Given eight identical standard dice (opposite sides sum to 7: 1-6;2-5 and 3-4).
(a)Construct a 2 cube, such that the sum of the pips on each side is the same.
(b) How many distinct solutions (rotations excluded) are there?
(c) Are there several possible target sums or just the one you have found?
I'm assuming that standard dice have a standard orientation: not just that 1 is opposite 6, etc., but also that, when looked at from a corner direction where 1, 2 and 3 are all visible, one of these two standards is used: 1, 2, 3 appear counterclockwise or 1, 2, 3 clockwise, consistently among the dice. The former, according to Wikipedia, is standard in the west, and the latter in Chinese dice. The sample diagrams below, use the western standard.
DIM cube(6, 8)
DATA 2453,3641,1562,1265,1463,2354
FOR i = 1 TO 6: READ neigh$(i): NEXT
FOR a = 1 TO 6
FOR b = 1 TO 6
FOR c = 1 TO 6
FOR d = 1 TO 6: PRINT a; b; c; d
cube(3, 3) = a
cube(3, 4) = b
cube(4, 4) = c
cube(4, 3) = d
facetot = a + b + c + d
FOR arot = 0 TO 3
cube(3, 2) = VAL(MID$(neigh$(a), arot + 1, 1))
cube(2, 3) = VAL(MID$(neigh$(a), (arot + 1) MOD 4 + 1, 1))
FOR brot = 0 TO 3
cube(2, 4) = VAL(MID$(neigh$(b), brot + 1, 1))
cube(3, 5) = VAL(MID$(neigh$(b), (brot + 1) MOD 4 + 1, 1))
FOR crot = 0 TO 3
cube(4, 5) = VAL(MID$(neigh$(c), crot + 1, 1))
cube(5, 4) = VAL(MID$(neigh$(c), (crot + 1) MOD 4 + 1, 1))
FOR drot = 0 TO 3
cube(5, 3) = VAL(MID$(neigh$(d), drot + 1, 1))
cube(4, 2) = VAL(MID$(neigh$(d), (drot + 1) MOD 4 + 1, 1))
FOR ap = 1 TO 6
FOR aprot = 0 TO 3
cube(3, 8) = ap
cube(1, 3) = VAL(MID$(neigh$(ap), aprot + 1, 1))
cube(3, 1) = VAL(MID$(neigh$(ap), (aprot + 1) MOD 4 + 1, 1))
cube(4, 1) = facetot - cube(3, 1) - cube(3, 2) - cube(4, 2)
cube(1, 4) = facetot - cube(1, 3) - cube(2, 3) - cube(2, 4)
IF cube(4, 1) > 0 AND cube(4, 1) < 7 THEN
IF cube(1, 4) > 0 AND cube(1, 4) < 7 THEN
FOR dp = 1 TO 6
FOR dprot = 0 TO 3
cube(4, 8) = dp
' cube(4, 1) = VAL(MID$(neigh$(dp), dprot + 1, 1))
cube(6, 3) = VAL(MID$(neigh$(dp), (dprot + 1) MOD 4 + 1, 1))
' cube(4, 1) = facetot - cube(3, 1) - cube(3, 2) - cube(4, 2)
cube(6, 4) = facetot - cube(5, 3) - cube(5, 4) - cube(6, 3)
IF cube(4, 1) = VAL(MID$(neigh$(dp), dprot + 1, 1)) THEN
IF cube(6, 4) > 0 AND cube(6, 4) < 7 THEN
bp = cube(1, 4)
cp = cube(6, 4)
FOR bprot = 0 TO 4
cube(3, 7) = VAL(MID$(neigh$(bp), bprot + 1, 1))
cube(3, 6) = VAL(MID$(neigh$(bp), (bprot + 1) MOD 4 + 1, 1))
IF cube(3, 7) > 0 AND cube(3, 7) < 7 THEN
IF cube(3, 6) > 0 AND cube(3, 6) < 7 THEN
FOR cprot = 0 TO 4
cube(4, 6) = VAL(MID$(neigh$(cp), cprot + 1, 1))
cube(4, 7) = VAL(MID$(neigh$(cp), (cprot + 1) MOD 4 + 1, 1))
IF cube(4, 7) > 0 AND cube(4, 7) < 7 THEN
IF cube(4, 6) > 0 AND cube(4, 6) < 7 THEN
IF cube(3, 5) + cube(3, 6) + cube(4, 5) + cube(4, 6) = facetot THEN
IF cube(3, 7) + cube(3, 8) + cube(4, 7) + cube(4, 8) = facetot THEN
GOSUB reportIt
END IF
END IF
END IF
END IF
NEXT
END IF
END IF
NEXT
END IF
END IF
NEXT
NEXT
END IF
END IF
NEXT
NEXT
NEXT
NEXT
NEXT
NEXT
NEXT
NEXT
NEXT
NEXT
PRINT ct
END
reportIt:
ct = ct + 1
IF INKEY$ > "" THEN
FOR row = 1 TO 6
FOR col = 1 TO 8
IF col > 2 AND col < 5 OR row > 2 AND row < 5 THEN
PRINT cube(row, col);
ELSE
PRINT " "; :
END IF
NEXT
PRINT
NEXT
PRINT
END IF
RETURN
finds 901728 raw solutions.
If there are no symmetric solutions, this represents 24 times the distinct solutions (24 rotations = 6 choices for top * 4 rotations about the vertical axis). And 901728 is a multiple of 24 so the answer to part b could be 901728/24 = 37572, but that divisibility might be a chance effect, meaning there really are more than just 37572 distinct solutions as the symmetric solutions need to be divided by a smaller number as they appear less than 24 times each.
In fact, a modified version of the program, which prints out to a file the first occurrence of each face total, apparently shows such symmetries:
3 1
2 3
2 3 1 1 2 2 3 1
1 3 6 1 3 2 1 4
2 2
2 3
9
1 3
4 2
3 2 1 1 4 1 2 2
2 3 6 2 4 1 5 1
2 1
4 3
10
1 2
2 3
3 3 1 1 2 3 1 2
1 1 3 3 2 1 3 2
2 1
3 2
8
4 1
4 2
5 2 1 1 4 2 3 1
1 3 6 3 1 4 5 2
2 5
3 1
11
2 4
4 2
4 2 1 1 4 1 5 1
3 3 6 4 1 6 4 2
2 2
6 2
12
2 5
4 2
4 2 1 1 4 3 6 1
4 3 6 5 1 5 4 2
2 4
1 6
13
2 6
4 2
4 2 1 1 4 2 4 1
5 3 6 6 2 6 3 6
2 3
4 5
14
3 5
4 3
6 2 1 2 6 3 6 2
4 3 6 6 2 4 6 1
2 3
5 5
15
6 5
4 1
4 2 1 3 5 3 6 2
5 5 6 6 2 6 5 3
3 3
6 4
16
6 6
4 1
5 2 1 4 2 4 5 4
5 5 6 6 5 6 5 3
3 4
6 4
17
5 5
4 4
6 2 1 5 6 6 4 3
5 5 6 6 3 3 5 6
3 5
4 6
18
5 4
5 5
6 4 1 6 4 5 6 3
5 4 6 6 5 5 4 6
5 4
4 6
19
4 5
5 6
6 6 4 4 5 6 4 5
4 4 6 6 5 4 6 5
5 4
6 5
20
As shown in the face totals, such totals can be anywhere from 8 through 20.
So part b remains unsolved.
For analysis, all 33 of the found solutions for a face total of 8 are shown here:
1 2
2 3
3 3 1 1 2 3 1 2
1 1 3 3 2 1 3 2
2 1
3 2
8
2 1
2 3
1 3 1 1 2 2 3 3
3 1 3 3 2 2 1 1
2 1
2 3
8
3 2
2 1
2 3 1 2 3 3 1 1
2 1 3 2 1 1 3 3
2 3
1 2
8
2 3
2 1
1 3 1 2 3 1 2 3
3 1 3 2 1 3 2 1
2 3
2 1
8
2 3
2 1
1 3 1 2 3 1 2 3
1 3 2 3 2 2 1 2
1 1
3 3
8
3 1
2 2
2 3 1 3 1 2 3 1
2 1 3 1 3 2 1 3
2 2
1 3
8
1 3
2 2
3 3 1 3 1 1 2 2
1 1 3 1 3 3 2 2
2 2
3 1
8
2 2
2 2
1 3 1 3 1 3 1 3
3 1 3 1 3 1 3 1
2 2
2 2
8
2 2
2 2
1 3 1 3 1 3 1 3
1 3 2 2 1 3 2 2
1 3
3 1
8
3 1
2 2
2 3 1 3 1 2 3 1
1 2 1 3 2 3 2 2
3 1
3 1
8
2 2
2 2
1 3 1 3 1 3 1 3
2 2 1 3 2 2 1 3
3 1
1 3
8
1 1
3 3
3 1 2 1 2 2 3 2
3 1 3 2 1 3 2 1
2 3
2 1
8
1 1
3 3
3 1 2 1 2 2 3 2
1 3 2 3 2 2 1 2
1 1
3 3
8
1 3
3 1
3 1 2 2 3 1 2 2
3 1 3 1 3 1 3 1
2 2
2 2
8
3 1
3 1
2 1 2 2 3 2 3 1
2 3 2 2 1 2 1 3
1 3
1 3
8
1 3
3 1
3 1 2 2 3 1 2 2
1 3 2 2 1 3 2 2
1 3
3 1
8
2 2
3 1
1 1 2 2 3 3 1 3
3 3 2 2 1 1 3 1
1 3
2 2
8
3 1
3 1
2 1 2 2 3 2 3 1
3 2 1 3 2 1 3 1
3 1
2 2
8
1 3
3 1
3 1 2 2 3 1 2 2
2 2 1 3 2 2 1 3
3 1
1 3
8
1 2
3 2
3 1 2 3 1 3 1 2
1 3 2 1 3 1 3 2
1 2
3 2
8
2 1
3 2
1 1 2 3 1 2 3 3
3 3 2 1 3 2 1 1
1 2
2 3
8
1 2
3 2
3 1 2 3 1 3 1 2
2 2 1 2 1 3 2 3
3 3
1 1
8
3 1
1 3
2 2 3 1 2 2 3 1
3 1 3 1 3 1 3 1
2 2
2 2
8
1 3
1 3
3 2 3 1 2 1 2 2
2 1 3 1 3 2 1 3
2 2
1 3
8
3 1
1 3
2 2 3 1 2 2 3 1
1 3 2 2 1 3 2 2
1 3
3 1
8
2 2
1 3
1 2 3 1 2 3 1 3
2 3 2 2 1 2 1 3
1 3
1 3
8
3 1
1 3
2 2 3 1 2 2 3 1
2 2 1 3 2 2 1 3
3 1
1 3
8
1 3
1 3
3 2 3 1 2 1 2 2
1 2 1 3 2 3 2 2
3 1
3 1
8
2 2
1 3
1 2 3 1 2 3 1 3
3 2 1 3 2 1 3 1
3 1
2 2
8
3 3
1 1
2 2 3 2 3 1 2 1
1 3 2 1 3 1 3 2
1 2
3 2
8
3 3
1 1
2 2 3 2 3 1 2 1
2 2 1 2 1 3 2 3
3 3
1 1
8
3 2
1 2
2 2 3 3 1 3 1 1
2 2 1 1 3 1 3 3
3 2
1 2
8
2 3
1 2
1 2 3 3 1 1 2 3
3 2 1 1 3 3 2 1
3 2
2 1
8
the total number of solutions, by face total, from 4 to 24 are:
0 0 0 0 33 648 6510 34920 79245 154464 350088 154464 79245 34920 6510 648 33 0 0 0 0
Note the symmetry as alternate solutions can be changed into one another by switching 1 with 6, 2 with 5, etc. so the 33 for face total 20 are just this process done on the 33 solutions for face total 8.
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Posted by Charlie
on 2013-11-18 21:28:31 |