The C code I wrote to calculate it is below. It finished nearly instantly. I was trying to determine how to write the rest of the code (could it fit in memory).
#include <stdio.h>
// Program to count how many 8 character combinations are necessary fo
+r brute force
// Constants to remove magic numbers
#define Z 25
int main (void) {
static const short int max[26] = {4, 3, 3, 4, 4, 4, 4, 3, 3, 2, 3,
+ 4, 3, 4, 4, 3, 1, 4, 5, 3, 4, 2, 3, 2, 3, 4}; // from STDERR of filt
+er.pl
short int have[26] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
+ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
int combo = 0;
int i, j, k, l, m, n, o, p;
for (i = 0; i <= Z; i++) {
if (++have[i] > max[i]) {
--have[i];
continue;
}
for (j = i; j <= Z; j++) {
if (++have[j] > max[j]) {
--have[j];
continue;
}
for (k = j; k <= Z; k++) {
if (++have[k] > max[k]) {
--have[k];
continue;
}
for (l = k; l <= Z; l++) {
if (++have[l] > max[l]) {
--have[l];
continue;
}
for (m = l; m <= Z; m++) {
if (++have[m] > max[m]) {
--have[m];
continue;
}
for (n = m; n <= Z; n++) {
if (++have[n] > max[n]) {
--have[n];
continue;
}
for (o = n; o <= Z; o++) {
if (++have[o] > max[o]) {
--have[o];
continue;
}
for (p = o; p <= Z; p++) {
if (++have[p] > max[p]) {
--have[p];
continue;
}
++combo;
--have[p];
}
--have[o];
}
--have[n];
}
--have[m];
}
--have[l];
}
--have[k];
}
--have[j];
}
--have[i];
}
printf("%i\n", combo);
return 0;
}
It also ended up being part of my final solution since I realized it was unnecessary to keep them in memory using a high water mark algorithm.
