12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061626364656667 |
- /*
- * Copyright (c) 1995 Colin Plumb. All rights reserved.
- * For licensing and other legal details, see the file legal.c.
- *
- * Compute the Jacobi symbol (small prime case only).
- */
- #include "bn.h"
- #include "jacobi.h"
- /*
- * For a small (usually prime, but not necessarily) prime p,
- * compute Jacobi(p,bn), which is -1, 0 or +1, using the following rules:
- * Jacobi(x, y) = Jacobi(x mod y, y)
- * Jacobi(0, y) = 0
- * Jacobi(1, y) = 1
- * Jacobi(2, y) = 0 if y is even, +1 if y is +/-1 mod 8, -1 if y = +/-3 mod 8
- * Jacobi(x1*x2, y) = Jacobi(x1, y) * Jacobi(x2, y) (used with x1 = 2 & x1 = 4)
- * If x and y are both odd, then
- * Jacobi(x, y) = Jacobi(y, x) * (-1 if x = y = 3 mod 4, +1 otherwise)
- */
- int
- bnJacobiQ(unsigned p, struct BigNum const *bn)
- {
- int j = 1;
- unsigned u = bnLSWord(bn);
- if (!(u & 1))
- return 0; /* Don't *do* that */
- /* First, get rid of factors of 2 in p */
- while ((p & 3) == 0)
- p >>= 2;
- if ((p & 1) == 0) {
- p >>= 1;
- if ((u ^ u>>1) & 2)
- j = -j; /* 3 (011) or 5 (101) mod 8 */
- }
- if (p == 1)
- return j;
- /* Then, apply quadratic reciprocity */
- if (p & u & 2) /* p = u = 3 (mod 4? */
- j = -j;
- /* And reduce u mod p */
- u = bnModQ(bn, p);
- /* Now compute Jacobi(u,p), u < p */
- while (u) {
- while ((u & 3) == 0)
- u >>= 2;
- if ((u & 1) == 0) {
- u >>= 1;
- if ((p ^ p>>1) & 2)
- j = -j; /* 3 (011) or 5 (101) mod 8 */
- }
- if (u == 1)
- return j;
- /* Now both u and p are odd, so use quadratic reciprocity */
- if (u < p) {
- unsigned t = u; u = p; p = t;
- if (u & p & 2) /* u = p = 3 (mod 4? */
- j = -j;
- }
- /* Now u >= p, so it can be reduced */
- u %= p;
- }
- return 0;
- }
|