/*
  Author:  Pate Williams (c) 1997

  Exercise 9.7
  "Suppose that Alice is using the Quisquater Scheme
  with n = 199543, b = 523, and v = 146152. Suppose
  that Olga has discovered that
  v ^ 465 101360 ^ b = v ^ 257 36056 ^ b mod n.
  Show how Olga can compute u." -Douglas R. Stinson-
  See "Cryptograhy: Theory and Practice" by Douglas
  R. Stinson page 304.
*/

#include <stdio.h>
#include "lip.h"

int main(void)
{
  long b = 523, n = 199543l, v = 146152l;
  long r1 = 456, r2 = 257, y1 = 101360l;
  long y2 = 36056l;
  verylong za = 0, zb = 0, zc = 0, zd = 0, zl = 0;
  verylong zn = 0, zt = 0, zu = 0, zv = 0;
  verylong zr1 = 0, zr2 = 0, zy1 = 0, zy2 = 0;

  zintoz(b, &zb);
  zintoz(n, &zn);
  zintoz(v, &zv);
  zintoz(r1, &zr1);
  zintoz(r2, &zr2);
  zintoz(y1, &zy1);
  zintoz(y2, &zy2);
  zsubmod(zr1, zr2, zb, &za);
  zinvmod(za, zb, &zt);
  zmul(za, zt, &zc);
  zsadd(zc, - 1l, &zd);
  zdiv(zd, zb, &zl, &zc);
  zdiv(zy1, zy2, &za, &zc);
  zexpmod(za, zt, zn, &zc);
  zexpmod(zv, zl, zn, &za);
  zmulmod(za, zc, zn, &zu);
  printf("b = %ld\n", b);
  printf("n = %ld\n", n);
  printf("v = %ld\n", v);
  printf("r_1 = %ld\n", r1);
  printf("r_2 = %ld\n", r2);
  printf("y_1 = %ld\n", y1);
  printf("y_2 = %ld\n", y2);
  printf("t = ");
  zwriteln(zt);
  printf("l = ");
  zwriteln(zl);
  printf("u = ");
  zwriteln(zu);
  zfree(&za);
  zfree(&zb);
  zfree(&zc);
  zfree(&zd);
  zfree(&zl);
  zfree(&zn);
  zfree(&zt);
  zfree(&zu);
  zfree(&zv);
  zfree(&zr1);
  zfree(&zr2);
  zfree(&zy1);
  zfree(&zy2);
  return 0;
}