Only seven positions need to be searched over with some pruning.
The solution appears to be unique subject to reflections and rotations.
- Pauli
Code:
#include <stdio.h>
/*
a b c
d e f g
h i j k l
m n o p
q r s
*/
int main() {
unsigned char usedbuf[120], *used;
for (int i=0; i<120; i++)
usedbuf[i] = 1;
used = usedbuf + 60;
for (int i=1; i<20; i++) used[i] = 0;
for (int a=1; a<20; a++) {
used[a] = 1;
for (int b=1; b<20; b++) {
if (used[b]) continue;
int c = 38 - a - b;
if (used[c] || c == b) continue;
used[b] = used[c] = 1;
for (int d=1; d<20; d++) {
if (used[d]) continue;
int h = 38 - a - d;
if (used[h] || d == h) continue;
used[d] = used[h] = 1;
for (int g=1; g<20; g++) {
if (used[g]) continue;
int l = 38 - c - g;
if (used[l] || l == g) continue;
used[l] = used[g] = 1;
for (int m=1; m<20; m++) {
if (used[m]) continue;
int q = 38 - m - h;
if (used[q] || m == q) continue;
used[m] = used[q] = 1;
for (int p=1; p<20; p++) {
if (used[p]) continue;
int s = 38 - p - l;
if (used[s] || p == s) continue;
int r = 38 - s - q;
if (used[r] || r == s || r == p) continue;
used[p] = used[s] = used[r] = 1;
for (int e=1; e<20; e++) {
if (used[e]) continue;
int f = 38 - d - e - g;
if (used[f] || e == f) continue;
int i = 38 - b - e - m;
if (used[i] || i == f || i == e) continue;
int k = 38 - b - f - p;
if (used[k] || k == i || k == f || k == e) continue;
used[e] = used[f] = used[i] = used[k] = 1;
int o = 38 - r - k - g;
if (used[o]) goto ex;
int n = 38 - i - d - r;
if (used[n] || n==o) goto ex;
int j = 38 - a - e - o - s;
if (used[j] || j==o || j==n) goto ex;
printf(" %3d %3d %3d\n", a, b, c);
printf(" %3d %3d %3d %3d\n", d, e, f, g);
printf("%3d %3d %3d %3d %3d\n", h, i, j, k, l);
printf(" %3d %3d %3d %3d\n", m, n, o, p);
printf(" %3d %3d %3d\n\n", q, r, s);
ex: used[e] = used[f] = used[i] = used[k] = 0;
}
used[p] = used[s] = used[r] = 0;
}
used[m] = used[q] = 0;
}
used[l] = used[g] = 0;
}
used[d] = used[h] = 0;
}
used[b] = used[c] = 0;
}
used[a] = 0;
}
}
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