org.bouncycastle.pqc.crypto.falcon.FalconKeyGen Maven / Gradle / Ivy
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The Bouncy Castle Crypto package is a Java implementation of cryptographic algorithms. This jar contains JCE provider and lightweight API for the Bouncy Castle Cryptography APIs for JDK 1.8 and up.
package org.bouncycastle.pqc.crypto.falcon;
class FalconKeyGen
{
FPREngine fpr;
FalconSmallPrimeList primes;
FalconFFT fft;
FalconCodec codec;
FalconVrfy vrfy;
FalconKeyGen()
{
this.fpr = new FPREngine();
this.primes = new FalconSmallPrimeList();
this.fft = new FalconFFT();
this.codec = new FalconCodec();
this.vrfy = new FalconVrfy();
}
private static int mkn(int logn)
{
return 1 << logn;
}
/*
* Reduce a small signed integer modulo a small prime. The source
* value x MUST be such that -p < x < p.
*/
int modp_set(int x, int p)
{
int w;
w = x;
w += p & -(w >>> 31);
return w;
}
/*
* Normalize a modular integer around 0.
*/
int modp_norm(int x, int p)
{
return (x - (p & (((x - ((p + 1) >>> 1)) >>> 31) - 1)));
}
/*
* Compute -1/p mod 2^31. This works for all odd integers p that fit
* on 31 bits.
*/
int modp_ninv31(int p)
{
int y;
y = 2 - p;
y *= 2 - p * y;
y *= 2 - p * y;
y *= 2 - p * y;
y *= 2 - p * y;
return 0x7FFFFFFF & -y;
}
/*
* Compute R = 2^31 mod p.
*/
int modp_R(int p)
{
/*
* Since 2^30 < p < 2^31, we know that 2^31 mod p is simply
* 2^31 - p.
*/
return (1 << 31) - p;
}
/*
* Addition modulo p.
*/
int modp_add(int a, int b, int p)
{
int d;
d = a + b - p;
d += p & -(d >>> 31);
return d;
}
/*
* Subtraction modulo p.
*/
int modp_sub(int a, int b, int p)
{
int d;
d = a - b;
d += p & -(d >>> 31);
return d;
}
/*
* Montgomery multiplication modulo p. The 'p0i' value is -1/p mod 2^31.
* It is required that p is an odd integer.
*/
int modp_montymul(int a, int b, int p, int p0i)
{
long z, w;
int d;
z = toUnsignedLong(a) * toUnsignedLong(b);
w = ((z * p0i) & toUnsignedLong(0x7FFFFFFF)) * p;
d = (int)((z + w) >>> 31) - p;
d += p & -(d >>> 31);
return d;
}
/*
* Compute R2 = 2^62 mod p.
*/
int modp_R2(int p, int p0i)
{
int z;
/*
* Compute z = 2^31 mod p (this is the value 1 in Montgomery
* representation), then double it with an addition.
*/
z = modp_R(p);
z = modp_add(z, z, p);
/*
* Square it five times to obtain 2^32 in Montgomery representation
* (i.e. 2^63 mod p).
*/
z = modp_montymul(z, z, p, p0i);
z = modp_montymul(z, z, p, p0i);
z = modp_montymul(z, z, p, p0i);
z = modp_montymul(z, z, p, p0i);
z = modp_montymul(z, z, p, p0i);
/*
* Halve the value mod p to get 2^62.
*/
z = (z + (p & -(z & 1))) >>> 1;
return z;
}
/*
* Compute 2^(31*x) modulo p. This works for integers x up to 2^11.
* p must be prime such that 2^30 < p < 2^31; p0i must be equal to
* -1/p mod 2^31; R2 must be equal to 2^62 mod p.
*/
int modp_Rx(int x, int p, int p0i, int R2)
{
int i;
int r, z;
/*
* 2^(31*x) = (2^31)*(2^(31*(x-1))); i.e. we want the Montgomery
* representation of (2^31)^e mod p, where e = x-1.
* R2 is 2^31 in Montgomery representation.
*/
x--;
r = R2;
z = modp_R(p);
for (i = 0; (1 << i) <= x; i++)
{
if ((x & (1 << i)) != 0)
{
z = modp_montymul(z, r, p, p0i);
}
r = modp_montymul(r, r, p, p0i);
}
return z;
}
/*
* Division modulo p. If the divisor (b) is 0, then 0 is returned.
* This function computes proper results only when p is prime.
* Parameters:
* a dividend
* b divisor
* p odd prime modulus
* p0i -1/p mod 2^31
* R 2^31 mod R
*/
int modp_div(int a, int b, int p, int p0i, int R)
{
int z, e;
int i;
e = p - 2;
z = R;
for (i = 30; i >= 0; i--)
{
int z2;
z = modp_montymul(z, z, p, p0i);
z2 = modp_montymul(z, b, p, p0i);
z ^= (z ^ z2) & -(int)((e >>> i) & 1);
}
/*
* The loop above just assumed that b was in Montgomery
* representation, i.e. really contained b*R; under that
* assumption, it returns 1/b in Montgomery representation,
* which is R/b. But we gave it b in normal representation,
* so the loop really returned R/(b/R) = R^2/b.
*
* We want a/b, so we need one Montgomery multiplication with a,
* which also remove one of the R factors, and another such
* multiplication to remove the second R factor.
*/
z = modp_montymul(z, 1, p, p0i);
return modp_montymul(a, z, p, p0i);
}
/*
* Bit-reversal index table.
*/
private short REV10[] = {
0, 512, 256, 768, 128, 640, 384, 896, 64, 576, 320, 832,
192, 704, 448, 960, 32, 544, 288, 800, 160, 672, 416, 928,
96, 608, 352, 864, 224, 736, 480, 992, 16, 528, 272, 784,
144, 656, 400, 912, 80, 592, 336, 848, 208, 720, 464, 976,
48, 560, 304, 816, 176, 688, 432, 944, 112, 624, 368, 880,
240, 752, 496, 1008, 8, 520, 264, 776, 136, 648, 392, 904,
72, 584, 328, 840, 200, 712, 456, 968, 40, 552, 296, 808,
168, 680, 424, 936, 104, 616, 360, 872, 232, 744, 488, 1000,
24, 536, 280, 792, 152, 664, 408, 920, 88, 600, 344, 856,
216, 728, 472, 984, 56, 568, 312, 824, 184, 696, 440, 952,
120, 632, 376, 888, 248, 760, 504, 1016, 4, 516, 260, 772,
132, 644, 388, 900, 68, 580, 324, 836, 196, 708, 452, 964,
36, 548, 292, 804, 164, 676, 420, 932, 100, 612, 356, 868,
228, 740, 484, 996, 20, 532, 276, 788, 148, 660, 404, 916,
84, 596, 340, 852, 212, 724, 468, 980, 52, 564, 308, 820,
180, 692, 436, 948, 116, 628, 372, 884, 244, 756, 500, 1012,
12, 524, 268, 780, 140, 652, 396, 908, 76, 588, 332, 844,
204, 716, 460, 972, 44, 556, 300, 812, 172, 684, 428, 940,
108, 620, 364, 876, 236, 748, 492, 1004, 28, 540, 284, 796,
156, 668, 412, 924, 92, 604, 348, 860, 220, 732, 476, 988,
60, 572, 316, 828, 188, 700, 444, 956, 124, 636, 380, 892,
252, 764, 508, 1020, 2, 514, 258, 770, 130, 642, 386, 898,
66, 578, 322, 834, 194, 706, 450, 962, 34, 546, 290, 802,
162, 674, 418, 930, 98, 610, 354, 866, 226, 738, 482, 994,
18, 530, 274, 786, 146, 658, 402, 914, 82, 594, 338, 850,
210, 722, 466, 978, 50, 562, 306, 818, 178, 690, 434, 946,
114, 626, 370, 882, 242, 754, 498, 1010, 10, 522, 266, 778,
138, 650, 394, 906, 74, 586, 330, 842, 202, 714, 458, 970,
42, 554, 298, 810, 170, 682, 426, 938, 106, 618, 362, 874,
234, 746, 490, 1002, 26, 538, 282, 794, 154, 666, 410, 922,
90, 602, 346, 858, 218, 730, 474, 986, 58, 570, 314, 826,
186, 698, 442, 954, 122, 634, 378, 890, 250, 762, 506, 1018,
6, 518, 262, 774, 134, 646, 390, 902, 70, 582, 326, 838,
198, 710, 454, 966, 38, 550, 294, 806, 166, 678, 422, 934,
102, 614, 358, 870, 230, 742, 486, 998, 22, 534, 278, 790,
150, 662, 406, 918, 86, 598, 342, 854, 214, 726, 470, 982,
54, 566, 310, 822, 182, 694, 438, 950, 118, 630, 374, 886,
246, 758, 502, 1014, 14, 526, 270, 782, 142, 654, 398, 910,
78, 590, 334, 846, 206, 718, 462, 974, 46, 558, 302, 814,
174, 686, 430, 942, 110, 622, 366, 878, 238, 750, 494, 1006,
30, 542, 286, 798, 158, 670, 414, 926, 94, 606, 350, 862,
222, 734, 478, 990, 62, 574, 318, 830, 190, 702, 446, 958,
126, 638, 382, 894, 254, 766, 510, 1022, 1, 513, 257, 769,
129, 641, 385, 897, 65, 577, 321, 833, 193, 705, 449, 961,
33, 545, 289, 801, 161, 673, 417, 929, 97, 609, 353, 865,
225, 737, 481, 993, 17, 529, 273, 785, 145, 657, 401, 913,
81, 593, 337, 849, 209, 721, 465, 977, 49, 561, 305, 817,
177, 689, 433, 945, 113, 625, 369, 881, 241, 753, 497, 1009,
9, 521, 265, 777, 137, 649, 393, 905, 73, 585, 329, 841,
201, 713, 457, 969, 41, 553, 297, 809, 169, 681, 425, 937,
105, 617, 361, 873, 233, 745, 489, 1001, 25, 537, 281, 793,
153, 665, 409, 921, 89, 601, 345, 857, 217, 729, 473, 985,
57, 569, 313, 825, 185, 697, 441, 953, 121, 633, 377, 889,
249, 761, 505, 1017, 5, 517, 261, 773, 133, 645, 389, 901,
69, 581, 325, 837, 197, 709, 453, 965, 37, 549, 293, 805,
165, 677, 421, 933, 101, 613, 357, 869, 229, 741, 485, 997,
21, 533, 277, 789, 149, 661, 405, 917, 85, 597, 341, 853,
213, 725, 469, 981, 53, 565, 309, 821, 181, 693, 437, 949,
117, 629, 373, 885, 245, 757, 501, 1013, 13, 525, 269, 781,
141, 653, 397, 909, 77, 589, 333, 845, 205, 717, 461, 973,
45, 557, 301, 813, 173, 685, 429, 941, 109, 621, 365, 877,
237, 749, 493, 1005, 29, 541, 285, 797, 157, 669, 413, 925,
93, 605, 349, 861, 221, 733, 477, 989, 61, 573, 317, 829,
189, 701, 445, 957, 125, 637, 381, 893, 253, 765, 509, 1021,
3, 515, 259, 771, 131, 643, 387, 899, 67, 579, 323, 835,
195, 707, 451, 963, 35, 547, 291, 803, 163, 675, 419, 931,
99, 611, 355, 867, 227, 739, 483, 995, 19, 531, 275, 787,
147, 659, 403, 915, 83, 595, 339, 851, 211, 723, 467, 979,
51, 563, 307, 819, 179, 691, 435, 947, 115, 627, 371, 883,
243, 755, 499, 1011, 11, 523, 267, 779, 139, 651, 395, 907,
75, 587, 331, 843, 203, 715, 459, 971, 43, 555, 299, 811,
171, 683, 427, 939, 107, 619, 363, 875, 235, 747, 491, 1003,
27, 539, 283, 795, 155, 667, 411, 923, 91, 603, 347, 859,
219, 731, 475, 987, 59, 571, 315, 827, 187, 699, 443, 955,
123, 635, 379, 891, 251, 763, 507, 1019, 7, 519, 263, 775,
135, 647, 391, 903, 71, 583, 327, 839, 199, 711, 455, 967,
39, 551, 295, 807, 167, 679, 423, 935, 103, 615, 359, 871,
231, 743, 487, 999, 23, 535, 279, 791, 151, 663, 407, 919,
87, 599, 343, 855, 215, 727, 471, 983, 55, 567, 311, 823,
183, 695, 439, 951, 119, 631, 375, 887, 247, 759, 503, 1015,
15, 527, 271, 783, 143, 655, 399, 911, 79, 591, 335, 847,
207, 719, 463, 975, 47, 559, 303, 815, 175, 687, 431, 943,
111, 623, 367, 879, 239, 751, 495, 1007, 31, 543, 287, 799,
159, 671, 415, 927, 95, 607, 351, 863, 223, 735, 479, 991,
63, 575, 319, 831, 191, 703, 447, 959, 127, 639, 383, 895,
255, 767, 511, 1023
};
/*
* Compute the roots for NTT and inverse NTT (binary case). Input
* parameter g is a primitive 2048-th root of 1 modulo p (i.e. g^1024 =
* -1 mod p). This fills gm[] and igm[] with powers of g and 1/g:
* gm[rev(i)] = g^i mod p
* igm[rev(i)] = (1/g)^i mod p
* where rev() is the "bit reversal" function over 10 bits. It fills
* the arrays only up to N = 2^logn values.
*
* The values stored in gm[] and igm[] are in Montgomery representation.
*
* p must be a prime such that p = 1 mod 2048.
*/
void modp_mkgm2(int[] srcgm, int gm, int[] srcigm, int igm, int logn,
int g, int p, int p0i)
{
int u, n;
int k;
int ig, x1, x2, R2;
n = mkn(logn);
/*
* We want g such that g^(2N) = 1 mod p, but the provided
* generator has order 2048. We must square it a few times.
*/
R2 = modp_R2(p, p0i);
g = modp_montymul(g, R2, p, p0i);
for (k = logn; k < 10; k++)
{
g = modp_montymul(g, g, p, p0i);
}
ig = modp_div(R2, g, p, p0i, modp_R(p));
k = 10 - logn;
x1 = x2 = modp_R(p);
for (u = 0; u < n; u++)
{
int v;
v = REV10[u << k];
srcgm[gm + v] = x1;
srcigm[igm + v] = x2;
x1 = modp_montymul(x1, g, p, p0i);
x2 = modp_montymul(x2, ig, p, p0i);
}
}
/*
* Compute the NTT over a polynomial (binary case). Polynomial elements
* are a[0], a[stride], a[2 * stride]...
*/
void modp_NTT2_ext(int[] srca, int a, int stride, int[] srcgm, int gm, int logn,
int p, int p0i)
{
int t, m, n;
if (logn == 0)
{
return;
}
n = mkn(logn);
t = n;
for (m = 1; m < n; m <<= 1)
{
int ht, u, v1;
ht = t >> 1;
for (u = 0, v1 = 0; u < m; u++, v1 += t)
{
int s;
int v;
int r1, r2;
s = srcgm[gm + m + u];
r1 = a + v1 * stride;
r2 = r1 + ht * stride;
for (v = 0; v < ht; v++, r1 += stride, r2 += stride)
{
int x, y;
x = srca[r1];
y = modp_montymul(srca[r2], s, p, p0i);
srca[r1] = modp_add(x, y, p);
srca[r2] = modp_sub(x, y, p);
}
}
t = ht;
}
}
/*
* Compute the inverse NTT over a polynomial (binary case).
*/
void modp_iNTT2_ext(int[] srca, int a, int stride, int[] srcigm, int igm, int logn,
int p, int p0i)
{
int t, m, n, k;
int ni;
int r;
if (logn == 0)
{
return;
}
n = mkn(logn);
t = 1;
for (m = n; m > 1; m >>= 1)
{
int hm, dt, u, v1;
hm = m >> 1;
dt = t << 1;
for (u = 0, v1 = 0; u < hm; u++, v1 += dt)
{
int s;
int v;
int r1, r2;
s = srcigm[igm + hm + u];
r1 = a + v1 * stride;
r2 = r1 + t * stride;
for (v = 0; v < t; v++, r1 += stride, r2 += stride)
{
int x, y;
x = srca[r1];
y = srca[r2];
srca[r1] = modp_add(x, y, p);
srca[r2] = modp_montymul(
modp_sub(x, y, p), s, p, p0i);
;
}
}
t = dt;
}
/*
* We need 1/n in Montgomery representation, i.e. R/n. Since
* 1 <= logn <= 10, R/n is an integer; morever, R/n <= 2^30 < p,
* thus a simple shift will do.
*/
ni = 1 << (31 - logn);
for (k = 0, r = a; k < n; k++, r += stride)
{
srca[r] = modp_montymul(srca[r], ni, p, p0i);
}
}
/*
* Simplified macros for NTT and iNTT (binary case) when the elements
* are consecutive in RAM.
*/
// #define modp_NTT2(a, gm, logn, p, p0i) modp_NTT2_ext(a, 1, gm, logn, p, p0i)
void modp_NTT2(int[] srca, int a, int[] srcgm, int gm, int logn, int p, int p0i)
{
modp_NTT2_ext(srca, a, 1, srcgm, gm, logn, p, p0i);
}
// #define modp_iNTT2(a, igm, logn, p, p0i) modp_iNTT2_ext(a, 1, igm, logn, p, p0i)
void modp_iNTT2(int[] srca, int a, int[] srcigm, int igm, int logn, int p, int p0i)
{
modp_iNTT2_ext(srca, a, 1, srcigm, igm, logn, p, p0i);
}
/*
* Given polynomial f in NTT representation modulo p, compute f' of degree
* less than N/2 such that f' = f0^2 - X*f1^2, where f0 and f1 are
* polynomials of degree less than N/2 such that f = f0(X^2) + X*f1(X^2).
*
* The new polynomial is written "in place" over the first N/2 elements
* of f.
*
* If applied logn times successively on a given polynomial, the resulting
* degree-0 polynomial is the resultant of f and X^N+1 modulo p.
*
* This function applies only to the binary case; it is invoked from
* solve_NTRU_binary_depth1().
*/
void modp_poly_rec_res(int[] srcf, int f, int logn,
int p, int p0i, int R2)
{
int hn, u;
hn = 1 << (logn - 1);
for (u = 0; u < hn; u++)
{
int w0, w1;
w0 = srcf[f + (u << 1) + 0];
w1 = srcf[f + (u << 1) + 1];
srcf[f + u] = modp_montymul(modp_montymul(w0, w1, p, p0i), R2, p, p0i);
}
}
/* ==================================================================== */
/*
* Custom bignum implementation.
*
* This is a very reduced set of functionalities. We need to do the
* following operations:
*
* - Rebuild the resultant and the polynomial coefficients from their
* values modulo small primes (of length 31 bits each).
*
* - Compute an extended GCD between the two computed resultants.
*
* - Extract top bits and add scaled values during the successive steps
* of Babai rounding.
*
* When rebuilding values using CRT, we must also recompute the product
* of the small prime factors. We always do it one small factor at a
* time, so the "complicated" operations can be done modulo the small
* prime with the modp_* functions. CRT coefficients (inverses) are
* precomputed.
*
* All values are positive until the last step: when the polynomial
* coefficients have been rebuilt, we normalize them around 0. But then,
* only additions and subtractions on the upper few bits are needed
* afterwards.
*
* We keep big integers as arrays of 31-bit words (in uint32_t values);
* the top bit of each uint32_t is kept equal to 0. Using 31-bit words
* makes it easier to keep track of carries. When negative values are
* used, two's complement is used.
*/
/*
* Subtract integer b from integer a. Both integers are supposed to have
* the same size. The carry (0 or 1) is returned. Source arrays a and b
* MUST be distinct.
*
* The operation is performed as described above if ctr = 1. If
* ctl = 0, the value a[] is unmodified, but all memory accesses are
* still performed, and the carry is computed and returned.
*/
int zint_sub(int[] srca, int a, int[] srcb, int b, int len,
int ctl)
{
int u;
int cc, m;
cc = 0;
m = -ctl;
for (u = 0; u < len; u++)
{
int aw, w;
aw = srca[a + u];
w = aw - srcb[b + u] - cc;
cc = w >>> 31;
aw ^= ((w & 0x7FFFFFFF) ^ aw) & m;
srca[a + u] = aw;
}
return cc;
}
/*
* Mutiply the provided big integer m with a small value x.
* This function assumes that x < 2^31. The carry word is returned.
*/
int zint_mul_small(int[] srcm, int m, int mlen, int x)
{
int u;
int cc;
cc = 0;
for (u = 0; u < mlen; u++)
{
long z;
z = toUnsignedLong(srcm[m + u]) * toUnsignedLong(x) + cc;
srcm[m + u] = (int)z & 0x7FFFFFFF;
cc = (int)(z >> 31);
}
return cc;
}
/*
* Reduce a big integer d modulo a small integer p.
* Rules:
* d is unsigned
* p is prime
* 2^30 < p < 2^31
* p0i = -(1/p) mod 2^31
* R2 = 2^62 mod p
*/
int zint_mod_small_unsigned(int[] srcd, int d, int dlen,
int p, int p0i, int R2)
{
int x;
int u;
/*
* Algorithm: we inject words one by one, starting with the high
* word. Each step is:
* - multiply x by 2^31
* - add new word
*/
x = 0;
u = dlen;
while (u-- > 0)
{
int w;
x = modp_montymul(x, R2, p, p0i);
w = srcd[d + u] - p;
w += p & -(w >>> 31);
x = modp_add(x, w, p);
}
return x;
}
/*
* Similar to zint_mod_small_unsigned(), except that d may be signed.
* Extra parameter is Rx = 2^(31*dlen) mod p.
*/
int zint_mod_small_signed(int[] srcd, int d, int dlen,
int p, int p0i, int R2, int Rx)
{
int z;
if (dlen == 0)
{
return 0;
}
z = zint_mod_small_unsigned(srcd, d, dlen, p, p0i, R2);
z = modp_sub(z, Rx & -(srcd[d + dlen - 1] >>> 30), p);
return z;
}
/*
* Add y*s to x. x and y initially have length 'len' words; the new x
* has length 'len+1' words. 's' must fit on 31 bits. x[] and y[] must
* not overlap.
*/
void zint_add_mul_small(int[] srcx, int x,
int[] srcy, int y, int len, int s)
{
int u;
int cc;
cc = 0;
for (u = 0; u < len; u++)
{
int xw, yw;
long z;
xw = srcx[x + u];
yw = srcy[y + u];
z = toUnsignedLong(yw) * toUnsignedLong(s) + toUnsignedLong(xw) + toUnsignedLong(cc);
srcx[x + u] = (int)z & 0x7FFFFFFF;
cc = (int)(z >>> 31);
}
srcx[x + len] = cc;
}
/*
* Normalize a modular integer around 0: if x > p/2, then x is replaced
* with x - p (signed encoding with two's complement); otherwise, x is
* untouched. The two integers x and p are encoded over the same length.
*/
void zint_norm_zero(int[] srcx, int x, int[] srcp, int p, int len)
{
int u;
int r, bb;
/*
* Compare x with p/2. We use the shifted version of p, and p
* is odd, so we really compare with (p-1)/2; we want to perform
* the subtraction if and only if x > (p-1)/2.
*/
r = 0;
bb = 0;
u = len;
while (u-- > 0)
{
int wx, wp, cc;
/*
* Get the two words to compare in wx and wp (both over
* 31 bits exactly).
*/
wx = srcx[x + u];
wp = (srcp[p + u] >>> 1) | (bb << 30);
bb = srcp[p + u] & 1;
/*
* We set cc to -1, 0 or 1, depending on whether wp is
* lower than, equal to, or greater than wx.
*/
cc = wp - wx;
cc = ((-cc) >>> 31) | -(cc >>> 31);
/*
* If r != 0 then it is either 1 or -1, and we keep its
* value. Otherwise, if r = 0, then we replace it with cc.
*/
r |= cc & ((r & 1) - 1);
}
/*
* At this point, r = -1, 0 or 1, depending on whether (p-1)/2
* is lower than, equal to, or greater than x. We thus want to
* do the subtraction only if r = -1.
*/
zint_sub(srcx, x, srcp, p, len, r >>> 31);
}
/*
* Rebuild integers from their RNS representation. There are 'num'
* integers, and each consists in 'xlen' words. 'xx' points at that
* first word of the first integer; subsequent integers are accessed
* by adding 'xstride' repeatedly.
*
* The words of an integer are the RNS representation of that integer,
* using the provided 'primes' are moduli. This function replaces
* each integer with its multi-word value (little-endian order).
*
* If "normalize_signed" is non-zero, then the returned value is
* normalized to the -m/2..m/2 interval (where m is the product of all
* small prime moduli); two's complement is used for negative values.
*/
void zint_rebuild_CRT(int[] srcxx, int xx, int xlen, int xstride,
int num, FalconSmallPrime[] primes, int normalize_signed,
int[] srctmp, int tmp)
{
int u;
int x;
srctmp[tmp + 0] = primes[0].p;
for (u = 1; u < xlen; u++)
{
/*
* At the entry of each loop iteration:
* - the first u words of each array have been
* reassembled;
* - the first u words of tmp[] contains the
* product of the prime moduli processed so far.
*
* We call 'q' the product of all previous primes.
*/
int p, p0i, s, R2;
int v;
p = primes[u].p;
s = primes[u].s;
p0i = modp_ninv31(p);
R2 = modp_R2(p, p0i);
for (v = 0, x = xx; v < num; v++, x += xstride)
{
int xp, xq, xr;
/*
* xp = the integer x modulo the prime p for this
* iteration
* xq = (x mod q) mod p
*/
xp = srcxx[x + u];
xq = zint_mod_small_unsigned(srcxx, x, u, p, p0i, R2);
/*
* New value is (x mod q) + q * (s * (xp - xq) mod p)
*/
xr = modp_montymul(s, modp_sub(xp, xq, p), p, p0i);
zint_add_mul_small(srcxx, x, srctmp, tmp, u, xr);
}
/*
* Update product of primes in tmp[].
*/
srctmp[tmp + u] = zint_mul_small(srctmp, tmp, u, p);
}
/*
* Normalize the reconstructed values around 0.
*/
if (normalize_signed != 0)
{
for (u = 0, x = xx; u < num; u++, x += xstride)
{
zint_norm_zero(srcxx, x, srctmp, tmp, xlen);
}
}
}
/*
* Negate a big integer conditionally: value a is replaced with -a if
* and only if ctl = 1. Control value ctl must be 0 or 1.
*/
void zint_negate(int[] srca, int a, int len, int ctl)
{
int u;
int cc, m;
/*
* If ctl = 1 then we flip the bits of a by XORing with
* 0x7FFFFFFF, and we add 1 to the value. If ctl = 0 then we XOR
* with 0 and add 0, which leaves the value unchanged.
*/
cc = ctl;
m = -ctl >>> 1;
for (u = 0; u < len; u++)
{
int aw;
aw = srca[a + u];
aw = (aw ^ m) + cc;
srca[a + u] = aw & 0x7FFFFFFF;
cc = aw >>> 31;
}
}
/*
* Replace a with (a*xa+b*xb)/(2^31) and b with (a*ya+b*yb)/(2^31).
* The low bits are dropped (the caller should compute the coefficients
* such that these dropped bits are all zeros). If either or both
* yields a negative value, then the value is negated.
*
* Returned value is:
* 0 both values were positive
* 1 new a had to be negated
* 2 new b had to be negated
* 3 both new a and new b had to be negated
*
* Coefficients xa, xb, ya and yb may use the full signed 32-bit range.
*/
int zint_co_reduce(int[] srca, int a, int[] srcb, int b, int len,
long xa, long xb, long ya, long yb)
{
int u;
long cca, ccb;
int nega, negb;
cca = 0;
ccb = 0;
for (u = 0; u < len; u++)
{
int wa, wb;
long za, zb;
wa = srca[a + u];
wb = srcb[b + u];
za = wa * xa + wb * xb + cca;
zb = wa * ya + wb * yb + ccb;
if (u > 0)
{
srca[a + u - 1] = (int)za & 0x7FFFFFFF;
srcb[b + u - 1] = (int)zb & 0x7FFFFFFF;
}
// cca = *(int64_t *)&za >> 31;
cca = za >> 31;
// ccb = *(int64_t *)&zb >> 31;
ccb = zb >> 31;
}
srca[a + len - 1] = (int)cca;
srcb[b + len - 1] = (int)ccb;
nega = (int)(cca >>> 63);
negb = (int)(ccb >>> 63);
zint_negate(srca, a, len, nega);
zint_negate(srcb, b, len, negb);
return nega | (negb << 1);
}
/*
* Finish modular reduction. Rules on input parameters:
*
* if neg = 1, then -m <= a < 0
* if neg = 0, then 0 <= a < 2*m
*
* If neg = 0, then the top word of a[] is allowed to use 32 bits.
*
* Modulus m must be odd.
*/
void zint_finish_mod(int[] srca, int a, int len, int[] srcm, int m, int neg)
{
int u;
int cc, xm, ym;
/*
* First pass: compare a (assumed nonnegative) with m. Note that
* if the top word uses 32 bits, subtracting m must yield a
* value less than 2^31 since a < 2*m.
*/
cc = 0;
for (u = 0; u < len; u++)
{
cc = (srca[a + u] - srcm[m + u] - cc) >>> 31;
}
/*
* If neg = 1 then we must add m (regardless of cc)
* If neg = 0 and cc = 0 then we must subtract m
* If neg = 0 and cc = 1 then we must do nothing
*
* In the loop below, we conditionally subtract either m or -m
* from a. Word xm is a word of m (if neg = 0) or -m (if neg = 1);
* but if neg = 0 and cc = 1, then ym = 0 and it forces mw to 0.
*/
xm = -neg >>> 1;
ym = -(neg | (1 - cc));
cc = neg;
for (u = 0; u < len; u++)
{
int aw, mw;
aw = srca[a + u];
mw = (srcm[m + u] ^ xm) & ym;
aw = aw - mw - cc;
srca[a + u] = aw & 0x7FFFFFFF;
cc = aw >>> 31;
}
}
/*
* Replace a with (a*xa+b*xb)/(2^31) mod m, and b with
* (a*ya+b*yb)/(2^31) mod m. Modulus m must be odd; m0i = -1/m[0] mod 2^31.
*/
void zint_co_reduce_mod(int[] srca, int a, int[] srcb, int b, int[] srcm, int m, int len,
int m0i, long xa, long xb, long ya, long yb)
{
int u;
long cca, ccb;
int fa, fb;
/*
* These are actually four combined Montgomery multiplications.
*/
cca = 0;
ccb = 0;
fa = ((srca[a + 0] * (int)xa + srcb[b + 0] * (int)xb) * m0i) & 0x7FFFFFFF;
fb = ((srca[a + 0] * (int)ya + srcb[b + 0] * (int)yb) * m0i) & 0x7FFFFFFF;
for (u = 0; u < len; u++)
{
int wa, wb;
long za, zb;
wa = srca[a + u];
wb = srcb[b + u];
za = wa * xa + wb * xb
+ srcm[m + u] * toUnsignedLong(fa) + cca;
zb = wa * ya + wb * yb
+ srcm[m + u] * toUnsignedLong(fb) + ccb;
if (u > 0)
{
srca[a + u - 1] = (int)za & 0x7FFFFFFF;
srcb[b + u - 1] = (int)zb & 0x7FFFFFFF;
}
cca = za >> 31;
ccb = zb >> 31;
}
srca[a + len - 1] = (int)cca;
srcb[b + len - 1] = (int)ccb;
/*
* At this point:
* -m <= a < 2*m
* -m <= b < 2*m
* (this is a case of Montgomery reduction)
* The top words of 'a' and 'b' may have a 32-th bit set.
* We want to add or subtract the modulus, as required.
*/
zint_finish_mod(srca, a, len, srcm, m, (int)(cca >>> 63));
zint_finish_mod(srcb, b, len, srcm, m, (int)(ccb >>> 63));
}
/*
* Compute a GCD between two positive big integers x and y. The two
* integers must be odd. Returned value is 1 if the GCD is 1, 0
* otherwise. When 1 is returned, arrays u and v are filled with values
* such that:
* 0 <= u <= y
* 0 <= v <= x
* x*u - y*v = 1
* x[] and y[] are unmodified. Both input values must have the same
* encoded length. Temporary array must be large enough to accommodate 4
* extra values of that length. Arrays u, v and tmp may not overlap with
* each other, or with either x or y.
*/
int zint_bezout(int[] srcu, int u, int[] srcv, int v,
int[] srcx, int x, int[] srcy, int y,
int len, int[] srctmp, int tmp)
{
/*
* Algorithm is an extended binary GCD. We maintain 6 values
* a, b, u0, u1, v0 and v1 with the following invariants:
*
* a = x*u0 - y*v0
* b = x*u1 - y*v1
* 0 <= a <= x
* 0 <= b <= y
* 0 <= u0 < y
* 0 <= v0 < x
* 0 <= u1 <= y
* 0 <= v1 < x
*
* Initial values are:
*
* a = x u0 = 1 v0 = 0
* b = y u1 = y v1 = x-1
*
* Each iteration reduces either a or b, and maintains the
* invariants. Algorithm stops when a = b, at which point their
* common value is GCD(a,b) and (u0,v0) (or (u1,v1)) contains
* the values (u,v) we want to return.
*
* The formal definition of the algorithm is a sequence of steps:
*
* - If a is even, then:
* a <- a/2
* u0 <- u0/2 mod y
* v0 <- v0/2 mod x
*
* - Otherwise, if b is even, then:
* b <- b/2
* u1 <- u1/2 mod y
* v1 <- v1/2 mod x
*
* - Otherwise, if a > b, then:
* a <- (a-b)/2
* u0 <- (u0-u1)/2 mod y
* v0 <- (v0-v1)/2 mod x
*
* - Otherwise:
* b <- (b-a)/2
* u1 <- (u1-u0)/2 mod y
* v1 <- (v1-v0)/2 mod y
*
* We can show that the operations above preserve the invariants:
*
* - If a is even, then u0 and v0 are either both even or both
* odd (since a = x*u0 - y*v0, and x and y are both odd).
* If u0 and v0 are both even, then (u0,v0) <- (u0/2,v0/2).
* Otherwise, (u0,v0) <- ((u0+y)/2,(v0+x)/2). Either way,
* the a = x*u0 - y*v0 invariant is preserved.
*
* - The same holds for the case where b is even.
*
* - If a and b are odd, and a > b, then:
*
* a-b = x*(u0-u1) - y*(v0-v1)
*
* In that situation, if u0 < u1, then x*(u0-u1) < 0, but
* a-b > 0; therefore, it must be that v0 < v1, and the
* first part of the update is: (u0,v0) <- (u0-u1+y,v0-v1+x),
* which preserves the invariants. Otherwise, if u0 > u1,
* then u0-u1 >= 1, thus x*(u0-u1) >= x. But a <= x and
* b >= 0, hence a-b <= x. It follows that, in that case,
* v0-v1 >= 0. The first part of the update is then:
* (u0,v0) <- (u0-u1,v0-v1), which again preserves the
* invariants.
*
* Either way, once the subtraction is done, the new value of
* a, which is the difference of two odd values, is even,
* and the remaining of this step is a subcase of the
* first algorithm case (i.e. when a is even).
*
* - If a and b are odd, and b > a, then the a similar
* argument holds.
*
* The values a and b start at x and y, respectively. Since x
* and y are odd, their GCD is odd, and it is easily seen that
* all steps conserve the GCD (GCD(a-b,b) = GCD(a, b);
* GCD(a/2,b) = GCD(a,b) if GCD(a,b) is odd). Moreover, either a
* or b is reduced by at least one bit at each iteration, so
* the algorithm necessarily converges on the case a = b, at
* which point the common value is the GCD.
*
* In the algorithm expressed above, when a = b, the fourth case
* applies, and sets b = 0. Since a contains the GCD of x and y,
* which are both odd, a must be odd, and subsequent iterations
* (if any) will simply divide b by 2 repeatedly, which has no
* consequence. Thus, the algorithm can run for more iterations
* than necessary; the final GCD will be in a, and the (u,v)
* coefficients will be (u0,v0).
*
*
* The presentation above is bit-by-bit. It can be sped up by
* noticing that all decisions are taken based on the low bits
* and high bits of a and b. We can extract the two top words
* and low word of each of a and b, and compute reduction
* parameters pa, pb, qa and qb such that the new values for
* a and b are:
* a' = (a*pa + b*pb) / (2^31)
* b' = (a*qa + b*qb) / (2^31)
* the two divisions being exact. The coefficients are obtained
* just from the extracted words, and may be slightly off, requiring
* an optional correction: if a' < 0, then we replace pa with -pa
* and pb with -pb. Each such step will reduce the total length
* (sum of lengths of a and b) by at least 30 bits at each
* iteration.
*/
int u0, u1, v0, v1, a, b;
int x0i, y0i;
int num, rc;
int j;
if (len == 0)
{
return 0;
}
/*
* u0 and v0 are the u and v result buffers; the four other
* values (u1, v1, a and b) are taken from tmp[].
*/
u0 = u;
v0 = v;
u1 = tmp;
v1 = u1 + len;
a = v1 + len;
b = a + len;
/*
* We'll need the Montgomery reduction coefficients.
*/
x0i = modp_ninv31(srcx[x + 0]);
y0i = modp_ninv31(srcy[y + 0]);
/*
* Initialize a, b, u0, u1, v0 and v1.
* a = x u0 = 1 v0 = 0
* b = y u1 = y v1 = x-1
* Note that x is odd, so computing x-1 is easy.
*/
// memcpy(a, x, len * sizeof *x);
System.arraycopy(srcx, x, srctmp, a, len);
// memcpy(b, y, len * sizeof *y);
System.arraycopy(srcy, y, srctmp, b, len);
// u0[0] = 1;
srcu[u0 + 0] = 1;
// memset(u0 + 1, 0, (len - 1) * sizeof *u0);
// memset(v0, 0, len * sizeof *v0);
srcv[v0 + 0] = 0;
for (int i = 1; i < len; i++)
{
srcu[u0 + i] = 0;
srcv[v0 + i] = 0;
}
// memcpy(u1, y, len * sizeof *u1);
System.arraycopy(srcy, y, srctmp, u1, len);
// memcpy(v1, x, len * sizeof *v1);
System.arraycopy(srcx, x, srctmp, v1, len);
// v1[0] --;
srctmp[v1 + 0]--;
/*
* Each input operand may be as large as 31*len bits, and we
* reduce the total length by at least 30 bits at each iteration.
*/
for (num = 62 * len + 30; num >= 30; num -= 30)
{
int c0, c1;
int a0, a1, b0, b1;
long a_hi, b_hi;
int a_lo, b_lo;
long pa, pb, qa, qb;
int i;
int r;
/*
* Extract the top words of a and b. If j is the highest
* index >= 1 such that a[j] != 0 or b[j] != 0, then we
* want (a[j] << 31) + a[j-1] and (b[j] << 31) + b[j-1].
* If a and b are down to one word each, then we use
* a[0] and b[0].
*/
c0 = -1;
c1 = -1;
a0 = 0;
a1 = 0;
b0 = 0;
b1 = 0;
j = len;
while (j-- > 0)
{
int aw, bw;
aw = srctmp[a + j];
bw = srctmp[b + j];
a0 ^= (a0 ^ aw) & c0;
a1 ^= (a1 ^ aw) & c1;
b0 ^= (b0 ^ bw) & c0;
b1 ^= (b1 ^ bw) & c1;
c1 = c0;
c0 &= (((aw | bw) + 0x7FFFFFFF) >>> 31) - 1;
}
/*
* If c1 = 0, then we grabbed two words for a and b.
* If c1 != 0 but c0 = 0, then we grabbed one word. It
* is not possible that c1 != 0 and c0 != 0, because that
* would mean that both integers are zero.
*/
a1 |= a0 & c1;
a0 &= ~c1;
b1 |= b0 & c1;
b0 &= ~c1;
a_hi = (toUnsignedLong(a0) << 31) + toUnsignedLong(a1);
b_hi = (toUnsignedLong(b0) << 31) + toUnsignedLong(b1);
a_lo = srctmp[a + 0];
b_lo = srctmp[b + 0];
/*
* Compute reduction factors:
*
* a' = a*pa + b*pb
* b' = a*qa + b*qb
*
* such that a' and b' are both multiple of 2^31, but are
* only marginally larger than a and b.
*/
pa = 1;
pb = 0;
qa = 0;
qb = 1;
for (i = 0; i < 31; i++)
{
/*
* At each iteration:
*
* a <- (a-b)/2 if: a is odd, b is odd, a_hi > b_hi
* b <- (b-a)/2 if: a is odd, b is odd, a_hi <= b_hi
* a <- a/2 if: a is even
* b <- b/2 if: a is odd, b is even
*
* We multiply a_lo and b_lo by 2 at each
* iteration, thus a division by 2 really is a
* non-multiplication by 2.
*/
int rt, oa, ob, cAB, cBA, cA;
long rz;
/*
* rt = 1 if a_hi > b_hi, 0 otherwise.
*/
rz = b_hi - a_hi;
rt = (int)((rz ^ ((a_hi ^ b_hi)
& (a_hi ^ rz))) >>> 63);
/*
* cAB = 1 if b must be subtracted from a
* cBA = 1 if a must be subtracted from b
* cA = 1 if a must be divided by 2
*
* Rules:
*
* cAB and cBA cannot both be 1.
* If a is not divided by 2, b is.
*/
oa = (a_lo >> i) & 1;
ob = (b_lo >> i) & 1;
cAB = oa & ob & rt;
cBA = oa & ob & ~rt;
cA = cAB | (oa ^ 1);
/*
* Conditional subtractions.
*/
a_lo -= b_lo & -cAB;
a_hi -= b_hi & -toUnsignedLong(cAB);
pa -= qa & -(long)cAB;
pb -= qb & -(long)cAB;
b_lo -= a_lo & -cBA;
b_hi -= a_hi & -toUnsignedLong(cBA);
qa -= pa & -(long)cBA;
qb -= pb & -(long)cBA;
/*
* Shifting.
*/
a_lo += a_lo & (cA - 1);
pa += pa & ((long)cA - 1);
pb += pb & ((long)cA - 1);
a_hi ^= (a_hi ^ (a_hi >> 1)) & -toUnsignedLong(cA);
b_lo += b_lo & -cA;
qa += qa & -(long)cA;
qb += qb & -(long)cA;
b_hi ^= (b_hi ^ (b_hi >> 1)) & (toUnsignedLong(cA) - 1);
}
/*
* Apply the computed parameters to our values. We
* may have to correct pa and pb depending on the
* returned value of zint_co_reduce() (when a and/or b
* had to be negated).
*/
r = zint_co_reduce(srctmp, a, srctmp, b, len, pa, pb, qa, qb);
pa -= (pa + pa) & -(long)(r & 1);
pb -= (pb + pb) & -(long)(r & 1);
qa -= (qa + qa) & -(long)(r >>> 1);
qb -= (qb + qb) & -(long)(r >>> 1);
zint_co_reduce_mod(srcu, u0, srctmp, u1, srcy, y, len, y0i, pa, pb, qa, qb);
zint_co_reduce_mod(srcv, v0, srctmp, v1, srcx, x, len, x0i, pa, pb, qa, qb);
}
/*
* At that point, array a[] should contain the GCD, and the
* results (u,v) should already be set. We check that the GCD
* is indeed 1. We also check that the two operands x and y
* are odd.
*/
rc = srctmp[a + 0] ^ 1;
for (j = 1; j < len; j++)
{
rc |= srctmp[a + j];
}
return ((1 - ((rc | -rc) >>> 31)) & srcx[x + 0] & srcy[y + 0]);
}
/*
* Add k*y*2^sc to x. The result is assumed to fit in the array of
* size xlen (truncation is applied if necessary).
* Scale factor 'sc' is provided as sch and scl, such that:
* sch = sc / 31
* scl = sc % 31
* xlen MUST NOT be lower than ylen.
*
* x[] and y[] are both signed integers, using two's complement for
* negative values.
*/
void zint_add_scaled_mul_small(int[] srcx, int x, int xlen,
int[] srcy, int y, int ylen, int k,
int sch, int scl)
{
int u;
int ysign, tw;
int cc;
if (ylen == 0)
{
return;
}
ysign = -(srcy[y + ylen - 1] >>> 30) >>> 1;
tw = 0;
cc = 0;
for (u = sch; u < xlen; u++)
{
int v;
int wy, wys, ccu;
long z;
/*
* Get the next word of y (scaled).
*/
v = u - sch;
wy = v < ylen ? srcy[y + v] : ysign;
wys = ((wy << scl) & 0x7FFFFFFF) | tw;
tw = wy >>> (31 - scl);
/*
* The expression below does not overflow.
*/
z = (toUnsignedLong(wys) * (long)k + toUnsignedLong(srcx[x + u]) + cc);
srcx[x + u] = (int)z & 0x7FFFFFFF;
/*
* Right-shifting the signed value z would yield
* implementation-defined results (arithmetic shift is
* not guaranteed). However, we can cast to unsigned,
* and get the next carry as an unsigned word. We can
* then convert it back to signed by using the guaranteed
* fact that 'int32_t' uses two's complement with no
* trap representation or padding bit, and with a layout
* compatible with that of 'uint32_t'.
*/
ccu = (int)(z >>> 31);
cc = ccu;
}
}
/*
* Subtract y*2^sc from x. The result is assumed to fit in the array of
* size xlen (truncation is applied if necessary).
* Scale factor 'sc' is provided as sch and scl, such that:
* sch = sc / 31
* scl = sc % 31
* xlen MUST NOT be lower than ylen.
*
* x[] and y[] are both signed integers, using two's complement for
* negative values.
*/
void zint_sub_scaled(int[] srcx, int x, int xlen,
int[] srcy, int y, int ylen, int sch, int scl)
{
int u;
int ysign, tw;
int cc;
if (ylen == 0)
{
return;
}
ysign = -(srcy[y + ylen - 1] >>> 30) >>> 1;
tw = 0;
cc = 0;
for (u = sch; u < xlen; u++)
{
int v;
int w, wy, wys;
/*
* Get the next word of y (scaled).
*/
v = u - sch;
wy = v < ylen ? srcy[y + v] : ysign;
wys = ((wy << scl) & 0x7FFFFFFF) | tw;
tw = wy >>> (31 - scl);
w = srcx[x + u] - wys - cc;
srcx[x + u] = w & 0x7FFFFFFF;
cc = w >>> 31;
}
}
/*
* Convert a one-word signed big integer into a signed value.
*/
int zint_one_to_plain(int[] srcx, int x)
{
int w;
w = srcx[x + 0];
w |= (w & 0x40000000) << 1;
return w;
}
/* ==================================================================== */
/*
* Convert a polynomial to floating-point values.
*
* Each coefficient has length flen words, and starts fstride words after
* the previous.
*
* IEEE-754 binary64 values can represent values in a finite range,
* roughly 2^(-1023) to 2^(+1023); thus, if coefficients are too large,
* they should be "trimmed" by pointing not to the lowest word of each,
* but upper.
*/
void poly_big_to_fp(FalconFPR[] srcd, int d, int[] srcf, int f, int flen, int fstride,
int logn)
{
int n, u;
n = mkn(logn);
if (flen == 0)
{
for (u = 0; u < n; u++)
{
srcd[d + u] = fpr.fpr_zero;
}
return;
}
for (u = 0; u < n; u++, f += fstride)
{
int v;
int neg, cc, xm;
FalconFPR x, fsc;
/*
* Get sign of the integer; if it is negative, then we
* will load its absolute value instead, and negate the
* result.
*/
neg = -(srcf[f + flen - 1] >>> 30);
xm = neg >>> 1;
cc = neg & 1;
x = fpr.fpr_zero;
fsc = fpr.fpr_one;
for (v = 0; v < flen; v++, fsc = fpr.fpr_mul(fsc, fpr.fpr_ptwo31))
{
int w;
w = (srcf[f + v] ^ xm) + cc;
cc = w >>> 31;
w &= 0x7FFFFFFF;
w -= (w << 1) & neg;
x = fpr.fpr_add(x, fpr.fpr_mul(fpr.fpr_of(w), fsc));
}
srcd[d + u] = x;
}
}
/*
* Convert a polynomial to small integers. Source values are supposed
* to be one-word integers, signed over 31 bits. Returned value is 0
* if any of the coefficients exceeds the provided limit (in absolute
* value), or 1 on success.
*
* This is not constant-time; this is not a problem here, because on
* any failure, the NTRU-solving process will be deemed to have failed
* and the (f,g) polynomials will be discarded.
*/
int poly_big_to_small(byte[] srcd, int d, int[] srcs, int s, int lim, int logn)
{
int n, u;
n = mkn(logn);
for (u = 0; u < n; u++)
{
int z;
z = zint_one_to_plain(srcs, s + u);
if (z < -lim || z > lim)
{
return 0;
}
srcd[d + u] = (byte)z;
}
return 1;
}
/*
* Subtract k*f from F, where F, f and k are polynomials modulo X^N+1.
* Coefficients of polynomial k are small integers (signed values in the
* -2^31..2^31 range) scaled by 2^sc. Value sc is provided as sch = sc / 31
* and scl = sc % 31.
*
* This function implements the basic quadratic multiplication algorithm,
* which is efficient in space (no extra buffer needed) but slow at
* high degree.
*/
void poly_sub_scaled(int[] srcF, int F, int Flen, int Fstride,
int[] srcf, int f, int flen, int fstride,
int[] srck, int k, int sch, int scl, int logn)
{
int n, u;
n = mkn(logn);
for (u = 0; u < n; u++)
{
int kf;
int v;
int x;
int y;
kf = -srck[k + u];
x = F + u * Fstride;
y = f;
for (v = 0; v < n; v++)
{
zint_add_scaled_mul_small(srcF,
x, Flen, srcf, y, flen, kf, sch, scl);
if (u + v == n - 1)
{
x = F;
kf = -kf;
}
else
{
x += Fstride;
}
y += fstride;
}
}
}
/*
* Subtract k*f from F. Coefficients of polynomial k are small integers
* (signed values in the -2^31..2^31 range) scaled by 2^sc. This function
* assumes that the degree is large, and integers relatively small.
* The value sc is provided as sch = sc / 31 and scl = sc % 31.
*/
void poly_sub_scaled_ntt(int[] srcF, int F, int Flen, int Fstride,
int[] srcf, int f, int flen, int fstride,
int[] srck, int k, int sch, int scl, int logn,
int[] srctmp, int tmp)
{
int gm, igm, fk, t1, x;
int y;
int n, u, tlen;
FalconSmallPrime[] primes;
n = mkn(logn);
tlen = flen + 1;
gm = tmp;
igm = gm + mkn(logn);
fk = igm + mkn(logn);
t1 = fk + n * tlen;
primes = this.primes.PRIMES;
/*
* Compute k*f in fk[], in RNS notation.
*/
for (u = 0; u < tlen; u++)
{
int p, p0i, R2, Rx;
int v;
p = primes[u].p;
p0i = modp_ninv31(p);
R2 = modp_R2(p, p0i);
Rx = modp_Rx(flen, p, p0i, R2);
modp_mkgm2(srctmp, gm, srctmp, igm, logn, primes[u].g, p, p0i);
for (v = 0; v < n; v++)
{
srctmp[t1 + v] = modp_set(srck[k + v], p);
}
modp_NTT2(srctmp, t1, srctmp, gm, logn, p, p0i);
for (v = 0, y = f, x = fk + u;
v < n; v++, y += fstride, x += tlen)
{
srctmp[x] = zint_mod_small_signed(srcf, y, flen, p, p0i, R2, Rx);
}
modp_NTT2_ext(srctmp, fk + u, tlen, srctmp, gm, logn, p, p0i);
for (v = 0, x = fk + u; v < n; v++, x += tlen)
{
srctmp[x] = modp_montymul(
modp_montymul(srctmp[t1 + v], srctmp[x], p, p0i), R2, p, p0i);
}
modp_iNTT2_ext(srctmp, fk + u, tlen, srctmp, igm, logn, p, p0i);
}
/*
* Rebuild k*f.
*/
zint_rebuild_CRT(srctmp, fk, tlen, tlen, n, primes, 1, srctmp, t1);
/*
* Subtract k*f, scaled, from F.
*/
for (u = 0, x = F, y = fk; u < n; u++, x += Fstride, y += tlen)
{
zint_sub_scaled(srcF, x, Flen, srctmp, y, tlen, sch, scl);
}
}
/* ==================================================================== */
/*
* Get a random 8-byte integer from a SHAKE-based RNG. This function
* ensures consistent interpretation of the SHAKE output so that
* the same values will be obtained over different platforms, in case
* a known seed is used.
*/
long get_rng_u64(SHAKE256 rng)
{
/*
* We enforce little-endian representation.
*/
byte[] tmp = new byte[8];
rng.inner_shake256_extract(tmp, 0, tmp.length);
return (tmp[0] & 0xffL)
| ((tmp[1] & 0xffL) << 8)
| ((tmp[2] & 0xffL) << 16)
| ((tmp[3] & 0xffL) << 24)
| ((tmp[4] & 0xffL) << 32)
| ((tmp[5] & 0xffL) << 40)
| ((tmp[6] & 0xffL) << 48)
| ((tmp[7] & 0xffL) << 56);
}
/*
* Table below incarnates a discrete Gaussian distribution:
* D(x) = exp(-(x^2)/(2*sigma^2))
* where sigma = 1.17*sqrt(q/(2*N)), q = 12289, and N = 1024.
* Element 0 of the table is P(x = 0).
* For k > 0, element k is P(x >= k+1 | x > 0).
* Probabilities are scaled up by 2^63.
*/
final long[] gauss_1024_12289 = {
1283868770400643928l, 6416574995475331444l, 4078260278032692663l,
2353523259288686585l, 1227179971273316331l, 575931623374121527l,
242543240509105209l, 91437049221049666l, 30799446349977173l,
9255276791179340l, 2478152334826140l, 590642893610164l,
125206034929641l, 23590435911403l, 3948334035941l,
586753615614l, 77391054539l, 9056793210l,
940121950l, 86539696l, 7062824l,
510971l, 32764l, 1862l,
94l, 4l, 0l
};
/*
* Generate a random value with a Gaussian distribution centered on 0.
* The RNG must be ready for extraction (already flipped).
*
* Distribution has standard deviation 1.17*sqrt(q/(2*N)). The
* precomputed table is for N = 1024. Since the sum of two independent
* values of standard deviation sigma has standard deviation
* sigma*sqrt(2), then we can just generate more values and add them
* together for lower dimensions.
*/
int mkgauss(SHAKE256 rng, int logn)
{
int u, g;
int val;
g = 1 << (10 - logn);
val = 0;
for (u = 0; u < g; u++)
{
/*
* Each iteration generates one value with the
* Gaussian distribution for N = 1024.
*
* We use two random 64-bit values. First value
* decides on whether the generated value is 0, and,
* if not, the sign of the value. Second random 64-bit
* word is used to generate the non-zero value.
*
* For constant-time code we have to read the complete
* table. This has negligible cost, compared with the
* remainder of the keygen process (solving the NTRU
* equation).
*/
long r;
int f, v, k, neg;
/*
* First value:
* - flag 'neg' is randomly selected to be 0 or 1.
* - flag 'f' is set to 1 if the generated value is zero,
* or set to 0 otherwise.
*/
r = get_rng_u64(rng);
neg = (int)(r >>> 63);
r &= ~(1l << 63);
f = (int)((r - gauss_1024_12289[0]) >>> 63);
/*
* We produce a new random 63-bit integer r, and go over
* the array, starting at index 1. We store in v the
* index of the first array element which is not greater
* than r, unless the flag f was already 1.
*/
v = 0;
r = get_rng_u64(rng);
r &= ~(1l << 63);
for (k = 1; k < gauss_1024_12289.length; k++)
{
int t;
t = (int)((r - gauss_1024_12289[k]) >>> 63) ^ 1;
v |= k & -(t & (f ^ 1));
f |= t;
}
/*
* We apply the sign ('neg' flag). If the value is zero,
* the sign has no effect.
*/
v = (v ^ -neg) + neg;
/*
* Generated value is added to val.
*/
val += v;
}
return val;
}
/*
* The MAX_BL_SMALL[] and MAX_BL_LARGE[] contain the lengths, in 31-bit
* words, of intermediate values in the computation:
*
* MAX_BL_SMALL[depth]: length for the input f and g at that depth
* MAX_BL_LARGE[depth]: length for the unreduced F and G at that depth
*
* Rules:
*
* - Within an array, values grow.
*
* - The 'SMALL' array must have an entry for maximum depth, corresponding
* to the size of values used in the binary GCD. There is no such value
* for the 'LARGE' array (the binary GCD yields already reduced
* coefficients).
*
* - MAX_BL_LARGE[depth] >= MAX_BL_SMALL[depth + 1].
*
* - Values must be large enough to handle the common cases, with some
* margins.
*
* - Values must not be "too large" either because we will convert some
* integers into floating-point values by considering the top 10 words,
* i.e. 310 bits; hence, for values of length more than 10 words, we
* should take care to have the length centered on the expected size.
*
* The following average lengths, in bits, have been measured on thousands
* of random keys (fg = max length of the absolute value of coefficients
* of f and g at that depth; FG = idem for the unreduced F and G; for the
* maximum depth, F and G are the output of binary GCD, multiplied by q;
* for each value, the average and standard deviation are provided).
*
* Binary case:
* depth: 10 fg: 6307.52 (24.48) FG: 6319.66 (24.51)
* depth: 9 fg: 3138.35 (12.25) FG: 9403.29 (27.55)
* depth: 8 fg: 1576.87 ( 7.49) FG: 4703.30 (14.77)
* depth: 7 fg: 794.17 ( 4.98) FG: 2361.84 ( 9.31)
* depth: 6 fg: 400.67 ( 3.10) FG: 1188.68 ( 6.04)
* depth: 5 fg: 202.22 ( 1.87) FG: 599.81 ( 3.87)
* depth: 4 fg: 101.62 ( 1.02) FG: 303.49 ( 2.38)
* depth: 3 fg: 50.37 ( 0.53) FG: 153.65 ( 1.39)
* depth: 2 fg: 24.07 ( 0.25) FG: 78.20 ( 0.73)
* depth: 1 fg: 10.99 ( 0.08) FG: 39.82 ( 0.41)
* depth: 0 fg: 4.00 ( 0.00) FG: 19.61 ( 0.49)
*
* Integers are actually represented either in binary notation over
* 31-bit words (signed, using two's complement), or in RNS, modulo
* many small primes. These small primes are close to, but slightly
* lower than, 2^31. Use of RNS loses less than two bits, even for
* the largest values.
*
* IMPORTANT: if these values are modified, then the temporary buffer
* sizes (FALCON_KEYGEN_TEMP_*, in inner.h) must be recomputed
* accordingly.
*/
final int[] MAX_BL_SMALL = {
1, 1, 2, 2, 4, 7, 14, 27, 53, 106, 209
};
final int[] MAX_BL_LARGE = {
2, 2, 5, 7, 12, 21, 40, 78, 157, 308
};
/*
* Average and standard deviation for the maximum size (in bits) of
* coefficients of (f,g), depending on depth. These values are used
* to compute bounds for Babai's reduction.
*/
final int[] bitlength_avg = {
4,
11,
24,
50,
102,
202,
401,
794,
1577,
3138,
6308
};
final int[] bitlength_std = {
0,
1,
1,
1,
1,
2,
4,
5,
8,
13,
25
};
/*
* Minimal recursion depth at which we rebuild intermediate values
* when reconstructing f and g.
*/
final int DEPTH_INT_FG = 4;
/*
* Compute squared norm of a short vector. Returned value is saturated to
* 2^32-1 if it is not lower than 2^31.
*/
int poly_small_sqnorm(byte[] srcf, int f, int logn)
{
int n, u;
int s, ng;
n = mkn(logn);
s = 0;
ng = 0;
for (u = 0; u < n; u++)
{
int z;
z = srcf[f + u];
s += (z * z);
ng |= s;
}
return s | -(ng >>> 31);
}
/*
* Convert a small vector to floating point.
*/
void poly_small_to_fp(FalconFPR[] srcx, int x, byte[] srcf, int f, int logn)
{
int n, u;
n = mkn(logn);
for (u = 0; u < n; u++)
{
srcx[x + u] = fpr.fpr_of(srcf[f + u]);
}
}
/*
* Input: f,g of degree N = 2^logn; 'depth' is used only to get their
* individual length.
*
* Output: f',g' of degree N/2, with the length for 'depth+1'.
*
* Values are in RNS; input and/or output may also be in NTT.
*/
void make_fg_step(int[] srcdata, int data, int logn, int depth,
int in_ntt, int out_ntt)
{
int n, hn, u;
int slen, tlen;
int fd, gd, fs, gs, gm, igm, t1;
FalconSmallPrime[] primes;
n = 1 << logn;
hn = n >> 1;
slen = MAX_BL_SMALL[depth];
tlen = MAX_BL_SMALL[depth + 1];
primes = this.primes.PRIMES;
/*
* Prepare room for the result.
*/
fd = data;
gd = fd + hn * tlen;
fs = gd + hn * tlen;
gs = fs + n * slen;
gm = gs + n * slen;
igm = gm + n;
t1 = igm + n;
// memmove(fs, data, 2 * n * slen * sizeof *data);
System.arraycopy(srcdata, data, srcdata, fs, 2 * n * slen);
/*
* First slen words: we use the input values directly, and apply
* inverse NTT as we go.
*/
for (u = 0; u < slen; u++)
{
int p, p0i, R2;
int v;
int x;
p = primes[u].p;
p0i = modp_ninv31(p);
R2 = modp_R2(p, p0i);
modp_mkgm2(srcdata, gm, srcdata, igm, logn, primes[u].g, p, p0i);
for (v = 0, x = fs + u; v < n; v++, x += slen)
{
srcdata[t1 + v] = srcdata[x];
}
if (in_ntt == 0)
{
modp_NTT2(srcdata, t1, srcdata, gm, logn, p, p0i);
}
for (v = 0, x = fd + u; v < hn; v++, x += tlen)
{
int w0, w1;
w0 = srcdata[t1 + (v << 1) + 0];
w1 = srcdata[t1 + (v << 1) + 1];
srcdata[x] = modp_montymul(
modp_montymul(w0, w1, p, p0i), R2, p, p0i);
}
if (in_ntt != 0)
{
modp_iNTT2_ext(srcdata, fs + u, slen, srcdata, igm, logn, p, p0i);
}
for (v = 0, x = gs + u; v < n; v++, x += slen)
{
srcdata[t1 + v] = srcdata[x];
}
if (in_ntt == 0)
{
modp_NTT2(srcdata, t1, srcdata, gm, logn, p, p0i);
}
for (v = 0, x = gd + u; v < hn; v++, x += tlen)
{
int w0, w1;
w0 = srcdata[t1 + (v << 1) + 0];
w1 = srcdata[t1 + (v << 1) + 1];
srcdata[x] = modp_montymul(
modp_montymul(w0, w1, p, p0i), R2, p, p0i);
}
if (in_ntt != 0)
{
modp_iNTT2_ext(srcdata, gs + u, slen, srcdata, igm, logn, p, p0i);
}
if (out_ntt == 0)
{
modp_iNTT2_ext(srcdata, fd + u, tlen, srcdata, igm, logn - 1, p, p0i);
modp_iNTT2_ext(srcdata, gd + u, tlen, srcdata, igm, logn - 1, p, p0i);
}
}
/*
* Since the fs and gs words have been de-NTTized, we can use the
* CRT to rebuild the values.
*/
zint_rebuild_CRT(srcdata, fs, slen, slen, n, primes, 1, srcdata, gm);
zint_rebuild_CRT(srcdata, gs, slen, slen, n, primes, 1, srcdata, gm);
/*
* Remaining words: use modular reductions to extract the values.
*/
for (u = slen; u < tlen; u++)
{
int p, p0i, R2, Rx;
int v;
int x;
p = primes[u].p;
p0i = modp_ninv31(p);
R2 = modp_R2(p, p0i);
Rx = modp_Rx(slen, p, p0i, R2);
modp_mkgm2(srcdata, gm, srcdata, igm, logn, primes[u].g, p, p0i);
for (v = 0, x = fs; v < n; v++, x += slen)
{
srcdata[t1 + v] = zint_mod_small_signed(srcdata, x, slen, p, p0i, R2, Rx);
}
modp_NTT2(srcdata, t1, srcdata, gm, logn, p, p0i);
for (v = 0, x = fd + u; v < hn; v++, x += tlen)
{
int w0, w1;
w0 = srcdata[t1 + (v << 1) + 0];
w1 = srcdata[t1 + (v << 1) + 1];
srcdata[x] = modp_montymul(
modp_montymul(w0, w1, p, p0i), R2, p, p0i);
}
for (v = 0, x = gs; v < n; v++, x += slen)
{
srcdata[t1 + v] = zint_mod_small_signed(srcdata, x, slen, p, p0i, R2, Rx);
}
modp_NTT2(srcdata, t1, srcdata, gm, logn, p, p0i);
for (v = 0, x = gd + u; v < hn; v++, x += tlen)
{
int w0, w1;
w0 = srcdata[t1 + (v << 1) + 0];
w1 = srcdata[t1 + (v << 1) + 1];
srcdata[x] = modp_montymul(
modp_montymul(w0, w1, p, p0i), R2, p, p0i);
}
if (out_ntt == 0)
{
modp_iNTT2_ext(srcdata, fd + u, tlen, srcdata, igm, logn - 1, p, p0i);
modp_iNTT2_ext(srcdata, gd + u, tlen, srcdata, igm, logn - 1, p, p0i);
}
}
}
/* d values are stored in the data[] array, at slen words per integer.
*
* Conditions:
* 0 <= depth <= logn
*
* Space use in data[]: enough room for any two successive values (f', g',
* f and g).
*/
void make_fg(int[] srcdata, int data, byte[] srcf, int f, byte[] srcg, int g,
int logn, int depth, int out_ntt)
{
int n, u;
int ft, gt, p0;
int d;
FalconSmallPrime[] primes;
n = mkn(logn);
ft = data;
gt = ft + n;
primes = this.primes.PRIMES;
p0 = primes[0].p;
for (u = 0; u < n; u++)
{
srcdata[ft + u] = modp_set(srcf[f + u], p0);
srcdata[gt + u] = modp_set(srcg[g + u], p0);
}
if (depth == 0 && out_ntt != 0)
{
int gm, igm;
int p, p0i;
p = primes[0].p;
p0i = modp_ninv31(p);
gm = gt + n;
igm = gm + n;
modp_mkgm2(srcdata, gm, srcdata, igm, logn, primes[0].g, p, p0i);
modp_NTT2(srcdata, ft, srcdata, gm, logn, p, p0i);
modp_NTT2(srcdata, gt, srcdata, gm, logn, p, p0i);
return;
}
for (d = 0; d < depth; d++)
{
make_fg_step(srcdata, data, logn - d, d,
d != 0 ? 1 : 0, ((d + 1) < depth || out_ntt != 0) ? 1 : 0);
}
}
/*
* Solving the NTRU equation, deepest level: compute the resultants of
* f and g with X^N+1, and use binary GCD. The F and G values are
* returned in tmp[].
*
* Returned value: 1 on success, 0 on error.
*/
int solve_NTRU_deepest(int logn_top,
byte[] srcf, int f, byte[] srcg, int g, int[] srctmp, int tmp)
{
int len;
int Fp, Gp, fp, gp, t1, q;
FalconSmallPrime[] primes;
len = MAX_BL_SMALL[logn_top];
primes = this.primes.PRIMES;
Fp = tmp;
Gp = Fp + len;
fp = Gp + len;
gp = fp + len;
t1 = gp + len;
make_fg(srctmp, fp, srcf, f, srcg, g, logn_top, logn_top, 0);
/*
* We use the CRT to rebuild the resultants as big integers.
* There are two such big integers. The resultants are always
* nonnegative.
*/
zint_rebuild_CRT(srctmp, fp, len, len, 2, primes, 0, srctmp, t1);
/*
* Apply the binary GCD. The zint_bezout() function works only
* if both inputs are odd.
*
* We can test on the result and return 0 because that would
* imply failure of the NTRU solving equation, and the (f,g)
* values will be abandoned in that case.
*/
if (zint_bezout(srctmp, Gp, srctmp, Fp, srctmp, fp, srctmp, gp, len, srctmp, t1) == 0)
{
return 0;
}
/*
* Multiply the two values by the target value q. Values must
* fit in the destination arrays.
* We can again test on the returned words: a non-zero output
* of zint_mul_small() means that we exceeded our array
* capacity, and that implies failure and rejection of (f,g).
*/
q = 12289;
if (zint_mul_small(srctmp, Fp, len, q) != 0
|| zint_mul_small(srctmp, Gp, len, q) != 0)
{
return 0;
}
return 1;
}
/*
* Solving the NTRU equation, intermediate level. Upon entry, the F and G
* from the previous level should be in the tmp[] array.
* This function MAY be invoked for the top-level (in which case depth = 0).
*
* Returned value: 1 on success, 0 on error.
*/
int solve_NTRU_intermediate(int logn_top,
byte[] srcf, int f, byte[] srcg, int g, int depth, int[] srctmp, int tmp)
{
/*
* In this function, 'logn' is the log2 of the degree for
* this step. If N = 2^logn, then:
* - the F and G values already in fk->tmp (from the deeper
* levels) have degree N/2;
* - this function should return F and G of degree N.
*/
int logn;
int n, hn, slen, dlen, llen, rlen, FGlen, u;
int Fd, Gd, Ft, Gt, ft, gt, t1;
FalconFPR[] rt1, rt2, rt3, rt4, rt5;
int scale_fg, minbl_fg, maxbl_fg, maxbl_FG, scale_k;
int x, y;
int[] k;
FalconSmallPrime[] primes;
logn = logn_top - depth;
n = 1 << logn;
hn = n >> 1;
/*
* slen = size for our input f and g; also size of the reduced
* F and G we return (degree N)
*
* dlen = size of the F and G obtained from the deeper level
* (degree N/2 or N/3)
*
* llen = size for intermediary F and G before reduction (degree N)
*
* We build our non-reduced F and G as two independent halves each,
* of degree N/2 (F = F0 + X*F1, G = G0 + X*G1).
*/
slen = MAX_BL_SMALL[depth];
dlen = MAX_BL_SMALL[depth + 1];
llen = MAX_BL_LARGE[depth];
primes = this.primes.PRIMES;
/*
* Fd and Gd are the F and G from the deeper level.
*/
Fd = tmp;
Gd = Fd + dlen * hn;
/*
* Compute the input f and g for this level. Note that we get f
* and g in RNS + NTT representation.
*/
ft = Gd + dlen * hn;
make_fg(srctmp, ft, srcf, f, srcg, g, logn_top, depth, 1);
/*
* Move the newly computed f and g to make room for our candidate
* F and G (unreduced).
*/
Ft = tmp;
Gt = Ft + n * llen;
t1 = Gt + n * llen;
// memmove(t1, ft, 2 * n * slen * sizeof *ft);
System.arraycopy(srctmp, ft, srctmp, t1, 2 * n * slen);
ft = t1;
gt = ft + slen * n;
t1 = gt + slen * n;
/*
* Move Fd and Gd _after_ f and g.
*/
// memmove(t1, Fd, 2 * hn * dlen * sizeof *Fd);
System.arraycopy(srctmp, Fd, srctmp, t1, 2 * hn * dlen);
Fd = t1;
Gd = Fd + hn * dlen;
/*
* We reduce Fd and Gd modulo all the small primes we will need,
* and store the values in Ft and Gt (only n/2 values in each).
*/
for (u = 0; u < llen; u++)
{
int p, p0i, R2, Rx;
int v;
int xs, ys, xd, yd;
p = primes[u].p;
p0i = modp_ninv31(p);
R2 = modp_R2(p, p0i);
Rx = modp_Rx(dlen, p, p0i, R2);
for (v = 0, xs = Fd, ys = Gd, xd = Ft + u, yd = Gt + u;
v < hn;
v++, xs += dlen, ys += dlen, xd += llen, yd += llen)
{
srctmp[xd] = zint_mod_small_signed(srctmp, xs, dlen, p, p0i, R2, Rx);
srctmp[yd] = zint_mod_small_signed(srctmp, ys, dlen, p, p0i, R2, Rx);
}
}
/*
* We do not need Fd and Gd after that point.
*/
/*
* Compute our F and G modulo sufficiently many small primes.
*/
for (u = 0; u < llen; u++)
{
int p, p0i, R2;
int gm, igm, fx, gx, Fp, Gp;
int v;
/*
* All computations are done modulo p.
*/
p = primes[u].p;
p0i = modp_ninv31(p);
R2 = modp_R2(p, p0i);
/*
* If we processed slen words, then f and g have been
* de-NTTized, and are in RNS; we can rebuild them.
*/
if (u == slen)
{
zint_rebuild_CRT(srctmp, ft, slen, slen, n, primes, 1, srctmp, t1);
zint_rebuild_CRT(srctmp, gt, slen, slen, n, primes, 1, srctmp, t1);
}
gm = t1;
igm = gm + n;
fx = igm + n;
gx = fx + n;
modp_mkgm2(srctmp, gm, srctmp, igm, logn, primes[u].g, p, p0i);
if (u < slen)
{
for (v = 0, x = ft + u, y = gt + u;
v < n; v++, x += slen, y += slen)
{
srctmp[fx + v] = srctmp[x];
srctmp[gx + v] = srctmp[y];
}
modp_iNTT2_ext(srctmp, ft + u, slen, srctmp, igm, logn, p, p0i);
modp_iNTT2_ext(srctmp, gt + u, slen, srctmp, igm, logn, p, p0i);
}
else
{
int Rx;
Rx = modp_Rx(slen, p, p0i, R2);
for (v = 0, x = ft, y = gt;
v < n; v++, x += slen, y += slen)
{
srctmp[fx + v] = zint_mod_small_signed(srctmp, x, slen,
p, p0i, R2, Rx);
srctmp[gx + v] = zint_mod_small_signed(srctmp, y, slen,
p, p0i, R2, Rx);
}
modp_NTT2(srctmp, fx, srctmp, gm, logn, p, p0i);
modp_NTT2(srctmp, gx, srctmp, gm, logn, p, p0i);
}
/*
* Get F' and G' modulo p and in NTT representation
* (they have degree n/2). These values were computed in
* a previous step, and stored in Ft and Gt.
*/
Fp = gx + n;
Gp = Fp + hn;
for (v = 0, x = Ft + u, y = Gt + u;
v < hn; v++, x += llen, y += llen)
{
srctmp[Fp + v] = srctmp[x];
srctmp[Gp + v] = srctmp[y];
}
modp_NTT2(srctmp, Fp, srctmp, gm, logn - 1, p, p0i);
modp_NTT2(srctmp, Gp, srctmp, gm, logn - 1, p, p0i);
/*
* Compute our F and G modulo p.
*
* General case:
*
* we divide degree by d = 2 or 3
* f'(x^d) = N(f)(x^d) = f * adj(f)
* g'(x^d) = N(g)(x^d) = g * adj(g)
* f'*G' - g'*F' = q
* F = F'(x^d) * adj(g)
* G = G'(x^d) * adj(f)
*
* We compute things in the NTT. We group roots of phi
* such that all roots x in a group share the same x^d.
* If the roots in a group are x_1, x_2... x_d, then:
*
* N(f)(x_1^d) = f(x_1)*f(x_2)*...*f(x_d)
*
* Thus, we have:
*
* G(x_1) = f(x_2)*f(x_3)*...*f(x_d)*G'(x_1^d)
* G(x_2) = f(x_1)*f(x_3)*...*f(x_d)*G'(x_1^d)
* ...
* G(x_d) = f(x_1)*f(x_2)*...*f(x_{d-1})*G'(x_1^d)
*
* In all cases, we can thus compute F and G in NTT
* representation by a few simple multiplications.
* Moreover, in our chosen NTT representation, roots
* from the same group are consecutive in RAM.
*/
for (v = 0, x = Ft + u, y = Gt + u; v < hn;
v++, x += (llen << 1), y += (llen << 1))
{
int ftA, ftB, gtA, gtB;
int mFp, mGp;
ftA = srctmp[fx + (v << 1) + 0];
ftB = srctmp[fx + (v << 1) + 1];
gtA = srctmp[gx + (v << 1) + 0];
gtB = srctmp[gx + (v << 1) + 1];
mFp = modp_montymul(srctmp[Fp + v], R2, p, p0i);
mGp = modp_montymul(srctmp[Gp + v], R2, p, p0i);
srctmp[x + 0] = modp_montymul(gtB, mFp, p, p0i);
srctmp[x + llen] = modp_montymul(gtA, mFp, p, p0i);
srctmp[y + 0] = modp_montymul(ftB, mGp, p, p0i);
srctmp[y + llen] = modp_montymul(ftA, mGp, p, p0i);
}
modp_iNTT2_ext(srctmp, Ft + u, llen, srctmp, igm, logn, p, p0i);
modp_iNTT2_ext(srctmp, Gt + u, llen, srctmp, igm, logn, p, p0i);
}
/*
* Rebuild F and G with the CRT.
*/
zint_rebuild_CRT(srctmp, Ft, llen, llen, n, primes, 1, srctmp, t1);
zint_rebuild_CRT(srctmp, Gt, llen, llen, n, primes, 1, srctmp, t1);
/*
* At that point, Ft, Gt, ft and gt are consecutive in RAM (in that
* order).
*/
/*
* Apply Babai reduction to bring back F and G to size slen.
*
* We use the FFT to compute successive approximations of the
* reduction coefficient. We first isolate the top bits of
* the coefficients of f and g, and convert them to floating
* point; with the FFT, we compute adj(f), adj(g), and
* 1/(f*adj(f)+g*adj(g)).
*
* Then, we repeatedly apply the following:
*
* - Get the top bits of the coefficients of F and G into
* floating point, and use the FFT to compute:
* (F*adj(f)+G*adj(g))/(f*adj(f)+g*adj(g))
*
* - Convert back that value into normal representation, and
* round it to the nearest integers, yielding a polynomial k.
* Proper scaling is applied to f, g, F and G so that the
* coefficients fit on 32 bits (signed).
*
* - Subtract k*f from F and k*g from G.
*
* Under normal conditions, this process reduces the size of F
* and G by some bits at each iteration. For constant-time
* operation, we do not want to measure the actual length of
* F and G; instead, we do the following:
*
* - f and g are converted to floating-point, with some scaling
* if necessary to keep values in the representable range.
*
* - For each iteration, we _assume_ a maximum size for F and G,
* and use the values at that size. If we overreach, then
* we get zeros, which is harmless: the resulting coefficients
* of k will be 0 and the value won't be reduced.
*
* - We conservatively assume that F and G will be reduced by
* at least 25 bits at each iteration.
*
* Even when reaching the bottom of the reduction, reduction
* coefficient will remain low. If it goes out-of-range, then
* something wrong occurred and the whole NTRU solving fails.
*/
/*
* Memory layout:
* - We need to compute and keep adj(f), adj(g), and
* 1/(f*adj(f)+g*adj(g)) (sizes N, N and N/2 fp numbers,
* respectively).
* - At each iteration we need two extra fp buffer (N fp values),
* and produce a k (N 32-bit words). k will be shared with one
* of the fp buffers.
* - To compute k*f and k*g efficiently (with the NTT), we need
* some extra room; we reuse the space of the temporary buffers.
*
* Arrays of 'fpr' are obtained from the temporary array itself.
* We ensure that the base is at a properly aligned offset (the
* source array tmp[] is supposed to be already aligned).
*/
rt1 = new FalconFPR[n];
rt2 = new FalconFPR[n];
rt3 = new FalconFPR[n];
rt4 = new FalconFPR[n];
rt5 = new FalconFPR[n >> 1];
k = new int[n];
/*
* Get f and g into rt3 and rt4 as floating-point approximations.
*
* We need to "scale down" the floating-point representation of
* coefficients when they are too big. We want to keep the value
* below 2^310 or so. Thus, when values are larger than 10 words,
* we consider only the top 10 words. Array lengths have been
* computed so that average maximum length will fall in the
* middle or the upper half of these top 10 words.
*/
rlen = (slen > 10) ? 10 : slen;
poly_big_to_fp(rt3, 0, srctmp, ft + slen - rlen, rlen, slen, logn);
poly_big_to_fp(rt4, 0, srctmp, gt + slen - rlen, rlen, slen, logn);
/*
* Values in rt3 and rt4 are downscaled by 2^(scale_fg).
*/
scale_fg = 31 * (slen - rlen);
/*
* Estimated boundaries for the maximum size (in bits) of the
* coefficients of (f,g). We use the measured average, and
* allow for a deviation of at most six times the standard
* deviation.
*/
minbl_fg = bitlength_avg[depth] - 6 * bitlength_std[depth];
maxbl_fg = bitlength_avg[depth] + 6 * bitlength_std[depth];
/*
* Compute 1/(f*adj(f)+g*adj(g)) in rt5. We also keep adj(f)
* and adj(g) in rt3 and rt4, respectively.
*/
fft.FFT(rt3, 0, logn);
fft.FFT(rt4, 0, logn);
fft.poly_invnorm2_fft(rt5, 0, rt3, 0, rt4, 0, logn);
fft.poly_adj_fft(rt3, 0, logn);
fft.poly_adj_fft(rt4, 0, logn);
/*
* Reduce F and G repeatedly.
*
* The expected maximum bit length of coefficients of F and G
* is kept in maxbl_FG, with the corresponding word length in
* FGlen.
*/
FGlen = llen;
maxbl_FG = 31 * llen;
/*
* Each reduction operation computes the reduction polynomial
* "k". We need that polynomial to have coefficients that fit
* on 32-bit signed integers, with some scaling; thus, we use
* a descending sequence of scaling values, down to zero.
*
* The size of the coefficients of k is (roughly) the difference
* between the size of the coefficients of (F,G) and the size
* of the coefficients of (f,g). Thus, the maximum size of the
* coefficients of k is, at the start, maxbl_FG - minbl_fg;
* this is our starting scale value for k.
*
* We need to estimate the size of (F,G) during the execution of
* the algorithm; we are allowed some overestimation but not too
* much (poly_big_to_fp() uses a 310-bit window). Generally
* speaking, after applying a reduction with k scaled to
* scale_k, the size of (F,G) will be size(f,g) + scale_k + dd,
* where 'dd' is a few bits to account for the fact that the
* reduction is never perfect (intuitively, dd is on the order
* of sqrt(N), so at most 5 bits; we here allow for 10 extra
* bits).
*
* The size of (f,g) is not known exactly, but maxbl_fg is an
* upper bound.
*/
scale_k = maxbl_FG - minbl_fg;
for (; ; )
{
int scale_FG, dc, new_maxbl_FG;
int scl, sch;
FalconFPR pdc, pt;
/*
* Convert current F and G into floating-point. We apply
* scaling if the current length is more than 10 words.
*/
rlen = (FGlen > 10) ? 10 : FGlen;
scale_FG = 31 * (int)(FGlen - rlen);
poly_big_to_fp(rt1, 0, srctmp, Ft + FGlen - rlen, rlen, llen, logn);
poly_big_to_fp(rt2, 0, srctmp, Gt + FGlen - rlen, rlen, llen, logn);
/*
* Compute (F*adj(f)+G*adj(g))/(f*adj(f)+g*adj(g)) in rt2.
*/
fft.FFT(rt1, 0, logn);
fft.FFT(rt2, 0, logn);
fft.poly_mul_fft(rt1, 0, rt3, 0, logn);
fft.poly_mul_fft(rt2, 0, rt4, 0, logn);
fft.poly_add(rt2, 0, rt1, 0, logn);
fft.poly_mul_autoadj_fft(rt2, 0, rt5, 0, logn);
fft.iFFT(rt2, 0, logn);
/*
* (f,g) are scaled by 'scale_fg', meaning that the
* numbers in rt3/rt4 should be multiplied by 2^(scale_fg)
* to have their true mathematical value.
*
* (F,G) are similarly scaled by 'scale_FG'. Therefore,
* the value we computed in rt2 is scaled by
* 'scale_FG-scale_fg'.
*
* We want that value to be scaled by 'scale_k', hence we
* apply a corrective scaling. After scaling, the values
* should fit in -2^31-1..+2^31-1.
*/
dc = scale_k - scale_FG + scale_fg;
/*
* We will need to multiply values by 2^(-dc). The value
* 'dc' is not secret, so we can compute 2^(-dc) with a
* non-constant-time process.
* (We could use ldexp(), but we prefer to avoid any
* dependency on libm. When using FP emulation, we could
* use our fpr_ldexp(), which is constant-time.)
*/
if (dc < 0)
{
dc = -dc;
pt = fpr.fpr_two;
}
else
{
pt = fpr.fpr_onehalf;
}
pdc = fpr.fpr_one;
while (dc != 0)
{
if ((dc & 1) != 0)
{
pdc = fpr.fpr_mul(pdc, pt);
}
dc >>= 1;
pt = fpr.fpr_sqr(pt);
}
for (u = 0; u < n; u++)
{
FalconFPR xv;
xv = fpr.fpr_mul(rt2[u], pdc);
/*
* Sometimes the values can be out-of-bounds if
* the algorithm fails; we must not call
* fpr_rint() (and cast to int32_t) if the value
* is not in-bounds. Note that the test does not
* break constant-time discipline, since any
* failure here implies that we discard the current
* secret key (f,g).
*/
if (!fpr.fpr_lt(fpr.fpr_mtwo31m1, xv)
|| !fpr.fpr_lt(xv, fpr.fpr_ptwo31m1))
{
return 0;
}
k[u] = (int)fpr.fpr_rint(xv);
}
/*
* Values in k[] are integers. They really are scaled
* down by maxbl_FG - minbl_fg bits.
*
* If we are at low depth, then we use the NTT to
* compute k*f and k*g.
*/
sch = (scale_k / 31);
scl = (scale_k % 31);
if (depth <= DEPTH_INT_FG)
{
poly_sub_scaled_ntt(srctmp, Ft, FGlen, llen, srctmp, ft, slen, slen,
k, 0, sch, scl, logn, srctmp, t1);
poly_sub_scaled_ntt(srctmp, Gt, FGlen, llen, srctmp, gt, slen, slen,
k, 0, sch, scl, logn, srctmp, t1);
}
else
{
poly_sub_scaled(srctmp, Ft, FGlen, llen, srctmp, ft, slen, slen,
k, 0, sch, scl, logn);
poly_sub_scaled(srctmp, Gt, FGlen, llen, srctmp, gt, slen, slen,
k, 0, sch, scl, logn);
}
/*
* We compute the new maximum size of (F,G), assuming that
* (f,g) has _maximal_ length (i.e. that reduction is
* "late" instead of "early". We also adjust FGlen
* accordingly.
*/
new_maxbl_FG = scale_k + maxbl_fg + 10;
if (new_maxbl_FG < maxbl_FG)
{
maxbl_FG = new_maxbl_FG;
if (FGlen * 31 >= maxbl_FG + 31)
{
FGlen--;
}
}
/*
* We suppose that scaling down achieves a reduction by
* at least 25 bits per iteration. We stop when we have
* done the loop with an unscaled k.
*/
if (scale_k <= 0)
{
break;
}
scale_k -= 25;
if (scale_k < 0)
{
scale_k = 0;
}
}
/*
* If (F,G) length was lowered below 'slen', then we must take
* care to re-extend the sign.
*/
if (FGlen < slen)
{
for (u = 0; u < n; u++, Ft += llen, Gt += llen)
{
int v;
int sw;
sw = -(srctmp[Ft + FGlen - 1] >>> 30) >>> 1;
for (v = FGlen; v < slen; v++)
{
srctmp[Ft + v] = sw;
}
sw = -(srctmp[Gt + FGlen - 1] >>> 30) >>> 1;
for (v = FGlen; v < slen; v++)
{
srctmp[Gt + v] = sw;
}
}
}
/*
* Compress encoding of all values to 'slen' words (this is the
* expected output format).
*/
for (u = 0, x = tmp, y = tmp;
u < (n << 1); u++, x += slen, y += llen)
{
// memmove(x, y, slen * sizeof *y);
System.arraycopy(srctmp, y, srctmp, x, slen);
}
return 1;
}
/*
* Solving the NTRU equation, binary case, depth = 1. Upon entry, the
* F and G from the previous level should be in the tmp[] array.
*
* Returned value: 1 on success, 0 on error.
*/
int solve_NTRU_binary_depth1(int logn_top,
byte[] srcf, int f, byte[] srcg, int g, int[] srctmp, int tmp)
{
/*
* The first half of this function is a copy of the corresponding
* part in solve_NTRU_intermediate(), for the reconstruction of
* the unreduced F and G. The second half (Babai reduction) is
* done differently, because the unreduced F and G fit in 53 bits
* of precision, allowing a much simpler process with lower RAM
* usage.
*/
int depth, logn;
int n_top, n, hn, slen, dlen, llen, u;
int Fd, Gd, Ft, Gt, ft, gt, t1;
FalconFPR[] rt1, rt2, rt3, rt4, rt5, rt6;
int x, y;
depth = 1;
n_top = 1 << logn_top;
logn = logn_top - depth;
n = 1 << logn;
hn = n >> 1;
/*
* Equations are:
*
* f' = f0^2 - X^2*f1^2
* g' = g0^2 - X^2*g1^2
* F' and G' are a solution to f'G' - g'F' = q (from deeper levels)
* F = F'*(g0 - X*g1)
* G = G'*(f0 - X*f1)
*
* f0, f1, g0, g1, f', g', F' and G' are all "compressed" to
* degree N/2 (their odd-indexed coefficients are all zero).
*/
/*
* slen = size for our input f and g; also size of the reduced
* F and G we return (degree N)
*
* dlen = size of the F and G obtained from the deeper level
* (degree N/2)
*
* llen = size for intermediary F and G before reduction (degree N)
*
* We build our non-reduced F and G as two independent halves each,
* of degree N/2 (F = F0 + X*F1, G = G0 + X*G1).
*/
slen = MAX_BL_SMALL[depth];
dlen = MAX_BL_SMALL[depth + 1];
llen = MAX_BL_LARGE[depth];
/*
* Fd and Gd are the F and G from the deeper level. Ft and Gt
* are the destination arrays for the unreduced F and G.
*/
Fd = tmp;
Gd = Fd + dlen * hn;
Ft = Gd + dlen * hn;
Gt = Ft + llen * n;
/*
* We reduce Fd and Gd modulo all the small primes we will need,
* and store the values in Ft and Gt.
*/
for (u = 0; u < llen; u++)
{
int p, p0i, R2, Rx;
int v;
int xs, ys, xd, yd;
p = this.primes.PRIMES[u].p;
p0i = modp_ninv31(p);
R2 = modp_R2(p, p0i);
Rx = modp_Rx(dlen, p, p0i, R2);
for (v = 0, xs = Fd, ys = Gd, xd = Ft + u, yd = Gt + u;
v < hn;
v++, xs += dlen, ys += dlen, xd += llen, yd += llen)
{
srctmp[xd] = zint_mod_small_signed(srctmp, xs, dlen, p, p0i, R2, Rx);
srctmp[yd] = zint_mod_small_signed(srctmp, ys, dlen, p, p0i, R2, Rx);
}
}
/*
* Now Fd and Gd are not needed anymore; we can squeeze them out.
*/
// memmove(tmp, Ft, llen * n * sizeof(uint32_t));
System.arraycopy(srctmp, Ft, srctmp, tmp, llen * n);
Ft = tmp;
// memmove(Ft + llen * n, Gt, llen * n * sizeof(uint32_t));
System.arraycopy(srctmp, Gt, srctmp, Ft + llen * n, llen * n);
Gt = Ft + llen * n;
ft = Gt + llen * n;
gt = ft + slen * n;
t1 = gt + slen * n;
/*
* Compute our F and G modulo sufficiently many small primes.
*/
for (u = 0; u < llen; u++)
{
int p, p0i, R2;
int gm, igm, fx, gx, Fp, Gp;
int e;
int v;
/*
* All computations are done modulo p.
*/
p = this.primes.PRIMES[u].p;
p0i = modp_ninv31(p);
R2 = modp_R2(p, p0i);
/*
* We recompute things from the source f and g, of full
* degree. However, we will need only the n first elements
* of the inverse NTT table (igm); the call to modp_mkgm()
* below will fill n_top elements in igm[] (thus overflowing
* into fx[]) but later code will overwrite these extra
* elements.
*/
gm = t1;
igm = gm + n_top;
fx = igm + n;
gx = fx + n_top;
modp_mkgm2(srctmp, gm, srctmp, igm, logn_top, this.primes.PRIMES[u].g, p, p0i);
/*
* Set ft and gt to f and g modulo p, respectively.
*/
for (v = 0; v < n_top; v++)
{
srctmp[fx + v] = modp_set(srcf[f + v], p);
srctmp[gx + v] = modp_set(srcg[g + v], p);
}
/*
* Convert to NTT and compute our f and g.
*/
modp_NTT2(srctmp, fx, srctmp, gm, logn_top, p, p0i);
modp_NTT2(srctmp, gx, srctmp, gm, logn_top, p, p0i);
for (e = logn_top; e > logn; e--)
{
modp_poly_rec_res(srctmp, fx, e, p, p0i, R2);
modp_poly_rec_res(srctmp, gx, e, p, p0i, R2);
}
/*
* From that point onward, we only need tables for
* degree n, so we can save some space.
*/
if (depth > 0)
{ /* always true */
// memmove(gm + n, igm, n * sizeof *igm);
System.arraycopy(srctmp, igm, srctmp, gm + n, n);
igm = gm + n;
// memmove(igm + n, fx, n * sizeof *ft);
System.arraycopy(srctmp, fx, srctmp, igm + n, n);
fx = igm + n;
// memmove(fx + n, gx, n * sizeof *gt);
System.arraycopy(srctmp, gx, srctmp, fx + n, n);
gx = fx + n;
}
/*
* Get F' and G' modulo p and in NTT representation
* (they have degree n/2). These values were computed
* in a previous step, and stored in Ft and Gt.
*/
Fp = gx + n;
Gp = Fp + hn;
for (v = 0, x = Ft + u, y = Gt + u;
v < hn; v++, x += llen, y += llen)
{
srctmp[Fp + v] = srctmp[x];
srctmp[Gp + v] = srctmp[y];
}
modp_NTT2(srctmp, Fp, srctmp, gm, logn - 1, p, p0i);
modp_NTT2(srctmp, Gp, srctmp, gm, logn - 1, p, p0i);
/*
* Compute our F and G modulo p.
*
* Equations are:
*
* f'(x^2) = N(f)(x^2) = f * adj(f)
* g'(x^2) = N(g)(x^2) = g * adj(g)
*
* f'*G' - g'*F' = q
*
* F = F'(x^2) * adj(g)
* G = G'(x^2) * adj(f)
*
* The NTT representation of f is f(w) for all w which
* are roots of phi. In the binary case, as well as in
* the ternary case for all depth except the deepest,
* these roots can be grouped in pairs (w,-w), and we
* then have:
*
* f(w) = adj(f)(-w)
* f(-w) = adj(f)(w)
*
* and w^2 is then a root for phi at the half-degree.
*
* At the deepest level in the ternary case, this still
* holds, in the following sense: the roots of x^2-x+1
* are (w,-w^2) (for w^3 = -1, and w != -1), and we
* have:
*
* f(w) = adj(f)(-w^2)
* f(-w^2) = adj(f)(w)
*
* In all case, we can thus compute F and G in NTT
* representation by a few simple multiplications.
* Moreover, the two roots for each pair are consecutive
* in our bit-reversal encoding.
*/
for (v = 0, x = Ft + u, y = Gt + u;
v < hn; v++, x += (llen << 1), y += (llen << 1))
{
int ftA, ftB, gtA, gtB;
int mFp, mGp;
ftA = srctmp[fx + (v << 1) + 0];
ftB = srctmp[fx + (v << 1) + 1];
gtA = srctmp[gx + (v << 1) + 0];
gtB = srctmp[gx + (v << 1) + 1];
mFp = modp_montymul(srctmp[Fp + v], R2, p, p0i);
mGp = modp_montymul(srctmp[Gp + v], R2, p, p0i);
srctmp[x + 0] = modp_montymul(gtB, mFp, p, p0i);
srctmp[x + llen] = modp_montymul(gtA, mFp, p, p0i);
srctmp[y + 0] = modp_montymul(ftB, mGp, p, p0i);
srctmp[y + llen] = modp_montymul(ftA, mGp, p, p0i);
}
modp_iNTT2_ext(srctmp, Ft + u, llen, srctmp, igm, logn, p, p0i);
modp_iNTT2_ext(srctmp, Gt + u, llen, srctmp, igm, logn, p, p0i);
/*
* Also save ft and gt (only up to size slen).
*/
if (u < slen)
{
modp_iNTT2(srctmp, fx, srctmp, igm, logn, p, p0i);
modp_iNTT2(srctmp, gx, srctmp, igm, logn, p, p0i);
for (v = 0, x = ft + u, y = gt + u;
v < n; v++, x += slen, y += slen)
{
srctmp[x] = srctmp[fx + v];
srctmp[y] = srctmp[gx + v];
}
}
}
/*
* Rebuild f, g, F and G with the CRT. Note that the elements of F
* and G are consecutive, and thus can be rebuilt in a single
* loop; similarly, the elements of f and g are consecutive.
*/
zint_rebuild_CRT(srctmp, Ft, llen, llen, n << 1, this.primes.PRIMES, 1, srctmp, t1);
zint_rebuild_CRT(srctmp, ft, slen, slen, n << 1, this.primes.PRIMES, 1, srctmp, t1);
/*
* Here starts the Babai reduction, specialized for depth = 1.
*
* Candidates F and G (from Ft and Gt), and base f and g (ft and gt),
* are converted to floating point. There is no scaling, and a
* single pass is sufficient.
*/
/*
* Convert F and G into floating point (rt1 and rt2).
*/
// rt1 = align_fpr(tmp, gt + slen * n);
rt1 = new FalconFPR[n];
rt2 = new FalconFPR[n];
poly_big_to_fp(rt1, 0, srctmp, Ft, llen, llen, logn);
poly_big_to_fp(rt2, 0, srctmp, Gt, llen, llen, logn);
/*
* Integer representation of F and G is no longer needed, we
* can remove it.
*/
// memmove(tmp, ft, 2 * slen * n * sizeof *ft);
System.arraycopy(srctmp, ft, srctmp, tmp, 2 * slen * n);
ft = tmp;
gt = ft + slen * n;
// rt3 = align_fpr(tmp, gt + slen * n);
// memmove(rt3, rt1, 2 * n * sizeof *rt1);
// rt1 = rt3;
// rt2 = rt1 + n;
rt3 = new FalconFPR[n];
rt4 = new FalconFPR[n];
/*
* Convert f and g into floating point (rt3 and rt4).
*/
poly_big_to_fp(rt3, 0, srctmp, ft, slen, slen, logn);
poly_big_to_fp(rt4, 0, srctmp, gt, slen, slen, logn);
/*
* Remove unneeded ft and gt. - not required as we have rt_ in separate array
*/
// memmove(tmp, rt1, 4 * n * sizeof *rt1);
// rt1 = (fpr *)tmp;
// rt2 = rt1 + n;
// rt3 = rt2 + n;
// rt4 = rt3 + n;
/*
* We now have:
* rt1 = F
* rt2 = G
* rt3 = f
* rt4 = g
* in that order in RAM. We convert all of them to FFT.
*/
fft.FFT(rt1, 0, logn);
fft.FFT(rt2, 0, logn);
fft.FFT(rt3, 0, logn);
fft.FFT(rt4, 0, logn);
/*
* Compute:
* rt5 = F*adj(f) + G*adj(g)
* rt6 = 1 / (f*adj(f) + g*adj(g))
* (Note that rt6 is half-length.)
*/
rt5 = new FalconFPR[n];
rt6 = new FalconFPR[n >> 1];
fft.poly_add_muladj_fft(rt5, 0, rt1, 0, rt2, 0, rt3, 0, rt4, 0, logn);
fft.poly_invnorm2_fft(rt6, 0, rt3, 0, rt4, 0, logn);
/*
* Compute:
* rt5 = (F*adj(f)+G*adj(g)) / (f*adj(f)+g*adj(g))
*/
fft.poly_mul_autoadj_fft(rt5, 0, rt6, 0, logn);
/*
* Compute k as the rounded version of rt5. Check that none of
* the values is larger than 2^63-1 (in absolute value)
* because that would make the fpr_rint() do something undefined;
* note that any out-of-bounds value here implies a failure and
* (f,g) will be discarded, so we can make a simple test.
*/
fft.iFFT(rt5, 0, logn);
for (u = 0; u < n; u++)
{
FalconFPR z;
z = rt5[u];
if (!fpr.fpr_lt(z, fpr.fpr_ptwo63m1) || !fpr.fpr_lt(fpr.fpr_mtwo63m1, z))
{
return 0;
}
rt5[u] = fpr.fpr_of(fpr.fpr_rint(z));
}
fft.FFT(rt5, 0, logn);
/*
* Subtract k*f from F, and k*g from G.
*/
fft.poly_mul_fft(rt3, 0, rt5, 0, logn);
fft.poly_mul_fft(rt4, 0, rt5, 0, logn);
fft.poly_sub(rt1, 0, rt3, 0, logn);
fft.poly_sub(rt2, 0, rt4, 0, logn);
fft.iFFT(rt1, 0, logn);
fft.iFFT(rt2, 0, logn);
/*
* Convert back F and G to integers, and return.
*/
Ft = tmp;
Gt = Ft + n;
// rt3 = align_fpr(tmp, Gt + n);
// memmove(rt3, rt1, 2 * n * sizeof *rt1);
// rt1 = rt3;
// rt2 = rt1 + n;
for (u = 0; u < n; u++)
{
srctmp[Ft + u] = (int)fpr.fpr_rint(rt1[u]);
srctmp[Gt + u] = (int)fpr.fpr_rint(rt2[u]);
}
return 1;
}
/*
* Solving the NTRU equation, top level. Upon entry, the F and G
* from the previous level should be in the tmp[] array.
*
* Returned value: 1 on success, 0 on error.
*/
int solve_NTRU_binary_depth0(int logn,
byte[] srcf, int f, byte[] srcg, int g, int[] srctmp, int tmp)
{
int n, hn, u;
int p, p0i, R2;
int Fp, Gp, t1, t2, t3, t4, t5;
int gm, igm, ft, gt;
int rt1, rt2, rt3;
n = 1 << logn;
hn = n >> 1;
/*
* Equations are:
*
* f' = f0^2 - X^2*f1^2
* g' = g0^2 - X^2*g1^2
* F' and G' are a solution to f'G' - g'F' = q (from deeper levels)
* F = F'*(g0 - X*g1)
* G = G'*(f0 - X*f1)
*
* f0, f1, g0, g1, f', g', F' and G' are all "compressed" to
* degree N/2 (their odd-indexed coefficients are all zero).
*
* Everything should fit in 31-bit integers, hence we can just use
* the first small prime p = 2147473409.
*/
p = this.primes.PRIMES[0].p;
p0i = modp_ninv31(p);
R2 = modp_R2(p, p0i);
Fp = tmp;
Gp = Fp + hn;
ft = Gp + hn;
gt = ft + n;
gm = gt + n;
igm = gm + n;
modp_mkgm2(srctmp, gm, srctmp, igm, logn, this.primes.PRIMES[0].g, p, p0i);
/*
* Convert F' anf G' in NTT representation.
*/
for (u = 0; u < hn; u++)
{
srctmp[Fp + u] = modp_set(zint_one_to_plain(srctmp, Fp + u), p);
srctmp[Gp + u] = modp_set(zint_one_to_plain(srctmp, Gp + u), p);
}
modp_NTT2(srctmp, Fp, srctmp, gm, logn - 1, p, p0i);
modp_NTT2(srctmp, Gp, srctmp, gm, logn - 1, p, p0i);
/*
* Load f and g and convert them to NTT representation.
*/
for (u = 0; u < n; u++)
{
srctmp[ft + u] = modp_set(srcf[f + u], p);
srctmp[gt + u] = modp_set(srcg[g + u], p);
}
modp_NTT2(srctmp, ft, srctmp, gm, logn, p, p0i);
modp_NTT2(srctmp, gt, srctmp, gm, logn, p, p0i);
/*
* Build the unreduced F,G in ft and gt.
*/
for (u = 0; u < n; u += 2)
{
int ftA, ftB, gtA, gtB;
int mFp, mGp;
ftA = srctmp[ft + u + 0];
ftB = srctmp[ft + u + 1];
gtA = srctmp[gt + u + 0];
gtB = srctmp[gt + u + 1];
mFp = modp_montymul(srctmp[Fp + (u >> 1)], R2, p, p0i);
mGp = modp_montymul(srctmp[Gp + (u >> 1)], R2, p, p0i);
srctmp[ft + u + 0] = modp_montymul(gtB, mFp, p, p0i);
srctmp[ft + u + 1] = modp_montymul(gtA, mFp, p, p0i);
srctmp[gt + u + 0] = modp_montymul(ftB, mGp, p, p0i);
srctmp[gt + u + 1] = modp_montymul(ftA, mGp, p, p0i);
}
modp_iNTT2(srctmp, ft, srctmp, igm, logn, p, p0i);
modp_iNTT2(srctmp, gt, srctmp, igm, logn, p, p0i);
Gp = Fp + n;
t1 = Gp + n;
// memmove(Fp, ft, 2 * n * sizeof *ft);
System.arraycopy(srctmp, ft, srctmp, Fp, 2 * n);
/*
* We now need to apply the Babai reduction. At that point,
* we have F and G in two n-word arrays.
*
* We can compute F*adj(f)+G*adj(g) and f*adj(f)+g*adj(g)
* modulo p, using the NTT. We still move memory around in
* order to save RAM.
*/
t2 = t1 + n;
t3 = t2 + n;
t4 = t3 + n;
t5 = t4 + n;
/*
* Compute the NTT tables in t1 and t2. We do not keep t2
* (we'll recompute it later on).
*/
modp_mkgm2(srctmp, t1, srctmp, t2, logn, this.primes.PRIMES[0].g, p, p0i);
/*
* Convert F and G to NTT.
*/
modp_NTT2(srctmp, Fp, srctmp, t1, logn, p, p0i);
modp_NTT2(srctmp, Gp, srctmp, t1, logn, p, p0i);
/*
* Load f and adj(f) in t4 and t5, and convert them to NTT
* representation.
*/
srctmp[t4 + 0] = srctmp[t5 + 0] = modp_set(srcf[f + 0], p);
for (u = 1; u < n; u++)
{
srctmp[t4 + u] = modp_set(srcf[f + u], p);
srctmp[t5 + n - u] = modp_set(-srcf[f + u], p);
}
modp_NTT2(srctmp, t4, srctmp, t1, logn, p, p0i);
modp_NTT2(srctmp, t5, srctmp, t1, logn, p, p0i);
/*
* Compute F*adj(f) in t2, and f*adj(f) in t3.
*/
for (u = 0; u < n; u++)
{
int w;
w = modp_montymul(srctmp[t5 + u], R2, p, p0i);
srctmp[t2 + u] = modp_montymul(w, srctmp[Fp + u], p, p0i);
srctmp[t3 + u] = modp_montymul(w, srctmp[t4 + u], p, p0i);
}
/*
* Load g and adj(g) in t4 and t5, and convert them to NTT
* representation.
*/
srctmp[t4 + 0] = srctmp[t5 + 0] = modp_set(srcg[g + 0], p);
for (u = 1; u < n; u++)
{
srctmp[t4 + u] = modp_set(srcg[g + u], p);
srctmp[t5 + n - u] = modp_set(-srcg[g + u], p);
}
modp_NTT2(srctmp, t4, srctmp, t1, logn, p, p0i);
modp_NTT2(srctmp, t5, srctmp, t1, logn, p, p0i);
/*
* Add G*adj(g) to t2, and g*adj(g) to t3.
*/
for (u = 0; u < n; u++)
{
int w;
w = modp_montymul(srctmp[t5 + u], R2, p, p0i);
srctmp[t2 + u] = modp_add(srctmp[t2 + u],
modp_montymul(w, srctmp[Gp + u], p, p0i), p);
srctmp[t3 + u] = modp_add(srctmp[t3 + u],
modp_montymul(w, srctmp[t4 + u], p, p0i), p);
}
/*
* Convert back t2 and t3 to normal representation (normalized
* around 0), and then
* move them to t1 and t2. We first need to recompute the
* inverse table for NTT.
*/
modp_mkgm2(srctmp, t1, srctmp, t4, logn, this.primes.PRIMES[0].g, p, p0i);
modp_iNTT2(srctmp, t2, srctmp, t4, logn, p, p0i);
modp_iNTT2(srctmp, t3, srctmp, t4, logn, p, p0i); // TODO fix binary_depth0 -> t1 value is wrong
for (u = 0; u < n; u++)
{
srctmp[t1 + u] = modp_norm(srctmp[t2 + u], p);
srctmp[t2 + u] = modp_norm(srctmp[t3 + u], p);
}
/*
* At that point, array contents are:
*
* F (NTT representation) (Fp)
* G (NTT representation) (Gp)
* F*adj(f)+G*adj(g) (t1)
* f*adj(f)+g*adj(g) (t2)
*
* We want to divide t1 by t2. The result is not integral; it
* must be rounded. We thus need to use the FFT.
*/
/*
* Get f*adj(f)+g*adj(g) in FFT representation. Since this
* polynomial is auto-adjoint, all its coordinates in FFT
* representation are actually real, so we can truncate off
* the imaginary parts.
*/
FalconFPR[]
tmp2 = new FalconFPR[3 * n];
// rt3 = align_fpr(tmp, t3);
rt1 = 0;
rt2 = rt1 + n;
rt3 = rt2 + n;
for (u = 0; u < n; u++)
{
tmp2[rt3 + u] = fpr.fpr_of(srctmp[t2 + u]);
}
fft.FFT(tmp2, rt3, logn);
// rt2 = align_fpr(tmp, t2);
// memmove(rt2, rt3, hn * sizeof *rt3);
System.arraycopy(tmp2, rt3, tmp2, rt2, hn);
/*
* Convert F*adj(f)+G*adj(g) in FFT representation.
*/
rt3 = rt2 + hn;
for (u = 0; u < n; u++)
{
tmp2[rt3 + u] = fpr.fpr_of(srctmp[t1 + u]);
}
fft.FFT(tmp2, rt3, logn);
/*
* Compute (F*adj(f)+G*adj(g))/(f*adj(f)+g*adj(g)) and get
* its rounded normal representation in t1.
*/
fft.poly_div_autoadj_fft(tmp2, rt3, tmp2, rt2, logn);
fft.iFFT(tmp2, rt3, logn);
for (u = 0; u < n; u++)
{
srctmp[t1 + u] = modp_set((int)fpr.fpr_rint(tmp2[rt3 + u]), p);
}
/*
* RAM contents are now:
*
* F (NTT representation) (Fp)
* G (NTT representation) (Gp)
* k (t1)
*
* We want to compute F-k*f, and G-k*g.
*/
t2 = t1 + n;
t3 = t2 + n;
t4 = t3 + n;
t5 = t4 + n;
modp_mkgm2(srctmp, t2, srctmp, t3, logn, this.primes.PRIMES[0].g, p, p0i);
for (u = 0; u < n; u++)
{
srctmp[t4 + u] = modp_set(srcf[f + u], p);
srctmp[t5 + u] = modp_set(srcg[g + u], p);
}
modp_NTT2(srctmp, t1, srctmp, t2, logn, p, p0i);
modp_NTT2(srctmp, t4, srctmp, t2, logn, p, p0i);
modp_NTT2(srctmp, t5, srctmp, t2, logn, p, p0i);
for (u = 0; u < n; u++)
{
int kw;
kw = modp_montymul(srctmp[t1 + u], R2, p, p0i);
srctmp[Fp + u] = modp_sub(srctmp[Fp + u],
modp_montymul(kw, srctmp[t4 + u], p, p0i), p);
srctmp[Gp + u] = modp_sub(srctmp[Gp + u],
modp_montymul(kw, srctmp[t5 + u], p, p0i), p);
}
modp_iNTT2(srctmp, Fp, srctmp, t3, logn, p, p0i);
modp_iNTT2(srctmp, Gp, srctmp, t3, logn, p, p0i);
for (u = 0; u < n; u++)
{
srctmp[Fp + u] = modp_norm(srctmp[Fp + u], p);
srctmp[Gp + u] = modp_norm(srctmp[Gp + u], p);
}
return 1;
}
/*
* Solve the NTRU equation. Returned value is 1 on success, 0 on error.
* G can be NULL, in which case that value is computed but not returned.
* If any of the coefficients of F and G exceeds lim (in absolute value),
* then 0 is returned.
*/
int solve_NTRU(int logn, byte[] srcF, int F, byte[] srcG, int G,
byte[] srcf, int f, byte[] srcg, int g, int lim, int[] srctmp, int tmp)
{
int n, u;
int ft, gt, Ft, Gt, gm;
int p, p0i, r;
FalconSmallPrime[] primes;
n = mkn(logn);
if (solve_NTRU_deepest(logn, srcf, f, srcg, g, srctmp, tmp) == 0)
{
return 0;
}
/*
* For logn <= 2, we need to use solve_NTRU_intermediate()
* directly, because coefficients are a bit too large and
* do not fit the hypotheses in solve_NTRU_binary_depth0().
*/
if (logn <= 2)
{
int depth;
depth = logn;
while (depth-- > 0)
{
if (solve_NTRU_intermediate(logn, srcf, f, srcg, g, depth, srctmp, tmp) == 0)
{
return 0;
}
}
}
else
{
int depth;
depth = logn;
while (depth-- > 2)
{
if (solve_NTRU_intermediate(logn, srcf, f, srcg, g, depth, srctmp, tmp) == 0)
{
return 0;
}
}
if (solve_NTRU_binary_depth1(logn, srcf, f, srcg, g, srctmp, tmp) == 0)
{
return 0;
}
if (solve_NTRU_binary_depth0(logn, srcf, f, srcg, g, srctmp, tmp) == 0)
{
return 0;
}
}
/*
* If no buffer has been provided for G, use a temporary one.
*/
if (srcG == null)
{
G = 0;
srcG = new byte[n];
}
/*
* Final F and G are in fk->tmp, one word per coefficient
* (signed value over 31 bits).
*/
if (poly_big_to_small(srcF, F, srctmp, tmp, lim, logn) == 0
|| poly_big_to_small(srcG, G, srctmp, tmp + n, lim, logn) == 0)
{
return 0;
}
/*
* Verify that the NTRU equation is fulfilled. Since all elements
* have short lengths, verifying modulo a small prime p works, and
* allows using the NTT.
*
* We put Gt[] first in tmp[], and process it first, so that it does
* not overlap with G[] in case we allocated it ourselves.
*/
Gt = tmp;
ft = Gt + n;
gt = ft + n;
Ft = gt + n;
gm = Ft + n;
primes = this.primes.PRIMES;
p = primes[0].p;
p0i = modp_ninv31(p);
modp_mkgm2(srctmp, gm, srctmp, tmp, logn, primes[0].g, p, p0i);
for (u = 0; u < n; u++)
{
srctmp[Gt + u] = modp_set(srcG[G + u], p);
}
for (u = 0; u < n; u++)
{
srctmp[ft + u] = modp_set(srcf[f + u], p);
srctmp[gt + u] = modp_set(srcg[g + u], p);
srctmp[Ft + u] = modp_set(srcF[F + u], p);
}
modp_NTT2(srctmp, ft, srctmp, gm, logn, p, p0i);
modp_NTT2(srctmp, gt, srctmp, gm, logn, p, p0i);
modp_NTT2(srctmp, Ft, srctmp, gm, logn, p, p0i);
modp_NTT2(srctmp, Gt, srctmp, gm, logn, p, p0i);
r = modp_montymul(12289, 1, p, p0i);
for (u = 0; u < n; u++)
{
int z;
z = modp_sub(modp_montymul(srctmp[ft + u], srctmp[Gt + u], p, p0i),
modp_montymul(srctmp[gt + u], srctmp[Ft + u], p, p0i), p);
if (z != r)
{
return 0;
}
}
return 1;
}
/*
* Generate a random polynomial with a Gaussian distribution. This function
* also makes sure that the resultant of the polynomial with phi is odd.
*/
void poly_small_mkgauss(SHAKE256 rng, byte[] srcf, int f, int logn)
{
int n, u;
int mod2;
n = mkn(logn);
mod2 = 0;
for (u = 0; u < n; u++)
{
int s;
for (; ; )
{
s = mkgauss(rng, logn);
/*
* We need the coefficient to fit within -127..+127;
* realistically, this is always the case except for
* the very low degrees (N = 2 or 4), for which there
* is no real security anyway.
*/
if (s < -127 || s > 127)
{
continue;
}
/*
* We need the sum of all coefficients to be 1; otherwise,
* the resultant of the polynomial with X^N+1 will be even,
* and the binary GCD will fail.
*/
if (u == n - 1)
{
if ((mod2 ^ (s & 1)) == 0)
{
continue;
}
}
else
{
mod2 ^= (s & 1);
}
srcf[f + u] = (byte)s;
break;
}
}
}
/* see falcon.h */
void keygen(SHAKE256 rng,
byte[] srcf, int f, byte[] srcg, int g, byte[] srcF, int F, byte[] srcG, int G, short[] srch, int h,
int logn)
{
/*
* Algorithm is the following:
*
* - Generate f and g with the Gaussian distribution.
*
* - If either Res(f,phi) or Res(g,phi) is even, try again.
*
* - If ||(f,g)|| is too large, try again.
*
* - If ||B~_{f,g}|| is too large, try again.
*
* - If f is not invertible mod phi mod q, try again.
*
* - Compute h = g/f mod phi mod q.
*
* - Solve the NTRU equation fG - gF = q; if the solving fails,
* try again. Usual failure condition is when Res(f,phi)
* and Res(g,phi) are not prime to each other.
*/
int n, u;
int[] itmp;
byte[] btmp;
short[] stmp;
FalconFPR[] ftmp;
int h2, tmp2;
SHAKE256 rc;
n = mkn(logn);
rc = rng;
/*
* We need to generate f and g randomly, until we find values
* such that the norm of (g,-f), and of the orthogonalized
* vector, are satisfying. The orthogonalized vector is:
* (q*adj(f)/(f*adj(f)+g*adj(g)), q*adj(g)/(f*adj(f)+g*adj(g)))
* (it is actually the (N+1)-th row of the Gram-Schmidt basis).
*
* In the binary case, coefficients of f and g are generated
* independently of each other, with a discrete Gaussian
* distribution of standard deviation 1.17*sqrt(q/(2*N)). Then,
* the two vectors have expected norm 1.17*sqrt(q), which is
* also our acceptance bound: we require both vectors to be no
* larger than that (this will be satisfied about 1/4th of the
* time, thus we expect sampling new (f,g) about 4 times for that
* step).
*
* We require that Res(f,phi) and Res(g,phi) are both odd (the
* NTRU equation solver requires it).
*/
for (; ; )
{
ftmp = new FalconFPR[3 * n];
int rt1, rt2, rt3;
FalconFPR bnorm;
int normf, normg, norm;
int lim;
/*
* The poly_small_mkgauss() function makes sure
* that the sum of coefficients is 1 modulo 2
* (i.e. the resultant of the polynomial with phi
* will be odd).
*/
poly_small_mkgauss(rc, srcf, f, logn);
poly_small_mkgauss(rc, srcg, g, logn);
/*
* Verify that all coefficients are within the bounds
* defined in max_fg_bits. This is the case with
* overwhelming probability; this guarantees that the
* key will be encodable with FALCON_COMP_TRIM.
*/
lim = 1 << (codec.max_fg_bits[logn] - 1);
for (u = 0; u < n; u++)
{
/*
* We can use non-CT tests since on any failure
* we will discard f and g.
*/
if (srcf[f + u] >= lim || srcf[f + u] <= -lim
|| srcg[g + u] >= lim || srcg[g + u] <= -lim)
{
lim = -1;
break;
}
}
if (lim < 0)
{
continue;
}
/*
* Bound is 1.17*sqrt(q). We compute the squared
* norms. With q = 12289, the squared bound is:
* (1.17^2)* 12289 = 16822.4121
* Since f and g are integral, the squared norm
* of (g,-f) is an integer.
*/
normf = poly_small_sqnorm(srcf, f, logn);
normg = poly_small_sqnorm(srcg, g, logn);
norm = (normf + normg) | -((normf | normg) >>> 31);
if ((norm & 0xffffffffL) >= 16823L)
{
continue;
}
/*
* We compute the orthogonalized vector norm.
*/
rt1 = 0;
rt2 = rt1 + n;
rt3 = rt2 + n;
poly_small_to_fp(ftmp, rt1, srcf, f, logn);
poly_small_to_fp(ftmp, rt2, srcg, g, logn);
fft.FFT(ftmp, rt1, logn);
fft.FFT(ftmp, rt2, logn);
fft.poly_invnorm2_fft(ftmp, rt3, ftmp, rt1, ftmp, rt2, logn);
fft.poly_adj_fft(ftmp, rt1, logn);
fft.poly_adj_fft(ftmp, rt2, logn);
fft.poly_mulconst(ftmp, rt1, fpr.fpr_q, logn);
fft.poly_mulconst(ftmp, rt2, fpr.fpr_q, logn);
fft.poly_mul_autoadj_fft(ftmp, rt1, ftmp, rt3, logn);
fft.poly_mul_autoadj_fft(ftmp, rt2, ftmp, rt3, logn);
fft.iFFT(ftmp, rt1, logn);
fft.iFFT(ftmp, rt2, logn);
bnorm = fpr.fpr_zero;
for (u = 0; u < n; u++)
{
bnorm = fpr.fpr_add(bnorm, fpr.fpr_sqr(ftmp[rt1 + u]));
bnorm = fpr.fpr_add(bnorm, fpr.fpr_sqr(ftmp[rt2 + u]));
}
if (!fpr.fpr_lt(bnorm, fpr.fpr_bnorm_max))
{
continue;
}
/*
* Compute public key h = g/f mod X^N+1 mod q. If this
* fails, we must restart.
*/
stmp = new short[2 * n];
if (srch == null)
{
h2 = 0;
srch = stmp;
tmp2 = h2 + n;
}
else
{
h2 = h;
tmp2 = 0;
}
if (vrfy.compute_public(srch, h2, srcf, f, srcg, g, logn, stmp, tmp2) == 0)
{
continue;
}
/*
* Solve the NTRU equation to get F and G.
*/
itmp = logn > 2 ? new int[28 * n] : new int[28 * n * 3];
lim = (1 << (codec.max_FG_bits[logn] - 1)) - 1;
if (solve_NTRU(logn, srcF, F, srcG, G, srcf, f, srcg, g, lim, itmp, 0) == 0)
{
continue;
}
/*
* Key pair is generated.
*/
break;
}
}
private long toUnsignedLong(int x)
{
return x & 0xffffffffL;
}
}