB Encode B2?", Hammer, Cryptologia, 3 (1), Jan. 1979, pp. 9-15. "Second Order Homophonic Ciphers", Hammer, Cryptologia, 12 (1) Jan. 1988, pp 11-20. ==> cryptology/Feynman.p <== What are the Feynman ciphers? ==> cryptology/Feynman.s <== When I was a graduate student at Caltech, Professor Feynman showed me three samples of code that he had been challenged with by a fellow scientist at Los Alamos and which he had not been able to crack. I also was unable to crack them. I posted them to Usenet and Jack C. Morrison of JPL cracked the first one. It is a simple transposition cipher: split the text into 5-column pieces, then read from lower right upward. What results are the opening lines of Chaucer's Canterbury Tales in Middle English. 1. Easier MEOTAIHSIBRTEWDGLGKNLANEA INOEEPEYSTNPEUOOEHRONLTIR OSDHEOTNPHGAAETOHSZOTTENT KEPADLYPHEODOWCFORRRNLCUE EEEOPGMRLHNNDFTOENEALKEHH EATTHNMESCNSHIRAETDAHLHEM TETRFSWEDOEOENEGFHETAEDGH RLNNGOAAEOCMTURRSLTDIDORE HNHEHNAYVTIERHEENECTRNVIO UOEHOTRNWSAYIFSNSHOEMRTRR EUAUUHOHOOHCDCHTEEISEVRLS KLIHIIAPCHRHSIHPSNWTOIISI SHHNWEMTIEYAFELNRENLEERYI PHBEROTEVPHNTYATIERTIHEEA WTWVHTASETHHSDNGEIEAYNHHH NNHTW 2. Harder XUKEXWSLZJUAXUNKIGWFSOZRAWURO RKXAOSLHROBXBTKCMUWDVPTFBLMKE FVWMUXTVTWUIDDJVZKBRMCWOIWYDX MLUFPVSHAGSVWUFWORCWUIDUJCNVT TBERTUNOJUZHVTWKORSVRZSVVFSQX OCMUWPYTRLGBMCYPOJCLRIYTVFCCM UWUFPOXCNMCIWMSKPXEDLYIQKDJWI WCJUMVRCJUMVRKXWURKPSEEIWZVXU LEIOETOOFWKBIUXPXUGOWLFPWUSCH 3. New Message WURVFXGJYTHEIZXSQXOBGSV RUDOOJXATBKTARVIXPYTMYA BMVUFXPXKUJVPLSDVTGNGOS IGLWURPKFCVGELLRNNGLPYT FVTPXAJOSCWRODORWNWSICL FKEMOTGJYCRRAOJVNTODVMN SQIVICRBICRUDCSKXYPDMDR OJUZICRVFWXIFPXIVVIEPYT DOIAVRBOOXWRAKPSZXTZKVR OSWCRCFVEESOLWKTOBXAUXV B Chris Cole Peregrine Systems uunet]peregrine]chris ==> cryptology/Voynich.p <== What are the ==> cryptology/Voynich.s <== The Voynich Manuscript is a manuscript that first surfaced in the court of Rudolf II (Holy Roman Emperor), who bought it for some large number of gold pieces (600?). Rudolf was interested in the occult, and the strange characters and bizarre illustrations suggested that it had some deep mystical/magical significance. After Rudolf's court broke up, the manuscript was sent to (if memory serves) Athanasius Kircher, with nobody on the list having been able to read it. It ended up in a chest of other manuscripts in the Villa Mondragone Ý?¨ in Italy, and was discovered there by Wilfred Voynich, a collector, in about 1910 or so. He took it to a linguist who wasn't a cryptanalyst, who identified it as a work by the 12th century monk Roger Bacon and produced extended bogus decryptions based on shorthand characters he saw in it. A great deal of effort by the best cryptanalysts in the country hasn't resulted in any breakthrough. William F. Friedman (arguably the best) thought it was written in an artificial language. I believe the manuscript is currently in the Beinecke Rare Book Collection at ÝHarvard?¨. Mary D'Imperio's paper is scholarly and detailed, and provides an excellent starting point for anyone who is interested in the subject. David Kahn's "The Codebreakers" has enough detail to tell you if you're interested; it also has one or more plates showing the script and some illustrations. I believe D'Imperio's monograph has been reprinted by Aegean Park Press. A number of people have published their own ideas about it, including Brumbaugh, without anybody agreeing. A recent publication from Aegean Park Press offers another decryption; I haven't seen that one. If you want *my* guess, it's a hoax made up by Edmund Kelley and an unnamed co-conspirator and sold to Rudolf through the reputation of John Dee (Queen Elizabeth I's astrologer). -- Jim Gillogly {hplabs, ihnp4}]sdcrdcf]randvax]jim jim@rand-unix.arpa I read "Labyrinths of Reason" by William Poundstone recently. I'm posting this to so many newsgroups in part to recommend this book, which, while of a popular nature, gives a good analysis of a wide variety of paradoxes and philosophical quandaries, and is a great read. Anyway, it mentions something called the Voynich manuscript, which is now at Yale University's Beinecke Rare Book and Manuscript Library. It's a real pity that I didn't know about this manuscript and go see it when I was at Yale. The Voynich manuscript is apparently very old. It is a 232-page illuminated manuscript written in a cipher that has never been cracked. (That's what Poundstone says - but see my hypothesis below.) If I may quote Poundstone's charming description, "Its author, subject matter, and meaning are unfathomed mysteries. No one even knows what language the text would be in if you deciphered it. Fanciful picutres of nude women, peculiar inventions, and nonexistent flora and fauna tantalize the would-be decipherer. Color sketches in the exacting style of a medieval herbal depict blossoms and spices that never spring from earth and constellations found in no sky. Plans for weird, otherworldly plumbing show nymphets frolicking in sitz baths connected with elbow-macaroni pipes. The manuscript has the eerie quality of a perfectly sensible book from an alternate universe." There is a picture of one page in Poundstone's book. It's written in a flowing script using "approximately 21 curlicued symbols," some of which are close to the Roman alphabet, but others of which supposedly resemble Cyrillic, Glagolitic, and Ethiopian. There is one tiny note in Middle High German, not necessarily by the original author, talking about the Herbal of Matthiolaus. Some astrology charts in the manuscript have the months labeled in Spanish. "What appears to be a cipher table on the first page has long faded into illegibility," and on the other hand, some scholars have guessed that a barely legible inscription on the *last* page is a key] It is said to have "languished for a long time at the Jesuit College of Mondragone in Frascati, Italy. Then in 1912 it was purchased by Wilfred M. Voynich, a Polish-born scientist and bibliophile... Voynich was the son-in-law of George Boole, the logician..." A letter written in 1666 claims that Holy Roman Emperor Rudolf II of Bohemia (1552-1612) bought the manuscript for 600 gold ducats. He may have bought it from Dr. John Dee, the famous astrologer. Rudolf thought the manuscript was written by Roger Bacon] ÝWouldn't it more likely have been written by Dee, out to make a fast ducat?¨ "Many of the most talented military code breakers of this century have tried to decipher it as a show of prowess. Herbert Yardley, the American code expert who solved the German cipher in WW1 and who cracked a Japanese diplomatic cipher without knowing the Japanese language, failed with the Voynich manuscript. So did John Manly, who unscrambled the Waberski cipher, and William Friedman, who defeated the Japanese "purple code" of the 1940's. Computers have been drafted into the effort in recent years, to no avail." Poundstone goes on to describe a kook, Newbold, who was apparently driven batty in his attempt to crack the manuscript. He then mentions that one Leo Levitov also claimed in 1987 to crack the cipher, saying that it was the text of a 12th-century cult of Isis worshipers, and that it describes a method of euthanasia by opening a vein in a warm bathtub, among other morbid matters. According to Levitov's translation the text begins: "ones treat the dying each the man lying deathly ill the one person who aches Isis each that dies treats the person" Poundstone rejects this translation. According to Poundstone, a William Bennett (see below) has analysed the text with a computer and finds that its entropy is less than any known European language, and closer to those of Polynesian languages. My wild hypothesis, on the basis solely of the evidence above, is this. Perhaps the text was meant to be RANDOM. Of course humans are lousy at generating random sequences. So I'm wondering how attempted random sequences (written in a weird alphabet) would compare statistically with the Voynich manuscript. Anyway, the only source Poundstone seems to cite, other than the manuscript itself, is Leo Levitov's "Solution of the Voynich Manuscript, A Liturgical Manual for the Endura Rite of the Cathari Heresy, the Cult of Isis," Laguna Hills, Calif., Aegean Park Press, 1987, and William Ralph Bennett Jr.'s "Scientific and Engineering Problem-Solving with the Computer," Englewood Cliffs, New Jersey, Prentice-Hall 1976. I will check the Bennett book; the other sounds hard to get ahold of] I would LOVE any further information about this bizarre puzzle. If anyone knows Bennett and can get samples of the Voynich manuscript in electronic form, I would LOVE to get my hands on it. Also, I would appreciate any information on: Voynich The Jesuit College of Mondragone Rudolf II The letter by Rudolf II (where is it? what does it say?) The attempts of Yardley, Friedman and Manly The Herbal of Matthiolaus and, just for the heck of it, the "Waberski cipher" and the "purple code"] This whole business sounds like a quagmire into which angels would fear to tread, but a fool like me finds it fascinating. -- sender's name lost (]?) To counter a few hypotheses that were suggested here: The Voynich Manuscript is certainly not strictly a polyalphabetic cipher like Vigenere or Beaufort or (the one usually called) Porta, because of the frequent repetitions of "words" at intervals that couldn't be multiples of any key length. I suppose one could imagine that it's an interrupted key Vig or something, but common elements appearing at places other than the beginnings of words would seem to rule that out. The I.C. is too high for a digraphic system like (an anachronistic) Playfair in any European language. One of the most interesting Voynich discoveries was made by Prescott Currier, who discovered that the two different "hands" (visually distinct handwriting) used different "dialects": that is, the frequencies for pages written in one hand are different from those written in the other. I confirmed this observation by running some correlation coefficients on the digraph matrices for the two kinds of pages. W. F. Friedman ("The Man Who Broke Purple") thought the Voynich was written in some artificial language. If it's not a hoax, I don't see any evidence to suggest he's wrong. My personal theory (yeah, I've offered too many of those lately) is that it was constructed by Edward Kelley, John Dee's scryer, with somebody else's help (to explain the second handwriting) -- perhaps Dee himself, although he's always struck me as a credulous dupe of Kelley rather than a co-conspirator (cf the Angelic language stuff). The best source I know for the Voynich is Mary D'Imperio's monograph "The Voynich Manuscript: An Elegant Enigma", which is available from Aegean Park Press. -- Jim Gillogly jim@rand.org Here's an update on the Voynich manuscript. This will concentrate on sources for information on the Voynich; later I will write a survey of what I have found out so far. I begin with some references to the case, kindly sent to me by Karl Kluge (the first three) and Micheal Roe (the rest). TITLE Thirty-five manuscripts : including the St. Blasien psalter, the Llangattock hours, the Gotha missal, the Roger Bacon (Voynich) cipher ms. Catalogue ; 100 35 manuscripts. CITATION New York, N.Y. : H.P. Kraus, Ý1962¨ 86 p., lxvii p. of plates, Ý1¨ leaf of plates : ill. (some col.), facsims. ; 36 cm. NOTES "30 years, 1932-1962" (Ý28¨ p.) in pocket. Includes indexes. SUBJECT Manuscripts Catalogs. Illumination of books and manuscripts Catalogs. AUTHOR Brumbaugh, Robert Sherrick, 1918- TITLE The most mysterious manuscript : the Voynich "Roger Bacon" cipher manuscript / edited by Robert S. Brumbaugh. CITATION Carbondale : Southern Illinois University Press, c1978. xii, 175 p. : ill. ; 22 cm. SUBJECT Bacon, Roger, 1214?-1294. Ciphers. AUTHOR D'Imperio, M. E. TITLE The Voynich manuscript : an elegant enigma / M. E. D'Imperio. CITATION Fort George E. Mead, Md. : National Security Agency/Central Security Service, 1978. ix, 140 p. : ill. ; 27 cm. NOTES Includes index. Bibliography: p. 124-131. SUBJECT Voynich manuscript. ÝNOTE: see alternate publisher below]¨ @book{Bennett76, author = "Bennett, William Ralph", title = "Scientific and Engineering Problem Solving with the Computer", address = "Englewood Cliffs, NJ", publisher = "Prentice-Hall", year = 1976} @book{dImperio78, author = "D'Imperio, M E", title = "The Voynich manuscript: An Elegant Enigma", publisher= "Aegean Park Press", year = 1978} @article{Friedman62, author = "Friedman, Elizebeth Smith", title = "``The Most Mysterious Manuscript'' Still Mysterious", booktitle = "Washington Post", month = "August 5", notes = "Section E", pages = "1,5", year = 1962} @book{Kahn67, author = "Kahn, David", title = "The Codebreakers", publisher = "Macmillan", year = "1967"} @article{Manly31, author = "Manly, John Matthews", title = "Roger Bacon and the Voynich MS", boooktitle = "Speculum VI", pages = "345--91", year = 1931} @article{ONeill44, author = "O'Neill, Hugh", title = "Botanical Remarks on the Voynich MS", journal = "Speculum XIX", pages = "p.126", year = 1944} @book{Poundstone88, author = "Poundstone, W.", title = "Labyrinths of Reason", publisher = "Doubleday", address = "New York", month = "November", year = 1988} @article{Zimanski70, author = "Zimanski, C.", title = "William Friedman and the Voynich Manuscript", journal = "Philological Quarterly", year = "1970"} @article{Guy91b, author = "Guy, J. B. M.", title = "Statistical Properties of Two Folios of the Voynich Manuscript", journal = "Cryptologia", volume = "XV", number = "4", pages = "pp. 207--218", month = "July", year = 1991} @article{Guy91a, author = "Guy, J. B. M.", title = "Letter to the Editor Re Voynich Manuscript", journal = "Cryptologia", volume = "XV", number = "3", pages = "pp. 161--166", year = 1991} This is by no means a complete list. It doesn't include Newbold's (largely discredited) work, nor work by Feely and Stong. In addition, there is the proposed decryption by Leo Levitov (also largely discredited): "Solution of the Voynich Manuscript: A Liturgical Manual for the Endura Rite of the Cathari Heresy, the Cult of Isis_, available from Aegean Park Press, P. O. Box 2837, Laguna Hills CA 92654-0837." According to Earl Boebert, this book is reviewed in Cryptologia XII, 1 (January 1988). I should add that Brumbaugh's book above gives a third, also largely discredited, decryption of the Voynich. According to smb@att.ulysses.com, Aegean Park Press does mail-order business and can be reached at the above address or at 714-586-8811 (an answering machine). Micheal Roe has explained how one get microfilms of the whole manuscript: "The Beinecke Rare Book Library, Yale University sells a microfilm of the manuscript. Their catalog number for the original is MS 408, ``The Voynich `Roger Bacon' Cipher MS''. You should write to them. The British Library Ýsic - should be Museum¨ has a photocopy of the MS donated to them by John Manly circa 1931. They apparently lost it until 12 March 1947, when it was entered in the catalogue (without cross-references under Voynich, Manly, Roger Bacon or any other useful keywords...) It appears as ``MS Facs 461: Positive rotographs of a Cipher MS (folios 1-56) acquired in 1912 by Wilfred M. Voynich in Southern Europe.' Correspondance between Newbold, Manly and various British Museum experts appears under ``MS Facs 439: Leaves of the Voynich MS, alleged to be in Roger Bacon's cypher, with correspondence and other pertinent material'' See John Manly's 1931 article in Speculum and Newbold's book for what the correspondance was about] There are also a number of press cuttings. Both of these in are in the manuscript collection, for which special permission is needed in addition to a normal British Library reader's pass." Also, Jim Gillogly has been extremely kind in making available part of the manuscript that was transcribed and keyed in by Mary D'Imperio (see above), using Prescott Currier's notation. It appears to consist of 166 of the total 232 pages. I hope to do some statistical studies on this, and I encourage others to do the same and let me know what they find] As Jim notes, the file is pub/jim/voynich.tar.Z and is available by anonymous ftp at rand.org. I've had a little trouble with this file at page 165, where I read "1650voynich 664" etc., with page 166 missing. If anyone else notes this let Jim or I know. Jim says he has confirmed by correlations between digraph matrices the discovery by Prescott Crurrier that the manuscript is written in two visibly distinct hands. These are marked "A" and "B" in the file voynich.tar.Z. Because of the possibility that the Voynich is nonsense, it would be interesting to compare the Voynich to the Codex Seraphinianus, which Kevin McCarty kindly reminded me of. He writes: "This is very odd. I know nothing of the Voynich manuscript, but I know of something which sounds very much like it and was created by an Italian artist, who it now seems was probably influenced by this work. It a book titled "Codex Seraphinianus", written in a very strange script. The title page contains only the book's title and the publisher's name: Abbeville Press, New York. The only clues in English (in *any* recognizable language) are some blurbs on the dust jacket that identify it as a modern work of art, and the copyright notice, in fine print, which reads "Library of Congress Cataloging in Publication Data Serafini, Luigi. Codex Seraphinianus. 1. Imaginary Languages. 2. Imaginary societies. 3. Encyclopedias and Dictionaries-- Miscellanea. I. Title. PN6381.S4 1983 818'.5407 83.-7076 ISBN 0-89659-428-9 First American Edition, 1983. Copyright (c) 1981 by Franco Maria Ricci. All rights reserved by Abbeville Press. No part of this book may be reproduced... without permission in writing from the publisher. Inquiries should be addressed to Abbeville Press, Inc., 505 Park Avenue, New York 10022. Printed and bound in Italy." The book is remarkable and bizarre. It *looks* like an encyclopedia for an imaginary world. Page after page of beautiful pictures of imaginary flora and fauna, with annotations and captions in a completely strange script. Machines, architecture, umm, 'situations', arcane diagrams, implements, an archeologist pointing at a Rosetta stone (with phony hieroglyphics), an article on penmanship (with unorthodox pens), and much more, finally ending with a brief index. The script in this work looks vaguely similar to the Voynich orthography shown in Poundstone's book (I just compared them); the alphabets look quite similar, but the Codex script is more cursive and less bookish than Voynich. It runs to about 200 pages, and probably ought to provide someone two things: - a possible explanation of what the Voynich manuscript is (a highly imaginative work of art) - a textual work which looks like it was inspired by it and might provide an interesting comparison for statistical study." I suppose it would be too much to hope that someone has already transcribed parts of the Codex, but nonetheless, if anyone has any in electronic form, I would love to have a copy for comparative statistics. Jacques Guy kindly summarized his analysis (in Cryptologia, see above) of the Voynich as follows: "I transcribed the two folios in Bennett's book and submitted them to letter-frequency counts, distinguishing word-initial, word-medial, word-final, isolated, line-initial, and line-final positions. I also submitted that transcription to Sukhotin's algorithm which, given a text written in an alphabetical system, identifies which symbols are vowels and which are consonants. The letter transcribed CT in Bennett's system came out as a consonant, the one transcribed CC as vowel. Now it so happens that CT is exactly the shape of the letter "t" in the Beneventan script (used in medieval Spain and Northern Italy), and CC is exactly the shape of "a" in that same script. I concluded that the author had a knowledge of that script, and that the values of CT and CC probably were "t" and "a". There's a lot more, but more shaky." By popular demand I've put a machine-readable copy of the Voynich Manuscript up for anonymous ftp: Host: rand.org File: pub/jim/voynich.tar.Z It uses Prescott Currier's notation, and was transcribed by Mary D'Imperio. If you use it in any analysis, be sure to give credit to D'Imperio, who put in a lot of effort to get it right. -- Jim Gillogly jim@rand.org This post is essentially a summary of the fruit of a short research quest at the local library. Brief description of the Voynich manuscript: The Voynich manuscript was bought (in about 1586) by the Holy Roman Emperor Rudolf II. He believed it to be the work of Roger Bacon an english 13th century philosopher. The manuscript consisted of about 200 pages with many illustrations. It is believed that the manuscript contains some secret scientific or magical knowledge since it is entirely written in secret writing (presumably in cipher). The Voynich Manuscript is often abbreviated "Voynich MS" in all of the books I have read on Voynich. This is done without explanation. I suppose it is just a convention started by the founding analysts of the manuscript to call it that. William R. Newbold, one of the original analysts of the Voynich MS after Voynich, claims to have arrived at a partial decipherment of the entire manuscript. His book The Cipher of Roger Bacon Ý2¨ contains a history of the unravelment of the cipher *and* keys to the cipher itself. As well as translations of several pages of the manuscript. Newbold derives his decipherment rules through a study of the medeival mind (which he is a leading scholar in) as well as the other writings of Roger Bacon. Says Newbold, ciphers in Roger Bacon's writings are not new, as Bacon discusses in other works the need for monks to use encipherment to protect their knowlege. Newbold includes many partial decipherments from the Voynich MS but most of them are presented in Latin only. Newbolds deciphering rules (from The Cipher of Roger Bacon Ý1¨) --------------------------------------------------------------- 1. Syllabification: Ýdouble all but the first and last letters of each word, and divide the product into biliteral groups or symbols.¨ 2. Translation: Ýtranslate these symbols into the alphabetic values¨ 3. Reversion: Ýchange the alphabetic values to the phonetic values, by use of the reversion alphabet¨ 4. Recomposition: Ý rearrange the letters in order, and thus recompose the true text¨. The text I copied this from failed to note step 0 which was: 0. Ignore. Ýignore the actual shape of every symbol and analyze only the (random?) properties of the direction of swirl and crosshatch patterns of the characters when viewed under a microscope. 14 distinct contruction patterns can be identified among the (much larger) set of symbols¨ John M. Manly in The Most Mysterious Manuscript Ý3¨, suggests that Newbold's method of decipherment is totally invalid. Manly goes on to show that it is not difficult to obtain *ANY DESIRABLE* message from the Voynich MS using Newbold's rules. He shows that after fifteen minutes deciphering a short sequence of letters he arrives at the plaintext message "Paris is lured into loving vestals..." and quips that he will furnish a continuation of the translation upon request] The reason I have spent so much time explaining Newbold's method is that Newbold presents the most convincing argument for how he arrived at his conclusions. Notwithstanding the fact that he invented the oija board of deciphering systems. Joseph Martin Feely, in his book on the Voynich MS Ý2¨ , claims to have found the key to deciphering at least one page of the Voynich MS. His entire book on the topic of the Voynich manuscript is devoted to the deciphering of the single page 78. Feely presents full tables of translation of the page 78 from its written form into latin (and english). It seems that Feely was using the exhaustive analysis method to determine the key. Feely suggests the following translation of (the first fiew lines of) page 78 of the Voynich MS: "the combined stream when well humidified, ramifies; afterward it is broken down smaller; afterward, at a distance, into the fore-bladder it comes Ý1¨. Then vesselled, it is after-a-while ruminated: well humidified it is clothed with veinlets Ý2¨. Thence after-a-bit they move down; tiny teats they provide (or live upon) in the outpimpling of the veinlets. They are impermiated; are thrown down below; they are ruminated; they are feminized with the tiny teats. .... " ... and so on for three more pages of "english plaintext". The descriptions by Feely say that this text is accompanied in the Voynich MS by an illustration that (he says) is unmistakably the internal female reproductive organs (I saw the plate myself and they DO look like fallopian tubes *AFTER* I read the explanation). The most informative work that I found (I feel) was "The Most Mysterious Manuscript". Of the five books on Voynich that I found, this was the only one that didn't claim to have found the key but was, rather, a collection of essays on the history of the Voynich MS and criticisms of various attempts by earlier scientists. It was also the *latest* book that I was able to consult, being published in 1978. My impression from the black and white plates of the Voynich MS I've seen, are that the illustrations are very weird when compared to other 'illuminated' manuscripts of this time. Particularly I would say that there is emphasis on the female nude that is unusual for the art of this period. I can't say that I myself believe the images to have ANYTHING to do with the text. My own conjecture is that the manuscript is a one-way encipherment. A cipher so clever that the inventor didn't even think of how it could be deciphered. Sorta like an /etc/passwd file. Bibliography ------------ 1. William R. Newbold. _The Cipher of Roger Bacon_Roland G Kent, ed. University of Pennsylvania Press, 1928. 2. Joseph Martin Feely. _Roger Bacon's Cipher: The Right Key Found_ Rochester N.Y.:Joseph Martin Feely, pub., 1943. 3. _The Most Mysterious Manuscript_ Robert S. Brumbaugh, ed. Southern Illinois Press, 1978 Unix filters are so wonderful. Massaging the machine readable file, we find: 4182 "words", of which 1284 are used more than once, 308 used 8+ times, 184 used 15+ times, 23 used 100+ times. Does this tell us anything about the language (if any) the text is written in? For those who may be interested, here are the 23 words used 100+ times: 121 2 115 4OFAE 114 4OFAM 155 4OFAN 195 4OFC89 162 4OFCC89 101 4OFCC9 189 89 111 8AE 492 8AM 134 8AN 156 8AR 248 OE 148 OR 111 S9 251 SC89 142 SC9 238 SOE 150 SOR 244 ZC89 116 ZC9 116 ZOE Could someone email the Voynich Ms. ref list that appeared here not very long ago? Thanks in advance... Also... I came across the following ref that is fun(?): The Voynich manuscript: an elegant enigma / M. E. D'Imperio Fort George E. Mead, Md. : National Security Agency(]) Central Security Service(?), 1978. ix, 140 p. : ill. ; 27 cm. The (?]) are mine... Sorry if this was already on the list, but the mention of the NSA (and what's the CSS?) made it jump out at me... -- Ron Carter ! rcarter@nyx.cs.du.edu rcarter GEnie 70707.3047 CIS Director ! Center for the Study of Creative Intelligence Denver, CO ! Knowledge is power. Knowledge to the people. Just say know. Distribution: na Organization: Wetware Diversions, San Francisco Keywords: From sci.archaeology: >From: jamie@cs.sfu.ca (Jamie Andrews) >Date: 16 Nov 91 00:49:08 GMT > > It seems like the person who would be most likely to solve >this Voynich manuscript cipher would have >(a) knowledge of the modern techniques for solving more complex > ciphers such as Playfairs and Vigineres; and >(b) knowledge of the possible contemporary and archaic languages > in which the plaintext could have been written. An extended discussion of the Voynich Manuscript may be found in the tape of the same name by Terence McKenna. I'm not sure who is currently publishing this particular McKenna tape but probably one of: Dolphin Tapes, POB 71, Big Sur, CA 93920 Sounds True, 1825 Pearl St., Boulder, CO 80302 Sound Photosynthesis, POB 2111, Mill Valley, CA 94942 The Spring 1988 issue of Gnosis magazine contained an article by McKenna giving some background of the Voynich Manuscipt and attempts to decipher it, and reviewing Leo Levitov's "Solution of the Voynich Manuscript" (published in 1987 by Aegean Park Press, POB 2837 Laguna Hills, CA 92654). Levitov's thesis is that the manuscript is the only surviving primary document of the Cathar faith (exterminated on the orders of the Pope in the Albigensian Crusade in the 1230s) and that it is in fact not encrypted material but rather is a highly polyglot form of Medieval Flemish with a large number of Old French and Old High German loan words, written in a special script. As far as I know Levitov's there has been no challenge to Levitov's claims so far. Michael Barlow, who had reviewed Levitov's book in Cryptologia, had sent me photocopies of the pages where much of the language was described (pp.21-31). I have just found them, and am looking at them now as I am typing this. Incidentally, I do not believe this has anything to do with cryptology proper, but the decipherment of texts in unknown languages. So if you are into cryptography proper, skip this. Looking at the "Voynich alphabet" pp.25-27, I made a list of the letters of the Voynich language as Levitov interprets them, and I added phonetic descriptions of the sounds I *think* Levitov meant to describe. Here it is: Letter# Phonetic Phonetic descriptions (IPA) in linguists' jargon: in plain English: 1 a low open, central unrounded a as in father e mid close, front, unrounded ay as in May O mid open, back, rounded aw as in law or o as in got (British pronunciation) 2 s unvoiced dental fricative s as in so 3 d voiced dental stop d 4 E mid, front, unrounded e as in wet 5 f unvoiced labiodental fricative f 6 i short, high open, front, i as in dim unrounded 7 i: long, high, front, unrounded ea as in weak 8 i:E (?) I can't make head nor tail of Levitov's explanations. Probably like "ei" in "weird" dragging along the "e": "weeeird"] (British pronunciation, with a silent "r") 9 C unvoiced palatal fricative ch in German ich 10 k unvoived velar stop k 11 l lateral, can't be more precise from description, probably like l in "loony" 12 m voiced bilabial nasal m 13 n voiced dental nasal n 14 r (?) cannot tell precisely from Scottish r? description Dutch r? 15 t no description; dental stop? t 16 t another form for #15 t 17 T (?) no description th as in this? th as in thick? 18 TE (?) again, no description or ET (?) 19 v voiced labiodental fricative v as in rave 20 v ditto, same as #19 ditto (By now, you will have guessed what my conclusion about Levitov's decipherment was) In the column headed "Phonetic (IPA)" I have used capital letters for lack of the special international phonetic symbols: E for the Greek letter "epsilon" O for the letter that looks like a mirror-image of "c" C for c-cedilla T for the Greek letter "theta" The colon (:) means that the sound represented by the preceding letter is long, e.g. "i:" is a long "i". The rest, #21 to 25, are not "letters" proper, but represent groups of two or more letters, just like #18 does. They are: 21 av 22a Ev 22b vE 23 CET 24 kET 25 sET That gives us a language with 6 vowels: a (#1), e (#1 again), O (#1 again), E (#4), i (#6), and i: (#7). Letter #8 is not a vowel, but a combination of two vowels: i: (#7) and probably E (#4). Levitov writes that the language is derived from Dutch. If so, it has lost the "oo" sound (English spelling; "oe" in Dutch spelling), and the three front rounded vowels of Dutch: u as in U ("you", polite), eu as in deur ("door"), u as in vlug ("quick"). Note that out of six vowels, three are confused under the same letter (#1), even though they sound very different from one another: a, e, O. Just imagine that you had no way of distinguishing between "last", "lest" and "lost" when writing in English, and you'll have a fair idea of the consequences. Let us look at the consonants now. I will put them in a matrix, with the points of articulation in one dimension, and the manner of articulation in the other (it's all standard procedure when analyzing a language). Brackets around a letter will mean that I could not tell where to place it exactly, and just took a guess. labial dental palatal velar nasal m n voiced stop d unvoiced stop t k voiced fricative v (T) unvoiced fricative f s C lateral l trill (?) (r) Note that there are only twelve consonant sounds. That is unheard of for a European language. No European language has so few consonant sounds. Spanish, which has very few sounds (only five vowels), has seventeen distinct consonants sounds, plus two semi-consonants. Dutch has from 18 to 20 consonants (depending on speakers, and how you analyze the sounds. Warning: I just counted them on the back of an envelope; I might have missed one or two). What is also extraordinary in Levitov's language is that it lacks a "g", and *BOTH* "b" and "p". I cannot think of one single language in the world that lacks both "b" and "p". Levitov also says that "m" occurs only word-finally, never at the beginning, nor in the middle of a word. That's true: the letter he says is an "m" is always word-final in the reproductions I have seen of the Voynich MS. But no language I know of behaves like that. All have an "m" (except one American Indian language, which is very famous for that, and the name of which escapes me right now), but, if there is a position where "m" never appears in some languages, that position is word-finally. Exactly the reverse of Levitov's language. What does Levitov say about the origin of the language? "The language was very much standardized. It was an application of a polyglot oral tongue into a literary language which would be understandable to people who did not understand Latin and to whom this language could be read." At first reading, I would dismiss it all as nonsense: "polyglot oral tongue" means nothing in linguistics terms. But Levitov is a medical doctor, so allowances must be made. The best meaning I can read into "polyglot oral tongue" is "a language that had never been written before and which had taken words from many different languages". That is perfectly reasonable: English for one, has done that. Half its vocabulary is Norman French, and some of the commonest words have non-Anglo-Saxon origins. "Sky", for instance, is a Danish word. So far, so good. Levitov continues: "The Voynich is actually a simple language because it follows set rules and has a very limited vocabulary.... There is a deliberate duality and plurality of words in the Voynich and much use of apostrophism". By "duality and plurality of words" Levitov means that the words are highly ambiguous, most words having two or more different meanings. I can only guess at what he means by apostrophism: running words together, leaving bits out, as we do in English: can not --> cannot --> can't, is not --> ain't. Time for a tutorial in the Voynich language as I could piece it together from Levitov's description. Because, according to Levitov, letter #1 represent 3 vowels sounds, I will represent it by just "a", but remember: it can be pronounced a, e, or o. But I will distinguish, as does Levitov, between the two letters which he says were both pronounced "v", using "v" for letter #20 and "w" for letter #21. Some vocabulary now. Some verbs first, which Levitov gives in the infinitive. In the Voynich language the infinitive of verbs ends in -en, just like in Dutch and in German. I have removed that grammatical ending in the list which follows, and given probable etymologies in parentheses (Levitov gives doesn't give any): ad = to aid, help ("aid") ak = to ache, pain ("ache") al = to ail ("ail") and = to undergo the "Endura" rite ("EndÝura¨", probably) d = to die ("dÝie¨") fad = to be for help (from f= for and ad=aid) fal = to fail ("fail") fil = to be for illness (from: f=for and il=ill) il = to be ill ("ill") k = to understand ("ken", Dutch and German "kennen" meaning "to know") l = to lie deathly ill, in extremis ("lie", "lay") s = to see ("see", Dutch "zien") t = to do, treat (German "tun" = to do) v = to will ("will" or Latin "volo" perhaps) vid = to be with death (from vi=with and d=die) vil = to want, wish, desire (German "willen") vis = to know ("wit", German "wissen", Dutch "weten") vit = to know (ditto) viT = to use (no idea, Latin "uti" perhaps?) vi = to be the way (Latin "via") eC = to be each ("each") ai:a = to eye, look at ("eye", "oog" in Dutch) en = to do (no idea) Example given by Levitov: enden "to do to death" made up of "en" (to do), "d" (to die) and "en" (infinitive ending). Well, to me, that's doing it the hard way. What's wrong with just "enden" = to end (German "enden", too]) More vocabulary: em = he or they (masculine) ("him") er = her or they (feminine) ("her") eT = it or they ("it" or perhaps "they" or Dutch "het") an = one ("one", Dutch "een") "There are no declensions of nouns or conjugation of verbs. Only the present tense is used" says Levitov. Examples: den = to die (infinitive) (d = die, -en = infinitive) deT = it/they die (d = die, eT = it/they) diteT = it does die (d = die, t = do, eT = it/they, with an "i" added to make it easier to pronounce, which is quite common and natural in languages) But Levitov contradicts himself immediately, giving another tense (known as present progressive in English grammar): dieT = it is dying But I may be unfair there, perhaps it is a compound: d = die, i = is ...-ing, eT = it/they. Plurals are formed by suffixing "s" in one part of the MS, "eT" in another: "ans" or "aneT" = ones. More: wians = we ones (wi = we, wie in Dutch, an = one, s = plural) vian = one way (vi = way, an = one) wia = one who (wi = who, a = one) va = one will (v = will, a = one) wa = who wi = who wieT = who, it (wi = who, eT = it) witeT = who does it (wi = who, t = do, eT = it/they) weT = who it is (wi = who, eT = it, then loss of "i", giving "weT") ker = she understands (k = understand, er =she) At this stage I would like to comment that we are here in the presence of a Germanic language which behaves very, very strangely in the way of the meanings of its compound words. For instance, "viden" (to be with death) is made up of the words for "with", "die" and the infinitive suffix. I am sure that Levitov here was thinking of a construction like German "mitkommen" which means "to come along" (to "withcome"). I suppose I could say "Bitte, sterben Sie mit" on the same model as "Bitte, kommen Sie mit" ("Come with me/us, please), thereby making up a verb "mitsterben", but that would mean "to die together with someone else", not "to be with death". Let us see how Levitov translates a whole sentence. Since he does not explain how he breaks up those compound words I have tried to do it using the vocabulary and grammar he provides in those pages. My tentative explanations are in parenthesis. TanvieT faditeT wan aTviteT anTviteT atwiteT aneT TanvieT = the one way (T = the (?), an = one, vi =way, eT = it) faditeT = doing for help (f = for, ad = aid, i = -ing, t = do, eT = it) wan = person (wi/wa = who, an = one) aTviteT = one that one knows (a = one, T = that, vit = know, eT = it. Here, Levitov adds one extra letter which is not in the text, getting "aTaviteT", which provide the second "one" of his translation) anTviteT = one that knows (an =one, T = that, vit = know, eT = it) atwiteT = one treats one who does it (a = one, t = do, wi = who, t = do, eT = it. Literally: "one does Ýone¨ who does it". The first "do" is translated as "treat", the second "one" is added in by Levitov: he added one letter, which gives him "atawiteT") aneT = ones (an = one, -eT = the plural ending) Levitov's translation of the above in better English: "the one way for helping a person who needs it, is to know one of the ones who do treat one". Need I say more? Does anyone still believe that Levitov's translations are worth anything? As an exercise, here is the last sentence on p.31, with its word-for-word translation by Levitov. I leave you to work it out, and to figure out what it might possibly mean. Good luck] tvieT nwn anvit fadan van aleC tvieT = do the ways nwn = not who does (but Levitov adds a letter to make it "nwen") anvit = one knows fadan = one for help van = one will aleC = each ail ==> cryptology/swiss.colony.p <== What are the 1987 Swiss Colony ciphers? ==> cryptology/swiss.colony.s <== Did anyone solve the 1987 'Crypto-gift' contest that was run by Swiss Colony? My friend and I worked on it for 4 months, but didn't get anywhere. My friend solved the 1986 puzzle in about a week and won $1000. I fear that we missed some clue that makes it incredibly easy to solve. I'm including the code, clues and a few notes for those of you so inclined to give it a shot. 197,333,318,511,824, 864,864,457,197,333, 824,769,372,769,864, 865,457,153,824,511,223,845,318, 489,953,234,769,703,489,845,703, 372,216,457,509,333,153,845,333, 511,864,621,611,769,707,153,333, 703,197,845,769,372,621,223,333, 197,845,489,953,223,769,216,223, 769,769,457,153,824,511,372,223, 769,824,824,216,865,845,153,769, 333,704,511,457,153,333,824,333, 953,372,621,234,953,234,865,703, 318,223,333,489,944,153,824,769, 318,457,234,845,318,223,372,769, 216,894,153,333,511,611, 769,704,511,153,372,621, 197,894,894,153,333,953, 234,845,318,223 CHRIS IS BACK WITH GOLD FOR YOU HIS RHYMES CONTAIN THE SECRET. YOU SCOUTS WHO'VE EARNED YOUR MERIT BADGE WILL QUICKLY LEARN TO READ IT. SO WHEN YOUR CHRISTMAS HAM'S ALL GONE AND YOU'RE READY FOR THE TUSSLE, BALL UP YOUR HAND INTO A FIST AND SHOW OUR MOUSE YOUR MUSCLE. PLEASE READ THESE CLUES WE LEAVE TO YOU BOTH FINE ONES AND THE COARSE; IF CARE IS USED TO HEED THEM ALL YOU'LL SUFFER NO REMORSE. Notes: The puzzle comes as a jigsaw that when assembled has the list of numbers. They are arranged as indicated on the puzzle, with commas. The lower right corner has a drawing of 'Secret Agent Chris Mouse'. He holds a box under his arm which looks like the box the puzzle comes in. The upper left corner has the words 'NEW 1987 $50,000 Puzzle'. The lower left corner is empty. The clues are printed on the entry form in upper case, with the punctuation as shown. Ed Rupp ...]ut-sally]oakhill]ed Motorola, Inc., Austin Tx. ==> decision/allais.p <== The Allais Paradox involves the choice between two alternatives: A. 89% chance of an unknown amount 10% chance of $1 million 1% chance of $1 million B. 89% chance of an unknown amount (the same amount as in A) 10% chance of $2.5 million 1% chance of nothing What is the rational choice? Does this choice remain the same if the unknown amount is $1 million? If it is nothing? ==> decision/allais.s <== This is "Allais' Paradox". Which choice is rational depends upon the subjective value of money. Many people are risk averse, and prefer the better chance of $1 million of option A. This choice is firm when the unknown amount is $1 million, but seems to waver as the amount falls to nothing. In the latter case, the risk averse person favors B because there is not much difference between 10% and 11%, but there is a big difference between $1 million and $2.5 million. Thus the choice between A and B depends upon the unknown amount, even though it is the same unknown amount independent of the choice. This violates the "independence axiom" that rational choice between two alternatives should depend only upon how those two alternatives differ. However, if the amounts involved in the problem are reduced to tens of dollars instead of millions of dollars, people's behavior tends to fall back in line with the axioms of rational choice. People tend to choose option B regardless of the unknown amount. Perhaps when presented with such huge numbers, people begin to calculate qualitatively. For example, if the unknown amount is $1 million the options are: A. a fortune, guaranteed B. a fortune, almost guaranteed a tiny chance of nothing Then the choice of A is rational. However, if the unknown amount is nothing, the options are: A. small chance of a fortune ($1 million) large chance of nothing B. small chance of a larger fortune ($2.5 million) large chance of nothing In this case, the choice of B is rational. The Allais Paradox then results from the limited ability to rationally calculate with such unusual quantities. The brain is not a calculator and rational calculations may rely on things like training, experience, and analogy, none of which would be help in this case. This hypothesis could be tested by studying the correlation between paradoxical behavior and "unusualness" of the amounts involved. If this explanation is correct, then the Paradox amounts to little more than the observation that the brain is an imperfect rational engine. ==> decision/division.p <== N-Person Fair Division If two people want to divide a pie but do not trust each other, they can still ensure that each gets a fair share by using the technique that one person cuts and the other person chooses. Generalize this technique to more than two people. Take care to ensure that no one can be cheated by a coalition of the others. ==> decision/division.s <== N-Person Fair Division Number the people from 1 to N. Person 1 cuts off a piece of the pie. Person 2 can either diminish the size of the cut off piece or pass. The same for persons 3 through N. The last person to touch the piece must take it and is removed from the process. Repeat this procedure with the remaining N - 1 people, until everyone has a piece. (cf. Luce and Raiffa, "Games and Decisions", Wiley, 1957, p. 366) There is a cute result in combinatorics called the Marriage Theorem. A village has n men and n women, such that for all 0 < k <= n and for any set of k men there are at least k women, each of whom is in love with at least one of the k men. All of the men are in love with all of the women :-}. The theorem asserts that there is a way to arrange the village into n monogamous couplings. The Marriage Theorem can be applied to the Fair Pie-Cutting Problem. One player cuts the pie into n pieces. Each of the players labels some non-null subset of the pieces as acceptable to him. For reasons given below he should "accept" each piece of size > 1/n, not just the best piece(s). The pie-cutter is required to "accept" all of the pieces. Given a set S of players let S' denote the set of pie-pieces acceptable to at least one player in S. Let t be the size of the largest set (T) of players satisfying !T! > !T'!. If there is no such set, the Marriage Theorem can be applied directly. Since the pie-cutter accepts every piece we know that t < n. Choose !T! - !T'! pieces at random from outside T', glue them together with the pieces in T' and let the players in T repeat the game with this smaller (t/n)-size pie. This is fair since they all rejected the other n-t pieces, so they believe this pie is larger than t/n. The remaining n-t players can each be assigned one of the remaining n-t pie-pieces without further ado due to the Marriage Theorem. (Otherwise the set T above was not maximal.) ==> decision/dowry.p <== Sultan's Dowry A sultan has granted a commoner a chance to marry one of his hundred daughters. The commoner will be presented the daughters one at a time. When a daughter is presented, the commoner will be told the daughter's dowry. The commoner has only one chance to accept or reject each daughter; he cannot return to a previously rejected daughter. The sultan's catch is that the commoner may only marry the daughter with the highest dowry. What is the commoner's best strategy assuming he knows nothing about the distribution of dowries? ==> decision/dowry.s <== Solution Since the commoner knows nothing about the distribution of the dowries, the best strategy is to wait until a certain number of daughters have been presented then pick the highest dowry thereafter. The exact number to skip is determined by the condition that the odds that the highest dowry has already been seen is just greater than the odds that it remains to be seen AND THAT IF IT IS SEEN IT WILL BE PICKED. This amounts to finding the smallest x such that: x/n > x/n * (1/(x+1) + ... + 1/(n-1)). Working out the math for n=100 and calculating the probability gives: The commoner should wait until he has seen 37 of the daughters, then pick the first daughter with a dowry that is bigger than any preceding dowry. With this strategy, his odds of choosing the daughter with the highest dowry are surprisingly high: about 37%. (cf. F. Mosteller, "Fifty Challenging Problems in Probability with Solutions", Addison-Wesley, 1965, #47; "Mathematical Plums", edited by Ross Honsberger, pp. 104-110) ==> decision/envelope.p <== Someone has prepared two envelopes containing money. One contains twice as much money as the other. You have decided to pick one envelope, but then the following argument occurs to you: Suppose my chosen envelope contains $X, then the other envelope either contains $X/2 or $2X. Both cases are equally likely, so my expectation if I take the other envelope is .5 * $X/2 + .5 * $2X = $1.25X, which is higher than my current $X, so I should change my mind and take the other envelope. But then I can apply the argument all over again. Something is wrong here] Where did I go wrong? In a variant of this problem, you are allowed to peek into the envelope you chose before finally settling on it. Suppose that when you peek you see $100. Should you switch now? ==> decision/envelope.s <== Let's follow the argument carefully, substituting real numbers for variables, to see where we went wrong. In the following, we will assume the envelopes contain $100 and $200. We will consider the two equally likely cases separately, then average the results. First, take the case that X=$100. "I have $100 in my hand. If I exchange I get $200. The value of the exchange is $200. The value from not exchanging is $100. Therefore, I gain $100 by exchanging." Second, take the case that X=$200. "I have $200 in my hand. If I exchange I get $100. The value of the exchange is $100. The value from not exchanging is $200. Therefore, I lose $100 by exchanging." Now, averaging the two cases, I see that the expected gain is zero. So where is the slip up? In one case, switching gets X/2 ($100), in the other case, switching gets 2X ($200), but X is different in the two cases, and I can't simply average the two different X's to get 1.25X. I can average the two numbers ($100 and $200) to get $150, the expected value of switching, which is also the expected value of not switching, but I cannot under any circumstances average X/2 and 2X. This is a classic case of confusing variables with constants. OK, so let's consider the case in which I looked into the envelope and found that it contained $100. This pins down what X is: a constant. Now the argument is that the odds of $50 is .5 and the odds of $200 is .5, so the expected value of switching is $125, so we should switch. However, the only way the odds of $50 could be .5 and the odds of $200 could be .5 is if all integer values are equally likely. But any probability distribution that is finite and equal for all integers would sum to infinity, not one as it must to be a probability distribution. Thus, the assumption of equal likelihood for all integer values is self-contradictory, and leads to the invalid proof that you should always switch. This is reminiscent of the plethora of proofs that 0=1; they always involve some illegitimate assumption, such as the validity of division by zero. Limiting the maximum value in the envelopes removes the self-contradiction and the argument for switching. Let's see how this works. Suppose all amounts up to $1 trillion were equally likely to be found in the first envelope, and all amounts beyond that would never appear. Then for small amounts one should indeed switch, but not for amounts above $500 billion. The strategy of always switching would pay off for most reasonable amounts but would lead to disastrous losses for large amounts, and the two would balance each other out. For those who would prefer to see this worked out in detail: Assume the smaller envelope is uniform on Ý$0,$M¨, for some value of $M. What is the expectation value of always switching? A quarter of the time $100 >= $M (i.e. 50% chance $X is in Ý$M/2,$M¨ and 50% chance the larger envelope is chosen). In this case the expected switching gain is -$50 (a loss). Thus overall the always switch policy has an expected (relative to $100) gain of (3/4)*$50 + (1/4)*(-$50) = $25. However the expected absolute gain (in terms of M) is: / M ! g f(g) dg, Ý where f(g) = (1/2)*UniformÝ0,M)(g) + /-M (1/2)*Uniform(-M,0¨(g). ¨ = 0. QED. OK, so always switching is not the optimal switching strategy. Surely there must be some strategy that takes advantage of the fact that we looked into the envelope and we know something we did not know before we looked. Well, if we know the maximum value $M that can be in the smaller envelope, then the optimal decision criterion is to switch if $100 < $M, otherwise stick. The reason for the stick case is straightforward. The reason for the switch case is due to the pdf of the smaller envelope being twice as high as that of the larger envelope over the range Ý0,$M). That is, the expected gain in switching is (2/3)*$100 + (1/3)*(-$50) = $50. What if we do not know the maximum value of the pdf? You can exploit the "test value" technique to improve your chances. The trick here is to pick a test value T. If the amount in the envelope is less than the test value, switch; if it is more, do not. This works in that if T happens to be in the range ÝM,2M¨ you will make the correct decision. Therefore, assuming the unknown pdf is uniform on Ý0,M¨, you are slightly better off with this technique. Of course, the pdf may not even be uniform, so the "test value" technique may not offer much of an advantage. If you are allowed to play the game repeatedly, you can estimate the pdf, but that is another story... ==> decision/exchange.p <== At one time, the Mexican and American dollars were devalued by 10 cents on each side of the border (i.e. a Mexican dollar was 90 cents in the US, and a US dollar was worth 90 cents in Mexico). A man walks into a bar on the American side of the border, orders 10 cents worth of beer, and tenders a Mexican dollar in change. He then walks across the border to Mexico, orders 10 cents worth of beer and tenders a US dollar in change. He continues this throughout the day, and ends up dead drunk with the original dollar in his pocket. Who pays for the drinks? ==> decision/exchange.s <== The man paid for all the drinks. But, you say, he ended up with the same amount of money that he started with] However, as he transported Mexican dollars into Mexico and US dollars into the US, he performed "economic work" by moving the currency to a location where it was in greater demand (and thus valued higher). The earnings from this work were spent on the drinks. Note that he can only continue to do this until the Mexican bar runs out of US dollars, or the US bar runs out of Mexican dollars, i.e., until he runs out of "work" to do. ==> decision/newcomb.p <== Newcomb's Problem A being put one thousand dollars in box A and either zero or one million dollars in box B and presents you with two choices: (1) Open box B only. (2) Open both box A and B. The being put money in box B only if it predicted you will choose option (1). The being put nothing in box B if it predicted you will do anything other than choose option (1) (including choosing option (2), flipping a coin, etc.). Assuming that you have never known the being to be wrong in predicting your actions, which option should you choose to maximize the amount of money you get? ==> decision/newcomb.s <== This is "Newcomb's Paradox". You are presented with two boxes: one certainly contains $1000 and the other might contain $1 million. You can either take one box or both. You cannot change what is in the boxes. Therefore, to maximize your gain you should take both boxes. However, it might be argued that you can change the probability that the $1 million is there. Since there is no way to change whether the million is in the box or not, what does it mean that you can change the probability that the million is in the box? It means that your choice is correlated with the state of the box. Events which proceed from a common cause are correlated. My mental states lead to my choice and, very probably, to the state of the box. Therefore my choice and the state of the box are highly correlated. In this sense, my choice changes the "probability" that the money is in the box. However, since your choice cannot change the state of the box, this correlation is irrelevant. The following argument might be made: your expected gain if you take both boxes is (nearly) $1000, whereas your expected gain if you take one box is (nearly) $1 million, therefore you should take one box. However, this argument is fallacious. In order to compute the expected gain, one would use the formulas: E(take one) = $0 * P(predict take both ! take one) + $1,000,000 * P(predict take one ! take one) E(take both) = $1,000 * P(predict take both ! take both) + $1,001,000 * P(predict take one ! take both) While you are given that P(do X ! predict X) is high, it is not given that P(predict X ! do X) is high. Indeed, specifying that P(predict X ! do X) is high would be equivalent to specifying that the being could use magic (or reverse causality) to fill the boxes. Therefore, the expected gain from either action cannot be determined from the information given. ==> decision/prisoners.p <== Three prisoners on death row are told that one of them has been chosen at random for execution the next day, but the other two are to be freed. One privately begs the warden to at least tell him the name of one other prisoner who will be freed. The warden relents: 'Susie will go free.' Horrified, the first prisoner says that because he is now one of only two remaining prisoners at risk, his chances of execution have risen from one-third to one-half] Should the warden have kept his mouth shut? ==> decision/prisoners.s <== Each prisoner had an equal chance of being the one chosen to be executed. So we have three cases: Prisoner executed: A B C Probability of this case: 1/3 1/3 1/3 Now, if A is to be executed, the warden will randomly choose either B or C, and tell A that name. When B or C is the one to be executed, there is only one prisoner other than A who will not be executed, and the warden will always give that name. So now we have: Prisoner executed: A A B C Name given to A: B C C B Probability: 1/6 1/6 1/3 1/3 We can calculate all this without knowing the warden's answer. When he tells us B will not be executed, we eliminate the middle two choices above. Now, among the two remaining cases, C is twice as likely as A to be the one executed. Thus, the probability that A will be executed is still 1/3, and C's chances are ==> decision/red.p <== I show you a shuffled deck of standard playing cards, one card at a time. At any point before I run out of cards, you must say "RED]". If the next card I show is red (i.e. diamonds or hearts), you win. We assume I the "dealer" don't have any control over what the order of cards is. The question is, what's the best strategy, and what is your probability of winning ? ==> decision/red.s <== If a deck has n cards, r red and b black, the best strategy wins with a probability of r/n. Thus, you can say "red" on the first card, the last card, or any other card you wish. Proof by induction on n. The statement is clearly true for one-card decks. Suppose it is true for n-card decks, and add a red card. I will even allow a nondeterministic strategy, meaning you say "red" on the first card with probability p. With probability 1-p, you watch the first card go by, and then apply the "optimal" strategy to the remaining n-card deck, since you now know its composition. The odds of winning are therefore: p * (r+1)/(n+1) + (1-p) * ((r+1)/(n+1) * r/n + b/(n+1) * (r+1)/n). After some algebra, this becomes (r+1)/(n+1) as expected. Adding a black card yields: p * r/(n+1) + (1-p) * (r/(n+1) * (r-1)/n + (b+1)/(n+1) * r/n). This becomes r/(n+1) as expected. ==> decision/rotating.table.p <== Four glasses are placed upside down in the four corners of a square rotating table. You wish to turn them all in the same direction, either all up or all down. You may do so by grasping any two glasses and, optionally, turning either over. There are two catches: you are blindfolded and the table is spun after each time you touch the glasses. How do you do it? ==> decision/rotating.table.s <== 1. Turn two adjacent glasses up. 2. Turn two diagonal glasses up. 3. Pull out two diagonal glasses. If one is down, turn it up and you're done. If not, turn one down and replace. 4. Take two adjacent glasses. Invert them both. 5. Take two diagonal glasses. Invert them both. References Probing the Rotating Table" W. T. Laaser and L. Ramshaw _The Mathematical Gardner_, Wadsworth International, Belmont CA 1981. ... we will see that such a procedure exists if and only if the parameters k and n satisfy the inequality k >= (1-1/p)n, where p is the largest prime factor of n. The paper mentions (without discussing) two other generalizations: more than two orientations of the glasses (Graham and Diaconis) and more symmetries in the table, e.g. those of a cube (Kim). ==> decision/stpetersburg.p <== What should you be willing to pay to play a game in which the payoff is calculated as follows: a coin is flipped until in comes up heads on the nth toss and the payoff is set at 2^n dollars? ==> decision/stpetersburg.s <== Classical decison theory says that you should be willing to pay any amount up to the expected value of the wager. Let's calculate the expected value: The probability of winning at step n is 2^-n, and the payoff at step n is 2^n, so the sum of the products of the probabilities and the payoffs is: E = sum over n (2^-n * 2^n) = sum over n (1) = infinity So you should be willing to pay any amount to play this game. This is called the "St. Petersburg Paradox." The classical solution to this problem was given by Bernoulli. He noted that people's desire for money is not linear in the amount of money involved. In other words, people do not desire $2 million twice as much as they desire $1 million. Suppose, for example, that people's desire for money is a logarithmic function of the amount of money. Then the expected VALUE of the game is: E = sum over n (2^-n * C * log(2^n)) = sum over n (2^-n * C' * n) = C'' Here the C's are constants that depend upon the risk aversion of the player, but at least the expected value is finite. However, it turns out that these constants are usually much higher than people are really willing to pay to play, and in fact it can be shown that any non-bounded utility function (map from amount of money to value of money) is prey to a generalization of the St. Petersburg paradox. So the classical solution of Bernoulli is only part of the story. The rest of the story lies in the observation that bankrolls are always finite, and this dramatically reduces the amount you are willing to bet in the St. Petersburg game. To figure out what would be a fair value to charge for playing the game we must know the bank's resources. Assume that the bank has 1 million dollars (1*K*K = 2^20). I cannot possibly win more than $1 million whether I toss 20 tails in a row or 2000. Therefore my expected amount of winning is E = sum n up to 20 (2^-n * 2^n) = sum n up to 20 (1) = $20 and my expected value of winning is E = sum n up to 20 (2^-n * C * log(2^n)) = some small number This is much more in keeping with what people would really pay to play the game. Incidentally, T.C. Fry suggested this change to the problem in 1928 (see W.W.R. Ball, Mathematical Recreations and Essays, N.Y.: Macmillan, 1960, pp. 44-45). The problem remains interesting when modified in this way, for the following reason. For a particular value of the bank's resources, let e denote the expected value of the player's winnings; and let p denote the probability that the player profits from the game, assuming the price of getting into the game is 0.8e (20% discount). Note that the expected value of the player's profit is 0.2e. Now let's vary the bank's resources and observe how e and p change. It will be seen that as e (and hence the expected value of the profit) increases, p diminishes. The more the game is to the player's advantage in terms of expected value of profit, the less likely it is that the player will come away with any profit at all. This is mildly counterintuitive. ==> decision/switch.p <== Switch? (The Monty Hall Problem) Two black marbles and a red marble are in a bag. You choose one marble from the bag without looking at it. Another person chooses a marble from the bag and it is black. You are given a chance to keep the marble you have or switch it with the one in the bag. If you want to end up with the red marble, is there an advantage to switching? What if the other person looked at the marbles remaining in the bag and purposefully selected a black one? ==> decision/switch.s <== Generalize the problem from three marbles to n marbles. If there are n marbles, your odds of having selected the red one are 1/n. After the other person selected a black one at random, your odds go up to 1/(n-1). There are n-2 marbles left in the bag, so your odds of selecting the red one by switching are 1/(n-2) times the odds that you did not already select it (n-2)/(n-1) or 1/(n-1), the same as the odds of already selecting it. Therefore there is no advantage to switching. If the person looked into the bag and selected a black one on purpose, then your odds of having selected the red one are not improved, so the odds of selecting the red one by switching are 1/(n-2) times (n-1)/n or (n-1)/n(n-2). This is (n-1)/(n-2) times better than the odds without switching, so you should switch. This is a clarified version of the Monty Hall "paradox": You are a participant on "Let's Make a Deal." Monty Hall shows you three closed doors. He tells you that two of the closed doors have a goat behind them and that one of the doors has a new car behind it. You pick one door, but before you open it, Monty opens one of the two remaining doors and shows that it hides a goat. He then offers you a chance to switch doors with the remaining closed door. Is it to your advantage to do so? The original Monty Hall problem (and solution) appears to be due to Steve Selvin, and appears in American Statistician, Feb 1975, V. 29, No. 1, p. 67 under the title ``A Problem in Probability.'' It should be of no surprise to readers of this group that he received several letters contesting the accuracy of his solution, so he responded two issues later (American Statistician, Aug 1975, V. 29, No. 3, p. 134). I extract a few words of interest, including a response from Monty Hall himself: ... The basis to my solution is that Monty Hall knows which box contains the prize and when he can open either of two boxes without exposing the prize, he chooses between them at random ... Benjamin King pointed out the critical assumptions about Monty Hall's behavior that are necessary to solve the problem, and emphasized that ``the prior distribution is not the only part of the probabilistic side of a decision problem that is subjective.'' Monty Hall wrote and expressed that he was not ``a student of statistics problems'' but ``the big hole in your argument is that once the first box is seen to be empty, the contestant cannot exchange his box.'' He continues to say, ``Oh, and incidentally, after one Ýbox¨ is seen to be empty, his chances are not 50/50 but remain what they were in the first place, one out of three. It just seems to the contestant that one box having been eliminated, he stands a better chance. Not so.'' I could not have said it better myself. The basic idea is that the Monty Hall problem is confusing for two reasons: first, there are hidden assumptions about Monty's motivation that cloud the issue in some peoples' minds; and second, novice probability students do not see that the opening of the door gave them any new information. Monty can have one of three basic motives: 1. He randomly opens doors. 2. He always opens the door he knows contains nothing. 3. He only opens a door when the contestant has picked the grand prize. These result in very different strategies: 1. No improvement when switching. 2. Double your odds by switching. 3. Don't switch] Most people, myself included, think that (2) is the intended interpretation of Monty's motive. A good way to see that Monty is giving you information by opening doors is to increase the number of doors from three to 100. If there are 100 doors, and Monty shows that 98 of them are empty, isn't it pretty clear that the chance the prize is behind the remaining door is 99/100? Reference (too numerous to mention, but this one should do): Leonard Gillman "The Car and the Goats" The American Mathematical Monthly, 99:1 (Jan 1992), pp. 3-7. ==> decision/truel.p <== A, B, and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 0.3, C's is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left. What should A's strategy be? ==> decision/truel.s <== This is problem 20 in Mosteller _Fifty Challenging Problems in Probability_ and it also appears (with an almost identical solution) on page 82 in Larsen & Marx _An Introduction to Probability and Its Applications_. Here's Mosteller's solution: A is naturally not feeling cheery about this enterprise. Having the first shot he sees that, if he hits C, B will then surely hit him, and so he is not going to shoot at C. If he shoots at B and misses him, then B clearly {I disagree; this is not at all clear]} shoots the more dangerous C first, and A gets one shot at B with probability 0.3 of succeeding. If he misses this time, the less said the better. On the other hand, suppose A hits B. Then C and A shoot alternately until one hits. A's chance of winning is (.5)(.3) + (.5)^2(.7)(.3) + (.5)^3(.7)^2(.3) + ... . Each term cooresponds to a sequence of misses by both C and A ending with a final hit by A. Summing the geometric series we get ... 3/13 < 3/10. Thus hitting B and finishing off with C has less probability of winning for A than just missing the first shot. So A fires his first shot into the ground and then tries to hit B with his next shot. C is out of luck. As much as I respect Mosteller, I have some serious problems with this solution. If we allow the option of firing into the ground, then if all fire into the ground with every shot, each will survive with probability 1. Now, the argument could be made that a certain strategy for X that both allows them to survive with probability 1 *and* gives less than a probability of survival of less than 1 for at least one of their foes would be preferred by X. However, if X pulls the trigger and actually hits someone what would the remaining person, say Y, do? If P(X hits)=1, clearly Y must try to hit X, since X firing at Y with intent to hit dominates any other strategy for X. If P(X hits)<1 and X fires at Y with intent to hit, then P(Y survives)<1 (since X could have hit Y). Thus, Y must insure that X can not follow this strategy by shooting back at X (thus insuring that P(X survives)<1). Therefore, I would conclude that the ideal strategy for all three players, assuming that they are rational and value survival above killing their enemies, would be to keep firing into the ground. If they don't value survival above killing their enemies (which is the only a priori assumption that I feel can be safely made in the absence of more information), then the problem can't be solved unless the function each player is trying to maximize is explicitly given. -- -- clong@remus.rutgers.edu (Chris Long) OK - I'll have a go at this. How about the payoff function being 1 if you win the "duel" (i.e. if at some point you are still standing and both the others have been shot) and 0 otherwise? This should ensure that an infinite sequence of deliberate misses is not to anyone's advantage. Furthermore, I don't think simple survival makes a realistic payoff function, since people with such a payoff function would not get involved in the fight in the first place] Ý I.e. I am presupposing a form of irrationality on the part of the fighters: they're only interested in survival if they win the duel. Come to think of it, this may be quite rational - spending the rest of my life firing a gun into the ground would be a very unattractive proposition to me :-) ¨ Now, denote each position in the game by the list of people left standing, in the order in which they get their turns (so the initial position is (A,B,C), and the position after A misses the first shot (B,C,A)). We need to know the value of each possible position for each person. By definition: valA(A) = 1 valB(A) = 0 valC(A) = 0 valA(B) = 0 valB(B) = 1 valC(B) = 0 valA(C) = 0 valB(C) = 0 valC(C) = 1 Consider the two player position (X,Y). An infinite sequence of misses has value zero to both players, and each player can ensure a positive payoff by trying to shoot the other player. So both players deliberately missing is a sub-optimal result for both players. The question is then whether both players should try to shoot the other first, or whether one should let the other take the first shot. Since having the first shot is always an advantage, given that some real shots are going to be fired, both players should try to shoot the other first. It is then easy to establish that: valA(A,B) = 3/10 valB(A,B) = 7/10 valC(A,B) = 0 valA(B,A) = 0 valB(B,A) = 1 valC(B,A) = 0 valA(B,C) = 0 valB(B,C) = 1 valC(B,C) = 0 valA(C,B) = 0 valB(C,B) = 5/10 valC(C,B) = 5/10 valA(C,A) = 3/13 valB(C,A) = 0 valC(C,A) = 10/13 valA(A,C) = 6/13 valB(A,C) = 0 valC(A,C) = 7/13 Now for the three player positions (A,B,C), (B,C,A) and (C,A,B). Again, the fact that an infinite sequence of misses is sub-optimal for all three players means that at least one player is going to decide to fire. However, it is less clear than in the 2 player case that any particular player is going to fire. In the 2 player case, each player knew that *if* it was sub-optimal for him to fire, then it was optimal for the other player to fire *at him* and that he would be at a disadvantage in the ensuing duel because of not having got the first shot. This is not necessarily true in the 3 player case. Consider the payoff to A in the position (A,B,C). If he shoots at B, his expected payoff is: 0.3*valA(C,A) + 0.7*valA(B,C,A) = 9/130 + 0.7*valA(B,C,A) If he shoots at C, his expected payoff is: 0.3*valA(B,A) + 0.7*valA(B,C,A) = 0.7*valA(B,C,A) And if he deliberately misses, his expected payoff is: valA(B,C,A) Since he tries to maximise his payoff, we can immediately eliminate shooting at C as a strategy - it is strictly dominated by shooting at B. So A's expected payoff is: valA(A,B,C) = MAX(valA(B,C,A), 9/130 + 0.7*valA(B,C,A)) A similar argument shows that C's expected payoffs in the (C,A,B) position are: For shooting at A: 0.5*valC(A,B,C) For shooting at B: 35/130 + 0.5*valC(A,B,C) For missing: valC(A,B,C) So C either shoots at B or deliberately misses, and: valC(C,A,B) = MAX(valC(A,B,C), 35/130 + 0.5*valC(A,B,C)) Each player can obtain a positive expected payoff by shooting at one of the other players, and it is known that an infinite sequence of misses will result in a zero payoff for all players. So it is known that some player's strategy must involve shooting at another player rather than deliberately missing. Now look at this from the point of view of player B. He knows that *if* it is sub-optimal for him to shoot at another player, then it is optimal for at least one of the other players to shoot. He also knows that if the other players choose to shoot, they will shoot *at him*. If he deliberately misses, therefore, the best that he can hope for is that they miss him and he is presented with the same situation again. This is clearly less good for him than getting his shot in first. So in position (B,C,A), he must shoot at another player rather than deliberately miss. B's expected payoffs are: For shooting at A: valB(C,B) = 5/10 For shooting at C: valB(A,B) = 7/10 So in position (B,C,A), B shoots at C for an expected payoff of 7/10. This gives us: valA(B,C,A) = 3/10 valB(B,C,A) = 7/10 valC(B,C,A) = 0 So valA(A,B,C) = MAX(3/10, 9/130 + 21/100) = 3/10, and A's best strategy is position (A,B,C) is to deliberately miss, giving us: valA(A,B,C) = 3/10 valB(A,B,C) = 7/10 valC(A,B,C) = 0 And finally, valC(C,A,B) = MAX(0, 35/130 + 0) = 7/26, and C's best strategy in position (C,A,B) is to shoot at B, giving us: valA(C,A,B) = 57/260 valB(C,A,B) = 133/260 valC(C,A,B) = 7/26 I suspect that, with this payoff function, all positions with 3 players can be resolved. For each player, we can establish that if their correct strategy is to fire at another player, then it is to fire at whichever of the other players is more dangerous. The most dangerous of the three players then finds that he has nothing to lose by firing at the second most dangerous. Questions: (a) In the general case, what are the optimal strategies for the other two players, possibly as functions of the hit probabilities and the cyclic order of the three players? (b) What happens in the 4 or more player case? -- David Seal ==> english/acronym.p <== What acronyms have become common words? ==> english/acronym.s <== The following is the list of acronyms which have become common nouns. An acronym is "a word formed from the initial letter or letters of each of the successive parts or major parts of a compound term" (Webster's Ninth). A common noun will occur uncapitalized in Webster's Ninth. Entries in the following table include the year in which they first entered the language (according to the Ninth), and the Merriam-Webster dictionary that first contains them. The following symbols are used: NI1 New International (1909) NI1+ New Words section of the New International (1931) NI2 New International Second Edition (1934) NI2+ Addendum section of the Second (1959, same as 1954) NI3 Third New International (1961) 9C Ninth New Collegiate (1983) 12W 12,000 Words (separately published addendum to the Third, 1986) asdic Anti-Submarine Detection Investigation Committee (1940, NI2+) dew Distant Early Warning (1953, 9C) dopa DihydrOxyPhenylAlanine (1917, NI3) fido Freaks + Irregulars + Defects + Oddities (1966, 9C) jato Jet-Assisted TakeOff (1947, NI2+) laser Light Amplification by Stimulated Emission of Radiation (1957, NI3) lidar LIght Detection And Ranging (1963, 9C) maser Microwave Amplification by Stimulated Emission of Radiation (1955, NI3) nitinol NIckel + TIn + Naval Ordinance Laboratory (1968, 9C) rad Radiation Absorbed Dose (1918, NI3) radar RAdio Detection And Ranging (ca. 1941, NI2+) rem Roentgen Equivalent Man (1947, NI3) rep Roentgen Equivalent Physical (1947, NI3) scuba Self-Contained Underwater Breathing Apparatus (1952, NI3) snafu Situation Normal -- All Fucked (Fouled) Up (ca. 1940, NI2+) sofar SOund Fixing And Ranging (1946, NI2+) sonar SOund NAvigation Ranging (1945, NI2+) tepa Tri-Ethylene Phosphor-Amide (1953, 9C) zip Zone Improvement Plan (1963, 9C) Below are blends that technically are also acronyms: alnico ALuminum + NIckel + CObalt (1935, NI2+) avgas AViation GASoline (1943, NI3) boff Box OFFice (1946, NI3) ceramal CERAMic ALloy (ca. 1948, NI2+) cermet CERamic METal (1948, NI2+) comsymp COMmunist SYMPathizer (ca. 1961, 9C) cyborg CYBernetic ORGanism (ca. 1962, 9C) dorper DORset horn + blackhead PERsian (1949, NI3) elhi ELementary school + HIgh school (1948, 9C) gox Gaseous OXygen (1959, 9C) hela HEnrietta LAcks (1953, 9C) kip KIlo- + Pound (1914, NI2) linac LINear ACcelerator (1950, 9C) loran LOng-RAnge Navigation (ca. 1932, NI2+) lox Liquid OXygen (1923, 9C) mascon MASs CONcentration (1968, 9C) maximin MAXImum + MINimum (1951, 9C) minimax MINImum + MAXimum (1918, 9C) modem MOdulator + DEModulator (ca. 1952, 9C) motocross MOTOr + CROSS-country (1951, 9C) napalm NAphthenic and PALMitic acids (1942, NI2+) parsec PARallax SECond (ca. 1913, NI1+) redox REDuction + OXidation (1828, NI2) selsyn SELf-SYNchronizing (1936, NI2+) shoran SHOrt-RAnge Navigation (ca. 1932, NI2+) silvex SILVa + EXterminator (1961, 9C) sitcom SITuation COMedy (1965, 9C) teleran TELEvision-RAdar Navigation (1946, NI2+) telex TELeprinter EXchange (ca. 1943, 9C) vidicon VIDeo + ICONoscope (1950, NI3) wilco WILl COmply (ca. 1938, NI3) Acronyms from other languages: agitprop AGITatsiya + PROPaganda (Russian, ca. 1926, NI2+) flak FLiegerAbwehrKanonen (German, 1938, NI2+) gestapo GEheime STAatsPOlizei (German, 1934, NI2+) gulag Glavnoe Upravlenie ispravitel'notrudovykh LAGerei (Russian, 1974, 9C) kolkhoz KOLlektivnoe KHOZyaistvo (Russian, 1921, NI2) moped MOtor + PEDal (Swedish, ca. 1955, 9C) sambo SAMozashchita Bez Oruzhiya (Russian, 1972, 9C) Selected near misses: athodyd Aero-THermODYnamic Duct (1945, NI2+) -- blend awol Absent WithOut Leave (1919, NI2+) -- usually capitalized benday BENjamin DAY (1903, NI1+) -- blend deet Di-Ethyl Tolumide (1962, 9C) -- pronunciation of D. E. T. echovirus Enteric Cytopathogenic Human Orphan VIRUS (1955, 9C) -- blend hi-fi HIgh FIdelity (1948, NI2+) -- hyphenated ibuprofen Iso-BUtyl PROpionic PHENyl (1969, 12W) -- PH pronounced f jaygee Junior Grade (1943, NI3) -- pronunciation of J. G. jayvee Junior Varsity (1937, NI3) -- pronunciation of J. V. jeep General Purpose (1940, NI2+) -- pronunciation of G. P. op-ed OPposite EDitorial (1970, 9C) -- hyphenated pj's PaJamas (1951, NI3) -- punctuated nazi NAtionalsoZIalist (German, 1930, NI2) -- shorten & alter nystatin New York STATe + -IN (1952, NI3) -- extraneous suffix reovirus Respiratory Enteric Orphan VIRUS (1959, 9C) -- blend sci-fi SCIence FIction (1955, 9C) -- hyphenated siloxane SILicon + OXygen + methANE (1922, NI3) -- blend tokamak TOroidskaja KAmera MAGneticheskaja (Russian, 1965, 9C) -- G pron. k tradevman TRAining DEVices MAN (ca. 1947, NI3) -- blend updo UPswept hairDO (1946, NI2+) -- blend veep Vice President (1940, NI2+) -- pronunciation of V. P. warfarin Wisconsin Alumni Research Foundation + coumARIN (ca. 1950, NI3) - blend yuppie Young Urban Professional + -PIE (1983, 9C) -- extraneous suffix Acronyms that should be in Webster's Ninth: biopic BIOgraphical PICture (12W) fifo First In, First Out (NI2+) lifo Last In, First Out (NI2+) nomic NO Metal In Composition (NI3) (John Bulten) quango QUAsi-Non Governmental Organization (12W) shazam Solomon Hercules Atlas Zeus Achilles Mercury (12W) tacan TACtical Air Navigation (12W) Supposed acronyms: posh Port Out, Starboard Home spiff Sales Productivity Incentive Fund tip To Insure (should be Ensure) Politeness (or Promptness) ==> english/ambiguous.p <== What word in the English language is the most ambiguous? What is the greatest number of parts of speech that a single word can be used for? ==> english/ambiguous.s <== In Webster's Ninth, "set" occupies 1.2 columns, has 25 vb entries, 11 vi entries, 23 noun entries, 7 adjective entries; "take" occupies 1.3 columns, has 19 vb entries, 8 vi entries, 4 noun entries. The word "like" occupies eight parts of speech: verb "Fruit flies like a banana." noun "He has his likes and dislikes." adjective "People of like tastes agree." adverb "The truth is more like this." conjunction "Time flies like an arrow." preposition "She cries like a woman." interjection "Like, man, that was far out." verbal auxiliary "So loud I like to fell out of bed." ==> english/antonym.p <== What words, when a single letter is added, reverse their meanings? Exclude words that are obtained by adding an "a-" to the beginning. ==> english/antonym.s <== e: fast -> feast, fiancee -> fiance h: treat -> threat r: fiend -> friend s: he -> she t: here -> there ==> english/behead.p <== Is there a sentence that remains a sentence when all its words are beheaded? ==> english/behead.s <== Show this bold Prussian that praises slaughter, slaughter brings rout. ==> english/capital.p <== What words change pronunciation when capitalized (e.g., polish -> Polish)? ==> english/capital.s <== A partial list is: askew august begin chile colon concord degas ewe (African language) herb job levy lima messier mobile natal nice polish rainier ravel reading tang (Chinese dynasty) tangier worms (Germany city) ==> english/charades.p <== A ....... surgeon was ....... to operate because he had ....... ==> english/charades.s <== A notable surgeon was not able to operate because he had no table. ==> english/contradictory.proverbs.p <== What are some proverbs that contradict one another? ==> english/contradictory.proverbs.s <== Beware of Greeks bearing gifts. Never look a gift horse in the mouth. Look before you leap. He who hesitates is lost. Nothing venture, nothing gain. Fools rush in where angels fear to tread. Seek and ye shall find. Curiosity killed the cat. Save for a rainy day. Tomorrow will take care of itself. Life is what we make it. What is to be will be. Too many cooks spoil the broth. Many hands make light work. One man's meat is another man's poison. Sauce for the goose is sauce for the gander. With age comes wisdom. Out of the mouths of babes and sucklings come all wise sayings. Bear ye one another's burdens. (Gal. 6:2) For every man shall bear his own burden. (Gal. 6:5) Great minds run in the same channel. Fools think alike. A rolling stone gathers no moss. A setting hen never lays. ==> english/contranym.p <== What words are their own antonym? ==> english/contranym.s <== In his 1989 book _Crazy English_, Richard Lederer calls such words contranyms and lists more than 35, although some are phrases instead of words. These can be divided into homographs (same spelling) and homophones (same pronunciation). A partial list of homographs: aught = all, nothing bill = invoice, money cleave = to separate, to join clip = cut apart, fasten together comprise = contain, compose dust = to remove, add fine particles fast = rapid, unmoving literally = actually, figuratively moot = debatable, not needing to be debated (already decided) note = promise to pay, money oversight = care, error peep = look quietly, beep peer = noble, companion put = lay, throw puzzle = pose problem, solve problem quantum = very small, very large (quantum leap) ravel = entangle, disentangle resign = to quit, to sign up again sanction = to approve of, to punish sanguine = murderous, optimistic scan = to examine closely, to glance at quickly set = fix, flow skin = to cover with, remove outer covering speak = express verbally, express nonverbally table = propose ÝBritish¨, set aside temper = calmness, passion trim = cut things off, put things on A very short list of homophones: aural, oral = heard, spoken fiance, fiancee = female betrothed, male betrothed raise, raze = erect, tear down A pair of French words which can be very confusing: La symetrie (symmetry) and L'asymetrie (asymmetry). Latin: immo = yes, no Possibilities: draw (curtains, open or close) (money, withdraw, accumulate interest) eke ==> english/element.p <== The name of what element ends in "h"? ==> english/element.s <== Bismuth. "The Elements" by Tom Lehrer Sung to the tune of "The Major-General's Song" from Gilbert & Sullivan's "The Pirates of Penzance": There's antimony, arsenic, aluminum, selenium And hydrogen and oxygen and nitrogen and rhenium And nickel, neodymium, neptunium, germanium And iron, americium, ruthenium, uranium, Europium, zirconium, lutetium, vanadium And lanthanum and osmium and astatine and radium And gold and protactinium and indium and gallium And iodine and thorium and thulium and thallium. There's yttrium, ytterbium, actinium, rubidium And boron, gadolinium, niobium, iridium And strontium and silicon and silver and samarium And BISMUTH, bromine, lithium, beryllium and barium. There's holmium and helium and hafnium and erbium And phosphorous and francium and fluorine and terbium And manganese and mercury, molybdenum, magnesium, Dysprosium and scandium and cerium and cesium And lead, praseodymium and platinum, plutonium, Palladium, promethium, potassium, polonium And tantalum, technetium, titanium, tellurium And cadmium and calcium and chromium and curium. There's sulfur, californium and fermium, berkelium And also mendelevium, einsteinium, nobelium And argon, krypton, neon, radon, xenon, zinc and rhodium And chlorine, carbon, cobalt, copper, tungsten, tin and sodium. These are the only ones of which the news has come to Ha'vard And there may be many others but they haven't been discavard. ==> english/equations.p <== Each equation below contains the initials of words that will make the phrase correct. Figure out the missing words. Lower case is used only to help the initials stand out better. Example: 26 = L. of the A. would be 26 = Letters of the Alphabet 1 = G. L. for M. K. 1 = S. C. in D. P. 1 = S. S. for a M. 1 = W. on a U. 2 = H. in a W. 2 = P. in a P. 3 = B. M., S. H. T. R.] 3 = D. of the C. 3 = W. M. 4 = Q. in a F. G. 4 = S. in a Y. 5 = D. in a Z. C. 5 = D. of the C. 5 = S. in the S. C. 5 = T. on a F. 6 = P. in a P. 6 = T. Z. in the U. S. 6 = of O. and a H. D. of the O. 7 = C. in a R. 7 = K. of F. in H. P. 7 = W. of the W. 8 = L. on a S. 8 = L. on an O. 8 = S. on a S. S. 9 = D. in a Z. C., with the S. C. 9 = L. of a C. 9 = P. in the S. S. 10 = L. I. B. 11 = P. on a C. T. 11 = P. on a F. T. 12 = D. of C. 12 = D. of J. 12 = S. of the Z. 12 = T. of I. 13 = B. D. 13 = S. on the A. F. 14 = D. in a F. 15 = M. on a D. M. C. 16 = O. in the P. 18 = H. on the G. C. 20 = C. in a P. 24 = B. B. B. in a P. 24 = B. B. to a C. 24 = H. in a D. 25 = Y. of M. for a S. A. 26 = L. of the A. 29 = D. in F. in a L. Y. 32 = D. F. at which W. F. 36 = I. on a Y. S. 40 = D. and N. of the G. F. 43 = B. in E. C. of N. 46 = C. in the H. B. 50 = W. to L. Y. L. 52 = W. in a Y. 54 = C. in a D. 57 = H. V. 64 = S. on a C. 76 = T. L. the B. P. 88 = C. in the S. 88 = P. K. 90 = D. in a R. A. 96 = T., by ? 100 = B. of B. on a W. 101 = D. 101 = a S. M. L. 200 = D. for P. G. in M. 206 = B. in the H. B. 365 = D. in a Y. 432 = P. in a H. 500 = M. in the I. F. H. 500 = S. in a R. 1000 = I. in N. Y. 1000 = W. that a P. is W. 1001 = A. N. 20000 = L. U. the S. ==> english/equations.s <== This puzzle originally was printed in "Games" magazine in 1981, by Will Shortz. Many people have added to it since then. 1 = G. L. for M. K. (1 giant leap for man kind) 1 = S. C. in D. P. (1 single calorie in diet pepsi) 1 = S. S. for a M. (1 small step for a man) 1 = W. on a U. (1 wheel on a unicycle) 2 = H. in a W. (2 halves in a whole) 2 = P. in a P. (2 peas in a pod) 3 = B. M., S. H. T. R.] (3 blind mice, see how they run]) 3 = D. of the C. (Days of the Condor -- movie) 3 = W. M. (3 wise men) 4 = Q. in a F. G. (4 quarters in a football game) 4 = S. in a Y. (4 seasons in a year) 5 = D. in a Z. C. (5 digits in a zip code) 5 = D. of the C. (Days of the Condor -- book) 5 = S. in the S. C. (stars in the Southern Cross) 5 = T. on a F. (5 toes on a foot) 6 = P. in a P. (6 pigs in a poke) 6 = T. Z. in the U. S. (time zones in the United States) 6 = of O. and a H. D. of the O. (6 of one and a half dozen of the other) 7 = C. in a R. (colors in a rainbow : ROYGBIV) 7 = K. of F. in H. P. (7 kinds of fruit in hawaiian punch) 7 = W. of the W. (7 wonders of the world) 8 = L. on a S. (legs on a spider) 8 = L. on an O. (8 legs on an octopus) 8 = S. on a S. S. (8 sides on a stop sign) 9 = D. in a Z. C., with the S. C. (digits in a zip code, with the street code) 9 = L. of a C. (9 lives of a cat) 9 = P. in the S. S. (9 planets in the solar system) 10 = L. I. B. (10 little indian boys) 11 = P. on a C. T. (11 players on a cricket team) 11 = P. on a F. T. (11 players on a football team) 12 = D. of C. (12 days of Christmas) 12 = D. of J. (disciples of Jesus) 12 = S. of the Z. (12 signs of the zodiac) 12 = T. of I. (12 tribes of Israel) 13 = B. D. (13 = baker's dozen) 13 = S. on the A. F. (13 stripes on the American flag) 14 = D. in a F. (14 days in a fortnight) 15 = M. on a D. M. C. (15 men on a dead man's chest) 16 = O. in the P. (ounces in the pound) 18 = H. on the G. C. (18 holes on the golf course) 20 = C. in a P. (20 cigarettes in a pack) 24 = B. B. B. in a P. (24 black birds baked in a pie) 24 = B. B. to a C. (24 beer bottles to a case) 24 = H. in a D. (24 hours in a day) 25 = Y. of M. for a S. A. (25 years of marriage for a silver anniversary) 26 = L. of the A. (letters of the alphabet) 29 = D. in F. in a L. Y. (29 days in Febuary in a leap year.) 32 = D. F. at which W. F. (32 degrees Fahrenheit at which water freezes) 36 = I. on a Y. S. (36 inches on a yard stick) 40 = D. and N. of the G. F. (40 days and nights of the great flood) 43 = B. in E. C. of N. (beans in each cup of Nescafe) 46 = C. in the H. B. (chromosomes in the human body) 50 = W. to L. Y. L. (50 ways to leave your lover) 52 = W. in a Y. (52 weeks in a year) 54 = C. in a D. (with the J.) (54 cards in a deck with the jokers) 57 = H. V. (57 heinz varieties) 64 = S. on a C. (64 squares on a checkerboard) 76 = T. L. the B. P. (76 trombones led the big parade) 88 = C. in the S. (constellations in the sky) 88 = P. K. (88 piano keys) 90 = D. in a R. A. (90 degrees in a right angle) 96 = T., by ? (96 Tears, by ?) 100 = B. of B. on a W. (100 bottles of beer on a wall) 101 = D. (101 dalmations) 101 = a S. M. L. (101, a silly millimeter longer) 200 = D. for P. G. in M. (200 dollars for passing go in monopoly) 206 = B. in the H. B. (206 bones in the human body) 365 = D. in a Y. (365 days in a year) 432 = P. in a H. (pints in a hogshead) 500 = M. in the I. F. H. (500 miles in the Indianapolis Five Hundred) 500 = S. in a R. (sheets in a ream) 1000 = I. in N. Y. (1000 islands in new york) 1000 = W. that a P. is W. (1000 words that a picture is worth) 1001 = A. N. (1001 arabian nights, as in tales of) 20000 = L. U. the S. (20000 leagues under the sea) ==> english/fossil.p <== What are some examples of idioms that include obsolete words? ==> english/fossil.s <== These are called fossil expresions -- words that have dropped out of common use but hang around in idioms. Not all of them are separate words, some are part of other words or have prefixes or suffixes attached. There are also words which have current meaning, but the meaning in the idiom is unrelated to it. idiom fossil meaning of fossil -------------------------------------------------- swashbuckler buckler small shield newfangled fangled siezed rank and file file column to and fro fro from gormless gorm attention hem and haw haw make the sound "haw" hem and haw hem make the sound "hem" hue and cry hue outcry kit and kaboodle kaboodle collection out of kilter kilter order kith and kin kith friends let or hinderance let hinderance footpad pad highwayman pratfall prat buttocks rank and file rank row raring to go raring enthusiastic ruthless ruth compassion short shrift shrift confession spick-and-span span chunk of wood spick-and-span spick nail (spike) swashbuckler swash bluster or stagger bank teller tell to count ==> english/frequency.p <== In the English language, what are the most frequently appearing: 1) letters overall? 2) letters BEGINNING words? 3) final letters? 4) digrams (ordered pairs of letters)? ==> english/frequency.s <== web2 = word list from Webster's Second Unabridged web2a = hyphenated words and phrases from Webster's Second Unabridged both = web2 + web2a net = several gigabytes of Usenet traffic 1) Most frequently appearing letters overall: web2: eiaorn tslcup mdhygb fvkwzx qj both: eairon tslcud pmhgyb fwvkzx qj net: etaoin srhldc umpfgy wbvkxj qz 2) Most frequently appearing letters BEGINNING words: web: spcaut mbdrhi eofgnl wvkjqz yx both: spcatb umdrhf eigowl nvkqjz yx net: taisow cmbphd frnelu gyjvkx qz 3) Most frequent final letters: web: eysndr ltacmg hkopif xwubzv jq both: eydsnr tlagcm hkpoiw fxbuzv jq net: estndr yolafg mhipuk cwxbvz jq 4) Most frequent digrams (ordered pairs of letters) web: er in ti on te al an at ic en is re ra le ri ro st ne ar ... both: er in te ti on an re al at le en ra ic ar st ri ro ed ne ... net: th he in er re an on at te es or en ar ha is ou it to st nd ... Program to compute this from word list in standard input: #include #include typedef struct { int count; char nameÝ3¨; } FREQ; FREQ allÝ256¨,initialÝ256¨,terminalÝ256¨,digramÝ65536¨; int compare(p,q) FREQ *p,*q; { return q->count - p->count; } void sort_and_print(freq,count,description) FREQ *freq; int count; char *description; { register FREQ *p; (void)qsort(freq,count,sizeof(*freq),compare); puts(description); for (p=freq;pcount) printf("%s %d\n",p->name,p->count); } main() { char sÝBUFSIZ¨; register char *p; register int i; while (gets(s)]=NULL) { if (islower(*s)) { initialÝ*s¨.count++; sprintf(initialÝ*s¨.name,"%c",*s); for (p=s;*p;p++) { if (isalpha(*p)) { allÝ*p¨.count++; sprintf(allÝ*p¨.name,"%c",*p); if (isalpha(pÝ1¨)) { i = pÝ0¨*256 + pÝ1¨; digramÝi¨.count++; sprintf(digramÝi¨.name,"%c%c",pÝ0¨,pÝ1¨); } } } terminalÝ*--p¨.count++; sprintf(terminalÝ*p¨.name,"%c",*p); } } sort_and_print(all,256,"overall character distribution: "); sort_and_print(initial,256,"initial character distribution: "); sort_and_print(terminal,256,"terminal character distribution: "); sort_and_print(digram,65536,"digram distribution: "); } ==> english/gry.p <== Find three completely different words ending in "gry." ==> english/gry.s <== Aside from "angry" and "hungry" and words derived therefrom, there is only one word ending with "-gry" in Webster's Third Unabridged: "aggry." However, this word is defective in that it is part of a phrase "aggry beads." The OED's usage examples all talk about "aggry beads." Moving to older dictionaries, we find that "gry" itself is a word in Webster's Second Unabridged (and the OED): gry, n. ÝL. gry, a trifle; Gr. gry, a grunt¨ 1. a measure equal to one-tenth of a line. ÝObs.¨ (Obs. = obsolete) 2. anything very small. ÝRare.¨ This is a list of 94 words, phrases and names ending in "gry": ÝExplanation of references is given at the end of the list.¨ aggry ÝOED:1:182; W2; W3¨ Agry Dagh (Mount Agry) ÝEB11¨ ahungry ÝOED:1:194; FW; W2¨ angry ÝOED; FW; W2; W3¨ anhungry ÝOED:1:332; W2¨ Badagry ÝJohnston; EB11¨ Ballingry ÝBartholomew:40; CLG:151; RD:164, pl.49¨ begry ÝOED:1:770,767¨ bewgry ÝOED:1:1160¨ bowgry ÝOED:1:1160¨ braggry ÝOED:1:1047¨ Bugry ÝTIG¨ Chockpugry ÝWorcester¨ Cogry ÝBBC¨ cony-gry ÝOED:2:956¨ conyngry ÝOED:2:956¨ Croftangry ÝDFC, as "Chrystal Croftangry"¨ dog-hungry ÝW2¨ Dshagry ÝStieler¨ Dzagry ÝAndree¨ eard-hungry ÝCED (see "yird"); CSD¨ Echanuggry ÝCentury:103-104, on inset map, Key 104 M 2¨ Egry ÝFrance; TIG¨ ever-angry ÝW2¨ fire-angry ÝW2¨ Gagry ÝEB11¨ gry (from Latin _gry_) ÝOED:4/2:475; W2¨ gry (from Romany _grai_) ÝW2¨ haegry ÝEDD (see "hagery")¨ half-angry ÝW2¨ hangry ÝOED:1:329¨ heart-angry ÝW2¨ heart-hungry ÝW2¨ higry pigry ÝOED:5/1:285¨ hogry ÝEDD (see "huggerie"); CSD¨ hogrymogry ÝEDD (see "huggerie"); CSD (as "hogry-mogry")¨ hongry ÝOED:5/1:459; EDD:3:282¨ huggrymuggry ÝEDD (see "huggerie"); CSD (as "huggry-muggry")¨ hungry ÝOED; FW; W2; W3¨ Hungry Bungry ÝDaily Illini, in ad for The Giraffe, Spring 1976¨ Jagry ÝEB11¨ kaingry ÝEDD (see "caingy")¨ land-hungry ÝW2¨ Langry ÝTIG; Times¨ Lisnagry ÝBartholomew:489¨ MacLoingry ÝPhillips (as "Flaithbhertach MacLoingry")¨ mad-angry ÝOED:6/2:14¨ mad-hungry ÝOED:6/2:14¨ magry ÝOED:6/2:36, 6/2:247-48¨ malgry ÝOED:6/2:247¨ Margry ÝIndians (see "Pierre Margry" in bibliog., v.2, p.1204)¨ maugry ÝOED:6/2:247-48¨ mawgry ÝOED:6/2:247¨ meagry ÝOED:6/2:267¨ meat-hungry ÝW2¨ menagry ÝOED (see "managery")¨ messagry ÝOED¨ overangry ÝRH1; RH2¨ Pelegry ÝCE (in main index as "Raymond de Pelegry")¨ Pingry ÝBio-Base; HPS:293-94, 120-21¨ podagry ÝOED; W2 (below the line)¨ Pongry ÝAndree (Supplement, p.572)¨ pottingry ÝOED:7/2:1195; Jamieson:3:532¨ puggry ÝOED:8/1:1573; FW; W2; W3¨ pugry ÝOED:8/1:1574¨ rungry ÝEDD:5:188¨ scavengry ÝOED (in 1715 quote under "scavengery")¨ Schtschigry ÝLG/1:2045; OSN:97¨ Seagry ÝTIG; EB11¨ Segry ÝJohnston; Andree¨ self-angry ÝW2¨ self-hungry ? Shchigry ÝCLG:1747; Johnson:594; OSN:97,206; Times:185,pl.45¨ shiggry ÝEDD¨ Shtchigry ÝLG/1:2045; LG/2:1701¨ Shtshigry ÝLipp¨ skugry ÝOED:9/2:156, 9/1:297; Jamieson:4:266¨ Sygry ÝAndree¨ Tangry ÝFrance¨ Tchangry ÝJohnson:594; LG/1:435,1117¨ Tchigry ÝJohnson:594¨ tear-angry ÝW2¨ tike-hungry ÝCSD¨ Tingry ÝFrance; EB11 (under "Princesse de Tingry")¨ toggry ÝSimmonds (as "Toggry", but all entries are capitalized)¨ ulgry ÝPartridge; Smith:24-25¨ unangry ÝW2¨ vergry ÝOED:12/1:123¨ Virgy ÝCLG:2090¨ Wirgy ÝCLG:2090; NAP:xxxix; Times:220, pl.62; WA:948¨ wind-angry. wind-hungry ÝW2¨ yeard-hungry ÝCED (see "yird")¨ yerd-hungry ÝCED (see "yird"); OED¨ yird-hungry ÝCED (see "yird")¨ Ymagry ÝOED:1:1009 (col. 3, 1st "boss" verb), (variant of "imagery")¨ This list was gathered from the following articles: George H. Scheetz. In Goodly Gree: With Goodwill. Word Ways 22:195 (Nov. 1989) Murray R. Pearce. Who's Flaithbhertach MacLoingry? Word Ways 23:6 (Feb. 1990) Harry B. Partridge. Gypsy Hobby Gry. Word Ways 23:9 (Feb. 1990) References: (Many references are of the form ÝSource:volume:page¨ or ÝSource:page¨.) Andree, Richard. Andrees Handatlas (index volume). 1925. Bartholomew, John. Gazetteer of the British Isles: Statistical and Topographical. 1887. BBC = BBC Pronouncing Dictionary of English Names. Bio-Base. (Microfiche) Detroit: Gale Research Company. 1980. CE = Catholic Encyclopedia. 1907. CED = Chambers English Dictionary. 1988. Century = "India, Northern Part." The Century Atlas of the World. 1897, 1898. CLG = The Colombia Lippincott Gazetteer of the World. L.E.Seltzer, ed. 1952. CSD = Chambers Scots Dictionary. 1971 reprint of 1911 edition. Daily Illini (University of Illinois at Urbana-Champaign). DFC = Dictionary of Fictional Characters. 1963. EB11 = Encyclopedia Britannica, 11th ed. EDD = The English Dialect Dictionary. Joseph Wright, ed. 1898. France = Map Index of France. G.H.Q. American Expeditionary Forces. 1918. FW = Funk & Wagnalls New Standard Dictionary of the English Language. 1943. HPS = The Handbook of Private Schools: An Annual Descriptive Survey of Independent Education, 66th ed. 1985. Indians = Handbook of American Indians North of Mexico. F. W. Hodge. 1912. Jamieson, John. An Etymological Dictionary of the Scottish Language. 1879-87. Johnston, Keith. Index Geographicus... 1864. LG/1 = Lippincott's Gazetteer of the World: A Complete Pronouncing Gazetteer or Geographical Dictionary of the World. 1888. LG/2 = Lippincott's New Gazetteer: ... 1906. Lipp = Lippincott's Pronouncing Gazetteer of the World. 1861, undated edition from late 1800's; 1902. NAP = Narodowy Atlas Polski. 1973-1978 ÝPolish language¨ OED = The Oxford English Dictionary. 1933. ÝForm: OED:volume/part number if applicable:page¨ OSN: U.S.S.R. Volume 6, S-T. Official Standard Names Approved by the United States Board on Geographic Names. Gazetteer #42, 2nd ed. June 1970. Partridge, Harry B. "Ad Memoriam Demetrii." Word Ways, 19 (Aug. 1986): 131. Phillips, Lawrence. Dictionary of Biographical Reference. 1889. RD = The Reader's Digest Complete Atlas of the British Isles, 1st ed. 1965. RH1 = Random House Dictionary of the English Language, Unabridged. 1966. RH2 = Random House Dictionary of the English Language, Second Edition Unabridged. 1987. Simmonds, P.L. Commercial Dictionary of Trade Products. 1883. Smith, John. The True Travels, Adventvres and Observations: London 1630. Stieler, Adolph. Stieler's Handatlas (index volume). 1925. TIG = The Times Index-Gazetteer of the World. 1965. Times = The Times Atlas of the World, 7th ed. 1985. W2 = Webster's New International Dictionary of the English Language, Second Edition, Unabridged. 1934. W3 = Webster's Third New International Dictionary of the English Language, Unabridged. 1961. WA = The World Atlas: Index-Gazetteer. Council of Ministires of the USSR, 1968. Worcester, J.E. Universal Gazetteer, Second Edition. 1823. Some words containing "gry" that do not end with "gry": agrypnia, agrypnotic, Gryllidae, gryllid, gryllus, Gryllus, grylloblattid, Gryllotalpa, gryllos, grypanian, Gryphaea, Gryll, Gryphaea, gryposis, grysbok, gryphon, Gryphosaurus, Grypotherium, grysbuck. Most of these are in Webster's Second also with one from Webster's Third Edition and one from the Random House Dictionary, Second Edition Unabridged. ==> english/homographs.p <== List all homographs (words that are spelled the same but pronounced differently) ==> english/homographs.s <== This list composed by Mark Brader Classes: A - All of the following "defects" absent B - Basic meanings are related C - Capitalization differs ("capitonyms") D - Different spellings also exist (US vs UK, hyphenation, etc.) E - Equal pronunciations also exist (US vs UK, regional, etc.) F - Foreign word, or may be distinguished with accent marks G - Gcontrived :-), coined, jargon, or other uncommon word N - Alleged, but I could not find support for this one in my dictionary and it is not familiar to me 3 - 3-way homograph 4 - 4-way homograph B abstract {corresponding noun and verb; henceforth abbreviated NV} B abuse {NV} B addict {NV} B advocate {NV} BG affect {alter; emotion} B affiliate {NV} B affix {NV} G agape {wide open; form of love} B aggregate {NV} G ai {sloth; ouch]} BE ally {NV} B alternate {NV} BD analyses {plural noun; singular verb (UK)} B animate {verb; adjective} A appropriate {take posession of; suitable} B approximate {verb; adjective} E are {form of to be; unit of area} B arithmetic {noun; adjective} B articulate {verb; adjective} 4DFG as {like; Roman coin; Persian card game; pl. of a} B aspirate {NV} B associate {NV} B attribute {NV} C august A axes {plural of ax; plural of axis} A bases {plural of base; plural of basis} A bass {~ fiddle; fishing for ~} N blessed A bow(ed) {~ and arrow; ~ to the king} E buffet {jostle; ~ lunch} B bustier {undergarment; more busty} B close {~ call; ~ the door} B closer {door ~; more close} B coagulate {NV} G coax {urge; coaxial cable} 3FG colon {":"; colonial farmer; Costa Rican monetary unit} B combat {NV} B combine {NV} A commune {take Communion; administrative district} A compact {closely arranged; treaty} B compound {NV} B compress {NV} B conduct {NV} B confect {NV} B confines {NV} B conflict {NV} B conglomerate {NV} B conjugate {NV} BE conserve {preserve; jam} A console {soothe; keyboard desk} B consort {NV} B construct {NV} B consummate {verb; adjective} N contact E content {what is contained; satisfied} B contest {NV} B contract {NV} B contrast {NV} N convent A converse {logic term; to talk} B convert {NV} B convict {NV} BE coordinate {NV} FG dame {woman; term in the game of Go} DE decameter {poetic line with 10 feet; 10 meters (US)} B defect {flaw; turn traitor} E defense {sports term; fortification} BE delegate {NV} B deliberate {adjective; verb} A desert {leave alone; Sahara ~} B desolate {adjective; verb} D dingy {dull; small boat} BE discharge {NV} N divers {plural diver; various} F do {perform; tonic note of scale} A does {~ the buck see the ~?} A dove {dived; pigeon} F dozen {12; stun (Scottish)} B drawer {one who draws; chest of ~s} B duplicate {NV} B elaborate {verb; adjective} A entrance {door; delight} BDE envelop {NV} N envelope N ergotism {logical reasoning; ergot poisoning} B escort {NV} N escrow B essay {piece of writing; try} B estimate {NV} CFG ewe {female sheep; African language} B excuse {NV} B exploit {NV} BF expose {NV} B ferment {NV} N fiasco {failure; bottle} BDE fillet {cut of meat/fish; band of ribbon/wood} G formal {ceremonious; methylal} DEG genet {civetlike animal; horselike animal} A gill {volume unit; organ in fish} A glower {sullen look; one that glows} B graduate {NV} F he {pronoun; Hebrew letter} CE herb {name; plant} A hinder {hamper; posterior} B house {NV} B import {NV} A incense {infuriate; perfume for burning} B increase {NV} B initiate {NV} B insert {NV} B insult {NV} B intern {NV} A intimate {~ relations; to suggest} A invalid {cripple; erroneous} B invite {NV} G is {form of to be; plural of i} B jagged {slashed or cut; having a zigzag edge} C Job BCF jubilate {rejoice; joyous song} CF junker/Junker 3A lather {suds; lath worker; lathe worker} A lead {~ pipe; ~ astray} B {past tense verb; adjective} BE legged {past tense verb; adjective} CF Lima B live {~ in peace; ~ audience} B lives {~ in peace; for all of our ~} D lower {to let down; frown} F manes {plural of mane; Roman gods} F mate {friend; type of tea} N mead A minute {60 seconds; tiny} B misconduct {NV} BE mobile {movable; wind-blown sculpture} B moderate {NV} EG molar {back tooth; chemical term} A moped {brooded; fun vehicle} B mouse {rodent; to hunt them} B mouth {NV} A mow {pile of hay; to cut down} B multiply {verb; adverb} A number {decimal ~; more numb} B object {thing; complain} E offense {sports term; attack} 3DG os {bone; esker; pl. of o} A overage {too old; surplus} BD paralyses {plural noun; singular verb (UK)} A pasty {pastelike; British meat pie} 3FG pate {head; food paste; porcelain paste for ceramics} A peaked {sharply pointed; unhealthy looking} A peer {equal; one who pees} B perfect {verb; adjective} G periodic {regularly occurring; ~ acids, HIO4 and related substances} B permit {NV} C Placer C polish A poll {head; group of students} B predicate {NV} N premise A present {current; Christmas ~} E primer {intro book/material (US); device for priming} B proceeds {goes; income} B produce {give rise to; fruits and vegetables} B progress {to move forward; work in ~} A project {planned undertaking; to throw forward} N prospect B protest {NV} A pussy {cat; infected} B putter/putting {golf club; one that puts} DG rabat {clerical garment; pottery piece used for polishing} DG rabbi {clerical garment; Jewish religious official} B ragged {teased; tattered} F re {pertaining to; 2nd note of scale} B read {present tense; past tense} C Reading F real {actual; former Spanish coin} B rebel {NV} B recess {NV} B recoil {NV} B record {NV} D recreate {relax; create again} 3BD redress {compensate; compensation; dress again} B refill {NV} B refund {NV} B refuse {NV} B regress {NV} B reject {NV} N repent {regret; creeping} B replay {NV} D represent {stand for; present again} B rerun {NV} D research {investigate; search again} A resent {be indignant; sent again} D reserve {hold back; serve again} D resign {quit; sign again} D resolve {settle dispute; solve again} D resort {vacation spot; sort again} F resume {work summary; restart} A river {watercourse; one who rives} F rose {flower; wine} DE routing {making a route for (US spelling); woodworking term} A row {a fight; ~,~,~ your boat} DF sake {purpose; Japanese drink} 3AF salve {ointment; salvage; hail]} N second B segment {NV} B separate {NV} A severer {cutter; more severe} 3AG sewer {one who sews; storm ~; head servant at table} A shower {one who shows; ~ stall} B syndicate {NV} A singer {one who singes; one who sings} A skied {past tense of ski; past tense of sky} A slaver {slave taker; drool} A slough {swamp; cast-off} A sow {~ seeds; female pig} A stingy {meager; able to sting} B subject {NV} A supply {in a supple way; ~ and demand} B survey {NV} B suspect {NV} N swinger {whopper; one that swings} CF tang {flavor; Chinese dynasty} A tarry {covered in tar; dawdle} A tear {~ down; shed a ~} A thou {you; slang for thousand} A thymic {of thyme; of thymus} A tier {one who ties; row or rank} B torment {NV} A tower {one who tows; leaning ~} B transfer {NV} B transplant {NV} B transport {NV} DG unionized {~ labor; ~ hydrogen} B upset {NV} G us {we; plural of u} B use {NV} A violist {viol player; viola player} A wind {~ the clock; north ~} CF worms A wound {injury; wrapped around} N yak {ox; laugh} ==> english/homophones.p <== What words have four or more spellings that sound alike? ==> english/homophones.s <== air, aire, are, ayr, ayer, e'er, ere, err, heir cense, cents, scents, sense eau, eaux, O, oh, owe ==> english/j.ending.p <== What words and names end in j? ==> english/j.ending.s <== Following is a compilation of words ending in j from various dictionaries. Capitalized words and words marked as foreign are included, but to keep the list to a managable size, personal and place names are excluded. aflaj plural of falaj (Cham) benj variant of bhang - hemp plant (NI2) bhimraj the rachet-tailed drongo (F&W) Bhumij branch of Munda tribes in India (NI3) Chuj a people of Northwestern Guatemala (NI3) esraj an Indian musical instrument with 3 or 4 strings (OED2) falaj a water channel as part of the ancient irrigation system of Oman (Cham) Funj variant of Fung - a people dominant in Sennar (NI3) gaj Omanese coin (NI2) genj a common type of cotton cloth in Sudan (F&W) gunj a grannery in India (NI2) hadj variant of hajj (NI3) haj variant of hajj (NI3) hajilij the bito - a small scrubby tree that grows in dry parts of Africa and Asia (NI2) hajj pilgimage to Mecca (NI3) hij obsolete form of hie or high (OED2) Jubaraj variant of Yuvaraja - the male heir to an Indian pricipality (OED2) kaleej variant of kalij (NI3) kalij any of several crested Indian pheasants (NI3) kankrej guzerat - a breed of Indian cattle (NI3) kharaj a tax on unbelievers (NI2) Khawarij plural of Kharijite - a member of the oldest religious sect of Islam (NI3) khiraj variant of kharaj (NI2) kilij a Turkish saber with a crescent shaped blade (RHD) kurunj variant of kurung - the Indian beech (NI2) Maharaj variant of Maharaja - East Indian prince (OED2) munj a tough Asiatic grass (NI3) naranj Maldive Island name for mancala - an Arabian board game (CD) pakhawaj a doubleheaded drum used in Indian music (OED2) raj rule (NI3) saj the Indian laurel (NI2) samaj Hindu religious society (NI3) sohmaj variant of samaj (NI2) somaj variant of samaj (NI2) svaraj variant of swaraj (F&W) swaraj local self-government in India (NI3) taj a tall conical cap worn by Moslems (NI3) tedj variant of tej (OED2) tej Ethiopian mead (OED2) Viraj in Hindu Mythology, the mysterious primeval being when differentiating itself into male and female (F&W) Yuvaraj same as Jubaraj (OED2) Yuveraj same as Jubaraj (OED2) Yuvraj same as Jubaraj (OED2) zij Persian astronomical tables (F&W) This list is almost certainly not complete. For example, on page 187 of Beyond Language, Dmitri Borgmann has "Udruj" in a word list. What reference he dug this word out of is unknown; the combined efforts of the NPL electronic mailing list could not produce the source of this word. So additions to this list will be welcomed by the author. REFERENCES CD - The Century Dictionary and Cyclopedia, 1911 Cham - Chambers English Dictionary, 1988 F&W - Funk & Wagnall's New Standard Dictionary of the English Language, 1941 NI2 - Webster's New International Dictionary, Second Edition, 1942 NI3 - Webster's Third New International Dictionary, 1981 OED2 - Oxford English Dictionary, Second Edition, 1989 RHD - Random House Dictionary of the English Language, 1966 --- Dan Tilque -- dant@logos.WR.TEK.COM ==> english/ladder.p <== Find the shortest word ladders stretching between the following pairs: hit - ace pig - sty four - five play - game green - grass wheat - bread order - chaos order - impel sixth - hubby speedy - comedy chasing - robbers effaces - cabaret griming - goblets vainest - injects vainest - infulae ==> english/ladder.s <== Using every unabridged dictionary available, the best yet found are: hit ait act ace pig peg seg sey sty four foud fond find fine five play blay bray bras baas bams gams game green grees greys grays grass wheat theat treat tread bread order older elder eider cider cides codes coles colls coals chals chaos order ormer armer ammer amper imper impel sixth sixty silty silly sally sably sabby nabby nubby hubby speedy speeds steeds steers sheers shyers sayers payers papers papery popery popely pomely comely comedy griming priming prising poising toising toiling coiling colling collins collies dollies doilies dailies bailies bailees bailers failers fablers gablers gabbers gibbers gibbets gobbets goblets chasing ceasing cessing messing massing masting marting martins martens martels cartels carpels carpers campers cambers combers cobbers combers robbers vainest fainest fairest sairest saidest saddest maddest middest mildest wildest wiliest winiest waniest caniest cantest contest confest confess confers conners canners fanners fawners pawners pawnees pawnces paunces jaunces jaunced jaunted saunted stunted stented stenned steined stained spained splined splines salines savines savings pavings parings earings enrings endings ondings ondines undines unlines unlives unwives unwires unwares unbares unbared unpared unpaged uncaged incaged incased incised incises incites indites indices indicts inducts indults insults insulas insulae infulae This is not another travelling salesman - it is merely finding the diameter of connected components of that graph. The simple algorithm for this is to do one depth first search from each word, resulting in an O(n*m) worst case algorithm (where n is the number of words, and m is the number of arcs). In practice, it is actually somewhat better, since the graph breaks down into many connected components. However, the diameters (and solutions) depend on what dictionary is used. Here are the results from various dictionaries: From /usr/dict/words (restricted to words all lower case alphabetical) (19,694 words): sixth - hubby (46 steps) From the official scrabble players dictionary (94,276 words): effaces - cabaret (57 steps) From the british official scrabble words (134,051 words): vainest - infulae (73 steps) From webster's ninth new collegiate dictionary (abridged) (78, 167 words): griming - goblets (56 steps) From all of the above, merged (180,676 words): vainest - injects (58 steps) To see the effect the dictionary has on paths, here are the lengths of the shortest paths these pairs, and for the ones mentioned in previous posts, for each dictionary (a - means that there is no path using only words from that dictionary): UDW OSPD OSW W9 ALL hit - ace 5 3 3 5 3 pig - sty - 5 4 5 4 four - five 6 6 5 7 5 play - game 8 7 7 8 7 green - grass 13 4 4 7 4 wheat - bread 6 6 6 6 6 sixth - hubby 46 9 9 - 9 effaces - cabaret - 57 - - 33 vainest - infulae - - 73 - 52 griming - goblets - 22 19 56 15 vainest - injects - - 72 - 58 ==> english/less.ness.p <== Find a word that forms two other words, unrelated in meaning, when "less" and "ness" are added. ==> english/less.ness.s <== base -> baseless, baseness light -> lightless, lightness sound -> soundless, soundness wit -> witless, witness ==> english/letter.rebus.p <== Define the letters of the alphabet using self-referential common phrases (e.g., "first of all" defines "a"). ==> english/letter.rebus.s <== A first of all, midday B fifth of bourbon, starting block C fifth of scotch D end of the world, back of my hand E end of the line, beginning of the end F starting friction, front G middle of the night, starting gate H end of the earth, top of the heap, middle of nowhere I next of kin J center of project K bottom of the deck, two of a kind L bottom of the barrel, starting line M top of my head N center of attention, final countdown, end run O second in command P bottom of the heap, the first of painters, starting point Q at the front of the queue, top quality R middle of the road, center of inertia S _Last of the Mohicans_, start of something big T top o' the morning, one's wit's end, bottom of my heart, last, central U second guess V center of gravity W end of the rainbow, top of the world X wax finish, climax Y top of your head, center of the cyclone, early years, final extremity Z led zeppelin ==> english/lipograms.p <== What books have been written without specific letters, vowels, etc.? ==> english/lipograms.s <== Such a book is called a lipogram. A novel-length example in English (omitting e) exists, titled _Gadsby_. Georges Perec wrote a French novel titled _La Disparition_ which does not contain the letter 'e', except in a few bits of text that the publisher had to include in or on the book somewhere -- such as the author's name :-). But these were all printed in red, making them somehow ``not count''. Perec also wrote another novel in which `e' was the only vowel. In _La Disparition_, unlike _Gadsby_, the lipogrammatic technique is reflected in the story. Objects disappear or become invisible. We know, however, more or less why the characters can't find things like eggs or even remember their names -- because the words for them can't be used. Amazingly, it's been ``translated'' into English (by Harry Mathews, I think). Another work which manages to Ýalmost¨ adhere to restrictive alphabetic rules while also remaining readable as well as providing amusement and literary satisfaction (though you have to like disjointed fiction) is _Alphabetical Africa_ by Walter Abish. The rules (which of course he doesn't explain, you can't help noticing most of them) have to do with initial letters of words. There are 52 chapters. In the first, all words begin with `a'; in the second, all words begin with either `a' or `b'; etc, until all words are allowed in chapter 26. Then in the second half, the letters are taken away one by one. It's remarkable when, for instance, you finally get `the' and realize how much or little you missed it; earlier, when `I' comes in, you feel something like the difference between third- and first-person narration. As one of the blurbs more or less says (I don't have it here to quote), reading this is like slowly taking a deep breath and letting it out again. ---- Mitch Marks mitchell@cs.uchicago.edu ==> english/multi.lingual.p <== What words in multiple languages are related in interesting ways? ==> english/multi.lingual.s <== Synonymous reversals: Dutch: nier (kidney), French: rein French: etats, English: state ==> english/near.palindrome.p <== What are some long near palindromes, i.e., words that except for one letter would be palindromes? ==> english/near.palindrome.s <== Here are the longest near palindromes in Webster's Ninth Collegiate: catalatic footstool red pepper detonated locofocos red spider dew-clawed nabataean retreater eisegesis possessor stargrass foolproof ratemeter webmember ==> english/palindromes.p <== What are some long palindromes? ==> english/palindromes.s <== The first words spoken were a palindrome: Madam, I'm Adam. or perhaps: Madam in Eden, I'm Adam. The response, of course, must have been: Eve Napolean's lament: Able was I ere I saw Elba. Has been improved with: Unremarkable was I ere I saw Elba, Kramer, nu? A fish is a: laminar animal Other palindromes in ascending length (drum roll please): Dennis sinned. Sir, I'm Iris. Sup not on pus. Name no one man. Naomi, did I moan? Enid and Edna dine. Revenge Meg? Never] No lemons, no melon. A Toyota's a Toyota. Ma is a nun, as I am. He harasses Sarah, eh? Niagara, O roar again] He lived as a devil, eh? Nurse, I spy gupsies, run] Sit on a potato pan, Otis] Slap a ham on Omaha, pals] A slut nixes sex in Tulsa. Rats live on no evil star. Ten animals I slam in a net. Go deliver a dare, vile dog. Was it a car or a cat I saw? Was it Eliot's toilet I saw? Al lets Della call Ed Stella. Draw, O Caeser, erase a coward. Did Eve salt an atlas? Eve did. No pinot noir on Orion to nip on. Naomi, sex at noon taxes] I moan. Evil I did dwell; lewd did I live. Yo, bad anaconda had no Canada boy . Egad] A base tone denotes a bad age. Satan, oscillate my metallic sonatas. Red dude kill lion. No ill-liked udder. I roamed under it as a tired, nude Maori. To Peru, named llama mall 'De Manure Pot'. Straw? No, too stupid a fad. I put soot on warts. Now, Ned, I am a maiden nun; Ned, I am a maiden won. Here we no got conical ill lilac in octogon ewer, eh? Salamander a ton now. Raw war won not, a Red Nam, alas. Fool] A dog lives sad a boxer, Rex. O bad ass evil god aloof] 'Tenor Octopus Night' netted a cadet tenth ginsu pot, coronet. Won total, I am a pro. Bali radar I labor. Pa, mail a tot now] Yo, boy] Trap gnus, nude. 'Kangaroo Rag' naked unsung party, O boy] Did I strap red nude, red rump, also slap murdered underparts? I did] Doc, note: I dissent. A fast never prevents a fatness. I diet on cod. So regards Rat's Lib: regrets no more hero monster gerbil stars' drag Eros. Degas, are we not drawn onward, we freer few, drawn onward to new eras aged? Garret, I ogle. Enemy democrats party; trap star comedy men, eel goiter rag. Sagas emit taxes, rat snot, or pastrami. I'm Arts, a proton star - sex at times a gas. Dr. Ana, Cataracts. Uranium enema smarts if fist rams, Amen] Emu in a rust car at a canard. T. Eliot, top bard, notes putrid tang emanating, is sad; I'd assign it a name: gnat dirt upset on drab pot toilet. Those wonderful proper names: Dennis, Nell, Edna, Leon, Nedra, Anita, Rolf, Nora, Alice, Carol, Leo, Jane, Reed, Dena, Dale, Basil, Rae, Penny, Lana, Dave, Denny, Lena, Ida, Bernadette, Ben, Ray, Lila, Nina, Jo, Ira, Mara, Sara, Mario, Jan, Ina, Lily, Arne, Bette, Dan, Reba, Diane, Lynn, Ed, Eva, Dana, Lynne, Pearl, Isabel, Ada, Ned, Dee, Rena, Joel, Lora, Cecil, Aaron, Flora, Tina, Arden, Noel, and Ellen sinned. A poem: Mood's mode] Pallas, I won] (Diaper pane, sold entire.) Melt till ever sere, hide it. Drown a more vile note; (Tar of rennet.) Ah, trowel, baton, eras ago. The reward? A "nisi." Two nag. Otary tastes putrid, yam was green. Odes up and on; stare we. Rats nod. Nap used one-erg saw. (May dirt upset satyr?) A toga now; 'tis in a drawer, eh? Togas are notable. (Worth a tenner for Ate`.) Tone liver. O Man, word-tied I. Here's revel] Little merit, Ned? Lose, Nap? Repaid now is all apedom's doom. -- Hubert Phillips: Headmaster's Palindromic List on his Memo Pad: Test on Erasmus Dr of Law Deliver soap Stop dynamo (OTC) Royal: phone no.? Tel: Law re Kate Race Ref. Football. Caps on for prep Is sofa sitable on? Pots- no tops XI--Staff over Knit up ties ('U') Sub-edit Nurse's order Ned (re paper) Canning is on test (snub slip-up) Eve's simple hot dish (crib) Birch (Sid) to help Miss Eve Pupil's buns Reaper den T-set: no sign in a/c Use it Red roses Put inkspot on stopper Run Tide Bus? Prof.--no space Rev off at six Caretaker (wall, etc.) Noel Bat is a fossil Too mand d*** pots Lab to offer one 'Noh' play-- Wal for duo? (I'd name Dr. O) or 'Pals Reviled'? See few owe fees (or demand IOU?) Sums are not set. -- Joyce Johnson (_New_Statesman_ competition in 1967. 126 words, 467 letters) Some word (not letter) palindromes: So patient a doctor to doctor a patient so. Girl, bathing on Bikini, eyeing boy, finds boy eyeing bikini on bathing girl. In German: Ein Neger mit Gazelle zagt im Regen nie. In Serbo-Croat: Ana voli Milovana. Ana nabra par banana. Imena Amen nema, a me mi. U pero soli i los o repu. Ako jad moli silom daj oka. Odano mati pita: a ti pitam, o nado? Evo sam iza padam mada pazim asove. v v v v A krt u razu mi laze no one zalim u zaru trka. Palindromes in other languages that are palindromes in English: Hebrew: aba or abba, English: dad German: tat, English: deed The timeless classic: A man, a plan, a canal; Panama? Has been improved by: A dog, a plan, a canal: pagoda] -- anonymous A man, a plan, a cat, a canal; Panama? -- Jim Saxe, plan file @ CMU, 9 October 1983 A man, a plan, a cat, a ham, a yak, a yam, a hat, a canal--Panama] -- Guy Jacobson, plan file @ CMU late 1983 A man, a plan, a caret, a ban, a myriad, a sum, a lac, a liar, a hoop, a pint, a catalpa, a gas, an oil, a bird, a yell, a vat, a caw, a pax, a wag, a tax, a nay, a ram, a cap, a yam, a gay, a tsar, a wall, a car, a luger, a ward, a bin, a woman, a vassal, a wolf, a tuna, a nit, a pall, a fret, a watt, a bay, a daub, a tan, a cab, a datum, a gall, a hat, a fag, a zap, a say, a jaw, a lay, a wet, a gallop, a tug, a trot, a trap, a tram, a torr, a caper, a top, a tonk, a toll, a ball, a fair, a sax, a minim, a tenor, a bass, a passer, a capital, a rut, an amen, a ted, a cabal, a tang, a sun, an ass, a maw, a sag, a jam, a dam, a sub, a salt, an axon, a sail, an ad, a wadi, a radian, a room, a rood, a rip, a tad, a pariah, a revel, a reel, a reed, a pool, a plug, a pin, a peek, a parabola, a dog, a pat, a cud, a nu, a fan, a pal, a rum, a nod, an eta, a lag, an eel, a batik, a mug, a mot, a nap, a maxim, a mood, a leek, a grub, a gob, a gel, a drab, a citadel, a total, a cedar, a tap, a gag, a rat, a manor, a bar, a gal, a cola, a pap, a yaw, a tab, a raj, a gab, a nag, a pagan, a bag, a jar, a bat, a way, a papa, a local, a gar, a baron, a mat, a rag, a gap, a tar, a decal, a tot, a led, a tic, a bard, a leg, a bog, a burg, a keel, a doom, a mix, a map, an atom, a gum, a kit, a baleen, a gala, a ten, a don, a mural, a pan, a faun, a ducat, a pagoda, a lob, a rap, a keep, a nip, a gulp, a loop, a deer, a leer, a lever, a hair, a pad, a tapir, a door, a moor, an aid, a raid, a wad, an alias, an ox, an atlas, a bus, a madam, a jag, a saw, a mass, an anus, a gnat, a lab, a cadet, an em, a natural, a tip, a caress, a pass, a baronet, a minimax, a sari, a fall, a ballot, a knot, a pot, a rep, a carrot, a mart, a part, a tort, a gut, a poll, a gateway, a law, a jay, a sap, a zag, a fat, a hall, a gamut, a dab, a can, a tabu, a day, a batt, a waterfall, a patina, a nut, a flow, a lass, a van, a mow, a nib, a draw, a regular, a call, a war, a stay, a gam, a yap, a cam, a ray, an ax, a tag, a wax, a paw, a cat, a valley, a drib, a lion, a saga, a plat, a catnip, a pooh, a rail, a calamus, a dairyman, a bater, a canal--Panama. --Dan Hoey, 'discovered' in 1984. Dan goes on to say "...a little work on the search algorithm could make it several times as long." The entire book _Satire: Veritas_ is a palindrome, it starts "Sir, I stra..." and ends "... Art, sir, is Satire: Veritas." References: Palindromes and Anagrams Howard W. Bergerson Dover Publications New York, 1973 ISBN 0-486-20664-5. The Oxford Guide to Word Games, chapter 11, titled "Palindromes" Tony Augarde ==> english/pangram.p <== A "pangram" is a sentence containing all 26 letters. What is the shortest pangram (measured by number of letters or words)? What is the shortest word list using all 26 letters in alphabetical order? In reverse alphabetical order? ==> english/pangram.s <== The single-letter words that have meanings unrelated to their letter shapes or sounds, position in the alphabet, etc. are: a - indefinite article; on; in; at; to; he; him; she; her; they; them; it; I; have; of; all c - 100; cocaine; programming language d - 500 e - base of natural logs; eccentricity; enlarging g - acceleration of gravity; general ability; $1000; general audience i - one; unit vector in x direction; personal pronoun; in; aye j - one; unit vector in y direction k - 1000; 1024; strikeout; unit vector in z direction l - 50; ell; elevated railroad m - 1000; em; pica; an antigen of human blood n - an indefinite number; en; an antigen of human blood o - oh q - quality of oscillatory circuit R - one of the three Rs; restricted audience t - t-shirt u - upper class v - five w - w particle x - unknown quantity; atmospherics; adults only y - unknown quantity; YMCA z - unknown quantity; buzzing sound; z particle It is therefore advisable to exclude single-letter words, with the possible exception of 'a'. As always, word acceptability varies with the dictionaries used. We use these: 9C - Merriam-Webster's Ninth New Collegiate Dictionary, 1986 NI3 - Merriam-Webster's Third New International Dictionary, 1961 NI2 - Merriam-Webster's New International Dictionary, Second Edition, 1935 OED - Oxford English Dictionary with Supplements, 1933 - 85 '+' indicates obsolete, dialectical, slang, or otherwise substandard word. Some exceptional pangrams: Using only words in 9C: Sympathizing would fix Quaker objectives. (5 words, 36 letters) Quick brown fox, jump over the lazy dogs. (8 words, 32 letters) Pack my box with five dozen liquor jugs. (8 words, 32 letters) Jackdaws love my big sphinx of quartz. (7 words, 31 letters) The five boxing wizards jump quickly. (6 words, 31 letters) How quickly daft jumping zebras vex. (6 words, 30 letters) Quartz glyph job vex'd cwm finks. (6 words?, 26 letters) Cwm, fjord-bank glyphs vext quiz. (6 words, 26 letters, Dmitri Borgmann) Using words in 9C and NI3: Veldt jynx grimps waqf zho buck. (6 words, 26 letters, Michael Jones) Using words in 9C, NI3 and NI3+: Squdgy fez, blank jimp crwth vox. (6 words, 26 letters, Claude Shannon) Using words in 9C, NI3, NI2 and NI2+: Phlegms fyrd wuz qvint jackbox. (5 words, 26 letters, Dmitri Borgmann) Some exceptional panalphabetic word lists: jackbox viewfinder phlegmy quartz (4 words, 31 letters, Mary Hazard) benzoxycamphors quick-flowing juventude (3 words, 36 letters, Darryl Francis) Some exceptional nearly panalphabetic isogrammatic word lists: blacksmith gunpowdery (2 words, 20 letters) humpbacks frowzy tingled (3 words, 22 letters) Some exceptional panalphabetic word lists with letters in alphabetical order: Using only words in 9C: a BCD ef ghi jack limn op querist ulva wax oyez (11 words, 37 letters) ABC defog hijack limn op querist ulva wax oyez (9 words, 38 letters) Using words in 9C and NI3: a BCD ef ghi jak limn op qres to uva wax oyez (12 words, 34 letters) ABC defy ghi jak limn op qres to uva wax oyez (11 words, 35 letters) ABC defy ghi jak limn opaquers turves wax oyez (9 words, 38 letters) scabicide afghani jokul manrope querist purview oxygenize (7 words, 51 letters) Using words in 9C, NI3 and NI3+: a BCD ef ghi jak limn op QRS to uva wax yez (12 words, 32 letters) ABC defy ghi jak limn op QRS to uva wax yez (11 words, 33 letters) ab cad ef ghi jak limn op qre stun vow ox yez (12 words, 34 letters) ABC defy ghi jak limn op querist uva wax yez (10 words, 35 letters) Using words in 9C, NI3, NI3+, NI2 and NI2+: ABC def ghi jak limn op qre struv wax yez (10 words, 32 letters) ABC def ghi jak limn opaquer struv wax yez (9 words, 34 letters) Using words in 9C, NI3, NI3+, NI2, NI2+ and the OED: ABC defog hij klam nop QRS tu vow XYZ (9 words, 29 letters, Jeff Grant) ABC def ghi jak limn op qres tu vow XYZ (10 words, 30 letters) ABC defog hij klam nop querist uvrow XYZ (8 words, 33 letters, Jeff Grant) ABC defyghe bij sklim nop querist uvrow XYZ (8 words, 36 letters) ABC defog hijack limnophil querist uvrow XYZ (7 words, 38 letters, Jeff Grant) Some exceptional panalphabetic word lists with letters in reverse alpha order: Using words from 9C: lazy ox ow vug tsar quip on milk jib hag fed cab a (13 words, 38 letters) lazy ox wave uts reequip on milk jihad gifted cabal (10 words, 42 letters) Using words from 9C and NI3: lazy ox ow vug tsar quip on milk jib hag fed caba (12 words, 38 letters) lazy ox wave uts roque pon milk jihad gifted caba (10 words, 40 letters) Using words from 9C, NI3 and NI2: zo yex wu vug tsar quip on milk jib hag fed caba (12 words, 37 letters) zo yex wave uts roque pon milk jihad gifted caba (10 words, 39 letters) All words are main entries in 9C except the following: 9C: ghi (at 'ghee') NI3: caba, fyrd, jak, opaquers, pon, qre(s), squdgy, uva NI3+: jimp, QRS (at 'QRS complex'), sklim, vox (at 'vox populi'), yez NI2: benzoxycamphors, jackbox, limnophil, quick-flowing, yex, zo NI2+: def, juventude, klam, quar, qvint, struv, tu, wuz OED: defyghe (at 'defy'), bij (at 'buy'), hij, nop, uvrow (at 'yuffrouw'), XYZ (at 'X') The first time I saw this pangram was in Gyles Brandeth's _The Joy of Lex_. It appeared there as: Waltz, nymph, for quick jigs vex Bud. (7 words, 28 letters, proper noun.) I always wondered why they didn't try modifying it as: Waltz, nymph, for quick jigs vex buds. (7 words, 29 letters, no proper noun.) However, why fast dances would irritate incipient flowers is beyond me, so I tried again: Waltz, dumb nymph, for quick jigs vex. (7 words, 29 letters, no proper noun, makes more sense.) However, sounds kind of sexist, and we can maybe chop off a letter and eliminate the sexism, although suffering some loss of sense: Waltz, bud nymph, for quick jigs vex. (7 words, 28 letters, no proper noun, makes less sense.) There are river nymphs and tree nymphs and mountain nymphs, so there can be nymphs of the aforementioned incipent flowers, right? Sense is a matter of opinion, so you can move the bud around or turn it into another imperative verb rather than a noun-as-adjective: Waltz, nymph, bud, for quick jigs vex. (7 words, 28 letters, no proper noun, sense is dubious.) ÝWe've all heard of budding youth, right?¨ Waltz, nymph, for quick bud jigs vex. (7 words, 28 letters, no proper noun, sense is dubious.) ÝYeah, we've all learned to dance a merry jig that looks like one of those infamous incipient flowers.¨ Dub waltz, nymph, for quick jigs vex. (7 words, 28 letters, no proper noun, came up with this on the spot and actually it looks pretty good]) ÝThe idea being that a nymph, being in control of the soundtrack for a TV sitcom, has to change the music to which a grandmother is listening, from something from Ireland to something from Strauss.¨ -- Stephen Joseph Smith It is fairly straightforward, if time-consuming, to search for minimal pangrams given a suitable lexicon, and the enclosed program does this. The run time is of the order of 20 MIPS-days if fed `Official Scrabble Words', a document nominally listing all sufficiently short words playable in tournament Scrabble in Britain. I also enclose a lexicon which will reproduce the OSW results much more quickly. The results are dominated by onomatopoeic interjections (`pst', `sh', etc.), and words borrowed from Welsh (`cwm', `crwth') and Arabic (`qat', `suq'). Other lexicons will contain a very different leavening of such words, and yield a very different set of pangrams. Readers are invited to form sentences (or, less challenging, newspaper headlines) from these pangrams. Few are amenable to this sort of thing. -- Steve Thomas -----cut-----here----- #include #include extern void *malloc (); extern void *realloc (); long getword (); #define MAXWORD 26 int listÝMAXWORD¨; int lp; struct list { struct list *next; char *word; }; struct word { long mask; struct list *list; } *w; int wp; int wsize; char wordbufÝBUFSIZ¨; char *letters = "qxjzvwfkbyhpmgcdultnoriase"; int cmp (ap, bp) struct word *ap, *bp; { char *p; long a = ap->mask, b = bp->mask; for (p = letters; *p; p++) { long m = 1L << (*p - 'a'); if ((a & m) ]= (b & m)) if ((a & m) ]= 0) return -1; else return 1; } return 0; } void * newmem (p, size) void *p; int size; { if (p) p = realloc (p, size); else p = malloc (size); if (p == NULL) { fprintf (stderr, "Out of memory\n"); exit (1); } return p; } char * dupstr (s) char *s; { char *p = newmem ((void *)NULL, strlen (s) + 1); strcpy (p, s); return p; } main (argc, argv) int argc; char **argv; { long m; int i, j; while ((m = getword (stdin)) ]= 0) { if (wp >= wsize) { wsize += 1000; w = newmem (w, wsize * sizeof (struct word)); } wÝwp¨.mask = m; wÝwp¨.list = newmem ((void *)NULL, sizeof (struct list)); wÝwp¨.list->word = dupstr (wordbuf); wÝwp¨.list->next = (struct list *)NULL; wp++; } qsort (w, wp, sizeof (struct word), cmp); for (i = 1, j = 1; j < wp; j++) { if (wÝj¨.mask == wÝi - 1¨.mask) { wÝj¨.list->next = wÝi - 1¨.list; wÝi - 1¨.list = wÝj¨.list; } else wÝi++¨ = wÝj¨; } wp = i; pangram (0L, 0, letters); exit (0); } pangram (sofar, min, lets) long sofar; int min; char *lets; { register int i; register long must; if (sofar == 0x3ffffff) { print (); return; } for (; *lets; lets++) if ((sofar & (1 << (*lets - 'a'))) == 0) break; must = 1 << (*lets - 'a'); for (i = min; i < wp; i++) { if (wÝi¨.mask & sofar) continue; if ((wÝi¨.mask & must) == 0) continue; listÝlp++¨ = i; pangram (wÝi¨.mask ! sofar, i + 1, lets); --lp; } } long getword (fp) FILE *fp; { long mask, m; char *p; char c; while (fgets (wordbuf, sizeof (wordbuf), fp) ]= NULL) { p = wordbuf; mask = 0L; while ((c = *p++) ]= '\0') { if (]islower (c)) break; m = 1L << (c - 'a'); if ((mask & m) ]= 0) break; mask != m; } if (c == '\n') pÝ-1¨ = c = '\0'; if (c == '\0' && mask) return mask; } return 0; } print () { int i; for (i = 0; i < lp; i++) { struct word *p = &wÝlistÝi¨¨; struct list *l; if (p->list->next == NULL) printf ("%s", p->list->word); else { printf ("("); for (l = p->list; l; l = l->next) { printf ("%s", l->word); if (l->next) printf (" "); } printf (")"); } if (i ]= lp - 1) printf (" "); } printf ("\n"); fflush (stdout); } -----and-----here----- ankh bad bag bald balk balks band bandh bang bank bap bard barf bark bed beds beg bend benj berk berks bez bhang bid big bight bilk bink bird birds birk bisk biz blad blag bland blank bled blend blight blimp blin blind blink blintz blip blitz block blond blunk blunks bod bods bog bok boks bold bond bong bonk bonks bop bord bords bosk box brad brank bred brink brinks brod brods brog brogh broghs bud bug bugs bulk bulks bump bumps bund bunds bung bungs bunk bunks burd burds burg burgh burghs burk burks burp busk by ch crwth cwm cwms dab dag dak damp dap deb debs debt deft delf delfs delft delph delphs depth derv dervs dhak dib dig dight dink dinks dirk disk div divs dob dobs dog dop dorp dowf drab draft drib dribs drip drop drub drubs drunk drunks dub dug dugs dung dunk dunks dup dusk dwarf dzo dzos fad fag falx fank fard fax fed fend fends fenks fez fib fid fig fight fink firk fisk fix fiz fjord fjords flab flag flak flank flap flax fled fleg flex flight flimp flip flisk flix flog flogs flong flongs flop flops flub flump flumps flung flunk flux fob fobs fog fogs fold folds folk folks fond fonds fop fops ford fords fork forks fox frab frank frap fremd fright friz frog frond frump frumps fub fud fug fugs fund funds funk funks fy fyrd fyrds gab gad gamb gamp gap gawk gawp ged geld gib gid gif gild gink gip gju gjus gled glib glid glift glitz glob globs glyph glyphs gob god gold golf golfs golp golps gonk gov govs gowd gowf gowfs gowk gowks graft graph grub grypt gub gubs gulf gulfs gulp gulph gulphs gunk gup gym gymp gyp gyps hadj hank hyp hyps jab jag jak jamb jap jark jerk jerks jib jibs jig jimp jink jinks jinx jird jirds jiz job jobs jog jogs jud juds jug jugs junk junks jynx kang kant kaw keb kebs ked kef kefs keg kelp kemb kemp kep kerb kerbs kerf kerfs kex khan khud khuds kid kids kif kifs kight kild kiln kilp kind kinds king kip klepht knag knight knob knobs knub knubs kob kobs kond kop kops kraft krantz kranz kvetch ky kynd kynds lav lev lez link luz lynx mawk nabk nth pad park pawk pax ped peg pegh peghs pelf pelfs penk perk perv pervs phang phiz phlox pig pight pix pleb plebs pled plight plink plod plongd plonk pluck plug plumb plumbs plunk ply pod polk polks pong pork pox poz prex prod prof prog pst pub pud pug pugh pulk pulks punk pyx qat qats qibla qiblas quark quiz sh skelf skid skrump skug sphinx spiv squawk st sunk suq swiz sylph tank thilk tyg vamp van vang vant veg veld velds veldt vend vends verb verbs vet vex vibs vild vint vly vox vug vugs vuln waltz wank welkt whack zag zap zarf zax zed zek zeks zel zig zigs zimb zimbs zing zings zip zips zit zurf zurfs ==> english/phonetic.letters.p <== What does "FUNEX" mean? ==> english/phonetic.letters.s <== FUNEX? (Have you any eggs?) SVFX. (Yes, we have eggs.) FUNEM? (Have you any ham?) SVFM. (Yes, we have ham.) FUMNX? (Have you ham and eggs?) S,S:VFM,VFX,VFMNX] (Yes, yes: we have ham, we have eggs, we have ham and eggs]) CD ED BD DUCKS? (See the itty bitty ducks?) MR NOT DUCKS] (Them are not ducks]) OSAR, CDEDBD WINGS? (Oh yes they are, see the itty bitty wings?) LILB MR DUCKS] (Well I'll be, them are ducks]) In Spanish: SOCKS. (Eso si que es.) ==> english/piglatin.p <== What words in pig latin also are words? ==> english/piglatin.s <== cess -> essay coke -> okay lawn -> onlay lout -> outlay lover -> overlay plover -> overplay plunder -> underplay sass -> assay stout -> outstay trash -> ashtray wear -> airway wonder -> underway ==> english/pleonasm.p <== What are some redundant terms that occur frequently (like "ABM missile")? ==> english/pleonasm.s <== 11.5% APR ABM missile ABS system AC current ACT tests AMOCO Oil Co. APL programming language ATM macine BASIC Code BBS System CAD design CNN news network DC current DMZ zone DOS operating system GMT time Geirangerfjorden (Fjord Fjord Fjord) HIV virus ISBN number ISDN network LCD display LED diode La Brea Tar Pits Los Altos Hills (The Hills Hills) MIDI Interface Mount Fujiyama (Mount Mountain) NATO organization NFS File System PCV valve PIN number RAM (or ROM) memory Ruidoso River (Noisy River River) SALT talks SAT test SCSI Interface SEATO organization VIN number floccinoccinihlipilification (from 4 latin words meaning "nothing") hoi polloi (a genuine bilingual redundancy) hot water heater ==> english/plurals/collision.p <== Two words, spelled and pronounced differently, have plurals spelled the same but pronounced differently. ==> english/plurals/collision.s <== axe and axis -> axes base and basis -> bases ellipse and ellipsis -> ellipses ==> english/plurals/doubtful.number.p <== A little word of doubtful number, a foe to rest and peaceful slumber. If you add an "s" to this, great is the metamorphosis. Plural is plural now no more, and sweet what bitter was before. What am I? ==> english/plurals/doubtful.number.s <== cares -> caress ==> english/plurals/drop.s.p <== What plural is formed by DROPPING the terminal "s" in a word? ==> english/plurals/drop.s.s <== necropolis -> necropoli ==> english/plurals/endings.p <== List a plural ending with each letter of the alphabet. ==> english/plurals/endings.s <== Legend 0 = plural formed (basically) by adding letter 1 = plural spelled differently from singular 2 = ditto, plural contains punctuation 3 = plural spelled the same as singular All entries are from Merriam-Webster's Ninth Collegiate Dictionary, except those marked "(NI3)", which are from the Third International. Entries in brackets are probable dictionary artifacts. A 0 VAS VASA B 1 SLUBBI SLEYB (NI3) C 0 CALPULLI CALPULLEC (NI3) D 2 GRANT-IN-AID GRANTS-IN-AID E 0 ALA ALAE F 1 SHARIF ASHRAF (NI3) G 0 AIRE AIRIG (NI3) H 0 LIRA LIROTH I 0 BAN BANI J 1 KHARIJITE KHAWARIJ (NI3) K 0 PULI PULIK L 1 ARMFUL ARMSFUL M 0 GOY GOYIM N 0 KRONE KRONEN O 2 DERRING-DO DERRINGS-DO (NI3) Ý1 MEO MIAO/MIXTECA MIXTECO/PAPIOPIO PAPIO/SUMU SUMO (NI3)¨ P 2 AIDE-DE-CAMP AIDES-DE-CAMP Q 3 QARAQALPAQ QARAQALPAQ (NI3) R 0 KRONE KRONER S 0 A AS T 0 MATZO MATZOT U 0 HALER HALERU V 3 TIV TIV (NI3) W 2 SON-IN-LAW SONS-IN-LAW Ý1 KWAPA QUAPAW (NI3)¨ X 0 EAU EAUX Y 0 GROSZ GROSZY Z 3 HERTZ HERTZ ==> english/plurals/french.p <== What English word, when spelled backwards, is its French plural? ==> english/plurals/french.s <== state/etats ==> english/plurals/man.p <== Words ending with "man" make their plurals by adding "s". ==> english/plurals/man.s <== caiman doberman German human leman ottoman pitman Pullman Roman shaman talisman ==> english/plurals/switch.first.p <== What plural is formed by switching the first two letters? ==> english/plurals/switch.first.s <== falaj -> aflaj (Chambers English Dictionary) ==> english/portmanteau.p <== What are some words formed by combining together parts of other words? ==> english/portmanteau.s <== Such words are called "Portmanteau" words. Here is a very incomplete list: beefalo beef, buffalo brunch breakfast, lunch chortle chuckle, snort fantabulous fantastic, fabulous flare flame, glare flounder flounce, founder glimmer gleam, shimmer glitz glamour, ritz liger lion, tiger motel motor, hotel smash smack, mash smog smoke, fog squiggle squirm, wiggle tangelo tangerine, pomelo tigon tiger, lion **** Unless noted otherwise, all words occur in Webster's Third New International Dictionary, Merriam-Webster, Springfield, MA, 1961. ==> english/potable.color.p <== Find words that are both beverages and colors. ==> english/potable.color.s <== burgundy champagne chartreuse chocolate claret cocoa coffee cream midori (Japanese for green. Does Japanese count?) rose wine ==> english/rare.trigraphs.p <== What trigraphs (three-letter combinations) occur in only one word? ==> english/rare.trigraphs.s <== Here is a list of all the trigraphs which occur exactly once in the union of _Official Scrabble Words_ (First Edition), the _Official Scrabble Players Dictionary_ and _Webster's Unabridged Dictionary (Second Edition)_, together with the words in which they occur. The definition of "word" is a problematic. For example, lots of words starting deoxy- contain the trigraph `eox', but no others do. Should `eox' be on the list? Common words are marked with a *. aae baaed adq*headquarter headquarters ajs svarajs aqs talaqs bks nabks bze subzero cda ducdame dph*headphone headphones dsf*handsful dts veldts dzu kudzu kudzus ekd*weekday weekdays evh evhoe evz evzone evzones exv sexvalent ezv*rendezvous fhu cliffhung fjo fjord fjords fsp*offspring offsprings gds smaragds ggp*eggplant eggplants gnb signboard signboards gnp*signpost signposted signposting signposts gnt sovereignty gty hogtying gza*zigzag zigzagged zigzagging zigzaggy zigzags hds camanachds hky droshky hlr kohlrabi kohlrabies kohlrabis hrj lehrjahre hyx asphyxia asphyxias asphyxiate asphyxies asphyxy itv mitvoth iwy skiwy ixg sixgun jds slojds jje hajjes jki pirojki pirojki jym jymold kky yukky ksg*thanksgiving kuz yakuza kvo mikvoth kyj*skyjack skyjacked skyjacker skyjackers skyjacking skyjackings skyjacks llj killjoy killjoys lmd filmdom filmdoms ltd*meltdown meltdowns lxe calxes lzy schmalzy mds fremds mfy comfy mhs ollamhs mky dumky mmm dwammming mpg*campground mss bremsstrahlung muo muon muonic muonium muoniums muons nhs sinhs njy benjy nuu continuum obg hobgoblin hobgoblins ojk pirojki okc*bookcase bookcases ovk sovkhoz sovkhozes sovkhozy pev*grapevine grapevines pfs dummkopfs php ephphatha pss topssmelt pyj pyjama pyjamaed pyjamas siq physique physiques slt juslted smk besmkes spb*raspberries raspberry spt claspt swy swythe syg*easygoing szy groszy tux*tux tuxedo tuxedoes tuxedos tuxes tvy outvying tzu tzuris ucd ducdame vho evhoe vkh sovkhoz sovkhozes sovkhozy sovkhos vly vly vns eevns voh evohe vun avuncular wcy gawcy wdu*sawdust sawdusted sawdusting sawdusts sawdusty wfr bowfront wft ewftes xeu exeunt xgl foxglove foxgloves xiw taxiway taxiways xls cacomixls xtd nextdoor xva sexvalent yks bashlyks yrf gyrfalcon gyrfalcons ysd paysd yxy asphyxy zhk pirozhki zow zowie zwo*buzzword buzzwords zzs*buzzsaw ==> english/records/pronunciation/silent.p <== What words have an exceptional number of silent letters? ==> english/records/pronunciation/silent.s <== longest sequence BROUGHAM (4, UGHA) for each letter AISLE, COMB, INDICT, HANDSOME, TWITCHED, HALFPENNY, GNOME, MYRRH, BUSINESS, MARIJUANA, KNOCK, TALK, MNEMONIC, AUTUMN, PEOPLE, PSYCHE, CINQCENTS, FORECASTLE, VISCOUNT, HAUTBOY, PLAQUE, FIVEPENCE, WRITE, TABLEAUX, PRAYER, RENDEZVOUS homophones, for each letter O(A)R, LAM(B), S(C)ENT, LE(D)GER, DO(E), WAF(F), REI(G)N, (H)OUR, WA(I)VE, HAJ(J)I, (K)NOT, HA(L)VE, PRIM(M)ER, DAM(N), J(O)UST, (P)SALTER, ?, CAR(R)IES, (S)CENT, TARO(T), B(U)Y, ?, T(W)O, ?, RE(Y), BIZ(Z) **** Unless noted otherwise, all words occur in Webster's Third New International Dictionary, Merriam-Webster, Springfield, MA, 1961. ==> english/records/pronunciation/spelling.p <== What words have exceptional ways to spell sounds? ==> english/records/pronunciation/spelling.s <== same spelling, different sound -OUGH (7) BOUGH, COUGH, DOUGH, HICCOUGH, LOUGH, ROUGH, THROUGH different spelling, same sound AIR (9) AIR, AIRE, ARE, AYR, AYER, E'ER, ERE, ERR, HEIR **** Unless noted otherwise, all words occur in Webster's Third New International Dictionary, Merriam-Webster, Springfield, MA, 1961. ==> english/records/pronunciation/syllable.p <== What words have an exceptional number of letters per syllable? ==> english/records/pronunciation/syllable.s <== longest for each number of syllables one SCRAUNCHED ÝSQUIRRELLED (11)¨ two SCRATCHBRUSHED (14) one, for each letter ARCHED, BROUGHAMS, CRAUNCHED, DRAUGHTS, EARTHED, FLINCHED, GROUCHED, HAUNCHED, ITCHED, JOUNCED, KNIGHTS, LAUNCHED, MOOCHED, NAUGHTS, OINKED, PREACHED, QUETCHED, REACHED, SCRAUNCHED, THOUGHTS, UMPHS, VOUCHED, WREATHED, XYSTS, YEARNED, ZOUAVES two, for each letter ARCHFIENDS, BREAKTHROUGHS, CLOTHESHORSE, DRAUGHTBOARDS, EARTHTONGUES, FLAMEPROOFED, GREATHEART, HAIRSBREADTHS, INTHRALLED, JUNETEENTHS, KNICKKNACKS, LIGHTWEIGHTS, MOOSETONGUES, NIGHTCLOTHES, OUTSTRETCHED, PLOUGHWRIGHTS, QUICKTHORNS, ROUGHSTRINGS, SCRATCHBRUSHED, THROATSTRAPS, UNSTRETCHED, VERSESMITHS, WHERETHROUGH, XANTHINES, YOURSELVES, ZEITGEISTS shortest for each number of syllables two AA three AREA (4) ÝO'IO (3)¨ four IEIE (4) five OXYOPIA (7) six ONIOMANIA ÝAMIOIDEI (8)¨ seven EPIDEMIOLOGY (12) ÝOMOHYOIDEI (10)¨ eight EPIZOOTIOLOGY nine EPIZOOTIOLOGICAL (16) ten EPIZOOTIOLOGICALLY twelve HUMUHUMUNUKUNUKUAPUAA (21) **** Unless noted otherwise, all words occur in Webster's Third New International Dictionary, Merriam-Webster, Springfield, MA, 1961. ==> english/records/spelling/longest.p <== What is the longest word in the English language? ==> english/records/spelling/longest.s <== The longest word to occur in both English and American "authoritative" unabridged dictionaries is "pneumonoultramicroscopicsilicovolcanoconiosis." The following is a brief citation history of this "word." New York Herald Tribune, February 23, 1935, p. 3 "Pneumonoultramicroscopicsilicovolcanokoniosis succeeded electrophotomicrographically as the longest word in the English language recognized by the National Puzzlers' League at the opening session of the organization's 103d semi-annual meeting held yesterday at the Hotel New Yorker. The puzzlers explained that the forty-five-letter word is the name of a special form of silicosis caused by ultra-microscopic particles of siliceous volcanic dust." Everett M. Smith (b. 1/1/1894), President of NPL and Radio News Editor of the Christian Science Monitor, cited the word at the convention. Smith was also President of the Yankee Puzzlers of Boston. It is not known whether Smith coined the word. "Bedside Manna. The Third Fun in Bed Book.", edited by Frank Scully, Simon and Schuster, New York, 1936, p. 87 "There's been a revival in interest in spelling, but Greg Hartswick, the cross word king and world's champion speller, is still in control of the situation. He'd never get any competition from us, that's sure, though pronouncing, let alone spelling, a 44 letter word like: Pneumonoultramicrosopicsilicovolkanakoniosis, a disease caused by ultra-microscopic particles of sandy volcanic dust might give even him laryngitis." It is likely that Scully, who resided in New York in February 1935, read the Herald Tribune article and slightly misremembered the word. Supplement to the Oxford English Dictionary, 1936 Both "-coniosis" and "-koniosis" are cited. "a factitious word alleged to mean 'a lung disease caused by the inhalation of very fine silica dust' but occurring chiefly as an instance of a very long word." Webster's first cite is "-koniosis" in the addendum to the Second Edition. The Third Edition changes the "-koniosis" to "-coniosis." I conjecture that this "word" was coined by word puzzlers, who then worked assiduously to get it into the major unabridged dictionaries (perhaps with a wink from the editors?) to put an end to the endless squabbling about what is the longest word. ==> english/records/spelling/most.p <== What word has the most variant spellings? ==> english/records/spelling/most.s <== catercorner There's eight spellings in Webster's Third. catercorner cater-cornered catacorner cata-cornered catty-corner catty-cornered kitty-corner kitty-cornered If you look in Random House, you will find one more which doesn't appear in Web3, but it only differs by a hyphen: cater-corner --- Dan Tilque -- dant@techbook.com ==> english/records/spelling/operations.on.words/deletion.p <== What exceptional words turn into other words by deletion of letters? ==> english/records/spelling/operations.on.words/deletion.s <== longest beheadable word P(REDETERMINATION) (16/15) longest for each letter (6-88,181,198,213,13-159,14-219,15-155,16-96,220, 17-85) APATHETICALLY, BLITHESOME, CHASTENING, DEMULSIFICATION, EMOTIONLESSNESS, FUTILITARIANISM, GASTRONOMICALLY, HEDRIOPHTHALMA, IDENTIFICATION, JUNCTIONAL, KINAESTHETIC, LIMITABLENESS, METHYLACETYLENE, NEOPALEOZOIC, OENANTHALDEHYDE, PREDETERMINATION, QUINTA, REVOLUTIONARILY, SELECTIVENESS, TREASONABLENESS, UPRAISER, VINDICATION, WHENCEFORWARD, XANTHOPHYLLITE, YOURSELVES, ZOOSPORIFEROUS longest beheadable down to a single letter PRESTATE (8) longest curtailable word (not a plural) (BULLETIN)G (9) longest curtailable down to a single letter LAMBASTES longest alternately beheadable and curtailable word ASHAMED (7) longest arbitrarily beheadable and curtailable (all subsequences words) SHADES (6) longest terminal ellision word D(EPILATION)S (11) longest letter subtraction down to a single letter STRANGLING, STRANGING, STANGING, STAGING, SAGING, AGING, GING, GIN, IN, I longest charitable word (subtract letter anywhere) PLEATS: LEATS,PEATS,PLATS,PLEAS,PLEAT shortest stingy word (no deletion possible) PRY (3) **** Unless noted otherwise, all words occur in Webster's Third New International Dictionary, Merriam-Webster, Springfield, MA, 1961. ==> english/records/spelling/operations.on.words/insertion.and.deletion.p <== What exceptional words turn into other words by both insertion and deletion of letters? ==> english/records/spelling/operations.on.words/insertion.and.deletion.s <== longest word both charitable and hospitable AMY: AM,AY,MY;GAMY,ARMY,AMOY,AMYL shortest word both stingy and hostile IMPETUOUS (9) **** Unless noted otherwise, all words occur in Webster's Third New International Dictionary, Merriam-Webster, Springfield, MA, 1961. ==> english/records/spelling/operations.on.words/insertion.p <== What exceptional words turn into other words by insertion of letters? ==> english/records/spelling/operations.on.words/insertion.s <== longest hydration (double reheadment) (D,R)EVOLUTIONIST (12/13) longest hospitable word (insert letter anywhere) CARES: SCARES, CHARES, CADRES, CARIES, CARETS, CARESS shortest hostile word (no deletion possible) SYZYGY (6) **** Unless noted otherwise, all words occur in Webster's Third New International Dictionary, Merriam-Webster, Springfield, MA, 1961. ==> english/records/spelling/operations.on.words/movement.p <== What exceptional words turn into other words by movement of letters? ==> english/records/spelling/operations.on.words/movement.s <== longest word allowing exchange of letters (metallege) CONSERVATIONAL, CONVERSATIONAL longest head-to-tail shift SPECULATION, PECULATIONS longest double head-to-tail shift STABLE-TABLES-ABLEST longest complete cyclic transposal ATE-TEA-EAT (3) **** Unless noted otherwise, all words occur in Webster's Third New International Dictionary, Merriam-Webster, Springfield, MA, 1961. ==> english/records/spelling/operations.on.words/substitution.p <== What exceptional words turn into other words by substitution of letters? ==> english/records/spelling/operations.on.words/substitution.s <== longest onalosi (substitution in every position possible) PASTERS: MASTERS,POSTERS,PALTERS,PASSERS,PASTORS,PASTELS,PASTERN shortest isolano (no substitution possible) ECRU longest word, all letters changed to other letters in minimum number of steps, yielding another word THUMBING-THUMPING-TRUMPING-TRAMPING- TRAPPING-CRAPPING-CRAPPIES-CRAPPOES longest word girders BADGER/SUNLIT, BUDLET/SANGIR (6) longest word with full vowel substitution CL(A,E,I,O,U)CKING (8) also Y D(A,E,I,O,U,Y)NE (4) longest words with vowel substitutions DESTRUCTIBILITIES, DISTRACTIBILITIES (17) longest word constant-letter-shifted to another PRIMERO-SULPHUR (7) arithmetical-letter-shifted DREAM-ETHER (5) constant-shift-with-transposal (shiftgrams) AEROPHANE-SILVERITE (9) longest word pair shifted one position on typewriter keyboard WAXIER-ESCORT (6) longest word pair confusable on a telephone keypad AMOUNTS-CONTOUR (7) **** Unless noted otherwise, all words occur in Webster's Third New International Dictionary, Merriam-Webster, Springfield, MA, 1961. ==> english/records/spelling/operations.on.words/transposition.p <== What exceptional words turn into other words by transposition of letters? ==> english/records/spelling/operations.on.words/transposition.s <== longest reversal DESSERTS,STRESSED (8) longest well-mixed transposal CINEMATOGRAPHER, MEGACHIROPTERAN (15) longest transposition list APERS, APRES, ASPER, PARES, PARSE, PEARS, PRASE, PRESA, RAPES, REAPS, SPARE, SPEAR (12) ANGRIEST, ANGRITES, ASTRINGE, GAIRTENS, GANISTER, GANTRIES, GRANITES, INGRATES, RANGIEST, TEARINGS (10) ÝSATING(ER), SIGNATE(R), TANGIER(S) (3)¨ ANORETICS, ATROSCINE, CANOTIERS, CERTOSINA, CONARITES, CREATIONS, REACTIONS, TRICOSANE (8) transposition with deletion, insertion, or substitution longest well-mixed transdeletion SONOLUMINESCENCES, UNECONOMICALNESSES (17/18) longest word transdeletable to a single letter CONCENTRATIONS-CONSTERNATION-CONTORNIATES-TRANSECTION- STENTORIAN-TRANSIENT-ENTRAINS-NASTIER-ASTERN-TEARS-SATE-TEA-AT-A (14) longest Baltimore transdeletion (word transdeletable on every letter) IDOLATERS: DELATORS, SOTERIAL, DILATERS, ASTEROID, STOLIDER, SOREDIAL, DILATORS, DIASTOLE, TAILORED (9) shortest word that cannot be transadded to another word SYZYGY (6) longest well-mixed transubstitution MICROELECTROPHORESIS, SPECTROCOLORIMETRIES (20) **** Unless noted otherwise, all words occur in Webster's Third New International Dictionary, Merriam-Webster, Springfield, MA, 1961. ==> english/records/spelling/operations.on.words/words.within.words.p <== What exceptional words contain other words? ==> english/records/spelling/operations.on.words/words.within.words.s <== longest non-trivial charade IN-DISC-RIM-IN-A-TI-ON (16) longest forward and reverse charade MAT-HE-MA-TI-CAL, LAC-IT-AM-EH-TAM longest snowball or rhopalic T-EM-PER-AMEN-TALLY (15) longest reverse rhopalic HETERO-TRANS-PLAN-TAT-IO-N (21) highest ratio of subwords/length (logogram) FIRESTONE: RE, TO, ON, NO, IF, FIR, IRE, RES, TON, ONE, NOT, RIF, FIRE, IRES, REST, TONE, FIRES, STONE, SERIF (20/9) longest charlinkade FORESTALL: FOREST, ALL; FORE, REST, TALL (9) longest alternade TRIENNIALLY: TINILY, RENAL (11) shortest three-letter-minimum word deletions PILGRIMAGE: RIM, GAG, PILE; GRIM, LAG, PIE **** Unless noted otherwise, all words occur in Webster's Third New International Dictionary, Merriam-Webster, Springfield, MA, 1961. ==> english/records/spelling/sets.of.words/nots.and.crosses.p <== What is the most number of letters that can be fit into a three by three grid of words, such that no letter is repeated in any row, column or diagonal? ==> english/records/spelling/sets.of.words/nots.and.crosses.s <== Games magazine ran a contest on this. The winner had 62: proxying! buckwash ! veldt ------------------------------------ stumbled! j ! zincography ------------------------------------ whack ! providently ! bumfs Here are some good tries: backsword !thumpingly ! fez ---------------------------------- vexingly ! q ! throwbacks = 61 ---------------------------------- thump ! beadworks ! jingly backsword ! thumpingly! vex ---------------------------------- vexingly ! q ! throwbacks = 60 ---------------------------------- thump ! bedrocks ! flying subjack !downrightly! fez ---------------------------------- novelwright! q ! backups = 59 ---------------------------------- pyx ! subface ! downright krafts ! exhuming ! blowzy ----------+-----------+----------- phylum ! j ! transfixed = 56 ----------+-----------+----------- vexing ! folkways ! chump klutz ! cymograph ! fend ----------+-----------+----------- exscind ! j ! kymograph = 54 ----------+-----------+----------- myograph !flunked ! vibs **** Unless noted otherwise, all words occur in Webster's Third New International Dictionary, Merriam-Webster, Springfield, MA, 1961. ==> english/records/spelling/sets.of.words/squares.p <== What are some exceptional word squares (square crosswords with no blanks)? ==> english/records/spelling/sets.of.words/squares.s <== Word squares are a particular example of a type of crossword known as "forms". They were more popular early in the 20th century than they are now, but people still like to compose and solve them. Forms appear every month in the _Enigma_, which is the monthly publication of the National Puzzlers' League. The membership fee is $13 for the first year, and information may be obtained from: David A. Rosen 207 East 27th St. #3K New York, NY 10016 All members have the option of choosing a nom de plume; for example, I go by the nom "Cubist". Another good place to find information on forms is in _Word Ways_, which is a quarterly journal of recreational linguistics: _Word Ways_ Spring Valley Road Morristown, NJ 07960 I'll have a paper appearing here at some point on the "support" of a form (which I'll discuss below). Word squares come in two flavors, regular and double. In regular word squares the words are the same across and down; in double word squares all words are different. The largest legitimate word square has order 9 (although Jeff Grant has come close to the 10), and what is considered to be the finest example was discovered by Eric Albert via computer search: necessism existence circumfer escarping sturnidae sempitern infidelic scenarize mergences All words appear in from Webster's New International Dictionary, Second Edition. It's the *only* single-source 9-square known, and its only flaw is that "Sturnidae" is a proper (capitalized) word. All words are also solid-form (no phrases, spaces, punctuation marks, etc.). Eric was using about 63,000 words when he discovered his square. Using about 78,500 9-letter words, I found an additional square: bortsches overtrust reparence trabeatae strestell creatural hunterite escalates steelless All are in the OED, except for "trabeatae", which is in NI2. This makes this square arguably the second-best ever discovered. All words are uncapitalized and solid-form, but it has the flaw of using more than one source. It is, however, the *only* known 9-square that uses only uncapitalized, solid-form dictionary words. There are about 2000 9-squares known, all of which were constructed by hand except for the two noted above. Almost all of these use very obscure sources of words. As a general rule of thumb, if you discover a new form via computer search, it is probably going to be of high quality, since it is hard to obtain computer-readable word lists that contain *really* obscure words. The largest known double word-squares are of order-8. They are considered to be about as hard to construct as a regular word square of order-9, and this is substantiated by the work I've done on the mathematics of form construction. The following fine example was constructed by Jeff Grant (see his article in _Word Ways_, Vol. 25 Num. 1, pp. 9-12): trattled hemerine apotomes metapore nailings aloisias tentmate assessed All are dictionary terms, but there are some weak entries, e.g. Aloisias: individuals named Aloisia, a feminine form of Aloysius occurring in the 16th and 17th century in parish registers of Hinton Charterhouse, England (The Oxford Dictionary of English Christian Names, 3rd Edition, E.G. Withycombe, 1977) Such words are, however, dear to the heart of logologists] For other examples of double squares see the article mentioned above. There are also many other types of forms. Some of the most common are pyramids, stars, and diamonds, and some come in regular and double varieties, and some are inherently double (e.g. rectangles). How hard is it to discover a square, anyway, and how many are there? As a data point, my program using the main (Air Force) entries in NI2 (26,332 words), found only seven 8x8 squares. This took an hour to run. They are: outtease appetite unabated acetated interact repeated repeated unweaned prenaris nopinene cadinene neomenia evenmete evenmete twigsome perscent apostate edentate toxicant pectosic pectosic teguexin ensconce bistered tindered emittent entresol entresol easement taconite antehall antehall rectoral amoebula amoebula anoxemia irenicum tearable tearable anaerobe tessular tessular seminist tincture entellus entellus cinnabar etiolate etiolate edentate esteemer deedless deedless tattlery declarer declared If the heuristic mathematics are worked out, the number of different words in your word-list before you'd expect to find a regular word square of order-n (the "support") is about e^{(n-1)/2}, where e ~ 15.7. For a double word square of order-n the support is about e^{n/2}. There is a simple algorithm which is more precise, and this gives a support of 75,641 for a regular 9-square, and a support of 272,976 for a double 9-square (using my 9-letter word list), which agrees well with reality. -- Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618 clong@remus.rutgers.edu ==> english/records/spelling/single.words.p <== What words have exceptional lengths, patterns ==> english/records/spelling/single.words.s <== Word Records from Webster's Third Spelling Letter Patterns Entire Word longest word trinitrophenylmethylnitramine (29,1) longest palindrome kinnikinnik (11,1) longest beginning with a palindrome adinida (7,1) longest beginning with b palindrome boob (4,1) longest beginning with c palindrome carac civic (5,2) longest beginning with d palindrome deified devoved (7,2) longest beginning with e palindrome ecce esse (4,2) longest beginning with f palindrome f (1,1) longest beginning with g palindrome goog (4,1) longest beginning with h palindrome hagigah halalah (7,2) longest beginning with i palindrome igigi imami (5,2) longest beginning with j palindrome j (1,1) longest beginning with k palindrome kinnikinnik (11,1) longest beginning with l palindrome lemel level lysyl (5,3) longest beginning with m palindrome malayalam (9,1) longest beginning with n palindrome nauruan (7,1) longest beginning with o palindrome oppo otto (4,2) longest beginning with p palindrome peeweep (7,1) longest beginning with q palindrome qazaq (5,1) longest beginning with r palindrome reviver rotator (7,2) longest beginning with s palindrome sawbwas seesees seities sememes (7,4) longest beginning with t palindrome terret tibbit tippit (6,3) longest beginning with u palindrome uku ulu utu (3,3) longest beginning with v palindrome vav (3,1) longest beginning with w palindrome waw wow (3,2) longest beginning with x palindrome x (1,1) longest beginning with y palindrome yaray (5,1) longest beginning with z palindrome z (1,1) longest with middle a palindrome halalah rotator (7,2) longest with middle b palindrome sawbwas (7,1) longest with middle c palindrome soccos succus (6,2) longest with middle d palindrome murdrum (7,1) longest with middle e palindrome sememes (7,1) longest with middle f palindrome deified (7,1) longest with middle g palindrome degged (6,1) longest with middle h palindrome aha ihi oho (3,3) longest with middle i palindrome hagigah reviver (7,2) longest with middle j palindrome kajak (5,1) longest with middle k palindrome kinnikinnik (11,1) longest with middle l palindrome hallah selles (6,2) longest with middle m palindrome sammas (6,1) longest with middle n palindrome adinida (7,1) longest with middle o palindrome devoved (7,1) longest with middle p palindrome tippit (6,1) longest with middle q palindrome q (1,1) longest with middle r palindrome nauruan (7,1) longest with middle s palindrome seesees (7,1) longest with middle t palindrome seities (7,1) longest with middle u palindrome alula arura (5,2) longest with middle v palindrome civic level rever tevet (5,4) longest with middle w palindrome peeweep (7,1) longest with middle x palindrome sexes (5,1) longest with middle y palindrome malayalam (9,1) longest with middle z palindrome kazak qazaq (5,2) longest tautonym tangantangan (12,1) longest beginning with a tautonym akeake atlatl (6,2) longest beginning with b tautonym bellabella (10,1) longest beginning with c tautonym caracara chowchow couscous (8,3) longest beginning with d tautonym dugdug dumdum (6,2) longest beginning with e tautonym ee (2,1) longest beginning with f tautonym froufrou (8,1) longest beginning with g tautonym ganggang greegree guitguit (8,3) longest beginning with h tautonym hotshots (8,0) ? longest beginning with i tautonym ipilipil (8,1) longest beginning with j tautonym juju (4,1) longest beginning with k tautonym kavakava kawakawa khuskhus kohekohe kouskous kukukuku (8,6) longest beginning with l tautonym lapulapu lavalava lomilomi (8,3) longest beginning with m tautonym mahimahi makomako matamata murumuru (8,4) longest beginning with n tautonym nagnag (6,1) longest beginning with o tautonym oo (2,1) longest beginning with p tautonym palapala pioupiou piripiri poroporo (8,4) longest beginning with q tautonym quiaquia (8,1) longest beginning with r tautonym riroriro (8,1) longest beginning with s tautonym sweeswee (8,1) longest beginning with t tautonym tangantangan (12,1) longest beginning with u tautonym ulaula (6,1) longest beginning with v tautonym valval verver (6,2) longest beginning with w tautonym wallawalla (10,1) longest beginning with x tautonym ? (0,0) ? longest beginning with y tautonym yariyari (8,1) longest beginning with z tautonym zoozoo (6,1) longest head 'n' tail einsteins muckamuck okeydokey overcover pungapung tarantara trinitrin (9,7) longest with middle a head 'n' tail muckamuck pungapung (9,2) longest with middle b head 'n' tail aba (3,1) longest with middle c head 'n' tail overcover (9,1) longest with middle d head 'n' tail okeydokey (9,1) longest with middle e head 'n' tail arear caeca (5,2) longest with middle f head 'n' tail efe ofo (3,2) longest with middle g head 'n' tail aggag algal edged magma (5,4) longest with middle h head 'n' tail outshouts (9,0) ? longest with middle i head 'n' tail trinitrin (9,1) longest with middle j head 'n' tail anjan (5,1) longest with middle k head 'n' tail arkar kokko (5,2) longest with middle l head 'n' tail ingling khalkha (7,2) longest with middle m head 'n' tail bamba bombo mamma pampa (5,4) longest with middle n head 'n' tail tarantara (9,1) longest with middle o head 'n' tail ingoing mesomes (7,2) longest with middle p head 'n' tail apa (3,1) longest with middle q head 'n' tail q (1,1) longest with middle r head 'n' tail adrad kurku ugrug verve (5,4) longest with middle s head 'n' tail hotshot (7,1) longest with middle t head 'n' tail einsteins (9,1) longest with middle u head 'n' tail mauma shush siusi veuve (5,4) longest with middle v head 'n' tail ava eve (3,2) longest with middle w head 'n' tail abwab (5,1) longest with middle x head 'n' tail manxman (7,1) longest with middle y head 'n' tail calycal (7,1) longest with middle z head 'n' tail z (1,1) Subset of Word longest internal palindrome kinnikinniks sensuousness sensuousnesses (11,3) longest internal tautonym anhydrohydroxyprogesterone anhydrohydroxyprogesterones kinnikinnick kinnikinnicks kinnikinnics kinnikinniks magnetophotophoresis methylethylpyridine micromicrofarad neuroneuronal trimethylethylene (10,11) longest repeated prefix kinnikinnick kinnikinnicks kinnikinnics kinnikinniks micromicrofarad neuroneuronal (10,6) most consecutive doubled letters bookkeeper bookkeeping (3,2) most doubled letters possessionlessness possessionlessnesses successlessness successlessnesses (4,4) longest two cadence humuhumunukunukuapuaa humuhumunukunukuapuaas (8,2) longest three cadence effervescence effervescences extendednesses neglectednesses pervertednesses redheadednesses reflectednesses unexpectednesses vallabhacharya vallabhacharyas (5,10) longest four cadence alveolopalatal coproporphyrinuria coproporphyrinurias distributivities gastroschisises humuhumunukunukuapuaa humuhumunukunukuapuaas inevitabilities roentgenometries somesthesises stresslessness stresslessnesses (4,12) longest five cadence indecipherablenesses recollectivenesses (4,2) Letter Counts Lipograms longest letters from first half hamamelidaceae (14,1) longest letters from second half nonsupports (11,0) ? longest without ab hydroxydesoxycorticosterone (27,1) longest without abcd philoprogenitivenesses (22,1) longest without a to h supposititiously (16,1) longest without a to k monotonously synonymously tumultuously voluptuously (12,4) longest without a to n prototropy zoosporous (10,2) longest without a to q susurrus (8,1) longest without a to s tutty (5,1) longest without e humuhumunukunukuapuaas macracanthorhynchiasis phonocardiographically prorhipidoglossomorpha supradiaphragmatically (22,5) longest without et humuhumunukunukuapuaas phonocardiographically prorhipidoglossomorpha (22,3) longest without eta coccidioidomycosis (18,1) longest without etai phyllospondylous (16,1) longest without etain chlorophyllous chromosomology chrysochlorous phyllomorphous polymorphously scolopophorous (14,6) longest without etains promorphology (13,1) Letter Choices Vowels longest all vowels aiee ieie (4,2) longest each vowel once entwicklungsroman (17,1) longest each vowel & y once cylindrocellular phosphuranylites ventriculography (16,3) shortest each vowel once eulogia eutocia eutopia isourea sequoia (7,5) shortest each vowel & y once oxyuridae (9,1) shortest vowels in order caesious (8,1) shortest vowels & y in order facetiously (11,1) longest vowels in order abstentious (11,1) longest vowels & y in order abstemiously (12,1) shortest vowels in reverse order muroidea (8,1) shortest vowels & y in reverse order ? (80,0) ? longest vowels in reverse order subcontinental (14,1) longest vowels & y in reverse order ? (0,0) ? longest one vowel strengths (9,1) longest two vowels schwartzbrots (13,1) longest containing a univocalic tathagatagarbhas (16,1) longest containing e univocalic strengthlessnesses (18,1) longest containing i univocalic instinctivistic (15,1) longest containing o univocalic loxolophodonts (14,1) longest containing u univocalic struldbrugs (11,1) longest containing y univocalic glycyls gypsyfy khlysts khlysty phytyls pyrryls qyrghyz rhythms styryls thymyls tyddyns (7,11) longest alternating vowel-consonant hypovitaminosises (17,1) longest alternating vowel-consonant excluding y aluminosilicates diketopiperazine epicoracohumeral (16,3) Consonants longest consonant string bergschrund bergschrunds catchphrase eschscholtzia eschscholtzias festschrift festschriften festschrifts goldschmidtine goldschmidtines goldschmidtite goldschmidtites lachsschinken lachsschinkens latchstring mischsprache mischsprachen nachschlag nachschlage nachschlags promptscript veldtschoen weltschmerz weltschmerzes (6,24) longest one consonant assessees coccaceae (9,2) longest two consonant nauseousnesses sensuousnesses (14,2) Isograms longest isogram dermatoglyphics (15,1) longest pair isogram scintillescent (14,1) longest trio isogram deeded (6,1) longest tetrad isogram kukukuku (8,1) longest polygram unprosperousnesses (18,0) ? longest pyramid chachalaca deadheaded disseisees evennesses keennesses kinnikinic rememberer sassanians sereneness sleeveless susurruses (10,11) most repeated letters dihydroxycholecalciferol hydroxydesoxycorticosterone hysterosalpingographies methyldihydromorphinone microspectrophotometrically octamethylpyrophosphoramide phosphatidylethanolamine pseudohermaphroditism tetrabromophenolphthalein tetraiodophenolphthalein trinitrophenylmethylnitramine (9,11) highest containing a repeated palaeacanthocephala tathagatagarbha tathagatagarbhas (6,3) highest containing b repeated bubbybush flibbertigibbet flibbertigibbets flibbertigibbety (4,4) highest containing c repeated chroococcaceae chroococcaceous circumcrescence circumcrescences echinococcic micrococcaceae (5,6) highest containing d repeated condiddled dadded deadheaded dendrodendritic diddered diddled diddledees didodecahedron disbudded dodded doddered doddled driddled dunderheaded dunderheadedness dunderheadednesses dyakisdodecahedral dyakisdodecahedron dyakisdodecahedrons fiddledeedee fiddleheaded granddaddy lepidodendrid lepidodendrids lepidodendroid muddleheaded muddleheadedness muddleheadednesses muddyheaded puddingheaded skedaddled woodshedded (4,32) highest containing e repeated ethylenediaminetetraacetate (7,1) highest containing f repeated chiffchaff chiffchaffs giffgaff giffgaffed giffgaffing giffgaffs riffraff (4,7) highest containing g repeated aggregating aggreging chugalugging gagging gaggling ganggang ganggangs gigging giggling gigglingly glugging goggling grigging grogging guggling lallygagging lollygagging zigzagging (4,18) highest containing h repeated ichthyophthiriasis ichthyophthirius ichthyophthiriuses rhamphorhynchid rhamphorhynchids rhamphorhynchoid rhamphorhynchus (4,7) highest containing i repeated dirigibilities discriminabilities distinguishabilities divisibilities ignitibilities indiscernibilities indiscerptibilities indistinguishability indivisibility infinitesimalities intelligibilities invincibilities (6,12) highest containing j repeated ajonjoli ajonjolis avijja avijjas djokjakarta gastrojejunal gastrojejunostomy hajj hajjes hajji hajjis jajman jajmani jajmans jajoba jejuna jejunal jejune jejunely jejuneness jejunenesses jejunities jejunity jejunostomies jejunostomy jejunum jeremejevite jeremejevites jimberjawed jimjams jinglejangle jinglejangles jinjili jinjilis jipijapa jipijapas jirajara jirajaras jiujitsu jiujitsus jiujutsu jiujutsus jogjakarta jojoba jujitsu jujitsus juju jujube jujubes jujus jujut highest containing k repeated kakkak kakkaks knickknack knickknackatories knickknackatory knickknackeries knickknackery knickknacky kukukuku kukukukus (4,10) highest containing l repeated allochlorophyll allochlorophylls alloplastically intellectualistically lillypillies lillypilly polysyllabically (5,7) highest containing m repeated dynamometamorphism hamamelidanthemum immunocompromised mammatocumulus mammectomies mammectomy mammiform mammilliform mammogram mammonism mammonisms mesembryanthemum mesembryanthemums meshummadim mohammedanism mohammedanisms muhammadanism muhammadanisms mummiform tetramethylammonium thermometamorphism zamzummim zamzummims (4,23) highest containing n repeated inconvenientness inconvenientnesses nannoplankton nannoplanktonic nondenominational nondenominationalism nonentanglement nonintervention noninterventionist syngenesiotransplantation unconvincingness unconvincingnesses (5,12) highest containing o repeated monogonoporous pseudomonocotyledonous (6,2) highest containing p repeated aplopappus haplopappus hyperleptoprosopic hyperleptoprosopy snippersnapper whippersnapper (4,6) highest containing q repeated qaraqalpaq qaraqalpaqs (3,2) highest containing r repeated ferriprotoporphyrin ferroprotoporphyrin (5,2) highest containing s repeated possessionlessnesses (9,1) highest containing t repeated ethylenediaminetetraacetate tetrasubstituted throttlebottom totipotentiality yttrotantalite (5,5) highest containing u repeated humuhumunukunukuapuaa humuhumunukunukuapuaas (9,2) highest containing v repeated overconservative ovoviviparity ovoviviparous ovoviviparously ovoviviparousness vulvovaginitis (3,6) highest containing w repeated bowwow bowwows powwow powwowed powwowing powwows swallowwort whillywhaw whillywhaws whitlowwort williwaw williwaws willowware willowweed willowworm willywaw willywaws (3,17) highest containing x repeated dextropropoxyphene executrix executrixes exlex exlexes exonarthex exotoxic exotoxin hexachlorocyclohexane hexahydroxy hexaxon hexoxide hydroxydeoxycorticosterone hydroxydesoxycorticosterone maxixe maxixes myxoxanthin oxyhexactine oxyhexaster paxwax paxwaxes paxywaxies paxywaxy saxifrax saxifraxes saxitoxin sextuplex xanthotoxin xanthoxenite xanthoxenites xanthoxylaceae xanthoxyletin xanthoxyletins xanthoxylin xanthoxylins xanthoxylum xanthoxylums (2,37) highest containing y repeated acetylphenylhydrazine acetylphenylhydrazines anhydrohydroxyprogesterone anhydrohydroxyprogesterones brachydactyly chylophylly cryptozygy cystopyelography cytophysiologically cytophysiology dacryocystorhinostomy dactylosymphysis dihydroxyphenylalanine dyssynergy glycolytically gypsyfy gypsyfying hydrodynamically hydronymy hydroxydeoxycorticosterone hydroxydesoxycorticosterone hydroxyethyl hydroxyethylation hydroxyethylations hydroxylysine hydroxymethyl hydroxymethylation highest containing z repeated pizzazz pizzazzes razzmatazz razzmatazzes (4,4) most different letters blepharoconjunctivitis pseudolamellibranchiata pseudolamellibranchiate psychogalvanometric (16,4) highest ratio length/letters kukukuku (400,1) highest ratio length/letters (no tautonyms) senselessnesses (375,1) lowest length 16 ratio length/letters ventriculography (106,1) lowest length 17 ratio length/letters entwicklungsroman hydrobasaluminite pterygomandibular (113,3) lowest length 18 ratio length/letters carboxyhemoglobins entwicklungsromane hyperglobulinemias psychogalvanometer ventriculographies (120,5) lowest length 19 ratio length/letters psychogalvanometric (118,1) lowest length 20 ratio length/letters brachycephalizations dimethyltubocurarine encephalomyocarditis magnetofluiddynamics moschellandsbergites (133,5) lowest length 21 ratio length/letters diphenylthiocarbazone pseudolamellibranchia sphygmomanometrically (140,3) lowest length 22 ratio length/letters blepharoconjunctivitis (137,1) lowest length 23 ratio length/letters pseudolamellibranchiata pseudolamellibranchiate (143,2) lowest length 24 ratio length/letters diphenylaminechlorarsine laryngotracheobronchitis meningoencephalomyelitis (171,3) lowest length 25 ratio length/letters spectroheliokinematograph (166,1) Letter Appearance longest narrow letters (ACEMNORSUVWXZ) erroneousnesses verrucosenesses (15,2) longest tall letters (BDFGHIJKLPQTY) lighttight lillypilly (10,2) longest vertical-symmetry letters (AHIMOTUVWXY) homotaxia thymomata (9,2) longest horizontal-symmetry letters (BCDEHIKOX) checkbook checkhook chookchie (9,3) highest ratio of dotted letters (IJ) jinjili (71,1) Typewriter longest top row proprietory proterotype rupturewort (11,3) longest middle row shakalshas (10,1) longest in order wettish (7,1) longest in reverse order bourree chapote chappie chappow gouttee (7,5) longest left hand tesseradecades (14,0) ? longest right hand hypolimnion kinnikinnik (11,2) longest alternating hands leucocytozoans (14,1) longest one finger deeded humhum hummum muhuhu muumuu (6,5) longest adjacent keys assessees redresser redresses seeresses sweeswees (9,5) Puzzle longest formed with chemical symbols nonrepresentationalism (22,1) longest formed with US postal codes convallarias (12,1) longest formed with compass points newnesses sweeswees (9,2) longest formed with piano notes cabbaged fabaceae fagaceae (8,3) Letter Order Alphabetical longest letters in order aegilops (8,1) longest letters in order with repeats aegilops (8,1) longest letters in reverse order sponged wronged (7,2) longest letters in reverse order with repeats trollied (8,1) longest roller-coaster decriminalizations provincializations (18,2) longest no letters in place trinitrophenylmethylnitramine (29,1) most letters in place abudefduf agammaglobulinemias archencephalon archetypical archetypically syngenesiotransplantation (5,6) most letters in place shifted cooperatively daughterlinesses definitivenesses gymnoplast gymnoplasts inoperative inoperativeness inopportunely intraoperatively neighborlinesses operatively postoperatively preoperatively undefendablenesses unoperative unspiritually (6,16) most consecutive letters in order consecutively bierstube bierstuben bierstubes gymnopaedia gymnopaedias gymnopaedic gymnopedia gymnopedias gymnophiona gymnoplast gymnoplasts klavierstuck limnopithecus limnoplankton limnoplanktonic overstudy overstuff overstuffed semnopithecus semnopitheque semnopitheques thamnophile thamnophiles thamnophiline thamnophilus thamnophis understudy (4,27) most consecutive letters in order aborticide aborticides abscinded absconded abscondence abscondences alimentotherapy aluminographies aluminography aluminotype aluminotypes ambuscade ambuscaded ambuscades helminthosporia helminthosporin helminthosporins helminthosporium helminthosporiums helminthosporoid laminograph laminographic laminographies laminography laminosioptes limnograph limnopithecus limnoplankton limnoplanktonic luminophor luminophors luminoscope opaquers reconstructive reconstructively most consecutive letters appropinquates appropinquations appropinquities equiponderates equiponderations perquisition perquisitions preconquest propinquities quadruplications sesquiterpenoid sesquiterpenoids (8,12) highest ratio of consecutive letters to length klompen (85,1) ==> english/repeat.p <== What is a sentence containing the most repeated words, without: using quotation marks, using proper names, using a language other than English, anything else distasteful. ==> english/repeat.s <== Five "had"s in a row: The parents were unable to conceive, so they hired someone else to be a surrogate. The parents had had a surrogate have their child. The parents had had had their child. The child had had no breakfast. The child whose parents had had had had had no breakfast. ==> english/repeated.words.p <== What is a sentence with the same word several times repeated? ==> english/repeated.words.s <== It is true for all that, that that that that that that signifies, is not the one to which I refer. Here are some steps to understanding the entire sentence: That is not the one to which I refer. That (that that that signifies) is not the one to which I refer. That that that that that that signifies, is not the one to which I refer. In Annamite: Ba ba ba ba. (Three ladies gave a box on the ear to the favorite of the Prince.) ==> english/rhyme.p <== What English words are hard to rhyme? "Rhyme is the identity in sound of an accented vowel in a word...and of all consonantal and vowel sounds following it; with a difference in the sound of the consonant immediately preceding the accented vowel." (From The Complete Rhyming Dictionary by Clement Wood). Appropriately Wood says a couple of pages later, "If a poet commences, 'October is the wildest month' he has estopped himself from any rhyme; since "month" has no rhyme in English." ==> english/rhyme.s <== NI3 = Merriam-Webster's Third New International Dictionary NI2 = Merriam-Webster's New International Dictionary, Second Edition RHD = Random House Unabridged Dictionary + means slang, foreign, obsolete, dialectical, etc. Word Rhyme Assonance --------------- --------------------------------------- -------------------- aitch brache (NI2+), taich (NI2+) naish angry unangry (NI2+) aggry angst lanx beards weirds breadth death bulb pulp carpet charpit chimney timne, polymny (NI2+) cusp wusp (NI2) bust depth stepped eighth faith else belts exit direxit (RHD+) sexist fiends teinds, piends filched hilched (NI3+), milched (NI2) zilch filth spilth, tilth fifth drift film pilm (NI3+) kiln fluxed luxed (NI3+), muxed (NI3+) ducked glimpsed rinsed gospel hostile gulf pulse jinxed outminxed (?) blinked leashed niched, tweesht (NI2+) liquid wicked mollusk smallest mouthed southed month grumph mulcts bulks mulched gulched (NI3+) bulged ninth pint oblige bides oomph sumph (NI3+) orange sporange pint jint (NI2+) bind poem phloem, proem pregnant regnant purple curple (NI3+), hirple (NI3+) puss schuss rhythm smitham scalds balds, caulds (NI3+), faulds (NI3+) scarce clairce (NI2), hairse (NI2+) cares sculpts gulps silver chilver (NI3+) sixth kicks spirit squiret (NI2+) tenth nth bent tsetse baronetcy, intermezzi, theetsee tuft yuft twelfth health widow kiddo width bridge window indo, lindo wolf bulls ==> english/self.ref.letters.p <== Construct a true sentence of the form: "This sentence contains _ a's, _ b's, _ c's, ...," where the numbers filling in the blanks are spelled out. ==> english/self.ref.letters.s <== A little history of the problem, culled from the pages of _Metamagical Themas_, Hofstadter's collection of his _Scientific American_ columns. First mention of it is in the Jan. '82 column, a followup to one on self- referential sentences. Lee Sallows opened the field with a sentence that began "Only the fool would take trouble to verify that his sentence was composed of ten a's ...." etc. Then in the addendum to the Jan.'83 column on viral sentences, Hofstadter quotes Sallows describing his Pangram Machine, "a clock-driven cascade of sixteen Johnson-counters," to tackle the problem. An early success was: "This pangram tallies five a's, one b, one c, two d's, twenty- eight e's, eight f's, six g's, eight h's, thirteen i's, one j, one k, three l's, two m's, eighteen n's, fifteen o's, two p's, one q, seven r's, twenty-five s's, twenty-two t's, four u's, four v's, nine w's, two x's, four y's, and one z." Sallows wagered ten guilders that no-one could create a perfect self- documenting sentence beginning, "This computer-generated pangram contains ...." within ten years. It was solved very quickly, after Sallows' challenge appeared in Dewdny's Oct. '84 SA column. Larry Tesler solved it by a method Hofstadter calls "Robinsonizing," which involves starting with an arbitrary set of values for each letter, getting the true values when the sentence is made, and plugging the new values back in, making a feedback loop. Eventually, you can zero in on a set of values that work. Tesler's sentence: This computer-generated pangram contains six a's, one b, three c's, three d's, thirty-seven e's, six f's, three g's, nine h's, twelve i's, one j, one k, two l's, three m's, twenty-two n's, thirteen o's, three p's, one q, fourteen r's, twenty-nine s's, twenty-four t's, five u's, six v's, seven w's, four x's, five y's, and one z. The method of solution (called "Robinsonizing," after the logician Raphael Robinson) is as follows: 1) Fix the count of a's. 2) Fix the count of b's. 3) Fix the count of c's. ... 26) Fix the count of z's. Then, if the sentence is still wrong, go back to step 1. Most attempts will fall into long loops (what Hofstadter calls attractive orbits), but with a good computer program, it's not too hard to find a Robinsonizing sequence that zeros in on a fixed set of values. The February and May 1992 _Word Ways_ have articles on this subject, titled "In Quest of a Pangram, (Part 1)" by Lee Sallows. It tells of his search for a self-referential pangram of the form, "This pangram contains _ a's, ..., and one z." (He built special hardware to search for them.) Two such pangrams given in the article are: This pangram lists four a's, one b, one c, two d's, twenty-nine e's, eight f's, three g's, five h's, eleven i's, one j, one k, three l's, two m's twenty-two n's, fifteen o's, two p's, one q, seven r's, twenty-six s's, nineteen t's, four u's, five v's, nine w's, two x's, four y's, and one z. This pangram contains four a's, one b, two c's, one d, thirty e's, six f's, five g's, seven h's, eleven i's, one j, one k, two l's, two m's eighteen n's, fifteen o's, two p's, one q, five r's, twenty-seven s's, eighteen t's, two u's, seven v's, eight w's, two x's, three y's, & one z. It also contains one in Dutch by Rudy Kousbroek: Dit pangram bevat vijf a's, twee b's, twee c's, drie d's, zesenveertig e's, vijf f's, vier g's, twee h's, vijftien i's, vier j's, een k, twee l's, twee m's, zeventien n's, een o, twee p's, een q, zeven r's, vierentwintig s's, zestien t's, een u, elf v's, acht w's, een x, een y, and zes z's. References: Dewdney, A.K. Scientific American, Oct. 1984, pp 18-22. Sallows, L.C.F. Abacus, Vol.2, No.3, Spring 1985, pp 22-40. Sallows, L.C.F. Word Ways, Feb. & May 1992 Hofstadter, D. Scientific American, Jan. 1982, pp 12-17. ==> english/self.ref.numbers.p <== What true sentence has the form: "There are _ 0's, _ 1's, _ 2's, ..., in this sentence"? ==> english/self.ref.numbers.s <== There are 1 0's, 7 1's, 3 2's, 2 3's, 1 4's, 1 5's, 1 6's, 2 7's, 1 8's, and 1 9's in this sentence. There are 1 0's, 11 1's, 2 2's, 1 3's, 1 4's, 1 5's, 1 6's, 1 7's, 1 8's and 1 9's in this sentence. ==> english/self.ref.words.p <== What sentence describes its own word, syllable and letter count? ==> english/self.ref.words.s <== This sentence contains ten words, eighteen syllables, and sixty-four letters. ==> english/sentence.p <== Find a sentence with words beginning with the letters of the alphabet, in order. ==> english/sentence.s <== After boxes containing dynamite exploded furiously generating hellish inferno jet killing laboring miners, novice operator, paralyzed, quickly refuses surgical treatment until veteran workers x-ray youth zealously. A big cuddly dog emitted fierce growls happily ignoring joyful kids licking minute nuts on pretty queer rotten smelly toadstalls underneath vampires who x-rayed young zombies. ==> english/snowball.p <== Construct the longest coherent sentence you can such that the nth word is n letters long. ==> english/snowball.s <== I do not know where family doctors acquired illegibly perplexing handwriting; nevertheless, extraordinary pharmaceutical intellectuality, counterbalancing indecipherability, transcendentalizes intercommunications' incomprehensibleness. ==> english/spoonerisms.p <== List some exceptional spoonerisms. ==> english/spoonerisms.s <== Original by Spooner himself: I am afraid you have tasted the whole worm, and must therefore take the next town drain. Some years ago in the Parliament, a certain member known for his quick and rapier wit, cut across a certain other member who was trying to make some bad joke. He called him a "Shining Wit" then apologized for making a Spoonerism. Another famous broadcast fluff was on the Canadian Broadcasting Corporation, which an announcer identified as the "Canadian Broadcorping Castration." Oh yes, another radio announcer one that has sort of crept into common English usage is "one swell foop". A friend of mine had just eaten dinner in the school cafeteria, and he didn't look very happy. Another of my friends said, "John, what's wrong?" Knowing exactly what he was saying, he said, "It's the bound grief I had for dinner]" A radio announcer, talking about a royal visit (or some such) said the visitor would be greeted with a "twenty one sun galoot". There are several fractured fables based on spoonerisms, such as: A king on a desert island was so beloved by his people, they decided to give him a very special gift for the anniversary of his coronation. So after much thought, they decided to make him a throne out of seashells, which were plentiful on the island. And when it was finished, they presented it to the king, who loved it. But he soon discovered it was very uncomfortable to sit on. So he told his subjects it was too special to use everyday (so as not to hurt their feelings) and put it in the attic of his palace (which was, of course, a hut like all the other dwellings on the island), planning to use it just for special occasions. But that night, it fell through the ceiling of his bedroom and landed on top of him, killing him instantly. And the moral of the story is: Those who live in grass houses shouldn't stow thrones] ==> english/states.p <== What long words have all bigrams either a postal state code or its reverse? ==> english/states.s <== 10 paramarine 10 indentment 10 cacocnemia 9 amendment 9 paramimia 9 paramenia 9 paralinin 9 paralalia 9 palilalia 9 palapalai 8 scalawag 8 memorial Disallowing reversals of state codes the longest common ones are: 8 malarial 7 malaria 6 scalar 6 marine 5 flaky Terry Donahue ==> english/telegrams.p <== Since telegrams cost by the word, phonetically similar messages can be cheaper. See if you can decipher these extreme cases: UTICA CHANSON MIGRATE INVENTION ANNUAL KNOBBY SORRY IN FACTUAL BEEN CLOVER. WEED LICHEN ICE CHEST FOREARM OTHER DISGUISE DELIMIT. CANCEL MYOCARDIA ITS INFORMAL FUNCTION. YEARN AFFIX, LOST UKASE, UGANDA JAIL, CONSERVE TENURES YACHT APPEAL. EYELET SHEILA INDIA HOUSE SHEILAS TURKEY. BOB STILT SEA, CANTANKEROUS BOAT, HUMUS GOAD IMMORTAL DECOS GUARD. MARY SINBAD SHEER TOURNEY AUGUSTA WIND NOCTURNE TOOTHBRUSH. WHINE YOSEMITE NAMES SOY CAN PHILATELIST. ALBEIT DETRACT UNIVERSE EDIFY MUSTAFA TICKET TICKET IN. ==> english/telegrams.s <== These are from an old "Games" magazine: UTICA CHANSON MIGRATE INVENTION ANNUAL KNOBBY SORRY IN FACTUAL BEEN CLOVER. You take a chance on my great invention and you'll not be sorry. In fact, you'll be in clover. WEED LICHEN ICE CHEST FOREARM OTHER DISGUISE DELIMIT. We'd like a nice chest for our mother; the sky's the limit. CANCEL MYOCARDIA ITS INFORMAL FUNCTION. Can't sell my ol' car dear; it's in for malfunction. YEARN AFFIX, LOST UKASE, UGANDA JAIL, CONSERVE TENURES YACHT APPEAL. You're in a fix. Lost your case. You goin' to jail. Can serve ten years. You ought to appeal. EYELET SHEILA INDIA HOUSE SHEILAS TURKEY. I let Sheila in their house; she lost her key. BOB STILT SEA, CANTANKEROUS BOAT, HUMUS GOAD IMMORTAL DECOS GUARD. Bob's still at sea; can't anchor his boat. You must go to him or tell the coast guard. MARY SINBAD SHEER TOURNEY AUGUSTA WIND NOCTURNE TOOTHBRUSH. Mary's in bed; she hurt her knee. A gust of wind knocked her into the brush. WHINE YOSEMITE NAMES SOY CAN PHILATELIST. Why don't you (why'n'ya) send me the names, so I can fill out a list. ALBEIT DETRACT UNIVERSE EDIFY MUSTAFA TICKET TICKET IN. I'll be at the track and I have a receipt if I must have a ticket to get in. ==> english/trivial.p <== Consider the free non-abelian group on the twenty-six letters of the alphabet with all relations of the form = , where and are homophones (i.e. they sound alike but are spelled differently). Show that every letter is trivial. For example, be = bee, so e is trivial. ==> english/trivial.s <== be = bee ==> e is trivial; ail = ale ==> i is trivial; week = weak ==> a is trivial; lie = lye ==> y is trivial; to = too ==> o is trivial; two = to ==> w is trivial; hour = our ==> h is trivial; faggot = fagot ==> g is trivial; bowl = boll ==> l is trivial; gell = jel ==> j is trivial; you = ewe ==> u is trivial; damn = dam ==> n is trivial; limb = limn ==> b is trivial; bass = base ==> s is trivial; cede = seed ==> c is trivial; knead = need ==> k is trivial; add = ad ==> d is trivial; awful = offal ==> f is trivial; gram = gramme ==> m is trivial; grip = grippe ==> p is trivial; cue = queue ==> q is trivial; carrel = carol ==> r is trivial; butt = but ==> t is trivial; lox = locks ==> x is trivial; tsar = czar ==> z is trivial; vlei = flay ==> v is trivial. For a related problem, see _The Jimmy's Book_ (_The American Mathematical Monthly_, Vol. 93, Num. 8 (Oct. 1986), p. 637): Consider the free group on twenty-six letters A, ..., Z. Mod out by the relation that defines two words to be equivalent if (a) one is a permutation of the other and (b) each appears as a legitimate English word in the dictionary. Identify the center of this group. -- clong@remus.rutgers.edu (Chris Long) ==> english/weird.p <== Make a sentence containing only words that violate the "i before e" rule. ==> english/weird.s <== From the May, 1990 _Word Ways_: That is IE - Or, Is That EI? by Paul Leopold Stockholm, Sweden "Seeing wherein neither weirdly-veiled sovereign deigned agreeing, their feisty heirs, leisurely eyeing eight heinous deity-freightened reindeer sleighs, counterfeited spontaneity, freeing rein (reveille, neighing]); forfeited obeisance, fleeing neighborhood. Kaleidoscopically-veined foreign heights being seized, either reigned, sleight surfeited, therein; reinvented skein-dyeing; reiteratedly inveighed, feigning weighty seismological reinforcement." The above passage appears in a book on the ecological conservation measures of the enlightened plutocracies of antiquity, Ancient Financier Aristocracies' Conscientious Scientific Species Policies, by Creighton Leigh Peirce and Keith Leiceister Reid. . . . Any beings decreeing such ogreish, albeit nonpareil, homogeneity must be nucleic protein-deficient from sauteing pharmacopoeial caffeine and codeine] From an 'fgrep cie /usr/dict/words', with similiar words removed. ancient coefficient concierge conscience conscientious deficient efficient financier glacier hacienda Muncie omniscient proficient science Societe(?) society species sufficient A search through Webster's on-line dictionary produced the following exceptions: Word: *cie* Possible matches are: 1. -facient 2. abortifacient 3. ancien regime 4. ancient 5. ancientry 6. boccie 7. cenospecies 8. christian science 9. coefficient 10. concierge 11. conscience 12. conscience money 13. conscientious 14. conscientious objector15. deficiency 16. deficiency disease 17. deficient 18. domestic science 19. earth science 20. ecospecies 21. efficiency 22. efficiency engineer 23. efficient 24. facies 25. fancier 26. financier 27. genospecies 28. geoscience 29. glacier 30. glacier theory 31. habeas corpus ad subjiciendum32. hacienda 33. inconscient 34. inefficiency 35. inefficient 36. insufficience 37. insufficiency 38. insufficient 39. international scientific vocabulary 40. library science 41. liquefacient 42. mental deficiency 43. mutafacient 44. natural science 45. nescience 46. omniscience 47. omniscient 48. physical science 49. political science 50. precieux 51. prescience 52. prescientific 53. prima facie 54. proficiency 55. proficient 56. pseudoscience 57. rubefacient 58. science 59. science fiction 60. scient 61. sciential 62. scientific 63. scientific method 64. scientism 65. scientist 66. scientistic 67. secret society 68. self-sufficiency 69. self-sufficient 70. social science 71. social scientist 72. societal 73. society 74. society verse 75. somnifacient 76. specie 77. species 78. stupefacient 79. sub specie aeternitatis80. subspecies 81. sufficiency 82. sufficient 83. sufficient condition 84. superficies 85. type species 86. unscientific 87. valenciennes 88. vers de societe ==> english/word.boundaries.p <== List some sentences that can be radically altered by changing word boundaries and punctuation. ==> english/word.boundaries.s <== Issues topping our mail: manslaughter. Is Sue stopping our mailman's laughter? The real ways I saw it. There always is a wit. You read evil tomes, Tim, at Ed's issue. "You're a devil, Tom]" estimated sis Sue. ==> english/word.torture.p <== What is the longest word all of whose contiguous subsequences are words? ==> english/word.torture.s <== This problem was discussed in _Word Ways_ in 1974-5. In August 1974, Ralph Beaman, in an article titled "Word Torture", offered the word SHADES, from which one obtains HADES, SHADE; ADES, HADE, SHAD; DES, ADE, HAD, SHA; ES, DE, AD, HA, SH; S, E, D, A, H. All of these are words given in Webster's Third. Since that time, a serious search has been launched for a seven-letter word. The near misses so far are: Date Person Word Missing Aug 74 Ralph Beaman GAMINES INES, GAMI, NES, INE Nov 74 Dmitri Borgmann ABASHED INE, NES, ABASHE, BASHE, ASHE (all in OED) May 75 David Robinson GUNITES GU, GUNIT (using Webster's Second) May 75 David Robinson ETAMINE ETAMI, TAMI (using Webster's Second) May 75 Ralph Beaman MORALES RAL (using Webster's Second) Aug 75 Tom Pulliam SHEAVES EAV (using Webster's Second) Webster's Second has been used for most of the attempts since it contains so many more words than Webster's Third. The seven-letter plateau remains to be achieved. ==> games/chess/knight.control.p <== How many knights does it take to attack or control the board? ==> games/chess/knight.control.s <== Fourteen knights are required to attack every square: 1 2 3 4 5 6 7 8 ___ ___ ___ ___ ___ ___ ___ ___ h ! ! ! ! ! ! ! ! ! --- --- --- --- --- --- --- --- g ! ! ! N ! N ! N ! N ! ! ! --- --- --- --- --- --- --- --- f ! ! ! ! ! ! ! ! ! --- --- --- --- --- --- --- --- e ! ! N ! N ! ! ! N ! N ! ! --- --- --- --- --- --- --- --- d ! ! ! ! ! ! ! ! ! --- --- --- --- --- --- --- --- c ! ! N ! N ! N ! N ! N ! N ! ! --- --- --- --- --- --- --- --- b ! ! ! ! ! ! ! ! ! --- --- --- --- --- --- --- --- a ! ! ! ! ! ! ! ! ! --- --- --- --- --- --- --- --- Three knights are needed to attack h1, g2, and a8; two more for b1, a2, and b3, and another two for h7, g8, and f7. The only alternative pattern is: 1 2 3 4 5 6 7 8 ___ ___ ___ ___ ___ ___ ___ ___ h ! ! ! ! ! ! ! ! ! --- --- --- --- --- --- --- --- g ! ! ! N ! ! ! N ! ! ! --- --- --- --- --- --- --- --- f ! ! ! N ! N ! N ! N ! ! ! --- --- --- --- --- --- --- --- e ! ! ! ! ! ! ! ! ! --- --- --- --- --- --- --- --- d ! ! ! N ! N ! N ! N ! ! ! --- --- --- --- --- --- --- --- c ! ! N ! N ! ! ! N ! N ! ! --- --- --- --- --- --- --- --- b ! ! ! ! ! ! ! ! ! --- --- --- --- --- --- --- --- a ! ! ! ! ! ! ! ! ! --- --- --- --- --- --- --- --- Twelve knights are needed to control (attack or occupy) the board: 1 2 3 4 5 6 7 8 ___ ___ ___ ___ ___ ___ ___ ___ a ! ! ! ! ! ! ! ! ! --- --- --- --- --- --- --- --- b ! ! ! N ! ! ! ! ! ! --- --- --- --- --- --- --- --- c ! ! ! N ! N ! ! N ! N ! ! --- --- --- --- --- --- --- --- d ! ! ! ! ! ! N ! ! ! --- --- --- --- --- --- --- --- e ! ! ! N ! ! ! ! ! ! --- --- --- --- --- --- --- --- f ! ! N ! N ! ! N ! N ! ! ! --- --- --- --- --- --- --- --- g ! ! ! ! ! ! N ! ! ! --- --- --- --- --- --- --- --- h ! ! ! ! ! ! ! ! ! --- --- --- --- --- --- --- --- Each knight can control at most one of the twelve squares a1, b1, b2, h1, g1, g2, a8, b8, b7, h8, g8, g7. This position is unique up to reflection. References Martin Gardner, _Mathematical Magic Show_. ==> games/chess/mutual.check.p <== What position is a stalemate for both sides and is reachable in a legal game (including the requirement to prevent check)? ==> games/chess/mutual.check.s <== Put the following configuration in one corner: ! ! x ! P x !B P P !K R B +--------- ("x" is a Black pawn), and the same with colors reversed in the h8 corner. --Noam D. Elkies (elkies@zariski.harvard.edu) Dept. of Mathematics, Harvard University ==> games/chess/mutual.stalemate.p <== What's the minimal number of pieces in a legal mutual stalemate? ==> games/chess/mutual.stalemate.s <== 6. W Kh8 e6 f7 h7 B Kf8 e7 W Kb1 B Ka3 b2 b3 b4 a4 W Kf1 B Kh1 Bg1 f2 f3 h2 ==> games/chess/queens.p <== How many ways can eight queens be placed so that they control the board? ==> games/chess/queens.s <== 92. The following program uses a backtracking algorithm to count positions: #include static int count = 0; void try(int row, int left, int right) { int poss, place; if (row == 0xFF) ++count; else { poss = ~(row!left!right) & 0xFF; while (poss ]= 0) { place = poss & -poss; try(row!place, (left!place)<<1, (right!place)>>1); poss &= ~place; } } } void main() { try(0,0,0); printf("There are %d solutions.\n", count); } -- Tony Lezard IS tony@mantis.co.uk OR tony%mantis.co.uk@uknet.ac.uk OR EVEN arl10@phx.cam.ac.uk if all else fails. ==> games/chess/size.of.game.tree.p <== How many different positions are there in the game tree of chess? ==> games/chess/size.of.game.tree.s <== Consider the following assignment of bit strings to square states: Square State Bit String ------ ----- --- ------ Empty 0 White Pawn 100 Black Pawn 101 White Rook 11111 Black Rook 11110 White Knight 11101 Black Knight 11100 White Bishop 11011 Black Bishop 11010 White Queen 110011 Black Queen 110010 White King 110001 Black King 110000 Record a position by listing the bit string for each of the 64 squares. For a position with all the pieces still on the board, this will take 164 bits. As pieces are captured, the number of bits needed goes down. As pawns promote, the number of bits go up. For positions where a King and Rook are in position to castle if castling is legal, we will need a bit to indicate if in fact castling is legal. Same for positions where an en-passant capture may be possible. I'm going to ignore these on the grounds that a more clever encoding of a position than the one that I am proposing could probably save as many bits as I need for these considerations, and thus conjecture that 164 bits is enough to encode a chess position. This gives an upper bound of 2^164 positions, or 2.3x10^49 positions. Jurg Nievergelt, of ETH Zurich, quoted the number 2^70 (or about 10^21) in e-mail, and referred to his paper "Information content of chess positions", ACM SIGART Newsletter 62, 13-14, April 1977, to be reprinted in "Machine Intelligence" (ed Michie), to appear 1990. Note that this latest estimate, 10^21, is not too intractable: 10^7 computers running at 10^7 positions per second could scan those in 10^7 seconds, which is less than 6 months. In fact, suppose there is a winning strategy in chess for white. Suppose further that the strategy starts from a strong book opening, proceeds through middle game with only moves that DT would pick using the singular extension technique, and finally ends in an endgame that DT can analyze completely. The book opening might take you ten moves into the game and DT has demonstarted its ability to analyze mates-in-20, so how many nodes would DT really have to visit? I suggest that by using external storage such a optical WORM memory, you could easily build up a transposition table for such a midgame. If DT did not find a mate, you could progressively expand the width of the search window and add to the table until it did. Of course there would be no guarantee of success, but the table built would be useful regardless. Also, you could change the book opening and add to the table. This project could continue indefinitely until finally it must solve the game (possibly using denser and denser storage media as technology advances). What do you think? ------- I think you are a little bit too optimistic about the feasibility. Solving mate-in-19 when the moves are forcing is one thing, but solving mate-in-19 when the moves are not forcing is another. Of course, human beings are no better at the latter task. But to solve the game in the way you described would seem to require the ability to handle the latter task. Anyway, we cannot really think about doing the sort of thing you described; DT is just a poor man's chess machine project (relatively speaking). --Hsu i dont think that you understand the numbers involved. the size of the tree is still VERY large compared to all the advances that you cite. (speed of DT, size of worms, endgame projects, etc) even starting a project will probably be a waste of time since the next advance will overtake it rather than augment it. (if you start on a journey to the stars today, you will be met there by humans) ken ==> games/cigarettes.p <== The game of cigarettes is played as follows: Two players take turns placing a cigarette on a circular table. The cigarettes can be placed upright (on end) or lying flat, but not so that it touches any other cigarette on the table. This continues until one person looses by not having a valid position on the table to place a cigarette. Is there a way for either of the players to guarantee a win? ==> games/cigarettes.s <== The first person wins by placing a cigarette at the center of the table, and then placing each of his cigarettes in a position symmetric (with respect to the center) to the place the second player just moved. If the second player could move, then symmetrically, so can the first player. ==> games/connect.four.p <== Is there a winning strategy for Connect Four? ==> games/connect.four.s <== An AI program has solved Connect Four for the standard 7 x 6 board. The conclusion: White wins, was confirmed by the brute force check made by James D. Allen, which has been published in rec.games.programmer. The program called VICTOR consists of a pure knowledge-based evaluation function which can give three values to a position: 1 won by white, 0 still unclear. -1 at least a draw for Black, This evaluation function is based on 9 strategic rules concerning the game, which all nine have been (mathematically) proven to be correct. This means that a claim made about the game-theoretical value of a position by VICTOR, is correct, although no search tree is built. If the result 1 or -1 is given, the program outputs a set of rules applied, indicating the way the result can be achieved. This way one evaluation can be used to play the game to the end without any extra calculation (unless the position was still unclear, of course). Using the evaluation function alone, it has been shown that Black can at least draw the game on any 6 x (2n) board. VICTOR found an easy strategy for these boardsizes, which can be taught to anyone within 5 minutes. Nevertheless, this strategy had not been encountered before by any humans, as far as I know. For 7 x (2n) boards a similar strategy was found, in case White does not start the game in the middle column. In these cases Black can therefore at least draw the game. Furthermore, VICTOR needed only to check a few dozen positions to show that Black can at least draw the game on the 7 x 4 board. Evaluation of a position on a 7 x 4 or 7 x 6 board costs between 0.01 and 10 CPU seconds on a Sun4. For the 7 x 6 board too many positions were unclear. For that reason a combination of Conspiracy-Number Search and Depth First Search was used to determine the game-theoretical value. This took several hundreds of hours on a Sun4. The main reason for the large amount of search needed, was the fact that in many variations, the win for White was very difficult to achieve. This caused many positions to be unclear for the evaluation function. Using the results of the search, a database will be constructed of roughly 500.000 positions with their game-theoretical value. Using this datebase, VICTOR can play against humans or other programs, winning all the time (playing White). The average move takes less than a second of calculation (search in the database or evaluation of the position by the evaluation function). Some variations are given below (columns and rows are numbered as is customary in chess): 1. d1, .. The only winning move. After 1. .., a1 wins 2. e1. Other second moves for White has not been checked yet. After 1. .., b1 wins 2. f1. Other second moves for White has not been checked yet. After 1. .., c1 wins 2. f1. Only 2 g1 has not been checked yet. All other second moves for White give Black at least a draw. After 1. .., d2 wins 2. d3. All other second moves for White give black at least a draw. A nice example of the difficulty White has to win: 1. d1, d2 2. d3, d4 3. d5, b1 4. b2] The first three moves for White are forced, while alternatives at the fourth moves of White are not checked yet. A variation which took much time to check and eventually turned out to be at least a draw for Black, was: 1. d1, c1 2. c2?, .. f1 wins, while c2 does not. 2. .., c3 Only move which gives Black the draw. 3. c4, .. White's best chance. 3. .., g1]] Only 3 .., d2 has not been checked completely, while all other third moves for Black have been shown to lose. The project has been described in my 'doctoraalscriptie' (Master thesis) which has been supervised by Prof.Dr H.J. van den Herik of the Rijksuniversiteit Limburg (The Netherlands). I will give more details if requested. Victor Allis. Vrije Universiteit van Amsterdam. The Netherlands. victor@cs.vu.nl ==> games/craps.p <== What are the odds in craps? ==> games/craps.s <== The game of craps: There is a person who rolls the two dice, and then there is the house. 1) On the first roll, if a 7 or 11 comes up, the roller wins. If a 2, 3, or 12 comes up the house wins. Anything else is a POINT, and more rolling is necessary, as per rule 2. 2) If a POINT appears on the first roll, keep rolling the dice. At each roll, if the POINT appears again, the roller wins. At each roll, if a 7 comes up, the house wins. Keep rolling until the POINT or a 7 comes up. Then there are the players, and they are allowed to place their bets with either the roller or with the house. ----- My computations: On the first roll, P.roller.trial(1) = 2/9, and P.house.trial(1) = 1/9. Let P(x) stand for the probability of a 4,5,6,8,9,10 appearing. Then on the second and onwards rolls, the probability is: Roller: --- (i - 2) P.roller.trial(i) = \ P(x) * ((5/6 - P(x)) * P(x) (i > 1) / --- x = 4,5,6,8,9,10 House: --- (i - 2) P.house.trial(i) = \ P(x) * ((5/6 - P(x)) * 1/6 (i > 1) / --- x = 4,5,6,8,9,10 Reasoning (roller): For the roller to win on the ith trial, a POINT should have appeared on the first trial (the first P(x) term), and the same POINT should appear on the ith trial (the last P(x) term). All the in between trials should come up with a number other than 7 or the POINT (hence the (5/6 - P(x)) term). Similar reasoning holds for the house. The numbers are: P.roller.trial(i) (i > 1) = (i-1) (i-1) (i-1) 1/72 * (27/36) + 2/81 * (26/36) + 25/648 * (25/36) P.house.trial(i) (i > 1) = (i-1) (i-1) (i-1) 2/72 * (27/36) + 3/81 * (26/36) + 30/648 * (25/36) ------------------------------------------------- The total probability comes to: P.roller = 2/9 + (1/18 + 4/45 + 25/198) = 0.4929292929292929.. P.house = 1/9 + (1/9 + 2/15 + 15/99) = 0.5070707070707070.. which is not even. =========================================================================== == Avinash Chopde (with standard disclaimer) abc@unhcs.unh.edu, abc@unh.unh.edu {.....}]uunet]unh]abc ==> games/crosswords/cryptic/clues.p <== What are some clues (indicators) used in cryptics? ==> games/crosswords/cryptic/clues.s <== The following list is derived from indicators used in a variety of crosswords: the letters in the left column are the letters being indicated; the right hand column is how these letters might be indicated in a clue. Caveat emptor: many of the entries in this list would be considered unsound in some puzzles (some of these unsound indicators are marked with a +). Entries marked * are used mostly in advanced cryptics. I would welcome corrections and additions to the list. ---------------------------------------------------------------------- a Austria a I a academician a accepted a ace a acre a active a adult a advanced a afternoon a aleph a alpha a amateur a ampere a an/ane a angstrom a answer a ante a are (metric) a articles - English a associate a atomic a ay a aye a before a blood group a bomb a effect a examination a fifty a film a first character a first class a first letter a five hundred a five thousand a good a high class a it a key + a level a midday a note + a one * a paper a road a strings (violin) a un a unit a violin string a vitamin a year aa motoring organisation ab able seaman ab hand ab rating ab sailor/salt/seaman ab tar abbe priest (Fr.) abe Lincoln abel first victim abel murder victim abel second child abel third man able can able expert abo native ac account ac accountant ac aircraftsman ac alternating current ac before Christ ac bill ac current aca accountant acas peacemakers acc account acc bill ace card ace champion ace expert ace one ace pilot ace service ace winner act decree act performance actor tree ad Christian era ad advertisement ad after date ad before the day ad contemporary ad in the modern age ad in the year of our Lord ad modern times ad notice ad now ad nowadays ad our time/era ad period ad present day ad promotion ad puff ad this era ad today adam first character adam first person adam number one add sum add tot aden port admin management admin running ado business ado difficulty ado fuss ado row ado trouble ae aged ae poet aet aged ag silver aga Muslim leader age (long) time age mature age period agent spy agm annual meeting agm meeting agm yearly meeting ai capital ai first class ai good ai high class ai main road ai sloth ail trouble ain own (Scot.) air appearance air display air song aire river ait island al Alabama al Alan al Albania al Albert al Capone al aluminium al gangster al one pound ala Alabama ala after the style of ala in the style of ala to the (Fr.) ala wings alas Alaska alb one pound ale beer aleph Hebrew letter all completely all everybody all everything alp mountain alp peak alph river alpha Greek letter alpha beginning alpha first character alpha first letter am America am American am I am am admitting am boasting am half day am hymns am in the morning am morning am self-confessed amen final word amen last word amer American ammo missiles amos bookmaker amp one member an I an articles - English an before an if (old word) an one * an un ana tales ane I ane one ane one ankh life symbol anne princess anon now ans answer ans brief reply ans collection ans short answer ant if it (old word) ant six-footer ant social worker ant soldier ant worker ape copy ape primate aq water ar arrive/arrival ar year of reign ara academician ara artist ara painter arab horse arc curve argo old ship aria song arm gun arm limb arm member arr arrive/arrival art contrivance art craft art cunning art painting art skill as Anglo-Saxon as ayes as ays as like as one's as specifically as when ash remains ash tree asia continent aside one fifteen asis existing state asp snake ass donkey asti wine ate goddess ate mischief athena goddess ation at one on atoll bikini au gold au to the (Fr.) aus Australia aux to the (Fr.) av bible av lived so long ave average ave greeting ave hail ave road ave way aver average avon county ay I ay agreement ay always ay ever ay yes aye I aye I say aye agreement aye always aye ever aye yes az Azed az scope, plenty of b Bach b Beethoven b Belgium b Brahms b Britain b British b a follower b bachelor b baron b bedbug b bee b bel b beta b beth b bishop b black b blood group b bloody b book * b born b boron b bowled b boy b breadth b inferior b key + b magnetic flux b note + b paper b second b second class b second letter b three hundred b three thousand b vitamin ba Bachelor of Arts ba airline ba bachelor ba barium ba degree ba graduate ba scholar bac airline bacon philosopher ban curse ban outlaw ban prohibition bar Inn bar lawyers bar prevent bar save barb horse bat fly-by-night bb bees bb books bb very black bc ancient times bc before Christ bc period bd beady bd bound bd cleric bd theologian be exist be live bea airline bear speculator bed in bed bee buzzer bee group of workers bee six-footer bee social worker bee worker bef Gort's men bess queen beta Greek letter beth Hebrew letter bi double bi two (double) bird prison bis two (twice) bit chewed bit piece biz business bk book bl British company bl lawman bl lawyer blue Conservative bm British Museum bm doctor bo American man boa snake board directors bob old shilling bp bishop br Britain br British br British Rail br bank rate br branch br bridge br brig br brother br brown * br lines/landline br railway(s) br trains br transport bra female support(er) bra support bra undergarment brass money brat child bren gun brer rabbit bridal old wedding bro brother bs bees bst summer time bull American policeman bull gold bus transport c Celsius/centigrade c Charles c Conservative c Cuba c about (approx.) c approx(imately) c around c cape c caput c carbon c caught c cedi c cent/centime c centi- c century c chapter c circa c club c cold c complex number c copyright c coulomb c electrical capacity c hundred c hundred thousand c key + c lot c many c note + c roughly c sea c see c speed of light c spring c tap c vitamin ca about (approx.) ca accountant ca approx(imately) ca calcium ca roughly cab transport cade conspirator cain first murderer cain killer cain murderer cal California cam river can able to + can is able to can prison can vessel cantuar archbishop cap chapter cap international car carat car transport carnation motor race cart transport cat jazz fan cato censor cattle neat cave warning cb Seabee (Amer.) cc county council cc seas cc small measure cc small quantity cc two hundred cd diplomat cd seedy ce Church of England ce church ce engineers ce this (Fr.) cent money cet this (Fr.) ch China ch Companion of Honour ch Switzerland ch award ch central heating ch champion ch chapter ch chief ch child ch church ch companion ch honour ch order cha tea chai gypsy woman chair president chal gypsy char daily che guerrilla che revolutionary cher dear (Fr.) chere dear (Fr.) chi Greek letter ci Channel Islands ci hundred and one cia secret service cia spies cid captain cid chief cid detectives cid police cid spanish hero cinc commander cl chlorine cl class cl clause cl gas - chlorine cl hundred and fifty co Colombia co business co care of co cobalt co commander co commanding officer co company co county co firm co gas - carbon monoxide co house co objector co officer cod fish cod swimmer col neck col pass cole old king colon stop com commander comb hairdresser composer scorer con Conservative con against con party con politician con study con swindle con trick cooler prison core decentralise corn naval commander cot bed cot house cow lower cow neat cr credit cr crown cr king cs Civil Service cs Czechoslovakia cs hundreds cs seas ct Connecticut ct carat ct caught ct cent/centime ct court ct small weight ct weight cu copper cu see you cue queue cure priest (Fr.) cutie pretty girl cv autobiography cy see why d (old) penny d Dee d Democrat d Deutsch d Germany d Schubert's works d copper d damn d date d daughter d day d dead/died d deci d degree d delete d delta d deserted d deuterium d diameter d diamond d differential operator d electrical flux d five hundred d four d four thousand d hundreds d key + d lot d many + d mark d note + d notice d number d strings (violin) d violin string d vitamin da American lawyer da District Attorney da agreement - foreign (Russ.) da dagger * da lawman da lawyer da yes (Russ.) dab expert dad father dad old man dail Irish house dam barrier dam mother dam restrain dame lady dan tribe das articles - German das the (Ger.) dc Washington dc Washington dc current dd cleric dd days dd divine dd doctor dd doctor of divinity dd theologian de from (Fr.) de of (Fr.) dean good man dec Christmas period dec last month decanter Tantalus' prisoner dee river deed indeed deed legal document deep in the main deep main deep sea deg degree del of the (Ital.) dela from the (Fr.) dela of the (Fr.) demi half den retreat den study der articles - German der the (Ger.) derby horse race des of the (Fr.) det detective di double di five hundred and one di princess di two (double) die articles - German die the (Ger.) dime 12.5 cents dior designer dis Hell dis Pluto dis underworld disc circle disc record disc ring dish pretty girl dit named dit reported dit said (Fr.) dit say (Fr.) diy amateur's department dk Denmark dm mark do (the) same do act do cheat do cook do ditto do note do party do work dodo double act doh note don fellow don nobleman don put on don university teacher down county dr dead reckoning dr doctor dr dram dr drawer dr healer drake bowler dt alcoholic state dt psychotic state du from the (Fr.) du of the (Fr.) dutch wife e Asian e Edward e Elizabeth e England/English e Spain e boat e bridge players e direction e east/eastern e eight e eight thousand e energy + e epsilon e eta e five e five thousand e key + e layer e logarithm base e low grade e note + e orient e oriental e point e quarter e strings (violin) e two hundred and fifty e two hundred and fifty thousand e universal set e violin string e vitamin ea East Africa ea each ea river * ea running water ea water ear listener ear organ ear spike * earp lawman eat Tanzania ebor archbishop ec London district ec city eccles Cakesville ed Edward ed editor ed journalist eden garden eden old Prime Minister eden paradise edison inventor edit censor ee ease eel fish eel swimmer eer always eer ever eer invariably eg for example eg for instance egg bomb egg cocktail egg encourage eight rowing boat ein number one (Ger.) el American railway el American railway el articles - Spanish el measure el printer's measure el small measure el the (Span.) eld old age eli priest eli prophet elia writer ell four feet ell length ell measure ely city ely city ely see em measure em printer's measure em small measure em small square em them en measure en printer's measure en small measure eng England/English ent otorhinolaryngology entry record eon age eon time ep record er Cockney girl er QE er difficulty er ever er hesitation er king er monarch er queen er royal badge era generation erasmus old scholar ere always ere before ergo so eric gradually erie lake err blunder err sin err wander erse Gaelic es French art es ease esp sixth sense esp telepathy est is (Fr.) et Egypt et alien et and (Fr.) et exotic et extraterrestrial et film eta Greek letter eta estimated time of arrival eta illegal army eta terrorists eton college eton educational establishment eton school etty artist eur continent eve first lady eve first mate eve lady eve woman ew bridge partners ew partnership ex former ex from ex late ex one time exe river eye I eye I say eye seer eyot island ezra pound f Fahrenheit f France f Friday f clef f farad f farthing f fathom f fellow f female f feminine f filly f fine f fluorine f folio f following f foot f force f forte f forty f frequency f gas - fluorine f hole f key + f loud f noisy f note + f strong f vitamin f woman fa football fa note fah note fal river fare China area (Far East) fast firm fe iron fed American detective ff folios ff followings ff fortissimo ff very loud ff very loud ff very strong ff very strong fig small illustration fir tree firm business firm company fist duke fl flourished fla Florida flower bloomer flu illness fo Foreign Office fo folio fob free on board foc free of charge fol folio fool dessert foot infantry for free on rail force police fore warning four rowing boat fr French fr father fr fragment fr franc fr frequently ft feet ft foot ft measure fz forced g Gauss g George g Germany g agent g clef g four hundred g gamma g gamut g gee g girl g gram(me) g grand (Amer.) g gravity g guinea g gulf g key + g man g midnight g note + g strings (violin) g suit g thousand g thousand g violin string g weight g-man American detective ga Georgia gab gift gad tribe gael Gaelic gal girl gar fish gat gun gate old goat gb Great Britain gb our islands gee horse gee little horse geiger counter gel jelly gen Genesis gen general gen information gen low down gent fellow george pilot gg Gee-gee gg horse gg little horse gi American soldier gi doughboy gi fighter gi government issue gi private gi serving man gi soldier gladiator old fighter glc capital authority gm counter go bargain go energy go in good condition go ready go success go traffic signal go work gotham New York gp doctor gr Greece gr Greek gr King George gr grain gr grammar gr gramme gr grouse gr king gr small weight gr weight grant general grass informer grist miller's corn grs gunmen gs general service gs general staff gt fast car gt sports car gu guinea gu old fiddle gue old fiddle h Dirac's constant h Hungary h Planck's constant h bomb h gas - hydrogen h hand h hard h heart h height h henry h horse h hospital h hot h hour h house h husband h hydrant h hydrogen h tap h two hundred h two hundred thousand h vitamin ha half ditch ha laugh ha this year haha ditch haha laugh hair net lock keeper ham (poor) actor han Chinese dynasty hand worker has bears haw hedge he (high) explosive he His Excellency he ambassador he excellency he gas - helium he governor he helium he legate he male he our man he the man head point hearth hard ground hebe goddess hehe laugh hen female hen layer her female her the woman's her woman hf half hg Dad's army hg mercury hh very hard hi greeting hi hello hic here in Rome hic this (Lat.) him male hm Her Majesty hm His Majesty hm king hm queen ho house hobo tramp hock prison hol short break hp hire purchase hp never-never hr hour hs here is ht high tension hun German hun barbarian i Italy i a i an/ane i ay i aye i che i electrical current i eye i first person i imaginary number i iodine i iota i island i line i lunchtime i number one i one i one i single i square root of -1 i straight line i un i unit i upright i yours truly ia Iowa iam I am iam admitting iam afternoon iam boasting iam early morning iam self-confessed ian Scot/Scotsman ians one answer ib in the same place ib same place ibid in the same place ic I see ic hundred ic in charge ic ninety nine ice diamond(s) ice hard water iceni old people id I had id I would id fish id genius id identification id instinct id same id that (Lat.) ida princess ide fish idem said ie that is (that's) if condition if provided/providing ign gin cocktail/sling ii eleven ii eyes ii two il Israel il articles - Italian il one pound il the (Ital.) ilb one pound ill I shall/will ill Illinois ill badly ill unwell im I am im admitting im boasting im self-confessed imp little devil imp mischevious child imp one member impi soldiers in at home in batting in elected in fashionable in favoured in in fashion in not out in playing in trendy in wearing ina princess inch island ind India ine oriental ing gin cocktail/sling inn local insect six-footer intens decimally io cry of triumph io joyful cry io maiden io ten io triumphant cry ioc dime (Amer.) iom Isle of Man iom island iom man ion number one returning iota Greek letter ious credit notes ious promises to pay ip trivial sum ir Iran ira illegal army ira terrorists ire anger ire rage irl Ireland is Iceland is ayes is ays is eyes is island is one's isis goddess isis river isle man ism doctrine ism theory iss exists ist first it Italian it sex appeal it the thing iv four iv ivy iv tea-time ive I have ix nine j Jack j Japan j heat j jay j joule j judge j justice j knave j one j pen j square root of -1 ja Jamaica ja agreement - foreign (Ger.) ja yes (Ger.) jack sailor/salt/seaman je I, being French je In Paris, I jo little woman jo sweetheart job bookmaker jock Scot/Scotsman joe American soldier jolly Marine jug prison k Kay k Khmer Republic k Kirkpatrick k Koechel k Mozart's works k Scarlatti's works k constant k kappa k kelvin k kilo k king k knight k monarch k potassium k thousand k twenty k twenty thousand k two hundred and fifty k vitamin ka double * ka genius * ka individuality key opener kg cagey kine cattle kine neat kine ox kish graphite kl kale km kilometre kn cayenne knee bender ko decisive blow ko kick off ko knock out ko stunner kr krypton kt knight kv cave (beware) kv cavy ky Kentucky l Labour l Liberal l Luxembourg l angle l apprentice l corner l el l elevated railway l ell l fifty l fifty thousand l half century l hand l inductance l inexperienced driver l la(m)bda l lake l lambert l latin l latitude l league l learner l learning l left l length l licentiate l line l lira/lire l litre l long l lumen l luminance l many + l money l new driver l novice l number plate l one pound l overhead railway l port l pound l pupil l railway l side l sovereign l student l trainee l tyro l vitamin la Los Angeles la Louisiana la articles - French la articles - Italian la articles - Spanish la look * la note la the (Fr.) la the (Ital.) la the (Span.) lab Labour lab laboratory lab party lab politician lab science centre lac aircraftsman lac hundred thousand lah note lakh hundred thousand lam beat lam pound lambda Greek letter lamed Hebrew letter lar Libya laud archbishop lay song lb one pound lb pound le articles - French le the (Fr.) lear king lee general leg limb leg member leg on leg support lei flowers lei wreath * lely artist les articles - French les the French let allow(ed) let hindrance let permit(ted) let service let with a tenant lewis gun lh left hand li fifty one lib Liberal lib book lib party lib politician limn old paint line shipping company ling fish ling swimmer lips mouthpiece lips speakers lit drunk lit loaded lit settled ll ells ll els ll fifty pounds ll fifty-fifty lner old railway lo look lo see loch lake log maths function log record los articles - Spanish loser fabulous hare lot large amount loti first item in sale lower cow lowing neat sound lp long playing lp record ls ells ls els lso orchestra lt Lieutenant lt officer lud old king lum chimney (Scot.) lv meal ticket m Bond's boss m French man m Malta m Monday m em m lot m maiden m maiden over m male m man m many + m mare m mark m married m masculine m mass m master m measure m member m meso- m meta- m meter (metre) m midday m mile m modulus m monsieur m month m moon m motorway m mu m noon m roof m small square m spymaster m thousand m vitamin ma Master of Arts ma academic ma degree ma educated man ma graduate ma master ma mother ma old woman ma scholar mab fairy queen mab queen mac Scot/Scotsman main deep main sea mal bad French mam mother man Friday man fellow man fighter man hand man husband man island man piece (chess) man soldier man worker manifesto show-ring mass Massachusetts mass service maxim gun may can mayo tree ring mb doctor mc master of ceremonies mc medal md doctor md one thousand five hundred me first person me note me number one men people men soldiers ment intended ment meant ment on purpose ment understood mer sea (Fr.) met New York opera met police mg magnesium mi main road mi motorway mi note mig aeroplane mill economist min half minute min thirty seconds ming Chinese dynasty ming china mini car miss Mississippi mm French men mm ems mm medal mn Merchant Navy mo Missouri mo doctor mo half minute mo month mo second mo short time mo time mo way of working mog cat mon Monday mon Scot/Scotsman moo low moo neat sound mos months moses lawgiver mot car test mot test moth fly-by-night mp fairly quiet mp member mp member of parliament mp military police mp mounted police mp mountie(s) mp policeman mp politician mp representative mph rate mph speed mps chemist mr mister ms ems ms handwriting ms manuscript ms paper ms text ms writing mss manuscripts mss papers mt hill mt mountain mtb torpedo boat mu Greek character mu Greek letter mum mother mum quiet mum silence mur wall (Fr.) mutt dog my gracious me n Avogadro's number n Norway n born n bridge players n direction n en n gas - nitrogen n half an em n indefinite number n knight n midday n name n neper n neuter n new n newton n ninety n ninety thousand n nitrogen n noon n north(ern) n note n noun n nu n number n point n pole + n quarter n unfavourable aspect n unknown number n unlimited number na North America na no (Scot.) na not (Scot.) na sodium nae no (Scot.) nae not (Scot.) nag horse nag little horse nat born nation people nation race national horse race nb nota bene nb note nco non-commissioned officer nco sergeant nd North Dakota nd no date ne Durham area ne Humberside ne Tyneside ne born ne bridge opponents ne gas - neon ne neon ne north-east ne not (old word) ne quarter neat cattle neat ox ned donkey ned little horse nee born nemo submariner ness head ness loch ness point net capture net fabric ney Marshal ng no good ni Northern Ireland ni Ulster ni nickel nib writer nick prison nie close (old word) nie near (old word) nil love nil nothing nitre chemical nitre fertiliser nj New Jersey nl not clear nl not far nl not permitted nn ens no New Orleans no indefinite number no not out no number no play (Jap.) no refusal noh play (Jap.) noi leading noi no one noi number one non no (Fr.) nose informer np North Pole np pole ns bridge partners ns ens ns not specific ns partnership nt Holy Writ nt New Testament nt Northern Territories nt book(s) nt books nt good book nt part of Bible nt preservationists nt scriptures nu Greek character nu Greek letter nu name unknown nu unidentified num miners number anaesthetic nun bluetit nurse shark nus dawn nus students' (union) nut teachers nv envy nw Merseyside nw bridge opponents nw north-west nw quarter ny Gotham ny New York o Ohio o around o aught o bald patch o ball o blob o blood group o cavity o cipher o circle o circuit o circular letter o dial o disc o duck o egg o eleven o eleven thousand o empty o examination o full moon o gas - oxygen o globe o gulf o hole o hollow o hoop o loop o love o naught o nil o no o nothing o nought o oh o omicron o opening o orb o ortho- o ought o owe o oxygen o pellet o ring o round o spangle oaks horse race oates conspirator ob old boy obe award obe honour obe order obit dead/died obit final message obit last word oc commander oc officer commanding odo bishop oe old English oe old Etonian offa old king og ogee og own goal og soccer blunder oic officer in charge ok acceptable ok all right ok approval ok authorisation ok correct ok fine ok okay old man captain ole cry of delight om award om honour om order omega Greek letter omega final letter omega final word omega last letter omitting skipping on about on acting on being broadcast on leg on on the menu on operating on performing on playing one lunchtime one undivided oo duck's eggs oo ohs oo owes oo spectacles oom Dutch uncle op operation op opposite op opus op out of print op work optic seer opus work or alternative or alternatively or before * or gold or yellow oral examination oral test ord no way ord ring road orion hunter orpheus classical musician os Ordinary Seaman os big os large letters os ohs os old style os outsize os owes os sailor os very large ost east (Ger.) ot Holy Writ ot Old Testament ot book(s) ot good book ot occupational therapy ot part of Bible ot scriptures ou Open University oui agreement - foreign (Fr.) oui yes (Fr.) ouija Franco-German agreement ous nothing to America ouse river out away out not at home out unfashionable over maiden ox bull ox neat oxonian dark blue oy oh, why oz ounce oz small weight oz weight oz wizard place p Celt p Kelt p Portugal p copper p four hundred p four hundred thousand p page p park(ing) p participle p pawn p pea p pedal p pee p peg p penny p phosphorous p pi p piano p pint p poise p power p president p prince p quiet p small change p soft p softly p vitamin pa Panama pa father pa old man pan god par standard para Brazilian para airborne soldier parent source pas dance pas step pate head pawn piece (chess) pawnbroker uncle pb lead pc copper pc policeman pe Peru pe gym pe physical education pe training peg tee pen author pen enclosure pen prison pen writer per by per for each per through pet cherished pet favourite pg paying guest ph local phi Greek letter pi Greek character pi Greek letter pi confusion * pi good pi religious pi upright pier mole pit Hell pitt old Prime Minister pl Poland pla mountain retreat pla port authority plato philosopher plo illegal army plo terrorists plot garden pm Prime Minister pm afternoon pm half day pm in the afternoon po Italian flower po Post Office po airman po palladium po pole po river polo Merchant of Venice poly college pony twenty five pounds pony twenty-five pounds pop father pop old man port left port wine pp pages pp peas pp pees pp pianissimo pp very quiet/soft(ly) pr Puerto Rico pr Romans/Roman people pr electoral system pr image building pr president pr price pr prince pr public relations pr two pra academician pra artist pra painter pres president prison bird pro expert pro for pro public relations officer prof academic provos terrorists ps footnote ps peas ps pees ps postscript ps second thoughts pt part pt physical training pt pint pt platinum pt point pt post town pt stop pt training pub local pv peavy q Celt q Kelt q Quebec q Queensland q boat q cue q electrical charge q farthing q five hundred q five hundred thousand q koppa q ninety q ninety thousand q quality q queen q query q question q queue q quintal q rational numbers qp kewpie (doll) qt cutie qt quart qt quiet qu quart qu queen quad prison que what (Fr.) qui who (Fr.) r Reaumur r Regina r Republican r Rex r Romania r are r arithmetic r castle r eighty r eighty thousand r hand r king r monarch r month r queen r radius r rain r rand r reading r real numbers r recipe r resistance r rho r right r river r road r rontgen unit r rook r royal r run r side r take * r writing ra Royal Academician ra Royal Academy ra Royal Artillery ra academician ra academy ra artillery ra artist ra big guns ra gunmen ra gunner(s) ra painter ra soldiers ra sun (god) rab Butler rac motoring organisation race people rada academy raf fliers raf service rage fashion rain waterfall ram butter ram music school ram sheep ran managed ran smuggled ras head ras prince rat art nouveau rat desert fighter rat desert(er) rat scab rat strikebreaker rate speed rc Roman Catholic rc church rd little way rd road rd way re Royal Engineer(s) re about re again re concerning re engineer(s) re note re over re religious education re sapper(s) re soldiers re touching rebecca Welsh riots rec recipe rec take red anarchist red bloody red cent * red communist red leftist red revolutionary red socialist regan princess regan wicked sister reine French queen reme engineers rene French man rep agent rep salesman rep traveller res old thing resh Hebrew letter ret soak rev priest rev vicar rex cat rg argy(-bargy) rh right hand ri Rhode Island ri king emperor rib wife rid clear rid free ring circle rio port rip final message rip last word river banker river flower river runner rly lines/landline rly railway rly transport rly way rm Marine rm Royal Marine(s) rm jolly rm old Irish magistrate rm resident magistrate rma Sandhurst rms mailboat rn Navy rn Royal Navy rn fleet rn sailors rn service ro right hand roc fabulous bird rock diamond rod fast car rod pole rod sports car roi French king roi king (Fr.) rom gypsy rose flower rosinante poor horse rot corruption rot decay rot rubbish rr Right Reverend rr Rolls Royce rr ars rr bishop rr car rsm non-commissioned officer rt arty rt right ru football ru rugby ruc Irish police rue street (Fr.) run manage run smuggle rur Capek's play rv bible ry landline ry line(s) ry little way ry rail ry railway ry transport ry way s Bach's works s God's s Sabbath s Saturday s Schmieder s Sweden s as s bend s bob s bridge players s direction s dollar s es s ess s has s his s is s largesse s old Bob s old shilling s paragon s part of collar s point s pole + s quarter s saint s second s seven s seven thousand s shilling s side s siemens s sister s snow s society s son s south(ern) s space s spade s square s stokes s sulphur s sun s us sa South Africa sa South America sa essay sa it sa sex appeal sa without date sad blue sailor able seaman saint paragon salt able seaman sam uncle sas soldiers sat Saturday satin dressmaker sc namely sc self-contained sc small capitals sc specifically sc that is sc viz scot fine scot tax scr scruple sculptor blockbuster sd South Dakota sd without a day fixed sdp nationalists se Home Counties se London Area se bridge opponents se quarter se south-east sea deep sea main sec dry sec second sec short time sec time see look seed children semi half semi house sent ecstatic set group set put seth fourth man sgt non-commissioned officer sgt sergeant sh hush sh quiet sh silence she (the) woman she female she lady she novel sherpa mountaineer si South Island si agreement - foreign (Span.) si note si silicon si yes (Ital.) si yes (Span.) sib relation sic thus side team silk dressmaker sin err sin evil sin without sin wrong sine maths function sine without sir knight sis sister sly tinker sm French king sm French queen sm king (Fr.) sm non-commissioned officer smith economist sn Essen sn partnership sn tin so ergo so note so therefore so thus so well soh note sol note sol sun-god som county some approx(imately) son issue sop soprano sos appeal sp South Pole sp childless sp odds sp pole sp species sp starting price sp without children spa spring spire Oxford dreamer spy agent spy mole square unfashionable sr old railway sr senior srn nurse ss German soldier ss Sunday School ss liner ss saints ss ship ss steamship st good man st hush st little way st paragon st road st saint st silence st stone st street st stumped st thoroughfare st way st weight stag speculator sten gun stet don't change it stir prison stop traffic signal sts saints sty filthy place stye eyesore su Soviet Union sub U-boat sub stand-in sub substitute sub warship supra over sure certain sw Cornwall sw Devon sw bridge opponents sw quarter sw south-west swift screecher swiss roll jammed cylinder sx Essex t Thailand t Tuesday t bandage t bar t bone t cart t cloth t cross t crossed t half dry t hundred and sixty t hundred and sixty thousand t junction t model + t peg t perfect letter t plate t rail t shirt t short time t square t tau t te t tea t tee t tesla t the t time t ton(ne) t tritium ta Territorial Army ta army ta cheers ta reserves ta soldiers ta terriers ta territorials ta thank you ta thanks ta volunteers tab label tace silence tag label tan beat tan brown tan maths function tar able seaman tar art nouveau tar sailor/salt/seaman tata Tosti's song tata goodbye tate gallery tau cross tay river tb torpedo boat td medal te Lawrence te note tea leaves tec detective ted Edward ted Heath tee peg teen old injury tees river tell archer temp secretary ten PM's address tene old injury tent wine ter three (triple) ter thrice test educational journal test examination test match teth Hebrew letter the article the articles - English ti note tic note tic spasm tic twitching tier row time father times daily timon misanthrope tin can tin cash tin money tin vessel tiny small tion empty container tit bird tit inferior horse tit poor horse tnt big banger tnt explosive tod fox todo commotion toe extremity toe member tom big bell tom cat tome book ton fashion ton hundred ton large amount ton weight tonne weight tor hell tor hill tor mountain tor point tor prominence tory Conservative tory party tory politician tp teepee tr Turkey tr transaction tr translation tram transport tree actor tres very (Fr.) tri three (triple) tri thrice troy ancient city troy old city try attempt try essay ts teas ts tees tt abstaining tt dry tt on the wagon tt race tt teas tt tees tt teetotal tt teetotaller tt thank you tu tradesmen tuck friar twelve eec two company u Conservative u Uruguay u Utah u about turn u acceptable u bend u boat u educational establishment u ewe u film u for all to see u high class u on view to all u posh u socially acceptable u suitable for children u superior u trap u tube u turn u union/Unionist u universal u university u upper class u uppish u upsilon u uranium u yew u you uc you see uk United Kingdom uk this country uk this island ule rubber ult last month um doubt um hesitation un United Nations un international un number one (Fr.) un one un one (dialect) un peacekeepers una number one (Ital.) unco very (Scot.) une number one (Fr.) uno international organisation uno number one (Ital.) up at university up excited up in court up mounted up riding up superior uq you queue ur ancient city ur hesitation ur old city ur primitive ur you are ure river uru Uruguay us America us American us as above us ewes us no good us transatlantic us undersecretary us use us useless us yews us you and me usa America use application use custom use employ(ment) use practice use practise ussr Soviet Union ut note ute half minute uu ewes uu use uu yews ux wife v Vatican v against v agent v bomb v day v five v look v neck v neckline v notch v opposing v see v sign v vanadium v vee v velocity v verb v verse v versus v very v victory v vide v volt v volume v win va Virginia vad nurse vale farewell vale goodbye vat tax vau Hebrew letter vb verb ve victory ver rev up very light vet surgeon vg for example vi half dozen vi six via old way vid see vid tanner/sixpence vide look vide see vin French wine vip big noise vip tanner/sixpence vir man/Roman vis viscount vj victory vo left hand vol book vol volume vy various years w Wednesday w Welsh w William w bridge players w direction w point w quarter w tungsten w watt w weak w west(ern) w whole numbers w wicket w width w wife w woman ward disadvantage (drawback) washington young feller we partnership we you and I wee little wee minor wee small who doctor wi Mayfair wi West Indies wi Westminster winner fabulous tortoise wise youth leaders wist knew (old word) women monstrous regiment woof bark wt small weight wt weight x Christ x PM's address x Xmas x across x body x chi x chromosome x cross x draw x ex,Exe x film x illiterate's signature x kiss x particle x ray x sign of love x sign of the times x spot marked x ten x ten thousand x thousand x times x unknown x vitamin x vote x wrong sign x xi xc ninety xi eleven xi side xi team xl excel xv side xv team y alloy y chromosome y level y measure y moth y one hundred and fifty y one hundred and fifty thousand y track y unknown y why y yard y year y yen y young y yttrium yard detectives yd measure ye the (old word) ye you (old word) yea agreement yew tree yr year yr your ys wise ys youth leaders yt that (old word) yu jade yule you will, say yy wise z Zambia z bar z bend z cedilla z final letter z integers z izzard z last character z last letter z omega z seven z seven thousand z sound of sleep z zed z zee z zero z zeta zo cross * zr Zaire zz (sound of) snoring ---------------------------------------------------------------------- -- Ross Beresford, ! Email (trusted): rberesfo@cix.compulink.co.uk 10 Wagtail Close, ! (work): ross@siesoft.co.uk Twyford, Reading, ! (under test): ross@dickens.demon.co.uk RG10 9ED, UK ! ==> games/crosswords/cryptic/double.p <== Each clue has two solutions, one for each diagram; one of the answers to 1ac. determines which solutions are for which diagram. All solutions are in Chamber's and Webster's Third except for one solution of each of 1dn, 3dn and 4dn, which can be found in Webster's 2nd. edition. ####################################################################### #1 !2 ! ! !3 !4 !5 #1 !2 ! ! !3 !4 !5 # # ! ! ! ! ! ! # ! ! ! ! ! ! # #----+----###########----#----#----#----+----###########----#----#----# #6 ! !7 ! ! # # #6 ! !7 ! ! # # # # ! ! ! ! # # # ! ! ! ! # # # #----#----#----######----#----#----#----#----#----######----#----#----# # # # #8 ! ! ! # # # #8 ! ! ! # # # # # ! ! ! # # # # ! ! ! # #----#----#----######----#----#----#----#----#----######----#----#----# #9 ! ! ! # # # #9 ! ! ! # # # # # ! ! ! # # # # ! ! ! # # # # #----#----#----######----#----#----#----#----#----######----#----#----# # # #10 ! ! ! ! # # #10 ! ! ! ! # # # # ! ! ! ! # # # ! ! ! ! # #----#----#----###########----+----#----#----#----###########----+----# #11 ! ! ! ! ! ! #11 ! ! ! ! ! ! # # ! ! ! ! ! ! # ! ! ! ! ! ! # ####################################################################### Ac. 1. What can have distinctive looking heads spaced about more prominently right. (7) 6. Vermin that can overrun fish and t'English tor perhaps. (5) 8. Old testament reversal - Adam's conclusion, start of sin. Felines initially with everything there. (4) 9. Black initiated cut, oozed out naturally. (4) 10. For instance, 11 with spleen dropping I count? (5) 11. Provoked explosion of grenade. (7) Dn. 1. Some of club taking part in theatrical function, for the equivalent of a fraction of a pound. (6) 2. Close-in light meter in one formation originally treated as limestone. (6) 3. Xingu River hombres having symmetrical shape. (5) 4. About sex-appeal measure - what waitresses should be? (6) 5. Old penny, least damaged, was preserved. (6) 7. IRA to harm ruling Englishman; extremes could be belonging to group. (5) ==> games/crosswords/cryptic/double.s <== +-+-+-+-+-+-+-+-+-+-+-+-+-+-+ !r e d c a p s!d e x t r a l! + + +-+-+ + + + + +-+-+ + + + !o t t e r!o!a!r o a c h!s!a! + + + +-+ + + + + + +-+ + + + !u!a!h!f a l l!a!z!m!t o m s! + + + +-+ + + + + + +-+ + + + !b l e d!r!i!t!c o o n!m!i!t! + + + +-+ + + + + + +-+ + + + !l!o!i r a t e!m!o!n o b l e! + + + +-+-+ + + + + +-+-+ + + !e n r a g e d!a n g e r e d! +-+-+-+-+-+-+-+-+-+-+-+-+-+-+ Notes. Left grid: Ac. 1. R + spaced (anag). 6. T'E tor (anag). 8. F-all. 9. B-led. 10. I-rate. Dn. 1. Ro-ub-le. 2. T.A.L. in one (anag). 4. it in pole. 5. anag of D+least. 7. anag of initial letters. Right grid: Ac. 1. D-extra-L. 6. 3 mngs. 8. OT (rev) + m-s. 9. initial letters. 10. No.-b(i)le. Dn. Dra-c-ma. 2. Zoo(m) in one (anag). 3. hidden. 4. SA (rev) + mile. 5. anag of D+least. 7. anag of final letters. -------------------------------------------------------------------- How I built it: it was hard] Basically, I started with a couple of word pairs which were easy to clue (e.g. enraged/angered - same meaning and anagrams of each other) and built a grid around them, trying to ensure corresponding words had something in common, either in meaning (their, among) or structure, (EtalON, EOzooN) and making sure that there was at least one word which could be used to distinguish the two grids (dextral). The clues were built in one of two ways: either the words had a common definition, and so a subsidiary indication which could refer to either was needed; or it was necessary to define each word in such a way that it was a subsidiary definition for all or part of the corresponding word, and deal with any remaining parts as before. I think the single hardest part was finding a definition of "interferometer" which could also be interpreted as "zoo" or "ozo". Roy rt@ukc.ac.uk ==> games/crosswords/cryptic/intro.p <== What are the rules for cluing cryptic crosswords? ==> games/crosswords/cryptic/intro.s <== This is a brief set of instructions for solving cryptic crossword puzzles for those of you who are intrigued by these puzzles, but haven't known how to begin solving them. For a more complete introduction, send a self-addressed, stamped envelope to The Atlantic Puzzler, 745 Boylston Street, Boston, Mass. 02116. The characteristic common to all cryptic crossword puzzles is the format of the clues. Each clue is a miniature word puzzle consisting of a straight definition of the answer and a cryptic definition of the answer. For example, Axle is poorly splined (7) yields SPINDLE. Axle is the straight definition. The cryptic definition (poorly splined) indicates an anagram of "splined". The number in parentheses is the number of letters in the answer. Punctuation and capitalization may be ignored in interpreting the clues. There are only eight categories of clues, as follows: 1. Anagram An anagram is a word formed by mixing up the letters of another word. An anagram clue is indicated by some word that means "mixed up", for example, out, crazy, bizarre, insane, etc. One or more words may contribute to the anagram. For example: Tim goes insane from selfishness (7) for EGOTISM (anagram of "Tim goes") 2. Double Definition A double definition is simply two definitions of the word. Most two-word clues are double definitions. For example: Release without charge (4) for FREE 3. Container A container clue indicates that something is to be put in (or wrapped around) something else. A container is indicated by phrases such as eaten by, contains, in, gobbles, etc. For example: In Missouri, consumed by fear (7) for AMONGST (MO = Missouri in ANGST = fear) 4. Hidden Word A hidden word is a word embedded in another word or words. It is indicated by phrases such as spot in, hides, at the heart of, covers, etc. For example: Worn spot in paper at typo (5) for RATTY (find ratty in "paper at typo") 5. Reversal A reversal is a definition of a word with the letters reversed. It is indicated by words such as back, reversed, up (for down clues), leftward (for across clues), etc. For example: Egad] Ray entirely reversed the lot of cloth (7) for YARDAGE ("Egad] Ray" reversed) 6. Homophone A homophone definition is a definition of a word that sounds the same as the answer, but is spelled differently. A homophone is indicated by words such as in audience, I hear, mouthed, verbally, etc. For example: Regrets prank, I hear (4) for RUES (the homophone is RUSE = prank) 7. Charade In a charade, the pieces of the word are "spelled" out in order. There are no auxiliary words that indicate a charade. For example: Excite a jerk extremist (7) for FANATIC (FAN = excite, A, TIC = jerk) 8. Deletion A deletion is a clue where you are instructed to remove a part of some word to make another word. For example, Times with poor wages (4) for AGES (with-poor WAGES, where with is abbreviated by W) Often the clue types are combined. Some common examples are 1) hidden word reversals where the answer is found backwards embedded in other words, and 2) containers or charades where the parts are anagrams. For example: Car shops have broken gear immersed in gasoline. (7) for GARAGES (RAGE = gear anagram in GAS = gasoline) All manner of common abbreviations, acronyms, and other symbology such as roman numerals are allowed. For example: c one hundred, cup, or centigrade vi six h hot s small ca california Two punctuation marks at the end of the clue have been reserved for special meaning. A question mark (?) indicates that the straight clue is not entirely straight (usually a pun). For example: I tie down mascara holder soundly? (7) for EYELASH (homophone of "I lash", mascara holder is a punning definition of EYELASH) An exclamation point (]) indicates that some part (usually all) of the clue overlaps. For example, the straight definition may also be the anagram indicator. Here is an example that entirely overlaps: A moped also has these] (6) for PEDALS (hidden word) Here, the entire clue indicates the hidden word, but the entire clue is also a straight definition of the answer. Give it a try] Cryptic crossword puzzles are a lot of fun. -- Steve Koehler ucsd.edu]telesoft]koehler telesoft]koehler@ucsd.edu koehler@telesoft.com ==> games/go-moku.p <== For a game of k in a row on an n x n board, for what values of k and n is there a win? Is (the largest such) k eventually constant or does it increase with n? ==> games/go-moku.s <== Berlekamp, Conway, and Guy's _Winning_Ways_ reports proof that the maximum k is between 4 and 7 inclusive, and it appears to be 5 or 6. They report: . 4-in-a-row is a draw on a 5x5 board (C. Y. Lee), but not on a 4x30 board (C. Lustenberger). . N-in-a-row is shown to be a draw on a NxN board for N>4, using a general pairing technique devised by A. W. Hales and R. I. Jewett. . 9-in-a-row is a draw even on an infinite board, a 1954 result of H. O. Pollak and C. E. Shannon. . More recently, the pseudonymous group T. G. L. Zetters showed that 8-in-a-row is a draw on an infinite board, and have made some progress on showing infinite 7-in-a-row to be a draw. Go-moku is 5-in-a-row played on a 19x19 go board. It is apparently a win for the first player, and so the Japanese have introduced several 'handicaps' for the first player (e.g., he must win with _exactly_ five: 6-in-a-row doesn't count), but apparently the game is still a win for the first player. None of these apparent results have been proven. ==> games/hi-q.p <== What is the quickest solution of the game Hi-Q (also called Solitair)? For those of you who aren't sure what the game looks like: 32 movable pegs ("+") are arranged on the following board such that only the middle position is empty ("-"). Just to be complete: the board consists of only these 33 positions. 1 2 3 4 5 6 7 1 + + + 2 + + + 3 + + + + + + + 4 + + + - + + + 5 + + + + + + + 6 + + + 7 + + + A piece moves on this board by jumping over one of its immediate neighboor (horizontally or vertically) into an empty space opposite. The peg that was jumped over, is hit and removed from the board. A move can contain multiple hits if you use the same peg to make the hits. You have to end with one peg exactly in the middle position (44). ==> games/hi-q.s <== 1: 46*44 2: 65*45 3: 57*55 4: 54*56 5: 52*54 6: 73*53 7: 43*63 8: 75*73*53 9: 35*55 10: 15*35 11: 23*43*63*65*45*25 12: 37*57*55*53 13: 31*33 14: 34*32 15: 51*31*33 16: 13*15*35 17: 36*34*32*52*54*34 18: 24*44 Found by Ernest Bergholt in 1912 and was proved to be minimal by John Beasley in 1964. References The Ins and Outs of Peg Solitaire John D Beasley Oxford U press, 1985 ISBN 0-19-853203-2 Winning Ways, Vol. 2, Ch. 23 Berlekamp, E.R. Academic Press, 1982 ISBN 01-12-091102-7 ==> games/jeopardy.p <== What are the highest, lowest, and most different scores contestants can achieve during a single game of Jeopardy? ==> games/jeopardy.s <== highest: $283,200.00, lowest: -$29,000.00, biggest difference: $309,700.00 (1) Our theoretical contestant has an itchy trigger finger, and rings in with an answer before either of his/her opponents. (2) The daily doubles (1 in the Jeopardy] round, 2 in the Double Jeopardy] round) all appear under an answer in the $100 or $200 rows. (3) All answers given by our contestant are (will be?) correct. Therefore: Round 1 (Jeopardy]): Max. score per category: $1500. For 6 categories - $100 for the DD, that's $8900. Our hero bets the farm and wins - score: $17,800. Round 2 (Double Jeopardy]): Max. score per category: $3000. Assume that the DDs are found last, in order. For 6 categories - $400 for both DDs, that's $17,600. Added to his/her winnings in Round 1, that's $35,400. After the 1st DD, where the whole thing is wagered, the contestant's score is $70,800. Then the whole amount is wagered again, yielding a total of $141,600. Round 3 (Final Jeopardy]): Our (very greedy] :) hero now bets the whole thing, to see just how much s/he can actually win. Assuming that his/her answer is right, the final amount would be $283,200. But the contestant can only take home $100,000; the rest is donated to charity. To calculate the lowest possible socre: -1500 x 6 = -9000 + 100 = -8900. On the Daily Double that appears in the 100 slot, you bet the maximum allowed, 500, and lose. So after the first round, you are at -9400. -3000 x 6 = -18000 + 400 = -17600 On the two Daily Doubles in the 200 slots, bet the maximum allowed, 1000. So after the second round you are at -9400 + -19600 = -29000. This is the lowest score you can achieve in Jeopardy before the Final Jeopardy round. The caveat here is that you *must* be the person sitting in the left-most seat (either a returning champion or the luckier of the three people who come in after a five-time champion "retires") at the beginning of the game, because otherwise you will not have control of the board when the first Daily Double comes along. ==> games/knight.tour.p <== For what board sizes is a knight's tour possible? ==> games/knight.tour.s <== A tour exists for boards of size 1x1, 3x4, 3xN with N >= 7, 4xN with N >= 5, and MxN with N >= M >= 5. In other words, for all rectangles except 1xN (excluding the trivial 1x1), 2xN, 3x3, 3x5, 3x6, 4x4. With the exception of 3x8 and 4xN, any even-sized board which allows a tour will also allow a closed (reentrant) tour. On an odd-sided board, there is one more square of one color than of the other. Every time a knight moves, it moves to a square of the other color than the one it is on. Therefore, on an odd-sided board, it must end the last move but one of the complete, reentrant tour on a square of the same color as that on which it started. It is then impossible to make the last move, for that move would end on a square of the same color as it begins on. Here is a solution for the 7x7 board (which is not reentrant). ------------------------------------ ! 17 ! 6 ! 33 ! 42 ! 15 ! 4 ! 25 ! ------------------------------------ ! 32 ! 47 ! 16 ! 5 ! 26 ! 35 ! 14 ! ------------------------------------ ! 7 ! 18 ! 43 ! 34 ! 41 ! 24 ! 3 ! ------------------------------------ ! 46 ! 31 ! 48 ! 27 ! 44 ! 13 ! 36 ! ------------------------------------ ! 19 ! 8 ! 45 ! 40 ! 49 ! 2 ! 23 ! ------------------------------------ ! 30 ! 39 ! 10 ! 21 ! 28 ! 37 ! 12 ! ------------------------------------ ! 9 ! 20 ! 29 ! 38 ! 11 ! 22 ! 1 ! ------------------------------------ Here is a solution for the 5x5 board (which is not reentrant). -------------------------- ! 5 ! 10 ! 15 ! 20 ! 3 ! -------------------------- ! 16 ! 21 ! 4 ! 9 ! 14 ! -------------------------- ! 11 ! 6 ! 25 ! 2 ! 19 ! -------------------------- ! 22 ! 17 ! 8 ! 13 ! 24 ! -------------------------- ! 7 ! 12 ! 23 ! 18 ! 1 ! -------------------------- Here is a reentrant 2x4x4 tour: 0 11 16 3 15 4 1 22 19 26 9 24 8 23 14 27 10 5 30 17 31 12 21 2 29 18 25 6 20 7 28 13 A reentrant 4x4x4 tour can be constructed by splicing two copies. It shouldn't be much more work now to completely solve the problem of which 3D rectangular boards allow tours. ==> games/nim.p <== Place 10 piles of 10 $1 bills in a row. A valid move is to reduce the last i>0 piles by the same amount j>0 for some i and j; a pile reduced to nothing is considered to have been removed. The loser is the player who picks up the last dollar, and they must forfeit half of what they picked up to the winner. 1) Who is the winner in Waldo Nim, the first or the second player? 2) How much more money than the loser can the winner obtain with best play on both parties? ==> games/nim.s <== For the particular game described we only need to consider positions for which the following condition holds for each pile: (number of bills in pile k) + k >= (number of piles) + 1 A GOOD position is defined as one in which this condition holds, with equality applying only to one pile P, and all piles following P having the same number of bills as P. ( So the initial position is GOOD, the special pile being the first. ) I now claim that if I leave you a GOOD position, and you make any move, I can move back to a GOOD position. Suppose there are n piles and the special pile is numbered (n-p+1) (so that the last p piles each contain p bills). (1) You take p bills from p or more piles; (a) If p = n, you have just taken the last bill and lost. (b) Otherwise I reduce pile (n-p) (which is now the last) to 1 bill. (2) You take p bills from r(p; I take (p-q) bills from (q+r-p) piles (b) q+r<=p; I take (p-q) bills from (q+r) piles Verifying that each of the resulting positions is GOOD is tedious but straightforward. It is left as an exercise for the reader. -- RobH ==> games/othello.p <== How good are computers at Othello? ==> games/othello.s <== The interesting game in which computers are undoubted masters of all they survey is Othello, where Kai-Fu Lee's (CMU) program "Bill" is so good it can only play itself to learn to get better. Bill has a fantastically correct and efficient evaluation function, that recently has been further improved by learning coefficients for additional terms made up of the pair-wise combination of the four old terms. This improved the quality of the play approximately as much as searching an extra two ply. Bill is so good it can beat lots of players with no search at all. Its 6 or 7 ply search sweeps aside all opposition (though Kai-Fu says that some very good players are now coming along in Japan, and he is not sure whether Bill would beat them). One interesting question remaining in Othello is the game theoretic value of the starting position. Bill's results seem to indicate that the first player has an advantage. It appears that, since Kai-Fu has published all his evaluation material, someone could build an Othello machine, and produce a constructive proof (as was done for Cubic) that it is a win for the first player. ==> games/risk.p <== What are the odds when tossing dice in Risk? ==> games/risk.s <== Attacker using 3 dice, Defender using 2: Probability that Attacker wins 2 = 2323 / 7776 Probability that Attacker wins 1 = 3724 / 7776 Probability that Attacker wins 0 = 1729 / 7776 Attacker using 3 dice, Defender using 1: Probability that Attacker wins 1 = 855 / 1296 Probability that Attacker wins 0 = 441 / 1296 Attacker using 2 dice, Defender using 2: Probability that Attacker wins 2 = 225 / 1296 Probability that Attacker wins 1 = 630 / 1296 Probability that Attacker wins 0 = 441 / 1296 Attacker using 2 dice, Defender using 1: Probability that Attacker wins 1 = 125 / 216 Probability that Attacker wins 0 = 91 / 216 Attacker using 1 dice, Defender using 2: Probability that Attacker wins 1 = 90 / 216 Probability that Attacker wins 0 = 126 / 216 Attacker using 1 dice, Defender using 1: Probability that Attacker wins 1 = 15 / 36 Probability that Attacker wins 0 = 21 / 36 ==> games/rubiks.clock.p <== How do you quickly solve Rubik's clock? ==> games/rubiks.clock.s <== Solution to Rubik's Clock The solution to Rubik's Clock is very simple and the clock can be "worked" in 10-20 seconds once the solution is known. In this description of how to solve the clock I will describe the different clocks as if they were on a map (e.g. N,NE,E,SE,S,SW,W,NW); this leaves the middle clock which I will just call M. To work the Rubik's clock choose one side to start from; it does not matter from which side you start. Your initial goal will be to align the N,S,E,W and M clocks. Use the following algorithm to do this: Ý1¨ Start with all buttons in the OUT position. Ý2¨ Choose a N,S,E,W clock that does not already have the same time as M (i.e. not aligned with M). Ý3¨ Push in the closest two buttons to the clock you chose in Ý2¨. Ý4¨ Using the knobs that are farest away from the clock you chose in Ý2¨ rotate the knob until M and the clock you chose are aligned. The time on the clocks at this point does not matter. Ý5¨ Go back to Ý1¨ until N,S,E,W and M are in alignment. Ý6¨ At this point N,S,E,W and M should all have the same time. Make sure all buttons are out and rotate any knob until N,S,E,W and M are pointing to 12 oclock. Now turn the puzzle over and repeat steps Ý1¨-Ý6¨ for this side. DO NOT turn any knobs other than the ones described in Ý1¨-Ý6¨. If you have done this correctly then on both sides of the puzzle N,S,E,W and M will all be pointing to 12. Now to align NE,SE,SW,NW. To finish the puzzle you only need to work from one side. Choose a side and use the following algorithm to align the corners: Ý1¨ Start with all buttons OUT on the side you're working from. Ý2¨ Choose a corner that is not aligned. Ý3¨ Press the button closest to that corner in. Ý4¨ Using any knob except for that corner's knob rotate all the clocks until they are in line with the corner clock. (Here "all the clocks" means N,S,E,W,M and any other clock that you have already aligned) There is no need at this point to return the clocks to 12 although if it is less confusing you can. Remember to return all buttons to their up position before you do so. Ý5¨ Return to Ý1¨ until all clocks are aligned. Ý6¨ With all buttons up rotate all the clocks to 12. ==> games/rubiks.cube.p <== What is known about bounds on solving Rubik's cube? ==> games/rubiks.cube.s <== The "official" world record was set by Minh Thai at the 1982 World Championships in Budapest Hungary, with a time of 22.95 seconds. Keep in mind mathematicians provided standardized dislocation patterns for the cubes to be randomized as much as possible. The fastest cube solvers from 19 different countries had 3 attempts each to solve the cube as quickly as possible. Minh and several others have unofficially solved the cube in times between 16 and 19 seconds. However, Minh averages around 25 to 26 seconds after 10 trials, and by best average of ten trials is about 27 seconds (although it is usually higher). Consider that in the World Championships 19 of the world's fastest cube solvers each solved the cube 3 times and no one solved the cube in less than 20 seconds... God's algorithm is the name given to an as yet (as far as I know) undiscovered method to solve the rubik's cube in the least number of moves; as apposed to using 'canned' moves. The known lower bound is 18 moves. This is established by looking at things backwards: suppose we can solve a position in N moves. Then by running the solution backwards, we can also go from the solved position to the position we started with in N moves. Now we count how many sequences of N moves there are from the starting position, making certain that we don't turn the same face twice in a row: N=0: 1 (empty) sequence; N=1: 18 sequences (6 faces can be turned, each in 3 different ways) N=2: 18*15 sequences (take any sequence of length 1, then turn any of the five faces which is not the last face turned, in any of 3 different ways); N=3: 18*15*15 sequences (take any sequence of length 2, then turn any of the five faces which is not the last face turned, in any of 3 different ways); : : N=i: 18*15^(i-1) sequences. So there are only 1 + 18 + 18*15 + 18*15^2 + ... + 18*15^(n-1) sequences of moves of length n or less. This sequence sums to (18/14)*(15^n - 1) + 1. Trying particular values of n, we find that there are about 8.4 * 10^18 sequences of length 16 or less, and about 1.3 times 10^20 sequences of length 17 or less. Since there are 2^10 * 3^7 * 8] * 12], or about 4.3 * 10^19, possible positions of the cube, we see that there simply aren't enough sequences of length 16 or less to reach every position from the starting position. So not every position can be solved in 16 or less moves - i.e. some positions require at least 17 moves. This can be improved to 18 moves by being a bit more careful about counting sequences which produce the same position. To do this, note that if you turn one face and then turn the opposite face, you get exactly the same result as if you'd done the two moves in the opposite order. When counting the number of essentially different sequences of N moves, therefore, we can split into two cases: (a) Last two moves were not of opposite faces. All such sequences can be obtained by taking a sequence of length N-1, choosing one of the 4 faces which is neither the face which was last turned nor the face opposite it, and choosing one of 3 possible ways to turn it. (If N=1, so that the sequence of length N-1 is empty and doesn't have a last move, we can choose any of the 6 faces.) (b) Last two moves were of opposite faces. All such sequences can be obtained by taking a sequence of length N-2, choosing one of the 2 opposite face pairs that doesn't include the last face turned, and turning each of the two faces in this pair (3*3 possibilities for how it was turned). (If N=2, so that the sequence of length N-2 is empty and doesn't have a last move, we can choose any of the 3 opposite face pairs.) This gives us a recurrence relation for the number X_N of sequences of length N: N=0: X_0 = 1 (the empty sequence) N=1: X_1 = 18 * X_0 = 18 N=2: X_2 = 12 * X_1 + 27 * X_0 = 243 N=3: X_3 = 12 * X_2 + 18 * X_1 = 3240 : : N=i: X_i = 12 * X_(i-1) + 18 * X_(i-2) If you do the calculations, you find that X_0 + X_1 + X_2 + ... + X_17 is about 2.0 * 10^19. So there are fewer essentially different sequences of moves of length 17 or less than there are positions of the cube, and so some positions require at least 18 moves. The upper bound of 50 moves is I believe due to Morwen Thistlethwaite, who developed a technique to solve the cube in a maximum of 50 moves. It involved a descent through a chain of subgroups of the full cube group, starting with the full cube group and ending with the trivial subgroup (i.e. the one containing the solved position only). Each stage involves a careful examination of the cube, essentially to work out which coset of the target subgroup it is in, followed by a table look-up to find a sequence to put it into that subgroup. Needless to say, it was not a fast technique] But it was fascinating to watch, because for the first three quarters or so of the solution, you couldn't really see anything happening - i.e. the position continued to appear random] If I remember correctly, one of the final subgroups in the chain was the subgroup generated by all the double twists of the faces - so near the end of the solution, you would suddenly notice that each face only had two colours on it. A few moves more and the solution was complete. Completely different from most cube solutions, in which you gradually see order return to chaos: with Morwen's solution, the order only re-appeared in the last 10-15 moves. With God's algorithm, of course, I would expect this effect to be even more pronounced: someone solving the cube with God's algorithm would probably look very much like a film of someone scrambling the cube, run in reverse] Finally, something I'd be curious to know in this context: consider the position in which every cubelet is in the right position, all the corner cubelets are in the correct orientation, and all the edge cubelets are "flipped" (i.e. the only change from the solved position is that every edge is flipped). What is the shortest sequence of moves known to get the cube into this position, or equivalently to solve it from this position? (I know of several sequences of 24 moves that do the trick.) The reason I'm interested in this particular position: it is the unique element of the centre of the cube group. As a consequence, I vaguely suspect (I'd hardly like to call it a conjecture :-) it may lie "opposite" the solved position in the cube graph - i.e. the graph with a vertex for each position of the cube and edges connecting positions that can be transformed into each other with a single move. If this is the case, then it is a good candidate to require the maximum possible number of moves in God's algorithm. -- David Seal dseal@armltd.co.uk To my knowledge, no one has ever demonstrated a specific cube position that takes 15 moves to solve. Furthermore, the lower bound is known to be greater than 15, due to a simple proof. The way we know the lower bound is by working backwards counting how many positions we can reach in a small number of moves from the solved position. If this is less than 43,252,003,274,489,856,000 (the total number of positions of Rubik's cube) then you need more than that number of moves to reach the other positions of the cube. Therefore, those positions will require more moves to solve. The answer depends on what we consider a move. There are three common definitions. The most restrictive is the QF metric, in which only a quarter-turn of a face is allowed as a single move. More common is the HF metric, in which a half-turn of a face is also counted as a single move. The most generous is the HS metric, in which a quarter- turn or half-turn of a central slice is also counted as a single move. These metrics are sometimes called the 12-generator, 18-generator, and 27-generator metrics, respectively, for the number of primitive moves. The definition does not affect which positions you can get to, or even how you get there, only how many moves we count for it. The answer is that even in the HS metric, the lower bound is 16, because at most 17,508,850,688,971,332,784 positions can be reached within 15 HS moves. In the HF metric, the lower bound is 18, because at most 19,973,266,111,335,481,264 positions can be reached within 17 HF moves. And in the QT metric, the lower bound is 21, because at most 39,812,499,178,877,773,072 positions can be reached within 20 QT moves. -- jjfink@skcla.monsanto.com writes: Lately in this conference I've noted several messages related to Rubik's Cube and Square 1. I've been an avid cube fanatic since 1981 and I've been gathering cube information since. Around Feb. 1990 I started to produce the Domain of the Cube Newsletter, which focuses on Rubik's Cube and all the cube variants produced to date. I include notes on unproduced prototype cubes which don't even exist, patent information, cube history (and prehistory), computer simulations of puzzles, etc. I'm up to the 4th issue. Anyways, if you're interested in other puzzles of the scramble by rotation type you may be interested in DOTC. It's available free to anyone interested. I am especially interested in contributing articles for the newsletter, e.g. ideas for new variants, God's Algorithm. Anyone ever write a Magic Dodecahedron simulation for a computer? Anyone understand Morwen Thistlethwaite's 50 move solution to Rubik's Cube? I'd love to hear from you. Drop me a SASE (say empire size) if you're interested in DOTC or if you would like to exchange notes on Rubik's Cube, Square 1 etc. I'm also interested in exchanging puzzle simulations, e.g. Rubik's Cube, Twisty Torus, NxNxN Simulations, etc, for Amiga and IBM computers. I've written several Rubik's Cube solving programs, and I'm trying to make the definitive puzzle solving engine. I'm also interested in AI programs for Rubik's Cube and the like. Ideal Toy put out the Rubik's Cube Newsletter, starting with issue #1 on May 1982. There were 4 issues in all, and I'm missing #3. I have: #1, May 1982 #2, Aug 1982 #3, Aug 1983 I am willing to trade photocopies with anyone to obtain #3. There was another sort of magazine, published in several languages called Rubik's Logic and Fantasy in space. I believe there were 8 issues in all. Unfortunately I don't have any of these] I'm willing to buy these off anyone interesting in selling. I would like to get the originals if at all possible... I'm also interested in buying any books on the cube or related puzzles. In particular I am _very_ interested in obtaining the following: Cube Games Don Taylor, Leanne Rylands Official Solution to Alexander's Star Adam Alexander The Amazing Pyraminx Dr. Ronald Turner-Smith The Winning Solution Minh Thai The Winning Solution to Rubik's Revenge Minh Thai Simple Solutions to Cubic Puzzles James G. Nourse I'm also interested in buying puzzles of the mechanical type. I'm still missing the Pyraminx Star (basically a Pyraminx with more tips on it), the Puck, and Hungarian Rings. If anyone out here is a fellow collector I'd like to hear from you. If you have a cube variant which you think is rare, or an idea for a cube variant we could swap notes. I'm in the middle of compiling an exhaustive library for computer simulations of puzzles. This includes simulations of all Uwe Meffert's puzzles which he prototyped but _never_ produced. In fact, I'm in the middle of working on a Pyraminx Hexagon solver. What? Never heard of it? Meffert did a lot of other puzzles which never were made. I invented some new "scramble by rotation" puzzles myself. My favourite creation is the Twisty Torus. It is a torus puzzle with segments (which slide around 360 degrees) with multiple rings around the circumference. The computer puzzle simulation library I'm forming will be described in depth in DOTC #4 (The Domain of the Cube Newsletter). So if you have any interesting computer puzzle programs please email me and tell me all about them] Also to the people interested in obtaining a subscription to DOTC, who are outside of Canada (which it seems is just about all of you]) please don't send U.S. or non-Canadian stamps (yeah, I know I said Self-Addressed Stamped Envelope before). Instead send me an international money order in Canadian funds for $6. I'll send you the first 4 issues (issue #4 is almost finished). Mark Longridge Address: 259 Thornton Rd N, Oshawa Ontario Canada, L1J 6T2 Email: mark.longridge@canrem.com One other thing, the six bucks is not for me to make any money. This is only to cover the cost of producing it and mailing it. I'm just trying to spread the word about DOTC and to encourage other mechanical puzzle lovers to share their ideas, books, programs and puzzles. Most of the programs I've written and/or collected are shareware for C64, Amiga and IBM. I have source for all my programs (all in C or Basic) and I am thinking of providing a disk with the 4th issue of DOTC. If the response is favourable I will continue to provide disks with DOTC. -- Mark Longridge writes: It may interest people to know that in the latest issue of "Cubism For Fun" % (# 28 that I just received yesterday) there is an article by Herbert Kociemba from Darmstadt. He describes a program that solves the cube. He states that until now he has found no configuration that required more than 21 turns to solve. He gives a 20 move manoeuvre to get at the "all edges flipped/ all corners twisted" position: DF^2U'B^2R^2B^2R^2LB'D'FD^2FB^2UF'RLU^2F' or in Varga's parlance: dofitabiribirilobadafodifobitofarolotifa Other things #28 contains are an analysis of Square 1, an article about triangular tilings by Martin Gardner, and a number of articles about other puzzles. -- % CFF is a newsletter published by the Dutch Cubusts Club NKC. Secretary: Anneke Treep Postbus 8295 6710 AG Ede The Netherlands Membership fee for 1992 is DFL 20 (about$ 11). -- -- dik t. winter References: E. C. Turner & K. F. Gold, "Rubik's Groups", American Mathematical Monthly, vol. 92 (1985), pp. 617-629. Cubelike Puzzles - What Are They and How Do You Solve Them? J.A. Eidswick A.M.M. March, 1986 Rubik's Revenge: The Group Theoretical Solution Mogens Esrom Larsen A.M.M. June-July, 1985 The Group of the Hungarian Magic Cube Chris Rowley Proceedings of the First Western Austrialian Conference on Algebra, 1982 Rubik's Cubic Compendium Erno Rubik, Tamas Varga, et al (Ed by David Singmaster) Oxford University Press, 1987 (Some chapters on mathematics of the cube.) David Singmaster, _Notes on Rubik's `Magic Cube'_ "Winning Ways" by Berlekamp, Elwyn R. Conway, John H. Guy, Richard K. Volume two, pages 760-768, 808, 809 ==> games/rubiks.magic.p <== How do you solve Rubik's Magic? ==> games/rubiks.magic.s <== The solution is in a 3x3 grid with a corner missing. +---+---+---+ +---+---+---+---+ ! 3 ! 5 ! 7 ! ! 1 ! 3 ! 5 ! 7 ! +---+---+---+ +---+---+---+---+ ! 1 ! 6 ! 8 ! ! 2 ! 4 ! 6 ! 8 ! +---+---+---+ +---+---+---+---+ ! 2 ! 4 ! Original Shape +---+---+ To get the 2x4 "standard" shape into this shape, follow this: 1. Lie it flat in front of you (4 going across). 2. Flip the pair (1,2) up and over on top of (3,4). 3. Flip the ONE square (2) up and over (1). ÝNote: if step 3 won't go, start over, but flip the entire original shape over (exposing the back).¨ 4. Flip the pair (2,4) up and over on top of (5,6). 5. Flip the pair (1,2) up and toward you on top of (blank,4). 6. Flip the ONE square (2) up and left on top of (1). 7. Flip the pair (2,4) up and toward you. Your puzzle won't be completely solved, but this is how to get the shape. Notice that 3,5,6,7,8 don't move. ==> games/scrabble.p <== What are some exceptional scrabble games? ==> games/scrabble.s <== The shortest scrabble game: The Scrabble Players News, Vol. XI No. 49, June 1983, contributed by Kyle Corbin of Raleigh, NC: ÝJ¨ J U S S O X ÝX¨U which can be done in 4 moves, JUS, SOX, ÝJ¨US, and ÝX¨U. In SPN Vol. XI, No. 52, December 1983, Alan Frank presented what he claimed is the shortest game where no blanks are used, also four moves: C WUD CUKES DEY S This was followed in SPN, Vol. XII No. 54, April 1984, by Terry Davis of Glasgow, KY: V V OÝX¨ ÝX¨U, which is three moves. He noted that the use of two blanks prevents such plays as VOLVOX. Unfortunately, it doesn't prevent SONOVOX. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Record for the highest scrabble score in a single turn (in a legal position): According to the Scrabble Players Newspaper (since renamed to Scrabble Players News) issue 44, p13, the highest score for one turn yet discovered, using the Official Scrabble Players Dictionary, 1st ed. (the 2nd edition is now in use in club and tournament play) and the Websters 9th New Collegiate Dictionary, was the following: d i s e q u i l i b r a t e D . . . . . . . e . . . . . . e . . . . . . . e . . . . . o m r a d i o a u t o g r a p(h)Y . . . . . . . . . . . w a s T . . . . . . . . . . b e . . h . . . . . . . . . . a . . g o . . . c o n j u n c t i v a L . . . . . . . . . . . . . n o . . . . . . . f i n i k i n G . . . . . . . a . . . (l) e i . . . . . . . d . s p e l t Z . . . . . . w e . . . . . . e . . . . . . r . . . . . . o r m e t h o x y f l u r a n e S for 1682 points. According to the May 1986 issue of GAMES, the highest known score achievable in one turn is 1,962 points. The word is BENZOXYCAMPHORS formed across the three triple-word scores on the bottom of the board. Apparently it was discovered by Darryl Francis, Ron Jerome, and Jeff Grant. As for other Scrabble trivia, the highest-scoring first move based on the Official Scrabble Players Dictionary is 120 points, with the words JUKEBOX, QUIZZED, SQUEEZE, or ZYMURGY. If Funk & Wagnall's New Standard Dictionary is used then ZYXOMMA, worth 130 points, can be formed. The highest-scoring game, based on Webster's Second and Third and on the Oxford English Dictionary, was devised by Ron Jerome and Ralph Beaman and totalled 4,142 points for the two players. The highest-scoring words in the game were BENZOXYCAMPHORS, VELVETEEN, and JACKPUDDINGHOOD. The following example of a SCRABBLE game produced a score of 2448 for one player and 1175 for the final word. It is taken from _Beyond Language_ (1967) by Dmitri Borgman (pp. 217-218). He credits this solution to Mrs. Josefa H. Byrne of San Francisco and implies that all words can be found in _Webster's Second Edition_. The two large words (multiplied by 27 as they span 3 triple word scores) are ZOOPSYCHOLOGIST (a psychologist who treats animals rather than humans) and PREJUDICATENESS (the condition or state of being decided beforehand). The asterisks (*) represent the blank tiles. (Please excuse any typo's). Board Player1 Player2 Z O O P S Y C H O L O G I S T ABILITY 76 ERI, YE 9 O N H A U R O W MAN, MI 10 EN 2 * R I B R O V E I FEN, FUN 14 MANIA 7 L T I K E G TABU 12 RIB 6 O L NEXT 11 AM 4 G I AX 9 END 6 I T IT, TIKE 10 LURE 6 * Y E LEND, LOGIC*AL 79 OO*LOGICAL 8 A R FUND, JUD 27 ATE, MA 7 L E N D M I ROVE 14 LO 2 E A Q DARE, DE 13 ES, ES, RE 6 W A X F E N U RE, ROW 14 IRE, IS, SO 7 E T A B U I A DARED, QUAD 22 ON 4 E N A M D A R E D WAX, WEE 27 WIG 9 P R E J U D I C A T E N E S S CHIT, HA 14 ON 2 PREJUDICATENESS, AN, MANIAC, QUADS, WEEP 911 OOP 8 ZOOPSYCHOLOGIST, HABILITY, TWIG, ZOOLOGICAL 1175 -------------------------------------- Total: 2438 93 F, N, V, T in loser's hand: +10 -10 -------------------------------------- Final Score: 2448 83 --------------------------------------------------------------------------- It is possible to form the following 14 7-letter OSPD words from the tiles: HUMANLY FATUOUS AMAZING EERIEST ROOFING TOILERS QUIXOTE JEWELRY CAPABLE PREVIEW BIDDERS HACKING OVATION DONATED ==> games/square-1.p <== Does anyone have any hints on how to solve the Square-1 puzzle? ==> games/square-1.s <== SHAPES 1. There are 29 different shapes for a side, counting reflections: 1 with 6 corners, 0 edges 3 with 5 corners, 2 edges 10 with 4 corners, 4 edges 10 with 3 corners, 6 edges 5 with 2 corners, 8 edges 2. Naturally, a surplus of corners on one side must be compensated by a deficit of corners on the other side. Thus there are 1*5 + 3*10 + C(10,2) = 5 + 30 + 55 = 90 distinct combinations of shapes, not counting the middle layer. 3. You can reach two squares from any other shape in at most 7 transforms, where a transform consists of (1) optionally twisting the top, (2) optionally twisting the bottom, and (3) flipping. 4. Each transform toggles the middle layer between Square and Kite, so you may need 8 transforms to reach a perfect cube. 5. The shapes with 4 corners and 4 edges on each side fall into four mutually separated classes. Side shapes can be assigned values: 0: Square, Mushroom, and Shield; 1: Left Fist and Left Paw; 2: Scallop, Kite, and Barrel; 3. Right Fist and Right Paw. The top and bottom's sum or difference, depending on how you look at them, is a constant. Notice that the side shapes with bilateral symmetry are those with even values. 6. To change this constant, and in particular to make it zero, you must attain a position that does not have 4 corners and 4 edges on each side. Almost any such position will do, but returning to 4 corners and 4 edges with the right constant is left to your ingenuity. 7. If the top and bottom are Squares but the middle is a Kite, just flip with the top and bottom 30deg out of phase and you will get a cube. COLORS 1. I do not know the most efficient way to restore the colors. What follows is my own suboptimal method. All flips keep the yellow stripe steady and flip the blue stripe. 2. You can permute the corners without changing the edges, so first get the edges right, then the corners. 3. This transformation sends the right top edge to the bottom and the left bottom edge to the top, leaving the other edges on the same side as they started: Twist top 30deg cl, flip, twist top 30deg ccl, twist bottom 150deg cl, flip, twist bottom 30deg cl, twist top 120deg cl, flip, twist top 30deg ccl, twist bottom 150deg cl, flip, twist bottom 30deg cl. Cl and ccl are defined looking directly at the face. With this transformation you can eventually get all the white edges on top. 4. Check the parity of the edge sequence on each side. If either is wrong, you need to fix it. Sorry -- I don't know how] (See any standard reference on combinatorics for an explanation of parity.) 5. The following transformation cyclically permutes ccl all the top edges but the right one and cl all the bottom edges but the left one. Apply the transformation in 3., and turn the whole cube 180deg. Repeat. This is a useful transformation, though not a cure-all. 6. Varying the transformation in 3. with other twists will produce other results. 7. The following transformation changes a cube into a Comet and Star: Flip to get Kite and Kite. Twist top and bottom cl 90deg and flip to get Barrel and Barrel. Twist top cl 30 and bottom cl 60 and flip to get Scallop and Scallop. Twist top cl 60 and bottom cl 120 and flip to get Comet and Star. The virtue of the Star is that it contains only corners, so that you can permute the corners without altering the edges. 8. To reach a Lemon and Star instead, replace the final bottom cl 120 with a bottom cl 60. In both these transformation the Star is on the bottom. 9. The following transformation cyclically permutes all but the bottom left rear. It sends the top left front to the bottom, and the bottom left front to the top. Go to Comet and Star. Twist star cl 60. Go to Lemon and Star -- you need not return all the way to the cube, but do it if you're unsure of yourself by following 7 backwards. Twist star cl 60. Return to cube by following 8 backwards. With this transformation you should be able to get all the white corners on top. 10. Check the parity of the corner sequences on both sides. If the bottom parity is wrong, here's how to fix it: Go to Lemon and Star. The colors on the Star will run WWGWWG. Twist it 180 and return to cube. 11. If the top parity is wrong, do the same thing, except that when you go from Scallop and Scallop to Lemon and Star, twist the top and bottom ccl instead of cl. The colors on the Star should now run GGWGGW. 12. Once the parity is right on both sides, the basic method is to go to Comet and Star, twist the star 120 cl (it will be WGWGWG), return to cube, twist one or both sides, go to Comet and Star, undo the star twist, return to cube, undo the side twists. With no side twists, this does nothing. If you twist the top, you will permute the top corners. If you twist the bottom, you will permute the bottom corners. Eventually you will get both the top and the bottom right. Don't forget to undo the side twists -- you need to have the edges in the right places. Happy twisting.... -- Col. G. L. Sicherman gls@windmill.att.COM ==> games/think.and.jump.p <== THINK & JUMP: FIRST THINK, THEN JUMP UNTIL YOU ARE LEFT WITH ONE PEG] O - O O - O / \ / \ / \ / \ O---O---O---O---O BOARD DESCRIPTION: To the right is a model of \ / \ / \ / \ / the Think & Jump board. The O---O---O---O---O---O O's represent holes which / \ / \ / \ / \ / \ / \ contain pegs. O---O---O---O---O---O---O \ / \ / \ / \ / \ / \ / O---O---O---O---O---O DIRECTIONS: To play this brain teaser, you begin / \ / \ / \ / \ by removing the center peg. Then, O---O---O---O---O moving any direction in the grid, \ / \ / \ / \ / jump over one peg at a time, O - O O - O removing the jumped peg - until only one peg is left. It's harder then it looks. But it's more fun than you can imagine. SKILL CHART: 10 pegs left - getting better 5 pegs left - true talent 1 peg left - you're a genius Manufactured by Pressman Toy Corporation, NY, NY. ==> games/think.and.jump.s <== Three-color the board in the obvious way. The initial configuration has 12 of each color, and each jump changes the parity of all three colors. Thus, it is impossible to achieve any position where the colors do not have the same parity; in particular, (1,0,0). If you remove the requirement that the initially-empty cell must be at the center, the game becomes solvable. The demonstration is left as an exercise. Karl Heuer rutgers]harvard]ima]haddock]karl karl@haddock.ima.isc.com Here is one way of reducing Think & Jump to two pegs. Long simplifies Balsley's scintillating snowflake solution: 1 U-S A - B C - D 2 H-U / \ / \ / \ / \ 3 V-T E---F---G---H---I 4 S-H \ / \ / \ / \ / 5 D-M J---K---L---M---N---O 6 F-S / \ / \ / \ / \ / \ / \ 7 Q-F P---Q---R---S---T---U---V 8 A-L \ / \ / \ / \ / \ / \ / 9 S-Q W---X---Y---Z---a---b 10 P-R / \ / \ / \ / \ 11 Z-N c---d---e---f---g 12 Y-K \ / \ / \ / \ / 13 h-Y h - i j - k 14 k-Z The board should now be in the snowflake pattern, i.e. look like o - * * - o / \ / \ / \ / \ *---o---*---o---* \ / \ / \ / \ / *---*---*---*---*---* / \ / \ / \ / \ / \ / \ o---o---o---o---o---o---o \ / \ / \ / \ / \ / \ / *---*---*---*---*---* / \ / \ / \ / \ *---o---*---o---* \ / \ / \ / \ / o - * * - o where o is empty and * is a peg. The top and bottom can now be reduced to single pegs individually. For example, we could continue 15 g-T 16 Y-a 17 i-Z 18 T-e 19 j-Y 20 b-Z 21 c-R 22 Z-X 23 W-Y 24 R-e which finishes the bottom. The top can be done in a similar manner. -- Chris Long ==> games/tictactoe.p <== In random tic-tac-toe, what is the probability that the first mover wins? ==> games/tictactoe.s <== Count cases. First assume that the game goes on even after a win. (Later figure out who won if each player gets a row of three.) Then there are 9]/5]4] possible final boards, of which 8*6]/2]4] - 2*6*4]/0]4] - 3*3*4]/0]4] - 1 = 98 have a row of three Xs. The first term is 8 rows times (6 choose 2) ways to put down the remaining 2 Xs. The second term is the number of ways X can have a diagonal row plus a horizontal or vertical row. The third term is the number of ways X can have a vertical and a horizontal row, and the 4th term is the number of ways X can have two diagonal rows. All the two-row configurations must be subtracted to avoid double-counting. There are 8*6]/1]5] = 48 ways O can get a row. There is no double- counting problem since only 4 Os are on the final board. There are 6*2*3]/2]1] = 36 ways that both players can have a row. (6 possible rows for X, each leaving 2 possible rows for O and (3 choose 2) ways to arrange the remaining row.) These cases need further consideration. There are 98 - 36 = 62 ways X can have a row but not O. There are 48 - 36 = 12 ways O can have a row but not X. There are 126 - 36 - 62 - 12 = 16 ways the game can be a tie. Now consider the 36 configurations in which each player has a row. Each such can be achieved in 5]4] = 2880 orders. There are 3*4]4] = 1728 ways that X's last move completes his row. In these cases O wins. There are 2*3*3]3] = 216 ways that Xs fourth move completes his row and Os row is already done in three moves. In these cases O also wins. Altogether, O wins 1728 + 216 = 1944 out of 2880 times in each of these 36 configurations. X wins the other 936 out of 2880. Altogether, the probability of X winning is ( 62 + 36*(936/2880) ) / 126. win: 737 / 1260 ( 0.5849206... ) lose: 121 / 420 ( 0.2880952... ) draw: 8 / 63 ( 0.1269841... ) 1000000 games: won 584865, lost 288240, tied 126895 Instead, how about just methodically having the program play every possible game, tallying up who wins? Wonderful idea, especially since there are only 9] ~ 1/3 million possible games. Of course some are identical because they end in fewer than 8 moves. It is clear that these should be counted multiple times since they are more probable than games that go longer. The result: 362880 games: won 212256, lost 104544, tied 46080 #include int boardÝ9¨; int N, move, won, lost, tied; int permÝ9¨ = { 0, 1, 2, 3, 4, 5, 6, 7, 8 }; int rowsÝ8¨Ý3¨ = { { 0, 1, 2 }, { 3, 4, 5 }, { 6, 7, 8 }, { 0, 3, 6 }, { 1, 4, 7 }, { 2, 5, 8 }, { 0, 4, 8 }, { 2, 4, 6 } }; main() { do { bzero((char *)board, sizeof board); for ( move=0; move<9; move++ ) { boardÝpermÝmove¨¨ = (move&1) ? 4 : 1; if ( move >= 4 && over() ) break; } if ( move == 9 ) tied++; #ifdef DEBUG printf("%1d%1d%1d\n%1d%1d%1d w %d, l %d, t %d\n%1d%1d%1d\n\n", boardÝ0¨, boardÝ1¨, boardÝ2¨, boardÝ3¨, boardÝ4¨, boardÝ5¨, won, lost, tied, boardÝ6¨, boardÝ7¨, boardÝ8¨); #endif N++; } while ( nextperm(perm, 9) ); printf("%d games: won %d, lost %d, tied %d\n", N, won, lost, tied); exit(0); } int s; int *row; over() { for ( row=rowsÝ0¨; row= 0 && cÝi¨ >= cÝi+1¨ ) i--; if ( i < 0 ) return 0; while ( cÝj¨ <= cÝi¨ ) j--; t = cÝi¨; cÝi¨ = cÝj¨; cÝj¨ = t; i++; j = n-1; while ( i < j ) { t = cÝi¨; cÝi¨ = cÝj¨; cÝj¨ = t; i++; j--; } return 1; } ==> geometry/K3,3.p <== Can three houses be connected to three utilities without the pipes crossing? _______ _______ _______ ! oil ! !water! ! gas ! !_____! !_____! !_____! _______ _______ _______ !HOUSE! !HOUSE! !HOUSE! ! one ! ! two ! !three! ==> geometry/K3,3.s <== The problem you describe is to draw a bipartite graph of 3 nodes connected in all ways to 3 nodes, all embedded in the plane. The graph is called K3,3. A famous theorem of Kuratowsky says that all graphs can be embedded in the plane, EXCEPT those containing K3,3 or K5 (the complete graph on 5 vertices, i.e., the graph with 5 nodes and 10 edges) as a subgraph. So your problem is a minimal example of a graph that cannot be embedded in the plane. The proofs that K5 and K3,3 are non-planar are really quite easy, and only depend on Euler's Theorem that F-E+V=2 for a planar graph. For K3,3 V is 6 and E is 9, so F would have to be 5. But each face has at least 4 edges, so E >= (F*4)/2 = 10, contradiction. For K5 V is 5 and E is 10, so F = 7. In this case each face has at least 3 edges, so E >= (F*3)/2 = 10.5, contradiction. The difficult part of Kuratowsky is the proof in the other direction] A quick, informal proof by contradiction without assuming Euler's Theorem: Using a map in which the houses are 1, 2, and 3 and the utilities are A, B, and C, there must be continuous lines that connect the buildings and divide the area into three sections bounded by the loops A-1-B-2-A, A-1-B-3-A, and A-2-B-3-A. (One of the areas is the infinite plane *around* whichever loop is the outer edge of the network.) C must be in one of these three areas; whichever area it is in, either 1, or 2, or 3, is *not* part of the loop that rings its area and hence is inaccessible to C. The usual quibble is to solve the puzzle by running one of the pipes underneath one of houses on its way to another house; the puzzle's instructions forbid crossing other *pipes*, but not crossing other *houses*. ==> geometry/bear.p <== If a hunter goes out his front door, goes 50 miles south, then goes 50 miles west, shoots a bear, goes 50 miles north and ends up in front of his house. What color was the bear? ==> geometry/bear.s <== The hunter's door is in one of two locations. One is a foot or so from the North Pole, facing north, such that his position in front of the door is precisely upon the North Pole. Since that's a ridiculous place to build a house and since bears do not roam within fifty miles of the pole, the bear is either imaginary or imported, and there is no telling what color it is. There is another place (actually a whole set) on earth from which one can go fifty miles south, fifty miles west, and fifty miles north and end up where one started. Consider the parallel of latitude close enough to the South Pole that the circumference of the earth at that latitude is 50/n miles, for some integer n. Take any point on that parallel of latitude and pick the point fifty miles north of it. Situate the hunter's front porch there. The hunter goes fifty miles south from his porch and is at a point we'll call A. He travels fifty miles west, going n times around the earth, and is at A again, where he shoots the bear. Fifty miles north from A he is back home. Since bears are not indigenous to the Antarctic, again the bear is either imaginary or imported and there is no telling what color it might be. ==> geometry/bisector.p <== If two angle bisectors of a triangle are equal, then the triangle is isosceles (more specifically, the sides opposite to the two angles being bisected are equal). ==> geometry/bisector.s <== The following proof is probably from Altshiller-Court's College Geometry, since that's where I first saw the problem. Let the triangle be ABC, with angle bisectors BE and CD. Let F be such that BEFD is a parallelagram. Let x = measure of angle CBE = angle DBE, y = measure of angle BCD = angle DCE, x' = measure of angle EFC, y' = measure of angle ECF. (You will probably want to draw a picture.) Suppose x > y. Consider the triangles EBC and DCB. Since BC = BC and BE = CD, we must have CE > BD. Now, since BD = EF, we have that CE > EF, so that x' > y'. Thus x+x' > y+y'. But, triangle FDC is isosceles, since DF = BE = DC, so x+x' = y+y', a contradiction. Similarly, we cannot have x < y. Therefore the base angles of ABC are equal, making ABC an isosceles triangle. QED ==> geometry/calendar.p <== Build a calendar from two sets of cubes. On the first set, spell the months with a letter on each face of three cubes. Use lowercase three-letter abbreviations for the names of all twelve months (e.g., "jan", "feb", "mar"). On the second set, number the days with a digit on each face of two cubes (e.g., "01", "02", etc.). ==> geometry/calendar.s <== First note that there are *nineteen* different letters in the month abbreviations (abcdef gjlmno prstuv y) so to get them all on the eighteen faces of 3 cubes, you know right away you're going to have to resort to trickery. So I wrote them all down and looked at which ones could be reversed to make another letter in the set. The only pair that jumped out at me was the d/p pair. Now I knew that it was at least feasible, as long as it wasn't necessary to duplicate any letters. Then I scanned the abbreviations to find ones that had a lot of common letters. The jan-jun-jul series looked like a good place to start: j a n u l was a good beginning but I realized right away that I had no room for duplicate letters and the second cube had both a and u so aug was going to be impossible. In fact I almost posted that answer. Then I realized that if Martin Gardner wrote about it, it must have a solution. :-) So I went back to the letter list. I don't put tails on my u's so it didn't strike me the first time through that n and u could be combined. Cube 1 Cube 2 Cube 3 j a n/u n/u l would let me get away with putting the g on the first cube to get aug, so I did. j a n/u g n/u l (1) Now came the fun part. The a was placed so I had to work around it for the other months that had an a in them (mar, apr, may). m a r d/p y (2) Now the d/p was placed so I had to work around that for sep and dec. This one was easy since they shared an e as well. d/p e s c (3) Now the e was placed so feb had to be worked in. f e b (4) The two months left (oct, nov) were far more complex. Not only did they have two "set" letters (c, n/u), there were two possible n/u's to be set with. That's why I left them for last. o t c n/u v (5) So now I had five pieces to fit together, so that no set would have more than six letters in it. Trial and error provided: j a n/u a b e g n/u l or, c d/p g r s m alphabetically: f l j y c d/p n/u m o e v t s n/u r o f b v t y Without some gimmick the days cannot be done. Because of the dates 11 and 22, there must be a 1 and a 2 on each cube. Thus there are 8 remaining spaces for the 8 remaining numbers, and because of 30, we put 3 and 0 on different cubes. I don't think the way you allocate the others matter. Now 6 numbers on each cube can produce at most 36 distinct pairs, and we need 31 distinct pairs to represent all possible dates. But since 3 each of {4,5,6,7,8,9} are on each cube, there are at least 9 representable numbers which can't be dates. Therefore there are at most 27 distinct numbers which are dates on the two cubes, and it can't be done. In particular, not all of {04,05,06,07,08,09} can be represented. The gimmick solution would be to represent the numbers in a stylised format (like say, on a digital clock or on a computer screen) such that the 6 can be turned upside down to be a 9. Then you can have 012 on both cubes, and three each of {3,4,5,6,7,8} on the other faces. Done. Example: 012468 012357 ==> geometry/circles.and.triangles.p <== Find the radius of the inscribed and circumscribed circles for a triangle. ==> geometry/circles.and.triangles.s <== Let a, b, and c be the sides of the triangle. Let s be the semiperimeter, i.e. s = (a + b + c) / 2. Let A be the area of the triangle, and let x be the radius of the incircle. Divide the triangle into three smaller triangles by drawing a line segment from each vertex to the incenter. The areas of the smaller triangles are ax/2, bx/2, and cx/2. Thus, A = ax/2 + bx/2 + cx/2, or A = sx. We use Heron's formula, which is A = sqrt(s(s-a)(s-b)(s-c)). This gives us x = sqrt((s-a)(s-b)(s-c)/s). The radius of the circumscribed circle is given by R = abc/4A. ==> geometry/coloring/cheese.cube.p <== A cube of cheese is divided into 27 subcubes. A mouse starts at one corner and eats through every subcube. Can it finish in the middle? ==> geometry/coloring/cheese.cube.s <== Give the subcubes a checkerboard-like coloring so that no two adjacent subcubes have the same color. If the corner subcubes are black, the cube will have 14 black subcubes and 13 white ones. The mouse always alternates colors and so must end in a black subcube. But the center subcube is white, so the mouse can't end there. ==> geometry/coloring/dominoes.p <== There is a chess board (of course with 64 squares). You are given 21 dominoes of size 3-by-1 (the size of an individual square on a chess board is 1-by-1). Which square on the chess board can you cut out so that the 21 dominoes exactly cover the remaining 63 squares? Or is it impossible? ==> geometry/coloring/dominoes.s <== !!!!!!!! !!!!!!!! !!!!!!!! ---***+* ---...+* ---*+O+* ---*+... ---*+*** There is only one way to remove a square, aside from rotations and reflections. To see that there is at most one way, do this: Label all the squares of the chessboard with A, B or C in sequence by rows starting from the top: ABCABCAB CABCABCA BCABCABC ABCABCAB CABCABCA BCABCABC ABCABCAB CABCABCA Every trimino must cover one A, one B and one C. There is one extra A square, so an A must be removed. Now label the board again by rows starting from the bottom: CABCABCA ABCABCAB BCABCABC CABCABCA ABCABCAB BCABCABC CABCABCA ABCABCAB The square removed must still be an A. The only squares that got marked with A both times are these: ........ ........ ..A..A.. ........ ........ ..A..A.. ........ ........ ==> geometry/construction/4.triangles.6.lines.p <== Can you construct 4 equilateral triangles with 6 toothpicks? ==> geometry/construction/4.triangles.6.lines.s <== Use the toothpicks as the edges of a tetrahedron. ==> geometry/construction/5.lines.with.4.points.p <== Arrange 10 points so that they form 5 rows of 4 each. ==> geometry/construction/5.lines.with.4.points.s <== Draw a 5 pointed star, put a point where any two lines meet. ==> geometry/construction/square.with.compass.p <== Construct a square with only a compass and a straight edge. ==> geometry/construction/square.with.compass.s <== Draw a circle (C1 at P1). Now draw a diameter D1 (intersects at P2 and P3). Set the compass larger than before. From points P2 and P3 draw another larger circle (C2 and C3). Where these two circles cross, draw a line (D2). This line should go the center of circle C1 at a rt angle to the original diameter line. This line should cross circle C1 at P4 and P5 Reset the compass to its original size. From P2 and P4 draw a circle (C4 and C5). These circles intersect at P6 and P1. Connect P6, P2, P1, P4 for a square. ==> geometry/cover.earth.p <== A thin membrane covers the surface of the earth. One square meter is added to the area of this membrane. How much is added to the radius and volume of this membrane? ==> geometry/cover.earth.s <== We know that V = (4/3)*pi*r^3 and A = 4*pi*r^2. We need to find out how much V increases if A increases by 1 m^2. dV / dr = 4 * pi * r^2 dA / dr = 8 * pi * r dV / dA = (dV / dr) / (dA / dr) = (4 * pi * r^2) / (8 * pi * r) = r/2 = 3,250,000 m If the area of the cover is increased by 1 square meter, then the volume it contains is increased by about 3.25 million cubic meters. We seem to be getting a lot of mileage out of such a small square of cotton. However, the new cover would not be very high above the surface of the planet -- about 6 nanometers (calculate dr/dA). ==> geometry/dissections/circle.p <== Can a circle be cut into similar pieces without point symmetry about the midpoint? Can it be done with a finite number of pieces? ==> geometry/dissections/circle.s <== Yes. Draw a circle inside the original circle, sharing a common point on the right. Now draw another circle inside the second, sharing a point at the left. Now draw another inside the third, sharing a point at the right. Continue in this way, coloring in every other region thus generated. Now, all the colored regions touch, so count this as one piece and the uncolored regions as a second piece. So the circle has been divided into two similar pieces and there is no point symmetry about the midpoint. Maybe it is cheating to call these single pieces, though. ==> geometry/dissections/hexagon.p <== Divide the hexagon into: 1) 3 indentical rhombuses. 2) 6 indentical kites(?). 3) 4 indentical trapezoids. 4) 8 indentcal shapes (any shape). 5) 12 identical shapes (any shape). ==> geometry/dissections/hexagon.s <== What is considered 'identical' for these questions? If mirror-image shapes are allowed, these are all pretty trivial. If not, the problems are rather more difficult... 1. Connect the center to every second vertex. 2. Connect the center to the midpoint of each side. 3. This is the hard one. If you allow mirror images, it's trivial: bisect the hexagon from vertex to vertex, then bisect with a perpendicular to that, from midpoint of side to midpoint of side. 4. This one's neat. Let the side length of the hexagon be 2 (WLOG). We can easily partition the hexagon into equilateral triangles with side 2 (6 of them), which can in turn be quartered into equilateral triangles with side 1. Thus, our original hexagon is partitioned into 24 unit equilateral triangles. Take the trapezoid formed by 3 of these little triangles. Place one such trapezoid on the inside of each face of the original hexagon, so that the long side of the trapezoid coincides with the side of the hexagon. This uses 6 trapezoids, and leaves a unit hexagon in the center as yet uncovered. Cover this little hexagon with two of the trapezoids. Voila. An 8-identical-trapezoid partition. 5. Easy. Do the rhombus partition in #1. Quarter each rhombus by connecting midpoints of opposite sides. This produces 12 small rhombi, each of which is equivalent to two adjacent small triangles as in #4. Except for #3, all of these partitions can be achieved by breaking up the hexagon into unit equilateral triangles, and then building these into the shapes desired. For #3, though, this would require (since there are 24 small triangles) trapezoids formed from 6 triangles each. The only trapezoid that can be built from 6 identical triangles is a parallelogram; I assume that the poster wouldn't have asked for a trapezoid if you could do it with a special case of trapezoid. At any rate, that parallelogram doesn't work. ==> geometry/dissections/square.70.p <== Since 1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2, can a 70x70 sqaure be dissected into 24 squares of size 1x1, 2x2, 3x3, etc.? ==> geometry/dissections/square.70.s <== Martin Gardner asked this in his Mathematical Games column in the September 1966 issue of Scientific American. William Cutler was the first of 24 readers who reduced the uncovered area to 49, using all but the 7x7 square. All the patterns were the same except for interchanging the squares of orders 17 and 18 and rearranging the squares of orders 1, ..., 6, 8, 9, and 10. Nobody proved that the solution is minimal. +----------------+-------------+----------------------+---------------------+ ! ! ! ! ! ! ! ! ! ! ! ! 11 ! ! ! ! ! ! ! ! ! 16 ! ! ! ! ! +-----+--+----+ 22 ! 21 ! ! ! ! 2! ! ! ! ! ! 5 +--+----+ ! ! ! ! ! ! ! ! +----------------+--+--+ 6 ! ! ! ! ! 3! ! ! ! ! ++-+-------+ ! ! ! !! ! ++--------------------+ ! !! 8 +----------------------++ ! ! 18 !! ! ! ! ! !! ! ! ! ! ++---------+ ! ! ! ! ! ! 20 ! ! ! 9 ! ! ! +------------------++ ! 23 ! ! ! !! ! ! ! ! ++----------+ ! ! ! ! ! +---++---------------+ ! ! ! ! !! ! ! 17 ! 10 ! ! 4 !! ! ! ! +---------------+-------+---++ ! ! +-+---------+---------------+ ! 15 ! ! ! ! ! ! ! ! ! ! ! 12 ! ! +------------------+-+ ! +-+-------------+ ! ! ! !1! ! ! ! +------------+-+ ! ! ! 24 ! ! ! ! ! ! ! ! ! 19 ! ! 13 ! 14 ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! +--------------------+-------------------------+--------------+-------------+ ==> geometry/dissections/square.five.p <== Can you dissect a square into 5 parts of equal area with just a straight edge? ==> geometry/dissections/square.five.s <== 1. Prove you can reflect points which lie on the sides of the square about the diagonals. 2. Construct two different rectangles whose vertices lie on the square and whose sides are parallel to the diagonals. 3. Construct points A, A', B, B' on one (extended) side of the square such that A/A' and B/B' are mirror image pairs with respect to another side of the square. 4. Construct the mirror image of the center of the square in one of the sides. 5. Divide the original square into 4 equal squares whose sides are parallel to the sides of the original square. 6. Divide one side of the square into 8 equal segments. 7. Construct a trapezoid in which one base is a square side and one base is 5/8 of the opposite square side. 8. Divide one side of the square into 5 equal segments. 9. Divide the square into 5 equal rectangles. ==> geometry/duck.and.fox.p <== A duck is swimming about in a circular pond. A ravenous fox (who cannot swim) is roaming the edges of the pond, waiting for the duck to come close. The fox can run faster than the duck can swim. In order to escape, the duck must swim to the edge of the pond before flying away. Assume that the duck can't fly until it has reached the edge of the pond. How much faster must the fox run that the duck swims in order to be always able to catch the duck? ==> geometry/duck.and.fox.s <== Assume the ratio of the fox's speed to the duck's is a, and the radius of the pond is r. The duck's best strategy is: 1. Swim around a circle of radius (r/a - delta) concentric with the pond until you are diametrically opposite the fox (you, the fox, and the center of the pond are colinear). 2. Swim a distance delta along a radial line toward the bank opposite the fox. 3. Observe which way the fox has started to run around the circle. Turn at a RIGHT ANGLE in the opposite direction (i.e. if you started swimming due south in step 2 and the fox started running to the east, i.e. clockwise around the pond, then start swimming due west). (Note: If at the beginning of step 3 the fox is still in the same location as at the start of step 2, i.e. directly opposite you, repeat step 2 instead of turning.) 4. While on your new course, keep track of the fox. If the fox slows down or reverses direction, so that you again become diametrically opposite the fox, go back to step 2. Otherwise continue in a straight line until you reach the bank. 5. Fly away. The duck should make delta as small as necessary in order to be able to escape the fox. The key to this strategy is that the duck initially follows a radial path away from the fox until the fox commits to running either clockwise or counterclockwise around the pond. The duck then turns onto a new course that intersects the circle at a point MORE than halfway around the circle from the fox's starting position. In fact, the duck swims along a tangent of the circle of radius r/a. Let theta = arc cos (1/a) then the duck swims a path of length r sin theta + delta but the fox has to run a path of length r*(pi + theta) - a*delta around the circle. In the limit as delta goes to 0, the duck will escape as long as r*(pi + theta) < a*r sin theta that is, pi + arc cos (1/a) - a * sqrt(a^2 - 1) < 0 Maximize a in the above: a = 4.6033388487517003525565820291030165130674... The fox can catch the duck as long as he can run about 4.6 times as fast as the duck can swim. "But wait," I hear you cry, "When the duck heads off to that spot 'more than halfway' around the circle, why doesn't the fox just double back? That way he'll reach that spot much quicker." That is why the duck's strategy has instructions to repeat step 2 under certain circumstances. Note that at the end of step 2, if the fox has started to run to head off the duck, say in a clockwise direction, he and the duck are now on the same side of some diameter of the circle. This continues to be true as long as both travel along their chosen paths at full speed. But if the fox were now to try to reach the duck's destination in a counterclockwise direction, then at some instant he and the duck must be on a diameter of the pond. At that instant, they have exactly returned to the situation that existed at the end of step 1, except that the duck is a little closer to the edge than she was before. That's why the duck always repeats step 2 if the fox is ever diametrically opposite her. Then the fox must commit again to go one way or the other. Every time the fox fails to commit, or reverses his commitment, the duck gets a distance delta closer to the edge. This is a losing strategy for the fox. The limiting ratio of velocities that this strategy works against cannot be improved by any other strategy, i.e., if the ratio of the duck's speed to the fox's speed is less than a then the duck cannot escape given the best fox strategy. Given a ratio R of speeds less than the above a, the fox is sure to catch the duck (or keep it in water indefinitely) by pursuing the following strategy: Do nothing so long as the duck is in a radius of R around the centre. As soon as it emerges from this circle, run at top speed around the circumference. If the duck is foolish enough not to position itself across from the center when it comes out of this circle, run "the short way around", otherwise run in either direction. To see this it is enough to verify that at the circumference of the circle of radius R, all straight lines connecting the duck to points on the circumference (in the smaller segment of the circle cut out by the tangent to the smaller circle) bear a ratio greater than R with the corresponding arc the fox must follow. That this is enough follows from the observation that the shortest curve from a point on a circle to a point on a larger concentric circle (shortest among all curves that don't intersect the interior of the smaller circle) is either a straight line or an arc of the smaller circle followed by a tangential straight line. ==> geometry/earth.band.p <== How much will a band around the equator rise above the surface if it is made one meter longer? ==> geometry/earth.band.s <== The formula for the circumference of a circle is 2 * pi * radius. Therefore, if you increase the circumference by 1 meter, you increase the radius by 1/(2 * pi) meters, or about 0.16 meters. ==> geometry/ham.sandwich.p <== Consider a ham sandwich, consisting of two pieces of bread and one of ham. Suppose the sandwich was dropped into a machine and spindled, torn and mutiliated. Is it still possible to divide the ham sandwich with a straight knife cut such that both the ham and the bread are divided in two parts of equal volume? ==> geometry/ham.sandwich.s <== Yes. There is a theorem in topology called the Ham Sandwich Theorem, which says: Given 3 (finite) volumes (each may be of any shape, and in several pieces), there is a plane that cuts each volume in half. One would learn about it typically in a first course in algebraic topology, or maybe in a course on introductory topology (if you studied the fundamental group). ==> geometry/hike.p <== You are hiking in a half-planar woods, exactly 1 mile from the edge, when you suddenly trip and lose your sense of direction. What's the shortest path that's guaranteed to take you out of the woods? Assume that you can navigate perfectly relative to your current location and (unknown) heading. ==> geometry/hike.s <== Go 2/sqrt(3) away from the starting point, turn 120 degrees and head 1/sqrt(3) along a tangent to the unit circle, then traverse an arc of length 7*pi/6 along this circle, then head off on a tangent 1 mile. This gives a minimum of sqrt(3) + 7*pi/6 + 1 = 6.397... It remains to prove this is the optimal answer. ==> geometry/hole.in.sphere.p <== Old Boniface he took his cheer, Then he bored a hole through a solid sphere, Clear through the center, straight and strong, And the hole was just six inches long. Now tell me, when the end was gained, What volume in the sphere remained? Sounds like I haven't told enough, But I have, and the answer isn't tough] ==> geometry/hole.in.sphere.s <== The volume of the leftover material is equal to the volume of a 6" sphere. First, lets look at the 2 dimensional equivalent of this problem. Two concentric circles where the chord of the outer circle that is tangent to the inner circle has length D. What is the area of the "doughnut" area between the circles? It is pi * (D/2)^2. The same area as a circle with that diameter. Proof: big circle radius is R little circle radius is r 2 2 area of donut = pi * R - pi * r 2 2 = pi * (R - r ) Draw a right triangle and apply the Pythagorean Theorem to see that 2 2 2 R - r = (D/2) so the area is 2 = pi * (D/2) Start with a sphere of radius R (where R > 6"), drill out the 6" high hole. We will now place this large "ring" on a plane. Next to it place a 6" high sphere. By Archemedes' theorem, it suffices to show that for any plane parallel to the base plane, the cross- sectional area of these two solids is the same. Take a general plane at height h above (or below) the center of the solids. The radius of the circle of intersection on the sphere is radius = srqt(3^2 - h^2) so the area is pi * ( 3^2 - h^2 ) For the ring, once again we are looking at the area between two concentric circles. The outer circle has radius sqrt(R^2 - h^2), The area of the outer circle is therefore pi (R^2 - h^2) The inner circle has radius sqrt(R^2 - 3^2). So the area of the inner circle is pi * ( R^2 - 3^2 ) the area of the doughnut is therefore pi(R^2 - h^2) - pi( R^2 - 3^2 ) = pi (R^2 - h^2 - R^2 + 3^2) = pi (3^2 - h^2) Therefore the areas are the same for every plane intersecting the solids. Therefore their volumes are the same. QED ==> geometry/ladders.p <== Two ladders form a rough X in an alley. The ladders are 11 and 13 meters long and they cross 4 meters off the ground. How wide is the alley? ==> geometry/ladders.s <== Ladders 1 and 2, denoted L1 and L2, respectively, will rest along two walls (taken to be perpendicular to the ground), and they will intersect at a point O = (a,s), a height s from the ground. Find the largest s such that this is possible. Then find the width of the alley, w = a+b, in terms of L1, L2, and s. This diagram is not to scale. B D !\ L1 L2 /! ! \ / ! BC = length of L1 ! \ / ! AD = length of L2 ! \ O / ! s = height of intersection x! \ / !y A = (0,0) ! /!\ ! AE = a ! m / ! \ n ! EC = b ! / !s \ ! AO = m ! / ! \ ! CO = n !/________!________\! (0,0) = A a E b C ----------------------------------------------------------------------------- Without loss of generality, let L2 >= L1. Observe that triangles AOB and DOC are similar. Let r be the ratio of similitude, so that x=ry. Consider right triangles CAB and ACD. By the Pythagorean theorem, L1^2 - x^2 = L2^2 - y^2. Substituting x=ry, this becomes y^2(1-r^2) = L2^2 - L1^2. Letting L= L2^2 -L1^2 (L>=0), and factoring, this becomes (*) y^2 (1+r)(1-r) = L Now, because parallel lines cut L1 (a transversal) in proportion, r = x/y = (L1-n)/n, and so L1/n = r+1. Now, x/s = L1/n = r+1, so ry = x = s(r+1). Solving for r, one obtains the formula r = s/(y-s). Substitute this into (*) to get (**) y^2 (y) (y-2s) = L (y-s)^2 NOTE: Observe that, since L>=0, it must be true that y-2s>=0. Now, (**) defines a fourth degree polynomial in y. It can be written in the form (by simply expanding (**)) (***) y^4 - 2s_y^3 - L_y^2 + 2sL_y - Ls^2 = 0 L1 and L2 are given, and so L is a constant. How large can s be? Given L, the value s=k is possible if and only if there exists a real solution, y', to (***), such that 2k <= y' < L2. Now that s has been chosen, L and s are constants, and (***) gives the desired value of y. (Make sure to choose the value satisfying 2s <= y' < L2. If the value of s is "admissible" (i.e., feasible), then there will exist exactly one such solution.) Now, w = sqrt(L2^2 - y^2), so this concludes the solution. L1 = 11, L2 = 13, s = 4. L = 13^2-11^2 = 48, so (***) becomes y^4 - 8_y^3 - 48_y^2 + 384_y - 768 = 0 Numerically find root y ~= 9.70940555, which yields w ~= 8.644504. ==> geometry/lattice/area.p <== Prove that the area of a triangle formed by three lattice points is integer/2. ==> geometry/lattice/area.s <== The formula for the area is A = ! x1*y2 + x2*y3 + x3*y1 - x1*y3 - x2*y1 - x3*y2 ! / 2 If the xi and yi are integers, A is of the form (integer/2) ==> geometry/lattice/equilateral.p <== Can an equlateral triangle have vertices at integer lattice points? ==> geometry/lattice/equilateral.s <== No. Suppose 2 of the vertices are (a,b) and (c,d), where a,b,c,d are integers. Then the 3rd vertex lies on the line defined by (x,y) = 1/2 (a+c,b+d) + t ((d-b)/(c-a),-1) (t any real number) and since the triangle is equilateral, we must have !!t ((d-b)/(c-a),-1)!! = sqrt(3)/2 !!(c,d)-(a,b)!! which yields t = +/- sqrt(3)/2 (c-a). Thus the 3rd vertex is 1/2 (a+c,b+d) +/- sqrt(3)/2 (d-b,a-c) which must be irrational in at least one coordinate. ==> geometry/rotation.p <== What is the smallest rotation that returns an object to its original state? ==> geometry/rotation.s <== 720 degrees. Objects are made of bosons (integer-spin particles) and fermions (half-odd-integer spin particles), and the wave function of a fermion changes sign upon being rotated by 360 degrees. To get it back to its original state you must rotate by another 360 degrees, for a total of 720 degrees. This fact is the basis of Fermi-Dirac statistics, the Pauli Exclusion Principle, electron orbits, chemistry, and life. Mathematically, this is due to the continuous double cover of SO(2) by SO(3), where SO(2) is the internal symmetry group of fermions and SO(3) is the group of rotations in three dimensional space. You can demonstrate this with a tray, which you hold in your right hand with the arm lowered, then rotate twice as you raise your arm and end up with the tray above your head, rotated twice about its vertical axis, but without having twisted your arm. Also, by attaching strings to a sphere, it is possible to see that a 360 degree rotation will entangle the strings, which another 360 degree rotation will disentangle. Hospitals have machines which take out your blood, centrifuge it to take out certain parts, then return it to your veins. Because of AIDS they must never let your blood touch the inside of the machine which has touched others' blood. So the inside is lined with a single piece of disposable branched plastic tubing. This tube must rotate rapidly in the centrifuge where several branches come out. Thus the tube should twist and tangle up the branches. But the machine untwists the branches as in the above discussion. At several hundred rounds per minute] References P. A. M. Dirac's "scissors demonstration" R. Penrose and W. Rindler Spinors and Space-time, vol. 1, p. 43 Cambridge University Press, 1984, R. Feynman and S. Weinberg Elementary Particles and the Laws of Physics, p. 29 Cambridge University Press, 1987 ==> geometry/smuggler.p <== Somewhere on the high sees smuggler S is attempting, without much luck, to outspeed coast guard G, whose boat can go faster than S's. G is one mile east of S when a heavy fog descends. It's so heavy that nobody can see or hear anything further than a few feet. Immediately after the fog descends, S changes course and attempts to escape at constant speed under a new, fixed course. Meanwhile, G has lost track of S. But G happens to know S's speed, that it is constant, and that S is sticking to some fixed heading, unknown to G. How does G catch S? G may change course and speed at will. He knows his own speed and course at all times. There is no wind, G does not have radio or radar, there is enough space for maneuvering, etc. ==> geometry/smuggler.s <== One way G can catch S is as follows (it is not the fastest way). G waits until he knows that S has traveled for one mile. At that time, both S and G are somewhere on a circle with radius one mile, and with its center at the original position of S. G then begins to travel with a velocity that has a radially outward component equal to that of S, and with a tangential component as large as possible, given G's own limitation of total speed. By doing so, G and S will always both be on an identical circle having its center at the original position of S. Because G has a tangential component whereas S does not, G will always catch S (actually, this is not proven until you solve the o.d.e. associated with the problem). If G can go at 40 mph and S goes at 20 mph, you can work out that it will take G at most 1h 49m 52s to catch S. On average, G will catch S in: ( -2pi + sqrt(3) ( exp(2pi/sqrt(3)) - 1 )) / 40pi hours, which is, 27 min and 17 sec. ==> geometry/table.in.corner.p <== Put a round table into a (perpendicular) corner so that the table top touches both walls and the feet are firmly on the ground. If there is a point on the perimeter of the table, in the quarter circle between the two points of contact, which is 10 cm from one wall and 5 cm from the other, what's the diameter of the table? ==> geometry/table.in.corner.s <== Consider the +X axis and the +Y axis to be the corner. The table has radius r which puts the center of the circle at (r,r) and makes the circle tangent to both axis. The equation of the circle (table's perimeter) is (x-r)^2 + (y-r)^2 = r^2 . This leads to r^2 - 2(x+y) + x^2 + y^2 = 0 Using x = 10, y = 5 we get the solutions 25 and 5. The former is the radius of the table. It's diameter is 50 cm. The latter number is the radius of a table that has a point which satisfies the conditions but is on the outside edge of the table. ==> geometry/tesseract.p <== If you suspend a cube by one corner and slice it in half with a horizontal plane through its centre of gravity, the section face is a hexagon. Now suspend a tesseract (a four dimensional hypercube) by one corner and slice it in half with a hyper-horizontal hyperplane through its centre of hypergravity. What is the shape of the section hyper-face? ==> geometry/tesseract.s <== The 4-cube is the set of all points in Ý-1,1¨^4 . The hyperplane { (x,y,z,w) : x + y + z + w = 0 } cuts the 4-cube in the desired manner. Now, { (.5,.5,-.5,-.5), (.5,-.5,.5,-.5), (.5,-.5,-.5,.5) } is an orthonormal basis for the hyperplane. Let (a,b,c) be a point on the hyperplane with respect to this basis. (a,b,c) is in the 4-cube if and only if !a! + !b! + !c! <= 2. The shape of the intersection is a regular octahedron. ==> geometry/tetrahedron.p <== Suppose you have a sphere of radius R and you have four planes that are all tangent to the sphere such that they form an arbitrary tetrahedron (it can be irregular). What is the ratio of the surface area of the tetrahedron to its volume? ==> geometry/tetrahedron.s <== For each face of the tetrahedron, construct a new tetrahedron with that face as the base and the center of the sphere as the fourth vertex. Now the original tetrahedron is divided into four smaller ones, each of height R. The volume of a tetrahedron is Ah/3 where A is the area of the base and h the height; in this case h=R. Combine the four tetrahedra algebraically to find that the volume of the original tetrahedron is R/3 times its surface area. ==> geometry/tiling/rational.sides.p <== A rectangular region R is divided into rectangular areas. Show that if each of the rectangles in the region has at least one side with rational length then the same can be said of R. ==> geometry/tiling/rational.sides.s <== "Fourteen proofs of a result about tiling a rectangle" (Stan Wagon) _The American Mathematical Monthly_, Aug-Sep 1987, Vol 94 #7. There was also a fifteenth proof published a few issues later, attributed to a (University of Kentucky?) student. ==> geometry/tiling/rectangles.with.squares.p <== Given two sorts of squares, (axa) and (bxb), what rectangles can be tiled? ==> geometry/tiling/rectangles.with.squares.s <== A rectangle can be tiled with (axa) and (bxb) squares, iff (i) gcd(a,b)=1 , and any of the following hold: either: both sides of the rectangle are multiples of a; or: both sides of the rectangle are multiples of b; or: one side is a multiple of (ab), and the other is any length EXCEPT one of a finite number of "bad" lengths: those numbers which are NOT positive integer combinations of a & b. { By Sylvester's theorem there are (a-1)(b-1)/2 of these, the largest being (a-1)(b-1)-1. } (ii) gcd(a,b) = d . Then merely apply (i) to the problem with a,b replaced by a/d, b/d and the rectangle lengths also divided by d. i.e. all cells must appear in (dxd) subsquares. ------ PROOF It is clear that (ii) follows from (i), and that simple constructions give the "if" part of (i). For the "only if" part, we prove that... (S) If one side of the rectangle is not divisible by a, and the other is not divisible by b, then the tiling is impossible. The results in (i) follow immediately from (S). To prove (S): ( Chakraborty-Hoey style ). ~~~~~~~~~~~~~~~~ Let the width of the rectangle be a NON-(a-multiple). Then the number of bxb squares starting (i.e. top edge) at row 1 must be a NON-a-multiple. Thus the number of bxb starting at row 2 must BE an a-multiple. Similarly for the number starting at rows 3,4,...,b . Then the number starting at row (b+1) must be a NON-a-multiple again. Similarly the number starting at rows (2b+1), (3b+1), (4b+1),... must all be non-a-multiples. So if the number of rows is NOT a multiple of b, (call it bx+r), then row (bx+1) must have a NON-a-multiple of bxb squares starting there, i.e. at least one, and there is no room left to squeeze it in. ÝQED¨ ---- A Rickard-style proof of (S) is ..BBB....BBWWW...WBBB....BBWWW...W(..etc) ~~~~~~~ also possible, by ..BBB....BBWWW...WBBB....BBWWW...W coloring the rectangle in ..BBB....BBWWW...WBBB....BBWWW...W vertical strips as shown here: <- a ->< b-a ><- a ->< b-a > Every square tile covers an a-multiple of black squares. But if the width is a NON-b-multiple, and the number of rows is a NON-a-multiple, then there are a NON-a-multiple of black squares in total. ÝQED¨ (Note: the coloring must have 1 column of blacks on the right, and any ==== spare columns of whites on the left.) =================== Bill Taylor. wft@math.canterbury.ac.nz >A Rickard-style proof of (S) is ..BBB....BBWWW...WBBB....BBWWW...W(..etc) > ~~~~~~~ also possible, by ..BBB....BBWWW...WBBB....BBWWW...W >coloring the rectangle in ..BBB....BBWWW...WBBB....BBWWW...W >vertical strips as shown here: <- a ->< b-a ><- a ->< b-a > > >Every square tile covers an a-multiple of black squares. But if the width >is a NON-b-multiple, and the number of rows is a NON-a-multiple, then there >are a NON-a-multiple of black squares in total. ÝQED¨ > >(Note: the coloring must have 1 column of blacks on the right, and any > ==== spare columns of whites on the left.) This statement of how to position the colouring isn't good enough, I'm afraid. Take a=4, b=7 and consider e.g. a 19x10 rectangle. Coloured your way, you get: BWWWBBBBWWWBBBBWWWB BWWWBBBBWWWBBBBWWWB ::::::::::::::::::: BWWWBBBBWWWBBBBWWWB BWWWBBBBWWWBBBBWWWB The result has 10*10=100 black squares in it, which *is* a multiple of a=4, despite the fact that 19 is not a multiple of 7 and 10 is not a multiple of 4. Of course, there is an alternative offset for the pattern that does give you the result you want: WWBBBBWWWBBBBWWWBBB WWBBBBWWWBBBBWWWBBB ::::::::::::::::::: WWBBBBWWWBBBBWWWBBB WWBBBBWWWBBBBWWWBBB To show this happens in general: because the width of the rectangle is a non-multiple of b, it is possible to position it on the pattern so that the leftmost column in the rectangle is white and the column just right of the right edge of the rectangle is black. Suppose N columns are black with this positioning. Then the rectangle contains N*H black cells, where H is the height of the rectangle. If we then shift the rectangle right by one, the number of black columns increases by 1 and it contains (N+1)*H black cells. The difference between these two numbers of black cells is H, which is not a multiple of a. Therefore N*H and (N+1)*H cannot both be multiples of a, and so one of these two positionings of the pattern will suit your purposes. David Seal dseal@armltd.co.uk ==> geometry/tiling/scaling.p <== A given rectangle can be entirely covered (i.e. concealed) by an appropriate arrangement of 25 disks of unit radius. Can the same rectangle be covered by 100 disks of 1/2 unit radius? ==> geometry/tiling/scaling.s <== Yes. The same configuration of circles, when every distance is reduced by half (including the diameters), will cover a similar rectangle whose sides are one half of the original one. The original rectangle is the union of four such rectangles. ==> geometry/tiling/seven.cubes.p <== Consider 7 cubes of equal size arranged as follows. Place 5 cubes so that they form a Swiss cross or a + (plus). ( 4 cubes on the sides and 1 in the middle). Now place one cube on top of the middle cube and the seventh below the middle cube, to effectively form a 3-dimensional swiss cross. Can a number of such blocks (of 7 cubes each) be arranged so that they are able to completely fill up a big cube (say 10 times the size of the small cubes)? It is all right if these blocks project out of the big cube, but there should be no holes or gaps. ==> geometry/tiling/seven.cubes.s <== Let n be a positive integer. Define the function f from Z^n to Z by f(x) = x_1+2x_2+3x_3+...+nx_n. For x in Z^n, say y is a neighbor of x if y and x differ by one in exactly one coordinate. Let S(x) be the set consisting of x and its 2n neighbors. It is easy to check that the values of f(y) for y in S(x) are congruent to 0,1,2,...,2n+1 (mod 2n+1) in some order. Using this, it is easy to check that every y in Z^n is a neighbor of one and only one x in Z^n such that f(x) is congruent to 0 (mod 2n+1). So Z^n can be tiled by clusters of the form S(x), where f(x) is congruent to 0 mod 2n+1. ==> group/group.01.p <== AEFHIKLMNTVWXYZ BCDGJOPQRSU ==> group/group.01.s <== AEFHIKLMNTVWXYZ drawn with straight lines BCDGJOPQRSU not drawn with straight lines ==> group/group.01a.p <== 147 0235689 ==> group/group.01a.s <== 147 drawn with straight lines 0235689 not drawn with straight lines ==> group/group.02.p <== ABEHIKMNOPTXZ CDFGJLQRSUVWY ==> group/group.02.s <== ABEHIKMNOPTXZ resembles Greek letter CDFGJLQRSUVWY does not resemble Greek letter ==> group/group.03.p <== BEJQXYZ DFGHLPRU KSTV CO AIW MN ==> group/group.03.s <== BEJQXYZ no state starting with this letter DFGHLPRU one state starting with this letter KSTV two states starting with this letter CO three states starting with this letter AIW four states starting with this letter five states starting with this letter six states starting with this letter seven states starting with this letter MN eight states starting with this letter ==> group/group.04.p <== BDO P ACGJLMNQRSUVWZ EFTY HIKX ==> group/group.04.s <== BDO no endpoint P one endpoint ACGJLMNQRSUVWZ two endpoints EFTY three endpoints HIKX four endpoints ==> group/group.05.p <== CEFGHIJKLMNSTUVWXYZ ADOPQR B ==> group/group.05.s <== CEFGHIJKLMNSTUVWXYZ no enclosed area ADOPQR one enclosed area B two enclosed areas ==> group/group.06.p <== BCEGKMQSW DFHIJLNOPRTUVXYZ ==> group/group.06.s <== BCEGKMQSW prime numbers DFHIJLNOPRTUVXYZ composi ==> induction/hanoi.p <== Is there an algorithom for solving the hanoi tower puzzle for any number of towers? Is there an equation for determining the minimum number of moves required to solve it, given a variable number of disks and towers? ==> induction/hanoi.s <== The best way of thinking of the Towers of Hanoi problem is inductively. To move n disks from post 1 to post 2, first move (n-1) disks from post 1 to post 3, then move disk n from post 1 to post 2, then move (n-1) disks from post 3 to post 2 (same procedure as moving (n-1) disks from post 1 to post 3). In order to figure out how to move (n-1) disks from post 1 to post 3, first move (n-2) disks . . . . As far as an algorithm which straightens out any legal position is concerned, the algorithm would go something like this: 1. Find the smallest disk. Call the post that it's on post 1. 2. Find the smallest disk which is not on post 1. This disk is, say, size k. (I am calling the smallest disk size 1 here.) Call the post that disk k is on post 2. Disks 1 through (k-1) are then all stacked up correctly on post 1 disk k is on top of post 2. This follow from the fact that the disks are in a legal postition. 3. Move disks 1 through (k-1) from post 1 to post 2, ignoring the other disks. This is just the standard Tower of Hanoi problem for (k-1) disks. 4. If the disks are not yet correctly arranged, repeat from step 1. In fact, this gives the straightening with the fewest number of moves. With 3 towers, for N number of disks, the formula for the minimum number of moves to complete the puzzle correctly is: (2^N) - 1 This bit of ancient folklore was invented by De Parville in 1884. ``In the great temple at Benares, says he, beneath the dome which marks the centre of the world, rests a brass plate in which are fixed three diamond needles, each a cubit high and as thick as the body of a bee. On one of these needles, at the creation, God placed sixty-four discs of pure gold, the largest disc resting on the brass plate, and the others getting smaller and smaller up to the top one. This is the Tower of Bramah. Day and night unceasingly the priests transfer the discs from one diamond needle to another according to the fixed and immutable laws of Bramah, which require that the priest on duty must not move more than one disc at a time and that he must place this disc on a needle so that there is no smaller disc below it. When the sixty-four discs shall have been thus transferred from the needle on which at the creation God placed them to one of the other needles, tower, temple, and Brahmins alike will crumble into dust, and with a thunderclap the world will vanish.'' (W W R Ball, MATHEMATICAL RECREATIONS AND ESSAYS, p. 304) This has been discussed by several authors, e.g. Er, Info Sci 42 (1987) 137-141. Graham, Knuth and Patashnik, _Concrete_Mathematics_. There are many papers claiming to solve this, and they are probably all correct but they rely on the unproven "Frame's conjecture". In particular, for the 4 peg case the conjecture states that an optimal solution begins by forming a substack of the k smallest discs, then moving the rest, and then moving those k again; k to be determined. Here is a extensible bc program that does the same work. The output format is not that great. We get 300 numbers as output. The first hundred represent N, the next 100 represent f(N) and the last hundred represent i, which is the number of discs to move to tmp1 using f(N). For convenience, I have here some values for N <= 48. Enjoy. Sharma N 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 f(N) 1 3 5 9 13 17 25 33 41 49 65 81 97 113 129 161 193 225 257 i 0 1 1 2 2 3 3 4 5 6 6 7 8 9 10 10 11 12 13 N 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 f(N) 289 321 385 449 513 577 641 705 769 897 1025 1153 1281 1409 1537 1665 i 14 15 15 16 17 18 19 20 21 21 22 23 24 25 26 27 N 36 37 38 39 40 41 42 43 44 45 46 47 48 f(N) 1793 2049 2305 2561 2817 3073 3329 3585 3841 4097 4609 5121 5633 i 28 28 29 30 31 32 33 34 35 36 36 37 38 /* This is the bc program that gives f(N) for 4 peg case */ w = 101; /* This represents the number of disks */ mÝ0¨ = 0; mÝ1¨ = 1; mÝ2¨ = 3; mÝ3¨ = 5; mÝ4¨ = 9; mÝ5¨ = 13; mÝ6¨ = 17; /* f(n) is the function that gives the min # of moves for 4 peg case */ define f(n) { return (mÝn¨); } /* g(n) is the function that fives the min # of moves for 3 peg case */ define g(n) { return (2^n - 1); } /* x(n) is the Optimization Routine */ define x(n) { auto j auto r auto i if(n == 1) return (1); j = f(1) + g(n-1); for(i = 2; i < n; i++) { r = f(i) + g(n-i); if(r < j) { j = r; d = i; } } return (j); } /* main program */ for(q = 4; q < w; q++) { t = x(q-1); mÝq¨ = 2 * t + 1; dÝq¨ = d; }; /*This for loop prints the number of discs from 1 <= n <= w*/ for(q = 1; q < w; q++) { q; } /*This for loop prints f(n) for 1 <= n <= w */ for(q = 1; q < w; q++) { mÝq¨; } /*This for loop prints i for 1 <= n <= w i represents the number of disks to be moved to tmp location using f(n) N-i-1 will be moved using g(n) */ for(q = 1; q < w; q++) { dÝq¨; } -- sharma@sharma.warren.mentorg.com ==> induction/n-sphere.p <== With what odds do three random points on an n-sphere form an acute triangle? ==> induction/n-sphere.s <== Select three points a, b, and c, randomly with respect to the surface of an n-sphere. These three points determine a fourth, x, which is the intersection of the sphere with the axis perpendicular to the abc plane. (Choose the pole nearest the plane.) I could have, just as easily, selected x, a distance d from x, and three points d units away from x. The distribution of d is not uniform, but that's ok. For every x and d, the three points abc form an acute triangle with probability pÝn-1¨. By induction, pÝn¨ = 1/4. ==> induction/paradox.p <== What simple property holds for the first 10,000 integers, then fails? ==> induction/paradox.s <== Consider the sequences defined by: s(1) = a; s(2) = b; s(n) = least integer such that s(n)/s(n-1) > s(n-1)/s(n-2). In other words, s(n) = 1+floor(s(n-1)^2/s(n-2)) for n >= 3. These sequences are similar in some ways to the classically-studied Pisot sequences. For example, if a = 1, b = 2, then we get the odd-indexed Fibonacci numbers. D. Boyd of UBC, an expert in Pisot sequences, pointed out the following. If we let a = 8, b = 55 in the definition above, then the resulting sequence s(n) appears to satisfy the following linear recurrence of order 4: s(n) = 6s(n-1) + 7s(n-2) - 5s(n-3) - 6s(n-4) Indeed, it does satisfy this linear recurrence for the first 11,056 terms. However, it fails at the 11,057th term] And s(11057) is a 9270 digit number. (The reason for this coincidence depends on a remarkable fact about the absolute values of the roots of the polynomial x^4 - 6x^3 - 7x^2 + 5x + 6.) ==> induction/party.p <== You're at a party. Any two (different) people at the party have exactly one friend in common (the friend is also at the party). Prove that there is at least one person at the party who is a friend of everyone else. Assume that the friendship relation is symmetric and not reflexive. ==> induction/party.s <== Here is an easy solution by induction. Let P be the set of people in the party, and n the size of P. If n=2 or 3, the result is trivial. Suppose now that n>3 and that the result is true for n-1. For any two distinct x,y in P, write x & y to mean that `x is a friend of y', and x ~& y to mean that `x is not a friend of y'. Take q in P. The hypothesis on the relation & is still satisfied on P-{q}; by induction, the result is thus true for P-{q}, and there is some p in P-{q} such that p & x for any x in P-{p,q}. We have two cases: a) p & q. Then the result holds for P with p. b) p ~& q. By hypothesis, there is a unique r in P-{p,q} such that p & r & q. For any x in P-{p,q}, if x & q, then p & x & q, and so x=r. Thus r is the unique friend of q. Now for any s in P-{q,r} there exists some x such that s & x & q, and so x=r. This means that r & s for any s in P-{q,r}, and as r & q, the results holds in P with r. The problem can also be solved by applying the spectral theory of graphs (see for instance Bollobas' excellent book, _Extremal Graph Theory_). The problem's condition is vacuous if there is only N=1 person at the "party", impossible if N=2 (If you aren't your own friend, nor I mine, somebody *else* must be our mutual friend), and trivial if N=3 (everybody must be everyone else's friend). Henceforth assume N>3. Let A,B be two friends, and C their mutual friend. Let a be the number of A's friends other than B and C, and likewise b,c. Each of A's friends is also friendly with exactly one other of A's friends, and with none of B and C's other friends (if A1,B1 are friends of A,B resp. and of each other then A1 and B have more than one mutual friend); likewise for B and C. Let M=N-(a+b+c+3) be the number of people not friendly with any of A,B,C. Each of them is friendly with exactly one of A's and one of B's friends; and each pair of a friend of A and a friend of B must have exactly one of them as a mutual friend. Thus M=ab; likewise M=ab=ac=bc. Thus either M and two of a,b,c vanish, or a=b=c=k (say), M=k^2, and N=k^3+3k+3. In the first case, say b=c=0; necessarily a is even, and A is a friend of everybody else at the party, each of whom is friendly with exactly one other person; clearly any such configuration (a graph of k/2+1 triangles with a common vertex) satisfies the problem's conditions). It remains to show that the second case is impossible. Since N=k^2+3k+3 does not depend on A,B,C, neither does k, and it quickly follows that the party's friendship graph is regular with reduced matrix Ý 0 k+2 0 ¨ Ý 1 1 k ¨ Ý 0 1 k+1 ¨ and eigenvalues k+2 and +-sqrt(k+1) and multiplicities 1,m1,m2 for some *integers* m1 and m2 such that (m1-m2)*sqrt(k+1)=-(k+2) (because the graph's matrix has trace zero). Thus sqrt(k+1) divides k+2 and k+1 divides (k+2)^2=(k+1)(k+3)+1 which is only possible if k=0, Q.E.D. ==> induction/roll.p <== An ordinary die is thrown until the running total of the throws first exceeds 12. What is the most likely final total that will be obtained? ==> induction/roll.s <== Claim: If you throw a die until the running total exceeds n>=5, a final outcome of n+1 is more likely than any other. Assume we throw an m for a total n+k>n+1, and assume m-k>=0. Now, it is just as likely to throw an m as an m-k+1, which means that the sum n+1 is just as likely as any other. Now consider the series of throws consisting of n-5 1's followed by a 6 and note that we cannot achieve more than an n+1 by changing the last die roll. Hence, a total of n+1 is more likely than any other. ==> induction/takeover.p <== After graduating from college, you have taken an important managing position in the prestigious financial firm of "Mary and Lee". You are responsable for all the decisions concerning take-over bids. Your immediate concern is whether to take over "Financial Data". There is no doubt that you will be successful if you are the first to bid and that this will be profitable for the firm and you in the long run. However, you know that there exist another n financial firms, similar to "Mary and Lee", that are also considering the possibility. Although you are likely to be the first one to move, you know that just after a take-over there is a lot of adjustment that needs to be done. In fact, for a period of time following any take-over the successful firm becomes a prime candidate for a take-over which will cost the job of whoever is responsable for take-overs. Among all financial firms it is common knowledge that the managers responsable for take-overs are rational and intelligent. What is your best response? ==> induction/takeover.s <== Assume the takeover is wise for n. The takeover is then unwise for n+1, as the other companies now find themselves in the same situation as you for n. If the decision is unwise for n, by similar reasoning it is wise to takeover FD for n+1. Now note that for n=1 the takeover decision is clearly unwise, hence by induction you should takeover FD iff n is even. ==> logic/29.p <== Three people check into a hotel. They pay $30 to the manager and go to their room. The manager finds out that the room rate is $25 and gives $5 to the bellboy to return. On the way to the room the bellboy reasons that $5 would be difficult to share among three people so he pockets $2 and gives $1 to each person. Now each person paid $10 and got back $1. So they paid $9 each, totalling $27. The bellboy has $2, totalling $29. Where is the remaining dollar? ==> logic/29.s <== Each person paid $9, totalling $27. The manager has $25 and the bellboy $2. The bellboy's $2 should be added to the manager's $25 or subtracted from the tenants' $27, not added to the tenants' $27. ==> logic/ages.p <== 1) Ten years from now Tim will be twice as old as Jane was when Mary was nine times as old as Tim. 2) Eight years ago, Mary was half as old as Jane will be when Jane is one year older than Tim will be at the time when Mary will be five times as old as Tim will be two years from now. 3) When Tim was one year old, Mary was three years older than Tim will be when Jane is three times as old as Mary was six years before the time when Jane was half as old as Tim will be when Mary will be ten years older than Mary was when Jane was one-third as old as Tim will be when Mary will be three times as old as she was when Jane was born. HOW OLD ARE THEY NOW? ==> logic/ages.s <== The solution: Tim is 3, Jane is 8, and Mary is 15. A little grumbling is in order here, as clue number 1 leads to the situation a year and a half ago, when Tim was 1 1/2, Jane was 6 1/2, and Mary was 13 1/2. This sort of problem is easy if you write down a set of equations. Let t be the year that Tim was born, j be the year that Jane was born, m be the year that Mary was born, and y be the current year. As indefinite years come up, let y1, y2, ... be the indefinite years. Then the equations are y + 10 - t = 2 (y1 - j) y1 - m = 9 (y1 - t) y - 8 - m = 1/2 (y2 - j) y2 - j = 1 + y3 - t y3 - m = 5 (y + 2 - t) t + 1 - m = 3 + y4 - t y4 - j = 3 (y5 - 6 - m) y5 - j = 1/2 (y6 - t) y6 - m = 10 + y7 - m y7 - j = 1/3 (y8 - t) y8 - m = 3 (j - m) t = y - 3 j = y - 8 m = y - 15 ==> logic/bookworm.p <== A bookworm eats from the first page of an encyclopedia to the last page. The bookworm eats in a straight line. The encyclopedia consists of ten 1000-page volumes. Not counting covers, title pages, etc., how many pages does the bookworm eat through? ==> logic/bookworm.s <== On a book shelf the first page of the first volume is on the "inside" __ __ B! ! ! !F A!1 !...........................!10!R C! ! ! !O K! ! ! !N ! ! ! !T ---------------------------------- so the bookworm eats only through the cover of the first volume, then 8 times 1000 pages of Volumes 2 - 9, then through the cover to the 1st page of Vol 10. He eats 8,000 pages. ==> logic/boxes.p <== Which Box Contains the Gold? Two boxes are labeled "A" and "B". A sign on box A says "The sign on box B is true and the gold is in box A". A sign on box B says "The sign on box A is false and the gold is in box A". Assuming there is gold in one of the boxes, which box contains the gold? ==> logic/boxes.s <== The problem cannot be solved with the information given. The sign on box A says "The sign on box B is true and the gold is in box A". The sign on box B says "The sign on box A is false and the gold is in box A". The following argument can be made: If the statement on box A is true, then the statement on box B is true, since that is what the statement on box A says. But the statement on box B states that the statement on box A is false, which contradicts the original assumption. Therefore, the statement on box A must be false. This implies that either the statement on box B is false or that the gold is in box B. If the statement on box B is false, then either the statement on box A is true (which it cannot be) or the gold is in box B. Either way, the gold is in box B. However, there is a hidden assumption in this argument: namely, that each statement must be either true or false. This assumption leads to paradoxes, for example, consider the statement: "This statement is false." If it is true, it is false; if it is false, it is true. The only way out of the paradox is to deny that the statement is either true or false and label it meaningless instead. Both of the statements on the boxes are therefore meaningless and nothing can be concluded from them. In general, statements about the truth of other statements lead to contradictions. Tarski invented metalanguages to avoid this problem. To avoid paradox, a statement about the truth of a statement in a language must be made in the metalanguage of the language. Common sense dictates that this problem cannot be solved with the information given. After all, how can we deduce which box contains the gold simply by reading statements written on the outside of the box? Suppose we deduce that the gold is in box B by whatever line of reasoning we choose. What is to stop us from simply putting the gold in box A, regardless of what we deduced? (cf. Smullyan, "What Is the Name of This Book?", Prentice-Hall, 1978, #70) ==> logic/calibans.will.p <== ---------------------------------------------- ! Caliban's Will by M.H. Newman ! ---------------------------------------------- When Caliban's will was opened it was found to contain the following clause: "I leave ten of my books to each of Low, Y.Y., and 'Critic,' who are to choose in a certain order. No person who has seen me in a green tie is to choose before Low. If Y.Y. was not in Oxford in 1920 the first chooser never lent me an umbrella. If Y.Y. or 'Critic' has second choice, 'Critic' comes before the one who first fell in love." Unfortunately Low, Y.Y., and 'Critic' could not remember any of the relevant facts; but the family solicitor pointed out that, assuming the problem to be properly constructed (i.e. assuming it to contain no statement superfluous to its solution) the relevant data and order could be inferred. What was the prescribed order of choosing; and who lent Caliban an umbrella? ==> logic/calibans.will.s <== Let T be "person who saw Caliban in a green tie." Let U be "person who lent Caliban an umbrella." Then the data are: (1) No T chooses before Low. (2) Either Y.Y. was in Oxford in March 1920 or the first chooser is not a U. (3) Either Low is second or Critic is not last. Consider first (3) If it could be shown that Low is first, then from (3), Critic is not last and therefore is second; i.e. the order is Low, Critic, Y.Y. Next (1) If both Critic and Y.Y. were T's would require Low first and (3) then gives the order Low-Critic-Y.Y., ie. (2) would be superfluous. Hence Critic and Y.Y. are not both T's. If neither Critic nor Y.Y. were a T, (1) would be trivially true for any ordering and therefore would give no information, i.e. would be superfluous. Hence just one of Y.Y. and Critic is a T. It follows that the only possible order in which Low is not first is: Not T, Low, T Now (2) First if Y.Y was in Oxford in March 1920, nothing follows from (2) about the order and (2) is superfluous. Hence Y.Y. was not in Oxford. If Low were a U he would not, by (2) come first, and so by (1) the order would be: Not T, Low, T i.e. (1) and (2) alone would fix an order, and (3) would be superfluous. Hence Low is not a U. It now follows, by the arguments just given for T's under (1) that just one of Y.Y. and Critic is a U. If the same one is the T and the U (2) follows from (1) (since Low is not a U); i.e (2) is superfluous. The situation is therefore: T's: just one of Y.Y. and Critic; not Low U's: the other one of Y.Y; not Low It now follows that "not T, Low, T" is impossible, for the "not T" is the "U" and therefore, by (2), is not first. Hence Low is first, and (3) gives the order: Low, Critic, Y.Y. Finally, Y.Y. is a T, and Critic is a U. For if Critic is a T, then by (1) Low precedes Critic and hence (3) allows only "Low, Critic, Y.Y"; (2) is superfluous. I.e. Critic (only) lent Caliban an umbrella. The problem is from _Problems Omnibus_ by Hubert Phillips, Arco Publications, London, 1960. Hubert Phillips was a noted puzzelist who contributed under his own name and the pseudonyms of "Caliban", "T.O. Hare", and "The Doc". ==> logic/camel.p <== An Arab sheikh tells his two sons that are to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower will win. The brothers, after wandering aimlessly for days, ask a wiseman for advise. After hearing the advice they jump on the camels and race as fast as they can to the city. What did the wiseman say? ==> logic/camel.s <== The wiseman tells them to switch camels. ==> logic/centrifuge.p <== You are a biochemist, working with a 12-slot centrifuge. This is a gadget that has 12 equally spaced slots around a central axis, in which you can place chemical samples you want centrifuged. When the machine is turned on, the samples whirl around the central axis and do their thing. To ensure that the samples are evenly mixed, they must be distributed in the 12 slots such that the centrifuge is balanced evenly. For example, if you wanted to mix 4 samples, you could place them in slots 12, 3, 6 and 9 (assuming the slots are numbered from 1 to 12 like a clock). Problem: Can you use the centrifuge to mix 5 samples? ==> logic/centrifuge.s <== The superposition of any two solutions is yet another solution, so given that the factors > 1 of 12 (2, 3, 4, 6, 12) are all solutions, the only thing to check about, for example, the proposed solution 2+3 is that not all ways of combining 2 & 3 would have centrifuge tubes from one subsolution occupying the slot for one of the tubes in another solution. For the case 2+3, there is no problem: Place 3 tubes, one in every 4th position, then place the 4th and 5th diametrically opposed (each will end up in a slot adjacent to one of the first 3 tubes). The obvious generalization is, what are the numbers of tubes that cannot be balanced? Observing that there are solutions for 2,3,4,5,6 tubes and that if X has a solution, 12-X has also one (obtained by swapping tubes and holes), it is obvious that 1 and 11 are the only cases without solutions. Here is how this problem is often solved in practice: A dummy tube is added to produce a total number of tubes that is easy to balance. For example, if you had to centrifuge just one sample, you'd add a second tube opposite it for balance. ==> logic/children.p <== A man walks into a bar, orders a drink, and starts chatting with the bartender. After a while, he learns that the bartender has three children. "How old are your children?" he asks. "Well," replies the bartender, "the product of their ages is 72." The man thinks for a moment and then says, "that's not enough information." "All right," continues the bartender, "if you go outside and look at the building number posted over the door to the bar, you'll see the sum of the ages." The man steps outside, and after a few moments he reenters and declares, "Still not enough]" The bartender smiles and says, "My youngest just loves strawberry ice cream." How old are the children? A variant of the problem is for the sum of the ages to be 13 and the product of the ages to be the number posted over the door. In this case, it is the oldest that loves ice cream. Then how old are they? ==> logic/children.s <== First, determine all the ways that three ages can multiply together to get 72: 72 1 1 (quite a feat for the bartender) 36 2 1 24 3 1 18 4 1 18 2 2 12 6 1 12 3 2 9 4 2 9 8 1 8 3 3 6 6 2 6 4 3 As the man says, that's not enough information; there are many possibilities. So the bartender tells him where to find the sum of the ages--the man now knows the sum even though we don't. Yet he still insists that there isn't enough info. This must mean that there are two permutations with the same sum; otherwise the man could have easily deduced the ages. The only pair of permutations with the same sum are 8 3 3 and 6 6 2, which both add up to 14 (the bar's address). Now the bartender mentions his "youngest"--telling us that there is one child who is younger than the other two. This is impossible with 8 3 3--there are two 3 year olds. Therefore the ages of the children are 6, 6, and 2. Pedants have objected that the problem is insoluble because there could be a youngest between two three year olds (even twins are not born exactly at the same time). However, the word "age" is frequently used to denote the number of years since birth. For example, I am the same age as my wife, even though technically she is a few months older than I am. And using the word "youngest" to mean "of lesser age" is also in keeping with common parlance. So I think the solution is fine as stated. In the sum-13 variant, the possibilities are: 11 1 1 10 2 1 9 3 1 9 2 2 8 4 1 8 3 2 7 5 1 7 4 2 7 3 3 6 6 1 6 5 2 6 4 3 The two that remain are 9 2 2 and 6 6 1 (both products equal 36). The final bit of info (oldest child) indicates that there is only one child with the highest age. This cancels out the 6 6 1 combination, leaving the childern with ages of 9, 2, and 2. ==> logic/condoms.p <== How can you have mutually safe sex with three women with only two condoms? ==> logic/condoms.s <== Use both condoms on the first woman. Take off the outer condom (turning it inside-out in the process) and set it aside. Use the inner condom alone on the second woman. Put the outer condom back on. Use it on the third woman. ==> logic/dell.p <== How can I solve logic puzzles (e.g., as published by Dell) automatically? ==> logic/dell.s <== #include #define EITHER if (SÝ1¨ = SÝ0¨, ] setjmp((S++)->jb)) { #define OR } else EITHER #define REJECT longjmp((--S)->jb, 1) #define END_EITHER } else REJECT; /* values in tmat: */ #define T_UNK 0 #define T_YES 1 #define T_NO 2 #define Val(t1,t2) (S->tmatÝt1¨Ýt2¨) #define CLASS(x) \ (((x) / NUM_ITEM) * NUM_ITEM) #define EVERY_TOKEN(x) \ (x = 0; x < TOT_TOKEN; x++) #define EVERY_ITEM(x, class) \ (x = CLASS(class); x < CLASS(class) + NUM_ITEM; x++) #define BEGIN \ struct state { \ char tmatÝTOT_TOKEN¨ÝTOT_TOKEN¨; \ jmp_buf jb; \ } StatesÝ100¨, *S = States; \ \ main() \ { \ int token; \ \ for EVERY_TOKEN(token) \ yes(token, token); \ EITHER /* Here is the problem-specific data */ #define NUM_ITEM 5 #define NUM_CLASS 6 #define TOT_TOKEN (NUM_ITEM * NUM_CLASS) #define HOUSE_0 0 #define HOUSE_1 1 #define HOUSE_2 2 #define HOUSE_3 3 #define HOUSE_4 4 #define ENGLISH 5 #define SPANISH 6 #define NORWEG 7 #define UKRAIN 8 #define JAPAN 9 #define GREEN 10 #define RED 11 #define IVORY 12 #define YELLOW 13 #define BLUE 14 #define COFFEE 15 #define TEA 16 #define MILK 17 #define OJUICE 18 #define WATER 19 #define DOG 20 #define SNAIL 21 #define FOX 22 #define HORSE 23 #define ZEBRA 24 #define OGOLD 25 #define PLAYER 26 #define CHESTER 27 #define LSTRIKE 28 #define PARLIA 29 char *namesݨ = { "HOUSE_0", "HOUSE_1", "HOUSE_2", "HOUSE_3", "HOUSE_4", "ENGLISH", "SPANISH", "NORWEG", "UKRAIN", "JAPAN", "GREEN", "RED", "IVORY", "YELLOW", "BLUE", "COFFEE", "TEA", "MILK", "OJUICE", "WATER", "DOG", "SNAIL", "FOX", "HORSE", "ZEBRA", "OGOLD", "PLAYER", "CHESTER", "LSTRIKE", "PARLIA", }; BEGIN yes(ENGLISH, RED); /* Clue 1 */ yes(SPANISH, DOG); /* Clue 2 */ yes(COFFEE, GREEN); /* Clue 3 */ yes(UKRAIN, TEA); /* Clue 4 */ EITHER /* Clue 5 */ yes(IVORY, HOUSE_0); yes(GREEN, HOUSE_1); OR yes(IVORY, HOUSE_1); yes(GREEN, HOUSE_2); OR yes(IVORY, HOUSE_2); yes(GREEN, HOUSE_3); OR yes(IVORY, HOUSE_3); yes(GREEN, HOUSE_4); END_EITHER yes(OGOLD, SNAIL); /* Clue 6 */ yes(PLAYER, YELLOW); /* Clue 7 */ yes(MILK, HOUSE_2); /* Clue 8 */ yes(NORWEG, HOUSE_0); /* Clue 9 */ EITHER /* Clue 10 */ yes(CHESTER, HOUSE_0); yes(FOX, HOUSE_1); OR yes(CHESTER, HOUSE_4); yes(FOX, HOUSE_3); OR yes(CHESTER, HOUSE_1); EITHER yes(FOX, HOUSE_0); OR yes(FOX, HOUSE_2); END_EITHER OR yes(CHESTER, HOUSE_2); EITHER yes(FOX, HOUSE_1); OR yes(FOX, HOUSE_3); END_EITHER OR yes(CHESTER, HOUSE_3); EITHER yes(FOX, HOUSE_2); OR yes(FOX, HOUSE_4); END_EITHER END_EITHER EITHER /* Clue 11 */ yes(PLAYER, HOUSE_0); yes(HORSE, HOUSE_1); OR yes(PLAYER, HOUSE_4); yes(HORSE, HOUSE_3); OR yes(PLAYER, HOUSE_1); EITHER yes(HORSE, HOUSE_0); OR yes(HORSE, HOUSE_2); END_EITHER OR yes(PLAYER, HOUSE_2); EITHER yes(HORSE, HOUSE_1); OR yes(HORSE, HOUSE_3); END_EITHER OR yes(PLAYER, HOUSE_3); EITHER yes(HORSE, HOUSE_2); OR yes(HORSE, HOUSE_4); END_EITHER END_EITHER yes(LSTRIKE, OJUICE); /* Clue 12 */ yes(JAPAN, PARLIA); /* Clue 13 */ EITHER /* Clue 14 */ yes(NORWEG, HOUSE_0); yes(BLUE, HOUSE_1); OR yes(NORWEG, HOUSE_4); yes(BLUE, HOUSE_3); OR yes(NORWEG, HOUSE_1); EITHER yes(BLUE, HOUSE_0); OR yes(BLUE, HOUSE_2); END_EITHER OR yes(NORWEG, HOUSE_2); EITHER yes(BLUE, HOUSE_1); OR yes(BLUE, HOUSE_3); END_EITHER OR yes(NORWEG, HOUSE_3); EITHER yes(BLUE, HOUSE_2); OR yes(BLUE, HOUSE_4); END_EITHER END_EITHER /* End of problem-specific data */ solveit(); OR printf("All solutions found\n"); exit(0); END_EITHER } no(a1, a2) { int non1, non2, token; if (Val(a1, a2) == T_YES) REJECT; else if (Val(a1, a2) == T_UNK) { Val(a1, a2) = T_NO; no(a2, a1); non1 = non2 = -1; for EVERY_ITEM(token, a1) if (Val(token, a2) ]= T_NO) if (non1 == -1) non1 = token; else break; if (non1 == -1) REJECT; else if (token == CLASS(a1) + NUM_ITEM) yes(non1, a2); for EVERY_TOKEN(token) if (Val(token, a1) == T_YES) no(a2, token); } } yes(a1, a2) { int token; if (Val(a1, a2) == T_NO) REJECT; else if (Val(a1, a2) == T_UNK) { Val(a1, a2) = T_YES; yes(a2, a1); for EVERY_ITEM(token, a1) if (token ]= a1) no(token, a2); for EVERY_TOKEN(token) if (Val(token, a1) == T_YES) yes(a2, token); else if (Val(token, a1) == T_NO) no(a2, token); } } solveit() { int token, tok2; for EVERY_TOKEN(token) for (tok2 = token; tok2 < TOT_TOKEN; tok2++) if (Val(token, tok2) == T_UNK) { EITHER yes(token, tok2); OR no(token, tok2); END_EITHER; return solveit(); } printf("Solution:\n"); for EVERY_ITEM(token, 0) { for (tok2 = NUM_ITEM; tok2 < TOT_TOKEN; tok2++) if (Val(token, tok2) == T_YES) printf("\t%s %s\n",namesÝtoken¨,namesÝtok2¨); printf("\n"); } REJECT; } --- james@crc.ricoh.com (James Allen) ==> logic/elimination.p <== 97 baseball teams participate in an annual state tournament. The way the champion is chosen for this tournament is by the same old elimination schedule. That is, the 97 teams are to be divided into pairs, and the two teams of each pair play against each other. After a team is eliminated from each pair, the winners would be again divided into pairs, etc. How many games must be played to determine a champion? ==> logic/elimination.s <== In order to determine a winner all but one team must lose. Therefore there must be at least 96 games. ==> logic/family.p <== Suppose that it is equally likely for a pregnancy to deliver a baby boy as it is to deliver a baby girl. Suppose that for a large society of people, every family continues to have children until they have a boy, then they stop having children. After 1,000 generations of families, what is the ratio of males to females? ==> logic/family.s <== The ratio will be 50-50 in both cases. We are not killing off any fetuses or babies, and half of all conceptions will be male, half female. When a family decides to stop does not affect this fact. ==> logic/flip.p <== How can a toss be called over the phone (without requiring trust)? ==> logic/flip.s <== A flips a coin. If the result is heads, A multiplies 2 90-digit prime numbers; if the result is tails, A multiplies 3 60-digit prime numbers. A tells B the result of the multiplication. B now calls either heads or tails and tells A. A then supplies B with the original numbers to verify the flip. ==> logic/friends.p <== Any group of 6 or more contains either 3 mutual friends or 3 mutual strangers. Prove it. ==> logic/friends.s <== Take a person X. Of the five other people, there must either be at least three acquaintances of X or at least three strangers of X. Assume wlog that X has three strangers A,B,C. Unless A,B,C is the required triad of acquaintances, they must include a pair of strangers, wlog A,B. Then X,A,B is the required triad of strangers, QED. ==> logic/hundred.p <== A sheet of paper has statements numbered from 1 to 100. Statement n says "exactly n of the statements on this sheet are false." Which statements are true and which are false? What if we replace "exactly" by "at least"? ==> logic/hundred.s <== It is tempting to argue as follows: Since only one statement can be true (they are mutually contradictory), therefore 99 are false. So, all are false except for statement 99. If replaced by "at least", and the "real" number of false statements is x, then statements x+1 to 100 will be false (since they falsely claim that there are more false statements than there actually are). So, 100-x are false, ie. x=100-x, so x=50. The first 50 statements are true, and statements 51 to 100 are false. However, there is a hidden and incorrect assumption in this argument. To see this, suppose that there is one statement on the sheet and it says "One statement is false" or "At least one statement is false," either way it implies "this statement is false," which is a familiar paradoxical statement. We have learned that this paradox arises because of the false assumption that all statements are either true or false. This is the hidden assumption in the above reasoning. If it is acknowledged that some of the statements on the page may be neither true nor false (i.e., meaningless), then nothing whatsoever can be concluded about which statements are true or false. This problem has been carefully contrived to appear to be solvable (like the vacuous statement "this statement is true"). By changing the numbers in some statements and changing "true" to "false," various circular forms of the liar's paradox can be constructed. From _Litton's Problematical Recreations_ ==> logic/inverter.p <== Can a digital logic circuit with two inverters invert N independent inputs? The circuit may contain any number of AND or OR gates. ==> logic/inverter.s <== It can be shown that N inverters can invert 2N-1 independent inputs, given an unlimited supply of AND and OR gates. The classic version of this puzzle is to invert 3 independent inputs using AND gates, OR gates, and only 2 inverters. So, start with N inverters. Replace 3 of them with 2. Keep doing that until you're down to 2 inverters. I was skeptical at first, because such a design requires so much feedback that I was sure the system would oscillate when switching between two particular states. But after writing a program to test every possible state change (32^2), it appears that this system settles after a maximum of 3 feedback logic iterations. I did not include gate delays in the simulation, however, which could increase the number of iterations before the system settles. In any case, it appears that the world needs only 2 inverters] :-) ==> logic/josephine.p <== The recent expedition to the lost city of Atlantis discovered scrolls attributted to the great poet, scholar, philosopher Josephine. They number eight in all, and here is the first. THE KINGDOM OF MAMAJORCA, WAS RULED BY QUEEN HENRIETTA I. IN MAMAJORCA WOMEN HAVE TO PASS AN EXTENSIVE LOGIC EXAM BEFORE THEY ARE ALLOWED TO GET MARRIED. QUEENS DO NOT HAVE TO TAKE THIS EXAM. ALL THE WOMEN IN MAMAJORCA ARE LOYAL TO THEIR QUEEN AND DO WHATEVER SHE TELLS THEM TO. THE QUEENS OF MAMAJORCA ARE TRUTHFUL. ALL SHOTS FIRED IN MAMAJORCA CAN BE HEARD IN EVERY HOUSE. ALL ABOVE FACTS ARE KNOWN TO BE COMMON KNOWLEDGE. HENRIETTA WAS WORRIED ABOUT THE INFIDELITY OF THE MARRIED MEN IN MAMAJORCA. SHE SUMMONED ALL THE WIVES TO THE TOWN SQUARE, AND MADE THE FOLLOWING ANNOUNCEMENT. "THERE IS AT LEAST ONE UNFAITHFUL HUSBAND IN MAMAJORCA. ALL WIVES KNOW WHICH HUSBANDS ARE UNFAITHFUL, BUT HAVE NO KNOWLEDGE ABOUT THE FIDELITY OF THEIR OWN HUSBAND. YOU ARE FORBIDDEN TO DISCUSS YOUR HUSBAND'S FAITHFULNESS WITH ANY OTHER WOMAN. IF YOU DISCOVER THAT YOUR HUSBAND IS UNFAITHFUL, YOU MUST SHOOT HIM AT PRECISELY MIDNIGHT OF THE DAY YOU FIND THAT OUT." THIRTY-NINE SILENT NIGHTS FOLLOWED THE QUEEN'S ANNOUNCEMENT. ON THE FORTIETH NIGHT, SHOTS WERE HEARD. QUEEN HENRIETTA I IS REVERED IN MAMAJORCAN HISTORY. As with all philosophers Josephine doesn't provide the question, but leaves it implicit in his document. So figure out the questions - there are two - and answer them. Here is Josephine's second scroll. QUEEN HENRIETTA I WAS SUCCEEDED BY DAUGHTER QUEEN HENRIETTA II. AFTER A WHILE HENRIETTA LIKE HER FAMOUS MOTHER BECAME WORRIED ABOUT THE INFIDELITY PROBLEM. SHE DECIDED TO ACT, AND SENT A LETTER TO HER SUBJECTS (WIVES) THAT CONTAINED THE EXACT WORDS OF HENRIETTA I'S FAMOUS SPEECH. SHE ADDED THAT THE LETTERS WERE GUARENTEED TO REACH ALL WIVES EVENTUALLY. QUEEN HENRIETTA II IS REMEMBERED AS A FOOLISH AND UNJUST QUEEN. What is the question and answer implied by this scroll? ==> logic/josephine.s <== The two questions for scroll #1 were: 1. How many husbands were shot on that fateful night? 2. Why is Queen Henrietta I revered in Mamajorca? The answers are: If there are n unfaithful husbands (UHs), every wife of an UH knows of n-1 UH's while every wife of a faithful husband knows of n UHs. Ýthis because everyone has perfect information about everything except the fidelity of their own husband¨. Now we do a simple induction: Assume that there is only one UH. Then all the wives but one know that there is just one UH, but the wife of the UH thinks that everyone is faithful. Upon hearing that "there is at least one UH", the wife realizes that the only husband it can be is her own, and so shoots him. Now, imagine that there are just two UH's. Each wife of an UH assumes that the situation is "only one UH in town" and so waits to hear the other wife (she knows who it is, of course) shoot her husband on the first night. When no one is shot, that can only be because her OWN husband was a second UH. The wife of the second UH makes the same deduction when no shot is fired the first night (she was waiting, and expecting the other to shoot, too). So they both figure it out after the first night, and shoot their husbands the second night. It is easy to tidy up the induction to show that the n UHs will all be shot just on the n'th midnight. The question for scroll #2 is: 3. Why is Queen Henrietta II not? The answer is: The problem now is that QHII didn't realize that it is *critical* that all of the wives, of faithful and UH's alike, to *BEGIN*AT*THE*SAME*MOMENT*. The uncertainty of having a particular wife's notice come a day or two late makes the whole logic path fall apart. That's why she's foolish. She is unjust, because some wives, honed and crack logicians all, remember, will *incorrectly* shoot faithful husbands. Let us imagine the situation with just a SINGLE UH in the whole country. And, wouldn't you know it, the notice to the wife of the UH just happens to be held up a day, whereas everyone else's arrived the first day. Now, all of the wives that got the notice the first day know that there is just one UH in the country. And they know that the wife of that UH will think that everyone is faithful, and so they'll expect her to figure it out and shoot her husband the first night. BUT SHE DIDN"T GET THE NOTICE THE FIRST NIGHT.... BUT THE OTHER WIVES HAVE NO WAY OF KNOWING THAT. So, the wife of the UH doesn't know that anything is going on and so (of course) doesn't do anything the first night. The next day she gets the notice, figures it all out, and her husband will be history come that midnight. BUT... *every* other wife thought that there should have been a shooting the first night, and since there wasn't there must have been an additional UH, and it can only have been _her_ husband. So on the second night **ALL** of the husbands are shot. Things are much more complicated if the mix of who gets the notice when is less simple than the one I mentioned above, but it is always wrong and/or tragic. NOTE: if the wives *know* that the country courier service (or however these things get delivered) is flaky, then they can avoid the massacre, but unless the wives exchange notes no one will ever be shot (since there is always a chance that rather than _your_ husband being an UH, you could reason that it might be that the wife of one of the UH's that you know about just hasn't gotten her copy of the scroll yet). I guess you could call this case "unjust", too, since the UH's evade punishment, despite the perfect logic of the wives. ==> logic/locks.and.boxes.p <== You want to send a valuable object to a friend. You have a box which is more than large enough to contain the object. You have several locks with keys. The box has a locking ring which is more than large enough to have a lock attached. But your friend does not have the key to any lock that you have. How do you do it? ==> logic/locks.and.boxes.s <== Attach a lock to the ring. Send it to her. She attaches her own lock and sends it back. You remove your lock and send it back to her. She removes her lock. ==> logic/mixing.p <== Start with a half cup of tea and a half cup of coffee. Take one tablespoon of the tea and mix it in with the coffee. Take one tablespoon of this mixture and mix it back in with the tea. Which of the two cups contains more of its original contents? ==> logic/mixing.s <== Mixing Liquids The two cups end up with the same volume of liquid they started with. The same amount of tea was moved to the coffee cup as coffee to the teacup. Therefore each cup contains the same amount of its original contents. ==> logic/number.p <== Mr. S. and Mr. P. are both perfect logicians, being able to correctly deduce any truth from any set of axioms. Two integers (not necessarily unique) are somehow chosen such that each is within some specified range. Mr. S. is given the sum of these two integers; Mr. P. is given the product of these two integers. After receiving these numbers, the two logicians do not have any communication at all except the following dialogue: <<1>> Mr. P.: I do not know the two numbers. <<2>> Mr. S.: I knew that you didn't know the two numbers. <<3>> Mr. P.: Now I know the two numbers. <<4>> Mr. S.: Now I know the two numbers. Given that the above statements are absolutely truthful, what are the two numbers? ==> logic/number.s <== The answer depends upon the ranges from which the numbers are chosen. The unique solution for the ranges Ý2,62¨ through Ý2,500+¨ is: SUM PRODUCT X Y 17 52 4 13 The unique solution for the ranges Ý3,94¨ through Ý3,500+¨ is: SUM PRODUCT X Y 29 208 13 16 There are no unique solutions for the ranges starting with 1, and there are no solutions for ranges starting with numbers above 3. A program to compute the possible pairs is included below. #include /* BEGINNING OF PROBLEM STATEMENT: Mr. S. and Mr. P. are both perfect logicians, being able to correctly deduce any truth from any set of axioms. Two integers (not necessarily unique) are somehow chosen such that each is within some specified range. Mr. S. is given the sum of these two integers; Mr. P. is given the product of these two integers. After receiving these numbers, the two logicians do not have any communication at all except the following dialogue: <<1>> Mr. P.: I do not know the two numbers. <<2>> Mr. S.: I knew that you didn't know the two numbers. <<3>> Mr. P.: Now I know the two numbers. <<4>> Mr. S.: Now I know the two numbers. Given that the above statements are absolutely truthful, what are the two numbers? END OF PROBLEM STATEMENT */ #define SMALLEST_MIN 1 #define LARGEST_MIN 10 #define SMALLEST_MAX 50 #define LARGEST_MAX 500 long PÝ(LARGEST_MAX + 1) * (LARGEST_MAX + 1)¨; /* products */ long SÝ(LARGEST_MAX + 1) + (LARGEST_MAX + 1)¨; /* sums */ find(long min, long max) { long i, j; /* * count factorizations in Pݨ * all PÝn¨ > 1 satisfy <<1>>. */ for(i = 0; i <= max * max; ++i) PÝi¨ = 0; for(i = min; i <= max; ++i) for(j = i; j <= max; ++j) ++PÝi * j¨; /* * decompose possible SUMs and check factorizations * all SÝn¨ == min - 1 satisfy <<2>>. */ for(i = min + min; i <= max + max; ++i) { for(j = i / 2; j >= min; --j) if(PÝj * (i - j)¨ < 2) break; SÝi¨ = j; } /* * decompose SUMs which satisfy <<2>> and see which products * they produce. All (PÝn¨ / 1000 == 1) satisfy <<3>>. */ for(i = min + min; i <= max + max; ++i) if(SÝi¨ == min - 1) for(j = i / 2; j >= min; --j) if(PÝj * (i - j)¨ > 1) PÝj * (i - j)¨ += 1000; /* * decompose SUMs which satisfy <<2>> again and see which products * satisfy <<3>>. Any (SÝn¨ == 999 + min) satisfies <<4>> */ for(i = min + min; i <= max + max; ++i) if(SÝi¨ == min - 1) for(j = i / 2; j >= min; --j) if(PÝj * (i - j)¨ / 1000 == 1) SÝi¨ += 1000; /* * find the answer(s) and print them */ printf("Ý%d,%d¨\n",min,max); for(i = min + min; i <= max + max; ++i) if(SÝi¨ == 999 + min) for(j = i / 2; j >= min; --j) if(PÝj * (i - j)¨ / 1000 == 1) printf("{ %d %d }: S = %d, P = %d\n", i - j, j, i, (i - j) * j); } main() { long min, max; for (min = SMALLEST_MIN; min <= LARGEST_MIN; min ++) for (max = SMALLEST_MAX; max <= LARGEST_MAX; max++) find(min,max); } ------------------------------------------------------------------------- = Jeff Kenton (617) 894-4508 = = jkenton@world.std.com = ------------------------------------------------------------------------- ==> logic/riddle.p <== Who makes it, has no need of it. Who buys it, has no use for it. Who uses it can neither see nor feel it. Tell me what a dozen rubber trees with thirty boughs on each might be? As I went over London Bridge I met my sister Jenny I broke her neck and drank her blood And left her standing empty It is said among my people that some things are improved by death. Tell me, what stinks while living, but in death, smells good? All right. Riddle me this: what goes through the door without pinching itself? What sits on the stove without burning itself? What sits on the table and is not ashamed? What work is it that the faster you work, the longer it is before you're done, and the slower you work, the sooner you're finished? Whilst I was engaged in sitting I spied the dead carrying the living. I know a word of letters three. Add two, and fewer there will be. I give you a group of three. One is sitting down, and will never get up. The second eats as much as is given to him, yet is always hungry. The third goes away and never returns. Whoever makes it, tells it not. Whoever takes it, knows it not. And whoever knows it wants it not. Two words, my answer is only two words. To keep me, you must give me. Sir, I bear a rhyme excelling In mystic force and magic spelling Celestial sprites elucidate All my own striving can't relate There is not wind enough to twirl That one red leaf, nearest of its clan, Which dances as often as dance it can. Half-way up the hill, I see thee at last Lying beneath me with thy sounds and sights -- A city in the twilight, dim and vast, With smoking roofs, soft bells, and gleaming lights. I am, in truth, a yellow fork From tables in the sky By inadvertent fingers dropped The awful cutlery. Of mansions never quite disclosed And never quite concealed The apparatus of the dark To ignorance revealed. Many-maned scud-thumper, Maker of worn wood, Shrub-ruster, Sky-mocker, Rave] Make me thy lyre, even as the forests are. What if my leaves fell like its own -- The tumult of thy mighty harmonies Will take from both a deep autumnal tone. This darksome burn, horseback brown, His rollock highroad roaring down, In coop and in comb the fleece of his foam Flutes and low to the body falls home. I've measured it from side to side, 'Tis three feet long and two feet wide. It is of compass small, and bare To thirsty suns and parching air. My love, when I gaze on thy beautiful face, Careering along, yet always in place -- The thought has often come into my mind If I ever shall see thy glorious behind. Then all thy feculent majesty recalls The nauseous mustiness of forsaken bowers, The leprous nudity of deserted halls -- The positive nastiness of sullied flowers. And I mark the colours, yellow and black, That fresco thy lithe, dictatorial thighs. When young, I am sweet in the sun. When middle-aged, I make you gay. When old, I am valued more than ever. I am always hungry, I must always be fed, The finger I lick Will soon turn red. All about, but cannot be seen, Can be captured, cannot be held, No throat, but can be heard. I am only useful When I am full, Yet I am always Full of holes. If you break me I do not stop working, If you touch me I may be snared, If you lose me Nothing will matter. If a man carried my burden He would break his back. I am not rich, But leave silver in my track. Until I am measured I am not known, Yet how you miss me When I have flown. I drive men mad For love of me, Easily beaten, Never free. When set loose I fly away, Never so cursed As when I go astray. I go around in circles But always straight ahead, Never complain No matter where I am led. Lighter than what I am made of, More of me is hidden Than is seen. I turn around once, What is out will not get in. I turn around again, What is in will not get out. Each morning I appear To lie at your feet, All day I will follow No matter how fast you run, Yet I nearly perish In the midday sun. Weight in my belly, Trees on my back, Nails in my ribs, Feet I do lack. Bright as diamonds, Loud as thunder, Never still, A thing of wonder. My life can be measured in hours, I serve by being devoured. Thin, I am quick Fat, I am slow Wind is my foe. To unravel me You need a simple key, No key that was made By locksmith's hand, But a key that only I Will understand. I am seen in the water If seen in the sky, I am in the rainbow, A jay's feather, And lapis lazuli. Glittering points That downward thrust, Sparkling spears That never rust. You heard me before, Yet you hear me again, Then I die, 'Till you call me again. Three lives have I. Gentle enough to soothe the skin, Light enough to caress the sky, Hard enough to crack rocks. You can see nothing else When you look in my face, I will look you in the eye And I will never lie. Lovely and round, I shine with pale light, grown in the darkness, A lady's delight. At the sound of me, men may dream Or stamp their feet At the sound of me, women may laugh Or sometimes weep When I am filled I can point the way, When I am empty Nothing moves me, I have two skins One without and one within. My tines be long, My tines be short My tines end ere My first report. What am I? ==> logic/riddle.s <== Who makes it, has no need of it. Who buys it, has no use for it. Who uses it can neither see nor feel it. coffin Tell me what a dozen rubber trees with thirty boughs on each might be? months of the year As I went over London Bridge I met my sister Jenny I broke her neck and drank her blood And left her standing empty gin It is said among my people that some things are improved by death. Tell me, what stinks while living, but in death, smells good? pig All right. Riddle me this: what goes through the door without pinching itself? What sits on the stove without burning itself? What sits on the table and is not ashamed? the sun What work is it that the faster you work, the longer it is before you're done, and the slower you work, the sooner you're finished? roasting meat on a spit Whilst I was engaged in sitting I spied the dead carrying the living. a ship I know a word of letters three. Add two, and fewer there will be. 'few' I give you a group of three. One is sitting down, and will never get up. The second eats as much as is given to him, yet is always hungry. The third goes away and never returns. stove, fire, and smoke Whoever makes it, tells it not. Whoever takes it, knows it not. And whoever knows it wants it not. counterfeit money Two words, my answer is only two words. To keep me, you must give me. your word Sir, I bear a rhyme excelling In mystic force and magic spelling Celestial sprites elucidate All my own striving can't relate ??? There is not wind enough to twirl That one red leaf, nearest of its clan, Which dances as often as dance it can. the sun, Samuel Taylor Coleridge Half-way up the hill, I see thee at last Lying beneath me with thy sounds and sights -- A city in the twilight, dim and vast, With smoking roofs, soft bells, and gleaming lights. the past, Longfellow I am, in truth, a yellow fork From tables in the sky By inadvertent fingers dropped The awful cutlery. Of mansions never quite disclosed And never quite concealed The apparatus of the dark To ignorance revealed. lightning, Emily Dickinson Many-maned scud-thumper, Maker of worn wood, Shrub-ruster, Sky-mocker, Rave] Portly pusher, Wind-slave. the ocean, John Updike Make me thy lyre, even as the forests are. What if my leaves fell like its own -- The tumult of thy mighty harmonies Will take from both a deep autumnal tone. the west wind, Percy Bysshe Shelley This darksome burn, horseback brown, His rollock highroad roaring down, In coop and in comb the fleece of his foam Flutes and low to the body falls home. river, Gerard Manley Hopkins I've measured it from side to side, 'Tis three feet long and two feet wide. It is of compass small, and bare To thirsty suns and parching air. the grave of a child, Wordsworth My love, when I gaze on thy beautiful face, Careering along, yet always in place -- The thought has often come into my mind If I ever shall see thy glorious behind. the moon, Sir Edmund Gosse Then all thy feculent majesty recalls The nauseous mustiness of forsaken bowers, The leprous nudity of deserted halls -- The positive nastiness of sullied flowers. And I mark the colours, yellow and black, That fresco thy lithe, dictatorial thighs. spider, Francis Saltus Saltus When young, I am sweet in the sun. When middle-aged, I make you gay. When old, I am valued more than ever. wine I am always hungry, I must always be fed, The finger I lick Will soon turn red. fire All about, but cannot be seen, Can be captured, cannot be held, No throat, but can be heard. wind I am only useful When I am full, Yet I am always Full of holes. sieve (or sponge) If you break me I do not stop working, If you touch me I may be snared, If you lose me Nothing will matter. heart If a man carried my burden He would break his back. I am not rich, But leave silver in my track. snail Until I am measured I am not known, Yet how you miss me When I have flown. time I drive men mad For love of me, Easily beaten, Never free. gold When set loose I fly away, Never so cursed As when I go astray. ? I go around in circles But always straight ahead, Never complain No matter where I am led. wagon wheel Lighter than what I am made of, More of me is hidden Than is seen. iceberg I turn around once, What is out will not get in. I turn around again, What is in will not get out. stopcock Each morning I appear To lie at your feet, All day I will follow No matter how fast you run, Yet I nearly perish In the midday sun. shadow Weight in my belly, Trees on my back, Nails in my ribs, Feet I do lack. ship Bright as diamonds, Loud as thunder, Never still, A thing of wonder. waterfall? (fireworks?) My life can be measured in hours, I serve by being devoured. Thin, I am quick Fat, I am slow Wind is my foe. candle To unravel me You need a simple key, No key that was made By locksmith's hand, But a key that only I Will understand. cipher I am seen in the water If seen in the sky, I am in the rainbow, A jay's feather, And lapis lazuli. blue Glittering points That downward thrust, Sparkling spears That never rust. icicle You heard me before, Yet you hear me again, Then I die, 'Till you call me again. echo Three lives have I. Gentle enough to soothe the skin, Light enough to caress the sky, Hard enough to crack rocks. water You can see nothing else When you look in my face, I will look you in the eye And I will never lie. your reflection Lovely and round, I shine with pale light, grown in the darkness, A lady's delight. pearl At the sound of me, men may dream Or stamp their feet At the sound of me, women may laugh Or sometimes weep music When I am filled I can point the way, When I am empty Nothing moves me, I have two skins One without and one within. sails? My tines be long, My tines be short My tines end ere My first report. What am I? lightning ==> logic/river.crossing.p <== Three humans, one big monkey and two small monkeys are to cross a river: a) Only humans and the big monkey can row the boat. b) At all times, the number of human on either side of the river must be GREATER OR EQUAL to the number of monkeys on THAT side. ( Or else the humans will be eaten by the monkeys]) ==> logic/river.crossing.s <== The three columns represent the left bank, the boat, and the right bank respectively. The < or > indicates the direction of motion of the boat. HHHMmm . . HHHm Mm> . HHHm m HHH mm HM Hm Hm HM mm HHH m HHHm . . HHHMmm ==> logic/ropes.p <== Two fifty foot ropes are suspended from a forty foot ceiling, about twenty feet apart. Armed with only a knife, how much of the rope can you steal? ==> logic/ropes.s <== Almost all of it. Tie the ropes together. Climb up one of them. Tie a loop in it as close as possible to the ceiling. Cut it below the loop. Run the rope through the loop and tie it to your waist. Climb the other rope (this may involve some swinging action). Pull the rope going through the loop tight and cut the other rope as close as possible to the ceiling. You will swing down on the rope through the loop. Lower yourself to the ground by letting out rope. Pull the rope through the loop. ==> logic/same.street.p <== Sally and Sue have a strong desire to date Sam. They all live on the same street yet neither Sally or Sue know where Sam lives. The houses on this street are numbered 1 to 99. Sally asks Sam "Is your house number a perfect square?". He answers. Then Sally asks "Is is greater than 50?". He answers again. Sally thinks she now knows the address of Sam's house and decides to visit. When she gets there, she finds out she is wrong. This is not surprising, considering Sam answered only the second question truthfully. Sue, unaware of Sally's conversation, asks Sam two questions. Sue asks "Is your house number a perfect cube?". He answers. She then asks "Is it greater than 25?". He answers again. Sue thinks she knows where Sam lives and decides to pay him a visit. She too is mistaken as Sam once again answered only the second question truthfully. If I tell you that Sam's number is less than Sue's or Sally's, and that the sum of their numbers is a perfect square multiplied by two, you should be able to figure out where all three of them live. ==> logic/same.street.s <== Sally and Sue have a strong desire to date Sam. They all live on the same street yet neither Sally or Sue know where Sam lives. The houses on this street are numbered 1 to 99. Sally asks Sam "Is your house number a perfect square?". He answers. Then Sally asks "Is is greater than 50?". He answers again. Sally thinks she now knows the address of Sam's house and decides to visit. Since Sally thinks that she has enough information, I deduce that Sam answered that his house number was a perfect square greater than 50. There are two of these {64,81} and Sally must live in one of them in order to have decided she knew where Sam lives. When she gets there, she finds out she is wrong. This is not surprising, considering Sam answered only the second question truthfully. So Sam's house number is greater than 50, but not a perfect square. Sue, unaware of Sally's conversation, asks Sam two questions. Sue asks "Is your house number a perfect cube?". He answers. She then asks "Is it greater than 25?". He answers again. Observation: perfect cubes greater than 25 are {27, 64}, less than 25 are {1,8}. Sue thinks she knows where Sam lives and decides to pay him a visit. She too is mistaken as Sam once again answered only the second question truthfully. Since Sam's house number is greater than 50, he told Sue that it was greater than 25 as well. Since Sue thought she knew which house was his, she must live in either of {27,64}. If I tell you that Sam's number is less than Sue's or Sally's, Since Sam's number is greater than 50, and Sue's is even bigger, she must live in 64. Assuming Sue and Sally are not roommates (although awkward social situations of this kind are not without precedent), Sally lives in 81. and that the sum of their numbers is a perfect square multiplied by two, you should be able to figure out where all three of them live. Sue + Sally + Sam = 2 p^2 for p an integer 64 + 81 + Sam = 2 p^2 Applying the constraint 50 < Sam < 64, looks like Sam = 55 (p = 10). In summary, Sam = 55 Sue = 64 Sally = 81 -- Tom Smith ==> logic/self.ref.p <== Find a number ABCDEFGHIJ such that A is the count of how many 0's are in the number, B is the number of 1's, and so on. ==> logic/self.ref.s <== 6210001000 For other numbers of digits: n=1: no sequence possible n=2: no sequence possible n=3: no sequence possible n=4: 1210, 2020 n=5: 21200 n=6: no sequence possible n=7: 3211000 n=8: 42101000 n=9: 521001000 n=10: 6210001000 n>10: (n-4), 2, 1, 0 * (n-7), 1, 0, 0, 0 No 1, 2, or 3 digit numbers are possible. Letting x_i be the ith digit, starting with 0, we see that (1) x_0 + ... + x_n = n+1 and (2) 0*x_0 + ... + n*x_n = n+1, where n+1 is the number of digits. I'll first prove that x_0 > n-3 if n>4. Assume not, then this implies that at least four of the x_i with i>0 are non-zero. But then we would have \sum_i i*x_i >= 10 by (2), impossible unless n=9, but it isn't possible in this case (51111100000 isn't valid). Now I'll prove that x_0 < n-1. x_0 clearly can't equal n; assume x_0 = n-1 ==> x_{n-1} = 1 by (2) if n>3. Now only one of the remaining x_i may be non-zero, and we must have that x_0 + ... + x_n = n+1, but since x_0 + x_{n-1} = n ==> the remaining x_i = 1 ==> by (2) that x_2 = 1. But this can't be, since x_{n-1} = 1 ==> x_1>0. Now assuming x_0 = n-2 we conclude that x_{n-2} = 1 by (2) if n>5 ==> x_1 + ... + x_{n-3} + x_{n-1} + x_n = 2 and 1*x_1 + ... + (n-3)*x_{n-3} + (n-1)*x_{n-1} + n*x_n = 3 ==> x_1=1 and x_2=1, contradiction. Case n>5: We have that x_0 = n-3 and if n>=7 ==> x_{n-3}=1 ==> x_1=2 and x_2=1 by (1) and (2). For the case n=6 we see that x_{n-3}=2 leads to an easy contradiction, and we get the same result. The cases n=4,5 are easy enough to handle, and lead to the two solutions above. -- -- clong@romulus.rutgers.edu (Chris Long) ==> logic/situation.puzzles.outtakes.p <== The following puzzles have been removed from my situation puzzles list, or never made it onto the list in the first place. There are a wide variety of reasons for the non-inclusion: some I think are obvious, some don't have enough of a story, some involve gimmicks that annoy me, some I think are riddles rather than situation puzzles, and some are so contrary to reality as to be unplayable. Basically, what it comes down to is that I don't like these enough to put them on my list. If you think of ways to make any of them more palatable to me, or to reorganize my entire list, or if you just want to chat, by all means contact me at zorn@apple.com. --jed e. hartman, 5/5/92 ----------------------------------------- Contra-reality puzzles, or, "That's not the way it works]" 2.10. A man is sitting in a train compartment. He sees a three- fingered hand through the compartment window, in the hallway of the train. He opens the compartment door and shoots the person with the three-fingered hand, but he goes free. (Michael Bernstein) 2.61. A man ran into a fire, and lived. A man stayed where there was no fire, and died. (Eric Wang original) 2.50. The pope is giving a speech. A man in the audience shoots the mayor who is behind the pope. (PRO) Date: 2 Feb 92 23:05:11 GMT In article <64023@netnews.upenn.edu>, weemba@libra (Matthew P Wiener) writes: >Here's Ýone¨ I made up years ago: "She stopped having sex. She died." 1.37. A holy man is dead in a room. (Perry Deess original) ----------------------------------------- Clocks, calendars, money, and other numerical trivia: 2.15. Two people are talking long distance on the phone; one is in an East-Coast state, the other is in a West-Coast state. The first asks the other "What time is it?", hears the answer, and says, "That's funny. It's the same time here]" (EMS) 2.19. A woman goes into a convenience store to buy a can of Coke. She pays for it with a $20 bill and receives $22 in change. (MI; partial MB wording) 2.20. A newspaper reported that Jacques Dubois finished first in the walking race held in Paris. The number of miles he walked was given as 62,137. The article was not in error. (AR, quoting Richard Fowell; MB wording) Organization: Penn State University Date: Tuesday, 4 Dec 1990 20:08:00 EST From: SCOTT MATTHEWS A man goes to a hardware store to buy a certain item. He asks the salesman how much this item costs to which he answers, "They are 3 for $1.00." The man say, "Okay I'll take 100," to which the salesman correctly replies, "That will be $1.00." The man pays $1.00 and leaves satisfied. What is the item. >"A man, his son, and his grandson had their first birthday together." (Matthew P Wiener original) ----------------------------------------- Just too weird and/or random and/or silly for me: 2.17. A woman walks up to a door and knocks. Another woman answers the door. The woman outside kills the woman inside. (GH) 2.59. A man is lying dead in a pool of blood and glass. (PRO) 2.60. The seals came up to do their show but immediately dove back into the water. (PRO) 2.58. A raft carrying passengers took a trip down a river. None of the passengers made it home alive. (CR; partial JM wording) ----------------------------------------- Confusing the map with the territory, or, call by reference: 2.22. In his own home a man watches as a woman dies, yet does nothing to save her. (MN) 2.39. King Henry VIII is lying at the bottom of the stairs with a gash across his face. (PRO) 2.40. A man travels to twenty countries and stays in each country for a month. During this time he never sees the light of day. (PRO) ----------------------------------------- How to prove your audience are sexists: 2.48. A boy and his father are injured in a car accident. Both are taken to a hospital. The father dies at arrival, but the boy lives and is taken to surgery. A grey-haired, bespectacled surgeon looks at the boy and says, "I cannot operate on this boy -- he's my son." (JV) 2.49. A husband coming home hears his wife call "Bill, don't kill me]". He walks in and finds his wife dead. Inside are a postman, a doctor, and a lawyer, none of whom the husband knows. The husband immediately realizes the postman killed his wife. (EMS; partial JM wording) ----------------------------------------- Need some work: 2.56. She said "I love you," and died. (EMS) Q. A woman gets up, drives to town, buys a gun, and shoots her husband. > >"He opens his mouth and she dies." (Ivan A Derzhanski) > >"He comes home, undresses, turns the light off and goes to bed. After a few > >minutes he springs up and says, `There's a corpse under my bed]'" (Ivan A Derzhanski) 2.34. A man is holding a box. Though he cannot see into it, he knows what's inside. (Eric Stephan original) ----------------------------------------- Miscellaneous others: 2.24. The telephone rang in the middle of the night and the woman woke up. When she answered it the caller hung up. The caller felt better. (Sasan Soltani) 2.27. A man called to a waiter in a restaurant, "There's a fly in my tea]" "I will bring you a fresh cup of tea," said the waiter. After a few moments, the man called out, "This is the same cup of tea]" How did he know? (PRO) 2.28. A man drives over a broken glass bottle. He travels the last 100 miles of the Sahara 5000 roadrace with a flat tire. (EMS) 2.35. A man was walking along some railroad tracks when he noticed that a train was coming. He ran toward the train before stepping aside. (RM) 2.41. A man puts a quarter down, and leaves. (PRO) 2.44. A dish moves, a scientist makes a discovery. (MN) 2.45. An Arab sheikh tells his two sons that are to race their camels to a distant city to see who will inherit his fortune. The one whose camel arrives last will win. The brothers, after wandering aimlessly for days, ask a wise man for advise. After hearing the advice they jump on the camels and race as fast as they can to their destination. (PRO) 2.46. Two children born in the same hospital, in the same hour, day, and year, have the same mother and father, but are not twins. (Sasan Soltani) 2.47. A couple will build a square house. In each wall they'll have a window, and each window will face north. (Sasan Soltani) The man who built it didn't use it, the man who used it didn't want it, etc. 2.52. A man pleads with his boss not to fly to Chicago. The boss goes anyway, and when he returns, he fires the man. (EMS) 2.53. On an archeological dig, the frozen remains of a man and woman are found. Immediately, the archeologists realize that the remains are those of Adam and Eve. (EMS) 2.54. A man carrying an attache case full of $20 bills falls on the way to the bank and is never seen again. (PRO) Q. A man sees his wife, and later kills her >From klkarp@remus.rutgers.edu Mon Dec 17 22:04:57 1990 Date: Mon, 17 Dec 90 22:07:25 EST (Karen Karp) 4) A guy is trapped in a room with a bed, a calender, a saw and a table. There are no windows or doors (except a vent to breathe if you get technical). How does the guy live and finally escape?? from Joe Kincaid: 6) A man is found dead at his work table. The investigating policeman looks the scene over and immediately declares it to be murder. From: Ivan A Derzhanski Date: Thu, 27 Feb 92 15:03:19 GMT Historical note: The oldest "situation puzzle" (well, kind of) I know of is described in the Maqamat of Al-Hariri. It is actually a puzzle for lawyers, and it goes like this: "A man (X) had a brother (B), and his wife had a brother (WB) too. All of them were free Muslims by birth. When X died, all he left went to WB; B got nothing. How can thbis be lawful?" Date: Sat, 17 Nov 1990 03:14:00 -0500 From: msb@sq.com (Mark Brader) By the way, this one reminds me of the Isaac Asimov story where an agent is shot and gives the dying clue "the blind man". I think that might have been the title, too, I don't remember. 1.35. A policeman follows a burglar into a bar. When he enters the bar he finds two similar-looking men, dressed alike, with the loot between them. After several minutes he arrests one of the men. (PRO, from "Which is Which?" by Isaac Asimov; partial JM wording) ==> logic/situation.puzzles.outtakes.s <== ----------------------------------------- Contra-reality puzzles, or, "That's not the way it works]" 2.10. A man is sitting in a train compartment. He sees a three- fingered hand through the compartment window, in the hallway of the train. He opens the compartment door and shoots the person with the three-fingered hand, but he goes free. (Michael Bernstein) 2.10. He's with a policeman, who's taking him to jail, and he uses the policeman's gun. He was convicted of his wife's murder; she had framed him for it somehow, involving cutting off two of her own fingers and mailing them to the police. Since he had already been convicted of her murder, he couldn't be tried twice for the same crime, and since he obviously hadn't actually been guilty before, he's set free. The main problem with this question is that as far as I know, that's NOT how the "double jeopardy" law works, and so it couldn't happen in real life. Nonetheless, it's a neat setup. ÝFurther info: >From ypay@leland.Stanford.EDU Wed Feb 26 12:06:34 1992 By Cheryl Balbes: Situation: A woman sees a man on a train eating an orange. She shoots and kills him. She is arrested and known to be sane and guilty but does not go to jail. Solution: The man and the woman were married. It was a terrible marriage - so terrible that he wanted revenge for it. So he cut off three of his fingers and burned down the house. The police arrived and arrested the woman for murder of her husband, citing the fingers as evidence of his death. She was tried and convicted and went to jail for most of her life. When she finally got out, she took a train ride and saw a man eating an orange. When he used the orange peeler, she could see that he was missing three fingers so she knew he was her husband. For ruining her life, she took out a gun and shot him dead. She was arrested and known to be guilty, but could not go to jail again for the same crime. Dan Cory¨ ÝIan Collier has a slightly variant answer: 30 years ago, the man and the woman (his wife) attempted to defraud the insurance company by faking her death. However he was found guilty of murdering his wife and given a long prison sentence (his wife remained hidden and made no attempt to prevent the conviction). The man, having just been freed from prison, summons his wife from a far city and shoots her. He is not punished, because he has already served the sentence.¨ 2.61. A man ran into a fire, and lived. A man stayed where there was no fire, and died. (Eric Wang original) 2.61. The two men were computer programmers working in a small room protected by a halon gas fire extinguisher system, when a fire broke out in an adjoining room. One of the programmers ran through the fire and escaped with only minor burns. The other one stayed in the room until the building's fire extinguishers kicked in, and died of oxygen starvation when the halon gas combined with all of the oxygen in the room. I'm told by Rolf Wilson that this really isn't how halon gas extinguishers work, so this puzzle should unfortunately be removed from the list. 2.50. The pope is giving a speech. A man in the audience shoots the mayor who is behind the pope. (PRO) 2.50. The pope has returned to the village where he began his priesthood fifty years earlier. He was late for the ceremony, so the mayor spoke first; he claimed to be the first person to give confession to the pope, fifty years earlier. When the pope arrived, he related that the first confession he had heard was that of the murder of a young woman. The man in the audience had a sister who was murdered at that time. The sanctity of the confessional is conveniently ignored. Date: 2 Feb 92 23:05:11 GMT In article <64023@netnews.upenn.edu>, weemba@libra (Matthew P Wiener) writes: >Here's Ýone¨ I made up years ago: "She stopped having sex. She died." A woman was a hemophiliac. She stayed pregnant whenever possible. When she stopped having sex, she had her first period, and bled to death. Ýjeh comments: there was a lot of debate on the Net about this, most of which tended to conclude that (a) menstrual fluid isn't blood; (b) most of the few female hemophiliacs die at birth; and (c) terminating a pregnancy in any conceivable (heh) way is likely to result in too much blood loss.¨ 1.37. A holy man is dead in a room. (Perry Deess original) 1.37. The man is a Moslem. He was caught stealing, and so his right hand was cut off. However, he's very devout, and thus isn't allowed to eat using his left hand; so he starved to death. It could be argued that this isn't very realistic; he could've gotten someone to feed him, or somehow eaten without using his hands at all. And I don't know if those rules of Islam really interact that way. Any real info would be appreciated. ÝIvan A Derzhanski writes: They are supposed to use their left hand. There is more than one tale in the _Arabian Nights_ about someone who has lost his right hand, typically as a punishment for theft (and typically unjustly). He conceals this fact (long sleeves and all that), until he is served a meal and the storyteller sees him eating with his left hand, something which is not exactly taboo, but is not the done thing either. Note that the punishment for theft is known to be loss of the right hand, not death by starvation. I would be more interested to know what a Hindu would do. Those people are much more careful about what is to be done with which hand. And, of course, there are many ways to lose a hand.¨ ----------------------------------------- Clocks, calendars, money, and other numerical trivia: 2.15. Two people are talking long distance on the phone; one is in an East-Coast state, the other is in a West-Coast state. The first asks the other "What time is it?", hears the answer, and says, "That's funny. It's the same time here]" (EMS) 2.15. One is in Eastern Oregon (in Mountain time), the other in Western Florida (in Central time), and it's daylight-savings changeover day at 1:30 AM. 2.15a. Variant answer: The east-coast state is in the USA, on Eastern Daylight time. The west-coast state is Western Australia. There's a twelve-hour time difference, so it's 8 o'clock in both places. (from Tim Lambert.) 2.19. A woman goes into a convenience store to buy a can of Coke. She pays for it with a $20 bill and receives $22 in change. (MI; partial MB wording) 2.19. It's in Canada; she pays in American money and receives change in Canadian money. Mark Brader points out that the amount of change can vary wildly depending on the price of the drink as well as both the official exchange rate and the actual exchange rate given. 2.20. A newspaper reported that Jacques Dubois finished first in the walking race held in Paris. The number of miles he walked was given as 62,137. The article was not in error. (AR, quoting Richard Fowell; MB wording) 2.20. The comma, in European numbers, is used the same way Americans use a decimal point. The man thus (Americans would say) walked 62.137 miles, or 100 km. Mark Brader points out that no newspaper in a country which uses the decimal comma would report the distance in miles; if anyone can think of a way around this problem, let me know. One possibility is to say that he walked 42,551 km (or whatever the actual length of a marathon is); but I don't know whether there are such things as walking marathons, or whether they'd be the same length in Europe as in America. Organization: Penn State University Date: Tuesday, 4 Dec 1990 20:08:00 EST From: SCOTT MATTHEWS A man goes to a hardware store to buy a certain item. He asks the salesman how much this item costs to which he answers, "They are 3 for $1.00." The man say, "Okay I'll take 100," to which the salesman correctly replies, "That will be $1.00." The man pays $1.00 and leaves satisfied. What is the item. ÝA: house numbers.¨ >"A man, his son, and his grandson had their first birthday together." (Matthew P Wiener original) David Grabiner's answer: Sweden had no leap-year days from 1748 to 1788 in order to catch up with the Gregorian calendar without creating excessive trouble. (In many other countries, people found that their loans suddenly became due eleven days earlier.) Thus, the grandfather was born in Sweden on February 29, 1744; the other two were born outside Sweden on February 29, 1768 and 1788, and returned to Sweden before their fourth birthdays. Ýjeh comments: this is, as Matt said in his original posting, an obscure-calendar puzzle; and the net indicated that the calendar story may not be true anyway.¨ ----------------------------------------- Just too weird and/or random and/or silly for me: 2.17. A woman walks up to a door and knocks. Another woman answers the door. The woman outside kills the woman inside. (GH) 2.17. The woman outside is a psychotic librarian. The woman inside has an extremely overdue book. 2.17a. Variant answer: The woman outside is married and lived at the home in question. She misplaced her key, and the door was answered by her husband's lover. Though this answer would allow the question to be in section 1, it's really a much-less-interesting version of #1.15, and it seems to me that it would be a fairly obvious answer. 2.59. A man is lying dead in a pool of blood and glass. (PRO) 2.59. The man caught a large fish and was so excited he went to a phone booth to call his wife. In trying to describe the size of the fish, he said, "It was THIS big]" and stretched his arms wide to indicate its length. His arms went through the sides of the phone booth, his wrists were sliced by broken glass, and he bled to death. ÝVariant from Bernd Wechner: A man makes a telephone call and dies. Answer: The man was ringing his wife, and learned from her that he had won the lottery. In jumping for joy he broke through the glass wall of the telephone booth and cut his wrists whereupon he bled to death.¨ 2.60. The seals came up to do their show but immediately dove back into the water. (PRO) 2.60. The seals were frightened by an audience of nuns, who, to the seals, looked like a herd of killer whales. 2.58. A raft carrying passengers took a trip down a river. None of the passengers made it home alive. (CR; partial JM wording) 2.58. The raft was floating down the Amazon river when it floated under a big tree. A snake was hanging down out of the tree, so the people pushed the entire raft away from the tree, where it capsized. The passengers were then eaten by piranha. ----------------------------------------- Confusing the map with the territory, or, call by reference: 2.22. In his own home a man watches as a woman dies, yet does nothing to save her. (MN) 2.22. He saw it happening on TV. 2.39. King Henry VIII is lying at the bottom of the stairs with a gash across his face. (PRO) 2.39. It is a painting of Henry VIII. 2.40. A man travels to twenty countries and stays in each country for a month. During this time he never sees the light of day. (PRO) 2.40. The man is a mummy, on tour to different museums throughout the world. Ýnote similarity of type to "ship at bottom of sea" and "husband who'd blown his brains out."¨ ----------------------------------------- How to prove your audience are sexists: 2.48. A boy and his father are injured in a car accident. Both are taken to a hospital. The father dies at arrival, but the boy lives and is taken to surgery. A grey-haired, bespectacled surgeon looks at the boy and says, "I cannot operate on this boy -- he's my son." (JV) 2.48. The surgeon is the boy's mother. As with #2.45, #2.46, and #2.47, I've frequently heard this as presented as a riddle; the attributions for these indicate the first person to tell them to me as mystery questions. 2.49. A husband coming home hears his wife call "Bill, don't kill me]". He walks in and finds his wife dead. Inside are a postman, a doctor, and a lawyer, none of whom the husband knows. The husband immediately realizes the postman killed his wife. (EMS; partial JM wording) 2.49. The postman is a man. The doctor and lawyer are women. ----------------------------------------- Need some work: 2.56. She said "I love you," and died. (EMS) 2.56. She was a circus performer who performed rope tricks. During one of them, she hung from the ceiling holding only a rope in her mouth. The other end of the rope was held by her husband. There's no motivation given for her choosing to do something so stupid; if anyone wants to twiddle this into a more reasonable question, please do. Q. A woman gets up, drives to town, buys a gun, and shoots her husband. A. The woman suspects her husband of cheating on her. She notes the mileage on the car each day. The previous night, hubby worked late at the office, but the mileage on the car is far greater than can be accounted for. (from Simon Travaglia) Ýjeh comments: This last could make a really nice puzzle, but has too many plotholes as it stands. Rework it sometime.¨ > >"He opens his mouth and she dies." (Ivan A Derzhanski) The male acrobat hangs from the ceiling and holds the female acrobat by his teeth. He drops her, and she breaks her spine. Ýjeh comments: again, this needs more of a real story for me to accept it.¨ > >"He comes home, undresses, turns the light off and goes to bed. After a few > >minutes he springs up and says, `There's a corpse under my bed]'" (Ivan A Derzhanski) He hears a watch tick under the bed. Why the watch has to be on the hand of someone (and if it is, he is obviously dead, because his breath is not heard) is left to the guessers' discretion. Ýjeh comments: see previous comment.¨ 2.34. A man is holding a box. Though he cannot see into it, he knows what's inside. (Eric Stephan original) 2.34. He's allergic to whatever's inside the box. Ýjeh comments: how is this different from just being able to smell it?¨ ----------------------------------------- Miscellaneous others: 2.24. The telephone rang in the middle of the night and the woman woke up. When she answered it the caller hung up. The caller felt better. (Sasan Soltani) 2.24. It was a husband calling from overseas to see that his wife arrived home all right. Hanging up before three seconds elapse results in no charge to the calling party. He could not call person-to-person because the local operators did not speak English. 2.27. A man called to a waiter in a restaurant, "There's a fly in my tea]" "I will bring you a fresh cup of tea," said the waiter. After a few moments, the man called out, "This is the same cup of tea]" How did he know? (PRO) 2.27. The man had already sugared his tea before sending it back. 2.28. A man drives over a broken glass bottle. He travels the last 100 miles of the Sahara 5000 roadrace with a flat tire. (EMS) 2.28. The flat tire is his spare. 2.35. A man was walking along some railroad tracks when he noticed that a train was coming. He ran toward the train before stepping aside. (RM) 2.35. The man was on a bridge, closer to the end the train was approaching from. 2.41. A man puts a quarter down, and leaves. (PRO) 2.41. The man has put a quarter of the cost of a new car into a down payment; he then drives away in the car. 2.44. A dish moves, a scientist makes a discovery. (MN) 2.44. The dish is a satellite dish. 2.45. An Arab sheikh tells his two sons that are to race their camels to a distant city to see who will inherit his fortune. The one whose camel arrives last will win. The brothers, after wandering aimlessly for days, ask a wise man for advise. After hearing the advice they jump on the camels and race as fast as they can to their destination. (PRO) 2.45. The wise man tells them to switch camels. 2.46. Two children born in the same hospital, in the same hour, day, and year, have the same mother and father, but are not twins. (Sasan Soltani) 2.46. The children are two of a set of triplets. Ýjeh wonders: is this fundamentally different from the people crowding under an umbrella or the black-painted town? Should all three be together on one list or the other?¨ 2.47. A couple will build a square house. In each wall they'll have a window, and each window will face north. (Sasan Soltani) 2.47. The house is at the south pole. This is much the same question as the age-old riddle asking what color a certain dead bear is. The man who built it didn't use it, the man who used it didn't want it, etc. (A: coffin.) Suggested as a story riddle by Ed Wagner. Ýjeh sez: If I included this, I'd feel obliged to include every riddle I've ever heard.¨ 2.52. A man pleads with his boss not to fly to Chicago. The boss goes anyway, and when he returns, he fires the man. (EMS) 2.52. The man was a night watchman who told his boss that last night he had a dream that the boss would die in a plane crash. The boss fired him for sleeping on the job. 2.53. On an archeological dig, the frozen remains of a man and woman are found. Immediately, the archeologists realize that the remains are those of Adam and Eve. (EMS) 2.53. The two bodies lacked what only Adam and Eve would lack -- bellybuttons. 2.54. A man carrying an attache case full of $20 bills falls on the way to the bank and is never seen again. (PRO) 2.54. The man falls off the river bank and drowns. Q. A man sees his wife, and later kills her A. The man sees his wife "performing" at a peep show. (from Simon Travaglia) >From klkarp@remus.rutgers.edu Mon Dec 17 22:04:57 1990 Date: Mon, 17 Dec 90 22:07:25 EST (Karen Karp) 4) A guy is trapped in a room with a bed, a calender, a saw and a table. There are no windows or doors (except a vent to breathe if you get technical). How does the guy live and finally escape?? 4) He eats dates from the calender, drinks water from the springs in the bed and saws the table in half, 2 halves make a whole and he escapes out the hole. from Joe Kincaid: 6) A man is found dead at his work table. The investigating policeman looks the scene over and immediately declares it to be murder. The man is a blind hemophiliac who *always* keeps his work table in precise order because of his condition(s). When he reached for his awl, it was turned upside-down and he impaled his hand on it. Being a hemophiliac, he bled to death. This couldn't happen by accident. Ýjeh comments: I suppose this *could* happen in real life; but it seems to me that it *could* happen by accident. Perhaps this is no less plausible than the "blind midget"-type puzzles; I'm ambivalent about it. But I'm leaving it out for now.¨ From: Ivan A Derzhanski Date: Thu, 27 Feb 92 15:03:19 GMT Historical note: The oldest "situation puzzle" (well, kind of) I know of is described in the Maqamat of Al-Hariri. It is actually a puzzle for lawyers, and it goes like this: "A man (X) had a brother (B), and his wife had a brother (WB) too. All of them were free Muslims by birth. When X died, all he left went to WB; B got nothing. How can thbis be lawful?" Solution: "X had married his son (S) to his mother-in-law (WM). S died, but left a son, who turned out to be a brother of X's wife (being a son of her mother, WM), but at the same time he is a grandson of X, and the grandson, as a direct descendant, has more rights to the heritage than the brother." Date: Sat, 17 Nov 1990 03:14:00 -0500 From: msb@sq.com (Mark Brader) By the way, this one reminds me of the Isaac Asimov story where an agent is shot and gives the dying clue "the blind man". I think that might have been the title, too, I don't remember. The solution: the cover role of the enemy agent who shot him was a repairman, and he got admission to the premises to fix a broken window blind. Ýjeh comments: reminds me of "The ÝCase of the?¨ Three Blind Mice," in which a dying man gasps that the one who killed him was "Mice..." Turns out not to be "my s...ister" or "my s...on" or the character named "Muyskins," but "my s...olicitor"; the guy is British.¨ 1.35. A policeman follows a burglar into a bar. When he enters the bar he finds two similar-looking men, dressed alike, with the loot between them. After several minutes he arrests one of the men. (PRO, from "Which is Which?" by Isaac Asimov; partial JM wording) 1.35. Both men were wearing glasses. The burglar, however, was wearing photosensitive sunglasses; the policeman noticed them changing shade and realized the man must have just entered. ==> logic/situation.puzzles.p <== Jed's List of Situation Puzzles History: original compilation 11/28/87 major revision 08/09/89 further additions 08/23/89 - 10/21/90 variants added to answer list 07/04/90 editing and renumbering 07/25/90 - 11/11/90 items removed; title changed 09/20/90 - 11/11/90 editing and additions 02/26/92 - 09/17/92 "A man lies dead in a room with fifty-three bicycles in front of him. What happened?" This is a list of what I refer to (for lack of a better name) as situation puzzles. In the game of situation puzzles, a situation like the one above is presented to a group of players, who must then try to find out more about the situation by asking further questions. The person who initially presented the situation can only answer "yes" or "no" to questions (or occasionally "irrelevant" or "doesn't matter"). My list has been divided into two sections. Section 1 consists of situation puzzles which are set in a realistic world; the situations could all actually occur. Section 2 consists of puzzles which involve double meanings for one or more words and those which could not possibly take place in reality as we know it, plus a few miscellaneous others. See the end of the list for more notes and comments. Section 1: "Realistic" situation puzzles. 1.1. In the middle of the ocean is a yacht. Several corpses are floating in the water nearby. (SJ) 1.2. A man is lying dead in a room. There is a large pile of gold and jewels on the floor, a chandelier attached to the ceiling, and a large open window. (DVS; partial JM wording) 1.3. A woman came home with a bag of groceries, got the mail, and walked into the house. On the way to the kitchen, she went through the living room and looked at her husband, who had blown his brains out. She then continued to the kitchen, put away the groceries, and made dinner. (partial JM wording) 1.4. A body is discovered in a park in Chicago in the middle of summer. It has a fractured skull and many other broken bones, but the cause of death was hypothermia. (MI, from _Hill Street Blues_) 1.5. A man lives on the twelfth floor of an apartment building. Every morning he takes the elevator down to the lobby and leaves the building. In the evening, he gets into the elevator, and, if there is someone else in the elevator -- or if it was raining that day -- he goes back to his floor directly. However, if there is nobody else in the elevator and it hasn't rained, he goes to the 10th floor and walks up two flights of stairs to his room. (MH) 1.6. A woman has incontrovertible proof in court that her husband was murdered by her sister. The judge declares, "This is the strangest case I've ever seen. Though it's a cut-and-dried case, this woman cannot be punished." (This is different from #1.43.) (MH) 1.7. A man walks into a bar and asks for a drink. The bartender pulls out a gun and points it at him. The man says, "Thank you," and walks out. (DVS) 1.8. A man is returning from Switzerland by train. If he had been in a non-smoking car he would have died. (DVS; MC wording) 1.9. A man goes into a restaurant, orders abalone, eats one bite, and kills himself. (TM and JM wording) 1.10. A man is found hanging in a locked room with a puddle of water under his feet. (This is different from #1.11.) 1.11. A man is dead in a puddle of blood and water on the floor of a locked room. (This is different from #1.10.) 1.12. A man is lying, dead, face down in the desert wearing a backpack. (This is different from #1.13, #2.11, and #2.12.) 1.13. A man is lying face down, dead, in the desert, with a match near his outstretched hand. (This is different from #1.12, #2.11, and #2.12.) (JH; partial JM wording) 1.14. A man is driving his car. He turns on the radio, listens for five minutes, turns around, goes home, and shoots his wife. (This is different from #1.15.) 1.15. A man driving his car turns on the radio. He then pulls over to the side of the road and shoots himself. (This is different from #1.14.) 1.16. Music stops and a woman dies. (DVS) 1.17. A man is dead in a room with a small pile of pieces of wood and sawdust in one corner. (from "Coroner's Inquest," by Marc Connelly) 1.18. A flash of light, a man dies. (ST original) 1.19. A rope breaks. A bell rings. A man dies. (KH) 1.20. A woman buys a new pair of shoes, goes to work, and dies. (DM) 1.21. A man is riding a subway. He meets a one-armed man, who pulls out a gun and shoots him. (SJ) 1.22. Two women are talking. One goes into the bathroom, comes out five minutes later, and kills the other. 1.23. A man is sitting in bed. He makes a phone call, saying nothing, and then goes to sleep. (SJ) 1.24. A man kills his wife, then goes inside his house and kills himself. (DH original, from "Nightmare in Yellow," by Fredric Brown) 1.25. Abel walks out of the ocean. Cain asks him who he is, and Abel answers. Cain kills Abel. (MWD original) 1.26. Two men enter a bar. They both order identical drinks. One lives; the other dies. (CR; partial JM wording) 1.27. Joe leaves his house, wearing a mask and carrying an empty sack. An hour later he returns. The sack is now full. He goes into a room and turns out the lights. (AL) 1.28. A man takes a two-week cruise to Mexico from the U.S. Shortly after he gets back, he takes a three-day cruise which doesn't stop at any other ports. He stays in his cabin all the time on both cruises. As a result, he makes $250,000. (MI, from "The Wager") 1.29. Hans and Fritz are German spies during World War II. They try to enter America, posing as returning tourists. Hans is immediately arrested. (JM) 1.30. Tim and Greg were talking. Tim said "The terror of flight." Greg said "The gloom of the grave." Greg was arrested. (MPW original, from "No Refuge Could Save," by Isaac Asimov) 1.31. A man is found dead in his parked car. Tire tracks lead up to the car and away. (SD) 1.32. A man dies in his own home. (ME original) 1.33. A woman in Paris in 1895 is waiting for her husband to come home. When he arrives, the house has burned to the ground and she's dead. (JM) 1.34. A man gets onto an elevator. When the elevator stops, he knows his wife is dead. (LA; partial KH wording) 1.35. Three men die. On the pavement are pieces of ice and broken glass. (JJ) 1.36. She lost her job when she invited them to dinner. (DS original) 1.37. A man is running along a corridor with a piece of paper in his hand. The lights flicker and the man drops to his knees and cries out, "Oh no]" (MP) 1.38. A car without a driver moves; a man dies. (EMS) 1.39. As I drive to work on my motorcycle, there is one corner which I go around at a certain speed whether it's rainy or sunny. If it's cloudy but not raining, however, I usually go faster. (SW original) 1.40. A woman throws something out a window and dies. (JM) 1.41. An avid birdwatcher sees an unexpected bird. Soon he's dead. (RSB original) 1.42. There are a carrot, a pile of pebbles, and a pipe lying together in the middle of a field. (PRO; partial JM wording) 1.43. Two brothers are involved in a murder. Though it's clear that one of them actually committed the crime, neither can be punished. (This is different from #1.6.) (from "Unreasonable Doubt," by Stanley Ellin) 1.44. An ordinary American citizen, with no passport, visits over thirty foreign countries in one day. He is welcomed in each country, and leaves each one of his own accord. (PRO) 1.45. If he'd turned on the light, he'd have lived. (JM) 1.46. A man is found dead on the floor in the living room. (ME original) 1.47. A man is found dead outside a large building with a hole in him. (JM, modified from PRO) 1.48. A man is found dead in an alley lying in a red pool with two sticks crossed near his head. (PRO) 1.49. A man lies dead next to a feather. (PRO) 1.50. There is blood on the ceiling of my bedroom. (MI original) 1.51. A man wakes up one night to get some water. He turns off the light and goes back to bed. The next morning he looks out the window, screams, and kills himself. (CR; KK wording) 1.52. She grabbed his ring, pulled on it, and dropped it. (JM, from _Math for Girls_) 1.53. A man sitting on a park bench reads a newspaper article headlined "Death at Sea" and knows a murder has been committed. 1.54. A man tries the new cologne his wife gave him for his birthday. He goes out to get some food, and is killed. (RW original) 1.55. A man in uniform stands on the beach of a tropical island. He takes out a cigarette, lights it, and begins smoking. He takes out a letter and begins reading it. The cigarette burns down between his fingers, but he doesn't throw it away. He cries. (RW) 1.56. A man went into a restaurant, had a large meal, and paid nothing for it. (JM original) 1.57. A married couple goes to a movie. During the movie the husband strangles the wife. He is able to get her body home without attracting attention. (from _Beyond the Easy Answer_) Section 2: Double meanings, fictional settings, and miscellaneous others. 2.1. A man shoots himself, and dies. (HL) (This is different from #2.2.) 2.2. A man walks into a room, shoots, and kills himself. (HL) (This is different from #2.1.) 2.3. Adults are holding children, waiting their turn. The children are handed (one at a time, usually) to a man, who holds them while a woman shoots them. If the child is crying, the man tries to stop the crying before the child is shot. (ML) 2.4. Hiking in the mountains, you walk past a large field and camp a few miles farther on, at a stream. It snows in the night, and the next day you find a cabin in the field with two dead bodies inside. (KL; KD and partial JM wording) 2.5. A man marries twenty women in his village but isn't charged with polygamy. 2.6. A man is alone on an island with no food and no water, yet he does not fear for his life. (MN) 2.7. Joe wants to go home, but he can't go home because the man in the mask is waiting for him. (AL wording) 2.8. A man is doing his job when his suit tears. Fifteen minutes later, he's dead. (RM) 2.9. A dead man lies near a pile of bricks and a beetle on top of a book. (MN) 2.10. At the bottom of the sea there lies a ship worth millons of dollars that will never be recovered. (TF original) 2.11. A man is found dead in the arctic with a pack on his back. (This is different from #1.12, #1.13, and #2.12.) (PRO) 2.12. There is a dead man lying in the desert next to a rock. (This is different from #1.12, #1.13, and #2.11.) (GH) 2.13. As a man jumps out of a window, he hears the telephone ring and regrets having jumped. (from "Some Days are Like That," by Bruce J. Balfour; partial JM wording) 2.14. Two people are playing cards. One looks around and realizes he's going to die. (JM original) 2.15. A man lies dead in a room with fifty-three bicycles in front of him. 2.16. A horse jumps over a tower and lands on a man, who disappears. (ES original) 2.17. A train pulls into a station, but none of the waiting passengers move. (MN) 2.18. A man pushes a car up to a hotel and tells the owner he's bankrupt. (DVS; partial AL and JM wording) 2.19. Three large people try to crowd under one small umbrella, but nobody gets wet. (CC) 2.20. A black man dressed all in black, wearing a black mask, stands at a crossroads in a totally black-painted town. All of the streetlights in town are broken. There is no moon. A black-painted car without headlights drives straight toward him, but turns in time and doesn't hit him. (AL and RM wording) 2.21. Bob and Carol and Ted and Alice all live in the same house. Bob and Carol go out to a movie, and when they return, Alice is lying dead on the floor in a puddle of water and glass. It is obvious that Ted killed her but Ted is not prosecuted or severely punished. 2.22. A man rides into town on Friday. He stays one night and leaves on Friday. (KK) 2.23. Bruce wins the race, but he gets no trophy. (EMS) 2.24. A woman opens an envelope and dyes. (AL) 2.25. A man was brought before a tribal chief, who asked him a question. If he had known the answer, he probably would have died. He didn't, and lived. (MWD original) 2.26. Two men are found dead outside of an igloo. (SK original) Attributions key: When I know who first told me the current version of a puzzle, I've put initials in parentheses after the puzzle statement; this is the key to those acknowledgments. The word "original" following an attribution means that, to the best of my knowledge, the cited person invented that puzzle. If a given puzzle isn't marked "original" but is attributed, that just means that's the first person I heard it from. I would appreciate it if attributions for originals were not removed; however, this list is hereby entered into the public domain, so do with it what you wish. LA == Laura Almasy RSB == Ranjit S. Bhatnagar CC == Chris Cole MC == Matt Crawford MWD == Matthew William Daly KD == Ken Duisenberg SD == Sylvia Dutcher ME == Marguerite Eisenstein TF == Thomas Freeman JH == Joaquin Hartman MH == Marcy Hartman KH == Karl Heuer GH == Geoff Hopcraft DH == David Huddleston MI == Mark Isaak SJ == Steve Jacquot JJ == J!rgen Jensen KK == Karen Karp SK == Shelby Kilmer KL == Ken Largman AL == Andy Latto HL == Howard Lazoff ML == Merlyn LeRoy RM == "Reaper Man" (real name unknown) TM == Ted McCabe JM == Jim Moskowitz DM == Damian Mulvena MN == Jan Mark Noworolski PRO == Peter R. Olpe (from his list) MP == Martin Pitwood CR == Charles Renert EMS == Ellen M. Sentovich (from her list) ES == Eric Stephan DS == Diana Stiefbold ST == Simon Travaglia DVS == David Van Stone RW == Randy Whitaker MPW == Matthew P Wiener SW == Steve Wilson (not sure of name) Special thanks to Jim Moskowitz, Karl Heuer, and Mark Brader, for a lot of discussion of small but important details and wording. Notes and comments: My outtakes list (items removed from this list for various reasons, most of which came down to the fact that I didn't like them) is now available from the r.p FAQ server. There are many possible wordings for most of the puzzles in this list. Most of them have what I consider the best wording of the variants I've heard; if you think there's a better way of putting one or more of them, or if you don't like my categorization of any of them, or if you have any other comments or suggestions, please drop me a note. If you know others not on this list, please send them to me. Of course, in telling a group of players one of these situations, you can add or remove details, either to make getting the answer harder or easier, or simply to throw in red herrings. I've made a few specific suggestions along these lines in the answer list, available in a separate file. Also in the answer list are variant problem statements and variant answers. --Jed Hartman zorn@apple.com (as of 9/92) ==> logic/situation.puzzles.s <== Answers to Jed's List of Situation Puzzles This is the list of answers to the puzzles in my situation puzzles list. See that list for more details. This document also contains variant setups and answers for some of the puzzles. Section 1: "Realistic" situation puzzles. 1.1. A bunch of people are on an ocean voyage in a yacht. One afternoon, they all decide to go swimming, so they put on swimsuits and dive off the side into the water. Unfortunately, they forget to set up a ladder on the side of the boat, so there's no way for them to climb back in, and they drown. 1.1a. Variant answer: The same situation, except that they set out a ladder which is just barely long enough. When they all dive into the water, the boat, without their weight, rises in the water until the ladder is just barely out of reach. (also from Steve Jacquot) 1.2. The room is the ballroom of an ocean liner which sank some time ago. The man ran out of air while diving in the wreck. 1.2a. Variant which puts this in section 2: same statement, ending with "a large window through which rays are coming." Answer: the rays are manta rays (this version tends to make people assume vampires are involved, unless they notice the awkwardness of the phrase involving rays). 1.3. The husband killed himself a while ago; it's his ashes in an urn on the mantelpiece that the wife looks at. It's debatable whether this belongs in section 2 for double meanings. 1.4. A poor peasant from somewhere in Europe wants desperately to get to the U.S. Not having money for airfare, he stows away in the landing gear compartment of a jet. He dies of hypothermia in mid-flight, and falls out when the landing gear compartment opens as the plane makes its final approach. 1.4a. Variant: A man is lying drowned in a dead forest. Answer: He's scuba diving when a firefighting plane lands nearby and fills its tanks with water, sucking him in with the water. He runs out of air while the plane is in flight; the plane then dumps its load of water, with him in it, onto a burning forest. (from Jim Moskowitz) 1.5. The man is a midget. He can't reach the upper elevator buttons, but he can ask people to push them for him. He can also push them with his umbrella. I've usually heard this stated with more details: "Every morning he wakes up, gets dressed, eats, goes to the elevator..." Ron Carter suggests a nice red herring: the man lives on the 13th floor of the building. 1.6. The sisters are Siamese twins. 1.6a. Variant: A man and his brother are in a bar drinking. They begin to argue (as always) and the brother won't get out of the man's face, shouting and cursing. The man, finally fed up, pulls out a pistol and blows his brother's brains out. He sits down to die. Answer: They are Siamese twins. In the original story, the argument started when one complained about the other's bad hygiene and bad breath. The shooter bled to death (from his brother's wounds) by the time the police arrived. (from Randy Whitaker, based on a 1987 _Weekly World News_ story) 1.7. The man has hiccups; the bartender scares them away by pulling a gun. 1.8. The man used to be blind; he's now returning from an eye operation which restored his sight. He's spent all his money on the operation, so when the train (which has no internal lighting) goes through a tunnel he at first thinks he's gone blind again and almost decides to kill himself. Fortunately, the light of the cigarettes people are smoking convinces him that he can still see. 1.8a. Variant: A man dies on a train he does not ordinarily catch. Answer: The man (a successful artist) has had an accident in which he injured his eyes. His head is bandaged and he has been warned not to remove the bandages under any circumstances lest the condition be irreversibly aggravated. He catches the train home from the hospital and cannot resist peeking. Seeing nothing at all (the same train-in-tunnel situation as above obtains, but without the glowing cigarettes this time), he assumes he is blinded and kills himself in grief. I like this version a lot, except that it makes much less sense that he'd be traveling alone. (from Bernd Wechner) 1.9. The man was in a ship that was wrecked on a desert island. When there was no food left, another passenger brought what he said was abalone but was really part of the man's wife (who had died in the wreck). The man suspects something fishy, so when they finally return to civilization, he orders abalone, realizes that what he ate before was his wife, and kills himself. 1.9a. Variant: same problem statement but with albatross instead of abalone. Answer: In this version, the man was in a lifeboat, with his wife, who died. He hallucinated an albatross landing in the boat which he caught and killed and ate; he thought that his wife had been washed overboard. When he actually eats albatross, he discovers that he had actually eaten his wife. 1.9b. Variant answer to 1.9a, with a slightly different problem statement: the man already knew that he had been eating human flesh. He asks the waiter in the restaurant what kind of soup is available, and the waiter responds, "Albatross soup." Thinking that "albatross soup" means "human soup," and sickened by the thought of such a society (place in a foreign country if necessary), he kills himself. (from Mike Neergaard) 1.10. He stood on a block of ice to hang himself. The fact that there's no furniture in the room can be added to the statement, but if it's mentioned in conjunction with the puddle of water the answer tends to be guessed more easily. 1.11. He stabbed himself with an icicle. 1.12. He jumped out of an airplane, but his parachute failed to open. Minor variant wording (from Joe Kincaid): he's on a mountain trail instead of in a desert. Minor variant wording (from Mike Reymond): he's got a ring in his hand (it came off of the ripcord). 1.12a. Silly variant: same problem statement, with the addition that one of the man's shoelaces is untied. Answer: He pulled his shoelace instead of the ripcord. 1.12b. Variant answer: The man was let loose in the desert with a pack full of poisoned food. He knows it's poisoned, and doesn't eat it -- he dies of hunger. (from Mike Neergaard) 1.13. He was with several others in a hot air balloon crossing the desert. The balloon was punctured and they began to lose altitude. They tossed all their non-essentials overboard, then their clothing and food, but were still going to crash in the middle of the desert. Finally, they drew matches to see who would jump over the side and save the others; this man lost. Minor variant wording: add that the man is nude. 1.14. The radio program is one of the call-up-somebody-and-ask-them-a- question contest shows; the announcer gives the phone number of the man's bedroom phone as the number he's calling, and a male voice answers. It's been suggested that such shows don't usually give the phone number being called; so instead the wife's name could be given as who's being called, and there could be appropriate background sounds when the other man answers the phone. 1.15. He worked as a DJ at a radio station. He decided to kill his wife, and so he put on a long record and quickly drove home and killed her, figuring he had a perfect alibi: he'd been at work. On the way back he turns on his show, only to discover that the record is skipping. 1.15a. Variant: The music stops and the man dies. Answer: The same, except it's a tape breaking instead of a record skipping. (from Michael Killianey) (See also #1.16, #1.19e, and #1.34a.) 1.16. The woman is a tightrope walker in a circus. Her act consists of walking the rope blindfolded, accompanied by music, without a net. The musician (organist, or calliopist, or pianist, or whatever) is supposed to stop playing when she reaches the end of the rope, telling her that it's safe to step off onto the platform. For unknown reasons (but with murderous intent), he stops the music early, and she steps off the rope to her death. 1.16a. Variant answer: The woman is a character in an opera, who "dies" at the end of her song. 1.16b. Variant answer: The "woman" is the dancing figure atop a music box, who "dies" when the box runs down. (Both of the above variants would probably require placing this puzzle in section 2 of the list.) 1.16c. Variant: Charlie died when the music stopped. Answer: Charlie was an insect sitting on a chair; the music playing was for the game Musical Chairs. (from Bob Philhower) (See also #1.15a, #1.19e, and #1.34a.) 1.17. The man is a blind midget, the shortest one in the circus. Another midget, jealous because he's not as short, has been sawing small pieces off of the first one's cane every night, so that every day he thinks he's taller. Since his only income is from being a circus midget, he decides to kill himself when he gets too tall. 1.17a. Slightly variant answer: Instead of sawing pieces off of the midget's cane, someone has sawed the legs off of his bed. He wakes up, stands up, and thinks he's grown during the night. 1.17b. Variant: A pile of sawdust, no net, a man dies. Answer: A midget is jealous of the clown who walks on stilts. He saws partway through the stilts; the clown walks along and falls and dies when they break. (from Peter R. Olpe) 1.17c. Rough sketch of variant: There were a mirror and a bottle on the table, and sawdust on the floor. He came in and dropped dead. Answer: He was a midget, but he wasn't aware of it, because the table used to be too high for him to see his reflection in the mirror, until someone shortened its legs. He was horrified by the discovery, and the shock killed him. (vaguely remembered by Ivan A Derzhanski, who adds that this would be best used as raw material for some elaboration. I agree; it's pretty implausible as is) 1.18. The man is a lion-tamer, posing for a photo with his lions. The lions react badly to the flash of the camera, and the man can't see properly, so he gets mauled. 1.18a. Variant: He couldn't find a chair, so he died. Answer: He was a lion-tamer. This one is kind of silly, but I like it, and it sounds possible to me (though I'm told a whip is more important than a chair to a lion-tamer). (from "Reaper Man," with Karl Heuer wording) 1.19. A blind man enjoys walking near a cliff, and uses the sound of a buoy to gauge his distance from the edge. One day the buoy's anchor rope breaks, allowing the buoy to drift away from the shore, and the man walks over the edge of the cliff. 1.19a. Variant: A bell rings. A man dies. A bell rings. Answer: A blind swimmer sets an alarm clock to tell him when and what direction to go to shore. The first bell is a buoy, which he mistakenly swims to, getting tired and drowning. Then the alarm clock goes off. In other variations, the first bell is a ship's bell, and/or the second bell is a hand-bell rung by a friend on shore at a pre-arranged time. 1.19b. Variant answer to 1.19a: The man falls off a belltower, pulling the bell-cord (perhaps he was climbing a steeple while hanging onto the rope), and dies. The second bell is one rung at his funeral. Could also be a variant on 1.19 (as suggested by Mike Neergaard): the bell-cord breaks when he falls (and there's no second bell involved). 1.19c. Variant answer to 1.19a: The man is a boxer. The first bell signals the start of a round; the second is either the end of the round or a funeral bell after he dies during the match. Could also be a variant on 1.19 (as suggested by Mike Neergaard): a boxing match in which the top rope breaks, tumbling a boxer to the floor (and he dies of a concussion). 1.19d. Variant: The wind stopped blowing and the man died. Answer: The sole survivor of a shipwreck reached a desert isle. Unfortunately, he was blind. Luckily, there was a freshwater spring on the island, and he rigged the ship's bell (which had drifted to the island also) at the spring's location. The bell rang in the wind, directing him to water. When he was becalmed for a week, he could not find water again, and so he died of thirst. (from Peter R. Olpe) 1.19e. Variant: The music stopped and the man died. Answer: Same as 1.19a, but the blind swimmer kept a portable transistor radio on the beach instead of a bell. When the batteries gave out, he got lost and drowned. (from Joe Kincaid) (See also #1.15a, #1.16, and #1.34a.) 1.20. The woman is the assistant to a (circus or sideshow) knife thrower. The new shoes have higher heels than she normally wears, so that the thrower misjudges his aim and one of his knives kills her during the show. 1.21. Several men were shipwrecked together. They agreed to survive by eating each other a piece at a time. Each of them in turn gave up an arm, but before they got to the last man, they were rescued. They all demanded that the last man live up to his end of the deal. Instead, he killed a bum and sent the bum's arm to the others in a box to "prove" that he had fulfilled the bargain. Later, one of them sees him on the subway, holding onto an overhead ring with the arm he supposedly cut off; the other realizes that the last man cheated, and kills him. 1.21a. Variant wording: A man sends a package to someone in Europe and gets a note back saying "Thank you. I received it." Answer: This is just a simpler version; the shipwreck situation is the same, and the man actually did send his own arm. 1.21b. Variant wording: Two men throw a box off of a cliff. Answer: Exactly the same situation as in 1.21a (one slight variation has a hand in the box instead of a whole arm), with the two men being two of the fellow passengers who had already lost their arms. 1.21c. Variant wording: A man in a Sherlock Holmes-style cape walks into a room, places a box on the table and leaves. Answer: In this one he's wearing the cape either to disguise the fact that he hasn't really cut off his arm/hand as required, or else simply in order to hide his now-missing limb. (from Joe Kincaid) 1.22. Both women are white; the one whose house this takes place in is single. A black friend of the other woman, the one who goes into the bathroom, was recently killed, reportedly by the KKK. The woman who goes into the bathroom discovers a bloodstained KKK robe in the other's laundry hamper, picks up a nail file from the medicine cabinet (or some other impromptu weapon), and goes out and kills the other. 1.22a. Variant: A man goes to hang his coat and realises he will die that day. Answer: The man (who is black) has car trouble and is in need of a telephone. He asks at the nearest house and on being invited in goes to hang his coat, whereupon he notices the white robes of the Ku Klux Klan in the closet. (from Bernd Wechner) 1.23. He is in a hotel, and is unable to sleep because the man in the adjacent room is snoring. He calls the room next door (from his own room number he can easily figure out his neighbor's, and from the room number, the telephone number). The snorer wakes up, answers the phone. The first man hangs up without saying anything and goes to sleep before the snorer gets back to sleep and starts snoring again. 1.23a. Slightly variant answer: It's a next-door neighbor in an apartment building who's snoring, rather than in a hotel. The caller thus knows his neighbor and the phone number. 1.24. It's the man's fiftieth birthday, and in celebration of this he plans to kill his wife, then take the money he's embezzled and move on to a new life in another state. His wife takes him out to dinner; afterward, on their front step, he kills her. He opens the door, dragging her body in with him, and all the lights suddenly turn on and a group of his friends shout "Surprise]" He kills himself. (Note that the whole first part, including the motive, isn't really necessary; it was just part of the original story.) 1.25. Abel is a prince of the island nation that he landed on. A cruel and warlike prince, he waged many land and naval battles along with his father the king. In one naval encounter, their ship sank, the king died, and the prince swam to a deserted island where he spent several months building a raft or small boat. In the meantime, a regent was appointed to the island nation, and he brought peace and prosperity. When Prince Abel returned to his kingdom, Cain (a native fisherman) realized that the peace of the land would only be maintained if Abel did not reascend to his throne, and killed the prince (with a piece of driftwood or some other impromptu weapon). 1.26. The drinks contain poisoned ice cubes; one man drinks slowly, giving them time to melt, while the other drinks quickly and thus doesn't get much of the poison. The fact that they drink at different speeds could be added to the statement, possibly along with red herrings such as saying that one of the men is big and burly and the other short and thin. 1.27. Joe is a kid who goes trick-or-treating for Halloween. 1.28. He's a smuggler. On the first cruise, someone brings the contraband to his cabin, and he hides it in an air conditioning duct. Returning to the U.S., he leaves without the contraband, and so passes through customs with no trouble. On the second trip, he has the same cabin on the same ship. Because it doesn't stop anywhere, he doesn't have to go through customs when he returns, so he gets the contraband off safely. 1.29. Hans and Fritz do everything right up until they're filling out a personal-information form and have to write down their birthdays. Fritz' birthday is, say, July 7, so he writes down 7/7/15. Hans, however, was born on, say, June 20, so he writes down 20/6/18 instead of what an American would write, 6/20/18. Note that this is only a problem because they *claim* to be returning Americans; as has been pointed out to me, there are lots of other nations which use the same date ordering. 1.30. Another WWII story. Greg is a German spy. His "friend" Tim is suspicious, so he plays a word-association game with him. When Tim says "The land of the free," Greg responds with "The home of the brave." Then Tim says "The terror of flight," and Greg says "The gloom of the grave." Any U.S. citizen knows the first verse of the national anthem, but only a spy would have memorized the third verse. (Why Tim knew the third verse is left as an exercise to the reader.) 1.31. The dead man was the driver in a hit-and-run acccident which paralyzed its victim. The victim did manage to get the license plate number of the car; now in a wheelchair, he eventually tracked down the driver and shot and killed him. 1.32. His home is a houseboat and he has run out of water while on an extended cruise. 1.32a. Variant wording: A man dies of thirst in his own home. This version goes more quickly because it gives more information; but it may be less likely to annoy people who think the original statement is too vague. 1.33. I'm told this is a true story. Windows in Paris at that time were apparently imperfectly flat; they could act as lenses. One particularly hot day, the sun shining in through such a window caused a woman's lingerie (which she was wearing at the time, awaiting her husband's return) to catch fire, and eventually the entire house caught and burned. 1.34. He's leaving a hospital after visiting his wife, who's on heavy life-support. When the power goes out, he knows she can't live without the life-support systems (he assumes that if the emergency backup generator were working, the elevator wouldn't lose power; this aspect isn't entirely satisfactory, so in a variant, the scene is at home rather than in a hospital). 1.34a. Variant: The music stops and a woman dies. Answer: The woman is confined in an iron lung, and the music is playing on her radio or stereo. The power goes out. (from Randy Whitaker) (See also #1.15a, #1.16, and #1.19e.) 1.35. A large man comes home to the penthouse apartment he shares with his beautiful young wife, taking the elevator up from the ground floor. He sees signs of lovemaking in the bedroom, and assumes that his wife is having an affair; her beau has presumably escaped down the stairs. The husband looks out the French windows and sees a good-looking man just leaving the main entrance of the building. The husband pushes the refrigerator out through the window onto the young man below. The husband dies of a heart attack from overexertion; the young man below dies from having a refrigerator fall on him; and the wife's boyfriend, who was hiding inside the refrigerator, also dies from the fall. 1.36. Let's say "she" is named Suzy, and "they" are named Harry and Jane. Harry is an elderly archaeologist who has found a very old skeleton, which he's dubbed "Jane" (a la "Lucy"). Suzy is a buyer for a museum; she's supposed to make some sort of purchase from Harry, so she invites him to have a business dinner with her (at a restaurant). When she calls to invite him, he keeps talking about "Jane," so Suzy assumes that Jane is his wife and says to bring her along. Harry, offended, calls Suzy's boss and complains; since Suzy should've known who Jane was, she gets fired. 1.37. The man is delivering a pardon, and the flicker of the lights indicates that the person to be pardoned has just been electrocuted. 1.38. The murderer sets the car on a slope above the hot dog stand where the victim works. He then wedges an ice block in the car to keep the brake pedal down, and puts the car in neutral, after which he flies to another city to avoid suspicion. It's a warm day; when the ice melts, the car rolls down the hill and strikes the hot dog man at his roadside stand, killing him. 1.39. There's a car wash on that corner. On rainy days, the rain reduces traction. On sunny days, water from the car wash has the same effect. If rain is threatening, though, the car wash gets little business and thus doesn't make the road wet, so I can take the corner faster. 1.40. The object she throws is a boomerang. It flies out, loops around, and comes back and hits her in the head, killing her. Boomerangs do not often return so close to the point from which they were thrown, but I believe it's possible for this to happen. 1.40a. Silly variant answer: She's in a submarine or spacecraft and throws a heavy object at the window, which breaks. 1.41. He is a passenger in an airplane and sees the bird get sucked into an engine at 20,000 feet. 1.42. They're the remains of a melted snowman. 1.43. One of the brothers (A) confesses to the murder. At his trial, his brother (B) is called as the only defense witness; B immediately confesses, in graphic detail, to having committed the crime. The defense lawyer refuses to have the trial stopped, and A is acquitted under the "reasonable doubt" clause. Immediately afterward, B goes on trial for the murder; A is called as the only defense witness and HE confesses. B is declared innocent; and though everyone knows that ONE of them did it, how can they tell who? Further, neither can be convicted of perjury until it's decided which of them did it... I don't know if that would actually work under our legal system, but someone else who heard the story said that his father was on the jury for a VERY similar case in New York some years ago. Mark Brader points out that the brothers might be convicted of conspiracy to commit perjury or to obstruct justice, or something of that kind. 1.44. He is a mail courier who delivers packages to the different foreign embassies in the United States. The land of an embassy belongs to the country of the embassy, not to the United States. 1.45. A man was shot during a robbery in his store one night. He staggered into the back room, where the telephone was, and called home, dialing by feel since he hadn't turned on the light. Once the call went through he gasped, "I'm at the store. I've been shot. Help]" or words to that effect. He set the phone down to await help, but none came; he'd treated the telephone pushbuttons like cash register numbers, when the arrangements of the numbers are upside down reflections of each other. The stranger he'd dialed had no way to know where "the store" was. 1.46. The dead man was playing Santa Claus, for whatever reason; he slipped while coming down the chimney and broke his neck. 1.46a. Variant answer: The dead man WAS Santa Claus. This moves the puzzle to section 2. 1.47. The man was struck by an object thrown from the roof of the Empire State Building. Originally I had the object being a penny, but several people suggested that a penny probably wouldn't be enough to penetrate someone's skull. Something aerodynamic and heavier, like a dart, was suggested, but I don't know how much mass would be required. 1.47a. Variant: A man is found dead outside a large marble building with three holes in him. Answer: The man was a paleontologist working with the Archaeological Research Institute. He was reviving a triceratops frozen in the ice age when it came to life and killed him. This couldn't possibly happen because triceratops didn't exist during the ice age. (from Peter R. Olpe) 1.48. The man died from eating a poisoned popsicle. 1.49. The man was a sword swallower in a carnival side-show. While he was practicing, someone tickled his throat with the feather, causing him to gag. 1.50. A mosquito bit me, and I swatted it when it later landed on my ceiling (so the blood is my own as well as the mosquito's). 1.51. The man is a lighthouse keeper. He turns off the light in the lighthouse and during the night a ship crashes on the rocks. Seeing this the next morning, the man realizes what he's done and commits suicide. 1.51a. Variant, similar to #1.15: The light goes out and a man dies. Answer: The lighthouse keeper uses his job as an alibi while he's elsewhere committing a crime, but the light goes out and a ship crashes, thereby disproving the alibi. The lighthouse keeper kills himself when he realizes his alibi is no good. (From Eric Wang) 1.51b. Variant answer to 1.51a: Someone else's alibi is disproven. (A man commits a heinous crime, claiming as his alibi that he was onboard a certain ship. When he learns that it was wrecked without reaching port safely, he realizes that his alibi is disproven and commits suicide to avoid being sent to prison.) (From Eric Wang) 1.52. They were skydiving. He broke his arm as he jumped from the plane by hitting it on the plane door; he couldn't reach his ripcord with his other arm. She pulled the ripcord for him. 1.52a. Sketch of variant answer: The ring was attached to the pin of a grenade that he was holding. Develop a situation from there. 1.53. The man is a travel agent. He had sold someone two tickets for an ocean voyage, one round-trip and one one-way. The last name of the man who bought the tickets is the same as the last name of the woman who "fell" overboard and drowned on the same voyage, which is the subject of the article he's reading. 1.54. The man is a beekeeper, and the bees attack en masse because they don't recognize his fragrance. Randy adds that this is based on something that actually happened to his grandfather, a beekeeper who was severely attacked by his bees when he used a new aftershave for the first time in 10 or 20 years. 1.55. He is a guard / attendant in a leper colony. The letter (to him) tells him that he has contracted the disease. The key is the cigarette burning down between his fingers -- leprosy is fairly unique in killing off sensory nerves without destroying motor ability. (Randy was told this by Gary Haas and Chris Englehard) 1.56. The man was a famous artist. A woman who collected autographs saw him dining; after he left the restaurant, she purchased the check that he used to pay for the meal from the restaurant manager. The check was therefore never cashed, so the artist never paid for the meal. 1.57. The movie is at a drive-in theatre. Section 2: Double meanings, fictional settings, and miscellaneous others. 2.1. The man is a heroin addict, and has contracted AIDS by using an infected needle. In despair, he shoots himself up with an overdose, thereby committing suicide. 2.2. The man walks into a casino and goes to the craps table. He bets all the money he owns, and shoots craps. Since he is now broke, he becomes despondent and commits suicide. 2.3. Kids getting their pictures taken with Santa. I see #2.1, #2.2, and #2.3 as different enough from each other to merit separate numbers, although they all rely on the same basic gimmick of alternate meanings of the word "shoot." 2.4. It's the cabin of an airplane that crashed there because of the snowstorm. 2.4a. Variant wording: A cabin, on the side of a mountain, locked from the inside, is opened, and 30 people are found dead inside. They had plenty of food and water. (from Ron Carter) 2.5. He's a priest; he is marrying them to other people, not to himself. 2.6. The "island" is a traffic island. 2.7. A baseball game is going on. The base-runner sees the catcher waiting at home plate with the ball, and so decides to stay at third base to avoid being tagged out. 2.7a. Variant: Two men are in a field. One is wearing a mask. The other man is running towards him to avoid him. Answer: the same, but the catcher isn't right at home plate; the runner is trying to get home before the catcher can. (from Hal Lowery, by way of Chris Riley) This phrasing would allow the puzzle to migrate to section 1, but I don't like it as much. 2.8. The man is an astronaut out on a space walk. 2.9. The man was an amateur mechanic, the book is a Volkswagen service manual, the beetle is a car, and the pile of bricks is what the car fell off of. 2.10. The Eagle landed in the Sea of Tranquility and will likely remain there for the foreseeable future. 2.11. It's a wolf pack; they've killed and eaten (most of) the man. 2.12. The dead man is Superman; the rock is Green Kryptonite. Invent a reasonable scenario from there. 2.13. This is a post-holocaust scenario of some kind; for whatever reason, the man believes himself to be the last human on earth. He doesn't want to live by himself, so he jumps, just before the telephone rings... (of course, it could be a computer calling, but he has no way of knowing). 2.14. The one who looks around sees his own reflection in the window (it's dark outside), but not his companion's. Thus, he realizes the other is a vampire, and that he's going to be killed by him. 2.15. The "bicycles" are Bicycle playing cards; the man was cheating at cards, and when the extra card was found, he was killed by the other players. 2.15a. Variant: There are 53 bees instead of 53 bicycles. Answer: The same (Bee is another brand of playing cards). 2.15b. Variant: There are 51 instead of 53. Answer: Someone saw the guy conceal a card, and proved the deck was defective by turning it up and pointing out the missing ace. Or, the game was bridge, and the others noticed the cheating when the deal didn't come out even. The man had palmed an ace during the shuffle and meant to put it in his own hand during the deal, but muffed it. (both answers from Mark Brader) 2.16. A chess game; knight takes pawn. 2.16a. Variant: It's the year 860 A.D., at Camelot. Two priests are sitting in the castle's chapel. The queen attacks the king. The two priests rise, shake hands, and leave the room. Answer: The two priests are playing chess; one of them just mated by moving his queen. (from Ellen M. Sentovich) 2.16b. Variant: A black leader dies in Africa. Answer: The black leader is a chess king, and the game was played in Africa. (from Erick Brethenoux) 2.17. It's a model train set. 2.17a. Variant: The Orient Express is derailed and a kitten plays nearby. Answer: The Orient Express is a model train which has been left running unattended. The kitten has playfully derailed it. (from Bernd Wechner) 2.18. It's a game of Monopoly. 2.19. The sun is shining; there's no rain. 2.20. It's daytime; the sun is out. 2.21. Alice is a goldfish; Ted is a cat. 2.21a. A very common variant uses the names Romeo and Juliet instead, to further mislead audiences. For example: Romeo is looking down on Juliet's dead body, which is on the floor surrounded by water and broken glass. (from Adam Carlson) 2.21b. Minor variant: Tom and Jean lay dead in a puddle of water with broken pieces of glass and a baseball nearby. Answer: Tom and Jean are both fish; it was a baseball, rather than a cat, that broke their tank. (from Mike Reymond) 2.22. Friday is a horse. 2.22a. Variant with the same basic gimmick: A woman comes home, sees Spaghetti on the wall and kills her husband. Answer: Spaghetti was the name of her pet dog. Her husband had it stuffed and mounted after it made a mess on his rug. (Simon Travaglia original) 2.23. Bruce is a horse. 2.24. Should be done orally; the envelope is an evelope of dye, and she's dying some cloth, but it sounds like "opens an envelope and dies" if said out loud. 2.25. The native chief asked him, "What is the third baseman's name in the Abbot and Costello routine 'Who's on First'?" The man, who had no idea, said "I don't know," the correct answer. However, he was a major smartass, so if he had known the answer he would have pointed out that What was the SECOND baseman's name. The chief, being quite humorless, would have executed him on the spot. This is fairly silly, but I like it too much to remove it from the list. 2.26. The men have gone spelunking and have taken an Igloo cooler with them so they can have a picnic down in the caves. They cleverly used dry ice to keep their beer cold, not realizing that as the dry ice sublimed (went from solid state to vapor state) it would push the lighter oxygen out of the cave and they would suffocate. ==> logic/smullyan/black.hat.p <== Three logicians, A, B, and C, are wearing hats, which they know are either black or white but not all white. A can see the hats of B and C; B can see the hats of A and C; C is blind. Each is asked in turn if they know the color of their own hat. The answers are: A: "No." B: "No." C: "Yes." What color is C's hat and how does she know? ==> logic/smullyan/black.hat.s <== A must see at least one black hat, or she would know that her hat is black since they are not all white. B also must see at least one black hat, and further, that hat had to be on C, otherwise she would know that her hat was black (since she knows A saw at least one black hat). So C knows that her hat is black, without even seeing the others' hats. ==> logic/smullyan/fork.three.men.p <== Three men stand at a fork in the road. One fork leads to Someplaceorother; the other fork leads to Nowheresville. One of these people always answers the truth to any yes/no question which is asked of him. The other always lies when asked any yes/no question. The third person randomly lies and tells the truth. Each man is known to the others, but not to you. What is the least number of yes/no questions you can ask of these men and pick the road to Someplaceorother? ==> logic/smullyan/fork.three.men.s <== It is clear that you must ask at least two questions, since you might be asking the first one of the randomizer and there is nothing you can tell from his answers. Start by asking A "Is B more likely to tell the truth than C?" If he answers "yes", then: If A is truthteller, B is randomizer, C is liar. If A is liar, B is randomizer, C is truthteller. If A is randomizer, C is truthteller or liar. If he answers "no", then: If A is truthteller, B is liar, C is randomizer. If A is liar, B is truthteller, C is randomizer. If A is randomizer, B is truthteller or liar. In either case, we now know somebody (C or B, respectively) who is either a truthteller or liar. Now, use the technique for finding information from a truthteller/liar, viz.: You ask him the following question: "If I were to ask a person of the opposite type to yourself if the left fork leads to Someplacerother, would he say yes?" If the person asked is a truthteller, he will tell you what a liar would say, which would be the wrong information. If the person asked is a liar, he will either tell you what a liar would say, or he will lie about what a truthteller would say. In either case, he will report the wrong information. If the answer is yes, take the right fork, if no take the left fork. ==> logic/smullyan/fork.two.men.p <== Two men stand at a fork in the road. One fork leads to Someplaceorother; the other fork leads to Nowheresville. One of these people always answers the truth to any yes/no question which is asked of him. The other always lies when asked any yes/no question. By asking one yes/no question, can you determine the road to Someplaceorother? ==> logic/smullyan/fork.two.men.s <== The question to ask is: "Will the other person say the right fork leads to Someplaceorother?" If the person asked says yes, then take the left fork, else take the right fork. If the person asked is the truthteller, then he correctly reports that the liar will misinform you about the right fork. If he is the liar, then he lies about what the truthteller will say. Either way, you should go the opposite direction from the way that the person asked says the other person will answer. The fact that there are two is a red herring - you only need one of either type. You ask him the following question: "If I were to ask a person of the opposite type to yourself if the left fork leads to Someplacerother, would he say yes?" If the person asked is a truthteller, he will tell you what a liar would say, which would be the wrong information. If the person asked is a liar, he will either tell you what a liar would say, or he will lie about what a truthteller would say. In either case, he will report the wrong information. If the answer is yes, take the right fork, if no take the left fork. This solution also removes the problem that the men may not know the other's identity. It is possible, of course, that the liars are malicious, and they will tell the truth if they figure out that you are trying to trick them. ==> logic/smullyan/integers.p <== Two logicians place cards on their foreheads so that what is written on the card is visible only to the other logician. Consecutive positive integers have been written on the cards. The following conversation ensues: A: "I don't know my number." B: "I don't know my number." A: "I don't know my number." B: "I don't know my number." ... n statements of ignorance later ... A or B: "I know my number." What is on the card and how does the logician know it? ==> logic/smullyan/integers.s <== If A saw 1, she would know that she had 2, and would say so. Therefore, A did not see 1. A says "I don't know my number." If B saw 2, she would know that she had 3, since she knows that A did not see 1, so B did not see 1 or 2. B says "I don't know my number." If A saw 3, she would know that she had 4, since she knows that B did not see 1 or 2, so A did not see 1, 2 or 3. A says "I don't know my number." If B saw 4, she would know that she had 5, since she knows that A did not see 1, 2 or 3, so B did not see 1, 2, 3 or 4. B says "I don't know my number." ... n statements of ignorance later ... If X saw n, she would know that she had n + 1, since she knows that ~X did not see 1 ... n - 1, so X did see n. X says "I know my number." And the number in n + 1. ==> logic/smullyan/liars.et.al.p <== Of a group of n men, some always lie, some never lie, and the rest sometimes lie. They each know which is which. You must determine the identity of each man by asking the least number of yes-or-no questions. ==> logic/smullyan/liars.et.al.s <== The real problem is to isolate the sometimes liars. Consider the case of three men: Ask man 1: "Does man 2 lie more than 3?" If the answer is yes, then man 2 cannot be the sometimes liar. Proof by analyzing the cases: Case 1: Man 2 is not the sometimes liar. Case 2: Man 2 is the sometimes liar, man 1 is the truth teller, and man 3 is the liar. Then man 1 would not say that man 2 lies more than man 3. Case 3: Man 2 is the sometimes liar, man 3 is the truth teller, and man 1 is the liar. Then man 1 would not say that man 2 lies more than man 3. QED. Similarly, if the answer is no, then man 3 cannot be the sometimes liar. Now ask the symmetric question of whichever man has been eliminated as the sometimes liar. The answer will now allow you to determine the identity of the sometimes liar. To determine the identity of the two remaining men, ask some question like "Does 1=1?" which is always true. This is not the only way to solve this problem. You could have asked the question which is always true (or false) second, which would now establish the identity of either the liar or the truth teller. Then ask the third question of this man to find out which of the other two is the sometimes liar. This problem requires three questions, whether or not they are yes-or-no questions. In order to identify all three men, you must identify the sometimes liar. You cannot identify the sometimes liar in one question since you may be asking it of the sometimes liar, and any answer from him conveys no information at all. Therefore at least two questions are necessary to identify the sometimes liar. Once the sometimes liar is identified, you still need one more question at least to identify the remaining men. Therefore, three questions are required. Suppose we have two truth-tellers, two liars, and two randomizers. The answer is 8. A proof follows. For brevity, "T" means truth-teller, "L" liar, "R" randomizer, "P" predictable (either T or L). Define a _pattern_ to be one of the C(6,2)=15 permutations of RRPPPP (each of which has C(4,2)=6 interpretations of the Ps as 2 Ts and 2 Ls). For any question Q, let ]Q denote the question "If I were to ask you Q, would you answer Yes?". Note that question ]Q directed toward any P will yield a truthful answer to question Q; in other words, a "Yes" answer to ]Q means that either Q is true or the respondent is an R, whereas "No" means that either Q is false or the respondent is an R. Ask #1, ]"Are both Rs in the set {#2, #3, #4}?". "No" implies that at most one of {#2, #3, #4} is an R. "Yes" implies that at most one of {#2, #5, #6} is an R. Without loss of generality, assume the former. Ask #2, ]"Is #3 an R?". "No" implies that #3 is a P. "Yes" implies that #4 is a P. Having identified someone as a P, there are at most C(5,2)=10 possible patterns, and hence at most 10*6=60 possible results. We can determine which one reflects reality with at most 6 more questions with a binary search. (At each step, bisect the set of possible answers, and ask the question ]"Is the correct pattern in the first subset?".) Now, let's show that it can't be done in 7. After asking your first two questions, renumber if necessary so that the first question was directed to #1 and the second to #2. (If you asked the same person twice, you're even worse off than in the analysis below.) You have no way to rule out the possibility that both are Rs, so pattern RRPPPP yields 6 possibilities. Of the four patterns RPRPPP RPPRPP RPPPRP RPPPPR, your first question gave no information and the second had one bit; so at best you can eliminate half of these 4*6 possibilities, leaving 12. Similarly for the four patterns PRRPPP PRPRPP PRPPRP PRPPPR there remain at least 12 possibilities. Of the remaining 6 patterns PPRRPP PPRPRP PPRPPR PPPRRP PPPRPR PPPPRR, your two bits of information can eliminate 3/4 of the 6*6, leaving 9. Thus, after two questions there are at least 6+12+12+9=39 arrangements that could have given the answers you heard; your five remaining questions have only 32 possible replies, so you can't distinguish them. ==> logic/smullyan/painted.heads.p <== While three logicians were sleeping under a tree, a malicious child painted their heads red. Upon waking, each logician spies the child's handiwork as it applied to the heads of the other two. Naturally they start laughing. Suddenly one falls silent. Why? ==> logic/smullyan/painted.heads.s <== The one who fell silent, presumably the quickest of the three, reasoned that his head must be painted also. The argument goes as follows. Let's call the quick one Q, and the other two D and S. Let's assume Q's head is untouched. Then D is laughing because S's head is painted, and vice versa. But eventually, D and S will realize that their head must be painted, because the other is laughing. So they will quit laughing as soon as they realize this. So, Q waits what he thinks is a reasonable amount of time for them to figure this out, and when they don't stop laughing, his worst fears are confirmed. He concludes that his assumption is invalid and he must be crowned in crimson too. ==> logic/smullyan/priest.p <== A priest takes confession of all the inhabitants in a small town. He discovers that in N married pairs in the town, one of the pair has committed adultery. Assume that the spouse of each adulterer does not know about the infidelity of his or her spouse, but that, since it is a small town, everyone knows about everyone else's infidelity. In other words, each spouse of an adulterer thinks there are N - 1 adulterers, but everyone else thinks there are N adulterers. The priest, who is an Old Testament type, decides that he should do something about the situation. He cannot break the confessional, but being an amateur logician of sorts, he hits upon a plan to do God's work. He announces in Mass one Sunday that the spouse of each adulterer has the moral obligation to punish his or her adulterous spouse by publicly denouncing them in church, and that he will make time during his next Sunday service for this, and continue to do so until all adulterers have been denounced. Is the priest correct? Will this result in every adulterer being denounced? ==> logic/smullyan/priest.s <== Yes. Let's start with the simple case that N = 1. The offended spouse reasons as follows: the priest knows there is at least one adulterer, but I don't know who this person is, and I would if it were anyone other than me, so it must be me. What happens if N = 2? On the first Sunday, the two offended spouses each calmly wait for the other to get up and condemn their spouses. When the other doesn't stand, they think: They do not think that they are a victim. But if they do not think they are victims, then they must think there are no adulterers, contrary to what the priest said. But everyone knows the priest speaks with the authority of God, so it is unthinkable that he is mistaken. The only remaining possibility is that they think there WAS another adulterer, and the only possibility is: MY SPOUSE] So, they know that they too must be victims. So on the next Sunday, they will get up. What if N = 3? On the first Sunday, each victim waits for the other two to get up. When they do not, they assume that they did not get up because they did not know about the other person (in other words, they hypothesize that each of the two other victims thought there was only one adulterer). However, each victim reasons, the two will now realize that they must be two victims, for the reasons given under the N = 2 case above. So they will get up next Sunday. This excuse lasts until the next Sunday, when still no one gets up, and now each victim realizes that either the priest was mistaken (unthinkable]) or there are really three victims, and I am ONE] So, on the third Sunday, all three get up. This reasoning can be repeated inductively to show that no one will do anything (except use up N - 1 excuses as to why no one got up) until the Nth Sunday, when all N victims will arise in unison. By the way, the rest of the town, which thinks there are N adulterers, is about to conclude that their perfectly innocent spouses have been unfaithful too. This includes the adulterous spouses, who are about to conclude that the door swings both ways. So the priest is playing a dangerous game. A movie plot in there somewhere? ==> logic/smullyan/stamps.p <== The moderator takes a set of 8 stamps, 4 red and 4 green, known to the logicians, and loosely affixes two to the forehead of each logician so that each logician can see all the other stamps except those 2 in the moderator's pocket and the two on her own head. He asks them in turn if they know the colors of their own stamps: A: "No" B: "No" C: "No" A: "No B: "Yes" What are the colors of her stamps, and what is the situation? ==> logic/smullyan/stamps.s <== B says: "Suppose I have red-red. A would have said on her second turn: 'I see that B has red-red. If I also have red-red, then all four reds would be used, and C would have realized that she had green-green. But C didn't, so I don't have red-red. Suppose I have green-green. In that case, C would have realized that if she had red-red, I would have seen four reds and I would have answered that I had green-green on my first turn. On the other hand, if she also has green-green Ýwe assume that A can see C; this line is only for completeness¨, then B would have seen four greens and she would have answered that she had two reds. So C would have realized that, if I have green-green and B has red-red, and if neither of us answered on our first turn, then she must have green-red. "'But she didn't. So I can't have green-green either, and if I can't have green-green or red-red, then I must have green-red.' So B continues: "But she (A) didn't say that she had green-red, so the supposition that I have red-red must be wrong. And as my logic applies to green-green as well, then I must have green-red." So B had green-red, and we don't know the distribution of the others certainly. (Actually, it is possible to take the last step first, and deduce that the person who answered YES must have a solution which would work if the greens and reds were switched -- red-green.) ==> logic/timezone.p <== Two people are talking long distance on the phone; one is in an East- Coast state, the other is in a West-Coast state. The first asks the other "What time is it?", hears the answer, and says, "That's funny. It's the same time here]" ==> logic/timezone.s <== One is in Eastern Oregon (in Mountain time), the other in Western Florida (in Central time), and it's daylight-savings changeover day at 1:30 AM. ==> logic/unexpected.p <== Swedish civil defense authorities announced that a civil defense drill would be held one day the following week, but the actual day would be a surprise. However, we can prove by induction that the drill cannot be held. Clearly, they cannot wait until Friday, since everyone will know it will be held that day. But if it cannot be held on Friday, then by induction it cannot be held on Thursday, Wednesday, or indeed on any day. What is wrong with this proof? ==> logic/unexpected.s <== This problem has generated a vast literature (see below). Several solutions of the paradox have been proposed, but as with most paradoxes there is no consensus on which solution is the "right" one. The earliest writers (O'Connor, Cohen, Alexander) see the announcement as simply a statement whose utterance refutes itself. If I tell you that I will have a surprise birthday party for you and then tell you all the details, including the exact time and place, then I destroy the surprise, refuting my statement that the birthday will be a surprise. Soon, however, it was noticed that the drill could occur (say on Wednesday), and still be a surprise. Thus the announcement is vindicated instead of being refuted. So a puzzle remains. One school of thought (Scriven, Shaw, Medlin, Fitch, Windt) interprets the announcement that the drill is unexpected as saying that the date of the drill cannot be deduced in advanced. This begs the question, deduced from which premises? Examination of the inductive argument shows that one of the premises used is the announcement itself, and in particular the fact that the drill is unexpected. Thus the word "unexpected" is defined circularly. Shaw and Medlin claim that this circularity is illegitimate and is the source of the paradox. Fitch uses Godelian techniques to produce a fully rigorous self-referential announcement, and shows that the resulting proposition is self-contradictory. However, none of these authors explain how it can be that this illegitimate or self-contradictory announcement nevertheless appears to be vindicated when the drill occurs. In other words, what they have shown is that under one interpretation of "surprise" the announcement is faulty, but their interpretation does not capture the intuition that the drill really is a surprise when it occurs and thus they are open to the charge that they have not captured the essence of the paradox. Another school of thought (Quine, Kaplan and Montague, Binkley, Harrison, Wright and Sudbury, McClelland, Chihara, Sorenson) interprets "surprise" in terms of "knowing" instead of "deducing." Quine claims that the victims of the drill cannot assert that on the eve of the last day they will "know" that the drill will occur on the next day. This blocks the inductive argument from the start, but Quine is not very explicit in showing what exactly is wrong with our strong intuition that everybody will "know" on the eve of the last day that the drill will occur on the following day. Later writers formalize the paradox using modal logic (a logic that attempts to represent propositions about knowing and believing) and suggest that various axioms about knowing are at fault, e.g., the axiom that if one knows something, then one knows that one knows it (the "KK axiom"). Sorenson, however, formulates three ingenious variations of the paradox that are independent of these doubtful axioms, and suggests instead that the problem is that the announcement involves a "blindspot": a statement that is true but which cannot be known by certain individuals even if they are presented with the statement. This idea was foreshadowed by O'Beirne and Binkley. Unfortunately, a full discussion of how this blocks the paradox is beyond the scope of this summary. Finally, there are two other approaches that deserve mention. Cargile interprets the paradox as a game between ideally rational agents and finds fault with the notion that ideally rational agents will arrive at the same conclusion independently of the situation they find themselves in. Olin interprets the paradox as an issue about justified belief: on the eve of the last day one cannot be justified in believing BOTH that the drill will occur on the next day AND that the drill will be a surprise even if both statements turn out to be true; hence the argument cannot proceed and the drill can be a surprise even on the last day. For those who wish to read some of the literature, good papers to start with are Bennett-Cargile and both papers of Sorenson. All of these provide overviews of previous work and point out some errors, and so it's helpful to read them before reading the original papers. For further reading on the "deducibility" side, Shaw, Medlin and Fitch are good representatives. Other papers that are definitely worth reading are Quine, Binkley, and Olin. D. O'Connor, "Pragmatic Paradoxes," Mind 57:358-9, 1948. L. Cohen, "Mr. O'Connor's 'Pragmatic Paradoxes,'" Mind 59:85-7, 1950. P. Alexander, "Pragmatic Paradoxes," Mind 59:536-8, 1950. M. Scriven, "Paradoxical Announcements," Mind 60:403-7, 1951. D. O'Connor, "Pragmatic Paradoxes and Fugitive Propositions," Mind 60:536-8, 1951 P. Weiss, "The Prediction Paradox," Mind 61:265ff, 1952. W. Quine, "On A So-Called Paradox," Mind 62:65-7, 1953. R. Shaw, "The Paradox of the Unexpected Examination," Mind 67:382-4, 1958. A. Lyon, "The Prediction Paradox," Mind 68:510-7, 1959. D. Kaplan and R. Montague, "A Paradox Regained," Notre Dame J Formal Logic 1:79-90, 1960. G. Nerlich, "Unexpected Examinations and Unprovable Statements," Mind 70:503-13, 1961. M. Gardner, "A New Prediction Paradox," Brit J Phil Sci 13:51, 1962. K. Popper, "A Comment on the New Prediction Paradox," Brit J Phil Sci 13:51, 1962. B. Medlin, "The Unexpected Examination," Am Phil Q 1:66-72, 1964. F. Fitch, "A Goedelized Formulation of the Prediction Paradox," Am Phil Q 1:161-4, 1964. R. Sharpe, "The Unexpected Examination," Mind 74:255, 1965. J. Chapman & R. Butler, "On Quine's So-Called 'Paradox,'" Mind 74:424-5, 1965. J. Bennett and J. Cargile, Reviews, J Symb Logic 30:101-3, 1965. J. Schoenberg, "A Note on the Logical Fallacy in the Paradox of the Unexpected Examination," Mind 75:125-7, 1966. J. Wright, "The Surprise Exam: Prediction on the Last Day Uncertain," Mind 76:115-7, 1967. J. Cargile, "The Surprise Test Paradox," J Phil 64:550-63, 1967. R. Binkley, "The Surprise Examination in Modal Logic," J Phil 65:127-36, 1968. C. Harrison, "The Unanticipated Examination in View of Kripke's Semantics for Modal Logic," in Philosophical Logic, J. Davis et al (ed.), Dordrecht, 1969. P. Windt, "The Liar in the Prediction Paradox," Am Phil Q 10:65-8, 1973. A. Ayer, "On a Supposed Antinomy," Mind 82:125-6, 1973. M. Edman, "The Prediction Paradox," Theoria 40:166-75, 1974. J. McClelland & C. Chihara, "The Surprise Examination Paradox," J Phil Logic 4:71-89, 1975. C. Wright and A. Sudbury, "The Paradox of the Unexpected Examination," Aust J Phil 55:41-58, 1977. I. Kvart, "The Paradox of the Surprise Examination," Logique et Analyse 337-344, 1978. R. Sorenson, "Recalcitrant Versions of the Prediction Paradox," Aust J Phil 69:355-62, 1982. D. Olin, "The Prediction Paradox Resolved," Phil Stud 44:225-33, 1983. R. Sorenson, "Conditional Blindspots and the Knowledge Squeeze: A Solution to the Prediction Paradox," Aust J Phil 62:126-35, 1984. C. Chihara, "Olin, Quine and the Surprise Examination," Phil Stud 47:191-9, 1985. R. Kirkham, "The Two Paradoxes of the Unexpected Hanging," Phil Stud 49:19-26, 1986. D. Olin, "The Prediction Paradox: Resolving Recalcitrant Variations," Aust J Phil 64:181-9, 1986. C. Janaway, "Knowing About Surprises: A Supposed Antinomy Revisited," Mind 98:391-410, 1989. -- tycchow@math.mit.edu. ==> logic/verger.p <== A very bright and sunny Day The Priest didst to the Verger say: "Last Monday met I strangers three None of which were known to Thee. I ask'd Them of Their Age combin'd which amounted twice to Thine] A Riddle now will I give Thee: Tell Me what Their Ages be]" So the Verger ask'd the Priest: "Give to Me a Clue at least]" "Keep Thy Mind and Ears awake, And see what Thou of this can make. Their Ages multiplied make plenty, Fifty and Ten Dozens Twenty." The Verger had a sleepless Night To try to get Their Ages right. "I almost found the Answer right. Please shed on it a little Light." "A little Clue I give to Thee, I'm older than all Strangers three." After but a little While The Verger answered with a Smile: "Inside my Head has rung a Bell. Now I know the answer well]" Now, the question is: How old i ==> logic/verger.s <== The puzzler tried to take the test; Intriguing rhymes he wished to best. But "Fifty and ten dozens twenty" made his headache pound aplenty. When he finally found some leisure, He took to task this witty treasure. "The product of the age must be Twenty-Four Hundred Fifty]" Knowing that, he took its primes, permuted them as many times as needed, til he found amounts equal to, by all accounts, twice the Verger's age, so that He would have that next day's spat. The reason for the lad's confusion was due to multiple solution] Hence he needed one more clue to give the answer back to you] Since only one could fit the bill, and then confirm the priest's age still, the eldest age of each solution by one could differ, with no coercion. <=(Sorry) Else, that last clue's revelation would not have brought information] With two, two, five, seven, and seven, construct three ages, another set of seven. Two sets of three yield sixty-four, Examine them, yet one time more. The eldest age of each would be forty-nine, and then, fifty] With lack of proper rhyme and meter, I've tried to be the first completor of this poem and a puzzle; my poetry, you'd try to muzzle] And lest you think my wit is thrifty, The answer, of course, must be fifty] If dispute, you wish to tender, note my addresss, as the sender] -- Kevin Nechodom ==> logic/weighing/balance.p <== You are given N balls and a balance scale and told that one ball is slightly heavier or lighter than the other identical ones. The scale lets you put the same number of balls on each side and observe which side (if either) is heavier. 1. What's the minimum # of weighings X (and way of doing them) that will always find the unique ball and whether it's heavy or light? 2. If you are told the unique ball is, in fact, heavier than the others, what's the minimum # of weighings Y to find it? ==> logic/weighing/balance.s <== Martin Gardner gave a neat solution to this problem. Assume that you are allowed N weighings. Write down the 3^N possible length N strings of the symbols '0', '1', and '2'. Eliminate the three such strings that consist of only one symbol repeated N times. For each string, find the first symbol that is different from the symbol preceeding it. Consider that pair of symbols. If that pair is not 01, 12, or 20, cross out that string. In other words, we only allow strings of the forms 0*01.*, 1*12.*, or 2*20.* ( using ed(1) regular expressions ). You will have (3^N-3)/2 strings left. This is how many balls you can handle in N weighings. Perform N weighings as follows: For weighing I, take all the balls that have a 0 in string position I, and weigh them against all the balls that have a 2 in string position I. If the side with the 0's in position I goes down, write down a 0. If the other side goes down, write down a 2. Otherwise, write down a 1. After the N weighings, you have written down an N symbol string. If your string matches the string on one of the balls, then that is the odd ball, and it is heavy. If none of them match, than change every 2 to a 0 in your string, and every 0 to a 2. You will then have a string that matches one of the balls, and that ball is lighter than the others. Note that if you only have to identify the odd ball, but don't have to determine if it is heavy or light, you can handle (3^N-3)/2+1 balls. Label the extra ball with a string of all 1's, and use the above method. Note also that you can handle (3^N-3)/2+1 balls if you *do* have to determine whether it is heavy or light, provided you have a single reference ball available, which you know has the correct weight. You do this by labelling the extra ball with a string of all 2s. This results in it being placed on the same side of the scales each time, and in that side of the scales having one more ball than the other each time. So you put the reference ball on the other side of the scales to the "all 2s" ball on each weighing. Proving that this works is straightforward, once you notice that the method of string construction makes sure that in each position, 1/3 of the strings have 0, 1/3 have 1, and 1/3 have 2, and that if a string occurs, then the string obtained by replacing each 0 with a 2 and each 2 with a 0 does not occur. ==> logic/weighing/box.p <== You have ten boxes; each contains nine balls. The balls in one box weigh 0.9 kg; the rest weigh 1.0 kg. You have one weighing on a scale to find the box containing the light balls. How do you do it? ==> logic/weighing/box.s <== Number the boxes 0-9. Take 0 balls from box 0, 1 ball from box 1, 2 balls from box 2, etc. Now weight all those balls and follow this table: If odd box is Weight is 0 45 kg 1 44.9 kg 2 44.8 kg 3 44.7 kg 4 44.6 kg 5 44.5 kg 6 44.4 kg 7 44.3 kg 8 44.2 kg 9 44.1 kg ==> logic/weighing/gummy.bears.p <== Real gummy drop bears have a mass of 10 grams, while imitation gummy drop bears have a mass of 9 grams. Spike has 7 cartons of gummy drop bears, 4 of which contain real gummy drop bears, the others imitation. Using a scale only once and the minimum number of gummy drop bears, how can Spike determine which cartons contain real gummy drop bears? ==> logic/weighing/gummy.bears.s <== Spike used 51 gummy drop bears: from the 7 boxes he took respectively 0, 1, 2, 4, 7, 13, and 24 bears. The notion is that each box of imitation bears will subtract its number of bears from the total "ideal" weight of 510 grams (1 gram of missing weight per bear), so Spike weighs the bears, subtracts the result from 510 to obtain a number N, and finds the unique combination of 3 numbers from the above list (since there are 3 "imitation" boxes) that sum to N. The trick is for the sums of all triples selected from the set S of numbers of bears to be unique. To accomplish this, I put numbers into S one at a time in ascending order, starting with the obvious choice, 0. (Why is this obvious? If I'd started with k > 0, then I could have improved on the resulting solution by subtracting k from each number) Each new number obviously had to be greater than any previous, because otherwise sums are not unique, but also the sums it made when paired with any previous number had to be distinct from all previous pairs (otherwise when this pair is combined with a third number you can't distinguish it from the other pair)--except for the last box, where we can ignore this point. And most obviously all the new triples had to be distinct from any old triples; it was easy to find what the new triples were by adding the newest number to each old sum of pairs. Now, in case you're curious, the possible weight deficits and their unique decompositions are: 3 = 0 + 1 + 2 5 = 0 + 1 + 4 6 = 0 + 2 + 4 7 = 1 + 2 + 4 8 = 0 + 1 + 7 9 = 0 + 2 + 7 10 = 1 + 2 + 7 11 = 0 + 4 + 7 12 = 1 + 4 + 7 13 = 2 + 4 + 7 14 = 0 + 1 + 13 15 = 0 + 2 + 13 16 = 1 + 2 + 13 17 = 0 + 4 + 13 18 = 1 + 4 + 13 19 = 2 + 4 + 13 20 = 0 + 7 + 13 21 = 1 + 7 + 13 22 = 2 + 7 + 13 24 = 4 + 7 + 13 25 = 0 + 1 + 24 26 = 0 + 2 + 24 27 = 1 + 2 + 24 28 = 0 + 4 + 24 29 = 1 + 4 + 24 30 = 2 + 4 + 24 31 = 0 + 7 + 24 32 = 1 + 7 + 24 33 = 2 + 7 + 24 35 = 4 + 7 + 24 37 = 0 + 13 + 24 38 = 1 + 13 + 24 39 = 2 + 13 + 24 41 = 4 + 13 + 24 44 = 7 + 13 + 24 Note that there had to be (7 choose 3) distinct values; they end up ranging from 3 to 44 inclusive with 7 numbers missing: 4, 23, 34, 36, 40, 42, and 43. -- David Karr (karr@cs.cornell.edu) ==> logic/weighing/weighings.p <== Some of the supervisors of Scandalvania's n mints are producing bogus coins. It would be easy to determine which mints are producing bogus coins but, alas, the only scale in the known world is located in Nastyville, which isn't on very friendly terms with Scandalville. In fact, Nastyville's king will only let you use the scale twice. Your job? You must determine which of the n mints are producing the bogus coins using only two weighings and the minimum number of coins (your king is rather parsimonious, to put it nicely). This is a true scale, i.e. it will tell you the weight of whatever you put on it. Good coins are known to have a weight of 1 ounce and it is also known that all the bogus mints (if any) produce coins that are light or heavy by the same amount. Some examples: if n=1 then we only need 1 coin, if n=2 then clearly 2 coins suffice, one from each mint. What are the solutions for n=3,4,5? What can be said for general n? ==> logic/weighing/weighings.s <== Oh gracious and wise king, I have solved this problem by first simplifying and then expanding. That is, consider the problem of being allowed only a single weighing. Stop reading right now if you want to think about it further. There are three possible outcomes for each mint (light, OK, heavy) which may be represented as (-1, 0, +1). Now, let each mint represent one place in base 3. Thus, the first mint is the ones place, the second the threes place, the third is the nines place and so on. The number of coins from each mint must equal the place. That is, we'll have 1 coin from mint 1, 3 from mint 2, 9 from mint 3, and, in general, 3^(n-1) from mint n. By weighing all coins at once, we will get a value between 1 + 3 + 9 + ... and -1 + -3 + -9 + ... In fact, we notice that that value will be unique for any mint outcomes. Thus, for the one weighing problem, we need sum for i=1 to n (3^(i-1)) which evaluates to (3^n - 1)/2 I'm fairly satisfied that this is a minimum for a single weighing. What does a second weighing give us? Well, we can divide the coins into two groups and use the same method. That is, if we have 5 mints, one weighing will be: 1 coin from mint 1 + 3 coins from mint 2 + 9 coins from mint 3 while the other weighing will be: 1 coin from mint 4 + 3 coins from mint 5 It's pretty plain that this gives us a total coinage of: 3^(n/2) - 1 for even n and, after some arithmetic agitation: 2 * 3^((n-1)/2) - 1 for odd n I think the flaw in this solution is that we don't know ahead of time the amount by which the coins are off weight. So if you weigh 1 coin from mint 1 together with 3 coins from mint 2 and the result is heavy by 3x units, you still don't know whether the bogus coins are from mint 3 (heavy by x units) or from mint 1 (heavy by 3x units). Note that we're not given the error amount, only the fact that is is equal for all bogus coins. Here is my partial solution: After considering the above, it would seem that on each of the two weighings we must include coins from all of the mints (except for the special cases of small n). So let ai (a sub i) be the number of coins from mint i on weighing 1 and bi be the number of coins from mint i on weighing 2. Let the error in the bogus coins have a value x, and let ci be a the counterfeit function: ci is 0 if mint i is good, 1 otherwise. Then Sum ai ci x = delta1 error on weighing 1 Sum bi ci x = delta2 error on weighing 2 Now the ratio of delta1 to delta2 will be rational regardless of the value of x, since x will factor out; let's call this ratio p over q (p and q relatively prime). We would like to choose { ai } and { bi } such that for any set of mints J, which will be a subset of { 1 , 2 , ... , n }, that Sum aj ( = Sum ai ci ) is relatively prime to Sum bj. If this is true then we can determine the error x; it will simply be delta1/p, which is equal to delta2/q. If the { ai } have been carefully chosen, we should be able to figure out the bogus mints from one of the weighings, provided that all subsets ( { { aj } over all J } ) have unique sums. This was the strategy proposed above, where is was suggested that ai = 3 ** (i-1) ; note that you can use base 2 instead of base 3 since all the errors have the same sign. Well, for the time being I'm stumped. This agrees with the analysis I've been fighting with. I actually came up with a pair of functions that "almost" works. So that the rest of you can save some time (in case you think the way I did): Weighing 1: 1 coin from each mint Weighing 2: 2^(k-1) coins from mint k, for 1...k...n (total 2^n - 1 coins) Consider the n mints to be one-bit-each -- bit set -> mint makes bogus coins. Then we can just state that we're trying to discover "K", where K is a number whose bit pattern _just_ describes the bogosity of each mint. OK - now, assuming we know 'x', and we only consider the *difference* of the weighing from what it should be, for weighing 1, the devaiation is just the Hamming weight of K -- that is the number of 1-bits in it -- that is, the number of bogosifying mints. For weighing 2, the deviation is just K] When the nth bit of K is set, then that mint contributes just 2^n to the deviation, and so the total deviation will just be K. So that set me in search of a lemma: given H(x) is the hamming weight of x, is f(x) = x / H(x) a 1-1 map integers into rationals? That is, if x/H(x) = y/H(y) can we conclude that x = y? The answer (weep) is NO. The lowest pair I could find are 402/603 (both give the ratio 100.5). Boy it sure looked like a good conjecture for a while] Sigh. There are two parts to the problem. First let us try to come up with a solution to finding the answer in 2 weighings - then worry about using the min. number of coins. Solutions are for GENERAL n. Let N = set of all mints, 1 to n. Card(N) = n. Let P = set of all bogus mints. Let Card(P) = p. Weighing I: Weigh n coins, 1 from each mint. Since each "good" coins weighs one ounce, let delta1 be the error in weighing. Since all bogus coins are identical, let delta1 be abs(error). If x is the weight by which one bogus coin differs from a good coin, delta1 = p * x. Weighing II: The coins to be weighed are composed thusly. Let a1 be the number of coins from mint 1, a2 # from mint2 .. and an from mint n. All ai's are distinct integers. Let A = Set of all ai's. Let delta2 = (abs.) error in weighing 2 = x * k where k is the number of coins that are bogus in weighing two. Or more formally k = sigma(ai) (over all i in P) Assuming p is not zero (from Weighing I - in that case go back and get beheaded for giving the king BAAAAAD advice), Let ratio = delta1/delta2 = p/k. Let IR = delta2/delta1 = k/p = inverse-ratio (for later proof). Let S(i) be the bag of all numbers generated by summing i distinct elements from A. Clearly there will be nCi (that n comb. i) elements in S(i). ÝA bag is a set that can have the same element occur more than once.¨ So S(1) = A and S(n) will have one element that is the sum of all the elements of A. Let R(i) = {x : For-all y in S(i), x = i/y} expressed as p/q (no common factors). (R is a bag too). Let R-A = Bag-Union(R(i) for 1>= i >=n). (can include same element twice) Choose A, such that all elements of R-A are DISTINCT, i.e. Bag(R-A) = Set(R-A). Let the sequence a1, a2, .. an, be an L-sequence if the above property is true. Or more simply, A is in L. ********************************************************************** CONJECTURE: The bogus mint problem is solved in two weighings if A is in L. Sketchy proof: R(1) = all possible ratios (= delta1/delta2) when p=1. R(i) = all possible ratio's when p=i. Since all possible combinations of bogus mints are reflected in R, just match the actual ratio with the generated table for n. ************************************************************************ A brief example. Say n=3. Skip to next line if you want. Let A=(2,3,7). p=1 possible ratios = 1/2 1/3 1/7 p=2 possible ratios = 2/5 2/9 1/5(2/10) p=3 possible ratios = 1/4(3/12) (lots of blood in Scandalvania). As all outcomes are distinct, and the actual ratio MUST be one of these, match it to the answer, and start sharpening the axe. Note that the minimum for n=3 is A=(0,1,3) possible ratios are p=1 infinity (delta2=0),1,1/3 p=2 2/1,2/3,1/2 p=3 3/4 ************************************************************************ All those with the determination to get this far are saying OK, OK how do we get A. I propose a solution that will generate A, to give you the answer in two weighings, but will not give you the optimal number of coins. Let a1=0 For i>=2 >=n ai = i*(a1 + a2 + ... + ai-1) + 1 ***************************************** * i-1 * * ai = i* ÝSigma(aj)¨ + 1 * ****Generator function G***** * j=1 * ***************************************** If A is L, all RATIO's are unique. Also all inverse-ratio's (1/ratio) are unique. I will prove that all inverse-ratio's (or IR's) are unique. Let A(k), be the series generated by the first k elements from eqn. G.(above) ************************************************************************ PROOF BY INDUCTION. A(1) = {0} is in L. A(2) = {0,1} is in L. ASSUME A(k) = {0,1, ..., ak} is in L. T.P.T. A(k+1) = {0,1, ..., ak, D) is in L where D is generated from G. We know that all IR's(inverse ratio's) from A(k) are distinct. Let K = set of all IR's of A(k). Since A(k+1) contains A(k), all IR's of A(k) will also be IR's of A(K+1). So for all P, such that (k+1) is not in P, we get a distinct IR. So consider cases when (k+1) is in P. p=1 (i.e. (k+1) = only bogus mint), IR = D ______________________________________________________________________ CONJECTURE: Highest IR for A(k) = max(K) = ak Proof: Since maxÝA(k)¨ = ak, for p'= 1, max IR = ak/1 = ak for p'= 2, max IR (max sum of 2 ai's)/2 = (ak + ak-1)/2 < ak (as ak>ak-1). for p'= i max IR sum of largest i elements of A(k) -------------------------------- i < i * ak/i = ak. So max. IR for A(k) is ak. ______________________________________________________________________ D > ak So for p=1 IR is distinct. Let Xim be the IR formed by choosing i elements from A(k+1). Note: We are choosing D and (i-1) elements from A(k). m is just an index to denote each distinct combination of (i-1) elemnts of A(i). ______________________________________________________________________ CONJECTURE : For p=j, all new IR's Xjm are limited to the range D/(j-1) > Xjm > D/j. Proof: Xjm = (D + {j-1 elements of A(k)})/j Clearly Xjm > D/j. To show: maxÝXjm¨ < D/(j-1) Note: a1 + a2 .. + ak < D/(k+1) maxÝXjm¨ = (D + ak + ak-1 + ... + a(k-j+1))/j < (D + D/(k+1))/j = D (k+2)/(k+1)j = ÝD/(j-1)¨ * alpha. alpha = (j-1)/(j) * (k+2)/(k+1) Since j <= k, (j-1)/j <= (k-1)/k < (k+1)/(k+2) IMPLIES alpha < 1. Conjecture proved. ______________________________________________________________________ CONJECTURE : For a given p, all newly generated IR's are distinct. Proof by contradiction: Assume this is not so. Implies (D + (p-1) elements of A(k))/p = (D + some other (p-1) elements of A(k))/p Implies SUMÝ(p-1) elements of A(k)¨ = SUMÝ some other (p-1) elements of A(k)¨ Implies SUMÝ(p-1) elements of A(k)¨/(p-1) = SUMÝsome other (p-1) elements¨/(p-1) Implies A(k) is NOT in L. Contra. Hence conjecture. ______________________________________________________________________ CONJECTURE: A(k+1) is in L. Since all newly generated IR's are distinct from each other, and all newly generated IR's are greater than previous IR's, A(k+1) is in L. ==> logic/zoo.p <== I took some nephews and nieces to the Zoo, and we halted at a cage marked Tovus Slithius, male and female. Beregovus Mimsius, male and female. Rathus Momus, male and female. Jabberwockius Vulgaris, male and female. The eight animals were asleep in a row, and the children began to guess which was which. "That one at the end is Mr Tove." "No, no] It's Mrs Jabberwock," and so on. I suggested that they should each write down the names in order from left to right, and offered a prize to the one who got most names right. As the four species were easily distinguished, no mistake would arise in pairing the animals; naturally a child who identified one animal as Mr Tove identified the other animal of the same species as Mrs Tove. The keeper, who consented to judge the lists, scrutinised them carefully. "Here's a queer thing. I take two of the lists, say, John's and Mary's. The animal which John supposes to be the animal which Mary supposes to be Mr Tove is the animal which Mary supposes to be the animal which John supposes to be Mrs Tove. It is just the same for every pair of lists, and for all four species. "Curiouser and curiouser] Each boy supposes Mr Tove to be the animal which he supposes to be Mr Tove; but each girl supposes Mr Tove to be the animal which she supposes to be Mrs Tove. And similarly for the oth- er animals. I mean, for instance, that the animal Mary calls Mr Tove is really Mrs Rathe, but the animal she calls Mrs Rathe is really Mrs Tove." "It seems a little involved," I said, "but I suppose it is a remarkable coincidence." "Very remarkable," replied Mr Dodgson (whom I had supposed to be the keeper) "and it could not have happened if you had brought any more children." How many nephews and nieces were there? Was the winner a boy or a girl? And how many names did the winner get right? Ýby Sir Arthur Eddington¨ ==> logic/zoo.s <== Given that there is at least one boy and one girl (John and Mary are mentioned) then the answer is that there were 3 nephews and 2 nieces, the winner was a boy who got 4 right. Number the animals 1 through 8, such that the females are even and the males are odd, with members of the same species consecutive; i.e. 1 is Mr. Tove, 2 Mrs. Tove, etc. Then each childs guesses can be represented by a permutation. I use the standard notation of a permutation as a set of orbits. For example: (1 3 5)(6 8) means 1 -> 3, 3 -> 5, 5 -> 1, 6 -> 8, 8 -> 6 and 2,4,7 are unchanged. Ý1¨ Let P be any childs guesses. Then P(mate(i)) = mate(P(i)). Ý2¨ If Q is another childs guesses, then ÝP,Q¨ = T, where ÝP,Q¨ is the commutator of P and Q (P composed with Q composed with P inverse composed with Q inverse) and T is the special permutation (1 2) (3 4) (5 6) (7 8) that just swaps each animal with its spouse. Ý3¨ If P represents a boy, then P*P = I (I use * for composition, and I for the identity permutation: (1)(2)(3)(4)(5)(6)(7)(8) Ý4¨ If P represents a girl, then P*P = T. Ý1¨ and Ý4¨ together mean that all girl's guesses must be of the form: (A B C D) (E F G H) where A and C are mates, as are B & D, E & F G & H. So without loss of generality let Mary = (1 3 2 4) (5 7 6 8) Without to much effort we see that the only possibilities for other girls "compatible" with Mary (I use compatible to mean the relation expressed in Ý2¨) are: g1: (1 5 2 6) (3 8 4 7) g2: (1 6 2 5) (3 7 4 8) g3: (1 7 2 8) (3 5 4 6) g4: (1 8 2 7) (3 6 4 5) Note that g1 is incompatible with g2 and g3 is incompatible with g4. Thus no 4 of Mary and g1-4 are mutually compatible. Thus there are at most three girls: Mary, g1 and g3 (without loss of generality) By Ý1¨ and Ý3¨, each boy must be represented as a product of transpostions and/or singletons: e.g. (1 3) (2 4) (5) (6) (7) (8) or (1) (2) (3 4) (5 8) (6 7). Let J represent John's guesses and consider J(1). If J(1) = 1, then J(2) = 2 (by Ý1¨) using Ý2¨ and Mary J(3) = 4, J(4) = 3, and g1 & J => J(5) = 6, J(6) = 5, & g3 & J => J(8) = 7 J(7) = 8 i.e. J = (1)(2)(3 4)(5 6)(7 8). But the ÝJ,Mary¨ <> T. In fact, we can see that J must have no fixed points, J(i) <> i for all i, since there is nothing special about i = 1. If J(1) = 2, then we get from Mary that J(3) = 3. contradiction. If J(1) = 3, then J(2) = 4, J(3) = 1, J(4) = 2 (from Mary) => J(5) = 7, J(6) = 8, J(7) = 5, J(8) = 6 => J = (1 3)(2 4)(5 7)(6 8) (from g1) But then J is incompatible with g3. A similar analysis shows that J(1) cannot be 4,5,6,7 or 8; i.e. no J can be compatible with all three girls. So without loss of generality, throw away g3. We have Mary = (1 3 2 4) (5 7 6 8) g1 = (1 5 2 6) (3 8 4 7) The following are the only possible boy guesses which are compatible with both of these: B1: (1)(2)(3 4)(5 6)(7)(8) B2: (1 2)(3)(4)(5)(6)(7 8) B3: (1 3)(2 4)(5 7)(6 8) B4: (1 4)(2 3)(5 8)(6 7) B5: (1 5)(2 6)(3 8)(4 7) B6: (1 6)(2 5)(3 7)(4 8) Note that B1 & B2 are incombatible, as are B3 & B4, B5 & B6, so at most three of them are mutually compatible. In fact, Mary, g1, B1, B3 and B5 are all mutually compatible (as are all the other possibilities you can get by choosing either B1 or B2, B3 or B4, B5 or B6. So if there are 2 girls there can be 3 boys, but no more, and we have already eliminated the case of 3 girls and 1 boy. The only other possibility to consider is whether there can be 4 or more boys and 1 girl. Suppose there are Mary and 4 boys. Each boy must map 1 to a different digit or they would not be mutually compatible. For example if b1 and b2 both map 1 to 3, then they both map 3 to 1 (since a boy's map consists of transpositions), so both b1*b2 and b2*b1 map 1 to 1. Furthermore, b1 and b2 cannot map 1 onto spouses. For example, if b1(1) = a and b is the spouse of a, then b1(2) = b. If b2(1) = b, then b2(2) = a. Then b1*b2(1) = b1(b) = 2 and b2*b1(1) = b2(a) = 2 (again using the fact that boys are all transpostions). Thus the four boys must be: B1: (1)(2)... or (1 2).... B2: (1 3)... or (1 4) ... B3: (1 5) ... or (1 6) ... B4: (1 7) ... or (1 8) ... Consider B4. The only permutation of the form (1 7)... which is compatible with Mary ( (1 3 2 4) (5 7 6 8) ) is: (1 7)(2 8)(3 5)(4 6) The only (1 8)... possibility is: (1 8)(2 7)(3 6)(4 5) Suppose B4 = (1 7)(2 8)(3 5)(4 6) If B3 starts (1 5), it must be (1 5)(2 6)(3 8)(4 7) to be compatible with B4. This is compatible with Mary also. Assuming this and B2 starts with (1 3) we get B2 = (1 3)(2 4)(5 8)(6 7) in order to be compatible with B4. But then B2*B3 and B3*B2 moth map 1 to 8. I.e. no B2 is mutually compatible with B3 & B4. Similarly if B2 starts with (1 4) it must be (1 4)(2 3)(5 7)(6 8) to work with B4, but this doesn't work with B3. Likewise B3 starting with (1 6) leads to no possible B2 and the identical reasoning eliminates B4 = (1 8)... So no B4 is possible] I.e at most 3 boys are mutually compatiblw with Mary, so 2 girls & 3 boys is optimal. Thus: Mary = (1 3 2 4) (5 7 6 8) Sue = (1 5 2 6) (3 8 4 7) John = (1)(2)(3 4)(5 6)(7)(8) Bob = (1 3)(2 4)(5 7)(6 8) Jim = (1 5)(2 6)(3 8)(4 7) is one optimal solution, with the winner being John (4 right: 1 2 7 & 8) ==> physics/balloon.p <== A helium-filled balloon is tied to the floor of a car that makes a sharp right turn. Does the balloon tilt while the turn is made? If so, which way? The windows are closed so there is no connection with the outside air. ==> physics/balloon.s <== Because of buoyancy, the helium balloon on the string will want to move in the direction opposite the effective gravitational field existing in the car. Thus, when the car turns the corner, the balloon will deflect towards the inside of the turn. ==> physics/bicycle.p <== A boy, a girl and a dog go for a 10 mile walk. The boy and girl can walk 2 mph and the dog can trot at 4 mph. They also have bicycle which only one of them can use at a time. When riding, the boy and girl can travel at 12 mph while the dog can peddle at 16 mph. What is the shortest time in which all three can complete the trip? ==> physics/bicycle.s <== First note that there's no apparent way to benefit from letting either the boy or girl ride the bike longer than the other. Any solution which gets the boy there faster, must involve him using the bike (forward) more; similarly for the girl. Thus the bike must go backwards more for it to remain within the 10-mile route. Thus the dog won't make it there in time. So the solution assumes they ride the bike for the same amount of time. Also note that there's no apparent way to benefit from letting any of the three arrive at the finish ahead of the others. If they do, they can probably take time out to help the others. So the solution assumes they all finish at the same time. The boy starts off on the bike, and travels 5.4 miles. At this point, he drops the bike and completes the rest of the trip on foot. The dog eventually reaches the bike, and takes it *backward* .8 miles (so the girl gets to it sooner) and then returns to trotting. Finally, the girl makes it to the bike and rides it to the end. The answer is 2.75 hours. The puzzle is in Vasek Chvatal, Linear Programming, W. H. Freeman & Co. The generalized problem (n people, 1 bike, different walking and riding speeds) is known as "The Bicycle Problem". A couple references are Masuda, S. (1970). "The bicycle problem," University of California, Berkeley: Operations Research Center Technical Report ORC 70-35. Chvatal, V. (1983). "On the bicycle problem," Discrete Applied Mathematics 5: pp. 165 - 173. As for the linear program which gives the lower bound of 2.75 hours, let tÝperson, mode, direction¨ by the amount of time "person" (boy, girl or dog) is travelling by "mode" (walk or bike) in "direction" (forward or backwards). Define TimeÝperson¨ to be the total time spent by person doing each of these four activities. The objective is to minimize the maximum of TÝperson¨, for person = boy, girl, dog, e.g. minimize T subject to T >= TÝboy¨, T >= TÝgirl¨, T >= TÝdog¨. Now just think of all the other linear constraints on the variables tÝx,y,z¨, such as everyone has to travel 10 miles, etc. In all, there are 8 contraints in 18 variables (including slack variables). Solving this program yields the lower bound. ==> physics/boy.girl.dog.p <== A boy, a girl and a dog are standing together on a long, straight road. Simulataneously, they all start walking in the same direction: The boy at 4 mph, the girl at 3 mph, and the dog trots back and forth between them at 10 mph. Assume all reversals of direction instantaneous. In one hour, where is the dog and in which direction is he facing? ==> physics/boy.girl.dog.s <== The dog's position and direction are indeterminate, other than that the dog must be between the boy and girl (endpoints included). To see this, simply time reverse the problem. No matter where the dog starts out, the three of them wind up together in one hour. This argument is not quite adequate. It is possible to construct problems where the orientation changes an infinite number of times initially, but for which there can be a definite result. This would be the case if the positions at time t are uniformly continuous in the positions at time s, s small. But suppose that at time a the dog is with the girl. Then the boy is at 4a, and the time it takes the dog to reach the boy is a/6, because the relative speed is 6 mph. So the time b at which the dog reaches the boy is proportional to a. A similar argument shows that the time the dog next reaches the girl is b + b/13, and is hence proportional to b. This makes the position of the dog at time (t > a) a periodic function of the logarithm of a, and thus does not approach a limit as a -> 0. ==> physics/brick.p <== What is the maximum overhang you can create with an infinite supply of bricks? ==> physics/brick.s <== You can create an infinite overhang. Let us reverse the problem: how far can brick 1 be from brick 0? Let us assume that the brick is of length 1. To determine the place of the center of mass a(n): a(1)=1/2 a(n)=1/nÝ(n-1)*a(n-1)+Ýa(n-1)+1/2¨¨=a(n-1)+1/(2n) Thus n 1 n 1 a(n)=Sum -- = 1/2 Sum - = 1/2 H(n) m=1 2m m=1 m Needless to say the limit for n->oo of half the Harmonic series is oo. ==> physics/cannonball.p <== A person in a boat drops a cannonball overboard; does the water level change? ==> physics/cannonball.s <== The cannonball in the boat displaces an amount of water equal to the MASS of the cannonball. The cannonball in the water displaces an amount of water equal to the VOLUME of the cannonball. Water is unable to support the level of salinity it would take to make it as dense as a cannonball, so the first amount is definitely more than the second amount, and the water level drops. ==> physics/dog.p <== A body of soldiers form a 50m-by-50m square ABCD on the parade ground. In a unit of time, they march forward 50m in formation to take up the position DCEF. The army's mascot, a small dog, is standing next to its handler at location A. When the B----C----E soldiers start marching, the dog ! ! ! forward--> begins to run around the moving A----D----F body in a clockwise direction, keeping as close to it as possible. When one unit of time has elapsed, the dog has made one complete circuit and has got back to its handler, who is now at location D. (We can assume the dog runs at a constant speed and does not delay when turning the corners.) How far does the dog travel? ==> physics/dog.s <== Let L be the side of the square, 50m, and let D be the distance the dog travels. Let v1 be the soldiers' marching speed and v2 be the speed of the dog. Then v1 = L / (1 time unit) and v2 = v1*D/L. Let t1, t2, t3, t4 be the time the dog takes to traverse each side of the square, in order. Find t1 through t4 in terms of L and D and solve t1+t2+t3+t4 = 1 time unit. While the dog runs along the back edge of the square in time t1, the soldiers advance a distance d=t1*v1, so the dog has to cover a distance sqrt(L^2 + (t1*v1)^2), which takes a time t1=sqrt(L^2 + (t1*v1)^2)/v2. Solving for t1 gives t1=L/sqrt(v2^2 - v1^2). The rest of the times are t2 = L/(v2-v1), t3 = t1, and t4 = L/(v2+v1). In t1+t2+t3+t4, eliminate v2 by using v2=v1*D/L and eliminate v1 by using v1=L/(1 time unit), obtaining 2 L (D + sqrt(D^2-L^2)) / (D^2 - L^2) = 1 which can be turned into D^4 - 4LD^3 - 2L^2D^2 + 4L^3D + 5L^4 = 0 which has a root D = 4.18113L = 209.056m. ==> physics/magnets.p <== You have two bars of iron. One is magnetic, the other is not. Without using any other instrument (thread, filings, other magnets, etc.), find out which is which. ==> physics/magnets.s <== Take the two bars, and put them together like a T, so that one bisects the other. ___________________ bar A ---> !___________________! ! ! ! ! ! ! ! ! bar B ------------> ! ! ! ! ! ! !_! If they stick together, then bar B is the magnet. If they don't, bar A is the magnet. (reasoning follows) Bar magnets are "dead" in their centers (ie there is no magnetic force, since the two poles cance out). So, if bar A is the magnet, then bar B won't stick to its center. However, bar magnets are quite "alive" at their edges (ie the magnetic force is concentrated). So, if bar B is the magnet, then bar A will stick nicely to its end. ==> physics/milk.and.coffee.p <== You are just served a hot cup of coffee and want it to be as hot as possible when you drink it some number of minutes later. Do you add milk when you get the cup or just before you drink it? ==> physics/milk.and.coffee.s <== Normalize your temperature scale so that 0 degrees = room temperature. Assume that the coffee cools at a rate proportional to the difference in temperature, and that the amount of milk is sufficiently small that the constant of proportinality is not changed when you add the milk. An early calculus homework problem is to compute that the temperature of the coffee decays exponentially with time, T(t) = exp(-ct) T0, where T0 = temperature at t=0. Let l = exp(-ct), where t is the duration of the experiment. Assume that the difference in specific heats of coffee and milk are negligible, so that if you add milk at temperature M to coffee at temperature C, you get a mix of temperature aM+bC, where a and b are constants between 0 and 1, with a+b=1. (Namely, a = the fraction of final volume that is milk, and b = fraction that is coffee.) If we let C denote the original coffee temperature and M the milk temperature, we see that Add milk later: aM + blC Add milk now: l(aM+bC) = laM+blC The difference is d=(1-l)aM. Since l<1 and a>0, we need to worry about whether M is positive or not. M>0: Warm milk. So d>0, and adding milk later is better. M=0: Room temp. So d=0, and it doesn't matter. M<0: Cold milk. So d<0, and adding milk now is better. Of course, if you wanted to be intuitive, the answer is obvious if you assume the coffee is already at room temperature and the milk is either scalding hot or subfreezing cold. Moral of the story: Always think of extreme cases when doing these puzzles. They are usually the key. Oh, by the way, if we are allowed to let the milk stand at room temperature, then let r = the corresponding exponential decay constant for your milk container. Add acclimated milk later: arM + blC We now have lots of cases, depending on whether rl: The milk pot is smaller than your coffee cup. (E.g., it's one of those tiny single-serving things.) M>0: The milk is warm. M<0: The milk is cold. Leaving out the analysis, I compute that you should... Add warm milk in large pots LATER. Add warm milk in small pots NOW. Add cold milk in large pots NOW. Add cold milk in small pots LATER. Of course, observe that the above summary holds for the case where the milk pot is allowed to acclimate; just treat the pot as of infinite size. ==> physics/mirror.p <== Why does a mirror appear to invert the left-right directions, but not up-down? ==> physics/mirror.s <== Mirrors invert front to back, not left to right. The popular misconception of the inversion is caused by the fact that a person when looking at another person expects him/her to face her/him, so with the left-hand side to the right. When facing oneself (in the mirror) one sees an 'uninverted' person. See Martin Gardner, ``Hexaflexagons and other mathematical diversions,'' University of Chicago Press 1988, Chapter 16. A letter by R.D. Tschigi and J.L. Taylor published in this book states that the fundamental reason is: ``Human beings are superficially and grossly bilaterally symmetrical, but subjectively and behaviorally they are relatively asymmetrical. The very fact that we can distinguish our right from our left side implies an asymettry of the perceiving system, as noted by Ernst Mach in 1900. We are thus, to a certain extent, an asymmetrical mind dwelling in a bilaterally symmetrical body, at least with respect to a casual visual inspection of our external form.'' Martin Gardner has also written the book ``The Amidextrous Universe.'' ==> physics/monkey.p <== Hanging over a pulley, there is a rope, with a weight at one end. At the other end hangs a monkey of equal weight. The rope weighs 4 ounces per foot. The combined ages of the monkey and it's mother is 4 years. The weight of the monkey is as many pounds as the mother is years old. The mother is twice as old as the monkey was when the mother was half as old as the monkey will be when the monkey is 3 times as old as the mother was when she was 3 times as old as the monkey. The weight of the rope and the weight is one-half as much again as the difference between the weight of the weight and the weight of the weight plus the weight of the monkey. How long is the rope? ==> physics/monkey.s <== The most difficult thing about this puzzle, as you probably expected, is translating all the convoluted problem statements into equations... the solution is pretty trivial after that. So... Let: m represent monkey M represent mother of monkey w represent weight r represent rope WÝx¨ = present weight of x (x is m, M, w, or r) AÝx,t¨ = age of x at time t (x is M or M, t is one of T1 thru T4) T1 = time at which mother is 3 times as old as monkey T2 = time at which monkey is 3 times as old as mother at T1 T3 = time at which mother is half as old as monkey at T2 T4 = present time For the ages, we have: AÝM,T1¨ = 3*AÝm,T1¨ AÝm,T2¨ = 3*AÝM,T1¨ = 9*AÝm,T1¨ AÝM,T3¨ = AÝm,T2¨/2 = 9*AÝm,T1¨/2 AÝm,T3¨ = AÝm,T1¨ + (T3-T1) = AÝm,T1¨ + (AÝM,T3¨-AÝM,T1¨) = AÝm,T1¨ + (9*AÝM,T1¨/2 - 3*AÝm,T1¨) = 5*AÝm,T1¨/2 AÝM,T4¨ = 2*AÝm,T3¨ = 5*AÝm,T1¨ AÝm,T4¨ = AÝm,T1¨ + (T4-T1) = AÝm,T1¨ + (AÝM,T4¨-AÝM,T1¨) = AÝm,T1¨ + (5*AÝm,T1¨ - 3*AÝm,T1¨) = 3*AÝm,T1¨ The present ages of monkey and mother sum to 4, so we have AÝm,T4¨ + AÝM,T4¨ = 4 3*AÝm,T1¨ + 5*AÝm,T1¨ = 4 8*AÝm,T1¨ = 4 AÝm,T1¨ = 1/2 Thus: AÝM,T4¨ = 5/2 AÝm,T4¨ = 3/2 Now for the weights, translating everything to ounces: Monkey's weight in lbs = mother's age in years, so: WÝm¨ = 16*5/2 = 40 Weight and monkey are same weight, so: WÝw¨ = WÝm¨ = 40 The last paragraph in the problem translates into: WÝr¨+WÝw¨ = (3/2)*((WÝw¨+WÝm¨)-WÝw¨) WÝr¨+ 40 = (3/2)*(( 40 + 40 )- 40 ) WÝr¨+ 40 = 60 WÝr¨ = 20 The rope weighs 4 ounces per foot, so its length is 5 feet. ==> physics/particle.p <== What is the longest time that a particle can take in travelling between two points if it never increases its acceleration along the way and reaches the second point with speed V? ==> physics/particle.s <== Assumptions: 1. x(0) = 0; x(T) = X 2. v(0) = 0; v(T) = V 3. d(a)/dt <= 0 Solution: a(t) = constant = A = V^2/2X which implies T = 2X/V. Proof: Consider assumptions as they apply to f(t) = A * t - v(t): 1. integral from 0 to T of f = 0 2. f(0) = f(T) = 0 3. d^2(f)/dt^2 <= 0 From the mean value theorem, f(t) = 0. QED. ==> physics/pole.in.barn.p <== Accelerate a pole of length l to a constant speed of 90% of the speed of light (.9c). Move this pole towards an open barn of length .9l (90% the length of the pole). Then, as soon as the pole is fully inside the barn, close the door. What do you see and what actually happens? ==> physics/pole.in.barn.s <== What the observer sees depends upon where the observer is, due to the finite speed of light. For definiteness, assume the forward end of the pole is marked "A" and the after end is marked "B". Let's also assume there is a light source inside the barn, and that the pole stops moving as soon as end "B" is inside the barn. An observer inside the barn next to the door will see the following sequence of events: 1. End "A" enters the barn and continues toward the back. 2. End "B" enters the barn and stops in front of the observer. 3. The door closes. 4. End "A" continues moving and penetrates the barn at the far end. 5. End "A" stops outside the barn. An observer at the other end of the barn will see: 1. End "A" enters the barn. 2. End "A" passes the observer and penetrates the back of the barn. 3. If the pole has markings on it, the observer will notice the part nearest him has stopped moving. However, both ends are still moving. 4. End "A" stops moving outside the barn. 5. End "B" continues moving until it enters the barn and then stops. 6. The door closes. After the observers have subtracted out the effects of the finite speed of light on what they see, both observers will agree on what happened: The pole entered the barn; the door closed so that the pole was completely contained within the barn; as the pole was being stopped it elongated and penetrated the back wall of the barn. Things are different if you are riding along with the pole. The pole is never inside the barn since it won't fit. End A of the pole penetrates the rear wall of the barn before the door is closed. If the wall of the barn is impenetrable, in all the above scenarios insert the wording "End A of the pole explodes" for "End A penetrates the barn." ==> physics/resistors.p <== What are the resistances between lattices of resistors in the shape of a: 1. Cube 2. Platonic solid 3. Hypercube 4. Plane sheet 5. Continuous sheet ==> physics/resistors.s <== 1. Cube The key idea is to observe that if you can show that two points in a circuit must be at the same potential, then you can connect them, and no current will flow through the connection and the overall properties of the circuit remain unchanged. In particular, for the cube, there are three resistors leaving the two "connection corners". Since the cube is completely symmetrical with respect to the three resistors, the far sides of the resistors may be connected together. And so we end up with: !---WWWWWW---! !---WWWWWW---! ! ! ! ! *--+---WWWWWW---+----+---WWWWWW---+---* ! ! ! ! !---WWWWWW---! !---WWWWWW---! 2. Platonic Solids Same idea for 8 12 and 20, since you use the symmetry to identify equi-potential points. The tetrahedron is hair more subtle: *---!---WWWWWW---!---* !\ /! W W W W W W W W W W W W ! \ / ! \ !! ! \ ! / \ W / \ W / <------- \ W / \!/ + By symmetry, the endpoints of the marked resistor are equi-potential. Hence they can be connected together, and so it becomes a simple: *---+---WWWWW---+----* ! ! +-WWW WWW-+ ! !-! ! !-WWW WWW-! 3. Hypercube Think of injecting a constant current I into the start vertex. It splits (by symmetry) into n equal currents in the n arms; the current of I/n then splits into I/n(n-1), which then splits into I/Ýn(n-1)(n-1)¨ and so on till the halfway point, when these currents start adding up. What is the voltage difference between the antipodal points? V = I x R; add up the voltages along any of the paths: n even: (n-2)/2 V = 2{I/n + I/(n(n-1)) + I/(n(n-1)(n-1)) + ... + I/(n(n-1) )} n odd: (n-3)/2 V = 2{I/n + I/(n(n-1)) + I/(n(n-1)(n-1)) + ... + I/(n(n-1) )} (n-1)/2 + I/(n(n-1) ) And R = V/I i.e. replace the Is in the above expression by 1s. For the 3-cube: R = 2{1/3} + 1/(3x2) = 5/6 ohm For the 4-cube: R = 2{1/4 + 1/(4x3)} = 2/3 ohm This formula yields the resistance from root to root of two (n-1)-ary trees of height n/2 with their end nodes identified (-when n is even; something similar when n is odd). Coincidentally, the 4-cube is such an animal and thus the answer 2/3 ohms is correct in that case. However, it does not provide the solution for n >= 5, as the hypercube does not have quite as many edges as were counted in the formula above. 4. The Infinite Plane For an infinite lattice: First inject a constant current I at a point; figure out the current flows (with heavy use of symmetry). Remove that current. Draw out a current I from the other point of interest (or inject a negative current) and figure out the flows (identical to earlier case, but displaced and in the other direction). By the principle of superposition, if you inject a current I into point a and take out a current I at point b at the same time, the currents in the paths are simply the sum of the currents obtained in the earlier two simpler cases. As in the n-cube, find the voltage between the points of interest, divide by I and voila'] As an illustration, in the adjacent points case: we have a current of I/4 in each of the four resistors: ^ ! ! v <--o--> -->o<-- ! ^ v ! (inject) (take out) And adding the currents, we have I/2 in the resistor connecting the two points. Therefore V=(1 ohm) x I/2 and effective resistance between the points = 1/2 ohm. You can (and showed how to) use symmetry to obtain the equivalent resistance of 1/2 between two adjacent nodes; but I doubt that symmetry alone will give you even the equivalent resistance of 2/pi between two diagonally adjacent nodes. ÝMore generally, the equivalent resistance between two nodes k diagonal units apart is (2/pi)(1+1/3+1/5+...+1/(2k-1)); that, plus symmetry and the known equivalent resistance between two adjacent nodes, is sufficient to derive all equivalent resistances in the lattice. 5. Continuous sheet Doesn't the resistance diverge in that case? The problem is that you can't inject current at a point. cf. "Random Walks and Electric Networks", by Doyle and Snell, published by the Mathematical Association of America. ==> physics/sail.p <== A sailor is in a sailboat on a river. The water (current) is flowing downriver at a velocity of 3 knots with respect to the land. The wind (air velocity) is zero, with respect to the land. The sailor wants to proceed downriver as quickly as possible, maximizing his downstream speed with respect to the land. Should he raise the sail, or not? ==> physics/sail.s <== Depends on the sail. If the boat is square-rigged, then not, since raising the sail will simply increase the air resistance. If the sailor has a fore-and-aft rig, then he should, since he can then tack into the wind. (Imagine the boat in still water with a 3-knot head wind). ==> physics/skid.p <== What is the fastest way to make a 90 degree turn on a slippery road? ==> physics/skid.s <== For higher speeds (measured at a small distance from the point of initiation of a sharp turn) the fastest way round is to "outside loop" - that is, steer away from the curve, and do a kidding 270. This technique is taught in advanced driving schools. References: M. Freeman and P. Palffy, American Journal of Physics, vol 50, p. 1098, 1982. P. Palffy and Unruh, American Journal of Physics, vol 49, p. 685, 1981. ==> physics/spheres.p <== Two spheres are the same size and weight, but one is hollow. They are made of uniform material, though of course not the same material. Without a minimum of apparatus, how can I tell which is hollow? ==> physics/spheres.s <== Since the balls have equal diameter and equal mass, their volume and density are also equal. However, the mass distribution is not equal, so they will have different moments of inertia - the hollow sphere has its mass concentrated at the outer edge, so its moment of inertia will be greater than the solid sphere. Applying a known torque and observing which sphere has the largest angular acceleration will determine which is which. An easy way to do this is to "race" the spheres down an inclined plane with enough friction to prevent the spheres from sliding. Then, by conservation of energy: mgh = 1/2 mv^2 + 1/2 Iw^2 Since the spheres are rolling without sliding, there is a relationship between velocity and angular velocity: w = v / r so mgh = 1/2 mv^2 + 1/2 I (v^2 / r^2) = 1/2 (m + I/r^2) v^2 and v^2 = 2mgh / (m + I / r^2) From this we can see that the sphere with larger moment of inertia (I) will have a smaller velocity when rolled from the same height, if mass and radius are equal with the other sphere. Thus the solid sphere will roll faster. ==> physics/wind.p <== Is a round-trip by airplane longer or shorter if there is wind blowing? ==> physics/wind.s <== It will take longer, by the ratio (s^2)/(s^2 - w^2) where s is the plane's speed, and w is the wind speed. The stronger the wind the longer it will take, up until the wind speed equals the planes speed, at which point the plane will never finish the trip. Math: s = plane's speed w = wind speed d = distance in one direction d / (s + w) = time to complete leg flying with the wind d / (s - w) = time to complete leg flying against the wind d / (s + w) + d / (s - w) = round trip time d / (s + w) + d / (s - w) = ratio of flying with wind to ------------------------- flying with no wind (bottom of d / s + d / s equation is top with w = 0) this simplifies to s^2 / (s^2 - w^2). ==> probability/amoeba.p <== A jar begins with one amoeba. Every minute, every amoeba turns into 0, 1, 2, or 3 amoebae with probability 25% for each case ( dies, does nothing, splits into 2, or splits into 3). What is the probability that the amoeba population eventually dies out? ==> probability/amoeba.s <== If p is the probability that a single amoeba's descendants will die out eventually, the probability that N amoebas' descendents will all die out eventually must be p^N, since each amoeba is independent of every other amoeba. Also, the probability that a single amoeba's descendants will die out must be independent of time when averaged over all the possibilities. At t=0, the probability is p, at t=1 the probability is 0.25(p^0+p^1+p^2+p^3), and these probabilities must be equal. Extinction probability p is a root of f(p)=p. In this case, p = sqrt(2)-1. The generating function for the sequence P(n,i), which gives the probability of i amoebas after n minutes, is f^n(x), where f^n(x) == f^(n-1) ( f(x) ), f^0(x) == x . That is, f^n is the nth composition of f with itself. Then f^n(0) gives the probability of 0 amoebas after n minutes, since f^n(0) = P(n,0). We then note that: f^(n+1)(x) = ( 1 + f^n(x) + (f^n(x))^2 + (f^n(x))^3 )/4 so that if f^(n+1)(0) -> f^n(0) we can solve the equation. The generating function also gives an expression for the expectation value of the number of amoebas after n minutes. This is d/dx(f^n(x)) evaluated at x=1. Using the chain rule we get f'(f^(n-1)(x))*d/dx(f^(n-1)(x)) and since f'(1) = 1.5 and f(1) = 1, we see that the result is just 1.5^n, as might be expected. ==> probability/apriori.p <== An urn contains one hundred white and black balls. You sample one hundred balls with replacement and they are all white. What is the probability that all the balls are white? ==> probability/apriori.s <== This question cannot be answered with the information given. In general, the following formula gives the conditional probability that all the balls are white given you have sampled one hundred balls and they are all white: P(100 white ! 100 white samples) = P(100 white samples ! 100 white) * P(100 white) ----------------------------------------------------------- sum(i=0 to 100) P(100 white samples ! i white) * P(i white) The probabilities P(i white) are needed to compute this formula. This does not seem helpful, since one of these (P(100 white)) is just what we are trying to compute. However, the following argument can be made: Before the experiment, all possible numbers of white balls from zero to one hundred are equally likely, so P(i white) = 1/101. Therefore, the odds that all 100 balls are white given 100 white samples is: P(100 white ! 100 white samples) = 1 / ( sum(i=0 to 100) (i/100)^100 ) = 63.6% This argument is fallacious, however, since we cannot assume that the urn was prepared so that all possible numbers of white balls from zero to one hundred are equally likely. In general, we need to know the P(i white) in order to calculate the P(100 white ! 100 white samples). Without this information, we cannot determine the answer. This leads to a general "problem": our judgments about the relative likelihood of things is based on past experience. Each experience allows us to adjust our likelihood judgment, based on prior probabilities. This is called Bayesian inference. However, if the prior probabilities are not known, then neither are the derived probabilities. But how are the prior probabilities determined? For example, if we are brains in the vat of a diabolical scientist, all of our prior experiences are illusions, and therefore all of our prior probabilities are wrong. All of our probability judgments indeed depend upon the assumption that we are not brains in a vat. If this assumption is wrong, all bets are off. ==> probability/cab.p <== A cab was involved in a hit and run accident at night. Two cab companies, the Green and the Blue, operate in the city. Here is some data: a) Although the two companies are equal in size, 85% of cab accidents in the city involve Green cabs and 15% involve Blue cabs. b) A witness identified the cab in this particular accident as Blue. The court tested the reliability of the witness under the same circumstances that existed on the night of the accident and concluded that the witness correctly identified each one of the two colors 80% of the time and failed 20% of the time. What is the probability that the cab involved in the accident was Blue rather than Green? If it looks like an obvious problem in statistics, then consider the following argument: The probability that the color of the cab was Blue is 80%] After all, the witness is correct 80% of the time, and this time he said it was Blue] What else need be considered? Nothing, right? If we look at Bayes theorem (pretty basic statistical theorem) we should get a much lower probability. But why should we consider statistical theorems when the problem appears so clear cut? Should we just accept the 80% figure as correct? ==> probability/cab.s <== The police tests don't apply directly, because according to the wording, the witness, given any mix of cabs, would get the right answer 80% of the time. Thus given a mix of 85% green and 15% blue cabs, he will say 20% of the green cabs and 80% of the blue cabs are blue. That's 20% of 85% plus 80% of 15%, or 17%+12% = 29% of all the cabs that the witness will say are blue. Of those, only 12/29 are actually blue. Thus P(cab is blue ! witness claims blue) = 12/29. That's just a little over 40%. Think of it this way... suppose you had a robot watching parts on a conveyor belt to spot defective parts, and suppose the robot made a correct determination only 50% of the time (I know, you should probably get rid of the robot...). If one out of a billion parts are defective, then to a very good approximation you'd expect half your parts to be rejected by the robot. That's 500 million per billion. But you wouldn't expect more than one of those to be genuinely defective. So given the mix of parts, a lot more than 50% of the REJECTED parts will be rejected by mistake (even though 50% of ALL the parts are correctly identified, and in particular, 50% of the defective parts are rejected). When the biases get so enormous, things starts getting quite a bit more in line with intuition. For a related real-life example of probability in the courtroom see People v. Collins, 68 Cal 2d319 (1968). ==> probability/coincidence.p <== Name some amazing coincidences. ==> probability/coincidence.s <== The answer to the question, "Who wrote the Bible," is, of course, Shakespeare. The King James Version was published in 1611. Shakespeare was 46 years old then (he turned 47 later in the year). Look up Psalm 46. Count 46 words from the beginning of the Psalm. You will find the word "Shake." Count 46 words from the end of the Psalm. You will find the word "Spear." An obvious coded message. QED. How many inches in the pole-to-pole diameter of the Earth? The answer is almost exactly 500,000,000 inches. Proof that the inch was defined by spacemen. ==> probability/coupon.p <== There is a free gift in my breakfast cereal. The manufacturers say that the gift comes in four different colours, and encourage one to collect all four (& so eat lots of their cereal). Assuming there is an equal chance of getting any one of the colours, what is the expected number of packets I must plough through to get all four? Can you generalise to n colours and/or unequal probabilities? ==> probability/coupon.s <== The problem is well known under the name of "COUPON COLLECTOR PROBLEM". The answer for n equally likely coupons is (1) C(n)=n*H(n) with H(n)=1+1/2+1/3+...+1/n. In the unequal probabilities case, with p_i the probability of coupon i, it becomes (2) C(n)=int_0^infty Ý1-prod_{i=1}^n (1-exp(-p_i*t))¨ dt which reduces to (1) when p_i=1/n (An easy exercise). In the equal probabilities case C(n)~n log(n). For a Zipf law, from (2), we have C(n)~n log^2(n). A related problem is the "BIRTHDAY PARADOX" familiar to people interested in hashing algorithms: With a party of 24 persons, you are likely (i.e. with probability >50%) to find two with the same birthday. The non equiprobable case was solved by: M. Klamkin and D. Newman, Extensions of the birthday surprise, J. Comb. Th. 3 (1967), 279-282. ==> probability/darts.p <== Peter throws two darts at a dartboard, aiming for the center. The second dart lands farther from the center than the first. If Peter now throws another dart at the board, aiming for the center, what is the probability that this third throw is also worse (i.e., farther from the center) than his first? Assume Peter's skilfulness is constant. ==> probability/darts.s <== Since the three darts are thrown independently, they each have a 1/3 chance of being the best throw. As long as the third dart is not the best throw, it will be worse than the first dart. Therefore the answer is 2/3. Ranking the three darts' results from A (best) to C (worst), there are, a priori, six equiprobable outcomes. possibility # 1 2 3 4 5 6 1st throw A A B B C C 2nd throw B C A C A B 3rd throw C B C A B A The information from the first two throws shows us that the first throw will not be the worst, nor the second throw the best. Thus possibilities 3, 5 and 6 are eliminated, leaving three equiprobable cases, 1, 2 and 4. Of these, 1 and 2 have the third throw worse than the first; 4 does not. Again the answer is 2/3. ==> probability/flips.p <== Consider a run of coin tosses: HHTHTTHTTTHTTTTHHHTHHHHHTHTTHT Define a success as a run of one H or T (as in THT or HTH). Use two different methods of sampling. The first method would consist of sampling the above data by taking 7 groups of three. This translates into the following sequences HHT, HTT, HTT, THT, TTT, HHH, and THH. In this sample there was one success, THT. The second method of sampling could be gotten by taking the groups in a serial sequence. Another way of explaining the method would be to take the tosses 1, 2, and 3 as the first set then tosses 2, 3, and 4 as the second set and so on to produce seven samples. The samples would be HHT, HTH, THT, HTT TTH, THT, and HTT, thus giving two success, HTH and THT. With these two methods what would be the expected value and the standard deviation for both methods? ==> probability/flips.s <== Case 1: Let us start with a simple case: Define success as T and consider sequences of length 1. In this case, the two sampling techniques are equivalent. Assuming that we are examining a truly random sequence of T and H. Thus if n groups of single sequences are considered or a sequence of length n is considered we will have the following statistics on the number of successes: Mean = n/2, and standard deviation (sd) = square_root(n)/2 Case 2: Define success as HT or TH: Here the two sampling techniques need to be distinguished: Sampling 1: Take "n" groups of two: Here probability of getting success in any group is 1/2 (TH and HT out of 4 possible cases). So the mean and the standard deviation is same as described in case 1. Sampling 2: Generate a sequence of length "n". It is easy to show that (n-1) samples are generated. The number of successes in this sequence is same as the number of T to H and H to T transitions. This problem is solved in John P. Robinson, Transition Count and Syndrome are Uncorrelated, IEEE Transactions on Information Theory, Jan 1988. I will just state the results: mean = (n-1)/2 and standard deviation = square_root(n-1)/2. In fact, if you want "n" samples in this case you need to generate length (n+1) sequence. Then the new mean and standard deviation are the same as described in Sampling 1. (replace n by n+1). The advantage in this sampling (i.e., sampling 2) is that you need only generate a sequence length of (n+1) to obtain n sample points as opposed to 2n (n groups of 2) in Sampling 1. OBSERVATION: The statistics has not changed. Case 3: Define success as THT and HTH. Sampling 1: This is a simple situation. Let us assume "n" samples need to be generated. Therefore, "n" groups of three are generated. The probability of a group being successful is 1/4 (THT and HTH out of 8 cases). Therefore mean and standard deviation are: mean= n/4 and sd= square_root(7n)/4 Sampling 2: This is not a simple situation. Let us generate a random sequence of length "n". There will be (n-2) samples in this case. Use an approach similar to that followed in the Jan 88 IEEE paper. I will just state the results. First we define a function or enumerator P(n,k) as the number of length "n" sequences that generate "k" successes. For example, P(4,1)= 4 (HHTH, HTHH, TTHT, and THTT are 4 possible length 4 sequences). I derived two generating functions g(x) and h(x) in order to enumerate P(n,k), they are compactly represented by the following matrix polynomial. _ _ _ _ _ _ ! g(x) ! ! 1 1 ! (n-3) ! 4 ! ! ! = ! ! ! ! ! h(x) ! ! 1 x ! !2+2x ! !_ _! !_ _! !_ _! The above is expressed as matrix generating function. It can be shown that P(n,k) is the coefficient of the x^k in the polynomial (g(x)+h(x)). For example, if n=4 we get (g(x)+h(x)) from the matrix generating function as (10+4x+2x^2). Clearly, P(4,1) (coefficient of x) is 4 and P(4,2)=2 ( There are two such sequences THTH, and HTHT). We can show that mean(k) = (n-2)/4 and sd= square_root(5n-12)/4 If we want to compare Sampling 1 with Sampling 2 then in Sampling 2 we need to generate "n" samples. This can be done by using sequences of length (n+2). Then our new statistics would be mean = n/4 (same as that in sampling 1) sd = square_root(5n-2)/4 (not the same as in sampling 1) sd in sampling 2 < sd in sampling 1. This difference can be explained by the fact that in serial sampling there is dependence on the past state. For example, if the past sample was TTT then we know that the next sample can't be a success. This was not the case in Case 1 or Case 2 (transition count). For example, in case 2 success was independent of previous sample. That is if my past sample was TT then my next sample can be TT or TH. The probability of success in Case 1 and Case 2 is not influenced by past history). Similar approach can be followed for higher dimensional cases. Here's a C program I had lying around that is relevant to the current discussion of coin-flipping. The algorithm is N^3 (for N flips) but it could certainly be improved. Compile with cc -o flip flip.c -lm -- Guy _________________ Cut here ___________________ #include #include char *progname; /* Program name */ #define NOT(c) (('H' + 'T') - (c)) /* flip.c -- a program to compute the expected waiting time for a sequence of coin flips, or the probabilty that one sequence of coin flips will occur before another. Guy Jacobson, 11/1/90 */ main (ac, av) int ac; char **av; { char *f1, *f2, *parseflips (); double compute (); progname = avÝ0¨; if (ac == 2) { f1 = parseflips (avÝ1¨); printf ("Expected number of flips until %s = %.1f\n", f1, compute (f1, NULL)); } else if (ac == 3) { f1 = parseflips (avÝ1¨); f2 = parseflips (avÝ2¨); if (strcmp (f1, f2) == 0) { printf ("Can't use the same flip sequence.\n"); exit (1); } printf ("Probability of flipping %s before %s = %.1f%%\n", avÝ1¨, avÝ2¨, compute (f1, f2) * 100.0); } else usage (); } char *parseflips (s) char *s; { char *f = s; while (*s) if (*s == 'H' !! *s == 'h') *s++ = 'H'; else if (*s == 'T' !! *s == 't') *s++ = 'T'; else usage (); return f; } usage () { printf ("usage: %s {HT}^n\n", progname); printf ("\tto get the expected waiting time, or\n"); printf ("usage: %s s1 s2\n\t(where s1, s2 in {HT}^n for some fixed n)\n", progname); printf ("\tto get the probability that s1 will occur before s2\n"); exit (1); } /* compute -- if f2 is non-null, compute the probability that flip sequence f1 will occur before f2. With null f2, compute the expected waiting time until f1 is flipped technique: Build a DFA to recognize (H+T)*f1 Ýor (H+T)*(f1+f2) when f2 is non-null¨. Randomly flipping coins is a Markov process on the graph of this DFA. We can solve for the probability that f1 precedes f2 or the expected waiting time for f1 by setting up a linear system of equations relating the values of these unknowns starting from each state of the DFA. Solve this linear system by Gaussian Elimination. */ typedef struct state { char *s; /* pointer to substring string matched */ int len; /* length of substring matched */ int backup; /* number of one of the two next states */ } state; double compute (f1, f2) char *f1, *f2; { double solvex0 (); int i, j, n1, n; state *dfa; int nstates; char *malloc (); n = n1 = strlen (f1); if (f2) n += strlen (f2); /* n + 1 states in the DFA */ dfa = (state *) malloc ((unsigned) ((n + 1) * sizeof (state))); if (]dfa) { printf ("Ouch, out of memory]\n"); exit (1); } /* set up the backbone of the DFA */ for (i = 0; i <= n; i++) { dfaÝi¨.s = (i <= n1) ? f1 : f2; dfaÝi¨.len = (i <= n1) ? i : i - n1; } /* for i not a final state, one next state of i is simply i+1 (this corresponds to another matching character of dfsÝi¨.s The other next state (the backup state) is now computed. It is the state whose substring matches the longest suffix with the last character changed */ for (i = 0; i <= n; i++) { dfaÝi¨.backup = 0; for (j = 1; j <= n; j++) if ((dfaÝj¨.len > dfaÝdfaÝi¨.backup¨.len) && dfaÝi¨.sÝdfaÝi¨.len¨ == NOT (dfaÝj¨.sÝdfaÝj¨.len - 1¨) && strncmp (dfaÝj¨.s, dfaÝi¨.s + dfaÝi¨.len - dfaÝj¨.len + 1, dfaÝj¨.len - 1) == 0) dfaÝi¨.backup = j; } /* our dfa has n + 1 states, so build a system n + 1 equations in n + 1 unknowns */ eqsystem (n + 1); for (i = 0; i < n; i++) if (i == n1) equation (1.0, n1, 0.0, 0, 0.0, 0, -1.0); else equation (1.0, i, -0.5, i + 1, -0.5, dfaÝi¨.backup, f2 ? 0.0 : -1.0); equation (1.0, n, 0.0, 0, 0.0, 0, 0.0); free (dfa); return solvex0 (); } /* a simple gaussian elimination equation solver */ double *m, **M; int rank; int neq = 0; /* create an n by n system of linear equations. allocate space for the matrix m, filled with zeroes and the dope vector M */ eqsystem (n) int n; { char *calloc (); int i; m = (double *) calloc (n * (n + 1), sizeof (double)); M = (double **) calloc (n, sizeof (double *)); if (]m !! ]M) { printf ("Ouch, out of memory]\n"); exit (1); } for (i = 0; i < n; i++) MÝi¨ = &mÝi * (n + 1)¨; rank = n; neq = 0; } /* add a new equation a * x_na + b * x_nb + c * x_nc + d = 0.0 (note that na, nb, and nc are not necessarily all distinct.) */ equation (a, na, b, nb, c, nc, d) double a, b, c, d; int na, nb, nc; { double *eq = MÝneq++¨; /* each row is an equation */ eqÝna + 1¨ += a; eqÝnb + 1¨ += b; eqÝnc + 1¨ += c; eqÝ0¨ = d; /* column zero holds the constant term */ } /* solve for the value of variable x_0. This will go nuts if therer are errors (for example, if m is singular) */ double solvex0 () { register i, j, jmax, k; register double max, val; register double *maxrow, *row; for (i = rank; i > 0; --i) { /* for each variable */ /* find pivot element--largest value in ith column*/ max = 0.0; for (j = 0; j < i; j++) if (fabs (MÝj¨Ýi¨) > fabs (max)) { max = MÝj¨Ýi¨; jmax = j; } /* swap pivot row with last row using dope vectors */ maxrow = MÝjmax¨; MÝjmax¨ = MÝi - 1¨; MÝi - 1¨ = maxrow; /* normalize pivot row */ max = 1.0 / max; for (k = 0; k <= i; k++) maxrowÝk¨ *= max; /* now eliminate variable i by subtracting multiples of pivot row */ for (j = 0; j < i - 1; j++) { row = MÝj¨; if (val = rowÝi¨) /* if variable i is in this eq */ for (k = 0; k <= i; k++) rowÝk¨ -= maxrowÝk¨ * val; } } /* the value of x0 is now in constant column of first row we only need x0, so no need to back-substitute */ val = -MÝ0¨Ý0¨; free (M); free (m); return val; } _________________________________________________________________ Guy Jacobson (201) 582-6558 AT&T Bell Laboratories uucp: {att,ucbvax}]ulysses]guy 600 Mountain Avenue internet: guy@ulysses.att.com Murray Hill NJ, 07974 ==> probability/flush.p <== Which set contains more flushes than the set of all possible hands? (1) Hands whose first card is an ace (2) Hands whose first card is the ace of spades (3) Hands with at least one ace (4) Hands with the ace of spades ==> probability/flush.s <== An arbitrary hand can have two aces but a flush hand can't. The average number of aces that appear in flush hands is the same as the average number of aces in arbitrary hands, but the aces are spread out more evenly for the flush hands, so set #3 contains a higher fraction of flushs. Aces of spades, on the other hand, are spread out the same way over possible hands as they are over flush hands, since there is only one of them in the deck. Whether or not a hand is flush is based solely on a comparison between different cards in the hand, so looking at just one card is necessarily uninformative. So the other sets contain the same fraction of flushes as the set of all possible hands. ==> probability/hospital.p <== A town has two hospitals, one big and one small. Every day the big hospital delivers 1000 babies and the small hospital delivers 100 babies. There's a 50/50 chance of male or female on each birth. Which hospital has a better chance of having the same number of boys as girls? ==> probability/hospital.s <== If there are 2N babies born, then the probability of an even split is (2N choose N) / (2 ** 2N) , where (2N choose N) = (2N)] / (N] * N]) . This is a DECREASING function. Think about it. If there are two babies born, then the probability of a split is 1/2 (just have the second baby be different from the first). With 2N babies, If there is a N,N-1 split in the first 2N-1, then there is a 1/2 chance of the last baby making it an even split. Otherwise there can be no even split. Therefore the probability is less than 1/2 overall for an even split. As N goes to infinity the probability of an even split approaches zero (although it is still the most likely event). ==> probability/icos.p <== The "house" rolls two 20-sided dice and the "player" rolls one 20-sided die. If the player rolls a number on his die between the two numbers the house rolled, then the player wins. Otherwise, the house wins (including ties). What are the probabilities of the player winning? ==> probability/icos.s <== It is easily seen that if any two of the three dice agree that the house wins. The probability that this does not happen is 19*18/(20*20). If the three numbers are different, the probability of winning is 1/3. So the chance of winning is 19*18/(20*20*3) = 3*19/200 = 57/200. ==> probability/intervals.p <== Given two random points x and y on the interval 0..1, what is the average size of the smallest of the three resulting intervals? ==> probability/intervals.s <== You could make a graph of the size of the smallest interval versus x and y. We know that the height of the graph is 0 along all the edges of the unit square where it is defined and on the line x=y. It achieves its maximum of 1/3 at (1/3,2/3) and (2/3,1/3). Assuming the graph has no curves or bizzare jags, the volume under the graph must be 1/9 and so the average height must also be 1/9. ==> probability/lights.p <== Waldo and Basil are exactly m blocks west and n blocks north from Central Park, and always go with the green light until they run out of options. Assuming that the probability of the light being green is 1/2 in each direction and that if the light is green in one direction it is red in the other, find the expected number of red lights that Waldo and Basil will encounter. ==> probability/lights.s <== Let E(m,n) be this number, and let (x)C(y) = x]/(y] (x-y)]). A model for this problem is the following nxm grid: ^ B---+---+---+ ... +---+---+---+ (m,0) ! ! ! ! ! ! ! ! ! N +---+---+---+ ... +---+---+---+ (m,1) <--W + E--> : : : : : : : : S +---+---+---+ ... +---+---+---+ (m,n-1) ! ! ! ! ! ! ! ! ! v +---+---+---+ ... +---+---+---E (m,n) where each + represents a traffic light. We can consider each traffic light to be a direction pointer, with an equal chance of pointing either east or south. IMHO, the best way to approach this problem is to ask: what is the probability that edge-light (x,y) will be the first red edge-light that the pedestrian encounters? This is easy to answer; since the only way to reach (x,y) is by going south x times and east y times, in any order, we see that there are (x+y)C(x) possible paths from (0,0) to (x,y). Since each of these has probability (1/2)^(x+y+1) of occuring, we see that the the probability we are looking for is (1/2)^(x+y+1)*(x+y)C(x). Multiplying this by the expected number of red lights that will be encountered from that point, (n-k+1)/2, we see that m-1 ----- \ E(m,n) = > ( 1/2 )^(n+k+1) * (n+k)C(n) * (m-k+1)/2 / ----- k=0 n-1 ----- \ + > ( 1/2 )^(m+k+1) * (m+k)C(m) * (n-k+1)/2 . / ----- k=0 Are we done? No] Putting on our Captain Clever Cap, we define n-1 ----- \ f(m,n) = > ( 1/2 )^k * (m+k)C(m) * k / ----- k=0 and n-1 ----- \ g(m,n) = > ( 1/2 )^k * (m+k)C(m) . / ----- k=0 Now, we know that n ----- \ f(m,n)/2 = > ( 1/2 )^k * (m+k-1)C(m) * (k-1) / ----- k=1 and since f(m,n)/2 = f(m,n) - f(m,n)/2, we get that n-1 ----- \ f(m,n)/2 = > ( 1/2 )^k * ( (m+k)C(m) * k - (m+k-1)C(m) * (k-1) ) / ----- k=1 - (1/2)^n * (m+n-1)C(m) * (n-1) n-2 ----- \ = > ( 1/2 )^(k+1) * (m+k)C(m) * (m+1) / ----- k=0 - (1/2)^n * (m+n-1)C(m) * (n-1) = (m+1)/2 * (g(m,n) - (1/2)^(n-1)*(m+n-1)C(m)) - (1/2)^n*(m+n-1)C(m)*(n-1) therefore f(m,n) = (m+1) * g(m,n) - (n+m) * (1/2)^(n-1) * (m+n-1)C(m) . Now, E(m,n) = (n+1) * (1/2)^(m+2) * g(m,n) - (1/2)^(m+2) * f(m,n) + (m+1) * (1/2)^(n+2) * g(n,m) - (1/2)^(n+2) * f(n,m) = (m+n) * (1/2)^(n+m+1) * (m+n)C(m) + (m-n) * (1/2)^(n+2) * g(n,m) + (n-m) * (1/2)^(m+2) * g(m,n) . Setting m=n in this formula, we see that E(n,n) = n * (1/2)^(2n) * (2n)C(n), and applying Stirling's theorem we get the beautiful asymptotic formula E(n,n) ~ sqrt(n/pi). ==> probability/lottery.p <== There n tickets in the lottery, k winners and m allowing you to pick another ticket. The problem is to determine the probability of winning the lottery when you start by picking 1 (one) ticket. A lottery has N balls in all, and you as a player can choose m numbers on each card, and the lottery authorities then choose n balls, define L(N,n,m,k) as the minimum number of cards you must purchase to ensure that at least one of your cards will have at least k numbers in common with the balls chosen in the lottery. ==> probability/lottery.s <== This relates to the problem of rolling a random number from 1 to 17 given a 20 sided die. You simply mark the numbers 18, 19, and 20 as "roll again". The probability of winning is, of course, k/(n-m) as for every k cases in which you get x "draw again"'s before winning, you get n-m-k similar cases where you get x "draw again"'s before losing. The example using dice makes it obvious, however. L(N,k,n,k) >= Ceiling((N-choose-k)/(n-choose-k)* (n-1-choose-k-1)/(N-1-choose-k-1)*L(N-1,k-1,n-1,k-1)) = Ceiling(N/n*L(N-1,k-1,n-1,k-1)) The correct way to see this is as follows: Suppose you have an (N,k,n,k) system of cards. Look at all of the cards that contain the number 1. They cover all ball sets that contain 1, and therefore these cards become an (N-1,k-1,n-1,k-1) covering upon deletion of the number 1. Therefore the number 1 must appear at least L(N-1,k-1,n-1,k-1). The same is true of all of the other numbers. There are N of them but n appear on each card. Thus we obtain the bound. ==> probability/particle.in.box.p <== A particle is bouncing randomly in a two-dimensional box. How far does it travel between bounces, on avergae? Suppose the particle is initially at some random position in the box and is traveling in a straight line in a random direction and rebounds normally at the edges. ==> probability/particle.in.box.s <== Let theta be the angle of the point's initial vector. After traveling a distance r, the point has moved r*cos(theta) horizontally and r*sin(theta) vertically, and thus has struck r*(sin(theta)+cos(theta))+O(1) walls. Hence the average distance between walls will be 1/(sin(theta)+cos(theta)). We now average this over all angles theta: 2/pi * intg from theta=0 to pi/2 (1/(sin(theta)+cos(theta))) dtheta which (in a computation which is left as an exercise) reduces to 2*sqrt(2)*ln(1+sqrt(2))/pi = 0.793515. ==> probability/pi.p <== Are the digits of pi random (i.e., can you make money betting on them)? ==> probability/pi.s <== No, the digits of pi are not truly random, therefore you can win money playing against a supercomputer that can calculate the digits of pi far beyond what we are currently capable of doing. This computer selects a position in the decimal expansion of pi -- say, at 10^100. Your job is to guess what the next digit or digit sequence is. Specifically, you have one dollar to bet. A bet on the next digit, if correct, returns 10 times the amount bet; a bet on the next two digits returns 100 times the amount bet, and so on. (The dollar may be divided in any fashion, so we could bet 1/3 or 1/10000 of a dollar.) You may place bets in any combination. The computer will tell you the position number, let you examine the digits up to that point, and calculate statistics for you. It is easy to set up strategies that might win, if the supercomputer doesn't know your strategy. For example, "Always bet on 7" might win, if you are lucky. Also, it is easy to set up bets that will always return a dollar. For example, if you bet a penny on every two-digit sequence, you are sure to win back your dollar. Also, there are strategies that might be winning, but we can't prove it. For example, it may be that a certain sequence of digits never occurs in pi, but we have no way of proving this. The problem is to find a strategy that you can prove will always win back more than a dollar. The assumption that the position is beyond the reach of calculation means that we must rely on general facts we know about the sequence of digits of pi, which is practically nil. But it is not completely nil, and the challenge is to find a strategy that will always win money. A theorem of Mahler (1953) states that for all integers p, q > 1, -42 !pi - p/q! > q This says that pi cannot have a rational approximation that is extremely tight. Now suppose that the computer picks position N. I know that the next 41 * N digits cannot be all zero. For if they were, then the first N digits, treated as a fraction with denominator 10^N, satisfies: !pi - p / 10^N! < 10^(-42 N) which contradicts Mahler's theorem. So, I split my dollar into 10^(41N) - 1 equal parts, and bet on each of the sequences of 41N digits, except the one with all zeroes. One of the bets is sure to win, so my total profit is about 10(^-41N) of a dollar] This strategy can be improved a number of ways, such as looking for other repeating patterns, or improvements to the bound of 42 -- but the earnings are so pathetic, it hardly seems worth the effort. Are there other winning strategies, not based on Mahler's theorem? I believe there are algorithms that generate 2N binary digits of pi, where the computations are separate for each block of N digits. Maybe from something like this, we can find a simple subsequence of the binary digits of pi which is always zero, or which has some simple pattern. ==> probability/random.walk.p <== Waldo has lost his car keys] He's not using a very efficient search; in fact, he's doing a random walk. He starts at 0, and moves 1 unit to the left or right, with equal probability. On the next step, he moves 2 units to the left or right, again with equal probability. For subsequent turns he follows the pattern 1, 2, 1, etc. His keys, in truth, were right under his nose at point 0. Assuming that he'll spot them the next time he sees them, what is the probability that poor Waldo will eventually return to 0? ==> probability/random.walk.s <== I can show the probability that Waldo returns to 0 is 1. Waldo's wanderings map to an integer grid in the plane as follows. Let (X_t,Y_t) be the cumulative sums of the length 1 and length 2 steps respectively taken by Waldo through time t. By looking only at even t, we get the ordinary random walk in the plane, which returns to the origin (0,0) with probability 1. In fact, landing at (2n, n) for any n will land Waldo on top of his keys too. There's no need to look at odd t. Similar considerations apply for step sizes of arbitrary (fixed) size. ==> probability/reactor.p <== There is a reactor in which a reaction is to take place. This reaction stops if an electron is present in the reactor. The reaction is started with 18 positrons; the idea being that one of these positrons would combine with any incoming electron (thus destroying both). Every second, exactly one particle enters the reactor. The probablity that this particle is an electron is 0.49 and that it is a positron is 0.51. What is the probablity that the reaction would go on for ever?? Note: Once the reaction stops, it cannot restart. ==> probability/reactor.s <== Let P(n) be the probability that, starting with n positrons, the reaction goes on forever. Clearly P'(n+1)=P'(0)*P'(n), where the ' indicates probabilistic complementation; also note that P'(n) = .51*P'(n+1) + .49*P'(n-1). Hence we have that P(1)=(P'(0))^2 and that P'(0) = .51*P'(1) ==> P'(0) equals 1 or 49/51. We thus get that either P'(18) = 1 or (49/51)^19 ==> P(18) = 0 or 1 - (49/51)^19. The answer is indeed the latter. A standard result in random walks (which can be easily derived using Markov chains) yields that if p>1/2 then the probability of reaching the absorbing state at +infinity as opposed to the absorbing state at -1 is 1-r^(-i), where r=p/(1-p) (p is the probability of moving from state n to state n-1, in our case .49) and i equals the starting location + 1. Therefore we have that P(18) = 1-(.49/.51)^19. ==> probability/roulette.p <== You are in a game of Russian roulette, but this time the gun (a 6 shooter revolver) has three bullets _in_a_row_ in three of the chambers. The barrel is spun only once. Each player then points the gun at his (her) head and pulls the trigger. If he (she) is still alive, the gun is passed to the other player who then points it at his (her) own head and pulls the trigger. The game stops when one player dies. Now to the point: would you rather be first or second to shoot? ==> probability/roulette.s <== All you need to consider are the six possible bullet configurations B B B E E E -> player 1 dies E B B B E E -> player 2 dies E E B B B E -> player 1 dies E E E B B B -> player 2 dies B E E E B B -> player 1 dies B B E E E B -> player 1 dies One therefore has a 2/3 probability of winning (and a 1/3 probability of dying) by shooting second. I for one would prefer this option. ==> probability/unfair.p <== Generate even odds from an unfair coin. For example, if you thought a coin was biased toward heads, how could you get the equivalent of a fair coin with several tosses of the unfair coin? ==> probability/unfair.s <== Toss twice. If both tosses give the same result, repeat this process (throw out the two tosses and start again). Otherwise, take the first of the two results. ==> series/series.01.p <== M, N, B, D, P ? ==> series/series.01.s <== T. If you say the sounds these letters make out loud, you will see that the next letter is T. ==> series/series.02.p <== H, H, L, B, B, C, N, O, F ? ==> series/series.02.s <== Answer 1: N, N, M, A The symbols for the elements. Answer 2: N, S, M, A The names of the elements. ==> series/series.03.p <== W, A, J, M, M, A, J? ==> series/series.03.s <== J, V, H, T, P, T, F, P, B, L. Presidents. ==> series/series.03a.p <== G, J, T, J, J, J, A, M, W, J, J, Z, M, F, J, ? ==> series/series.03a.s <== G, J, T, J, J, J, A, M, W, J, J, Z, M, F, J, A. Presidents' first names. ==> series/series.03b.p <== A, J, B, C, G, T, C, V, J, T, D, F, K, B, H, ? ==> series/series.03b.s <== A, J, B, C, G, T, C, V, J, T, D, F, K, B, H, J. Vice Presidents. ==> series/series.03c.p <== M, A, M, D, E, L, R, H, ? ==> series/series.03c.s <== M, A, M, D, E, L, R, H, A. Presidents' wives' first names. ==> series/series.04.p <== A, E, H, I, K, L, ? ==> series/series.04.s <== M, N, O, P, U, W. Letters in the Hawaiian alphabet. ==> series/series.05.p <== A B C D E F G H? ==> series/series.05.s <== M. The names of the cross-streets travelling west on (say) Commonwealth Avenue from Boston Garden: Arlington, Berkeley, Clarendon, Dartmouth, Exeter, Fairfield, Gloucester, Hereford, Massachusetts Ave. ==> series/series.06.p <== Z, O, T, T, F, F, S, S, E, N? ==> series/series.06.s <== T. The name of the integers starting with zero. ==> series/series.06a.p <== F, S, T, F, F, S, ? ==> series/series.06a.s <== The words "first", "second", "third", etc. The same as the previous from this point on. ==> series/series.07.p <== 1, 1 1, 2 1, 1 2 1 1, ... What is the pattern and asymptotics of this series? ==> series/series.07.s <== Each line is derived from the last by the transformation (for example) ... z z z x x y y y ... -> ... 3 z 2 x 3 y ... John Horton Conway analyzed this in "The Weird and Wonderful Chemistry of Audioactive Decay" (T M Cover & B Gopinath (eds) OPEN PROBLEMS IN COMMUNICATION AND COMPUTATION, Springer-Verlag (1987)). You can also find his most complete FRACTRAN paper in this collection. First, he points out that under this sequence, you frequently get adjacent subsequences XY which cannot influence each other in any future derivation of the sequence rule. The smallest such are called "atoms" or "elements". As Conway claims to have proved, there are 92 atoms which show up eventually in every sequence, no matter what the starting value (besides <> and <22>), and always in the same non-zero limiting proportions. Conway named them after some other list of 92 atoms. As a puzzle, see if you can recreate the list from the following, in decreasing atomic number: U Pa Th Ac Ra Fr Rn Ho.AT Po Bi Pm.PB Tl Hg Au Pt Ir Os Re Ge.Ca.W Ta HF.Pa.H.Ca.W Lu Yb Tm ER.Ca.Co HO.Pm Dy Tb Ho.GD EU.Ca.Co Sm PM.Ca.Zn Nd Pr Ce LA.H.Ca.Co Ba Cs Xe I Ho.TE Eu.Ca.SB Pm.SN In Cd Ag Pd Rh Ho.RU Eu.Ca.TC Mo Nb Er.ZR Y.H.Ca.Tc SR.U Rb Kr Br Se As GE.Na Ho.GA Eu.Ca.Ac.H.Ca.ZN Cu Ni Zn.CO Fe Mn CR.Si V Ti Sc Ho.Pa.H.CA.Co K Ar Cl S P Ho.SI Al Mg Pm.NA Ne F O N C B Be Ge.Ca.LI He Hf.Pa.H.Ca.Li Uranium is 3, Protactinium is 13, etc. Rn => Ho.AT means the following: Radon forms a string that consists of two atoms, Holmium on the left, and Astatine on the right. I capitalize the symbol for At to remind you that Astatine, and not Holmium, is one less than Radon in atomic number. As a check, against you or me making a mistake, Hf is 111xx, Nd is 111xxx, In and Ni are 111xxxxx, K is 111x, and H is 22. Next see if you can at least prove that any atom other than Hydrogen, eventually (and always thereafter) forms strings containing all 92 atoms. The grand Conway theorem here is that every string eventually forms (within a universal time limit) strings containing all the 92 atoms in certain specific non-zero limiting proportions, and that digits N greater than 3 are eventually restricted to one of two atomic patterns (ie, abc...N and def...N for some {1,2,3} sequences abc... and def...), which Conway calls isotopes of Np and Pu. (For N=2, these are He and Li), and that these transuranic atoms have a zero limiting proportion. The longest lived exotic element is Methuselum (2233322211N) which takes about 25 applications to reduce to the periodic table. -Matthew P Wiener (weemba@libra.wistar.upenn.edu) Conway gives many results on the ultimate behavior of strings under this transformation: for example, taking the sequence derived from 1 (or any other string except 2 2), the limit of the ratio of length of the (n+1)th term to the length of the nth term as n->infinity is a fixed constant, namely 1.30357726903429639125709911215255189073070250465940... This number is from Ilan Vardi, "Computational Recreations in Mathematica", Addison Wesley 1991, page 13. Another sequence that is related but not nearly as interesting is: 1, 11, 21, 1112, 3112, 211213, 312213, 212223, 114213, 31121314, 41122314, 31221324, 21322314, and 21322314 generates itself, so we have a cycle. ==> series/series.08a.p <== G, L, M, B, C, L, M, C, F, S, ? ==> series/series.08a.s <== Army officer ranks, descending. ==> series/series.08b.p <== A, V, R, R, C, C, L, L, L, E, ? ==> series/series.08b.s <== Navy officer ranks, descending. ==> series/series.09a.p <== S, M, S, S, S, C, P, P, P, ? ==> series/series.09a.s <== Army non-comm ranks, descending. ==> series/series.09b.p <== M, S, C, P, P, P, S, S, S, ? ==> series/series.09b.s <== Navy non-comm ranks, descending. ==> series/series.10.p <== D, P, N, G, C, M, M, S, ? ==> series/series.10.s <== N, V, N, N, R. States in Constitution ratification order. ==> series/series.11.p <== R O Y G B ? ==> series/series.11.s <== V. Colors. ==> series/series.12.p <== A, T, G, C, L, ? ==> series/series.12.s <== V, L, S, S, C, A, P. Zodiacal signs. ==> series/series.13.p <== M, V, E, M, J, S, ? ==> series/series.13.s <== U, N, P. Names of the Planets. ==> series/series.14.p <== A, B, D, O, P, ? ==> series/series.14.s <== Q, R. Only letters with an inside as printed. ==> series/series.14a.p <== A, B, D, E, G, O, P, ? ==> series/series.14a.s <== Q. Letters with cursive insides. ==> series/series.15.p <== A, E, F, H, I, ? ==> series/series.15.s <== L, M, N, O, S, U. Letters whose English names start with vowels. ==> series/series.16.p <== A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, X, Y? ==> series/series.16.s <== Z. Letters whose English names have one syllable. ==> series/series.17.p <== T, P, O, F, O, F, N, T, S, F, T, F, E, N, S, N? ==> series/series.17.s <== T, T, T, E, T, E. Digits of Pi. ==> series/series.18.p <== 10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, ___ , 100, 121, 10000 ==> series/series.18.s <== 10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, 31 , 100, 121, 10000 Sixteen in base n for n=16, 15, ..., 2. ==> series/series.19.p <== 1 01 01011 0101101011011 0101101011011010110101101101011011 etc. Each string is formed from the previous string by substituting '01' for '1' and '011' for '0' simultaneously at each occurance. Notice that each string is an initial substring of the previous string so that we may consider them all as initial substrings of an infinite string. The puzzle then is, given n, determine if the nth digit is 0 or 1 without having to construct all the previous digits. That is, give a non-recursive formula for the nth digit. ==> series/series.19.s <== Let G equal the limit string generated by the above process and define the string F by FÝ0¨ = "0", FÝn¨ = "1" if n = floor(phi*m) for some positive integer m, FÝn¨ = "0" if n = floor(phi^2*m) for some positive integer m, where floor(x) is the greatest integer =< x and phi = (1 + \/5)/2; I claim that F = G. I will try to motivate my solution. Let gÝ0¨="0" and define gÝn+1¨ to be the string that results from replacing "0" in gÝn¨ with "01" and "1" with "011"; furthermore, let s(n) and t(n) be the number of "0"'s and "1"'s in gÝn¨, respectively. Note that we have the following recursive formulas : s(n+1) = s(n) + t(n) and t(n+1) = s(n) + 2t(n). I claim that s(n) = Fib(2n-1) and t(n) = Fib(2n), where Fib(m) is the mth Fibonacci number (defined by Fib(-1) = 1, Fib(0) = 0, Fib(n+1) = Fib(n) + Fib(n-1) for n>=0); this is easily established by induction. Now noting that Fib(2n)/Fib(2n-1) -> phi as n -> infinity, we see that if the density of the "0"'s and "1"'s exists, they must be be 1/phi^2 and 1/phi, respectively. What is the simplest generating sequence which has this property? Answer: the one given above. Proof: We start with Beatty's Theorem: if a and b are positive irrational numbers such that 1/a + 1/b = 1, then every positive integer has a representation of the form floor(am) or floor(bm) (m a positive integer), and this representation is unique. This shows that F is well-defined. I now claim that Lemma: If S(n) and T(n) (yes, two more functions; apparently today's the day that functions have their picnic) represent the number of "0"'s and "1"'s in the initial string of F of length n, then S(n) = ceil(n/phi^2) and T(n) = floor(n/phi) (ceil(x) is the smallest integer >= x). Proof of lemma: using the identity phi^2 = phi + 1 we see that S(n) + T(n) = n, hence for a given n either S(n) = S(n-1) + 1 or T(n) = T(n-1) + 1. Now note that if FÝn-1¨="1" ==> n-1 = floor(phi*m) for some positive integer m and since phi*m-1 < floor(phi*m) < phi*m ==> m-1/phi < (n-1)/phi < m ==> T(n) = T(n-1) + 1. To finish, note that if FÝn-1¨="0" ==> n-1 = floor(phi^2*m) for some positive integer m and since phi^2*m-1 < floor(phi^2*m) < phi^2*m ==> m-1/phi^2 < (n-1)/phi^2 < m ==> S(n) = S(n-1) + 1. Q.E.D. I will now show that F is invariant under the operation of replacing "0" with "01" and "1" with "011"; it will then follow that F=G. Note that this is equivalent to showing that FÝ2S(n) + 3T(n)¨ = "0", FÝ2S(n) + 3T(n) + 1¨ = "1", and that if n = Ýphi*m¨ for some positive integer m, then FÝ2S(n) + 3T(n) + 2¨ = "1". One could waste hours trying to prove some fiendish identities; watch how I sidestep this trap. For the first part, note that by the above lemma FÝ2S(n) + 3T(n)¨ = FÝ2*ceil(n/phi^2) + 3*floor(n/phi)¨ = FÝ2n + floor(n/phi)¨ = FÝ2n + floor(n*phi-n)¨ = FÝfloor(phi*n+n)¨ = FÝfloor(phi^2*n)¨ ==> FÝ2S(n) + 3T(n)¨ = "0". For the second, it is easy to see that since phi^2>2, if FÝm¨="0" ==> FÝm¨="1" hence the first part implies the second part. Finally, note that if n = Ýphi*m¨ for some positive integer m, then FÝ2S(n) + 3T(n) + 3¨ = FÝ2S(n+1) + 3T(n+1)¨ = "0", hence by the same reasoning as above FÝ2S(n) + 3T(n) + 2¨ = "1". Q.E.D. -- clong@remus.rutgers.edu (Chris Long) ==> series/series.20.p <== 1 2 5 16 64 312 1812 12288 ==> series/series.20.s <== ANSWER: 95616 The sum of factorial(k)*factorial(n-k) for k=0,...,n. ==> series/series.21.p <== 5, 6, 5, 6, 5, 5, 7, 5, ? ==> series/series.21.s <== The number of letters in the ordinal numbers. First 5 Second 6 Third 5 Fourth 6 Fifth 5 Sixth 5 Seventh 7 Eighth 6 Ninth 5 etc. ==> series/series.22.p <== 3 1 1 0 3 7 5 5 2 ? ==> series/series.22.s <== ANSWER: 4 The digits of pi expressed in base eight. ==> series/series.23.p <== 22 22 30 13 13 16 16 28 28 11 ? ==> series/series.23.s <== ANSWER: 15 The birthdays of the Presidents of the United States. ==> series/series.24.p <== What is the next letter in the sequence: W, I, T, N, L, I, T? ==> series/series.24.s <== S. First letters of words in question. ==> series/series.25.p <== 1 3 4 9 10 12 13 27 28 30 31 36 37 39 40 ? ==> series/series.25.s <== 1 3 4 9 10 12 13 27 28 30 31 36 37 39 40 ... i in binary, treated as a base 3 number and converted to decimal. ==> series/series.26.p <== 1 3 2 6 7 5 4 12 13 15 14 10 11 9 8 24 25 27 26 ? ==> series/series.26.s <== 1 3 2 6 7 5 4 12 13 15 14 10 11 9 8 24 25 27 26 ... Take i in binary, for each 1 bit (in i, not changed) flip the next bit. This can also be phrased in reversing sequences of numbers. More simply, just the integers in reflective-Gray-code order. ==> series/series.27.p <== 0 1 1 2 1 2 1 3 2 2 1 3 1 2 2 4 1 3 1 3 2 2 1 4 2 ? ==> series/series.27.s <== 0 1 1 2 1 2 1 3 2 2 1 3 1 2 2 4 1 3 1 3 2 2 1 4 2 ... Number of factors in prime factorization of i. ==> series/series.28.p <== 0 2 3 4 5 5 7 6 6 7 11 7 13 9 8 8 17 8 19 9 10 13 23 9 10 ? ==> series/series.28.s <== 0 2 3 4 5 5 7 6 6 7 11 7 13 9 8 8 17 8 19 9 10 13 23 9 10 ... Sum of factors in prime factorization of i. ==> series/series.29.p <== 1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1 2 2 3 2 3 3 4 2 3 3 4 3 4 ? ==> series/series.29.s <== 1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1 2 2 3 2 3 3 4 2 3 3 4 3 4 ... The number of 1s in the binary expansion of n. ==> series/series.30.p <== I I T Y W I M W Y B M A D ==> series/series.30.s <== ? (first letters of "If I tell you what it means will you buy me a drink?") ==> series/series.31.p <== 6 2 5 5 4 5 6 3 7 ==> series/series.31.s <== 6. The number of segments on a standard calculator display it takes to represent the digits starting with 0. _ _ _ _ _ _ _ _ ! ! ! _! _! !_! !_ !_ ! !_! !_! !_! ! !_ _! ! _! !_! ! !_! _! ==> series/series.32.p <== 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 ==> series/series.32.s <== 0 -> 1 01 -> 10 0110 -> 1001 01101001 -> 10010110 Recursively append the inverse. This sequence is known as the Morse-Thue sequence. It can be defined non-recursively as the nth term is the mod 2 count of 1s in n written in binary: 0->0 1->1 10->1 11->0 100->1 101->0 110->0 111->1 etc. Reference: Dekking, et. al., "Folds] I,II,III" The Mathematical Intelligencer, v4,#3,#4,#4. ==> series/series.33.p <== 2 12 360 75600 ==> series/series.33.s <== 2 = 2^1 12 = 2^2 * 3^1 360 = 2^3 * 3^2 * 5^1 75600 = 2^4 * 3^3 * 5^2 * 7^1 174636000 = 2^5 * 3^4 * 5^3 * 7^2 * 11^1 ==> series/series.34.p <== 3 5 4 4 3 5 5 4 3 ==> series/series.34.s <== The number of letters in the English words for the counting numbers. ==> series/series.35.p <== 1 2 3 2 1 2 3 4 2 1 2 3 4 2 2 3 ==> series/series.35.s <== The number of letters in the Roman numeral representation of the numbers. ==> trivia/area.codes.p <== When looking at a map of the distribution of telephone area codes for North America, it appears that they are randomly distributed. I am doubtful that this is the case, however. Does anyone know how the area codes were/are chosen? ==> trivia/area.codes.s <== Originally, back in the middle 1950's when direct dialing of long distance calls first became possible, the idea was to assign area codes with the 'shortest' dialing time required to the larger cities. Touch tone dialing was very rare. Most dialed calls were with 'rotary' dials. Area codes like 212, 213, 312 and 313 took very little time to dial (while waiting for the dial to return to normal) as opposed, for example, to 809, 908, 709, etc ... So the 'quickest to dial' area codes were assigned to the places which would probably receive the most direct dialed calls, i.e. New York City got 212, Chicago got 312, Los Angeles got 213, etc ... Washington, DC got 202, which is a little longer to dial than 212, but much shorter than others. In order of size and estimated amount of telephone traffic, the numbers got larger: San Fransisco got 415, which is sort of in the middle, and Miami got 305, etc. At the other end of the spectrum came places like Hawaii (it only got statehood as of about 1958) with 808, Puerto Rico with 809, Newfounland with 709, etc. The original (and still in use until about 1993) plan is that area codes have a certain construction to the numbers: The first digit will be 2 through 9. The second digit will always be 0 or 1. The third digit will be 1 through 9. Three digit numbers with two zeros will be special codes, ie. 700, 800 or 900. Three digit numbers with two ones are for special local codes, i.e. 411 for local directory assistance, 611 for repairs, etc. Three digit codes ending in '10', i.e. 410, 510, 610, 710, 810, 910 were 'area codes' for the AT&T (and later on Western Union) TWX network. This rule has been mostly abolished, however 610 is still Canadian TWX, and 910 is still used by Western Union TWX. Gradually the '10' codes are being converted to regular area codes. We are running out of possible combinations of numbers using the above rules, and it is estimated that beginning in 1993-94, area codes will begin looking like regular telephone prefix codes, with numbers other than 0 or 1 as the second digit. I hope this gives you a basic idea. There were other rules at one time such as not having an area code with zero in the second digit in the same state as a code with one in the second digit, etc .. but after the initial assignment of numbers back almost forty years ago, some of those rules were dropped when it became apparent they were not flexible enough. Patrick Townson TELECOM Digest Moderator -- Patrick Townson patrick@chinet.chi.il.us / ptownson@eecs.nwu.edu / US Mail: 60690-1570 FIDO: 115/743 / AT&T Mail: 529-6378 (]ptownson) / MCI Mail: 222-4956 ==> trivia/eskimo.snow.p <== How many words do the Eskimo have for snow? ==> trivia/eskimo.snow.s <== Couple of weeks ago, someone named D.K. Holm in the Boston Phoenix came up with the list, drawn from the Inupiat Eskimo Dictionary by Webster and Zibell, and from Thibert's English-Eskimo Eskimo-English Dictionary. The words may remind you of generated passwords. Eskimo English Eskimo English ---------------------------------+---------------------------- apun snow ! pukak sugar snow apingaut first snowfall ! pokaktok salt-like snow aput spread-out snow ! miulik sleet kanik frost ! massak snow mixed with water kanigruak frost on a ! auksalak melting snow living surface ! aniuk snow for melting ayak snow on clothes ! into water kannik snowflake ! akillukkak soft snow nutagak powder snow ! milik very soft snow aniu packed snow ! mitailak soft snow covering an aniuvak snowbank ! opening in an ice floe natigvik snowdrift ! sillik hard, crusty snow kimaugruk snowdrift that ! kiksrukak glazed snow in a thaw blocks something ! mauya snow that can be perksertok drifting snow ! broken through akelrorak newly drifting snow ! katiksunik light snow mavsa snowdrift overhead ! katiksugnik light snow deep enough and about to fall ! for walking kaiyuglak rippled surface ! apuuak snow patch of snow ! sisuuk avalanche =*= ==> trivia/federal.reserve.p <== What is the pattern to this list: Boston, MA New York, NY Philadelphia, PA Cleveland, OH Richmond, VA Atlanta, GA Chicago, IL St. Louis, MO Minneapolis, MN Kansas City, MO Dallas, TX San Francisco, CA ==> trivia/federal.reserve.s <== Each of the cities is a location for a Federal Reserve. The cities are listed in alphabetical order based on the letter that represents each city on a dollar bill. ==> trivia/jokes.self-referential.p <== What are some self-referential jokes? ==> trivia/jokes.self-referential.s <== Q: What is alive, green, lives all over the world, and has seventeen legs? A: Grass. I lied about the legs. The two rules for success are: 1. Never tell them everything you know.