Commit 80d89e6a authored by Bodo Möller's avatar Bodo Möller
Browse files

Sign-related fixes (and tests). 

BN_mod_exp_mont does not work properly yet if modulus m
is negative (we want computations to be carried out
modulo |m|).
parent bc5f2740

CHANGES

+4 −0
Original line number Diff line number Diff line
@@ -3,6 +3,10 @@ 


 Changes between 0.9.6 and 0.9.7  [xx XXX 2000]



  *) BN_div bugfix: If the result is 0, the sign (res->neg) must not be

     set.

     [Bodo Moeller]



  *) Changed the LHASH code to use prototypes for callbacks, and created

     macros to declare and implement thin (optionally static) functions

     that provide type-safety and avoid function pointer casting for the

crypto/bn/bn_div.c

+2 −0
Original line number Diff line number Diff line
@@ -241,6 +241,8 @@ int BN_div(BIGNUM *dv, BIGNUM *rm, const BIGNUM *num, const BIGNUM *divisor, 
		}

	else

		res->top--;

	if (res->top == 0)

		res->neg = 0;

	resp--;



	for (i=0; i<loop-1; i++)

crypto/bn/bn_sqrt.c

+14 −15
Original line number Diff line number Diff line
@@ -133,21 +133,16 @@ 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;

		}

	/* 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

		/* 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.

		 *     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;
@@ -159,16 +154,16 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) 
	

	if (e == 2)

		{

		/* p == 5  (mod 8)

		/* |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),

		 *      b := (2*a)^((|p|-5)/8),

		 *      i := (2*a)*b^2

		 * we have

		 *     i^2 = (2*a)^((1 + (p-5)/4)*2)

		 *     i^2 = (2*a)^((1 + (|p|-5)/4)*2)

		 *         = (2*a)^((p-1)/2)

		 *         = -1;

		 * so if we set
@@ -195,7 +190,7 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) 
		/* t := 2*a */

		if (!BN_mod_lshift1_quick(t, a, p)) goto end;



		/* b := (2*a)^((p-5)/8) */

		/* 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;
@@ -218,6 +213,8 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) 
	

	/* 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

		{
@@ -240,7 +237,7 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) 
				if (!BN_set_word(y, i)) goto end;

			}

		

		r = BN_kronecker(y, p, ctx);

		r = BN_kronecker(y, q, ctx); /* here 'q' is |p| */

		if (r < -1) goto end;

		if (r == 0)

			{
@@ -262,6 +259,8 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx) 
		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. */

crypto/bn/bntest.c

+14 −2
Original line number Diff line number Diff line
@@ -907,6 +907,7 @@ int test_kron(BIO *bp, BN_CTX *ctx) 
	 * works.) */



	if (!BN_generate_prime(b, 512, 0, NULL, NULL, genprime_cb, NULL)) goto err;

	b->neg = rand_neg();

	putc('\n', stderr);



	for (i = 0; i < num0; i++)
@@ -914,12 +915,17 @@ int test_kron(BIO *bp, BN_CTX *ctx) 
		if (!BN_bntest_rand(a, 512, 0, 0)) goto err;

		a->neg = rand_neg();



		/* t := (b-1)/2  (note that b is odd) */

		/* t := (|b|-1)/2  (note that b is odd) */

		if (!BN_copy(t, b)) goto err;

		t->neg = 0;

		if (!BN_sub_word(t, 1)) goto err;

		if (!BN_rshift1(t, t)) goto err;

		/* r := a^t mod b */

		if (!BN_mod_exp(r, a, t, b, ctx)) goto err;

		/* FIXME: Using BN_mod_exp (Montgomery variant) leads to

		 * incorrect results if  b  is negative ("Legendre symbol

		 * computation failed").

		 * We want computations to be carried out modulo |b|. */

		if (!BN_mod_exp_simple(r, a, t, b, ctx)) goto err;



		if (BN_is_word(r, 1))

			legendre = 1;
@@ -938,6 +944,9 @@ int test_kron(BIO *bp, BN_CTX *ctx) 
		

		kronecker = BN_kronecker(a, b, ctx);

		if (kronecker < -1) goto err;

		/* we actually need BN_kronecker(a, |b|) */

		if (a->neg && b->neg)

			kronecker = -kronecker;

		

		if (legendre != kronecker)

			{
@@ -991,6 +1000,7 @@ int test_sqrt(BIO *bp, BN_CTX *ctx) 
			if (!BN_generate_prime(p, 256, 0, a, r, genprime_cb, NULL)) goto err;

			putc('\n', stderr);

			}

		p->neg = rand_neg();



		for (j = 0; j < num2; j++)

			{
@@ -1003,6 +1013,8 @@ int test_sqrt(BIO *bp, BN_CTX *ctx) 
			if (!BN_nnmod(a, a, p, ctx)) goto err;

			if (!BN_mod_sqr(a, a, p, ctx)) goto err;

			if (!BN_mul(a, a, r, ctx)) goto err;

			if (rand_neg())

				if (!BN_sub(a, a, p)) goto err;



			if (!BN_mod_sqrt(r, a, p, ctx)) goto err;

			if (!BN_mod_sqr(r, r, p, ctx)) goto err;