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Semi-Minimalist Painting (Posted on 2006-03-13) Difficulty: 4 of 5
A semi-minimalist painter created a work which consisted of a 6 x 6 array of small colored squares. Each small square contained just one color.

At the art gallery, six girl students were examining the painting. Each girl chose to report on exactly one horizontal row of small squares, by assigning a different number to each color in that row. The six row patterns, in the original order, were

121341
112213
123221
121222
122113
122134
The girls did not consult one another, so a given digit in one row does not necessarily represent the same color as the same digit in a different row.

Another group of six girls did the same process, but this time for the columns, rather than the rows. The column patterns they came up with look like this (but the array below shows the columns in no particular order):

1  1  1  1  1  1
2  1  2  2  1  2
3  2  3  2  2  2
4  3  2  2  3  1
3  2  1  2  4  2
2  4  1  3  3  2
Remember: the rows in the first table are shown in the correct order, but the columns in the second table are shown randomly. Outside of the particular row, for the first table, or column for the second table, do not expect the same digit-to-color coding scheme.

There were more green squares than any other color. How many squares were painted green?

  Submitted by Charlie    
Rating: 3.2000 (5 votes)
Solution: (Hide)
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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
please explain solution processCynthia2006-06-29 00:47:36
re: Where is this from?Charlie2006-03-17 07:56:25
QuestionWhere is this from?Tristan2006-03-14 20:46:01
re(3): Solutiontomarken2006-03-14 14:59:17
re(2): Solutiontomarken2006-03-14 14:45:52
re: SolutionMagda2006-03-14 14:16:27
re: Solutiontomarken2006-03-14 12:01:23
re: SolutionSoumitra Pal2006-03-14 11:58:14
SolutionSoumitra Pal2006-03-14 11:53:58
SolutionSolutiontomarken2006-03-13 16:14:45
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