index d5caf5136f1248007dc6bb6f215665351cc055a9..2c6fb2379a99328add8f52434e3c50efcd2d39d9 100644 (file)
@@ -56,7 +56,7 @@
* [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
@@ -240,131 +240,216 @@ BIGNUM *BN_mod_inverse(BIGNUM *in,
/* From  B = a mod |n|,  A = |n|  it follows that
*
*      0 <= B < A,
-        *      sign*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) <= 400))
{
-               BIGNUM *tmp;
+               /* Binary inversion algorithm; requires odd modulus.
+                * This is faster than the general algorithm if the modulus
+                * is sufficiently small. */
+               int shift;
+
+               while (!BN_is_zero(B))
+                       {
+                       /*
+                        *      0 < B < 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,
-                * (*)  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))
-                       {
-                       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), but  A  may no longer be larger than  B.
+                        * Both  A  and  B  are odd.
+                        * The following computations ensure that
+                        *
+                        *      0 =< B < A = |n|,
+                        * (1) -sign*X*a  ==  B   (mod |n|),
+                        * (2)  sign*Y*a  ==  A   (mod |n|)
+                        */
+                       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 (less efficient than binary inversion) */
+
+               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
-                *       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_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 (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.
*/

@@ -378,7 +463,7 @@ BIGNUM *BN_mod_inverse(BIGNUM *in,
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;
}