pigweed / third_party / boringssl / boringssl / 95c29f3cd1f6c08c6c0927868683392eea727ccb / . / crypto / bn / kronecker.c

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* software must display the following acknowledgment: | |

* "This product includes software developed by the OpenSSL Project | |

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* 6. Redistributions of any form whatsoever must retain the following | |

* acknowledgment: | |

* "This product includes software developed by the OpenSSL Project | |

* for use in the OpenSSL Toolkit (http://www.openssl.org/)" | |

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* | |

* This product includes cryptographic software written by Eric Young | |

* (eay@cryptsoft.com). This product includes software written by Tim | |

* Hudson (tjh@cryptsoft.com). */ | |

#include <openssl/bn.h> | |

#include "internal.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; | |

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_CTX_start(ctx); | |

A = BN_CTX_get(ctx); | |

B = BN_CTX_get(ctx); | |

if (B == NULL) { | |

goto end; | |

} | |

if (!BN_copy(A, a) || | |

!BN_copy(B, b)) { | |

goto end; | |

} | |

/* Kronecker symbol, imlemented 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++; | |

} | |

if (!BN_rshift(B, B, i)) { | |

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++; | |

} | |

if (!BN_rshift(A, A, i)) { | |

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|) */ | |

if (!BN_nnmod(B, B, A, ctx)) { | |

ret = -2; | |

goto end; | |

} | |

tmp = A; | |

A = B; | |

B = tmp; | |

tmp->neg = 0; | |

} | |

end: | |

BN_CTX_end(ctx); | |

return ret; | |

} |