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Devilish Cubes (Posted on 2009-07-27) Difficulty: 3 of 5
I once was presented with a set of 4 cubes each having a different net configuration.

The object was to place the cubes in a line such that no face within any of the four rectangles so formed contained the same colour within that rectangle.
Then too, no colour was orthogonally adjacent to a face of the same colour in an adjoining rectangle.
Lastly the end faces of the composite prism were both different to each other and not repeated within the rectangular faces.

A B C D
4 2 1 6
3 3 3 3
5 2 3 1 4 6 4 2 2 2 4 4
4 5 5 5
j k l m
A A A A A A
       
       
       
Caution: This is not an on-line interactive presentation.

The table above shows a set of cubic nets. Replace j, k, l & m with an appropriate label A, B, C or D to indicate the net used.

Additionally add just one digit to each cell in compliance with those nets so that each row along with the end values contains unique values.

See The Solution Submitted by brianjn    
Rating: 2.0000 (1 votes)

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Solution computer solution | Comment 1 of 2

There seem to be 24 basic solutions, barring trivial variations.

In the below table of the 24, cube A is always shown in the second column (column k), but could be exchanged with the cube to its right (column l) as the left and right faces are hidden. Also the visible faces of the entire set could be rotated 1, 2 or 3 positions. Also, the entire structure could be flipped left and right, as cube A is always presented in the sequence 2543, rather than the 3452 it would take upon this flipping. Thus each solution presented would have 4*2*2-1 = 15 trivial variations.

 BADC    DACB    DABC    BADC    DACB    DABC    CABD    CABD    BADC    DACB
642531  652341  652431  642351  632541  632451  132456  152436  632451  642531
 3542    4523    4532    3524    2543    2534    4532    2534    2534    3542
 2435    3452    3425    2453    5432    5423    5423    3425    5423    2435
 5324    2345    2354    5342    4325    4352    2354    4352    4352    5324
 
 DABC    BADC    DACB    DABC    BADC    DACB    DABC    BADC    DACB    DABC
642351  632541  652431  652341  652431  642351  642531  652341  632451  632541
 3524    2543    4532    4523    4532    3524    3542    4523    2534    2543
 2453    5432    3425    3452    3425    2453    2435    3452    5423    5432
 5342    4325    2354    2345    2354    5342    5324    2345    4352    4325
 
 BACD    CADB    BACD    CADB
152436  132456  132456  152436
 2534    4532    4532    2534
 3425    5423    5423    3425
 4352    2354    2354    4352
 
 


 The program assumes that 6 and 1 are the end colors as these are not present on cube A.
 
 DECLARE FUNCTION orient$ (cube!, orientation!)
 DATA 233445,436251,232154,434652
 
 DIM SHARED c$(4)
 
 CLS
 
 FOR cube = 1 TO 4
  READ c$(cube)
 NEXT
 
 FOR o1 = 0 TO 11 STEP 4 ' reduced to eliminate trivial variations
  cb$(1) = orient$(1, o1)
 FOR o2 = 0 TO 23
  cb$(2) = orient$(2, o2)
 FOR o3 = 0 TO 23
  cb$(3) = orient$(3, o3)
 FOR o4 = 0 TO 23
  cb$(4) = orient$(4, o4)
 
 good = 1
 REDIM col(6): colUsed = 0
 FOR i = 1 TO 4
  s$(i) = ""
  FOR j = 1 TO 4
   IF INSTR(s$(i), MID$(cb$(j), i, 1)) > 0 THEN good = 0: EXIT FOR
   s$(i) = s$(i) + MID$(cb$(j), i, 1)
   IF i > 1 THEN
     IF RIGHT$(s$(i), 1) = MID$(s$(i - 1), j, 1) THEN good = 0: EXIT FOR
     IF i = 4 THEN
       IF RIGHT$(s$(i), 1) = MID$(s$(1), j, 1) THEN good = 0: EXIT FOR
     END IF
   END IF
   cNum = VAL(MID$(cb$(j), i, 1))
   IF col(cNum) = 0 THEN
     col(cNum) = 1
     colUsed = colUsed + 1
     IF colUsed > 4 THEN good = 0: EXIT FOR
   END IF
  NEXT
  IF good = 0 THEN EXIT FOR
 NEXT
 IF good THEN
   FOR e1 = 2 TO 4
   FOR e2 = 2 TO 4
    IF e1 <> e2 THEN
     IF MID$(cb$(e1), 5, 1) = "6" AND MID$(cb$(e2), 6, 1) = "1" OR MID$(cb$(e1), 5, 1) = "1" AND MID$(cb$(e2), 6, 1) = "6" THEN
       n2 = 9 - e1 - e2
       c0 = (solCt MOD 10) * 8 + 1
       r0 = (solCt \ 10) * 8 + 1
       LOCATE r0, c0 + 1
       PRINT MID$("ABCD", e1, 1) + MID$("ABCD", 1, 1) + MID$("ABCD", n2, 1) + MID$("ABCD", e2, 1);
       LOCATE r0 + 1, c0
       PRINT MID$(cb$(e1), 5, 1) + MID$(cb$(e1), 1, 1) + MID$(cb$(1), 1, 1) + MID$(cb$(n2), 1, 1) + MID$(cb$(e2), 1, 1) + MID$(cb$(e2), 6, 1);
       LOCATE r0 + 2, c0 + 1
       PRINT MID$(cb$(e1), 2, 1) + MID$(cb$(1), 2, 1) + MID$(cb$(n2), 2, 1) + MID$(cb$(e2), 2, 1);
       LOCATE r0 + 3, c0 + 1
       PRINT MID$(cb$(e1), 3, 1) + MID$(cb$(1), 3, 1) + MID$(cb$(n2), 3, 1) + MID$(cb$(e2), 3, 1);
       LOCATE r0 + 4, c0 + 1
       PRINT MID$(cb$(e1), 4, 1) + MID$(cb$(1), 4, 1) + MID$(cb$(n2), 4, 1) + MID$(cb$(e2), 4, 1);
       solCt = solCt + 1
     END IF
    END IF
   NEXT
   NEXT
 
 END IF
 
 
 NEXT
 NEXT
 NEXT
 NEXT
 
 FUNCTION orient$ (cube, orientation)
  cu$ = c$(cube)
  e1 = orientation \ 4 + 1
  e2 = orientation \ 4 + 4: IF e2 > 6 THEN e2 = e2 - 6
  SELECT CASE e1
   CASE 1
    seq$ = "2356"
   CASE 2
    seq$ = "1643"
   CASE 3
    seq$ = "1245"
   CASE 4
    seq$ = "2653"
   CASE 5
    seq$ = "1346"
   CASE 6
    seq$ = "1542"
  END SELECT
  rot = orientation MOD 4
  s$ = MID$(seq$, rot + 1) + LEFT$(seq$, rot)
  o$ = ""
  FOR i = 1 TO 4
    o$ = o$ + MID$(cu$, VAL(MID$(s$, i, 1)), 1)
  NEXT
  o$ = o$ + MID$(cu$, e1, 1) + MID$(cu$, e2, 1)
  orient$ = o$
 END FUNCTION
 
Note the orientations for cube 1 (cube A) are limited to every fourth one to eliminate rotations from the basic orientation and to under 12  so that only the faces with the 2 or one of the 3's can face the left end cap.


  Posted by Charlie on 2009-07-27 16:28:53
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