4f22f40507
Reviewed-by: Richard Levitte <levitte@openssl.org>
140 lines
3.2 KiB
C
140 lines
3.2 KiB
C
/*
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* Copyright 2000-2016 The OpenSSL Project Authors. All Rights Reserved.
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*
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* Licensed under the OpenSSL license (the "License"). You may not use
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* this file except in compliance with the License. You can obtain a copy
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* in the file LICENSE in the source distribution or at
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* https://www.openssl.org/source/license.html
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*/
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#include "internal/cryptlib.h"
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#include "bn_lcl.h"
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/* least significant word */
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#define BN_lsw(n) (((n)->top == 0) ? (BN_ULONG) 0 : (n)->d[0])
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/* Returns -2 for errors because both -1 and 0 are valid results. */
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int BN_kronecker(const BIGNUM *a, const BIGNUM *b, BN_CTX *ctx)
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{
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int i;
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int ret = -2; /* avoid 'uninitialized' warning */
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int err = 0;
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BIGNUM *A, *B, *tmp;
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/*-
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* In 'tab', only odd-indexed entries are relevant:
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* For any odd BIGNUM n,
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* tab[BN_lsw(n) & 7]
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* is $(-1)^{(n^2-1)/8}$ (using TeX notation).
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* Note that the sign of n does not matter.
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*/
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static const int tab[8] = { 0, 1, 0, -1, 0, -1, 0, 1 };
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bn_check_top(a);
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bn_check_top(b);
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BN_CTX_start(ctx);
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A = BN_CTX_get(ctx);
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B = BN_CTX_get(ctx);
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if (B == NULL)
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goto end;
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err = !BN_copy(A, a);
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if (err)
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goto end;
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err = !BN_copy(B, b);
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if (err)
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goto end;
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/*
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* Kronecker symbol, implemented according to Henri Cohen,
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* "A Course in Computational Algebraic Number Theory"
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* (algorithm 1.4.10).
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*/
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/* Cohen's step 1: */
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if (BN_is_zero(B)) {
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ret = BN_abs_is_word(A, 1);
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goto end;
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}
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/* Cohen's step 2: */
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if (!BN_is_odd(A) && !BN_is_odd(B)) {
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ret = 0;
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goto end;
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}
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/* now B is non-zero */
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i = 0;
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while (!BN_is_bit_set(B, i))
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i++;
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err = !BN_rshift(B, B, i);
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if (err)
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goto end;
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if (i & 1) {
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/* i is odd */
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/* (thus B was even, thus A must be odd!) */
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/* set 'ret' to $(-1)^{(A^2-1)/8}$ */
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ret = tab[BN_lsw(A) & 7];
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} else {
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/* i is even */
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ret = 1;
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}
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if (B->neg) {
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B->neg = 0;
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if (A->neg)
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ret = -ret;
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}
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/*
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* now B is positive and odd, so what remains to be done is to compute
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* the Jacobi symbol (A/B) and multiply it by 'ret'
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*/
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while (1) {
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/* Cohen's step 3: */
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/* B is positive and odd */
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if (BN_is_zero(A)) {
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ret = BN_is_one(B) ? ret : 0;
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goto end;
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}
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/* now A is non-zero */
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i = 0;
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while (!BN_is_bit_set(A, i))
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i++;
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err = !BN_rshift(A, A, i);
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if (err)
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goto end;
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if (i & 1) {
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/* i is odd */
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/* multiply 'ret' by $(-1)^{(B^2-1)/8}$ */
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ret = ret * tab[BN_lsw(B) & 7];
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}
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/* Cohen's step 4: */
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/* multiply 'ret' by $(-1)^{(A-1)(B-1)/4}$ */
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if ((A->neg ? ~BN_lsw(A) : BN_lsw(A)) & BN_lsw(B) & 2)
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ret = -ret;
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/* (A, B) := (B mod |A|, |A|) */
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err = !BN_nnmod(B, B, A, ctx);
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if (err)
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goto end;
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tmp = A;
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A = B;
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B = tmp;
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tmp->neg = 0;
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}
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end:
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BN_CTX_end(ctx);
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if (err)
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return -2;
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else
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return ret;
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}
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