1 /*
2 * Copyright 2000-2016 The OpenSSL Project Authors. All Rights Reserved.
3 *
4 * Licensed under the Apache License 2.0 (the "License"). You may not use
5 * this file except in compliance with the License. You can obtain a copy
6 * in the file LICENSE in the source distribution or at
7 * https://www.openssl.org/source/license.html
8 */
9
10 #include "internal/cryptlib.h"
11 #include "bn_local.h"
12
13 /* least significant word */
14 #define BN_lsw(n) (((n)->top == 0) ? (BN_ULONG) 0 : (n)->d[0])
15
16 /* Returns -2 for errors because both -1 and 0 are valid results. */
BN_kronecker(const BIGNUM * a,const BIGNUM * b,BN_CTX * ctx)17 int BN_kronecker(const BIGNUM *a, const BIGNUM *b, BN_CTX *ctx)
18 {
19 int i;
20 int ret = -2; /* avoid 'uninitialized' warning */
21 int err = 0;
22 BIGNUM *A, *B, *tmp;
23 /*-
24 * In 'tab', only odd-indexed entries are relevant:
25 * For any odd BIGNUM n,
26 * tab[BN_lsw(n) & 7]
27 * is $(-1)^{(n^2-1)/8}$ (using TeX notation).
28 * Note that the sign of n does not matter.
29 */
30 static const int tab[8] = { 0, 1, 0, -1, 0, -1, 0, 1 };
31
32 bn_check_top(a);
33 bn_check_top(b);
34
35 BN_CTX_start(ctx);
36 A = BN_CTX_get(ctx);
37 B = BN_CTX_get(ctx);
38 if (B == NULL)
39 goto end;
40
41 err = !BN_copy(A, a);
42 if (err)
43 goto end;
44 err = !BN_copy(B, b);
45 if (err)
46 goto end;
47
48 /*
49 * Kronecker symbol, implemented according to Henri Cohen,
50 * "A Course in Computational Algebraic Number Theory"
51 * (algorithm 1.4.10).
52 */
53
54 /* Cohen's step 1: */
55
56 if (BN_is_zero(B)) {
57 ret = BN_abs_is_word(A, 1);
58 goto end;
59 }
60
61 /* Cohen's step 2: */
62
63 if (!BN_is_odd(A) && !BN_is_odd(B)) {
64 ret = 0;
65 goto end;
66 }
67
68 /* now B is non-zero */
69 i = 0;
70 while (!BN_is_bit_set(B, i))
71 i++;
72 err = !BN_rshift(B, B, i);
73 if (err)
74 goto end;
75 if (i & 1) {
76 /* i is odd */
77 /* (thus B was even, thus A must be odd!) */
78
79 /* set 'ret' to $(-1)^{(A^2-1)/8}$ */
80 ret = tab[BN_lsw(A) & 7];
81 } else {
82 /* i is even */
83 ret = 1;
84 }
85
86 if (B->neg) {
87 B->neg = 0;
88 if (A->neg)
89 ret = -ret;
90 }
91
92 /*
93 * now B is positive and odd, so what remains to be done is to compute
94 * the Jacobi symbol (A/B) and multiply it by 'ret'
95 */
96
97 while (1) {
98 /* Cohen's step 3: */
99
100 /* B is positive and odd */
101
102 if (BN_is_zero(A)) {
103 ret = BN_is_one(B) ? ret : 0;
104 goto end;
105 }
106
107 /* now A is non-zero */
108 i = 0;
109 while (!BN_is_bit_set(A, i))
110 i++;
111 err = !BN_rshift(A, A, i);
112 if (err)
113 goto end;
114 if (i & 1) {
115 /* i is odd */
116 /* multiply 'ret' by $(-1)^{(B^2-1)/8}$ */
117 ret = ret * tab[BN_lsw(B) & 7];
118 }
119
120 /* Cohen's step 4: */
121 /* multiply 'ret' by $(-1)^{(A-1)(B-1)/4}$ */
122 if ((A->neg ? ~BN_lsw(A) : BN_lsw(A)) & BN_lsw(B) & 2)
123 ret = -ret;
124
125 /* (A, B) := (B mod |A|, |A|) */
126 err = !BN_nnmod(B, B, A, ctx);
127 if (err)
128 goto end;
129 tmp = A;
130 A = B;
131 B = tmp;
132 tmp->neg = 0;
133 }
134 end:
135 BN_CTX_end(ctx);
136 if (err)
137 return -2;
138 else
139 return ret;
140 }
141
