From eb8dd9dca1228af0cd132f515509051ecfabf6f6 Mon Sep 17 00:00:00 2001 From: cvs2svn Date: Mon, 14 Apr 2025 17:32:06 +0000 Subject: This commit was manufactured by cvs2git to create tag 'tb_20250414'. --- src/lib/libcrypto/bn/bn_mod_sqrt.c | 723 ------------------------------------- 1 file changed, 723 deletions(-) delete mode 100644 src/lib/libcrypto/bn/bn_mod_sqrt.c (limited to 'src/lib/libcrypto/bn/bn_mod_sqrt.c') diff --git a/src/lib/libcrypto/bn/bn_mod_sqrt.c b/src/lib/libcrypto/bn/bn_mod_sqrt.c deleted file mode 100644 index 280002cc48..0000000000 --- a/src/lib/libcrypto/bn/bn_mod_sqrt.c +++ /dev/null @@ -1,723 +0,0 @@ -/* $OpenBSD: bn_mod_sqrt.c,v 1.3 2023/08/03 18:53:55 tb Exp $ */ - -/* - * Copyright (c) 2022 Theo Buehler - * - * Permission to use, copy, modify, and distribute this software for any - * purpose with or without fee is hereby granted, provided that the above - * copyright notice and this permission notice appear in all copies. - * - * THE SOFTWARE IS PROVIDED "AS IS" AND THE AUTHOR DISCLAIMS ALL WARRANTIES - * WITH REGARD TO THIS SOFTWARE INCLUDING ALL IMPLIED WARRANTIES OF - * MERCHANTABILITY AND FITNESS. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR - * ANY SPECIAL, DIRECT, INDIRECT, OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES - * WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN - * ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF - * OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS SOFTWARE. - */ - -#include - -#include "bn_local.h" - -/* - * Tonelli-Shanks according to H. Cohen "A Course in Computational Algebraic - * Number Theory", Section 1.5.1, Springer GTM volume 138, Berlin, 1996. - * - * Under the assumption that p is prime and a is a quadratic residue, we know: - * - * a^[(p-1)/2] = 1 (mod p). (*) - * - * To find a square root of a (mod p), we handle three cases of increasing - * complexity. In the first two cases, we can compute a square root using an - * explicit formula, thus avoiding the probabilistic nature of Tonelli-Shanks. - * - * 1. p = 3 (mod 4). - * - * Set n = (p+1)/4. Then 2n = 1 + (p-1)/2 and (*) shows that x = a^n (mod p) - * is a square root of a: x^2 = a^(2n) = a * a^[(p-1)/2] = a (mod p). - * - * 2. p = 5 (mod 8). - * - * This uses a simplification due to Atkin. By Theorem 1.4.7 and 1.4.9, the - * Kronecker symbol (2/p) evaluates to (-1)^[(p^2-1)/8]. From p = 5 (mod 8) - * we get (p^2-1)/8 = 1 (mod 2), so (2/p) = -1, and thus - * - * 2^[(p-1)/2] = -1 (mod p). (**) - * - * Set b = (2a)^[(p-5)/8]. With (p-1)/2 = 2 + (p-5)/2, (*) and (**) show - * - * i = 2 a b^2 is a square root of -1 (mod p). - * - * Indeed, i^2 = 2^2 a^2 b^4 = 2^[(p-1)/2] a^[(p-1)/2] = -1 (mod p). Because - * of (i-1)^2 = -2i (mod p) and i (-i) = 1 (mod p), a square root of a is - * - * x = a b (i-1) - * - * as x^2 = a^2 b^2 (-2i) = a (2 a b^2) (-i) = a (mod p). - * - * 3. p = 1 (mod 8). - * - * This is the Tonelli-Shanks algorithm. For a prime p, the multiplicative - * group of GF(p) is cyclic of order p - 1 = 2^s q, with odd q. Denote its - * 2-Sylow subgroup by S. It is cyclic of order 2^s. The squares in S have - * order dividing 2^(s-1). They are the even powers of any generator z of S. - * If a is a quadratic residue, 1 = a^[(p-1)/2] = (a^q)^[2^(s-1)], so b = a^q - * is a square in S. Therefore there is an integer k such that b z^(2k) = 1. - * Set x = a^[(q+1)/2] z^k, and find x^2 = a (mod p). - * - * The problem is thus reduced to finding a generator z of the 2-Sylow - * subgroup S of GF(p)* and finding k. An iterative constructions avoids - * the need for an explicit k, a generator is found by a randomized search. - * - * While we do not actually know that p is a prime number, we can still apply - * the formulas in cases 1 and 2 and verify that we have indeed found a square - * root of p. Similarly, in case 3, we can try to find a quadratic non-residue, - * which will fail for example if p is a square. The iterative construction - * may or may not find a candidate square root which we can then validate. - */ - -/* - * Handle the cases where p is 2, p isn't odd or p is one. Since BN_mod_sqrt() - * can run on untrusted data, a primality check is too expensive. Also treat - * the obvious cases where a is 0 or 1. - */ - -static int -bn_mod_sqrt_trivial_cases(int *done, BIGNUM *out_sqrt, const BIGNUM *a, - const BIGNUM *p, BN_CTX *ctx) -{ - *done = 1; - - if (BN_abs_is_word(p, 2)) - return BN_set_word(out_sqrt, BN_is_odd(a)); - - if (!BN_is_odd(p) || BN_abs_is_word(p, 1)) { - BNerror(BN_R_P_IS_NOT_PRIME); - return 0; - } - - if (BN_is_zero(a) || BN_is_one(a)) - return BN_set_word(out_sqrt, BN_is_one(a)); - - *done = 0; - - return 1; -} - -/* - * Case 1. We know that (a/p) = 1 and that p = 3 (mod 4). - */ - -static int -bn_mod_sqrt_p_is_3_mod_4(BIGNUM *out_sqrt, const BIGNUM *a, const BIGNUM *p, - BN_CTX *ctx) -{ - BIGNUM *n; - int ret = 0; - - BN_CTX_start(ctx); - - if ((n = BN_CTX_get(ctx)) == NULL) - goto err; - - /* Calculate n = (|p| + 1) / 4. */ - if (!BN_uadd(n, p, BN_value_one())) - goto err; - if (!BN_rshift(n, n, 2)) - goto err; - - /* By case 1 above, out_sqrt = a^n is a square root of a (mod p). */ - if (!BN_mod_exp_ct(out_sqrt, a, n, p, ctx)) - goto err; - - ret = 1; - - err: - BN_CTX_end(ctx); - - return ret; -} - -/* - * Case 2. We know that (a/p) = 1 and that p = 5 (mod 8). - */ - -static int -bn_mod_sqrt_p_is_5_mod_8(BIGNUM *out_sqrt, const BIGNUM *a, const BIGNUM *p, - BN_CTX *ctx) -{ - BIGNUM *b, *i, *n, *tmp; - int ret = 0; - - BN_CTX_start(ctx); - - if ((b = BN_CTX_get(ctx)) == NULL) - goto err; - if ((i = BN_CTX_get(ctx)) == NULL) - goto err; - if ((n = BN_CTX_get(ctx)) == NULL) - goto err; - if ((tmp = BN_CTX_get(ctx)) == NULL) - goto err; - - /* Calculate n = (|p| - 5) / 8. Since p = 5 (mod 8), simply shift. */ - if (!BN_rshift(n, p, 3)) - goto err; - BN_set_negative(n, 0); - - /* Compute tmp = 2a (mod p) for later use. */ - if (!BN_mod_lshift1(tmp, a, p, ctx)) - goto err; - - /* Calculate b = (2a)^n (mod p). */ - if (!BN_mod_exp_ct(b, tmp, n, p, ctx)) - goto err; - - /* Calculate i = 2 a b^2 (mod p). */ - if (!BN_mod_sqr(i, b, p, ctx)) - goto err; - if (!BN_mod_mul(i, tmp, i, p, ctx)) - goto err; - - /* A square root is out_sqrt = a b (i-1) (mod p). */ - if (!BN_sub_word(i, 1)) - goto err; - if (!BN_mod_mul(out_sqrt, a, b, p, ctx)) - goto err; - if (!BN_mod_mul(out_sqrt, out_sqrt, i, p, ctx)) - goto err; - - ret = 1; - - err: - BN_CTX_end(ctx); - - return ret; -} - -/* - * Case 3. We know that (a/p) = 1 and that p = 1 (mod 8). - */ - -/* - * Simple helper. To find a generator of the 2-Sylow subgroup of GF(p)*, we - * need to find a quadratic non-residue of p, i.e., n such that (n/p) = -1. - */ - -static int -bn_mod_sqrt_n_is_non_residue(int *is_non_residue, const BIGNUM *n, - const BIGNUM *p, BN_CTX *ctx) -{ - switch (BN_kronecker(n, p, ctx)) { - case -1: - *is_non_residue = 1; - return 1; - case 1: - *is_non_residue = 0; - return 1; - case 0: - /* n divides p, so ... */ - BNerror(BN_R_P_IS_NOT_PRIME); - return 0; - default: - return 0; - } -} - -/* - * The following is the only non-deterministic part preparing Tonelli-Shanks. - * - * If we find n such that (n/p) = -1, then n^q (mod p) is a generator of the - * 2-Sylow subgroup of GF(p)*. To find such n, first try some small numbers, - * then random ones. - */ - -static int -bn_mod_sqrt_find_sylow_generator(BIGNUM *out_generator, const BIGNUM *p, - const BIGNUM *q, BN_CTX *ctx) -{ - BIGNUM *n, *p_abs; - int i, is_non_residue; - int ret = 0; - - BN_CTX_start(ctx); - - if ((n = BN_CTX_get(ctx)) == NULL) - goto err; - if ((p_abs = BN_CTX_get(ctx)) == NULL) - goto err; - - for (i = 2; i < 32; i++) { - if (!BN_set_word(n, i)) - goto err; - if (!bn_mod_sqrt_n_is_non_residue(&is_non_residue, n, p, ctx)) - goto err; - if (is_non_residue) - goto found; - } - - if (!bn_copy(p_abs, p)) - goto err; - BN_set_negative(p_abs, 0); - - for (i = 0; i < 128; i++) { - if (!bn_rand_interval(n, 32, p_abs)) - goto err; - if (!bn_mod_sqrt_n_is_non_residue(&is_non_residue, n, p, ctx)) - goto err; - if (is_non_residue) - goto found; - } - - /* - * The probability to get here is < 2^(-128) for prime p. For squares - * it is easy: for p = 1369 = 37^2 this happens in ~3% of runs. - */ - - BNerror(BN_R_TOO_MANY_ITERATIONS); - goto err; - - found: - /* - * If p is prime, n^q generates the 2-Sylow subgroup S of GF(p)*. - */ - - if (!BN_mod_exp_ct(out_generator, n, q, p, ctx)) - goto err; - - /* Sanity: p is not necessarily prime, so we could have found 0 or 1. */ - if (BN_is_zero(out_generator) || BN_is_one(out_generator)) { - BNerror(BN_R_P_IS_NOT_PRIME); - goto err; - } - - ret = 1; - - err: - BN_CTX_end(ctx); - - return ret; -} - -/* - * Initialization step for Tonelli-Shanks. - * - * In the end, b = a^q (mod p) and x = a^[(q+1)/2] (mod p). Cohen optimizes this - * to minimize taking powers of a. This is a bit confusing and distracting, so - * factor this into a separate function. - */ - -static int -bn_mod_sqrt_tonelli_shanks_initialize(BIGNUM *b, BIGNUM *x, const BIGNUM *a, - const BIGNUM *p, const BIGNUM *q, BN_CTX *ctx) -{ - BIGNUM *k; - int ret = 0; - - BN_CTX_start(ctx); - - if ((k = BN_CTX_get(ctx)) == NULL) - goto err; - - /* k = (q-1)/2. Since q is odd, we can shift. */ - if (!BN_rshift1(k, q)) - goto err; - - /* x = a^[(q-1)/2] (mod p). */ - if (!BN_mod_exp_ct(x, a, k, p, ctx)) - goto err; - - /* b = ax^2 = a^q (mod p). */ - if (!BN_mod_sqr(b, x, p, ctx)) - goto err; - if (!BN_mod_mul(b, a, b, p, ctx)) - goto err; - - /* x = ax = a^[(q+1)/2] (mod p). */ - if (!BN_mod_mul(x, a, x, p, ctx)) - goto err; - - ret = 1; - - err: - BN_CTX_end(ctx); - - return ret; -} - -/* - * Find smallest exponent m such that b^(2^m) = 1 (mod p). Assuming that a - * is a quadratic residue and p is a prime, we know that 1 <= m < r. - */ - -static int -bn_mod_sqrt_tonelli_shanks_find_exponent(int *out_exponent, const BIGNUM *b, - const BIGNUM *p, int r, BN_CTX *ctx) -{ - BIGNUM *x; - int m; - int ret = 0; - - BN_CTX_start(ctx); - - if ((x = BN_CTX_get(ctx)) == NULL) - goto err; - - /* - * If r <= 1, the Tonelli-Shanks iteration should have terminated as - * r == 1 implies b == 1. - */ - if (r <= 1) { - BNerror(BN_R_P_IS_NOT_PRIME); - goto err; - } - - /* - * Sanity check to ensure taking squares actually does something: - * If b is 1, the Tonelli-Shanks iteration should have terminated. - * If b is 0, something's very wrong, in particular p can't be prime. - */ - if (BN_is_zero(b) || BN_is_one(b)) { - BNerror(BN_R_P_IS_NOT_PRIME); - goto err; - } - - if (!bn_copy(x, b)) - goto err; - - for (m = 1; m < r; m++) { - if (!BN_mod_sqr(x, x, p, ctx)) - goto err; - if (BN_is_one(x)) - break; - } - - if (m >= r) { - /* This means a is not a quadratic residue. As (a/p) = 1, ... */ - BNerror(BN_R_P_IS_NOT_PRIME); - goto err; - } - - *out_exponent = m; - - ret = 1; - - err: - BN_CTX_end(ctx); - - return ret; -} - -/* - * The update step. With the minimal m such that b^(2^m) = 1 (mod m), - * set t = y^[2^(r-m-1)] (mod p) and update x = xt, y = t^2, b = by. - * This preserves the loop invariants a b = x^2, y^[2^(r-1)] = -1 and - * b^[2^(r-1)] = 1. - */ - -static int -bn_mod_sqrt_tonelli_shanks_update(BIGNUM *b, BIGNUM *x, BIGNUM *y, - const BIGNUM *p, int m, int r, BN_CTX *ctx) -{ - BIGNUM *t; - int ret = 0; - - BN_CTX_start(ctx); - - if ((t = BN_CTX_get(ctx)) == NULL) - goto err; - - /* t = y^[2^(r-m-1)] (mod p). */ - if (!BN_set_bit(t, r - m - 1)) - goto err; - if (!BN_mod_exp_ct(t, y, t, p, ctx)) - goto err; - - /* x = xt (mod p). */ - if (!BN_mod_mul(x, x, t, p, ctx)) - goto err; - - /* y = t^2 = y^[2^(r-m)] (mod p). */ - if (!BN_mod_sqr(y, t, p, ctx)) - goto err; - - /* b = by (mod p). */ - if (!BN_mod_mul(b, b, y, p, ctx)) - goto err; - - ret = 1; - - err: - BN_CTX_end(ctx); - - return ret; -} - -static int -bn_mod_sqrt_p_is_1_mod_8(BIGNUM *out_sqrt, const BIGNUM *a, const BIGNUM *p, - BN_CTX *ctx) -{ - BIGNUM *b, *q, *x, *y; - int e, m, r; - int ret = 0; - - BN_CTX_start(ctx); - - if ((b = BN_CTX_get(ctx)) == NULL) - goto err; - if ((q = BN_CTX_get(ctx)) == NULL) - goto err; - if ((x = BN_CTX_get(ctx)) == NULL) - goto err; - if ((y = BN_CTX_get(ctx)) == NULL) - goto err; - - /* - * Factor p - 1 = 2^e q with odd q. Since p = 1 (mod 8), we know e >= 3. - */ - - e = 1; - while (!BN_is_bit_set(p, e)) - e++; - if (!BN_rshift(q, p, e)) - goto err; - - if (!bn_mod_sqrt_find_sylow_generator(y, p, q, ctx)) - goto err; - - /* - * Set b = a^q (mod p) and x = a^[(q+1)/2] (mod p). - */ - if (!bn_mod_sqrt_tonelli_shanks_initialize(b, x, a, p, q, ctx)) - goto err; - - /* - * The Tonelli-Shanks iteration. Starting with r = e, the following loop - * invariants hold at the start of the loop. - * - * a b = x^2 (mod p) - * y^[2^(r-1)] = -1 (mod p) - * b^[2^(r-1)] = 1 (mod p) - * - * In particular, if b = 1 (mod p), x is a square root of a. - * - * Since p - 1 = 2^e q, we have 2^(e-1) q = (p - 1) / 2, so in the first - * iteration this follows from (a/p) = 1, (n/p) = -1, y = n^q, b = a^q. - * - * In subsequent iterations, t = y^[2^(r-m-1)], where m is the smallest - * m such that b^(2^m) = 1. With x = xt (mod p) and b = bt^2 (mod p) the - * first invariant is preserved, the second and third follow from - * y = t^2 (mod p) and r = m as well as the choice of m. - * - * Finally, r is strictly decreasing in each iteration. If p is prime, - * let S be the 2-Sylow subgroup of GF(p)*. We can prove the algorithm - * stops: Let S_r be the subgroup of S consisting of elements of order - * dividing 2^r. Then S_r = and b is in S_(r-1). The S_r form a - * descending filtration of S and when r = 1, then b = 1. - */ - - for (r = e; r >= 1; r = m) { - /* - * Termination condition. If b == 1 then x is a square root. - */ - if (BN_is_one(b)) - goto done; - - /* Find smallest exponent 1 <= m < r such that b^(2^m) == 1. */ - if (!bn_mod_sqrt_tonelli_shanks_find_exponent(&m, b, p, r, ctx)) - goto err; - - /* - * With t = y^[2^(r-m-1)], update x = xt, y = t^2, b = by. - */ - if (!bn_mod_sqrt_tonelli_shanks_update(b, x, y, p, m, r, ctx)) - goto err; - - /* - * Sanity check to make sure we don't loop indefinitely. - * bn_mod_sqrt_tonelli_shanks_find_exponent() ensures m < r. - */ - if (r <= m) - goto err; - } - - /* - * If p is prime, we should not get here. - */ - - BNerror(BN_R_NOT_A_SQUARE); - goto err; - - done: - if (!bn_copy(out_sqrt, x)) - goto err; - - ret = 1; - - err: - BN_CTX_end(ctx); - - return ret; -} - -/* - * Choose the smaller of sqrt and |p| - sqrt. - */ - -static int -bn_mod_sqrt_normalize(BIGNUM *sqrt, const BIGNUM *p, BN_CTX *ctx) -{ - BIGNUM *x; - int ret = 0; - - BN_CTX_start(ctx); - - if ((x = BN_CTX_get(ctx)) == NULL) - goto err; - - if (!BN_lshift1(x, sqrt)) - goto err; - - if (BN_ucmp(x, p) > 0) { - if (!BN_usub(sqrt, p, sqrt)) - goto err; - } - - ret = 1; - - err: - BN_CTX_end(ctx); - - return ret; -} - -/* - * Verify that a = (sqrt_a)^2 (mod p). Requires that a is reduced (mod p). - */ - -static int -bn_mod_sqrt_verify(const BIGNUM *a, const BIGNUM *sqrt_a, const BIGNUM *p, - BN_CTX *ctx) -{ - BIGNUM *x; - int ret = 0; - - BN_CTX_start(ctx); - - if ((x = BN_CTX_get(ctx)) == NULL) - goto err; - - if (!BN_mod_sqr(x, sqrt_a, p, ctx)) - goto err; - - if (BN_cmp(x, a) != 0) { - BNerror(BN_R_NOT_A_SQUARE); - goto err; - } - - ret = 1; - - err: - BN_CTX_end(ctx); - - return ret; -} - -static int -bn_mod_sqrt_internal(BIGNUM *out_sqrt, const BIGNUM *a, const BIGNUM *p, - BN_CTX *ctx) -{ - BIGNUM *a_mod_p, *sqrt; - BN_ULONG lsw; - int done; - int kronecker; - int ret = 0; - - BN_CTX_start(ctx); - - if ((a_mod_p = BN_CTX_get(ctx)) == NULL) - goto err; - if ((sqrt = BN_CTX_get(ctx)) == NULL) - goto err; - - if (!BN_nnmod(a_mod_p, a, p, ctx)) - goto err; - - if (!bn_mod_sqrt_trivial_cases(&done, sqrt, a_mod_p, p, ctx)) - goto err; - if (done) - goto verify; - - /* - * Make sure that the Kronecker symbol (a/p) == 1. In case p is prime - * this is equivalent to a having a square root (mod p). The cost of - * BN_kronecker() is O(log^2(n)). This is small compared to the cost - * O(log^4(n)) of Tonelli-Shanks. - */ - - if ((kronecker = BN_kronecker(a_mod_p, p, ctx)) == -2) - goto err; - if (kronecker <= 0) { - /* This error is only accurate if p is known to be a prime. */ - BNerror(BN_R_NOT_A_SQUARE); - goto err; - } - - lsw = BN_lsw(p); - - if (lsw % 4 == 3) { - if (!bn_mod_sqrt_p_is_3_mod_4(sqrt, a_mod_p, p, ctx)) - goto err; - } else if (lsw % 8 == 5) { - if (!bn_mod_sqrt_p_is_5_mod_8(sqrt, a_mod_p, p, ctx)) - goto err; - } else if (lsw % 8 == 1) { - if (!bn_mod_sqrt_p_is_1_mod_8(sqrt, a_mod_p, p, ctx)) - goto err; - } else { - /* Impossible to hit since the trivial cases ensure p is odd. */ - BNerror(BN_R_P_IS_NOT_PRIME); - goto err; - } - - if (!bn_mod_sqrt_normalize(sqrt, p, ctx)) - goto err; - - verify: - if (!bn_mod_sqrt_verify(a_mod_p, sqrt, p, ctx)) - goto err; - - if (!bn_copy(out_sqrt, sqrt)) - goto err; - - ret = 1; - - err: - BN_CTX_end(ctx); - - return ret; -} - -BIGNUM * -BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) -{ - BIGNUM *out_sqrt; - - if ((out_sqrt = in) == NULL) - out_sqrt = BN_new(); - if (out_sqrt == NULL) - goto err; - - if (!bn_mod_sqrt_internal(out_sqrt, a, p, ctx)) - goto err; - - return out_sqrt; - - err: - if (out_sqrt != in) - BN_free(out_sqrt); - - return NULL; -} -LCRYPTO_ALIAS(BN_mod_sqrt); -- cgit v1.2.3-55-g6feb