aboutsummaryrefslogtreecommitdiffstats
path: root/contrib/libs/openssl/crypto/bn/bn_kron.c
blob: dcd0bc72ff8b92061265397630ce5e502045814c (plain) (blame)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
/*
 * 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_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;
}