mirror of
https://github.com/openssl/openssl.git
synced 2024-12-15 06:01:37 +08:00
141 lines
3.2 KiB
C
141 lines
3.2 KiB
C
/*
|
|
* Copyright 2000-2016 The OpenSSL Project Authors. All Rights Reserved.
|
|
*
|
|
* Licensed under the OpenSSL license (the "License"). You may not use
|
|
* this file except in compliance with the License. You can obtain a copy
|
|
* in the file LICENSE in the source distribution or at
|
|
* https://www.openssl.org/source/license.html
|
|
*/
|
|
|
|
#include "internal/cryptlib.h"
|
|
#include "bn_lcl.h"
|
|
|
|
/* least significant word */
|
|
#define BN_lsw(n) (((n)->top == 0) ? (BN_ULONG) 0 : (n)->d[0])
|
|
|
|
/* Returns -2 for errors because both -1 and 0 are valid results. */
|
|
int BN_kronecker(const BIGNUM *a, const BIGNUM *b, BN_CTX *ctx)
|
|
{
|
|
int i;
|
|
int ret = -2; /* avoid 'uninitialized' warning */
|
|
int err = 0;
|
|
BIGNUM *A, *B, *tmp;
|
|
/*-
|
|
* In 'tab', only odd-indexed entries are relevant:
|
|
* For any odd BIGNUM n,
|
|
* tab[BN_lsw(n) & 7]
|
|
* is $(-1)^{(n^2-1)/8}$ (using TeX notation).
|
|
* Note that the sign of n does not matter.
|
|
*/
|
|
static const int tab[8] = { 0, 1, 0, -1, 0, -1, 0, 1 };
|
|
|
|
bn_check_top(a);
|
|
bn_check_top(b);
|
|
|
|
BN_CTX_start(ctx);
|
|
A = BN_CTX_get(ctx);
|
|
B = BN_CTX_get(ctx);
|
|
if (B == NULL)
|
|
goto end;
|
|
|
|
err = !BN_copy(A, a);
|
|
if (err)
|
|
goto end;
|
|
err = !BN_copy(B, b);
|
|
if (err)
|
|
goto end;
|
|
|
|
/*
|
|
* Kronecker symbol, implemented according to Henri Cohen,
|
|
* "A Course in Computational Algebraic Number Theory"
|
|
* (algorithm 1.4.10).
|
|
*/
|
|
|
|
/* Cohen's step 1: */
|
|
|
|
if (BN_is_zero(B)) {
|
|
ret = BN_abs_is_word(A, 1);
|
|
goto end;
|
|
}
|
|
|
|
/* Cohen's step 2: */
|
|
|
|
if (!BN_is_odd(A) && !BN_is_odd(B)) {
|
|
ret = 0;
|
|
goto end;
|
|
}
|
|
|
|
/* now B is non-zero */
|
|
i = 0;
|
|
while (!BN_is_bit_set(B, i))
|
|
i++;
|
|
err = !BN_rshift(B, B, i);
|
|
if (err)
|
|
goto end;
|
|
if (i & 1) {
|
|
/* i is odd */
|
|
/* (thus B was even, thus A must be odd!) */
|
|
|
|
/* set 'ret' to $(-1)^{(A^2-1)/8}$ */
|
|
ret = tab[BN_lsw(A) & 7];
|
|
} else {
|
|
/* i is even */
|
|
ret = 1;
|
|
}
|
|
|
|
if (B->neg) {
|
|
B->neg = 0;
|
|
if (A->neg)
|
|
ret = -ret;
|
|
}
|
|
|
|
/*
|
|
* now B is positive and odd, so what remains to be done is to compute
|
|
* the Jacobi symbol (A/B) and multiply it by 'ret'
|
|
*/
|
|
|
|
while (1) {
|
|
/* Cohen's step 3: */
|
|
|
|
/* B is positive and odd */
|
|
|
|
if (BN_is_zero(A)) {
|
|
ret = BN_is_one(B) ? ret : 0;
|
|
goto end;
|
|
}
|
|
|
|
/* now A is non-zero */
|
|
i = 0;
|
|
while (!BN_is_bit_set(A, i))
|
|
i++;
|
|
err = !BN_rshift(A, A, i);
|
|
if (err)
|
|
goto end;
|
|
if (i & 1) {
|
|
/* i is odd */
|
|
/* multiply 'ret' by $(-1)^{(B^2-1)/8}$ */
|
|
ret = ret * tab[BN_lsw(B) & 7];
|
|
}
|
|
|
|
/* Cohen's step 4: */
|
|
/* multiply 'ret' by $(-1)^{(A-1)(B-1)/4}$ */
|
|
if ((A->neg ? ~BN_lsw(A) : BN_lsw(A)) & BN_lsw(B) & 2)
|
|
ret = -ret;
|
|
|
|
/* (A, B) := (B mod |A|, |A|) */
|
|
err = !BN_nnmod(B, B, A, ctx);
|
|
if (err)
|
|
goto end;
|
|
tmp = A;
|
|
A = B;
|
|
B = tmp;
|
|
tmp->neg = 0;
|
|
}
|
|
end:
|
|
BN_CTX_end(ctx);
|
|
if (err)
|
|
return -2;
|
|
else
|
|
return ret;
|
|
}
|