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Octo Vertices (Posted on 2008-12-30) Difficulty: 3 of 5
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.

See The Solution Submitted by brianjn    
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Solution computer solution | Comment 2 of 3 |

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
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