/src/openssl30/crypto/bn/bn_sqrt.c

Source (jump to first uncovered line)
/*
 * Copyright 2000-2022 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"

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, otherwise an error or
 * an incorrect "result" will be returned.
 */
{
    BIGNUM *ret = in;
    int err = 1;
    int r;
    BIGNUM *A, *b, *q, *t, *x, *y;
    int e, i, j;
    int used_ctx = 0;

    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;
        }

        ERR_raise(ERR_LIB_BN, BN_R_P_IS_NOT_PRIME);
        return NULL;
    }

    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);
    used_ctx = 1;
    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) {
        /*-
         * 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.
         */
        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) {
        /*-
         * |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,
         * Subject: Square Roots and Cognate Matters modulo p=8n+5.
         * URL: https://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind9211&L=NMBRTHRY&P=4026
         * November 1992.)
         */

        /* 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 {
            if (!BN_priv_rand_ex(y, BN_num_bits(p), 0, 0, 0, ctx))
                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 */
            ERR_raise(ERR_LIB_BN, 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.
         */
        ERR_raise(ERR_LIB_BN, 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)) {
        ERR_raise(ERR_LIB_BN, BN_R_P_IS_NOT_PRIME);
        goto end;
    }

    /*-
     * 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.
     */

    /* 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) {
        /*-
         * 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.
         */

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

        /* Find the smallest i, 0 < i < e, such that b^(2^i) = 1. */
        for (i = 1; i < e; i++) {
            if (i == 1) {
                if (!BN_mod_sqr(t, b, p, ctx))
                    goto end;

            } else {
                if (!BN_mod_mul(t, t, t, p, ctx))
                    goto end;
            }
            if (BN_is_one(t))
                break;
        }
        /* If not found, a is not a square or p is not prime. */
        if (i >= e) {
            ERR_raise(ERR_LIB_BN, BN_R_NOT_A_SQUARE);
            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;
    }

 vrfy:
    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)) {
            ERR_raise(ERR_LIB_BN, BN_R_NOT_A_SQUARE);
            err = 1;
        }
    }

 end:
    if (err) {
        if (ret != in)
            BN_clear_free(ret);
        ret = NULL;
    }
    if (used_ctx)
        BN_CTX_end(ctx);
    bn_check_top(ret);
    return ret;
}

Coverage Report

Created: 2023-09-25 06:45

Line	Count	Source (jump to first uncovered line)
1		/*
2		* Copyright 2000-2022 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
13		BIGNUM BN_mod_sqrt(BIGNUM in, const BIGNUM a, const BIGNUM p, BN_CTX *ctx)
14		/*
15		* Returns 'ret' such that ret^2 == a (mod p), using the Tonelli/Shanks
16		* algorithm (cf. Henri Cohen, "A Course in Algebraic Computational Number
17		* Theory", algorithm 1.5.1). 'p' must be prime, otherwise an error or
18		* an incorrect "result" will be returned.
19		*/
20	61.4k	{
21	61.4k	BIGNUM *ret = in;
22	61.4k	int err = 1;
23	61.4k	int r;
24	61.4k	BIGNUM A, b, q, t, x, y;
25	61.4k	int e, i, j;
26	61.4k	int used_ctx = 0;
27
28	61.4k	if (!BN_is_odd(p) \|\| BN_abs_is_word(p, 1)) {
29	0	if (BN_abs_is_word(p, 2)) {
30	0	if (ret == NULL)
31	0	ret = BN_new();
32	0	if (ret == NULL)
33	0	goto end;
34	0	if (!BN_set_word(ret, BN_is_bit_set(a, 0))) {
35	0	if (ret != in)
36	0	BN_free(ret);
37	0	return NULL;
38	0	}
39	0	bn_check_top(ret);
40	0	return ret;
41	0	}
42
43	0	ERR_raise(ERR_LIB_BN, BN_R_P_IS_NOT_PRIME);
44	0	return NULL;
45	0	}
46
47	61.4k	if (BN_is_zero(a) \|\| BN_is_one(a)) {
48	167	if (ret == NULL)
49	0	ret = BN_new();
50	167	if (ret == NULL)
51	0	goto end;
52	167	if (!BN_set_word(ret, BN_is_one(a))) {
53	0	if (ret != in)
54	0	BN_free(ret);
55	0	return NULL;
56	0	}
57	167	bn_check_top(ret);
58	167	return ret;
59	167	}
60
61	61.2k	BN_CTX_start(ctx);
62	61.2k	used_ctx = 1;
63	61.2k	A = BN_CTX_get(ctx);
64	61.2k	b = BN_CTX_get(ctx);
65	61.2k	q = BN_CTX_get(ctx);
66	61.2k	t = BN_CTX_get(ctx);
67	61.2k	x = BN_CTX_get(ctx);
68	61.2k	y = BN_CTX_get(ctx);
69	61.2k	if (y == NULL)
70	0	goto end;
71
72	61.2k	if (ret == NULL)
73	0	ret = BN_new();
74	61.2k	if (ret == NULL)
75	0	goto end;
76
77		/* A = a mod p */
78	61.2k	if (!BN_nnmod(A, a, p, ctx))
79	0	goto end;
80
81		/* now write \|p\| - 1 as 2^eq where q is odd /
82	61.2k	e = 1;
83	1.62M	while (!BN_is_bit_set(p, e))
84	1.56M	e++;
85		/* we'll set q later (if needed) */
86
87	61.2k	if (e == 1) {
88		/*-
89		* The easy case: (\|p\|-1)/2 is odd, so 2 has an inverse
90		* modulo (\|p\|-1)/2, and square roots can be computed
91		* directly by modular exponentiation.
92		* We have
93		* 2 * (\|p\|+1)/4 == 1 (mod (\|p\|-1)/2),
94		* so we can use exponent (\|p\|+1)/4, i.e. (\|p\|-3)/4 + 1.
95		*/
96	40.7k	if (!BN_rshift(q, p, 2))
97	0	goto end;
98	40.7k	q->neg = 0;
99	40.7k	if (!BN_add_word(q, 1))
100	0	goto end;
101	40.7k	if (!BN_mod_exp(ret, A, q, p, ctx))
102	0	goto end;
103	40.7k	err = 0;
104	40.7k	goto vrfy;
105	40.7k	}
106
107	20.4k	if (e == 2) {
108		/*-
109		* \|p\| == 5 (mod 8)
110		*
111		* In this case 2 is always a non-square since
112		* Legendre(2,p) = (-1)^((p^2-1)/8) for any odd prime.
113		* So if a really is a square, then 2*a is a non-square.
114		* Thus for
115		* b := (2*a)^((\|p\|-5)/8),
116		* i := (2a)b^2
117		* we have
118		* i^2 = (2a)^((1 + (\|p\|-5)/4)2)
119		* = (2*a)^((p-1)/2)
120		* = -1;
121		* so if we set
122		* x := ab(i-1),
123		* then
124		* x^2 = a^2 * b^2 * (i^2 - 2*i + 1)
125		* = a^2 * b^2 * (-2*i)
126		* = a(-i)(2ab^2)
127		* = a(-i)i
128		* = a.
129		*
130		* (This is due to A.O.L. Atkin,
131		* Subject: Square Roots and Cognate Matters modulo p=8n+5.
132		* URL: https://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind9211&L=NMBRTHRY&P=4026
133		* November 1992.)
134		*/
135
136		/* t := 2a /
137	2.71k	if (!BN_mod_lshift1_quick(t, A, p))
138	0	goto end;
139
140		/* b := (2a)^((\|p\|-5)/8) /
141	2.71k	if (!BN_rshift(q, p, 3))
142	0	goto end;
143	2.71k	q->neg = 0;
144	2.71k	if (!BN_mod_exp(b, t, q, p, ctx))
145	0	goto end;
146
147		/* y := b^2 */
148	2.71k	if (!BN_mod_sqr(y, b, p, ctx))
149	0	goto end;
150
151		/* t := (2a)b^2 - 1 */
152	2.71k	if (!BN_mod_mul(t, t, y, p, ctx))
153	0	goto end;
154	2.71k	if (!BN_sub_word(t, 1))
155	0	goto end;
156
157		/* x = abt */
158	2.71k	if (!BN_mod_mul(x, A, b, p, ctx))
159	0	goto end;
160	2.71k	if (!BN_mod_mul(x, x, t, p, ctx))
161	0	goto end;
162
163	2.71k	if (!BN_copy(ret, x))
164	0	goto end;
165	2.71k	err = 0;
166	2.71k	goto vrfy;
167	2.71k	}
168
169		/*
170		* e > 2, so we really have to use the Tonelli/Shanks algorithm. First,
171		* find some y that is not a square.
172		*/
173	17.7k	if (!BN_copy(q, p))
174	0	goto end; /* use 'q' as temp */
175	17.7k	q->neg = 0;
176	17.7k	i = 2;
177	175k	do {
178		/*
179		* For efficiency, try small numbers first; if this fails, try random
180		* numbers.
181		*/
182	175k	if (i < 22) {
183	169k	if (!BN_set_word(y, i))
184	0	goto end;
185	169k	} else {
186	6.54k	if (!BN_priv_rand_ex(y, BN_num_bits(p), 0, 0, 0, ctx))
187	0	goto end;
188	6.54k	if (BN_ucmp(y, p) >= 0) {
189	2.52k	if (!(p->neg ? BN_add : BN_sub) (y, y, p))
190	0	goto end;
191	2.52k	}
192		/* now 0 <= y < \|p\| */
193	6.54k	if (BN_is_zero(y))
194	0	if (!BN_set_word(y, i))
195	0	goto end;
196	6.54k	}
197
198	175k	r = BN_kronecker(y, q, ctx); /* here 'q' is \|p\| */
199	175k	if (r < -1)
200	0	goto end;
201	175k	if (r == 0) {
202		/* m divides p */
203	163	ERR_raise(ERR_LIB_BN, BN_R_P_IS_NOT_PRIME);
204	163	goto end;
205	163	}
206	175k	}
207	175k	while (r == 1 && ++i < 82);
208
209	17.5k	if (r != -1) {
210		/*
211		* Many rounds and still no non-square -- this is more likely a bug
212		* than just bad luck. Even if p is not prime, we should have found
213		* some y such that r == -1.
214		*/
215	107	ERR_raise(ERR_LIB_BN, BN_R_TOO_MANY_ITERATIONS);
216	107	goto end;
217	107	}
218
219		/* Here's our actual 'q': */
220	17.4k	if (!BN_rshift(q, q, e))
221	0	goto end;
222
223		/*
224		* Now that we have some non-square, we can find an element of order 2^e
225		* by computing its q'th power.
226		*/
227	17.4k	if (!BN_mod_exp(y, y, q, p, ctx))
228	0	goto end;
229	17.4k	if (BN_is_one(y)) {
230	0	ERR_raise(ERR_LIB_BN, BN_R_P_IS_NOT_PRIME);
231	0	goto end;
232	0	}
233
234		/*-
235		* Now we know that (if p is indeed prime) there is an integer
236		* k, 0 <= k < 2^e, such that
237		*
238		* a^q * y^k == 1 (mod p).
239		*
240		* As a^q is a square and y is not, k must be even.
241		* q+1 is even, too, so there is an element
242		*
243		* X := a^((q+1)/2) * y^(k/2),
244		*
245		* and it satisfies
246		*
247		* X^2 = a^q * a * y^k
248		* = a,
249		*
250		* so it is the square root that we are looking for.
251		*/
252
253		/* t := (q-1)/2 (note that q is odd) */
254	17.4k	if (!BN_rshift1(t, q))
255	0	goto end;
256
257		/* x := a^((q-1)/2) */
258	17.4k	if (BN_is_zero(t)) { /* special case: p = 2^e + 1 */
259	14	if (!BN_nnmod(t, A, p, ctx))
260	0	goto end;
261	14	if (BN_is_zero(t)) {
262		/* special case: a == 0 (mod p) */
263	0	BN_zero(ret);
264	0	err = 0;
265	0	goto end;
266	14	} else if (!BN_one(x))
267	0	goto end;
268	17.4k	} else {
269	17.4k	if (!BN_mod_exp(x, A, t, p, ctx))
270	0	goto end;
271	17.4k	if (BN_is_zero(x)) {
272		/* special case: a == 0 (mod p) */
273	0	BN_zero(ret);
274	0	err = 0;
275	0	goto end;
276	0	}
277	17.4k	}
278
279		/* b := ax^2 (= a^q) /
280	17.4k	if (!BN_mod_sqr(b, x, p, ctx))
281	0	goto end;
282	17.4k	if (!BN_mod_mul(b, b, A, p, ctx))
283	0	goto end;
284
285		/* x := ax (= a^((q+1)/2)) /
286	17.4k	if (!BN_mod_mul(x, x, A, p, ctx))
287	0	goto end;
288
289	654k	while (1) {
290		/*-
291		* Now b is a^q * y^k for some even k (0 <= k < 2^E
292		* where E refers to the original value of e, which we
293		* don't keep in a variable), and x is a^((q+1)/2) * y^(k/2).
294		*
295		* We have a*b = x^2,
296		* y^2^(e-1) = -1,
297		* b^2^(e-1) = 1.
298		*/
299
300	654k	if (BN_is_one(b)) {
301	14.2k	if (!BN_copy(ret, x))
302	0	goto end;
303	14.2k	err = 0;
304	14.2k	goto vrfy;
305	14.2k	}
306
307		/* Find the smallest i, 0 < i < e, such that b^(2^i) = 1. */
308	30.2M	for (i = 1; i < e; i++) {
309	30.2M	if (i == 1) {
310	640k	if (!BN_mod_sqr(t, b, p, ctx))
311	0	goto end;
312
313	29.6M	} else {
314	29.6M	if (!BN_mod_mul(t, t, t, p, ctx))
315	0	goto end;
316	29.6M	}
317	30.2M	if (BN_is_one(t))
318	637k	break;
319	30.2M	}
320		/* If not found, a is not a square or p is not prime. */
321	640k	if (i >= e) {
322	3.20k	ERR_raise(ERR_LIB_BN, BN_R_NOT_A_SQUARE);
323	3.20k	goto end;
324	3.20k	}
325
326		/* t := y^2^(e - i - 1) */
327	637k	if (!BN_copy(t, y))
328	0	goto end;
329	1.27M	for (j = e - i - 1; j > 0; j--) {
330	638k	if (!BN_mod_sqr(t, t, p, ctx))
331	0	goto end;
332	638k	}
333	637k	if (!BN_mod_mul(y, t, t, p, ctx))
334	0	goto end;
335	637k	if (!BN_mod_mul(x, x, t, p, ctx))
336	0	goto end;
337	637k	if (!BN_mod_mul(b, b, y, p, ctx))
338	0	goto end;
339	637k	e = i;
340	637k	}
341
342	57.7k	vrfy:
343	57.7k	if (!err) {
344		/*
345		* verify the result -- the input might have been not a square (test
346		* added in 0.9.8)
347		*/
348
349	57.7k	if (!BN_mod_sqr(x, ret, p, ctx))
350	0	err = 1;
351
352	57.7k	if (!err && 0 != BN_cmp(x, A)) {
353	15.0k	ERR_raise(ERR_LIB_BN, BN_R_NOT_A_SQUARE);
354	15.0k	err = 1;
355	15.0k	}
356	57.7k	}
357
358	61.2k	end:
359	61.2k	if (err) {
360	18.5k	if (ret != in)
361	0	BN_clear_free(ret);
362	18.5k	ret = NULL;
363	18.5k	}
364	61.2k	if (used_ctx)
365	61.2k	BN_CTX_end(ctx);
366	61.2k	bn_check_top(ret);
367	61.2k	return ret;
368	57.7k	}