+/*
+ * Copyright (c) 2018 Thomas Pornin <pornin@bolet.org>
+ *
+ * Permission is hereby granted, free of charge, to any person obtaining
+ * a copy of this software and associated documentation files (the
+ * "Software"), to deal in the Software without restriction, including
+ * without limitation the rights to use, copy, modify, merge, publish,
+ * distribute, sublicense, and/or sell copies of the Software, and to
+ * permit persons to whom the Software is furnished to do so, subject to
+ * the following conditions:
+ *
+ * The above copyright notice and this permission notice shall be
+ * included in all copies or substantial portions of the Software.
+ *
+ * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
+ * EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
+ * MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND
+ * NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS
+ * BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN
+ * ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
+ * CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
+ * SOFTWARE.
+ */
+
+#include "inner.h"
+
+/* see bearssl_rsa.h */
+size_t
+br_rsa_i15_compute_privexp(void *d,
+ const br_rsa_private_key *sk, uint32_t e)
+{
+ /*
+ * We want to invert e modulo phi = (p-1)(q-1). This first
+ * requires computing phi, which is easy since we have the factors
+ * p and q in the private key structure.
+ *
+ * Since p = 3 mod 4 and q = 3 mod 4, phi/4 is an odd integer.
+ * We could invert e modulo phi/4 then patch the result to
+ * modulo phi, but this would involve assembling three modulus-wide
+ * values (phi/4, 1 and e) and calling moddiv, that requires
+ * three more temporaries, for a total of six big integers, or
+ * slightly more than 3 kB of stack space for RSA-4096. This
+ * exceeds our stack requirements.
+ *
+ * Instead, we first use one step of the extended GCD:
+ *
+ * - We compute phi = k*e + r (Euclidean division of phi by e).
+ * If public exponent e is correct, then r != 0 (e must be
+ * invertible modulo phi). We also have k != 0 since we
+ * enforce non-ridiculously-small factors.
+ *
+ * - We find small u, v such that u*e - v*r = 1 (using a
+ * binary GCD; we can arrange for u < r and v < e, i.e. all
+ * values fit on 32 bits).
+ *
+ * - Solution is: d = u + v*k
+ * This last computation is exact: since u < r and v < e,
+ * the above implies d < r + e*((phi-r)/e) = phi
+ */
+
+ uint16_t tmp[4 * ((BR_MAX_RSA_FACTOR + 14) / 15) + 12];
+ uint16_t *p, *q, *k, *m, *z, *phi;
+ const unsigned char *pbuf, *qbuf;
+ size_t plen, qlen, u, len, dlen;
+ uint32_t r, a, b, u0, v0, u1, v1, he, hr;
+ int i;
+
+ /*
+ * Check that e is correct.
+ */
+ if (e < 3 || (e & 1) == 0) {
+ return 0;
+ }
+
+ /*
+ * Check lengths of p and q, and that they are both odd.
+ */
+ pbuf = sk->p;
+ plen = sk->plen;
+ while (plen > 0 && *pbuf == 0) {
+ pbuf ++;
+ plen --;
+ }
+ if (plen < 5 || plen > (BR_MAX_RSA_FACTOR / 8)
+ || (pbuf[plen - 1] & 1) != 1)
+ {
+ return 0;
+ }
+ qbuf = sk->q;
+ qlen = sk->qlen;
+ while (qlen > 0 && *qbuf == 0) {
+ qbuf ++;
+ qlen --;
+ }
+ if (qlen < 5 || qlen > (BR_MAX_RSA_FACTOR / 8)
+ || (qbuf[qlen - 1] & 1) != 1)
+ {
+ return 0;
+ }
+
+ /*
+ * Output length is that of the modulus.
+ */
+ dlen = (sk->n_bitlen + 7) >> 3;
+ if (d == NULL) {
+ return dlen;
+ }
+
+ p = tmp;
+ br_i15_decode(p, pbuf, plen);
+ plen = (p[0] + 15) >> 4;
+ q = p + 1 + plen;
+ br_i15_decode(q, qbuf, qlen);
+ qlen = (q[0] + 15) >> 4;
+
+ /*
+ * Compute phi = (p-1)*(q-1), then move it over p-1 and q-1 (that
+ * we do not need anymore). The mulacc function sets the announced
+ * bit length of t to be the sum of the announced bit lengths of
+ * p-1 and q-1, which is usually exact but may overshoot by one 1
+ * bit in some cases; we readjust it to its true length.
+ */
+ p[1] --;
+ q[1] --;
+ phi = q + 1 + qlen;
+ br_i15_zero(phi, p[0]);
+ br_i15_mulacc(phi, p, q);
+ len = (phi[0] + 15) >> 4;
+ memmove(tmp, phi, (1 + len) * sizeof *phi);
+ phi = tmp;
+ phi[0] = br_i15_bit_length(phi + 1, len);
+ len = (phi[0] + 15) >> 4;
+
+ /*
+ * Divide phi by public exponent e. The final remainder r must be
+ * non-zero (otherwise, the key is invalid). The quotient is k,
+ * which we write over phi, since we don't need phi after that.
+ */
+ r = 0;
+ for (u = len; u >= 1; u --) {
+ /*
+ * Upon entry, r < e, and phi[u] < 2^15; hence,
+ * hi:lo < e*2^15. Thus, the produced word k[u]
+ * must be lower than 2^15, and the new remainder r
+ * is lower than e.
+ */
+ uint32_t hi, lo;
+
+ hi = r >> 17;
+ lo = (r << 15) + phi[u];
+ phi[u] = br_divrem(hi, lo, e, &r);
+ }
+ if (r == 0) {
+ return 0;
+ }
+ k = phi;
+
+ /*
+ * Compute u and v such that u*e - v*r = GCD(e,r). We use
+ * a binary GCD algorithm, with 6 extra integers a, b,
+ * u0, u1, v0 and v1. Initial values are:
+ * a = e u0 = 1 v0 = 0
+ * b = r u1 = r v1 = e-1
+ * The following invariants are maintained:
+ * a = u0*e - v0*r
+ * b = u1*e - v1*r
+ * 0 < a <= e
+ * 0 < b <= r
+ * 0 <= u0 <= r
+ * 0 <= v0 <= e
+ * 0 <= u1 <= r
+ * 0 <= v1 <= e
+ *
+ * At each iteration, we reduce either a or b by one bit, and
+ * adjust u0, u1, v0 and v1 to maintain the invariants:
+ * - if a is even, then a <- a/2
+ * - otherwise, if b is even, then b <- b/2
+ * - otherwise, if a > b, then a <- (a-b)/2
+ * - otherwise, if b > a, then b <- (b-a)/2
+ * Algorithm stops when a = b. At that point, the common value
+ * is the GCD of e and r; it must be 1 (otherwise, the private
+ * key or public exponent is not valid). The (u0,v0) or (u1,v1)
+ * pairs are the solution we are looking for.
+ *
+ * Since either a or b is reduced by at least 1 bit at each
+ * iteration, 62 iterations are enough to reach the end
+ * condition.
+ *
+ * To maintain the invariants, we must compute the same operations
+ * on the u* and v* values that we do on a and b:
+ * - When a is divided by 2, u0 and v0 must be divided by 2.
+ * - When b is divided by 2, u1 and v1 must be divided by 2.
+ * - When b is subtracted from a, u1 and v1 are subtracted from
+ * u0 and v0, respectively.
+ * - When a is subtracted from b, u0 and v0 are subtracted from
+ * u1 and v1, respectively.
+ *
+ * However, we want to keep the u* and v* values in their proper
+ * ranges. The following remarks apply:
+ *
+ * - When a is divided by 2, then a is even. Therefore:
+ *
+ * * If r is odd, then u0 and v0 must have the same parity;
+ * if they are both odd, then adding r to u0 and e to v0
+ * makes them both even, and the division by 2 brings them
+ * back to the proper range.
+ *
+ * * If r is even, then u0 must be even; if v0 is odd, then
+ * adding r to u0 and e to v0 makes them both even, and the
+ * division by 2 brings them back to the proper range.
+ *
+ * Thus, all we need to do is to look at the parity of v0,
+ * and add (r,e) to (u0,v0) when v0 is odd. In order to avoid
+ * a 32-bit overflow, we can add ((r+1)/2,(e/2)+1) after the
+ * division (r+1 does not overflow since r < e; and (e/2)+1
+ * is equal to (e+1)/2 since e is odd).
+ *
+ * - When we subtract b from a, three cases may occur:
+ *
+ * * u1 <= u0 and v1 <= v0: just do the subtractions
+ *
+ * * u1 > u0 and v1 > v0: compute:
+ * (u0, v0) <- (u0 + r - u1, v0 + e - v1)
+ *
+ * * u1 <= u0 and v1 > v0: compute:
+ * (u0, v0) <- (u0 + r - u1, v0 + e - v1)
+ *
+ * The fourth case (u1 > u0 and v1 <= v0) is not possible
+ * because it would contradict "b < a" (which is the reason
+ * why we subtract b from a).
+ *
+ * The tricky case is the third one: from the equations, it
+ * seems that u0 may go out of range. However, the invariants
+ * and ranges of other values imply that, in that case, the
+ * new u0 does not actually exceed the range.
+ *
+ * We can thus handle the subtraction by adding (r,e) based
+ * solely on the comparison between v0 and v1.
+ */
+ a = e;
+ b = r;
+ u0 = 1;
+ v0 = 0;
+ u1 = r;
+ v1 = e - 1;
+ hr = (r + 1) >> 1;
+ he = (e >> 1) + 1;
+ for (i = 0; i < 62; i ++) {
+ uint32_t oa, ob, agtb, bgta;
+ uint32_t sab, sba, da, db;
+ uint32_t ctl;
+
+ oa = a & 1; /* 1 if a is odd */
+ ob = b & 1; /* 1 if b is odd */
+ agtb = GT(a, b); /* 1 if a > b */
+ bgta = GT(b, a); /* 1 if b > a */
+
+ sab = oa & ob & agtb; /* 1 if a <- a-b */
+ sba = oa & ob & bgta; /* 1 if b <- b-a */
+
+ /* a <- a-b, u0 <- u0-u1, v0 <- v0-v1 */
+ ctl = GT(v1, v0);
+ a -= b & -sab;
+ u0 -= (u1 - (r & -ctl)) & -sab;
+ v0 -= (v1 - (e & -ctl)) & -sab;
+
+ /* b <- b-a, u1 <- u1-u0 mod r, v1 <- v1-v0 mod e */
+ ctl = GT(v0, v1);
+ b -= a & -sba;
+ u1 -= (u0 - (r & -ctl)) & -sba;
+ v1 -= (v0 - (e & -ctl)) & -sba;
+
+ da = NOT(oa) | sab; /* 1 if a <- a/2 */
+ db = (oa & NOT(ob)) | sba; /* 1 if b <- b/2 */
+
+ /* a <- a/2, u0 <- u0/2, v0 <- v0/2 */
+ ctl = v0 & 1;
+ a ^= (a ^ (a >> 1)) & -da;
+ u0 ^= (u0 ^ ((u0 >> 1) + (hr & -ctl))) & -da;
+ v0 ^= (v0 ^ ((v0 >> 1) + (he & -ctl))) & -da;
+
+ /* b <- b/2, u1 <- u1/2 mod r, v1 <- v1/2 mod e */
+ ctl = v1 & 1;
+ b ^= (b ^ (b >> 1)) & -db;
+ u1 ^= (u1 ^ ((u1 >> 1) + (hr & -ctl))) & -db;
+ v1 ^= (v1 ^ ((v1 >> 1) + (he & -ctl))) & -db;
+ }
+
+ /*
+ * Check that the GCD is indeed 1. If not, then the key is invalid
+ * (and there's no harm in leaking that piece of information).
+ */
+ if (a != 1) {
+ return 0;
+ }
+
+ /*
+ * Now we have u0*e - v0*r = 1. Let's compute the result as:
+ * d = u0 + v0*k
+ * We still have k in the tmp[] array, and its announced bit
+ * length is that of phi.
+ */
+ m = k + 1 + len;
+ m[0] = (2 << 4) + 2; /* bit length is 32 bits, encoded */
+ m[1] = v0 & 0x7FFF;
+ m[2] = (v0 >> 15) & 0x7FFF;
+ m[3] = v0 >> 30;
+ z = m + 4;
+ br_i15_zero(z, k[0]);
+ z[1] = u0 & 0x7FFF;
+ z[2] = (u0 >> 15) & 0x7FFF;
+ z[3] = u0 >> 30;
+ br_i15_mulacc(z, k, m);
+
+ /*
+ * Encode the result.
+ */
+ br_i15_encode(d, dlen, z);
+ return dlen;
+}
projects / BearSSL / commitdiff