/src/openssl/crypto/bn/bn_gcd.c

Source (jump to first uncovered line)
/*
 * Copyright 1995-2024 The OpenSSL Project Authors. All Rights Reserved.
 *
 * Licensed under the Apache License 2.0 (the "License").  You may not use
 * this file except in compliance with the License.  You can obtain a copy
 * in the file LICENSE in the source distribution or at
 * https://www.openssl.org/source/license.html
 */

#include "internal/cryptlib.h"
#include "bn_local.h"
#include "internal/constant_time.h"

/*
 * bn_mod_inverse_no_branch is a special version of BN_mod_inverse. It does
 * not contain branches that may leak sensitive information.
 *
 * This is a static function, we ensure all callers in this file pass valid
 * arguments: all passed pointers here are non-NULL.
 */
static ossl_inline
BIGNUM *bn_mod_inverse_no_branch(BIGNUM *in,
                                 const BIGNUM *a, const BIGNUM *n,
                                 BN_CTX *ctx, int *pnoinv)
{
    BIGNUM *A, *B, *X, *Y, *M, *D, *T, *R = NULL;
    BIGNUM *ret = NULL;
    int sign;

    bn_check_top(a);
    bn_check_top(n);

    BN_CTX_start(ctx);
    A = BN_CTX_get(ctx);
    B = BN_CTX_get(ctx);
    X = BN_CTX_get(ctx);
    D = BN_CTX_get(ctx);
    M = BN_CTX_get(ctx);
    Y = BN_CTX_get(ctx);
    T = BN_CTX_get(ctx);
    if (T == NULL)
        goto err;

    if (in == NULL)
        R = BN_new();
    else
        R = in;
    if (R == NULL)
        goto err;

    if (!BN_one(X))
        goto err;
    BN_zero(Y);
    if (BN_copy(B, a) == NULL)
        goto err;
    if (BN_copy(A, n) == NULL)
        goto err;
    A->neg = 0;

    if (B->neg || (BN_ucmp(B, A) >= 0)) {
        /*
         * Turn BN_FLG_CONSTTIME flag on, so that when BN_div is invoked,
         * BN_div_no_branch will be called eventually.
         */
         {
            BIGNUM local_B;
            bn_init(&local_B);
            BN_with_flags(&local_B, B, BN_FLG_CONSTTIME);
            if (!BN_nnmod(B, &local_B, A, ctx))
                goto err;
            /* Ensure local_B goes out of scope before any further use of B */
        }
    }
    sign = -1;
    /*-
     * 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|).
     */

    while (!BN_is_zero(B)) {
        BIGNUM *tmp;

        /*-
         *      0 < B < A,
         * (*) -sign*X*a  ==  B   (mod |n|),
         *      sign*Y*a  ==  A   (mod |n|)
         */

        /*
         * Turn BN_FLG_CONSTTIME flag on, so that when BN_div is invoked,
         * BN_div_no_branch will be called eventually.
         */
        {
            BIGNUM local_A;
            bn_init(&local_A);
            BN_with_flags(&local_A, A, BN_FLG_CONSTTIME);

            /* (D, M) := (A/B, A%B) ... */
            if (!BN_div(D, M, &local_A, B, ctx))
                goto err;
            /* Ensure local_A goes out of scope before any further use of A */
        }

        /*-
         * 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.
         */

        if (!BN_mul(tmp, D, X, ctx))
            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;
    }

    /*-
     * The while loop (Euclid's algorithm) ends when
     *      A == gcd(a,n);
     * we have
     *       sign*Y*a  ==  A  (mod |n|),
     * where  Y  is non-negative.
     */

    if (sign < 0) {
        if (!BN_sub(Y, n, Y))
            goto err;
    }
    /* Now  Y*a  ==  A  (mod |n|).  */

    if (BN_is_one(A)) {
        /* Y*a == 1  (mod |n|) */
        if (!Y->neg && BN_ucmp(Y, n) < 0) {
            if (!BN_copy(R, Y))
                goto err;
        } else {
            if (!BN_nnmod(R, Y, n, ctx))
                goto err;
        }
    } else {
        *pnoinv = 1;
        /* caller sets the BN_R_NO_INVERSE error */
        goto err;
    }

    ret = R;
    *pnoinv = 0;

 err:
    if ((ret == NULL) && (in == NULL))
        BN_free(R);
    BN_CTX_end(ctx);
    bn_check_top(ret);
    return ret;
}

/*
 * This is an internal function, we assume all callers pass valid arguments:
 * all pointers passed here are assumed non-NULL.
 */
BIGNUM *int_bn_mod_inverse(BIGNUM *in,
                           const BIGNUM *a, const BIGNUM *n, BN_CTX *ctx,
                           int *pnoinv)
{
    BIGNUM *A, *B, *X, *Y, *M, *D, *T, *R = NULL;
    BIGNUM *ret = NULL;
    int sign;

    /* This is invalid input so we don't worry about constant time here */
    if (BN_abs_is_word(n, 1) || BN_is_zero(n)) {
        *pnoinv = 1;
        return NULL;
    }

    *pnoinv = 0;

    if ((BN_get_flags(a, BN_FLG_CONSTTIME) != 0)
        || (BN_get_flags(n, BN_FLG_CONSTTIME) != 0)) {
        return bn_mod_inverse_no_branch(in, a, n, ctx, pnoinv);
    }

    bn_check_top(a);
    bn_check_top(n);

    BN_CTX_start(ctx);
    A = BN_CTX_get(ctx);
    B = BN_CTX_get(ctx);
    X = BN_CTX_get(ctx);
    D = BN_CTX_get(ctx);
    M = BN_CTX_get(ctx);
    Y = BN_CTX_get(ctx);
    T = BN_CTX_get(ctx);
    if (T == NULL)
        goto err;

    if (in == NULL)
        R = BN_new();
    else
        R = in;
    if (R == NULL)
        goto err;

    if (!BN_one(X))
        goto err;
    BN_zero(Y);
    if (BN_copy(B, a) == NULL)
        goto err;
    if (BN_copy(A, n) == NULL)
        goto err;
    A->neg = 0;
    if (B->neg || (BN_ucmp(B, A) >= 0)) {
        if (!BN_nnmod(B, B, A, ctx))
            goto err;
    }
    sign = -1;
    /*-
     * 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|).
     */

    if (BN_is_odd(n) && (BN_num_bits(n) <= 2048)) {
        /*
         * 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 systems, 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;
            }

            /*
             * 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 {
        /* general inversion algorithm */

        while (!BN_is_zero(B)) {
            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)) {
                if (!BN_one(D))
                    goto err;
                if (!BN_sub(M, A, B))
                    goto err;
            } else if (BN_num_bits(A) == BN_num_bits(B) + 1) {
                /* A/B is 1, 2, or 3 */
                if (!BN_lshift1(T, B))
                    goto err;
                if (BN_ucmp(A, T) < 0) {
                    /* A < 2*B, so D=1 */
                    if (!BN_one(D))
                        goto err;
                    if (!BN_sub(M, A, B))
                        goto err;
                } else {
                    /* 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)) {
                    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;
            }

            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|),
     * where  Y  is non-negative.
     */

    if (sign < 0) {
        if (!BN_sub(Y, n, Y))
            goto err;
    }
    /* Now  Y*a  ==  A  (mod |n|).  */

    if (BN_is_one(A)) {
        /* Y*a == 1  (mod |n|) */
        if (!Y->neg && BN_ucmp(Y, n) < 0) {
            if (!BN_copy(R, Y))
                goto err;
        } else {
            if (!BN_nnmod(R, Y, n, ctx))
                goto err;
        }
    } else {
        *pnoinv = 1;
        goto err;
    }
    ret = R;
 err:
    if ((ret == NULL) && (in == NULL))
        BN_free(R);
    BN_CTX_end(ctx);
    bn_check_top(ret);
    return ret;
}

/* solves ax == 1 (mod n) */
BIGNUM *BN_mod_inverse(BIGNUM *in,
                       const BIGNUM *a, const BIGNUM *n, BN_CTX *ctx)
{
    BN_CTX *new_ctx = NULL;
    BIGNUM *rv;
    int noinv = 0;

    if (ctx == NULL) {
        ctx = new_ctx = BN_CTX_new_ex(NULL);
        if (ctx == NULL) {
            ERR_raise(ERR_LIB_BN, ERR_R_BN_LIB);
            return NULL;
        }
    }

    rv = int_bn_mod_inverse(in, a, n, ctx, &noinv);
    if (noinv)
        ERR_raise(ERR_LIB_BN, BN_R_NO_INVERSE);
    BN_CTX_free(new_ctx);
    return rv;
}

/*
 * The numbers a and b are coprime if the only positive integer that is a
 * divisor of both of them is 1.
 * i.e. gcd(a,b) = 1.
 *
 * Coprimes have the property: b has a multiplicative inverse modulo a
 * i.e there is some value x such that bx = 1 (mod a).
 *
 * Testing the modulo inverse is currently much faster than the constant
 * time version of BN_gcd().
 */
int BN_are_coprime(BIGNUM *a, const BIGNUM *b, BN_CTX *ctx)
{
    int ret = 0;
    BIGNUM *tmp;

    BN_CTX_start(ctx);
    tmp = BN_CTX_get(ctx);
    if (tmp == NULL)
        goto end;

    ERR_set_mark();
    BN_set_flags(a, BN_FLG_CONSTTIME);
    ret = (BN_mod_inverse(tmp, a, b, ctx) != NULL);
    /* Clear any errors (an error is returned if there is no inverse) */
    ERR_pop_to_mark();
end:
    BN_CTX_end(ctx);
    return ret;
}

/*-
 * This function is based on the constant-time GCD work by Bernstein and Yang:
 * https://eprint.iacr.org/2019/266
 * Generalized fast GCD function to allow even inputs.
 * The algorithm first finds the shared powers of 2 between
 * the inputs, and removes them, reducing at least one of the
 * inputs to an odd value. Then it proceeds to calculate the GCD.
 * Before returning the resulting GCD, we take care of adding
 * back the powers of two removed at the beginning.
 * Note 1: we assume the bit length of both inputs is public information,
 * since access to top potentially leaks this information.
 */
int BN_gcd(BIGNUM *r, const BIGNUM *in_a, const BIGNUM *in_b, BN_CTX *ctx)
{
    BIGNUM *g, *temp = NULL;
    BN_ULONG pow2_numbits, pow2_numbits_temp, pow2_condition_mask, pow2_flag;
    int i, j, top, rlen, glen, m, delta = 1, cond = 0, pow2_shifts, ret = 0;

    /* Note 2: zero input corner cases are not constant-time since they are
     * handled immediately. An attacker can run an attack under this
     * assumption without the need of side-channel information. */
    if (BN_is_zero(in_b)) {
        ret = BN_copy(r, in_a) != NULL;
        r->neg = 0;
        return ret;
    }
    if (BN_is_zero(in_a)) {
        ret = BN_copy(r, in_b) != NULL;
        r->neg = 0;
        return ret;
    }

    bn_check_top(in_a);
    bn_check_top(in_b);

    BN_CTX_start(ctx);
    temp = BN_CTX_get(ctx);
    g = BN_CTX_get(ctx);

    /* make r != 0, g != 0 even, so BN_rshift is not a potential nop */
    if (g == NULL
        || !BN_lshift1(g, in_b)
        || !BN_lshift1(r, in_a))
        goto err;

    /* find shared powers of two, i.e. "shifts" >= 1 */
    pow2_flag = 1;
    pow2_shifts = 0;
    pow2_numbits = 0;
    for (i = 0; i < r->dmax && i < g->dmax; i++) {
        pow2_numbits_temp = r->d[i] | g->d[i];
        pow2_condition_mask = constant_time_is_zero_bn(pow2_flag);
        pow2_flag &= constant_time_is_zero_bn(pow2_numbits_temp);
        pow2_shifts += pow2_flag;
        pow2_numbits = constant_time_select_bn(pow2_condition_mask,
                                               pow2_numbits, pow2_numbits_temp);
    }
    pow2_numbits = ~pow2_numbits;
    pow2_shifts *= BN_BITS2;
    pow2_flag = 1;
    for (j = 0; j < BN_BITS2; j++) {
        pow2_flag &= pow2_numbits;
        pow2_shifts += pow2_flag;
        pow2_numbits >>= 1;
    }

    /* subtract shared powers of two; shifts >= 1 */
    if (!BN_rshift(r, r, pow2_shifts)
        || !BN_rshift(g, g, pow2_shifts))
        goto err;

    /* expand to biggest nword, with room for a possible extra word */
    top = 1 + ((r->top >= g->top) ? r->top : g->top);
    if (bn_wexpand(r, top) == NULL
        || bn_wexpand(g, top) == NULL
        || bn_wexpand(temp, top) == NULL)
        goto err;

    /* re arrange inputs s.t. r is odd */
    BN_consttime_swap((~r->d[0]) & 1, r, g, top);

    /* compute the number of iterations */
    rlen = BN_num_bits(r);
    glen = BN_num_bits(g);
    m = 4 + 3 * ((rlen >= glen) ? rlen : glen);

    for (i = 0; i < m; i++) {
        /* conditionally flip signs if delta is positive and g is odd */
        cond = ((unsigned int)-delta >> (8 * sizeof(delta) - 1)) & g->d[0] & 1
            /* make sure g->top > 0 (i.e. if top == 0 then g == 0 always) */
            & (~((unsigned int)(g->top - 1) >> (sizeof(g->top) * 8 - 1)));
        delta = (-cond & -delta) | ((cond - 1) & delta);
        r->neg ^= cond;
        /* swap */
        BN_consttime_swap(cond, r, g, top);

        /* elimination step */
        delta++;
        if (!BN_add(temp, g, r))
            goto err;
        BN_consttime_swap(g->d[0] & 1 /* g is odd */
                /* make sure g->top > 0 (i.e. if top == 0 then g == 0 always) */
                & (~((unsigned int)(g->top - 1) >> (sizeof(g->top) * 8 - 1))),
                g, temp, top);
        if (!BN_rshift1(g, g))
            goto err;
    }

    /* remove possible negative sign */
    r->neg = 0;
    /* add powers of 2 removed, then correct the artificial shift */
    if (!BN_lshift(r, r, pow2_shifts)
        || !BN_rshift1(r, r))
        goto err;

    ret = 1;

 err:
    BN_CTX_end(ctx);
    bn_check_top(r);
    return ret;
}

Coverage Report

Created: 2025-06-22 06:56

Line	Count	Source (jump to first uncovered line)
1		/*
2		* Copyright 1995-2024 The OpenSSL Project Authors. All Rights Reserved.
3		*
4		* Licensed under the Apache License 2.0 (the "License"). You may not use
5		* this file except in compliance with the License. You can obtain a copy
6		* in the file LICENSE in the source distribution or at
7		* https://www.openssl.org/source/license.html
8		*/
9
10		#include "internal/cryptlib.h"
11		#include "bn_local.h"
12		#include "internal/constant_time.h"
13
14		/*
15		* bn_mod_inverse_no_branch is a special version of BN_mod_inverse. It does
16		* not contain branches that may leak sensitive information.
17		*
18		* This is a static function, we ensure all callers in this file pass valid
19		* arguments: all passed pointers here are non-NULL.
20		*/
21		static ossl_inline
22		BIGNUM bn_mod_inverse_no_branch(BIGNUM in,
23		const BIGNUM a, const BIGNUM n,
24		BN_CTX ctx, int pnoinv)
25	0	{
26	0	BIGNUM A, B, X, Y, M, D, T, R = NULL;
27	0	BIGNUM *ret = NULL;
28	0	int sign;
29
30	0	bn_check_top(a);
31	0	bn_check_top(n);
32
33	0	BN_CTX_start(ctx);
34	0	A = BN_CTX_get(ctx);
35	0	B = BN_CTX_get(ctx);
36	0	X = BN_CTX_get(ctx);
37	0	D = BN_CTX_get(ctx);
38	0	M = BN_CTX_get(ctx);
39	0	Y = BN_CTX_get(ctx);
40	0	T = BN_CTX_get(ctx);
41	0	if (T == NULL)
42	0	goto err;
43
44	0	if (in == NULL)
45	0	R = BN_new();
46	0	else
47	0	R = in;
48	0	if (R == NULL)
49	0	goto err;
50
51	0	if (!BN_one(X))
52	0	goto err;
53	0	BN_zero(Y);
54	0	if (BN_copy(B, a) == NULL)
55	0	goto err;
56	0	if (BN_copy(A, n) == NULL)
57	0	goto err;
58	0	A->neg = 0;
59
60	0	if (B->neg \|\| (BN_ucmp(B, A) >= 0)) {
61		/*
62		* Turn BN_FLG_CONSTTIME flag on, so that when BN_div is invoked,
63		* BN_div_no_branch will be called eventually.
64		*/
65	0	{
66	0	BIGNUM local_B;
67	0	bn_init(&local_B);
68	0	BN_with_flags(&local_B, B, BN_FLG_CONSTTIME);
69	0	if (!BN_nnmod(B, &local_B, A, ctx))
70	0	goto err;
71		/* Ensure local_B goes out of scope before any further use of B */
72	0	}
73	0	}
74	0	sign = -1;
75		/*-
76		* From B = a mod \|n\|, A = \|n\| it follows that
77		*
78		* 0 <= B < A,
79		* -signXa == B (mod \|n\|),
80		* signYa == A (mod \|n\|).
81		*/
82
83	0	while (!BN_is_zero(B)) {
84	0	BIGNUM *tmp;
85
86		/*-
87		* 0 < B < A,
88		* () -signX*a == B (mod \|n\|),
89		* signYa == A (mod \|n\|)
90		*/
91
92		/*
93		* Turn BN_FLG_CONSTTIME flag on, so that when BN_div is invoked,
94		* BN_div_no_branch will be called eventually.
95		*/
96	0	{
97	0	BIGNUM local_A;
98	0	bn_init(&local_A);
99	0	BN_with_flags(&local_A, A, BN_FLG_CONSTTIME);
100
101		/* (D, M) := (A/B, A%B) ... */
102	0	if (!BN_div(D, M, &local_A, B, ctx))
103	0	goto err;
104		/* Ensure local_A goes out of scope before any further use of A */
105	0	}
106
107		/*-
108		* Now
109		* A = D*B + M;
110		* thus we have
111		* (*) signYa == DB + M (mod \|n\|).
112		*/
113
114	0	tmp = A; /* keep the BIGNUM object, the value does not
115		* matter */
116
117		/* (A, B) := (B, A mod B) ... */
118	0	A = B;
119	0	B = M;
120		/* ... so we have 0 <= B < A again */
121
122		/*-
123		* Since the former M is now B and the former B is now A,
124		* (**) translates into
125		* signYa == D*A + B (mod \|n\|),
126		* i.e.
127		* signYa - D*A == B (mod \|n\|).
128		* Similarly, (*) translates into
129		* -signXa == A (mod \|n\|).
130		*
131		* Thus,
132		* signYa + DsignX*a == B (mod \|n\|),
133		* i.e.
134		* sign(Y + DX)*a == B (mod \|n\|).
135		*
136		* So if we set (X, Y, sign) := (Y + D*X, X, -sign), we arrive back at
137		* -signXa == B (mod \|n\|),
138		* signYa == A (mod \|n\|).
139		* Note that X and Y stay non-negative all the time.
140		*/
141
142	0	if (!BN_mul(tmp, D, X, ctx))
143	0	goto err;
144	0	if (!BN_add(tmp, tmp, Y))
145	0	goto err;
146
147	0	M = Y; /* keep the BIGNUM object, the value does not
148		* matter */
149	0	Y = X;
150	0	X = tmp;
151	0	sign = -sign;
152	0	}
153
154		/*-
155		* The while loop (Euclid's algorithm) ends when
156		* A == gcd(a,n);
157		* we have
158		* signYa == A (mod \|n\|),
159		* where Y is non-negative.
160		*/
161
162	0	if (sign < 0) {
163	0	if (!BN_sub(Y, n, Y))
164	0	goto err;
165	0	}
166		/* Now Ya == A (mod \|n\|). /
167
168	0	if (BN_is_one(A)) {
169		/* Ya == 1 (mod \|n\|) /
170	0	if (!Y->neg && BN_ucmp(Y, n) < 0) {
171	0	if (!BN_copy(R, Y))
172	0	goto err;
173	0	} else {
174	0	if (!BN_nnmod(R, Y, n, ctx))
175	0	goto err;
176	0	}
177	0	} else {
178	0	*pnoinv = 1;
179		/* caller sets the BN_R_NO_INVERSE error */
180	0	goto err;
181	0	}
182
183	0	ret = R;
184	0	*pnoinv = 0;
185
186	0	err:
187	0	if ((ret == NULL) && (in == NULL))
188	0	BN_free(R);
189	0	BN_CTX_end(ctx);
190	0	bn_check_top(ret);
191	0	return ret;
192	0	}
193
194		/*
195		* This is an internal function, we assume all callers pass valid arguments:
196		* all pointers passed here are assumed non-NULL.
197		*/
198		BIGNUM int_bn_mod_inverse(BIGNUM in,
199		const BIGNUM a, const BIGNUM n, BN_CTX *ctx,
200		int *pnoinv)
201	0	{
202	0	BIGNUM A, B, X, Y, M, D, T, R = NULL;
203	0	BIGNUM *ret = NULL;
204	0	int sign;
205
206		/* This is invalid input so we don't worry about constant time here */
207	0	if (BN_abs_is_word(n, 1) \|\| BN_is_zero(n)) {
208	0	*pnoinv = 1;
209	0	return NULL;
210	0	}
211
212	0	*pnoinv = 0;
213
214	0	if ((BN_get_flags(a, BN_FLG_CONSTTIME) != 0)
215	0	\|\| (BN_get_flags(n, BN_FLG_CONSTTIME) != 0)) {
216	0	return bn_mod_inverse_no_branch(in, a, n, ctx, pnoinv);
217	0	}
218
219	0	bn_check_top(a);
220	0	bn_check_top(n);
221
222	0	BN_CTX_start(ctx);
223	0	A = BN_CTX_get(ctx);
224	0	B = BN_CTX_get(ctx);
225	0	X = BN_CTX_get(ctx);
226	0	D = BN_CTX_get(ctx);
227	0	M = BN_CTX_get(ctx);
228	0	Y = BN_CTX_get(ctx);
229	0	T = BN_CTX_get(ctx);
230	0	if (T == NULL)
231	0	goto err;
232
233	0	if (in == NULL)
234	0	R = BN_new();
235	0	else
236	0	R = in;
237	0	if (R == NULL)
238	0	goto err;
239
240	0	if (!BN_one(X))
241	0	goto err;
242	0	BN_zero(Y);
243	0	if (BN_copy(B, a) == NULL)
244	0	goto err;
245	0	if (BN_copy(A, n) == NULL)
246	0	goto err;
247	0	A->neg = 0;
248	0	if (B->neg \|\| (BN_ucmp(B, A) >= 0)) {
249	0	if (!BN_nnmod(B, B, A, ctx))
250	0	goto err;
251	0	}
252	0	sign = -1;
253		/*-
254		* From B = a mod \|n\|, A = \|n\| it follows that
255		*
256		* 0 <= B < A,
257		* -signXa == B (mod \|n\|),
258		* signYa == A (mod \|n\|).
259		*/
260
261	0	if (BN_is_odd(n) && (BN_num_bits(n) <= 2048)) {
262		/*
263		* Binary inversion algorithm; requires odd modulus. This is faster
264		* than the general algorithm if the modulus is sufficiently small
265		* (about 400 .. 500 bits on 32-bit systems, but much more on 64-bit
266		* systems)
267		*/
268	0	int shift;
269
270	0	while (!BN_is_zero(B)) {
271		/*-
272		* 0 < B < \|n\|,
273		* 0 < A <= \|n\|,
274		* (1) -signXa == B (mod \|n\|),
275		* (2) signYa == A (mod \|n\|)
276		*/
277
278		/*
279		* Now divide B by the maximum possible power of two in the
280		* integers, and divide X by the same value mod \|n\|. When we're
281		* done, (1) still holds.
282		*/
283	0	shift = 0;
284	0	while (!BN_is_bit_set(B, shift)) { /* note that 0 < B */
285	0	shift++;
286
287	0	if (BN_is_odd(X)) {
288	0	if (!BN_uadd(X, X, n))
289	0	goto err;
290	0	}
291		/*
292		* now X is even, so we can easily divide it by two
293		*/
294	0	if (!BN_rshift1(X, X))
295	0	goto err;
296	0	}
297	0	if (shift > 0) {
298	0	if (!BN_rshift(B, B, shift))
299	0	goto err;
300	0	}
301
302		/*
303		* Same for A and Y. Afterwards, (2) still holds.
304		*/
305	0	shift = 0;
306	0	while (!BN_is_bit_set(A, shift)) { /* note that 0 < A */
307	0	shift++;
308
309	0	if (BN_is_odd(Y)) {
310	0	if (!BN_uadd(Y, Y, n))
311	0	goto err;
312	0	}
313		/* now Y is even */
314	0	if (!BN_rshift1(Y, Y))
315	0	goto err;
316	0	}
317	0	if (shift > 0) {
318	0	if (!BN_rshift(A, A, shift))
319	0	goto err;
320	0	}
321
322		/*-
323		* We still have (1) and (2).
324		* Both A and B are odd.
325		* The following computations ensure that
326		*
327		* 0 <= B < \|n\|,
328		* 0 < A < \|n\|,
329		* (1) -signXa == B (mod \|n\|),
330		* (2) signYa == A (mod \|n\|),
331		*
332		* and that either A or B is even in the next iteration.
333		*/
334	0	if (BN_ucmp(B, A) >= 0) {
335		/* -sign(X + Y)a == B - A (mod \|n\|) */
336	0	if (!BN_uadd(X, X, Y))
337	0	goto err;
338		/*
339		* NB: we could use BN_mod_add_quick(X, X, Y, n), but that
340		* actually makes the algorithm slower
341		*/
342	0	if (!BN_usub(B, B, A))
343	0	goto err;
344	0	} else {
345		/* sign(X + Y)a == A - B (mod \|n\|) */
346	0	if (!BN_uadd(Y, Y, X))
347	0	goto err;
348		/*
349		* as above, BN_mod_add_quick(Y, Y, X, n) would slow things down
350		*/
351	0	if (!BN_usub(A, A, B))
352	0	goto err;
353	0	}
354	0	}
355	0	} else {
356		/* general inversion algorithm */
357
358	0	while (!BN_is_zero(B)) {
359	0	BIGNUM *tmp;
360
361		/*-
362		* 0 < B < A,
363		* () -signX*a == B (mod \|n\|),
364		* signYa == A (mod \|n\|)
365		*/
366
367		/* (D, M) := (A/B, A%B) ... */
368	0	if (BN_num_bits(A) == BN_num_bits(B)) {
369	0	if (!BN_one(D))
370	0	goto err;
371	0	if (!BN_sub(M, A, B))
372	0	goto err;
373	0	} else if (BN_num_bits(A) == BN_num_bits(B) + 1) {
374		/* A/B is 1, 2, or 3 */
375	0	if (!BN_lshift1(T, B))
376	0	goto err;
377	0	if (BN_ucmp(A, T) < 0) {
378		/* A < 2B, so D=1 /
379	0	if (!BN_one(D))
380	0	goto err;
381	0	if (!BN_sub(M, A, B))
382	0	goto err;
383	0	} else {
384		/* A >= 2B, so D=2 or D=3 /
385	0	if (!BN_sub(M, A, T))
386	0	goto err;
387	0	if (!BN_add(D, T, B))
388	0	goto err; /* use D (:= 3B) as temp /
389	0	if (BN_ucmp(A, D) < 0) {
390		/* A < 3B, so D=2 /
391	0	if (!BN_set_word(D, 2))
392	0	goto err;
393		/*
394		* M (= A - 2*B) already has the correct value
395		*/
396	0	} else {
397		/* only D=3 remains */
398	0	if (!BN_set_word(D, 3))
399	0	goto err;
400		/*
401		* currently M = A - 2B, but we need M = A - 3B
402		*/
403	0	if (!BN_sub(M, M, B))
404	0	goto err;
405	0	}
406	0	}
407	0	} else {
408	0	if (!BN_div(D, M, A, B, ctx))
409	0	goto err;
410	0	}
411
412		/*-
413		* Now
414		* A = D*B + M;
415		* thus we have
416		* (*) signYa == DB + M (mod \|n\|).
417		*/
418
419	0	tmp = A; /* keep the BIGNUM object, the value does not matter */
420
421		/* (A, B) := (B, A mod B) ... */
422	0	A = B;
423	0	B = M;
424		/* ... so we have 0 <= B < A again */
425
426		/*-
427		* Since the former M is now B and the former B is now A,
428		* (**) translates into
429		* signYa == D*A + B (mod \|n\|),
430		* i.e.
431		* signYa - D*A == B (mod \|n\|).
432		* Similarly, (*) translates into
433		* -signXa == A (mod \|n\|).
434		*
435		* Thus,
436		* signYa + DsignX*a == B (mod \|n\|),
437		* i.e.
438		* sign(Y + DX)*a == B (mod \|n\|).
439		*
440		* So if we set (X, Y, sign) := (Y + D*X, X, -sign), we arrive back at
441		* -signXa == B (mod \|n\|),
442		* signYa == A (mod \|n\|).
443		* Note that X and Y stay non-negative all the time.
444		*/
445
446		/*
447		* most of the time D is very small, so we can optimize tmp := D*X+Y
448		*/
449	0	if (BN_is_one(D)) {
450	0	if (!BN_add(tmp, X, Y))
451	0	goto err;
452	0	} else {
453	0	if (BN_is_word(D, 2)) {
454	0	if (!BN_lshift1(tmp, X))
455	0	goto err;
456	0	} else if (BN_is_word(D, 4)) {
457	0	if (!BN_lshift(tmp, X, 2))
458	0	goto err;
459	0	} else if (D->top == 1) {
460	0	if (!BN_copy(tmp, X))
461	0	goto err;
462	0	if (!BN_mul_word(tmp, D->d[0]))
463	0	goto err;
464	0	} else {
465	0	if (!BN_mul(tmp, D, X, ctx))
466	0	goto err;
467	0	}
468	0	if (!BN_add(tmp, tmp, Y))
469	0	goto err;
470	0	}
471
472	0	M = Y; /* keep the BIGNUM object, the value does not matter */
473	0	Y = X;
474	0	X = tmp;
475	0	sign = -sign;
476	0	}
477	0	}
478
479		/*-
480		* The while loop (Euclid's algorithm) ends when
481		* A == gcd(a,n);
482		* we have
483		* signYa == A (mod \|n\|),
484		* where Y is non-negative.
485		*/
486
487	0	if (sign < 0) {
488	0	if (!BN_sub(Y, n, Y))
489	0	goto err;
490	0	}
491		/* Now Ya == A (mod \|n\|). /
492
493	0	if (BN_is_one(A)) {
494		/* Ya == 1 (mod \|n\|) /
495	0	if (!Y->neg && BN_ucmp(Y, n) < 0) {
496	0	if (!BN_copy(R, Y))
497	0	goto err;
498	0	} else {
499	0	if (!BN_nnmod(R, Y, n, ctx))
500	0	goto err;
501	0	}
502	0	} else {
503	0	*pnoinv = 1;
504	0	goto err;
505	0	}
506	0	ret = R;
507	0	err:
508	0	if ((ret == NULL) && (in == NULL))
509	0	BN_free(R);
510	0	BN_CTX_end(ctx);
511	0	bn_check_top(ret);
512	0	return ret;
513	0	}
514
515		/* solves ax == 1 (mod n) */
516		BIGNUM BN_mod_inverse(BIGNUM in,
517		const BIGNUM a, const BIGNUM n, BN_CTX *ctx)
518	0	{
519	0	BN_CTX *new_ctx = NULL;
520	0	BIGNUM *rv;
521	0	int noinv = 0;
522
523	0	if (ctx == NULL) {
524	0	ctx = new_ctx = BN_CTX_new_ex(NULL);
525	0	if (ctx == NULL) {
526	0	ERR_raise(ERR_LIB_BN, ERR_R_BN_LIB);
527	0	return NULL;
528	0	}
529	0	}
530
531	0	rv = int_bn_mod_inverse(in, a, n, ctx, &noinv);
532	0	if (noinv)
533	0	ERR_raise(ERR_LIB_BN, BN_R_NO_INVERSE);
534	0	BN_CTX_free(new_ctx);
535	0	return rv;
536	0	}
537
538		/*
539		* The numbers a and b are coprime if the only positive integer that is a
540		* divisor of both of them is 1.
541		* i.e. gcd(a,b) = 1.
542		*
543		* Coprimes have the property: b has a multiplicative inverse modulo a
544		* i.e there is some value x such that bx = 1 (mod a).
545		*
546		* Testing the modulo inverse is currently much faster than the constant
547		* time version of BN_gcd().
548		*/
549		int BN_are_coprime(BIGNUM a, const BIGNUM b, BN_CTX *ctx)
550	0	{
551	0	int ret = 0;
552	0	BIGNUM *tmp;
553
554	0	BN_CTX_start(ctx);
555	0	tmp = BN_CTX_get(ctx);
556	0	if (tmp == NULL)
557	0	goto end;
558
559	0	ERR_set_mark();
560	0	BN_set_flags(a, BN_FLG_CONSTTIME);
561	0	ret = (BN_mod_inverse(tmp, a, b, ctx) != NULL);
562		/* Clear any errors (an error is returned if there is no inverse) */
563	0	ERR_pop_to_mark();
564	0	end:
565	0	BN_CTX_end(ctx);
566	0	return ret;
567	0	}
568
569		/*-
570		* This function is based on the constant-time GCD work by Bernstein and Yang:
571		* https://eprint.iacr.org/2019/266
572		* Generalized fast GCD function to allow even inputs.
573		* The algorithm first finds the shared powers of 2 between
574		* the inputs, and removes them, reducing at least one of the
575		* inputs to an odd value. Then it proceeds to calculate the GCD.
576		* Before returning the resulting GCD, we take care of adding
577		* back the powers of two removed at the beginning.
578		* Note 1: we assume the bit length of both inputs is public information,
579		* since access to top potentially leaks this information.
580		*/
581		int BN_gcd(BIGNUM r, const BIGNUM in_a, const BIGNUM in_b, BN_CTX ctx)
582	0	{
583	0	BIGNUM g, temp = NULL;
584	0	BN_ULONG pow2_numbits, pow2_numbits_temp, pow2_condition_mask, pow2_flag;
585	0	int i, j, top, rlen, glen, m, delta = 1, cond = 0, pow2_shifts, ret = 0;
586
587		/* Note 2: zero input corner cases are not constant-time since they are
588		* handled immediately. An attacker can run an attack under this
589		* assumption without the need of side-channel information. */
590	0	if (BN_is_zero(in_b)) {
591	0	ret = BN_copy(r, in_a) != NULL;
592	0	r->neg = 0;
593	0	return ret;
594	0	}
595	0	if (BN_is_zero(in_a)) {
596	0	ret = BN_copy(r, in_b) != NULL;
597	0	r->neg = 0;
598	0	return ret;
599	0	}
600
601	0	bn_check_top(in_a);
602	0	bn_check_top(in_b);
603
604	0	BN_CTX_start(ctx);
605	0	temp = BN_CTX_get(ctx);
606	0	g = BN_CTX_get(ctx);
607
608		/* make r != 0, g != 0 even, so BN_rshift is not a potential nop */
609	0	if (g == NULL
610	0	\|\| !BN_lshift1(g, in_b)
611	0	\|\| !BN_lshift1(r, in_a))
612	0	goto err;
613
614		/* find shared powers of two, i.e. "shifts" >= 1 */
615	0	pow2_flag = 1;
616	0	pow2_shifts = 0;
617	0	pow2_numbits = 0;
618	0	for (i = 0; i < r->dmax && i < g->dmax; i++) {
619	0	pow2_numbits_temp = r->d[i] \| g->d[i];
620	0	pow2_condition_mask = constant_time_is_zero_bn(pow2_flag);
621	0	pow2_flag &= constant_time_is_zero_bn(pow2_numbits_temp);
622	0	pow2_shifts += pow2_flag;
623	0	pow2_numbits = constant_time_select_bn(pow2_condition_mask,
624	0	pow2_numbits, pow2_numbits_temp);
625	0	}
626	0	pow2_numbits = ~pow2_numbits;
627	0	pow2_shifts *= BN_BITS2;
628	0	pow2_flag = 1;
629	0	for (j = 0; j < BN_BITS2; j++) {
630	0	pow2_flag &= pow2_numbits;
631	0	pow2_shifts += pow2_flag;
632	0	pow2_numbits >>= 1;
633	0	}
634
635		/* subtract shared powers of two; shifts >= 1 */
636	0	if (!BN_rshift(r, r, pow2_shifts)
637	0	\|\| !BN_rshift(g, g, pow2_shifts))
638	0	goto err;
639
640		/* expand to biggest nword, with room for a possible extra word */
641	0	top = 1 + ((r->top >= g->top) ? r->top : g->top);
642	0	if (bn_wexpand(r, top) == NULL
643	0	\|\| bn_wexpand(g, top) == NULL
644	0	\|\| bn_wexpand(temp, top) == NULL)
645	0	goto err;
646
647		/* re arrange inputs s.t. r is odd */
648	0	BN_consttime_swap((~r->d[0]) & 1, r, g, top);
649
650		/* compute the number of iterations */
651	0	rlen = BN_num_bits(r);
652	0	glen = BN_num_bits(g);
653	0	m = 4 + 3 * ((rlen >= glen) ? rlen : glen);
654
655	0	for (i = 0; i < m; i++) {
656		/* conditionally flip signs if delta is positive and g is odd */
657	0	cond = ((unsigned int)-delta >> (8 * sizeof(delta) - 1)) & g->d[0] & 1
658		/* make sure g->top > 0 (i.e. if top == 0 then g == 0 always) */
659	0	& (~((unsigned int)(g->top - 1) >> (sizeof(g->top) * 8 - 1)));
660	0	delta = (-cond & -delta) \| ((cond - 1) & delta);
661	0	r->neg ^= cond;
662		/* swap */
663	0	BN_consttime_swap(cond, r, g, top);
664
665		/* elimination step */
666	0	delta++;
667	0	if (!BN_add(temp, g, r))
668	0	goto err;
669	0	BN_consttime_swap(g->d[0] & 1 /* g is odd */
670		/* make sure g->top > 0 (i.e. if top == 0 then g == 0 always) */
671	0	& (~((unsigned int)(g->top - 1) >> (sizeof(g->top) * 8 - 1))),
672	0	g, temp, top);
673	0	if (!BN_rshift1(g, g))
674	0	goto err;
675	0	}
676
677		/* remove possible negative sign */
678	0	r->neg = 0;
679		/* add powers of 2 removed, then correct the artificial shift */
680	0	if (!BN_lshift(r, r, pow2_shifts)
681	0	\|\| !BN_rshift1(r, r))
682	0	goto err;
683
684	0	ret = 1;
685
686	0	err:
687	0	BN_CTX_end(ctx);
688	0	bn_check_top(r);
689	0	return ret;
690	0	}