/src/openssl111/crypto/bn/bn_gcd.c

Source (jump to first uncovered line)
/*
 * Copyright 1995-2022 The OpenSSL Project Authors. All Rights Reserved.
 *
 * Licensed under the OpenSSL license (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"

/*
 * 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();
        if (ctx == NULL) {
            BNerr(BN_F_BN_MOD_INVERSE, ERR_R_MALLOC_FAILURE);
            return NULL;
        }
    }

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

/*-
 * 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 mask = 0;
    int i, j, top, rlen, glen, m, bit = 1, delta = 1, cond = 0, shifts = 0, 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 */
    for (i = 0; i < r->dmax && i < g->dmax; i++) {
        mask = ~(r->d[i] | g->d[i]);
        for (j = 0; j < BN_BITS2; j++) {
            bit &= mask;
            shifts += bit;
            mask >>= 1;
        }
    }

    /* subtract shared powers of two; shifts >= 1 */
    if (!BN_rshift(r, r, shifts)
        || !BN_rshift(g, g, 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 = (-delta >> (8 * sizeof(delta) - 1)) & g->d[0] & 1
            /* make sure g->top > 0 (i.e. if top == 0 then g == 0 always) */
            & (~((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) */
                & (~((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, shifts)
        || !BN_rshift1(r, r))
        goto err;

    ret = 1;

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

Coverage Report

Created: 2023-06-08 06:41

Line	Count	Source (jump to first uncovered line)
1		/*
2		* Copyright 1995-2022 The OpenSSL Project Authors. All Rights Reserved.
3		*
4		* Licensed under the OpenSSL license (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
13		/*
14		* bn_mod_inverse_no_branch is a special version of BN_mod_inverse. It does
15		* not contain branches that may leak sensitive information.
16		*
17		* This is a static function, we ensure all callers in this file pass valid
18		* arguments: all passed pointers here are non-NULL.
19		*/
20		static ossl_inline
21		BIGNUM bn_mod_inverse_no_branch(BIGNUM in,
22		const BIGNUM a, const BIGNUM n,
23		BN_CTX ctx, int pnoinv)
24	0	{
25	0	BIGNUM A, B, X, Y, M, D, T, R = NULL;
26	0	BIGNUM *ret = NULL;
27	0	int sign;
28
29	0	bn_check_top(a);
30	0	bn_check_top(n);
31
32	0	BN_CTX_start(ctx);
33	0	A = BN_CTX_get(ctx);
34	0	B = BN_CTX_get(ctx);
35	0	X = BN_CTX_get(ctx);
36	0	D = BN_CTX_get(ctx);
37	0	M = BN_CTX_get(ctx);
38	0	Y = BN_CTX_get(ctx);
39	0	T = BN_CTX_get(ctx);
40	0	if (T == NULL)
41	0	goto err;
42
43	0	if (in == NULL)
44	0	R = BN_new();
45	0	else
46	0	R = in;
47	0	if (R == NULL)
48	0	goto err;
49
50	0	if (!BN_one(X))
51	0	goto err;
52	0	BN_zero(Y);
53	0	if (BN_copy(B, a) == NULL)
54	0	goto err;
55	0	if (BN_copy(A, n) == NULL)
56	0	goto err;
57	0	A->neg = 0;
58
59	0	if (B->neg \|\| (BN_ucmp(B, A) >= 0)) {
60		/*
61		* Turn BN_FLG_CONSTTIME flag on, so that when BN_div is invoked,
62		* BN_div_no_branch will be called eventually.
63		*/
64	0	{
65	0	BIGNUM local_B;
66	0	bn_init(&local_B);
67	0	BN_with_flags(&local_B, B, BN_FLG_CONSTTIME);
68	0	if (!BN_nnmod(B, &local_B, A, ctx))
69	0	goto err;
70		/* Ensure local_B goes out of scope before any further use of B */
71	0	}
72	0	}
73	0	sign = -1;
74		/*-
75		* From B = a mod \|n\|, A = \|n\| it follows that
76		*
77		* 0 <= B < A,
78		* -signXa == B (mod \|n\|),
79		* signYa == A (mod \|n\|).
80		*/
81
82	0	while (!BN_is_zero(B)) {
83	0	BIGNUM *tmp;
84
85		/*-
86		* 0 < B < A,
87		* () -signX*a == B (mod \|n\|),
88		* signYa == A (mod \|n\|)
89		*/
90
91		/*
92		* Turn BN_FLG_CONSTTIME flag on, so that when BN_div is invoked,
93		* BN_div_no_branch will be called eventually.
94		*/
95	0	{
96	0	BIGNUM local_A;
97	0	bn_init(&local_A);
98	0	BN_with_flags(&local_A, A, BN_FLG_CONSTTIME);
99
100		/* (D, M) := (A/B, A%B) ... */
101	0	if (!BN_div(D, M, &local_A, B, ctx))
102	0	goto err;
103		/* Ensure local_A goes out of scope before any further use of A */
104	0	}
105
106		/*-
107		* Now
108		* A = D*B + M;
109		* thus we have
110		* (*) signYa == DB + M (mod \|n\|).
111		*/
112
113	0	tmp = A; /* keep the BIGNUM object, the value does not
114		* matter */
115
116		/* (A, B) := (B, A mod B) ... */
117	0	A = B;
118	0	B = M;
119		/* ... so we have 0 <= B < A again */
120
121		/*-
122		* Since the former M is now B and the former B is now A,
123		* (**) translates into
124		* signYa == D*A + B (mod \|n\|),
125		* i.e.
126		* signYa - D*A == B (mod \|n\|).
127		* Similarly, (*) translates into
128		* -signXa == A (mod \|n\|).
129		*
130		* Thus,
131		* signYa + DsignX*a == B (mod \|n\|),
132		* i.e.
133		* sign(Y + DX)*a == B (mod \|n\|).
134		*
135		* So if we set (X, Y, sign) := (Y + D*X, X, -sign), we arrive back at
136		* -signXa == B (mod \|n\|),
137		* signYa == A (mod \|n\|).
138		* Note that X and Y stay non-negative all the time.
139		*/
140
141	0	if (!BN_mul(tmp, D, X, ctx))
142	0	goto err;
143	0	if (!BN_add(tmp, tmp, Y))
144	0	goto err;
145
146	0	M = Y; /* keep the BIGNUM object, the value does not
147		* matter */
148	0	Y = X;
149	0	X = tmp;
150	0	sign = -sign;
151	0	}
152
153		/*-
154		* The while loop (Euclid's algorithm) ends when
155		* A == gcd(a,n);
156		* we have
157		* signYa == A (mod \|n\|),
158		* where Y is non-negative.
159		*/
160
161	0	if (sign < 0) {
162	0	if (!BN_sub(Y, n, Y))
163	0	goto err;
164	0	}
165		/* Now Ya == A (mod \|n\|). /
166
167	0	if (BN_is_one(A)) {
168		/* Ya == 1 (mod \|n\|) /
169	0	if (!Y->neg && BN_ucmp(Y, n) < 0) {
170	0	if (!BN_copy(R, Y))
171	0	goto err;
172	0	} else {
173	0	if (!BN_nnmod(R, Y, n, ctx))
174	0	goto err;
175	0	}
176	0	} else {
177	0	*pnoinv = 1;
178		/* caller sets the BN_R_NO_INVERSE error */
179	0	goto err;
180	0	}
181
182	0	ret = R;
183	0	*pnoinv = 0;
184
185	0	err:
186	0	if ((ret == NULL) && (in == NULL))
187	0	BN_free(R);
188	0	BN_CTX_end(ctx);
189	0	bn_check_top(ret);
190	0	return ret;
191	0	}
192
193		/*
194		* This is an internal function, we assume all callers pass valid arguments:
195		* all pointers passed here are assumed non-NULL.
196		*/
197		BIGNUM int_bn_mod_inverse(BIGNUM in,
198		const BIGNUM a, const BIGNUM n, BN_CTX *ctx,
199		int *pnoinv)
200	42.7k	{
201	42.7k	BIGNUM A, B, X, Y, M, D, T, R = NULL;
202	42.7k	BIGNUM *ret = NULL;
203	42.7k	int sign;
204
205		/* This is invalid input so we don't worry about constant time here */
206	42.7k	if (BN_abs_is_word(n, 1) \|\| BN_is_zero(n)) {
207	0	*pnoinv = 1;
208	0	return NULL;
209	0	}
210
211	42.7k	*pnoinv = 0;
212
213	42.7k	if ((BN_get_flags(a, BN_FLG_CONSTTIME) != 0)
214	42.7k	\|\| (BN_get_flags(n, BN_FLG_CONSTTIME) != 0)) {
215	0	return bn_mod_inverse_no_branch(in, a, n, ctx, pnoinv);
216	0	}
217
218	42.7k	bn_check_top(a);
219	42.7k	bn_check_top(n);
220
221	42.7k	BN_CTX_start(ctx);
222	42.7k	A = BN_CTX_get(ctx);
223	42.7k	B = BN_CTX_get(ctx);
224	42.7k	X = BN_CTX_get(ctx);
225	42.7k	D = BN_CTX_get(ctx);
226	42.7k	M = BN_CTX_get(ctx);
227	42.7k	Y = BN_CTX_get(ctx);
228	42.7k	T = BN_CTX_get(ctx);
229	42.7k	if (T == NULL)
230	0	goto err;
231
232	42.7k	if (in == NULL)
233	0	R = BN_new();
234	42.7k	else
235	42.7k	R = in;
236	42.7k	if (R == NULL)
237	0	goto err;
238
239	42.7k	if (!BN_one(X))
240	0	goto err;
241	42.7k	BN_zero(Y);
242	42.7k	if (BN_copy(B, a) == NULL)
243	0	goto err;
244	42.7k	if (BN_copy(A, n) == NULL)
245	0	goto err;
246	42.7k	A->neg = 0;
247	42.7k	if (B->neg \|\| (BN_ucmp(B, A) >= 0)) {
248	42.7k	if (!BN_nnmod(B, B, A, ctx))
249	0	goto err;
250	42.7k	}
251	42.7k	sign = -1;
252		/*-
253		* From B = a mod \|n\|, A = \|n\| it follows that
254		*
255		* 0 <= B < A,
256		* -signXa == B (mod \|n\|),
257		* signYa == A (mod \|n\|).
258		*/
259
260	42.7k	if (BN_is_odd(n) && (BN_num_bits(n) <= 2048)) {
261		/*
262		* Binary inversion algorithm; requires odd modulus. This is faster
263		* than the general algorithm if the modulus is sufficiently small
264		* (about 400 .. 500 bits on 32-bit systems, but much more on 64-bit
265		* systems)
266		*/
267	42.7k	int shift;
268
269	2.70M	while (!BN_is_zero(B)) {
270		/*-
271		* 0 < B < \|n\|,
272		* 0 < A <= \|n\|,
273		* (1) -signXa == B (mod \|n\|),
274		* (2) signYa == A (mod \|n\|)
275		*/
276
277		/*
278		* Now divide B by the maximum possible power of two in the
279		* integers, and divide X by the same value mod \|n\|. When we're
280		* done, (1) still holds.
281		*/
282	2.66M	shift = 0;
283	3.62M	while (!BN_is_bit_set(B, shift)) { /* note that 0 < B */
284	960k	shift++;
285
286	960k	if (BN_is_odd(X)) {
287	510k	if (!BN_uadd(X, X, n))
288	0	goto err;
289	510k	}
290		/*
291		* now X is even, so we can easily divide it by two
292		*/
293	960k	if (!BN_rshift1(X, X))
294	0	goto err;
295	960k	}
296	2.66M	if (shift > 0) {
297	924k	if (!BN_rshift(B, B, shift))
298	0	goto err;
299	924k	}
300
301		/*
302		* Same for A and Y. Afterwards, (2) still holds.
303		*/
304	2.66M	shift = 0;
305	4.39M	while (!BN_is_bit_set(A, shift)) { /* note that 0 < A */
306	1.72M	shift++;
307
308	1.72M	if (BN_is_odd(Y)) {
309	934k	if (!BN_uadd(Y, Y, n))
310	0	goto err;
311	934k	}
312		/* now Y is even */
313	1.72M	if (!BN_rshift1(Y, Y))
314	0	goto err;
315	1.72M	}
316	2.66M	if (shift > 0) {
317	1.70M	if (!BN_rshift(A, A, shift))
318	0	goto err;
319	1.70M	}
320
321		/*-
322		* We still have (1) and (2).
323		* Both A and B are odd.
324		* The following computations ensure that
325		*
326		* 0 <= B < \|n\|,
327		* 0 < A < \|n\|,
328		* (1) -signXa == B (mod \|n\|),
329		* (2) signYa == A (mod \|n\|),
330		*
331		* and that either A or B is even in the next iteration.
332		*/
333	2.66M	if (BN_ucmp(B, A) >= 0) {
334		/* -sign(X + Y)a == B - A (mod \|n\|) */
335	953k	if (!BN_uadd(X, X, Y))
336	0	goto err;
337		/*
338		* NB: we could use BN_mod_add_quick(X, X, Y, n), but that
339		* actually makes the algorithm slower
340		*/
341	953k	if (!BN_usub(B, B, A))
342	0	goto err;
343	1.70M	} else {
344		/* sign(X + Y)a == A - B (mod \|n\|) */
345	1.70M	if (!BN_uadd(Y, Y, X))
346	0	goto err;
347		/*
348		* as above, BN_mod_add_quick(Y, Y, X, n) would slow things down
349		*/
350	1.70M	if (!BN_usub(A, A, B))
351	0	goto err;
352	1.70M	}
353	2.66M	}
354	42.7k	} else {
355		/* general inversion algorithm */
356
357	0	while (!BN_is_zero(B)) {
358	0	BIGNUM *tmp;
359
360		/*-
361		* 0 < B < A,
362		* () -signX*a == B (mod \|n\|),
363		* signYa == A (mod \|n\|)
364		*/
365
366		/* (D, M) := (A/B, A%B) ... */
367	0	if (BN_num_bits(A) == BN_num_bits(B)) {
368	0	if (!BN_one(D))
369	0	goto err;
370	0	if (!BN_sub(M, A, B))
371	0	goto err;
372	0	} else if (BN_num_bits(A) == BN_num_bits(B) + 1) {
373		/* A/B is 1, 2, or 3 */
374	0	if (!BN_lshift1(T, B))
375	0	goto err;
376	0	if (BN_ucmp(A, T) < 0) {
377		/* A < 2B, so D=1 /
378	0	if (!BN_one(D))
379	0	goto err;
380	0	if (!BN_sub(M, A, B))
381	0	goto err;
382	0	} else {
383		/* A >= 2B, so D=2 or D=3 /
384	0	if (!BN_sub(M, A, T))
385	0	goto err;
386	0	if (!BN_add(D, T, B))
387	0	goto err; /* use D (:= 3B) as temp /
388	0	if (BN_ucmp(A, D) < 0) {
389		/* A < 3B, so D=2 /
390	0	if (!BN_set_word(D, 2))
391	0	goto err;
392		/*
393		* M (= A - 2*B) already has the correct value
394		*/
395	0	} else {
396		/* only D=3 remains */
397	0	if (!BN_set_word(D, 3))
398	0	goto err;
399		/*
400		* currently M = A - 2B, but we need M = A - 3B
401		*/
402	0	if (!BN_sub(M, M, B))
403	0	goto err;
404	0	}
405	0	}
406	0	} else {
407	0	if (!BN_div(D, M, A, B, ctx))
408	0	goto err;
409	0	}
410
411		/*-
412		* Now
413		* A = D*B + M;
414		* thus we have
415		* (*) signYa == DB + M (mod \|n\|).
416		*/
417
418	0	tmp = A; /* keep the BIGNUM object, the value does not matter */
419
420		/* (A, B) := (B, A mod B) ... */
421	0	A = B;
422	0	B = M;
423		/* ... so we have 0 <= B < A again */
424
425		/*-
426		* Since the former M is now B and the former B is now A,
427		* (**) translates into
428		* signYa == D*A + B (mod \|n\|),
429		* i.e.
430		* signYa - D*A == B (mod \|n\|).
431		* Similarly, (*) translates into
432		* -signXa == A (mod \|n\|).
433		*
434		* Thus,
435		* signYa + DsignX*a == B (mod \|n\|),
436		* i.e.
437		* sign(Y + DX)*a == B (mod \|n\|).
438		*
439		* So if we set (X, Y, sign) := (Y + D*X, X, -sign), we arrive back at
440		* -signXa == B (mod \|n\|),
441		* signYa == A (mod \|n\|).
442		* Note that X and Y stay non-negative all the time.
443		*/
444
445		/*
446		* most of the time D is very small, so we can optimize tmp := D*X+Y
447		*/
448	0	if (BN_is_one(D)) {
449	0	if (!BN_add(tmp, X, Y))
450	0	goto err;
451	0	} else {
452	0	if (BN_is_word(D, 2)) {
453	0	if (!BN_lshift1(tmp, X))
454	0	goto err;
455	0	} else if (BN_is_word(D, 4)) {
456	0	if (!BN_lshift(tmp, X, 2))
457	0	goto err;
458	0	} else if (D->top == 1) {
459	0	if (!BN_copy(tmp, X))
460	0	goto err;
461	0	if (!BN_mul_word(tmp, D->d[0]))
462	0	goto err;
463	0	} else {
464	0	if (!BN_mul(tmp, D, X, ctx))
465	0	goto err;
466	0	}
467	0	if (!BN_add(tmp, tmp, Y))
468	0	goto err;
469	0	}
470
471	0	M = Y; /* keep the BIGNUM object, the value does not matter */
472	0	Y = X;
473	0	X = tmp;
474	0	sign = -sign;
475	0	}
476	0	}
477
478		/*-
479		* The while loop (Euclid's algorithm) ends when
480		* A == gcd(a,n);
481		* we have
482		* signYa == A (mod \|n\|),
483		* where Y is non-negative.
484		*/
485
486	42.7k	if (sign < 0) {
487	42.7k	if (!BN_sub(Y, n, Y))
488	0	goto err;
489	42.7k	}
490		/* Now Ya == A (mod \|n\|). /
491
492	42.7k	if (BN_is_one(A)) {
493		/* Ya == 1 (mod \|n\|) /
494	42.7k	if (!Y->neg && BN_ucmp(Y, n) < 0) {
495	1.78k	if (!BN_copy(R, Y))
496	0	goto err;
497	40.9k	} else {
498	40.9k	if (!BN_nnmod(R, Y, n, ctx))
499	0	goto err;
500	40.9k	}
501	42.7k	} else {
502	0	*pnoinv = 1;
503	0	goto err;
504	0	}
505	42.7k	ret = R;
506	42.7k	err:
507	42.7k	if ((ret == NULL) && (in == NULL))
508	0	BN_free(R);
509	42.7k	BN_CTX_end(ctx);
510	42.7k	bn_check_top(ret);
511	42.7k	return ret;
512	42.7k	}
513
514		/* solves ax == 1 (mod n) */
515		BIGNUM BN_mod_inverse(BIGNUM in,
516		const BIGNUM a, const BIGNUM n, BN_CTX *ctx)
517	42.7k	{
518	42.7k	BN_CTX *new_ctx = NULL;
519	42.7k	BIGNUM *rv;
520	42.7k	int noinv = 0;
521
522	42.7k	if (ctx == NULL) {
523	0	ctx = new_ctx = BN_CTX_new();
524	0	if (ctx == NULL) {
525	0	BNerr(BN_F_BN_MOD_INVERSE, ERR_R_MALLOC_FAILURE);
526	0	return NULL;
527	0	}
528	0	}
529
530	42.7k	rv = int_bn_mod_inverse(in, a, n, ctx, &noinv);
531	42.7k	if (noinv)
532	42.7k	BNerr(BN_F_BN_MOD_INVERSE, BN_R_NO_INVERSE);
533	42.7k	BN_CTX_free(new_ctx);
534	42.7k	return rv;
535	42.7k	}
536
537		/*-
538		* This function is based on the constant-time GCD work by Bernstein and Yang:
539		* https://eprint.iacr.org/2019/266
540		* Generalized fast GCD function to allow even inputs.
541		* The algorithm first finds the shared powers of 2 between
542		* the inputs, and removes them, reducing at least one of the
543		* inputs to an odd value. Then it proceeds to calculate the GCD.
544		* Before returning the resulting GCD, we take care of adding
545		* back the powers of two removed at the beginning.
546		* Note 1: we assume the bit length of both inputs is public information,
547		* since access to top potentially leaks this information.
548		*/
549		int BN_gcd(BIGNUM r, const BIGNUM in_a, const BIGNUM in_b, BN_CTX ctx)
550	0	{
551	0	BIGNUM g, temp = NULL;
552	0	BN_ULONG mask = 0;
553	0	int i, j, top, rlen, glen, m, bit = 1, delta = 1, cond = 0, shifts = 0, ret = 0;
554
555		/* Note 2: zero input corner cases are not constant-time since they are
556		* handled immediately. An attacker can run an attack under this
557		* assumption without the need of side-channel information. */
558	0	if (BN_is_zero(in_b)) {
559	0	ret = BN_copy(r, in_a) != NULL;
560	0	r->neg = 0;
561	0	return ret;
562	0	}
563	0	if (BN_is_zero(in_a)) {
564	0	ret = BN_copy(r, in_b) != NULL;
565	0	r->neg = 0;
566	0	return ret;
567	0	}
568
569	0	bn_check_top(in_a);
570	0	bn_check_top(in_b);
571
572	0	BN_CTX_start(ctx);
573	0	temp = BN_CTX_get(ctx);
574	0	g = BN_CTX_get(ctx);
575
576		/* make r != 0, g != 0 even, so BN_rshift is not a potential nop */
577	0	if (g == NULL
578	0	\|\| !BN_lshift1(g, in_b)
579	0	\|\| !BN_lshift1(r, in_a))
580	0	goto err;
581
582		/* find shared powers of two, i.e. "shifts" >= 1 */
583	0	for (i = 0; i < r->dmax && i < g->dmax; i++) {
584	0	mask = ~(r->d[i] \| g->d[i]);
585	0	for (j = 0; j < BN_BITS2; j++) {
586	0	bit &= mask;
587	0	shifts += bit;
588	0	mask >>= 1;
589	0	}
590	0	}
591
592		/* subtract shared powers of two; shifts >= 1 */
593	0	if (!BN_rshift(r, r, shifts)
594	0	\|\| !BN_rshift(g, g, shifts))
595	0	goto err;
596
597		/* expand to biggest nword, with room for a possible extra word */
598	0	top = 1 + ((r->top >= g->top) ? r->top : g->top);
599	0	if (bn_wexpand(r, top) == NULL
600	0	\|\| bn_wexpand(g, top) == NULL
601	0	\|\| bn_wexpand(temp, top) == NULL)
602	0	goto err;
603
604		/* re arrange inputs s.t. r is odd */
605	0	BN_consttime_swap((~r->d[0]) & 1, r, g, top);
606
607		/* compute the number of iterations */
608	0	rlen = BN_num_bits(r);
609	0	glen = BN_num_bits(g);
610	0	m = 4 + 3 * ((rlen >= glen) ? rlen : glen);
611
612	0	for (i = 0; i < m; i++) {
613		/* conditionally flip signs if delta is positive and g is odd */
614	0	cond = (-delta >> (8 * sizeof(delta) - 1)) & g->d[0] & 1
615		/* make sure g->top > 0 (i.e. if top == 0 then g == 0 always) */
616	0	& (~((g->top - 1) >> (sizeof(g->top) * 8 - 1)));
617	0	delta = (-cond & -delta) \| ((cond - 1) & delta);
618	0	r->neg ^= cond;
619		/* swap */
620	0	BN_consttime_swap(cond, r, g, top);
621
622		/* elimination step */
623	0	delta++;
624	0	if (!BN_add(temp, g, r))
625	0	goto err;
626	0	BN_consttime_swap(g->d[0] & 1 /* g is odd */
627		/* make sure g->top > 0 (i.e. if top == 0 then g == 0 always) */
628	0	& (~((g->top - 1) >> (sizeof(g->top) * 8 - 1))),
629	0	g, temp, top);
630	0	if (!BN_rshift1(g, g))
631	0	goto err;
632	0	}
633
634		/* remove possible negative sign */
635	0	r->neg = 0;
636		/* add powers of 2 removed, then correct the artificial shift */
637	0	if (!BN_lshift(r, r, shifts)
638	0	\|\| !BN_rshift1(r, r))
639	0	goto err;
640
641	0	ret = 1;
642
643	0	err:
644	0	BN_CTX_end(ctx);
645	0	bn_check_top(r);
646	0	return ret;
647	0	}