Find Star of David 26 Solutions

Star6.c

// Given the Star of David, arrange the numbers 1-12 (there are 12 intersections) so that
// each line (there are 6 lines and 4 intersections per line) adds up to 26.
// 
// Intersections on array are numbered TDLR, from a0 to a11.

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

// prototypes
int compute(int*);
void display(int *);


#define NVALUES 12

int compute(int *a)
{
    return  (a[0]  + a[3]  + a[6]  + a[10] == 26) &&
    (a[0]  + a[2]  + a[5]  + a[7]  == 26) &&
    (a[1]  + a[2]  + a[3]  + a[4]  == 26) &&
    (a[7]  + a[8]  + a[9]  + a[10] == 26) &&
    (a[1]  + a[5]  + a[8]  + a[11] == 26) &&
    (a[11] + a[9]  + a[6]  + a[4]  == 26);
}


void display(int *a)
{
    printf("[ ");
    for (int i = 0; i < NVALUES; i++) {
        printf("%d%c", a[i], (i < NVALUES-1 ? ',' : ' '));
    }
    printf("]\n");
}

int main(int argc, char *argv[]) 
{
    // seed random numbers
    srand48(time(0));
    
    int a[NVALUES], i, count, solutions;
    
    // initialize array with values 1-12
    for(i = 0; i < NVALUES; i++) {
        a[i] = i + 1;
    }
    
    count = 0;
    solutions = 0;
    
    while (solutions < 10) {
        count++;        
        
        // shuffle
        for(i = 0; i < NVALUES-1; i++) {
            int r = (int)(drand48() * (NVALUES-i));
            int t = a[i]; a[i] = a[i+r]; a[i+r] = t;
        }
        
        if (compute(a)) {
            printf("Took %d iterations to find ", count);
            display(a);
            count = 0;
            solutions++;
        }
    }
}