openssl/crypto/bn/bn_sqrt.c

/* crypto/bn/bn_mod.c */
/* 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.
 *
 * Redistribution and use in source and binary forms, with or without
 * modification, are permitted provided that the following conditions
 * are met:
 *
 * 1. Redistributions of source code must retain the above copyright
 *    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
 *    the documentation and/or other materials provided with the
 *    distribution.
 *
 * 3. All advertising materials mentioning features or use of this
 *    software must display the following acknowledgment:
 *    "This product includes software developed by the OpenSSL Project
 *    for use in the OpenSSL Toolkit. (http://www.openssl.org/)"
 *
 * 4. The names "OpenSSL Toolkit" and "OpenSSL Project" must not be used to
 *    endorse or promote products derived from this software without
 *    prior written permission. For written permission, please contact
 *    openssl-core@openssl.org.
 *
 * 5. Products derived from this software may not be called "OpenSSL"
 *    nor may "OpenSSL" appear in their names without prior written
 *    permission of the OpenSSL Project.
 *
 * 6. Redistributions of any form whatsoever must retain the following
 *    acknowledgment:
 *    "This product includes software developed by the OpenSSL Project
 *    for use in the OpenSSL Toolkit (http://www.openssl.org/)"
 *
 * THIS SOFTWARE IS PROVIDED BY THE OpenSSL PROJECT ``AS IS'' AND ANY
 * EXPRESSED OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR
 * PURPOSE ARE DISCLAIMED.  IN NO EVENT SHALL THE OpenSSL PROJECT OR
 * ITS CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL,
 * SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT
 * NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES;
 * LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT,
 * STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
 * ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED
 * OF THE POSSIBILITY OF SUCH DAMAGE.
 * ====================================================================
 *
 * This product includes cryptographic software written by Eric Young
 * (eay@cryptsoft.com).  This product includes software written by Tim
 * Hudson (tjh@cryptsoft.com).
 *
 */

#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 *ret = in;
	int err = 1;
	int r;
	BIGNUM *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)))
				{
				BN_free(ret);
				return NULL;
				}
			return ret;
			}

		BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
		return(NULL);
		}

#if 0 /* if BN_mod_sqrt is used with correct input, this just wastes time */
	r = BN_kronecker(a, p, ctx);
	if (r < -1) return NULL;
	if (r == -1)
		{
		BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
		return(NULL);
		}
#endif

	BN_CTX_start(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;

	/* now write  |p| - 1  as  2^e*q  where  q  is odd */
	e = 1;
	while (!BN_is_bit_set(p, e))
		e++;
	if (!BN_rshift(q, p, e)) goto end;
	q->neg = 0;

	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.  (q+1)/2.
		 */
		if (!BN_add_word(q,1)) goto end;
		if (!BN_rshift1(q,q)) goto end;
		if (!BN_mod_exp(ret, a, q, p, ctx)) goto end;
		err = 0;
		goto end;
		}
	
	/* e > 1, so we really have to use the Tonelli/Shanks algorithm.
	 * First, find some  y  that is not a square. */
	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, p, ctx);
		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;
		}


	/* 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) */
			if (!BN_zero(ret)) goto end;
			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) */
			if (!BN_zero(ret)) goto end;
			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 end;
			}


		/* 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;
		}

 end:
	if (err)
		{
		if (ret != NULL && ret != in)
			{
			BN_clear_free(ret);
			}
		ret = NULL;
		}
	BN_CTX_end(ctx);
	return ret;
	}