The pattern of colors, using letters this time and the same coding scheme for the whole table, is:
abacda
bbaabc
ca aac
bcbccc
a aac
accab
The blank spaces could be either d or a fifth color, e.
The sequence of columns from the original table of columns is 352614.
There are 13 a's, 7 b's, 11 c's, 1 d and 4 blanks. Even if the blanks are all d's, there are more a's than any other letter, and so a represents green and there were 13 green squares on the painting.
DECLARE SUB permute (a$)
DATA 1,2,1,3,4,1
DATA 1,1,2,2,1,3
DATA 1,2,3,2,2,1
DATA 1,2,1,2,2,2
DATA 1,2,2,1,1,3
DATA 1,2,2,1,3,4
DATA 1,2,3,4,3,2
DATA 1,1,2,3,2,4
DATA 1,2,3,2,1,1
DATA 1,2,2,2,2,3
DATA 1,1,2,3,4,3
DATA 1,2,2,1,2,2
DIM SHARED colList$
DIM SHARED byRow(6, 6)
DIM SHARED byCol(6, 6)
DIM SHARED currRow, currCol
colList$ = "123456"
FOR r = 1 TO 6
FOR c = 1 TO 6
READ byRow(r, c)
NEXT
NEXT
FOR c = 1 TO 6
FOR r = 1 TO 6
READ byCol(r, c)
NEXT
NEXT
rc = 1: cc = 1
h$ = colList$
DO
currRow = 1: currCol = 1: nextLet = 1
REDIM SHARED board$(6, 6)
FOR c = 1 TO 6
board$(1, c) = MID$("abacda", c, 1)
NEXT
good = 1
didSome = 1
DO
didSome2 = didSome
didSome = 0
FOR c = 1 TO 6
cType = VAL(MID$(colList$, c, 1))
FOR r = 1 TO 6
IF board$(r, c) > "" THEN
FOR r2 = 1 TO 6
IF byCol(r2, cType) = byCol(r, cType) AND r2 <> r THEN
IF board$(r2, c) > "" AND board$(r2, c) <> board$(r, c) THEN
good = 0: EXIT FOR
END IF
IF board$(r2, c) = "" THEN didSome = 1
board$(r2, c) = board$(r, c)
END IF
IF byCol(r2, cType) <> byCol(r, cType) AND r2 <> r THEN
IF board$(r2, c) > "" AND board$(r2, c) = board$(r, c) THEN
good = 0: EXIT FOR
END IF
END IF
NEXT
IF good = 0 THEN EXIT FOR
END IF
NEXT
IF good = 0 THEN EXIT FOR
NEXT
IF good = 0 THEN EXIT DO
FOR r = 1 TO 6
FOR c = 1 TO 6
IF board$(r, c) > "" THEN
FOR c2 = 1 TO 6
IF byRow(r, c2) = byRow(r, c) AND c2 <> c THEN
IF board$(r, c2) > "" AND board$(r, c2) <> board$(r, c) THEN
good = 0: EXIT FOR
END IF
IF board$(r, c2) = "" THEN didSome = 1
board$(r, c2) = board$(r, c)
END IF
IF byRow(r, c2) <> byRow(r, c) AND c2 <> c THEN
IF board$(r, c2) > "" AND board$(r, c2) = board$(r, c) THEN
good = 0: EXIT FOR
END IF
END IF
NEXT
IF good = 0 THEN EXIT FOR
END IF
NEXT
IF good = 0 THEN EXIT FOR
NEXT
IF good = 0 THEN EXIT DO
LOOP UNTIL didSome2 = 0
IF good THEN
REDIM lct(5)
FOR r = 1 TO 6
FOR c = 1 TO 6
board$(r, c) = LEFT$(board$(r, c) + " ", 1)
PRINT board$(r, c);
s = INSTR(" abcd", board$(r, c)) - 1
lct(s) = lct(s) + 1
NEXT
PRINT
NEXT
ct = ct + 1
FOR i = 0 TO 4: PRINT lct(i); : NEXT
PRINT "------"; ct; colList$
END IF
permute colList$
LOOP UNTIL colList$ = h$
The permute subroutine is shown elsewhere on this site.
The puzzle is a rewording of Enigma 1375: Patterns of Colours; New Scientist, 21 January 2006.
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