* [including the GNU Public Licence.]
*/
/* ====================================================================
- * Copyright (c) 1998-2000 The OpenSSL Project. All rights reserved.
+ * Copyright (c) 1998-2001 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
ret=1;
err:
BN_CTX_end(ctx);
+ bn_check_top(r);
return(ret);
}
{
if (!BN_lshift(a,a,shifts)) goto err;
}
+ bn_check_top(a);
return(a);
err:
return(NULL);
/* From B = a mod |n|, A = |n| it follows that
*
* 0 <= B < A,
- * X*a == B (mod |n|),
- * -sign*Y*a == A (mod |n|).
+ * -sign*X*a == B (mod |n|),
+ * sign*Y*a == A (mod |n|).
*/
- while (!BN_is_zero(B))
+ if (BN_is_odd(n) && (BN_num_bits(n) <= (BN_BITS <= 32 ? 450 : 2048)))
{
- BIGNUM *tmp;
+ /* Binary inversion algorithm; requires odd modulus.
+ * This is faster than the general algorithm if the modulus
+ * is sufficiently small (about 400 .. 500 bits on 32-bit
+ * sytems, but much more on 64-bit systems) */
+ int shift;
+
+ while (!BN_is_zero(B))
+ {
+ /*
+ * 0 < B < |n|,
+ * 0 < A <= |n|,
+ * (1) -sign*X*a == B (mod |n|),
+ * (2) sign*Y*a == A (mod |n|)
+ */
+
+ /* Now divide B by the maximum possible power of two in the integers,
+ * and divide X by the same value mod |n|.
+ * When we're done, (1) still holds. */
+ shift = 0;
+ while (!BN_is_bit_set(B, shift)) /* note that 0 < B */
+ {
+ shift++;
+
+ if (BN_is_odd(X))
+ {
+ if (!BN_uadd(X, X, n)) goto err;
+ }
+ /* now X is even, so we can easily divide it by two */
+ if (!BN_rshift1(X, X)) goto err;
+ }
+ if (shift > 0)
+ {
+ if (!BN_rshift(B, B, shift)) goto err;
+ }
- /*
- * 0 < B < A,
- * (*) X*a == B (mod |n|),
- * -sign*Y*a == A (mod |n|)
- */
- /* (D, M) := (A/B, A%B) ... */
- if (BN_num_bits(A) == BN_num_bits(B))
- {
- if (!BN_one(D)) goto err;
- if (!BN_sub(M,A,B)) goto err;
+ /* Same for A and Y. Afterwards, (2) still holds. */
+ shift = 0;
+ while (!BN_is_bit_set(A, shift)) /* note that 0 < A */
+ {
+ shift++;
+
+ if (BN_is_odd(Y))
+ {
+ if (!BN_uadd(Y, Y, n)) goto err;
+ }
+ /* now Y is even */
+ if (!BN_rshift1(Y, Y)) goto err;
+ }
+ if (shift > 0)
+ {
+ if (!BN_rshift(A, A, shift)) goto err;
+ }
+
+
+ /* We still have (1) and (2).
+ * Both A and B are odd.
+ * The following computations ensure that
+ *
+ * 0 <= B < |n|,
+ * 0 < A < |n|,
+ * (1) -sign*X*a == B (mod |n|),
+ * (2) sign*Y*a == A (mod |n|),
+ *
+ * and that either A or B is even in the next iteration.
+ */
+ if (BN_ucmp(B, A) >= 0)
+ {
+ /* -sign*(X + Y)*a == B - A (mod |n|) */
+ if (!BN_uadd(X, X, Y)) goto err;
+ /* NB: we could use BN_mod_add_quick(X, X, Y, n), but that
+ * actually makes the algorithm slower */
+ if (!BN_usub(B, B, A)) goto err;
+ }
+ else
+ {
+ /* sign*(X + Y)*a == A - B (mod |n|) */
+ if (!BN_uadd(Y, Y, X)) goto err;
+ /* as above, BN_mod_add_quick(Y, Y, X, n) would slow things down */
+ if (!BN_usub(A, A, B)) goto err;
+ }
}
- else if (BN_num_bits(A) == BN_num_bits(B) + 1)
+ }
+ else
+ {
+ /* general inversion algorithm */
+
+ while (!BN_is_zero(B))
{
- /* A/B is 1, 2, or 3 */
- if (!BN_lshift1(T,B)) goto err;
- if (BN_ucmp(A,T) < 0)
+ BIGNUM *tmp;
+
+ /*
+ * 0 < B < A,
+ * (*) -sign*X*a == B (mod |n|),
+ * sign*Y*a == A (mod |n|)
+ */
+
+ /* (D, M) := (A/B, A%B) ... */
+ if (BN_num_bits(A) == BN_num_bits(B))
{
- /* A < 2*B, so D=1 */
if (!BN_one(D)) goto err;
if (!BN_sub(M,A,B)) goto err;
}
- else
+ else if (BN_num_bits(A) == BN_num_bits(B) + 1)
{
- /* A >= 2*B, so D=2 or D=3 */
- if (!BN_sub(M,A,T)) goto err;
- if (!BN_add(D,T,B)) goto err; /* use D (:= 3*B) as temp */
- if (BN_ucmp(A,D) < 0)
+ /* A/B is 1, 2, or 3 */
+ if (!BN_lshift1(T,B)) goto err;
+ if (BN_ucmp(A,T) < 0)
{
- /* A < 3*B, so D=2 */
- if (!BN_set_word(D,2)) goto err;
- /* M (= A - 2*B) already has the correct value */
+ /* A < 2*B, so D=1 */
+ if (!BN_one(D)) goto err;
+ if (!BN_sub(M,A,B)) goto err;
}
else
{
- /* only D=3 remains */
- if (!BN_set_word(D,3)) goto err;
- /* currently M = A - 2*B, but we need M = A - 3*B */
- if (!BN_sub(M,M,B)) goto err;
+ /* A >= 2*B, so D=2 or D=3 */
+ if (!BN_sub(M,A,T)) goto err;
+ if (!BN_add(D,T,B)) goto err; /* use D (:= 3*B) as temp */
+ if (BN_ucmp(A,D) < 0)
+ {
+ /* A < 3*B, so D=2 */
+ if (!BN_set_word(D,2)) goto err;
+ /* M (= A - 2*B) already has the correct value */
+ }
+ else
+ {
+ /* only D=3 remains */
+ if (!BN_set_word(D,3)) goto err;
+ /* currently M = A - 2*B, but we need M = A - 3*B */
+ if (!BN_sub(M,M,B)) goto err;
+ }
}
}
- }
- else
- {
- if (!BN_div(D,M,A,B,ctx)) goto err;
- }
-
- /* Now
- * A = D*B + M;
- * thus we have
- * (**) -sign*Y*a == D*B + M (mod |n|).
- */
-
- tmp=A; /* keep the BIGNUM object, the value does not matter */
-
- /* (A, B) := (B, A mod B) ... */
- A=B;
- B=M;
- /* ... so we have 0 <= B < A again */
-
- /* Since the former M is now B and the former B is now A,
- * (**) translates into
- * -sign*Y*a == D*A + B (mod |n|),
- * i.e.
- * -sign*Y*a - D*A == B (mod |n|).
- * Similarly, (*) translates into
- * X*a == A (mod |n|).
- *
- * Thus,
- * -sign*Y*a - D*X*a == B (mod |n|),
- * i.e.
- * -sign*(Y + D*X)*a == B (mod |n|).
- *
- * So if we set (X, Y, sign) := (Y + D*X, X, -sign), we arrive back at
- * X*a == B (mod |n|),
- * -sign*Y*a == A (mod |n|).
- * Note that X and Y stay non-negative all the time.
- */
-
- /* most of the time D is very small, so we can optimize tmp := D*X+Y */
- if (BN_is_one(D))
- {
- if (!BN_add(tmp,X,Y)) goto err;
- }
- else
- {
- if (BN_is_word(D,2))
- {
- if (!BN_lshift1(tmp,X)) goto err;
- }
- else if (BN_is_word(D,4))
+ else
{
- if (!BN_lshift(tmp,X,2)) goto err;
+ if (!BN_div(D,M,A,B,ctx)) goto err;
}
- else if (D->top == 1)
+
+ /* Now
+ * A = D*B + M;
+ * thus we have
+ * (**) sign*Y*a == D*B + M (mod |n|).
+ */
+
+ tmp=A; /* keep the BIGNUM object, the value does not matter */
+
+ /* (A, B) := (B, A mod B) ... */
+ A=B;
+ B=M;
+ /* ... so we have 0 <= B < A again */
+
+ /* Since the former M is now B and the former B is now A,
+ * (**) translates into
+ * sign*Y*a == D*A + B (mod |n|),
+ * i.e.
+ * sign*Y*a - D*A == B (mod |n|).
+ * Similarly, (*) translates into
+ * -sign*X*a == A (mod |n|).
+ *
+ * Thus,
+ * sign*Y*a + D*sign*X*a == B (mod |n|),
+ * i.e.
+ * sign*(Y + D*X)*a == B (mod |n|).
+ *
+ * So if we set (X, Y, sign) := (Y + D*X, X, -sign), we arrive back at
+ * -sign*X*a == B (mod |n|),
+ * sign*Y*a == A (mod |n|).
+ * Note that X and Y stay non-negative all the time.
+ */
+
+ /* most of the time D is very small, so we can optimize tmp := D*X+Y */
+ if (BN_is_one(D))
{
- if (!BN_copy(tmp,X)) goto err;
- if (!BN_mul_word(tmp,D->d[0])) goto err;
+ if (!BN_add(tmp,X,Y)) goto err;
}
else
{
- if (!BN_mul(tmp,D,X,ctx)) goto err;
+ if (BN_is_word(D,2))
+ {
+ if (!BN_lshift1(tmp,X)) goto err;
+ }
+ else if (BN_is_word(D,4))
+ {
+ if (!BN_lshift(tmp,X,2)) goto err;
+ }
+ else if (D->top == 1)
+ {
+ if (!BN_copy(tmp,X)) goto err;
+ if (!BN_mul_word(tmp,D->d[0])) goto err;
+ }
+ else
+ {
+ if (!BN_mul(tmp,D,X,ctx)) goto err;
+ }
+ if (!BN_add(tmp,tmp,Y)) goto err;
}
- if (!BN_add(tmp,tmp,Y)) goto err;
+
+ M=Y; /* keep the BIGNUM object, the value does not matter */
+ Y=X;
+ X=tmp;
+ sign = -sign;
}
-
- M=Y; /* keep the BIGNUM object, the value does not matter */
- Y=X;
- X=tmp;
- sign = -sign;
}
-
+
/*
- * The while loop ends when
+ * The while loop (Euclid's algorithm) ends when
* A == gcd(a,n);
* we have
- * -sign*Y*a == A (mod |n|),
+ * sign*Y*a == A (mod |n|),
* where Y is non-negative.
*/
if (BN_is_one(A))
{
/* Y*a == 1 (mod |n|) */
- if (BN_ucmp(Y,n) < 0)
+ if (!Y->neg && BN_ucmp(Y,n) < 0)
{
if (!BN_copy(R,Y)) goto err;
}
err:
if ((ret == NULL) && (in == NULL)) BN_free(R);
BN_CTX_end(ctx);
+ bn_check_top(ret);
return(ret);
}
projects / openssl.git / blobdiff