F I D D L E by Doug Beeferman Systematic ways to solve selected levels NOTATION: @ - standard unit square X - center of rotation unit square --> When I say "2-edge", "3-edge", etc., I am referring to a piece's edge that has length 2, 3, etc. --> The First (and only!) Fundamental Rule of Fiddle is as follows: When a piece is rotated, its center of rotation does not move. LEVEL E, "Jagged edge": The first ten levels of Fiddle rely heavily on visualization, and the remainder rely heavily on logic. But the two aren't unrelated. Look at piece (1). Our goal is to construct a solid square- no holes are to be left unfilled. A region of (1) encloses a small "bay"- that is, it surrounds a region of unfilled squares on three sides. Clearly this "bay" must be filled for a solid square to be constructed, regardless of what we know of the other pieces. It must be filled by another piece, or possibly by more than one. Look around at the other pieces and determine which piece can singlehandedly fill (1)'s bay. Only (2) can. Now, on the side of (1) opposite of where you've placed (2) there's an empty region of 2 unit squares length. There are two possibilities: (a) this side of (1) is an EDGE of the 11x11 square, or (b) this side of (1) is not an edge, and there is at least one more piece in this direction. Inspection will tell you that (b) is true, because if this side were an edge then the only piece that could fill the 2-square void (piece (6)) would cause the edge to have length 13, which exceeds the desired 11. Piece (4) is the only remaining piece that fits nicely next to this side of (1). After (4) is placed, the correct positions of (6) and (5) are visually apparent. Then the final piece, (3), can be placed. -------------------- LEVEL M, "One good turn": 1= @@@X 3= @ ... @ @ @@@ @@@ @ @ @ @ @@@ @@X@ @ @ @@ The area that piece (1) encloses must be filled, and the only piece that can fill it is (3). A corollary to the First Fundamental Rule is that the proper position for a piece that is to be rotated into another can be found by considering how far apart the centers of rotation are to be. For our rotation of (3) into (1), we want the end result to look like: 111X <-- C(1) 222221 If the center of (3) can be placed 3 squares to the 211121 left and 4 squares below the center of (1), it can 222111 be rotated to fill the hole as shown. Manipulate X <-- C(3) (3) so that its center is in this position and 2 rotate as necessary. 2222 After (1) and (3) are joined, the remaining pieces can quickly be put into place to complete the square. -------------------- LEVEL T, "Spacemaker": 1= @ 2= @ 3= @@@ 4= @ @ ... @@ @X@ X @X@ @X The outer piece (6) has dimensions of the completed square, so the solution entails filling it exactly. Piece (5) must enter the square somehow. Its center of rotation must be within the square in the solution, so by the First Fundamental Rule it must be within the square before it is rotated into place. It can only enter the square by way of (6)'s 2-square void at the upper left. After (6) is rotated into the square it will fill the two upper, undivided quadrants. The divided quadrants below have to be filled, therefore, with pieces 1 through 4. Two solid 3x3 boxes need to made; (2) joins with (4) to form such a box, and (3) joins with (1) to form another. To move (2) from the upper-right quadrant of (6) to its destination in the lower left, rotate (6) clockwise so that (2) can first be moved to the lower right quadrant. Follow a similar procedure to join (2) with (4), and again to join (3) with (1), rotating pieces 1 through 4 as necessary to fit inside the 3x3 quadrants. Soon you'll be able to rotate (5) into place. -------------------- LEVEL W, "Holey Piece": This level is one of three in the game in which the final positions of the pieces are clear from the beginning. Pieces (1) and (2) both have holes to fill. Since the holes are completely within the pieces (they are "lakes" rather than "bays"), filling them can only be accomplished by rotating the "mainland" pieces to surround the "island" pieces. Which do we fill first, (1) or (2)? These pieces have to ultimately attach to each other (it's clear how) and we know that a piece cannot slide if it has an embedded "island"; therefore, when whatever we choose to be the second piece is filled it must be attached to the other piece. Filling requires rotation, and we should notice that (2) cannot be rotated at all when they're attached; (1) can. We should therefore fill (2) first. (I won't describe that here.) Filling (1) subsequently is a matter of recognizing that all three holes are the same. Placing one of the islands correctly in the middle (surrounded on three sides by (2)) will not restrict (1)'s freedom to rotate. Do this before bringing (1) over. Then face (1)'s back (its 15-edge) to the center. Fitting the other two islands should then be easy. -------------------- LEVEL Y, "Revenge of Jigsaw": (Warning! Read this at your own risk!) 1= @ 2= @@ 3= @@ 4= @@@ 5= @ @ 6= @ 7= @ 8= @ 9= @@ @@@ @@@ @@@ @@@@ @@@ @@@@@ @@@ @@@ @@@@ @@@ @@ @@@ @@ @@@ @@ @@ @@@@ @ @ @ @ @ @@@@ @@@ @@@@ This is essentially a small jigsaw puzzle. If there were a thousand pieces instead of nine, how would you go about solving it? With a square end result in mind, you'd probably find the corner pieces and edge pieces and isolate them from the "center" pieces (i.e., the pieces which are to be completely surrounded by others.) Among these nine pieces there is a well-defined number of corner pieces: four. Let's find them first. It can be shown (write me if you want to know precisely how) that only pieces (2), (3), (4), and (7) can be corners. All of the remaining pieces might be edge pieces-- one or some of them might instead be in the center. The position (upper left, upper right, etc.) we choose for the first corner is arbitrary-- the corners can be rotated to assume any of the four positions. Piece (3) is peculiar among the four corners in that it has an edge length that exceeds all edge lengths on all other corners-- 4. Its other edge length is 2. The completed square is to be 9 by 9; both sides upon which (3) is incident must total 9 in length. There are edge pieces available with lengths 2, 3, and 4, and there are corner pieces available (other than (3)) with lengths 2 and 3. For the edge of (3) with length 2, it is clear that it must be next to an edge of length 4, because anything less would not allow for a total side with length as high as 9. (2+3+3<9, 2+2+3<9, 2+2+2<9). Both pieces (1) and (8) have edges of length 4 that can be placed next to the 2-edge of (3), but only (8) in turn allows a corner with edge length 3 ((4) or (7)) to complete the edge. So (8) goes next to the 2-edge corner (3). The corner that goes next to (8) to complete the edge must be (7), because putting (4) there leaves a 1x2 void that cannot be filled by any other piece. Now let's return to the _other_ side of the corner (3)-- that is, the 4-edge side. Adjacent to this must be a 3-edge-- the only 2-edge piece that fits (9) does not allow for a corner piece (there are two remaining at this point, and we know that they have to be (2) and (4)) to complete the 9-side. The only piece that works is (5). The corner (2) then completes the side. The placement of the final corner (4) is now obvious. Pieces (1) and (6) fit interchangably between the corners (2,4) and (7,4). But, as it turns out, piece 9 is the sole center piece, and it can only reside there if (1) is placed between (7) and (4), and (6) between (2) and (4). Now that the exact destinations of the pieces are known, the matter of getting them there is not that hard. Rotation might be necessary on whatever you choose to be the last piece placement. Only one of the pieces can rotate nicely into its final position. Be careful! If you start with (3) in the lower left, the completed arrangement will look roughly like: 7 1 4 8 9 6 3 5 2 As the startup configuration might have led you to believe, the solution is a 3 by 3 arrangement of the pieces. How are you supposed to figure this all out in 8 minutes? You're not. I would have felt bad had I left it out, though. ------------------- LEVEL Z, "Nightmare": Just kidding! This level's worth too much for me to reveal the answer. Clue: start with the "what must go where" strategy. It's actually not nearly as mightmarish as Y. Write me if you want the systematic solution. 12-Jan-1992