bn_gcd.c 16.9 KB
Newer Older
R
Rich Salz 已提交
1 2
/*
 * Copyright 1995-2016 The OpenSSL Project Authors. All Rights Reserved.
B
Bodo Möller 已提交
3
 *
R
Rich Salz 已提交
4 5 6 7
 * 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
B
Bodo Möller 已提交
8
 */
9

10
#include "internal/cryptlib.h"
11 12 13
#include "bn_lcl.h"

static BIGNUM *euclid(BIGNUM *a, BIGNUM *b);
14

15
int BN_gcd(BIGNUM *r, const BIGNUM *in_a, const BIGNUM *in_b, BN_CTX *ctx)
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52
{
    BIGNUM *a, *b, *t;
    int ret = 0;

    bn_check_top(in_a);
    bn_check_top(in_b);

    BN_CTX_start(ctx);
    a = BN_CTX_get(ctx);
    b = BN_CTX_get(ctx);
    if (a == NULL || b == NULL)
        goto err;

    if (BN_copy(a, in_a) == NULL)
        goto err;
    if (BN_copy(b, in_b) == NULL)
        goto err;
    a->neg = 0;
    b->neg = 0;

    if (BN_cmp(a, b) < 0) {
        t = a;
        a = b;
        b = t;
    }
    t = euclid(a, b);
    if (t == NULL)
        goto err;

    if (BN_copy(r, t) == NULL)
        goto err;
    ret = 1;
 err:
    BN_CTX_end(ctx);
    bn_check_top(r);
    return (ret);
}
53

U
Ulf Möller 已提交
54
static BIGNUM *euclid(BIGNUM *a, BIGNUM *b)
55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117
{
    BIGNUM *t;
    int shifts = 0;

    bn_check_top(a);
    bn_check_top(b);

    /* 0 <= b <= a */
    while (!BN_is_zero(b)) {
        /* 0 < b <= a */

        if (BN_is_odd(a)) {
            if (BN_is_odd(b)) {
                if (!BN_sub(a, a, b))
                    goto err;
                if (!BN_rshift1(a, a))
                    goto err;
                if (BN_cmp(a, b) < 0) {
                    t = a;
                    a = b;
                    b = t;
                }
            } else {            /* a odd - b even */

                if (!BN_rshift1(b, b))
                    goto err;
                if (BN_cmp(a, b) < 0) {
                    t = a;
                    a = b;
                    b = t;
                }
            }
        } else {                /* a is even */

            if (BN_is_odd(b)) {
                if (!BN_rshift1(a, a))
                    goto err;
                if (BN_cmp(a, b) < 0) {
                    t = a;
                    a = b;
                    b = t;
                }
            } else {            /* a even - b even */

                if (!BN_rshift1(a, a))
                    goto err;
                if (!BN_rshift1(b, b))
                    goto err;
                shifts++;
            }
        }
        /* 0 <= b <= a */
    }

    if (shifts) {
        if (!BN_lshift(a, a, shifts))
            goto err;
    }
    bn_check_top(a);
    return (a);
 err:
    return (NULL);
}
B
Bodo Möller 已提交
118

119
/* solves ax == 1 (mod n) */
A
Andy Polyakov 已提交
120
static BIGNUM *BN_mod_inverse_no_branch(BIGNUM *in,
121 122
                                        const BIGNUM *a, const BIGNUM *n,
                                        BN_CTX *ctx);
123

124
BIGNUM *BN_mod_inverse(BIGNUM *in,
125 126 127 128 129 130 131 132 133
                       const BIGNUM *a, const BIGNUM *n, BN_CTX *ctx)
{
    BIGNUM *rv;
    int noinv;
    rv = int_bn_mod_inverse(in, a, n, ctx, &noinv);
    if (noinv)
        BNerr(BN_F_BN_MOD_INVERSE, BN_R_NO_INVERSE);
    return rv;
}
134 135

BIGNUM *int_bn_mod_inverse(BIGNUM *in,
136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183
                           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;

    if (pnoinv)
        *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);
    }

    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;

    BN_one(X);
    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;
184 185 186 187 188 189 190
    /*-
     * 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|).
     */
191

R
Rich Salz 已提交
192
    if (BN_is_odd(n) && (BN_num_bits(n) <= 2048)) {
193 194 195
        /*
         * Binary inversion algorithm; requires odd modulus. This is faster
         * than the general algorithm if the modulus is sufficiently small
F
FdaSilvaYY 已提交
196
         * (about 400 .. 500 bits on 32-bit systems, but much more on 64-bit
197 198 199 200 201
         * systems)
         */
        int shift;

        while (!BN_is_zero(B)) {
202 203 204 205 206 207
            /*-
             *      0 < B < |n|,
             *      0 < A <= |n|,
             * (1) -sign*X*a  ==  B   (mod |n|),
             * (2)  sign*Y*a  ==  A   (mod |n|)
             */
208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252

            /*
             * 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;
            }

253 254 255 256 257 258 259 260 261 262 263 264
            /*-
             * 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.
             */
265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292
            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;

293 294 295 296 297
            /*-
             *      0 < B < A,
             * (*) -sign*X*a  ==  B   (mod |n|),
             *      sign*Y*a  ==  A   (mod |n|)
             */
298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343

            /* (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;
            }

344 345 346 347 348 349
            /*-
             * Now
             *      A = D*B + M;
             * thus we have
             * (**)  sign*Y*a  ==  D*B + M   (mod |n|).
             */
350 351 352 353 354 355 356 357 358

            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 */

359 360 361 362 363 364 365 366 367 368 369 370 371 372
            /*-
             * 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|).
             *
F
FdaSilvaYY 已提交
373
             * So if we set  (X, Y, sign) := (Y + D*X, X, -sign), we arrive back at
374 375 376 377
             *      -sign*X*a  ==  B   (mod |n|),
             *       sign*Y*a  ==  A   (mod |n|).
             * Note that  X  and  Y  stay non-negative all the time.
             */
378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413

            /*
             * 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;
        }
    }

414 415 416 417 418 419 420
    /*-
     * 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.
     */
421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453

    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 {
        if (pnoinv)
            *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);
}

/*
 * BN_mod_inverse_no_branch is a special version of BN_mod_inverse. It does
 * not contain branches that may leak sensitive information.
454
 */
A
Andy Polyakov 已提交
455
static BIGNUM *BN_mod_inverse_no_branch(BIGNUM *in,
456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496
                                        const BIGNUM *a, const BIGNUM *n,
                                        BN_CTX *ctx)
{
    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;

    BN_one(X);
    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.
         */
497 498
         {
            BIGNUM local_B;
R
Rich Salz 已提交
499
            bn_init(&local_B);
500 501 502 503 504
            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 */
        }
505 506
    }
    sign = -1;
507 508 509 510 511 512 513
    /*-
     * 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|).
     */
514 515 516 517

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

518 519 520 521 522
        /*-
         *      0 < B < A,
         * (*) -sign*X*a  ==  B   (mod |n|),
         *      sign*Y*a  ==  A   (mod |n|)
         */
523 524 525 526 527

        /*
         * Turn BN_FLG_CONSTTIME flag on, so that when BN_div is invoked,
         * BN_div_no_branch will be called eventually.
         */
528 529
        {
            BIGNUM local_A;
R
Rich Salz 已提交
530
            bn_init(&local_A);
531
            BN_with_flags(&local_A, A, BN_FLG_CONSTTIME);
532

533 534 535 536 537
            /* (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 */
        }
538

539 540 541 542 543 544
        /*-
         * Now
         *      A = D*B + M;
         * thus we have
         * (**)  sign*Y*a  ==  D*B + M   (mod |n|).
         */
545 546 547 548 549 550 551 552 553

        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 */

554 555 556 557 558 559 560 561 562 563 564 565 566 567
        /*-
         * 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|).
         *
F
FdaSilvaYY 已提交
568
         * So if we set  (X, Y, sign) := (Y + D*X, X, -sign), we arrive back at
569 570 571 572
         *      -sign*X*a  ==  B   (mod |n|),
         *       sign*Y*a  ==  A   (mod |n|).
         * Note that  X  and  Y  stay non-negative all the time.
         */
573 574 575 576 577 578 579 580 581 582 583 584 585

        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;
    }

586 587 588 589 590 591 592
    /*-
     * 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.
     */
593 594 595 596 597 598 599 600 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620

    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 {
        BNerr(BN_F_BN_MOD_INVERSE_NO_BRANCH, BN_R_NO_INVERSE);
        goto err;
    }
    ret = R;
 err:
    if ((ret == NULL) && (in == NULL))
        BN_free(R);
    BN_CTX_end(ctx);
    bn_check_top(ret);
    return (ret);
}