bn_sqrt.c 9.2 KB
Newer Older
1
/*
M
Matt Caswell 已提交
2
 * Copyright 2000-2018 The OpenSSL Project Authors. All Rights Reserved.
B
BN_sqrt  
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
BN_sqrt  
Bodo Möller 已提交
8 9
 */

10
#include "internal/cryptlib.h"
B
BN_sqrt  
Bodo Möller 已提交
11 12
#include "bn_lcl.h"

13 14 15 16 17
BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
/*
 * Returns 'ret' such that ret^2 == a (mod p), using the Tonelli/Shanks
 * algorithm (cf. Henri Cohen, "A Course in Algebraic Computational Number
 * Theory", algorithm 1.5.1). 'p' must be prime!
B
BN_sqrt  
Bodo Möller 已提交
18
 */
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41
{
    BIGNUM *ret = in;
    int err = 1;
    int r;
    BIGNUM *A, *b, *q, *t, *x, *y;
    int e, i, j;

    if (!BN_is_odd(p) || BN_abs_is_word(p, 1)) {
        if (BN_abs_is_word(p, 2)) {
            if (ret == NULL)
                ret = BN_new();
            if (ret == NULL)
                goto end;
            if (!BN_set_word(ret, BN_is_bit_set(a, 0))) {
                if (ret != in)
                    BN_free(ret);
                return NULL;
            }
            bn_check_top(ret);
            return ret;
        }

        BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
K
KaoruToda 已提交
42
        return NULL;
43 44 45 46 47 48 49 50 51 52 53 54 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
    }

    if (BN_is_zero(a) || BN_is_one(a)) {
        if (ret == NULL)
            ret = BN_new();
        if (ret == NULL)
            goto end;
        if (!BN_set_word(ret, BN_is_one(a))) {
            if (ret != in)
                BN_free(ret);
            return NULL;
        }
        bn_check_top(ret);
        return ret;
    }

    BN_CTX_start(ctx);
    A = BN_CTX_get(ctx);
    b = BN_CTX_get(ctx);
    q = BN_CTX_get(ctx);
    t = BN_CTX_get(ctx);
    x = BN_CTX_get(ctx);
    y = BN_CTX_get(ctx);
    if (y == NULL)
        goto end;

    if (ret == NULL)
        ret = BN_new();
    if (ret == NULL)
        goto end;

    /* A = a mod p */
    if (!BN_nnmod(A, a, p, ctx))
        goto end;

    /* now write  |p| - 1  as  2^e*q  where  q  is odd */
    e = 1;
    while (!BN_is_bit_set(p, e))
        e++;
    /* we'll set  q  later (if needed) */

    if (e == 1) {
85 86 87 88 89 90 91 92
        /*-
         * The easy case:  (|p|-1)/2  is odd, so 2 has an inverse
         * modulo  (|p|-1)/2,  and square roots can be computed
         * directly by modular exponentiation.
         * We have
         *     2 * (|p|+1)/4 == 1   (mod (|p|-1)/2),
         * so we can use exponent  (|p|+1)/4,  i.e.  (|p|-3)/4 + 1.
         */
93 94 95 96 97 98 99 100 101 102 103 104
        if (!BN_rshift(q, p, 2))
            goto end;
        q->neg = 0;
        if (!BN_add_word(q, 1))
            goto end;
        if (!BN_mod_exp(ret, A, q, p, ctx))
            goto end;
        err = 0;
        goto vrfy;
    }

    if (e == 2) {
M
Matt Caswell 已提交
105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130
        /*-
         * |p| == 5  (mod 8)
         *
         * In this case  2  is always a non-square since
         * Legendre(2,p) = (-1)^((p^2-1)/8)  for any odd prime.
         * So if  a  really is a square, then  2*a  is a non-square.
         * Thus for
         *      b := (2*a)^((|p|-5)/8),
         *      i := (2*a)*b^2
         * we have
         *     i^2 = (2*a)^((1 + (|p|-5)/4)*2)
         *         = (2*a)^((p-1)/2)
         *         = -1;
         * so if we set
         *      x := a*b*(i-1),
         * then
         *     x^2 = a^2 * b^2 * (i^2 - 2*i + 1)
         *         = a^2 * b^2 * (-2*i)
         *         = a*(-i)*(2*a*b^2)
         *         = a*(-i)*i
         *         = a.
         *
         * (This is due to A.O.L. Atkin,
         * <URL: http://listserv.nodak.edu/scripts/wa.exe?A2=ind9211&L=nmbrthry&O=T&P=562>,
         * November 1992.)
         */
131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181

        /* t := 2*a */
        if (!BN_mod_lshift1_quick(t, A, p))
            goto end;

        /* b := (2*a)^((|p|-5)/8) */
        if (!BN_rshift(q, p, 3))
            goto end;
        q->neg = 0;
        if (!BN_mod_exp(b, t, q, p, ctx))
            goto end;

        /* y := b^2 */
        if (!BN_mod_sqr(y, b, p, ctx))
            goto end;

        /* t := (2*a)*b^2 - 1 */
        if (!BN_mod_mul(t, t, y, p, ctx))
            goto end;
        if (!BN_sub_word(t, 1))
            goto end;

        /* x = a*b*t */
        if (!BN_mod_mul(x, A, b, p, ctx))
            goto end;
        if (!BN_mod_mul(x, x, t, p, ctx))
            goto end;

        if (!BN_copy(ret, x))
            goto end;
        err = 0;
        goto vrfy;
    }

    /*
     * e > 2, so we really have to use the Tonelli/Shanks algorithm. First,
     * find some y that is not a square.
     */
    if (!BN_copy(q, p))
        goto end;               /* use 'q' as temp */
    q->neg = 0;
    i = 2;
    do {
        /*
         * For efficiency, try small numbers first; if this fails, try random
         * numbers.
         */
        if (i < 22) {
            if (!BN_set_word(y, i))
                goto end;
        } else {
182
            if (!BN_priv_rand(y, BN_num_bits(p), 0, 0))
183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229
                goto end;
            if (BN_ucmp(y, p) >= 0) {
                if (!(p->neg ? BN_add : BN_sub) (y, y, p))
                    goto end;
            }
            /* now 0 <= y < |p| */
            if (BN_is_zero(y))
                if (!BN_set_word(y, i))
                    goto end;
        }

        r = BN_kronecker(y, q, ctx); /* here 'q' is |p| */
        if (r < -1)
            goto end;
        if (r == 0) {
            /* m divides p */
            BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
            goto end;
        }
    }
    while (r == 1 && ++i < 82);

    if (r != -1) {
        /*
         * Many rounds and still no non-square -- this is more likely a bug
         * than just bad luck. Even if p is not prime, we should have found
         * some y such that r == -1.
         */
        BNerr(BN_F_BN_MOD_SQRT, BN_R_TOO_MANY_ITERATIONS);
        goto end;
    }

    /* Here's our actual 'q': */
    if (!BN_rshift(q, q, e))
        goto end;

    /*
     * Now that we have some non-square, we can find an element of order 2^e
     * by computing its q'th power.
     */
    if (!BN_mod_exp(y, y, q, p, ctx))
        goto end;
    if (BN_is_one(y)) {
        BNerr(BN_F_BN_MOD_SQRT, BN_R_P_IS_NOT_PRIME);
        goto end;
    }

230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247
    /*-
     * Now we know that (if  p  is indeed prime) there is an integer
     * k,  0 <= k < 2^e,  such that
     *
     *      a^q * y^k == 1   (mod p).
     *
     * As  a^q  is a square and  y  is not,  k  must be even.
     * q+1  is even, too, so there is an element
     *
     *     X := a^((q+1)/2) * y^(k/2),
     *
     * and it satisfies
     *
     *     X^2 = a^q * a     * y^k
     *         = a,
     *
     * so it is the square root that we are looking for.
     */
248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285

    /* t := (q-1)/2  (note that  q  is odd) */
    if (!BN_rshift1(t, q))
        goto end;

    /* x := a^((q-1)/2) */
    if (BN_is_zero(t)) {        /* special case: p = 2^e + 1 */
        if (!BN_nnmod(t, A, p, ctx))
            goto end;
        if (BN_is_zero(t)) {
            /* special case: a == 0  (mod p) */
            BN_zero(ret);
            err = 0;
            goto end;
        } else if (!BN_one(x))
            goto end;
    } else {
        if (!BN_mod_exp(x, A, t, p, ctx))
            goto end;
        if (BN_is_zero(x)) {
            /* special case: a == 0  (mod p) */
            BN_zero(ret);
            err = 0;
            goto end;
        }
    }

    /* b := a*x^2  (= a^q) */
    if (!BN_mod_sqr(b, x, p, ctx))
        goto end;
    if (!BN_mod_mul(b, b, A, p, ctx))
        goto end;

    /* x := a*x    (= a^((q+1)/2)) */
    if (!BN_mod_mul(x, x, A, p, ctx))
        goto end;

    while (1) {
286 287 288 289 290 291 292 293 294
        /*-
         * Now  b  is  a^q * y^k  for some even  k  (0 <= k < 2^E
         * where  E  refers to the original value of  e,  which we
         * don't keep in a variable),  and  x  is  a^((q+1)/2) * y^(k/2).
         *
         * We have  a*b = x^2,
         *    y^2^(e-1) = -1,
         *    b^2^(e-1) = 1.
         */
295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331

        if (BN_is_one(b)) {
            if (!BN_copy(ret, x))
                goto end;
            err = 0;
            goto vrfy;
        }

        /* find smallest  i  such that  b^(2^i) = 1 */
        i = 1;
        if (!BN_mod_sqr(t, b, p, ctx))
            goto end;
        while (!BN_is_one(t)) {
            i++;
            if (i == e) {
                BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
                goto end;
            }
            if (!BN_mod_mul(t, t, t, p, ctx))
                goto end;
        }

        /* t := y^2^(e - i - 1) */
        if (!BN_copy(t, y))
            goto end;
        for (j = e - i - 1; j > 0; j--) {
            if (!BN_mod_sqr(t, t, p, ctx))
                goto end;
        }
        if (!BN_mod_mul(y, t, t, p, ctx))
            goto end;
        if (!BN_mod_mul(x, x, t, p, ctx))
            goto end;
        if (!BN_mod_mul(b, b, y, p, ctx))
            goto end;
        e = i;
    }
B
BN_sqrt  
Bodo Möller 已提交
332

333
 vrfy:
334 335 336 337 338 339 340 341 342 343 344 345 346 347
    if (!err) {
        /*
         * verify the result -- the input might have been not a square (test
         * added in 0.9.8)
         */

        if (!BN_mod_sqr(x, ret, p, ctx))
            err = 1;

        if (!err && 0 != BN_cmp(x, A)) {
            BNerr(BN_F_BN_MOD_SQRT, BN_R_NOT_A_SQUARE);
            err = 1;
        }
    }
348

B
BN_sqrt  
Bodo Möller 已提交
349
 end:
350
    if (err) {
R
Rich Salz 已提交
351
        if (ret != in)
352 353 354 355 356 357 358
            BN_clear_free(ret);
        ret = NULL;
    }
    BN_CTX_end(ctx);
    bn_check_top(ret);
    return ret;
}