diff options
Diffstat (limited to 'openssl/crypto/bn/bn_sqrt.c')
| -rw-r--r-- | openssl/crypto/bn/bn_sqrt.c | 678 |
1 files changed, 347 insertions, 331 deletions
diff --git a/openssl/crypto/bn/bn_sqrt.c b/openssl/crypto/bn/bn_sqrt.c index 6beaf9e5e..232af99a2 100644 --- a/openssl/crypto/bn/bn_sqrt.c +++ b/openssl/crypto/bn/bn_sqrt.c @@ -1,6 +1,8 @@ /* crypto/bn/bn_sqrt.c */ -/* Written by Lenka Fibikova <fibikova@exp-math.uni-essen.de> - * and Bodo Moeller for the OpenSSL project. */ +/* + * Written by Lenka Fibikova <fibikova@exp-math.uni-essen.de> and Bodo + * Moeller for the OpenSSL project. + */ /* ==================================================================== * Copyright (c) 1998-2000 The OpenSSL Project. All rights reserved. * @@ -9,7 +11,7 @@ * are met: * * 1. Redistributions of source code must retain the above copyright - * notice, this list of conditions and the following disclaimer. + * notice, this list of conditions and the following disclaimer. * * 2. Redistributions in binary form must reproduce the above copyright * notice, this list of conditions and the following disclaimer in @@ -58,336 +60,350 @@ #include "cryptlib.h" #include "bn_lcl.h" - -BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) -/* Returns 'ret' such that - * ret^2 == a (mod p), - * using the Tonelli/Shanks algorithm (cf. Henri Cohen, "A Course - * in Algebraic Computational Number Theory", algorithm 1.5.1). - * 'p' must be prime! +BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) +/* + * Returns 'ret' such that ret^2 == a (mod p), using the Tonelli/Shanks + * algorithm (cf. Henri Cohen, "A Course in Algebraic Computational Number + * Theory", algorithm 1.5.1). 'p' must be prime! */ - { - BIGNUM *ret = in; - int err = 1; - int r; - BIGNUM *A, *b, *q, *t, *x, *y; - int e, i, j; - - if (!BN_is_odd(p) || BN_abs_is_word(p, 1)) - { - if (BN_abs_is_word(p, 2)) - { - if (ret == NULL) - ret = BN_new(); - if (ret == NULL) - goto end; - if (!BN_set_word(ret, BN_is_bit_set(a, 0))) - { - if (ret != in) - BN_free(ret); - return NULL; - } - bn_check_top(ret); - return ret; - } - - BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME); - return(NULL); - } - - if (BN_is_zero(a) || BN_is_one(a)) - { - if (ret == NULL) - ret = BN_new(); - if (ret == NULL) - goto end; - if (!BN_set_word(ret, BN_is_one(a))) - { - if (ret != in) - BN_free(ret); - return NULL; - } - bn_check_top(ret); - return ret; - } - - BN_CTX_start(ctx); - A = BN_CTX_get(ctx); - b = BN_CTX_get(ctx); - q = BN_CTX_get(ctx); - t = BN_CTX_get(ctx); - x = BN_CTX_get(ctx); - y = BN_CTX_get(ctx); - if (y == NULL) goto end; - - if (ret == NULL) - ret = BN_new(); - if (ret == NULL) goto end; - - /* A = a mod p */ - if (!BN_nnmod(A, a, p, ctx)) goto end; - - /* now write |p| - 1 as 2^e*q where q is odd */ - e = 1; - while (!BN_is_bit_set(p, e)) - e++; - /* we'll set q later (if needed) */ - - if (e == 1) - { - /* The easy case: (|p|-1)/2 is odd, so 2 has an inverse - * modulo (|p|-1)/2, and square roots can be computed - * directly by modular exponentiation. - * We have - * 2 * (|p|+1)/4 == 1 (mod (|p|-1)/2), - * so we can use exponent (|p|+1)/4, i.e. (|p|-3)/4 + 1. - */ - if (!BN_rshift(q, p, 2)) goto end; - q->neg = 0; - if (!BN_add_word(q, 1)) goto end; - if (!BN_mod_exp(ret, A, q, p, ctx)) goto end; - err = 0; - goto vrfy; - } - - if (e == 2) - { - /* |p| == 5 (mod 8) - * - * In this case 2 is always a non-square since - * Legendre(2,p) = (-1)^((p^2-1)/8) for any odd prime. - * So if a really is a square, then 2*a is a non-square. - * Thus for - * b := (2*a)^((|p|-5)/8), - * i := (2*a)*b^2 - * we have - * i^2 = (2*a)^((1 + (|p|-5)/4)*2) - * = (2*a)^((p-1)/2) - * = -1; - * so if we set - * x := a*b*(i-1), - * then - * x^2 = a^2 * b^2 * (i^2 - 2*i + 1) - * = a^2 * b^2 * (-2*i) - * = a*(-i)*(2*a*b^2) - * = a*(-i)*i - * = a. - * - * (This is due to A.O.L. Atkin, - * <URL: http://listserv.nodak.edu/scripts/wa.exe?A2=ind9211&L=nmbrthry&O=T&P=562>, - * November 1992.) - */ - - /* t := 2*a */ - if (!BN_mod_lshift1_quick(t, A, p)) goto end; - - /* b := (2*a)^((|p|-5)/8) */ - if (!BN_rshift(q, p, 3)) goto end; - q->neg = 0; - if (!BN_mod_exp(b, t, q, p, ctx)) goto end; - - /* y := b^2 */ - if (!BN_mod_sqr(y, b, p, ctx)) goto end; - - /* t := (2*a)*b^2 - 1*/ - if (!BN_mod_mul(t, t, y, p, ctx)) goto end; - if (!BN_sub_word(t, 1)) goto end; - - /* x = a*b*t */ - if (!BN_mod_mul(x, A, b, p, ctx)) goto end; - if (!BN_mod_mul(x, x, t, p, ctx)) goto end; - - if (!BN_copy(ret, x)) goto end; - err = 0; - goto vrfy; - } - - /* e > 2, so we really have to use the Tonelli/Shanks algorithm. - * First, find some y that is not a square. */ - if (!BN_copy(q, p)) goto end; /* use 'q' as temp */ - q->neg = 0; - i = 2; - do - { - /* For efficiency, try small numbers first; - * if this fails, try random numbers. - */ - if (i < 22) - { - if (!BN_set_word(y, i)) goto end; - } - else - { - if (!BN_pseudo_rand(y, BN_num_bits(p), 0, 0)) goto end; - if (BN_ucmp(y, p) >= 0) - { - if (!(p->neg ? BN_add : BN_sub)(y, y, p)) goto end; - } - /* now 0 <= y < |p| */ - if (BN_is_zero(y)) - if (!BN_set_word(y, i)) goto end; - } - - r = BN_kronecker(y, q, ctx); /* here 'q' is |p| */ - if (r < -1) goto end; - if (r == 0) - { - /* m divides p */ - BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME); - goto end; - } - } - while (r == 1 && ++i < 82); - - if (r != -1) - { - /* Many rounds and still no non-square -- this is more likely - * a bug than just bad luck. - * Even if p is not prime, we should have found some y - * such that r == -1. - */ - BNerr(BN_F_BN_MOD_SQRT, BN_R_TOO_MANY_ITERATIONS); - goto end; - } - - /* Here's our actual 'q': */ - if (!BN_rshift(q, q, e)) goto end; - - /* Now that we have some non-square, we can find an element - * of order 2^e by computing its q'th power. */ - if (!BN_mod_exp(y, y, q, p, ctx)) goto end; - if (BN_is_one(y)) - { - BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME); - goto end; - } - - /* Now we know that (if p is indeed prime) there is an integer - * k, 0 <= k < 2^e, such that - * - * a^q * y^k == 1 (mod p). - * - * As a^q is a square and y is not, k must be even. - * q+1 is even, too, so there is an element - * - * X := a^((q+1)/2) * y^(k/2), - * - * and it satisfies - * - * X^2 = a^q * a * y^k - * = a, - * - * so it is the square root that we are looking for. - */ - - /* t := (q-1)/2 (note that q is odd) */ - if (!BN_rshift1(t, q)) goto end; - - /* x := a^((q-1)/2) */ - if (BN_is_zero(t)) /* special case: p = 2^e + 1 */ - { - if (!BN_nnmod(t, A, p, ctx)) goto end; - if (BN_is_zero(t)) - { - /* special case: a == 0 (mod p) */ - BN_zero(ret); - err = 0; - goto end; - } - else - if (!BN_one(x)) goto end; - } - else - { - if (!BN_mod_exp(x, A, t, p, ctx)) goto end; - if (BN_is_zero(x)) - { - /* special case: a == 0 (mod p) */ - BN_zero(ret); - err = 0; - goto end; - } - } - - /* b := a*x^2 (= a^q) */ - if (!BN_mod_sqr(b, x, p, ctx)) goto end; - if (!BN_mod_mul(b, b, A, p, ctx)) goto end; - - /* x := a*x (= a^((q+1)/2)) */ - if (!BN_mod_mul(x, x, A, p, ctx)) goto end; - - while (1) - { - /* Now b is a^q * y^k for some even k (0 <= k < 2^E - * where E refers to the original value of e, which we - * don't keep in a variable), and x is a^((q+1)/2) * y^(k/2). - * - * We have a*b = x^2, - * y^2^(e-1) = -1, - * b^2^(e-1) = 1. - */ - - if (BN_is_one(b)) - { - if (!BN_copy(ret, x)) goto end; - err = 0; - goto vrfy; - } - - - /* find smallest i such that b^(2^i) = 1 */ - i = 1; - if (!BN_mod_sqr(t, b, p, ctx)) goto end; - while (!BN_is_one(t)) - { - i++; - if (i == e) - { - BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE); - goto end; - } - if (!BN_mod_mul(t, t, t, p, ctx)) goto end; - } - - - /* t := y^2^(e - i - 1) */ - if (!BN_copy(t, y)) goto end; - for (j = e - i - 1; j > 0; j--) - { - if (!BN_mod_sqr(t, t, p, ctx)) goto end; - } - if (!BN_mod_mul(y, t, t, p, ctx)) goto end; - if (!BN_mod_mul(x, x, t, p, ctx)) goto end; - if (!BN_mod_mul(b, b, y, p, ctx)) goto end; - e = i; - } +{ + BIGNUM *ret = in; + int err = 1; + int r; + BIGNUM *A, *b, *q, *t, *x, *y; + int e, i, j; + + if (!BN_is_odd(p) || BN_abs_is_word(p, 1)) { + if (BN_abs_is_word(p, 2)) { + if (ret == NULL) + ret = BN_new(); + if (ret == NULL) + goto end; + if (!BN_set_word(ret, BN_is_bit_set(a, 0))) { + if (ret != in) + BN_free(ret); + return NULL; + } + bn_check_top(ret); + return ret; + } + + BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME); + return (NULL); + } + + if (BN_is_zero(a) || BN_is_one(a)) { + if (ret == NULL) + ret = BN_new(); + if (ret == NULL) + goto end; + if (!BN_set_word(ret, BN_is_one(a))) { + if (ret != in) + BN_free(ret); + return NULL; + } + bn_check_top(ret); + return ret; + } + + BN_CTX_start(ctx); + A = BN_CTX_get(ctx); + b = BN_CTX_get(ctx); + q = BN_CTX_get(ctx); + t = BN_CTX_get(ctx); + x = BN_CTX_get(ctx); + y = BN_CTX_get(ctx); + if (y == NULL) + goto end; + + if (ret == NULL) + ret = BN_new(); + if (ret == NULL) + goto end; + + /* A = a mod p */ + if (!BN_nnmod(A, a, p, ctx)) + goto end; + + /* now write |p| - 1 as 2^e*q where q is odd */ + e = 1; + while (!BN_is_bit_set(p, e)) + e++; + /* we'll set q later (if needed) */ + + if (e == 1) { + /*- + * The easy case: (|p|-1)/2 is odd, so 2 has an inverse + * modulo (|p|-1)/2, and square roots can be computed + * directly by modular exponentiation. + * We have + * 2 * (|p|+1)/4 == 1 (mod (|p|-1)/2), + * so we can use exponent (|p|+1)/4, i.e. (|p|-3)/4 + 1. + */ + if (!BN_rshift(q, p, 2)) + goto end; + q->neg = 0; + if (!BN_add_word(q, 1)) + goto end; + if (!BN_mod_exp(ret, A, q, p, ctx)) + goto end; + err = 0; + goto vrfy; + } + + if (e == 2) { + /*- + * |p| == 5 (mod 8) + * + * In this case 2 is always a non-square since + * Legendre(2,p) = (-1)^((p^2-1)/8) for any odd prime. + * So if a really is a square, then 2*a is a non-square. + * Thus for + * b := (2*a)^((|p|-5)/8), + * i := (2*a)*b^2 + * we have + * i^2 = (2*a)^((1 + (|p|-5)/4)*2) + * = (2*a)^((p-1)/2) + * = -1; + * so if we set + * x := a*b*(i-1), + * then + * x^2 = a^2 * b^2 * (i^2 - 2*i + 1) + * = a^2 * b^2 * (-2*i) + * = a*(-i)*(2*a*b^2) + * = a*(-i)*i + * = a. + * + * (This is due to A.O.L. Atkin, + * <URL: http://listserv.nodak.edu/scripts/wa.exe?A2=ind9211&L=nmbrthry&O=T&P=562>, + * November 1992.) + */ + + /* t := 2*a */ + if (!BN_mod_lshift1_quick(t, A, p)) + goto end; + + /* b := (2*a)^((|p|-5)/8) */ + if (!BN_rshift(q, p, 3)) + goto end; + q->neg = 0; + if (!BN_mod_exp(b, t, q, p, ctx)) + goto end; + + /* y := b^2 */ + if (!BN_mod_sqr(y, b, p, ctx)) + goto end; + + /* t := (2*a)*b^2 - 1 */ + if (!BN_mod_mul(t, t, y, p, ctx)) + goto end; + if (!BN_sub_word(t, 1)) + goto end; + + /* x = a*b*t */ + if (!BN_mod_mul(x, A, b, p, ctx)) + goto end; + if (!BN_mod_mul(x, x, t, p, ctx)) + goto end; + + if (!BN_copy(ret, x)) + goto end; + err = 0; + goto vrfy; + } + + /* + * e > 2, so we really have to use the Tonelli/Shanks algorithm. First, + * find some y that is not a square. + */ + if (!BN_copy(q, p)) + goto end; /* use 'q' as temp */ + q->neg = 0; + i = 2; + do { + /* + * For efficiency, try small numbers first; if this fails, try random + * numbers. + */ + if (i < 22) { + if (!BN_set_word(y, i)) + goto end; + } else { + if (!BN_pseudo_rand(y, BN_num_bits(p), 0, 0)) + goto end; + if (BN_ucmp(y, p) >= 0) { + if (!(p->neg ? BN_add : BN_sub) (y, y, p)) + goto end; + } + /* now 0 <= y < |p| */ + if (BN_is_zero(y)) + if (!BN_set_word(y, i)) + goto end; + } + + r = BN_kronecker(y, q, ctx); /* here 'q' is |p| */ + if (r < -1) + goto end; + if (r == 0) { + /* m divides p */ + BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME); + goto end; + } + } + while (r == 1 && ++i < 82); + + if (r != -1) { + /* + * Many rounds and still no non-square -- this is more likely a bug + * than just bad luck. Even if p is not prime, we should have found + * some y such that r == -1. + */ + BNerr(BN_F_BN_MOD_SQRT, BN_R_TOO_MANY_ITERATIONS); + goto end; + } + + /* Here's our actual 'q': */ + if (!BN_rshift(q, q, e)) + goto end; + + /* + * Now that we have some non-square, we can find an element of order 2^e + * by computing its q'th power. + */ + if (!BN_mod_exp(y, y, q, p, ctx)) + goto end; + if (BN_is_one(y)) { + BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME); + goto end; + } + + /*- + * Now we know that (if p is indeed prime) there is an integer + * k, 0 <= k < 2^e, such that + * + * a^q * y^k == 1 (mod p). + * + * As a^q is a square and y is not, k must be even. + * q+1 is even, too, so there is an element + * + * X := a^((q+1)/2) * y^(k/2), + * + * and it satisfies + * + * X^2 = a^q * a * y^k + * = a, + * + * so it is the square root that we are looking for. + */ + + /* t := (q-1)/2 (note that q is odd) */ + if (!BN_rshift1(t, q)) + goto end; + + /* x := a^((q-1)/2) */ + if (BN_is_zero(t)) { /* special case: p = 2^e + 1 */ + if (!BN_nnmod(t, A, p, ctx)) + goto end; + if (BN_is_zero(t)) { + /* special case: a == 0 (mod p) */ + BN_zero(ret); + err = 0; + goto end; + } else if (!BN_one(x)) + goto end; + } else { + if (!BN_mod_exp(x, A, t, p, ctx)) + goto end; + if (BN_is_zero(x)) { + /* special case: a == 0 (mod p) */ + BN_zero(ret); + err = 0; + goto end; + } + } + + /* b := a*x^2 (= a^q) */ + if (!BN_mod_sqr(b, x, p, ctx)) + goto end; + if (!BN_mod_mul(b, b, A, p, ctx)) + goto end; + + /* x := a*x (= a^((q+1)/2)) */ + if (!BN_mod_mul(x, x, A, p, ctx)) + goto end; + + while (1) { + /*- + * Now b is a^q * y^k for some even k (0 <= k < 2^E + * where E refers to the original value of e, which we + * don't keep in a variable), and x is a^((q+1)/2) * y^(k/2). + * + * We have a*b = x^2, + * y^2^(e-1) = -1, + * b^2^(e-1) = 1. + */ + + if (BN_is_one(b)) { + if (!BN_copy(ret, x)) + goto end; + err = 0; + goto vrfy; + } + + /* find smallest i such that b^(2^i) = 1 */ + i = 1; + if (!BN_mod_sqr(t, b, p, ctx)) + goto end; + while (!BN_is_one(t)) { + i++; + if (i == e) { + BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE); + goto end; + } + if (!BN_mod_mul(t, t, t, p, ctx)) + goto end; + } + + /* t := y^2^(e - i - 1) */ + if (!BN_copy(t, y)) + goto end; + for (j = e - i - 1; j > 0; j--) { + if (!BN_mod_sqr(t, t, p, ctx)) + goto end; + } + if (!BN_mod_mul(y, t, t, p, ctx)) + goto end; + if (!BN_mod_mul(x, x, t, p, ctx)) + goto end; + if (!BN_mod_mul(b, b, y, p, ctx)) + goto end; + e = i; + } vrfy: - if (!err) - { - /* verify the result -- the input might have been not a square - * (test added in 0.9.8) */ - - if (!BN_mod_sqr(x, ret, p, ctx)) - err = 1; - - if (!err && 0 != BN_cmp(x, A)) - { - BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE); - err = 1; - } - } + if (!err) { + /* + * verify the result -- the input might have been not a square (test + * added in 0.9.8) + */ + + if (!BN_mod_sqr(x, ret, p, ctx)) + err = 1; + + if (!err && 0 != BN_cmp(x, A)) { + BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE); + err = 1; + } + } end: - if (err) - { - if (ret != NULL && ret != in) - { - BN_clear_free(ret); - } - ret = NULL; - } - BN_CTX_end(ctx); - bn_check_top(ret); - return ret; - } + if (err) { + if (ret != NULL && ret != in) { + BN_clear_free(ret); + } + ret = NULL; + } + BN_CTX_end(ctx); + bn_check_top(ret); + return ret; +} |