/* Given a set of consecutive characters, find a hash expression that
   would produce the desired symbols.

   Result:
   ('A' ^ 91) * 22 % 54 = ' '
   ('B' ^ 91) * 22 % 54 = '\n'
   ('C' ^ 91) * 22 % 54 = '*'
   ('D' ^ 91) * 22 % 54 = '"'
*/

#include<stdio.h>

/* List of consecutive unused characters from prime_cut2.c */
static const char *kKeySets[3] =
{
   "5678",
   "ABCD",
   "wxyz",
};

/* Hash function. */
static int Hash(char input, int p, int q, int r)
{
   return (input ^ r) * p % q;
}

/* Try hashing a set of keys to the desired symbols. */
static void TryKeySet(const char *keys)
{
   int p, q, r, v, coverage;
   const char *k;

   for(p = 0; p < 1000; p++)
   {
      for(q = 1; q < 1000; q++)
      {
         for(r = 0; r < 256; r++)
         {
            coverage = 0;
            for(k = keys; *k; k++)
            {
               v = Hash(*k, p, q, r);
               if( v == ' ' ) coverage |= 1;
               if( v == '\n' ) coverage |= 2;
               if( v == '"' ) coverage |= 4;
               if( v == '*' ) coverage |= 8;
            }
            if( coverage == 15 )
            {
               printf("keys = %s, p = %d, q = %d, r = %d:", keys, p, q, r);
               for(k = keys; *k; k++)
                  printf("\t%d->%d", *k, Hash(*k, p, q, r));
               putchar('\n');
            }
         }
      }
   }
}

int main(void)
{
   int k;

   for(k = 0; k < 3; k++)
      TryKeySet(kKeySets[k]);
   return 0;
}