/*
  Author:  Pate Williams (c) 1997

  Exercise III.1.1
  "In certain computer bullentin-boards systems it is
  customary, if you want to post a message that may
  offend some people (e.g. a dirty joke), to encipher
  the letters (but not the blanks or punctuation) by
  a translation C = P + b mod 26. Using frequency
  analysis decrypt the following ciphertext:
  Jr frjrq n fzvyr ba n ubefr'f nff, naq n lrne
  yngre vg jnf ryrpgrq Cerfvqrag!" -Neal Koblitz-
  See "A Course in Number Theory and Cryptography"
  by Neal Koblitz second edition page 61.
*/

#include <stdio.h>
#include <string.h>

struct code {long alpha, count;};

int main(void)
{
  char cipher[3][25] = {"JRFRJRQNFZVYRBANUBEFRF",
                        "NFFNAQNLRNEYNGREVGJNFR",
                        "YRPGRQCERFVQRAG"};
  long b, count = 0, i, j, p;
  long ciphertext[128], frequency[26] = {0};
  struct code c[26], t;

  for (i = 0; i < 3; i++) {
    for (j = 0; j < strlen(cipher[i]); j++) {
      ciphertext[count] = cipher[i][j] - 'A';
      frequency[ciphertext[count++]]++;
    }
  }
  for (i = 0; i < 26; i++) {
    c[i].alpha = i;
    c[i].count = frequency[i];
  }
  /* sort the code array into descending order */
  for (i = 0; i < 25; i++)
    for (j = i + 1; j < 26; j++)
      if (c[i].count < c[j].count)
        t = c[i], c[i] = c[j], c[j] = t;
  b = (c[0].alpha - 4) % 26;
  if (b < 0) b += 26;
  printf("the ciphertext is as follows:\n");
  for (i = 0; i < count; i++) {
    printf("%c", ciphertext[i] + 'A');
    if ((i + 1) % 25 == 0) printf("\n");
  }
  printf("\n");
  printf("the plaintext is as follows:\n");
  for (i = 0; i < count; i++) {
    p = (ciphertext[i] - b) % 26;
    if (p < 0) p += 26;
    printf("%c", p + 'A');
    if ((i + 1) % 25 == 0) printf("\n");
  }
  printf("\n");
  return 0;
}