In the following octahedral net the vertices A, B, C, D, E and F, and are to be assigned distinctive values from 1 through 9.


M, N, O, P, Q, R, S and T are the values of the respective sums of the three vertices which form the respective surfaces to which each is assigned.

M + N = O + P = Q + R = S + T

Find unique sets of values¹ for A, B, C, D, E and F such that the values of M through T (but not necessarily in that order) form series which increment by 2 of which there are 7.

Background statistics:
                                   Unique        Actual
                M+N=   Interval     Sets2       Solutions
                 21       1           1            96
                 23       1           2            96
                 25       1           2            96
                 26       2           2            96
                 27       1           5           480
                 29       1           4           288 
                 30       2           3           144
                 31       1           8           288
                 33       1           6           480
                 34       2           2            96
                 35       1           2            96
                 37       1           2            96
                 39       1           1            96
                                     40          2448 

Note:
1. For M+N=21,
          1, 2, 3, 6, 4, 5
          1, 2, 3, 6, 5, 4
      and 1, 2, 5, 3, 4, 6  
   are the first 3 of 96 solutions.  The 96 values do not
   discriminate amongst vertex rotation or reflections. 
2. Each unique set configures the octahedron in more ways than one.

The pairs of rows below contain the same set of integers within the pair, and there are indeed 7 pairs. The reversal of E and F changes only the handedness of the octahedron as E and F are opposite vertices.

 A  B  C  D  E  F            M   N   O   P   Q   R  S   T   
 1  4  9  2  3  7            18   8  20  6  12  14  10  16
 1  4  9  2  7  3            14  12  16  10  8  18   6  20
 1  6  9  2  3  5            16  10  20  6  12  14   8  18
 1  6  9  2  5  3            14  12  18  8  10  16   6  20
 1  6  9  2  5  7            18  12  22  8  14  16  10  20
 1  6  9  2  7  5            16  14  20  10 12  18   8  22
 1  5  9  3  4  8            20  10  22  8  14  16  12  18
 1  5  9  3  8  4            16  14  18  12  10  20  8  22
 1  7  9  3  4  6            18  12  22  8  14  16  10  20
 1  7  9  3  6  4            16  14  20  10  12 18   8  22
 1  7  9  3  6  8            20  14  24  10  16 18  12  22
 1  7  9  3  8  6            18  16  22  12  14 20  10  24
 1  8  9  4  5  7            20  14  24  10  16 18  12  22
 1  8  9  4  7  5            18  16  22  12  14 20  10  24

The program as initially written contains a bug that actually had no effect. It assumes vertex A has the digit 1, but in actual fact, there's no initial assurance that a 1 is part of every solution:

DECLARE SUB permute (a$)
CLS
a = 1
FOR c = 2 TO 9
 used(c) = 1
 FOR d = 2 TO 6
   IF used(d) = 0 THEN
     used(d) = 1
 FOR x = d + 1 TO 9
   IF used(x) = 0 THEN
     used(x) = 1
     s1$ = LTRIM$(STR$(x))
 FOR y = x + 1 TO 9
   IF used(y) = 0 THEN
     used(y) = 1
     s2$ = s1$ + LTRIM$(STR$(y))
 FOR z = y + 1 TO 9
   IF used(z) = 0 THEN
     used(z) = 1
     s3$ = s2$ + LTRIM$(STR$(z))

     h$ = s3$
     DO
       b = VAL(MID$(s3$, 1, 1))
       e = VAL(MID$(s3$, 2, 1))
       f = VAL(MID$(s3$, 3, 1))
       m = c + d + f: n = a + b + e
       o = c + f + b: p = a + d + e
       q = a + b + f: r = e + d + c
       s = f + d + a: t = b + c + e
       v(1) = m
       v(2) = n
       v(3) = o
       v(4) = p
       v(5) = q
       v(6) = r
       v(7) = s
       v(8) = t
       DO
        done = 1
        FOR i = 1 TO 7
         IF v(i) > v(i + 1) THEN SWAP v(i), v(i + 1): done = 0
        NEXT
       LOOP UNTIL done
       FOR i = 1 TO 7
         IF v(i + 1) - v(i) <> 2 THEN done = 0: EXIT FOR
       NEXT
       IF done THEN PRINT a; b; c; d; e; f, m; n; o; p; q; r; s; t

       permute s3$
     LOOP UNTIL h$ = s3$

     used(z) = 0
   END IF
 NEXT
     used(y) = 0
   END IF
 NEXT
     used(x) = 0
   END IF
 NEXT
     used(d) = 0
   END IF
 NEXT d
 used(c) = 0
NEXT c

END

A modified version, varying A from 2 to 4, without allowing for any 1's, verifies there is no solution without a 1.

Posted by Charlie on 2008-12-30 17:45:22