Commit bac68541 authored by Bodo Möller's avatar Bodo Möller
Browse files

Faster BN_mod_sqrt algorithm for p == 5 (8).

parent 1a4d6400

CHANGES

+1 −1
Original line number Diff line number Diff line
@@ -32,7 +32,7 @@ 
     [Richard Levitte]



  *) New function BN_mod_sqrt for computing square roots modulo a prime

     (Tonelli-Shanks algorithm).

     (Tonelli-Shanks algorithm unless  p == 3 (mod 4)  or  p == 5 (mod 8)).

     [Lenka Fibikova <fibikova@exp-math.uni-essen.de>, Bodo Moeller]



  *) Store verify_result within SSL_SESSION also for client side to

crypto/bn/bn_sqrt.c

+79 −5
Original line number Diff line number Diff line
@@ -93,6 +93,20 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) 
		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)))

			{

			BN_free(ret);

			return NULL;

			}

		return ret;

		}



#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;
@@ -119,6 +133,8 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) 
	e = 1;

	while (!BN_is_bit_set(p, e))

		e++;

	if (e > 2)

		/* we don't need this  q  if  e = 1 or 2 */

		if (!BN_rshift(q, p, e)) goto end;

	q->neg = 0;
@@ -129,16 +145,74 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) 
		 * 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.

		 * so we can use exponent  (p+1)/4,  i.e.  (p-3)/4 + 1.

		 */

		if (!BN_rshift(q, p, 2)) goto end;

		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.

	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.)

		 */



		/* make sure that  a  is reduced modulo p */

		if (a->neg || BN_ucmp(a, p) >= 0)

			{

			if (!BN_nnmod(x, a, p, ctx)) goto end;

			a = x; /* use x as temporary variable */

			}



		/* 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;

		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; /* cannot become negative */



		/* 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 end;

		}

	

	/* e > 2, so we really have to use the Tonelli/Shanks algorithm.

	 * First, find some  y  that is not a square. */

	i = 2;

	do