1 /* $OpenBSD: bn_isqrt.c,v 1.10 2023/06/04 17:28:35 tb Exp $ */
2 /*
3 * Copyright (c) 2022 Theo Buehler <tb@openbsd.org>
4 *
5 * Permission to use, copy, modify, and distribute this software for any
6 * purpose with or without fee is hereby granted, provided that the above
7 * copyright notice and this permission notice appear in all copies.
8 *
9 * THE SOFTWARE IS PROVIDED "AS IS" AND THE AUTHOR DISCLAIMS ALL WARRANTIES
10 * WITH REGARD TO THIS SOFTWARE INCLUDING ALL IMPLIED WARRANTIES OF
11 * MERCHANTABILITY AND FITNESS. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR
12 * ANY SPECIAL, DIRECT, INDIRECT, OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES
13 * WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN
14 * ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF
15 * OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS SOFTWARE.
16 */
17
18 #include <stddef.h>
19 #include <stdint.h>
20
21 #include <openssl/bn.h>
22 #include <openssl/err.h>
23
24 #include "bn_local.h"
25 #include "crypto_internal.h"
26
27 /*
28 * Calculate integer square root of |n| using a variant of Newton's method.
29 *
30 * Returns the integer square root of |n| in the caller-provided |out_sqrt|;
31 * |*out_perfect| is set to 1 if and only if |n| is a perfect square.
32 * One of |out_sqrt| and |out_perfect| can be NULL; |in_ctx| can be NULL.
33 *
34 * Returns 0 on error, 1 on success.
35 *
36 * Adapted from pure Python describing cpython's math.isqrt(), without bothering
37 * with any of the optimizations in the C code. A correctness proof is here:
38 * https://github.com/mdickinson/snippets/blob/master/proofs/isqrt/src/isqrt.lean
39 * The comments in the Python code also give a rather detailed proof.
40 */
41
42 int
bn_isqrt(BIGNUM * out_sqrt,int * out_perfect,const BIGNUM * n,BN_CTX * in_ctx)43 bn_isqrt(BIGNUM *out_sqrt, int *out_perfect, const BIGNUM *n, BN_CTX *in_ctx)
44 {
45 BN_CTX *ctx = NULL;
46 BIGNUM *a, *b;
47 int c, d, e, s;
48 int cmp, perfect;
49 int ret = 0;
50
51 if (out_perfect == NULL && out_sqrt == NULL) {
52 BNerror(ERR_R_PASSED_NULL_PARAMETER);
53 goto err;
54 }
55
56 if (BN_is_negative(n)) {
57 BNerror(BN_R_INVALID_RANGE);
58 goto err;
59 }
60
61 if ((ctx = in_ctx) == NULL)
62 ctx = BN_CTX_new();
63 if (ctx == NULL)
64 goto err;
65
66 BN_CTX_start(ctx);
67
68 if ((a = BN_CTX_get(ctx)) == NULL)
69 goto err;
70 if ((b = BN_CTX_get(ctx)) == NULL)
71 goto err;
72
73 if (BN_is_zero(n)) {
74 perfect = 1;
75 BN_zero(a);
76 goto done;
77 }
78
79 if (!BN_one(a))
80 goto err;
81
82 c = (BN_num_bits(n) - 1) / 2;
83 d = 0;
84
85 /* Calculate s = floor(log(c)). */
86 if (!BN_set_word(b, c))
87 goto err;
88 s = BN_num_bits(b) - 1;
89
90 /*
91 * By definition, the loop below is run <= floor(log(log(n))) times.
92 * Comments in the cpython code establish the loop invariant that
93 *
94 * (a - 1)^2 < n / 4^(c - d) < (a + 1)^2
95 *
96 * holds true in every iteration. Once this is proved via induction,
97 * correctness of the algorithm is easy.
98 *
99 * Roughly speaking, A = (a << (d - e)) is used for one Newton step
100 * "a = (A >> 1) + (m >> 1) / A" approximating m = (n >> 2 * (c - d)).
101 */
102
103 for (; s >= 0; s--) {
104 e = d;
105 d = c >> s;
106
107 if (!BN_rshift(b, n, 2 * c - d - e + 1))
108 goto err;
109
110 if (!BN_div_ct(b, NULL, b, a, ctx))
111 goto err;
112
113 if (!BN_lshift(a, a, d - e - 1))
114 goto err;
115
116 if (!BN_add(a, a, b))
117 goto err;
118 }
119
120 /*
121 * The loop invariant implies that either a or a - 1 is isqrt(n).
122 * Figure out which one it is. The invariant also implies that for
123 * a perfect square n, a must be the square root.
124 */
125
126 if (!BN_sqr(b, a, ctx))
127 goto err;
128
129 /* If a^2 > n, we must have isqrt(n) == a - 1. */
130 if ((cmp = BN_cmp(b, n)) > 0) {
131 if (!BN_sub_word(a, 1))
132 goto err;
133 }
134
135 perfect = cmp == 0;
136
137 done:
138 if (out_perfect != NULL)
139 *out_perfect = perfect;
140
141 if (out_sqrt != NULL) {
142 if (!bn_copy(out_sqrt, a))
143 goto err;
144 }
145
146 ret = 1;
147
148 err:
149 BN_CTX_end(ctx);
150
151 if (ctx != in_ctx)
152 BN_CTX_free(ctx);
153
154 return ret;
155 }
156
157 /*
158 * is_square_mod_N[r % N] indicates whether r % N has a square root modulo N.
159 * The tables are generated in regress/lib/libcrypto/bn/bn_isqrt.c.
160 */
161
162 const uint8_t is_square_mod_11[] = {
163 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0,
164 };
165 CTASSERT(sizeof(is_square_mod_11) == 11);
166
167 const uint8_t is_square_mod_63[] = {
168 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0,
169 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0,
170 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0,
171 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
172 };
173 CTASSERT(sizeof(is_square_mod_63) == 63);
174
175 const uint8_t is_square_mod_64[] = {
176 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
177 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
178 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
179 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
180 };
181 CTASSERT(sizeof(is_square_mod_64) == 64);
182
183 const uint8_t is_square_mod_65[] = {
184 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0,
185 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0,
186 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0,
187 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0,
188 1,
189 };
190 CTASSERT(sizeof(is_square_mod_65) == 65);
191
192 /*
193 * Determine whether n is a perfect square or not.
194 *
195 * Returns 1 on success and 0 on error. In case of success, |*out_perfect| is
196 * set to 1 if and only if |n| is a perfect square.
197 */
198
199 int
bn_is_perfect_square(int * out_perfect,const BIGNUM * n,BN_CTX * ctx)200 bn_is_perfect_square(int *out_perfect, const BIGNUM *n, BN_CTX *ctx)
201 {
202 BN_ULONG r;
203
204 *out_perfect = 0;
205
206 if (BN_is_negative(n))
207 return 1;
208
209 /*
210 * Before performing an expensive bn_isqrt() operation, weed out many
211 * obvious non-squares. See H. Cohen, "A course in computational
212 * algebraic number theory", Algorithm 1.7.3.
213 *
214 * The idea is that a square remains a square when reduced modulo any
215 * number. The moduli are chosen in such a way that a non-square has
216 * probability < 1% of passing the four table lookups.
217 */
218
219 /* n % 64 */
220 r = BN_lsw(n) & 0x3f;
221
222 if (!is_square_mod_64[r % 64])
223 return 1;
224
225 if ((r = BN_mod_word(n, 11 * 63 * 65)) == (BN_ULONG)-1)
226 return 0;
227
228 if (!is_square_mod_63[r % 63] ||
229 !is_square_mod_65[r % 65] ||
230 !is_square_mod_11[r % 11])
231 return 1;
232
233 return bn_isqrt(NULL, out_perfect, n, ctx);
234 }
235