===========================================================================
 BBS: The Abacus * HST/DS * Potterville, MI
Date: 03-07-93 (16:58)             Number: 19
From: EARL MONTGOMERY              Refer#: NONE
  To: RICH GELDREICH                Recvd: NO  
Subj: Screen 13 Compression          Conf: (35) Quick Basi
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' Rich, I finally found a copy of one of your Screen 13
' compression routines. It was dated 24 Dec, 1991. It
' uses the Huffman routine. After studying it I decided to
' try to out do you! <smile>. Well I did to a certain extent.
' If you put the same graphics in yours that I have in mine,
' yours decompresses on my 386 in 48 seconds. Mine in 32.
' Your compression was 18209 bytes. Mine was 15181 bytes.
' The big problem with mine is that if A$ gets close to
' 16000 bytes I get an OUT of String Space error. Shouldn't
' I be able to get close to 32767 with a variable string with
' a $ sign after it? I would appreciate any comments you might
' have.
DIM b$(10)
DEFLNG A-Z: SCREEN 13: COLOR 4
REM Graphic portion
LINE (80, 80)-(300, 150), RND * 255, BF
LINE (0, 0)-(200, 2), , BF
CIRCLE (100, 100), 100
PAINT (102, 102), 234, 4
LINE (0, 199)-(160, 199), 5
LINE (161, 199)-(319, 199), 4
FOR x = 0 TO 14
LINE (RND * 319, RND * 199)-(RND * 319, RND * 199), RND * 255
NEXT
DEF SEG = &HA000
REM Reading how many times a value is repeated and the value
REM Builing A$
FOR x = 0 TO 63999
p = PEEK(x)
IF PEEK(x + 1) = p THEN cnt = cnt + 1: tempcnt = cnt:  ELSE cnt = 1
IF cnt = 1 THEN a$ = a$ + STR$(tempcnt) + STR$(p): tempcnt = 1
NEXT
OPEN "compress.bin" FOR OUTPUT AS #1
PRINT #1, a$
CLOSE : CLS
OPEN "compress.bin" FOR INPUT AS #1
INPUT #1, a$
CLOSE : cnt = 0
REM This part is a little tricky. I am looking for a space in A$. Then
REM finding the value to the left of it which gives me the marker value
REM in the routine SetPix:
step1:
cnt = cnt + 1
b$ = LEFT$(a$, cnt)
IF RIGHT$(b$, 1) = " " THEN a = VAL(LEFT$(b$, cnt)): marker = marker + a: a$ = R
IGHT$(a$, LEN(a$) - cnt): cnt = 0: GOTO step2
GOTO step1
REM Looking for another space then finding the value to the left of it
REM which gives me the color value (B) which I use in setpix.
step2:
cnt = cnt + 1
IF LEN(a$) = 1 THEN b = VAL(a$): GOTO setpix
b$ = LEFT$(a$, cnt)
IF RIGHT$(b$, 1) = " " THEN b = VAL(LEFT$(b$, cnt)): a$ = RIGHT$(a$, LEN(a$) - c
nt): cnt = 0: GOTO setpix
GOTO step2
REM Marker1 is initially 0 but after the loop equals marker.
REM The next time around marker will increase in value.
REM The IF x=64000 then 999 routine is a little Brute Force but at the
REM time I couldn't figure a neat way to exit.
REM After the loop we start the whole process again until x=64000
setpix:
FOR x = marker1 TO marker
POKE x, b
IF x = 64000 THEN 999
NEXT
marker1 = marker: GOTO step1
999 GOTO 999
Earl

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