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4f22f405 RS |
1 | /* |
2 | * Copyright 2000-2016 The OpenSSL Project Authors. All Rights Reserved. | |
265592b9 | 3 | * |
4f22f405 RS |
4 | * Licensed under the OpenSSL license (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 | |
265592b9 BM |
8 | */ |
9 | ||
b39fc560 | 10 | #include "internal/cryptlib.h" |
265592b9 BM |
11 | #include "bn_lcl.h" |
12 | ||
265592b9 BM |
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. */ | |
17 | int BN_kronecker(const BIGNUM *a, const BIGNUM *b, BN_CTX *ctx) | |
0f113f3e MC |
18 | { |
19 | int i; | |
20 | int ret = -2; /* avoid 'uninitialized' warning */ | |
21 | int err = 0; | |
22 | BIGNUM *A, *B, *tmp; | |
35a1cc90 MC |
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 | */ | |
0f113f3e MC |
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 | /* | |
0d4fb843 | 49 | * Kronecker symbol, implemented according to Henri Cohen, |
0f113f3e MC |
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 | } |