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 Octo Vertices (Posted on 2008-12-30)
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 No Rating

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 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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