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