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authorBodo Möller <bodo@openssl.org>2000-12-06 12:25:33 +0000
committerBodo Möller <bodo@openssl.org>2000-12-06 12:25:33 +0000
commitbac685417a0890642bada45bfaf1e44de9c92beb (patch)
treea82d422df98406d98c90bc2ab218656054d27c98 /crypto/bn/bn_sqrt.c
parent1a4d6400ae84ebaaadbd46227d22a7f2fa3ac607 (diff)
Faster BN_mod_sqrt algorithm for p == 5 (8).
Diffstat (limited to 'crypto/bn/bn_sqrt.c')
-rw-r--r--crypto/bn/bn_sqrt.c84
1 files changed, 79 insertions, 5 deletions
diff --git a/crypto/bn/bn_sqrt.c b/crypto/bn/bn_sqrt.c
index 2a72c189cb..a54d9d2919 100644
--- a/crypto/bn/bn_sqrt.c
+++ b/crypto/bn/bn_sqrt.c
@@ -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;
@@ -119,7 +133,9 @@ 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 (!BN_rshift(q, p, e)) goto end;
+ 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;
if (e == 1)
@@ -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_add_word(q,1)) goto end;
- if (!BN_rshift1(q,q)) goto end;
+ if (!BN_rshift(q, p, 2)) goto end;
+ if (!BN_add_word(q, 1)) 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