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/*
* Copyright 2000-2016 The OpenSSL Project Authors. All Rights Reserved.
*
* Licensed under the Apache License 2.0 (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_local.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;
}
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