/** * Copyright (c) 2013-2014 Tomas Dzetkulic * Copyright (c) 2013-2014 Pavol Rusnak * Copyright (c) 2015 Jochen Hoenicke * Copyright (c) 2016 Alex Beregszaszi * * Permission is hereby granted, free of charge, to any person obtaining * a copy of this software and associated documentation files (the "Software"), * to deal in the Software without restriction, including without limitation * the rights to use, copy, modify, merge, publish, distribute, sublicense, * and/or sell copies of the Software, and to permit persons to whom the * Software is furnished to do so, subject to the following conditions: * * The above copyright notice and this permission notice shall be included * in all copies or substantial portions of the Software. * * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS * OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, * FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL * THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES * OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, * ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR * OTHER DEALINGS IN THE SOFTWARE. */ #include #include #include #include "bignum.h" /* big number library */ /* The structure bignum256 is an array of nine 32-bit values, which * are digits in base 2^30 representation. I.e. the number * bignum256 a; * represents the value * sum_{i=0}^8 a.val[i] * 2^{30 i}. * * The number is *normalized* iff every digit is < 2^30. * * As the name suggests, a bignum256 is intended to represent a 256 * bit number, but it can represent 270 bits. Numbers are usually * reduced using a prime, either the group order or the field prime. * The reduction is often partly done by bn_fast_mod, and similarly * implicitly in bn_multiply. A *partly reduced number* is a * normalized number between 0 (inclusive) and 2*prime (exclusive). * * A partly reduced number can be fully reduced by calling bn_mod. * Only a fully reduced number is guaranteed to fit in 256 bit. * * All functions assume that the prime in question is slightly smaller * than 2^256. In particular it must be between 2^256-2^224 and * 2^256 and it must be a prime number. */ inline uint32_t read_be(const uint8_t *data) { return (((uint32_t)data[0]) << 24) | (((uint32_t)data[1]) << 16) | (((uint32_t)data[2]) << 8) | (((uint32_t)data[3])); } inline void write_be(uint8_t *data, uint32_t x) { data[0] = x >> 24; data[1] = x >> 16; data[2] = x >> 8; data[3] = x; } inline uint32_t read_le(const uint8_t *data) { return (((uint32_t)data[3]) << 24) | (((uint32_t)data[2]) << 16) | (((uint32_t)data[1]) << 8) | (((uint32_t)data[0])); } inline void write_le(uint8_t *data, uint32_t x) { data[3] = x >> 24; data[2] = x >> 16; data[1] = x >> 8; data[0] = x; } // convert a raw bigendian 256 bit value into a normalized bignum. // out_number is partly reduced (since it fits in 256 bit). void bn_read_be(const uint8_t *in_number, bignum256 *out_number) { int i; uint32_t temp = 0; for (i = 0; i < 8; i++) { // invariant: temp = (in_number % 2^(32i)) >> 30i // get next limb = (in_number % 2^(32(i+1))) >> 32i uint32_t limb = read_be(in_number + (7 - i) * 4); // temp = (in_number % 2^(32(i+1))) << 30i temp |= limb << (2*i); // store 30 bits into val[i] out_number->val[i]= temp & 0x3FFFFFFF; // prepare temp for next round temp = limb >> (30 - 2*i); } out_number->val[8] = temp; } // convert a normalized bignum to a raw bigendian 256 bit number. // in_number must be fully reduced. void bn_write_be(const bignum256 *in_number, uint8_t *out_number) { int i; uint32_t temp = in_number->val[8] << 16; for (i = 0; i < 8; i++) { // invariant: temp = (in_number >> 30*(8-i)) << (16 + 2i) uint32_t limb = in_number->val[7 - i]; temp |= limb >> (14 - 2*i); write_be(out_number + i * 4, temp); temp = limb << (18 + 2*i); } } // convert a raw little endian 256 bit value into a normalized bignum. // out_number is partly reduced (since it fits in 256 bit). void bn_read_le(const uint8_t *in_number, bignum256 *out_number) { int i; uint32_t temp = 0; for (i = 0; i < 8; i++) { // invariant: temp = (in_number % 2^(32i)) >> 30i // get next limb = (in_number % 2^(32(i+1))) >> 32i uint32_t limb = read_le(in_number + i * 4); // temp = (in_number % 2^(32(i+1))) << 30i temp |= limb << (2*i); // store 30 bits into val[i] out_number->val[i]= temp & 0x3FFFFFFF; // prepare temp for next round temp = limb >> (30 - 2*i); } out_number->val[8] = temp; } // convert a normalized bignum to a raw little endian 256 bit number. // in_number must be fully reduced. void bn_write_le(const bignum256 *in_number, uint8_t *out_number) { int i; uint32_t temp = in_number->val[8] << 16; for (i = 0; i < 8; i++) { // invariant: temp = (in_number >> 30*(8-i)) << (16 + 2i) uint32_t limb = in_number->val[7 - i]; temp |= limb >> (14 - 2*i); write_le(out_number + (7 - i) * 4, temp); temp = limb << (18 + 2*i); } } void bn_read_uint32(uint32_t in_number, bignum256 *out_number) { out_number->val[0] = in_number & 0x3FFFFFFF; out_number->val[1] = in_number >> 30; out_number->val[2] = 0; out_number->val[3] = 0; out_number->val[4] = 0; out_number->val[5] = 0; out_number->val[6] = 0; out_number->val[7] = 0; out_number->val[8] = 0; } void bn_read_uint64(uint64_t in_number, bignum256 *out_number) { out_number->val[0] = in_number & 0x3FFFFFFF; out_number->val[1] = (in_number >>= 30) & 0x3FFFFFFF; out_number->val[2] = in_number >>= 30; out_number->val[3] = 0; out_number->val[4] = 0; out_number->val[5] = 0; out_number->val[6] = 0; out_number->val[7] = 0; out_number->val[8] = 0; } // a must be normalized int bn_bitcount(const bignum256 *a) { int i; for (i = 8; i >= 0; i--) { int tmp = a->val[i]; if (tmp != 0) { return i * 30 + (32 - __builtin_clz(tmp)); } } return 0; } #define DIGITS 78 // log10(2 ^ 256) unsigned int bn_digitcount(const bignum256 *a) { bignum256 val; memcpy(&val, a, sizeof(bignum256)); unsigned int digits = 1; for (unsigned int i = 0; i < DIGITS; i += 3) { uint32_t limb; bn_divmod1000(&val, &limb); if (limb >= 100) { digits = i + 3; } else if (limb >= 10) { digits = i + 2; } else if (limb >= 1) { digits = i + 1; } } return digits; } // sets a bignum to zero. void bn_zero(bignum256 *a) { int i; for (i = 0; i < 9; i++) { a->val[i] = 0; } } // sets a bignum to one. void bn_one(bignum256 *a) { a->val[0] = 1; a->val[1] = 0; a->val[2] = 0; a->val[3] = 0; a->val[4] = 0; a->val[5] = 0; a->val[6] = 0; a->val[7] = 0; a->val[8] = 0; } // checks that a bignum is zero. // a must be normalized // function is constant time (on some architectures, in particular ARM). int bn_is_zero(const bignum256 *a) { int i; uint32_t result = 0; for (i = 0; i < 9; i++) { result |= a->val[i]; } return !result; } // Check whether a < b // a and b must be normalized // function is constant time (on some architectures, in particular ARM). int bn_is_less(const bignum256 *a, const bignum256 *b) { int i; uint32_t res1 = 0; uint32_t res2 = 0; for (i = 8; i >= 0; i--) { res1 = (res1 << 1) | (a->val[i] < b->val[i]); res2 = (res2 << 1) | (a->val[i] > b->val[i]); } return res1 > res2; } // Check whether a == b // a and b must be normalized // function is constant time (on some architectures, in particular ARM). int bn_is_equal(const bignum256 *a, const bignum256 *b) { int i; uint32_t result = 0; for (i = 0; i < 9; i++) { result |= (a->val[i] ^ b->val[i]); } return !result; } // Assigns res = cond ? truecase : falsecase // assumes that cond is either 0 or 1. // function is constant time. void bn_cmov(bignum256 *res, int cond, const bignum256 *truecase, const bignum256 *falsecase) { int i; uint32_t tmask = (uint32_t) -cond; uint32_t fmask = ~tmask; assert (cond == 1 || cond == 0); for (i = 0; i < 9; i++) { res->val[i] = (truecase->val[i] & tmask) | (falsecase->val[i] & fmask); } } // shift number to the left, i.e multiply it by 2. // a must be normalized. The result is normalized but not reduced. void bn_lshift(bignum256 *a) { int i; for (i = 8; i > 0; i--) { a->val[i] = ((a->val[i] << 1) & 0x3FFFFFFF) | ((a->val[i - 1] & 0x20000000) >> 29); } a->val[0] = (a->val[0] << 1) & 0x3FFFFFFF; } // shift number to the right, i.e divide by 2 while rounding down. // a must be normalized. The result is normalized. void bn_rshift(bignum256 *a) { int i; for (i = 0; i < 8; i++) { a->val[i] = (a->val[i] >> 1) | ((a->val[i + 1] & 1) << 29); } a->val[8] >>= 1; } // sets bit in bignum void bn_setbit(bignum256 *a, uint8_t bit) { a->val[bit / 30] |= (1 << (bit % 30)); } // clears bit in bignum void bn_clearbit(bignum256 *a, uint8_t bit) { a->val[bit / 30] &= ~(1 << (bit % 30)); } // tests bit in bignum uint32_t bn_testbit(bignum256 *a, uint8_t bit) { return a->val[bit / 30] & (1 << (bit % 30)); } // a = b ^ c void bn_xor(bignum256 *a, const bignum256 *b, const bignum256 *c) { int i; for (i = 0; i < 9; i++) { a->val[i] = b->val[i] ^ c->val[i]; } } // multiply x by 1/2 modulo prime. // it computes x = (x & 1) ? (x + prime) >> 1 : x >> 1. // assumes x is normalized. // if x was partly reduced, it is also partly reduced on exit. // function is constant time. void bn_mult_half(bignum256 * x, const bignum256 *prime) { int j; uint32_t xodd = -(x->val[0] & 1); // compute x = x/2 mod prime // if x is odd compute (x+prime)/2 uint32_t tmp1 = (x->val[0] + (prime->val[0] & xodd)) >> 1; for (j = 0; j < 8; j++) { uint32_t tmp2 = (x->val[j+1] + (prime->val[j+1] & xodd)); tmp1 += (tmp2 & 1) << 29; x->val[j] = tmp1 & 0x3fffffff; tmp1 >>= 30; tmp1 += tmp2 >> 1; } x->val[8] = tmp1; } // multiply x by k modulo prime. // assumes x is normalized, 0 <= k <= 4. // guarantees x is partly reduced. void bn_mult_k(bignum256 *x, uint8_t k, const bignum256 *prime) { int j; for (j = 0; j < 9; j++) { x->val[j] = k * x->val[j]; } bn_fast_mod(x, prime); } // compute x = x mod prime by computing x >= prime ? x - prime : x. // assumes x partly reduced, guarantees x fully reduced. void bn_mod(bignum256 *x, const bignum256 *prime) { const int flag = bn_is_less(x, prime); // x < prime bignum256 temp; bn_subtract(x, prime, &temp); // temp = x - prime bn_cmov(x, flag, x, &temp); } // auxiliary function for multiplication. // compute k * x as a 540 bit number in base 2^30 (normalized). // assumes that k and x are normalized. void bn_multiply_long(const bignum256 *k, const bignum256 *x, uint32_t res[18]) { int i, j; uint64_t temp = 0; // compute lower half of long multiplication for (i = 0; i < 9; i++) { for (j = 0; j <= i; j++) { // no overflow, since 9*2^60 < 2^64 temp += k->val[j] * (uint64_t)x->val[i - j]; } res[i] = temp & 0x3FFFFFFFu; temp >>= 30; } // compute upper half for (; i < 17; i++) { for (j = i - 8; j < 9 ; j++) { // no overflow, since 9*2^60 < 2^64 temp += k->val[j] * (uint64_t)x->val[i - j]; } res[i] = temp & 0x3FFFFFFFu; temp >>= 30; } res[17] = temp; } // auxiliary function for multiplication. // reduces res modulo prime. // assumes res normalized, res < 2^(30(i-7)) * 2 * prime // guarantees res normalized, res < 2^(30(i-8)) * 2 * prime void bn_multiply_reduce_step(uint32_t res[18], const bignum256 *prime, uint32_t i) { // let k = i-8. // on entry: // 0 <= res < 2^(30k + 31) * prime // estimate coef = (res / prime / 2^30k) // by coef = res / 2^(30k + 256) rounded down // 0 <= coef < 2^31 // subtract (coef * 2^(30k) * prime) from res // note that we unrolled the first iteration uint32_t j; uint32_t coef = (res[i] >> 16) + (res[i + 1] << 14); uint64_t temp = 0x2000000000000000ull + res[i - 8] - prime->val[0] * (uint64_t)coef; assert (coef < 0x80000000u); res[i - 8] = temp & 0x3FFFFFFF; for (j = 1; j < 9; j++) { temp >>= 30; // Note: coeff * prime->val[j] <= (2^31-1) * (2^30-1) // Hence, this addition will not underflow. temp += 0x1FFFFFFF80000000ull + res[i - 8 + j] - prime->val[j] * (uint64_t)coef; res[i - 8 + j] = temp & 0x3FFFFFFF; // 0 <= temp < 2^61 + 2^30 } temp >>= 30; temp += 0x1FFFFFFF80000000ull + res[i - 8 + j]; res[i - 8 + j] = temp & 0x3FFFFFFF; // we rely on the fact that prime > 2^256 - 2^224 // res = oldres - coef*2^(30k) * prime; // and // coef * 2^(30k + 256) <= oldres < (coef+1) * 2^(30k + 256) // Hence, 0 <= res < 2^30k (2^256 + coef * (2^256 - prime)) // < 2^30k (2^256 + 2^31 * 2^224) // < 2^30k (2 * prime) } // auxiliary function for multiplication. // reduces x = res modulo prime. // assumes res normalized, res < 2^270 * 2 * prime // guarantees x partly reduced, i.e., x < 2 * prime void bn_multiply_reduce(bignum256 *x, uint32_t res[18], const bignum256 *prime) { int i; // res = k * x is a normalized number (every limb < 2^30) // 0 <= res < 2^270 * 2 * prime. for (i = 16; i >= 8; i--) { bn_multiply_reduce_step(res, prime, i); assert(res[i + 1] == 0); } // store the result for (i = 0; i < 9; i++) { x->val[i] = res[i]; } } // Compute x := k * x (mod prime) // both inputs must be smaller than 180 * prime. // result is partly reduced (0 <= x < 2 * prime) // This only works for primes between 2^256-2^224 and 2^256. void bn_multiply(const bignum256 *k, bignum256 *x, const bignum256 *prime) { uint32_t res[18] = {0}; bn_multiply_long(k, x, res); bn_multiply_reduce(x, res, prime); explicit_bzero(res, sizeof(res)); } // partly reduce x modulo prime // input x does not have to be normalized. // x can be any number that fits. // prime must be between (2^256 - 2^224) and 2^256 // result is partly reduced, smaller than 2*prime void bn_fast_mod(bignum256 *x, const bignum256 *prime) { int j; uint32_t coef; uint64_t temp; coef = x->val[8] >> 16; // substract (coef * prime) from x // note that we unrolled the first iteration temp = 0x2000000000000000ull + x->val[0] - prime->val[0] * (uint64_t)coef; x->val[0] = temp & 0x3FFFFFFF; for (j = 1; j < 9; j++) { temp >>= 30; temp += 0x1FFFFFFF80000000ull + x->val[j] - prime->val[j] * (uint64_t)coef; x->val[j] = temp & 0x3FFFFFFF; } } // square root of x = x^((p+1)/4) // http://en.wikipedia.org/wiki/Quadratic_residue#Prime_or_prime_power_modulus // assumes x is normalized but not necessarily reduced. // guarantees x is reduced void bn_sqrt(bignum256 *x, const bignum256 *prime) { // this method compute x^1/2 = x^(prime+1)/4 uint32_t i, j, limb; bignum256 res, p; bn_one(&res); // compute p = (prime+1)/4 memcpy(&p, prime, sizeof(bignum256)); bn_addi(&p, 1); bn_rshift(&p); bn_rshift(&p); for (i = 0; i < 9; i++) { // invariants: // x = old(x)^(2^(i*30)) // res = old(x)^(p % 2^(i*30)) // get the i-th limb of prime - 2 limb = p.val[i]; for (j = 0; j < 30; j++) { // invariants: // x = old(x)^(2^(i*30+j)) // res = old(x)^(p % 2^(i*30+j)) // limb = (p % 2^(i*30+30)) / 2^(i*30+j) if (i == 8 && limb == 0) break; if (limb & 1) { bn_multiply(x, &res, prime); } limb >>= 1; bn_multiply(x, x, prime); } } bn_mod(&res, prime); memcpy(x, &res, sizeof(bignum256)); explicit_bzero(&res, sizeof(res)); explicit_bzero(&p, sizeof(p)); } #if ! USE_INVERSE_FAST // in field G_prime, small but slow void bn_inverse(bignum256 *x, const bignum256 *prime) { // this method compute x^-1 = x^(prime-2) uint32_t i, j, limb; bignum256 res; bn_one(&res); for (i = 0; i < 9; i++) { // invariants: // x = old(x)^(2^(i*30)) // res = old(x)^((prime-2) % 2^(i*30)) // get the i-th limb of prime - 2 limb = prime->val[i]; // this is not enough in general but fine for secp256k1 & nist256p1 because prime->val[0] > 1 if (i == 0) limb -= 2; for (j = 0; j < 30; j++) { // invariants: // x = old(x)^(2^(i*30+j)) // res = old(x)^((prime-2) % 2^(i*30+j)) // limb = ((prime-2) % 2^(i*30+30)) / 2^(i*30+j) // early abort when only zero bits follow if (i == 8 && limb == 0) break; if (limb & 1) { bn_multiply(x, &res, prime); } limb >>= 1; bn_multiply(x, x, prime); } } bn_mod(&res, prime); memcpy(x, &res, sizeof(bignum256)); } #else // in field G_prime, big and complicated but fast // the input must not be 0 mod prime. // the result is smaller than prime void bn_inverse(bignum256 *x, const bignum256 *prime) { int i, j, k, cmp; struct combo { uint32_t a[9]; int len1; } us, vr, *odd, *even; uint32_t pp[8]; uint32_t temp32; uint64_t temp; // The algorithm is based on Schroeppel et. al. "Almost Modular Inverse" // algorithm. We keep four values u,v,r,s in the combo registers // us and vr. us stores u in the first len1 limbs (little endian) // and s in the last 9-len1 limbs (big endian). vr stores v and r. // This is because both u*s and v*r are guaranteed to fit in 8 limbs, so // their components are guaranteed to fit in 9. During the algorithm, // the length of u and v shrinks while r and s grow. // u,v,r,s correspond to F,G,B,C in Schroeppel's algorithm. // reduce x modulo prime. This is necessary as it has to fit in 8 limbs. bn_fast_mod(x, prime); bn_mod(x, prime); // convert x and prime to 8x32 bit limb form temp32 = prime->val[0]; for (i = 0; i < 8; i++) { temp32 |= prime->val[i + 1] << (30-2*i); us.a[i] = pp[i] = temp32; temp32 = prime->val[i + 1] >> (2+2*i); } temp32 = x->val[0]; for (i = 0; i < 8; i++) { temp32 |= x->val[i + 1] << (30-2*i); vr.a[i] = temp32; temp32 = x->val[i + 1] >> (2+2*i); } us.len1 = 8; vr.len1 = 8; // set s = 1 and r = 0 us.a[8] = 1; vr.a[8] = 0; // set k = 0. k = 0; // only one of the numbers u,v can be even at any time. We // let even point to that number and odd to the other. // Initially the prime u is guaranteed to be odd. odd = &us; even = &vr; // u = prime, v = x // r = 0 , s = 1 // k = 0 for (;;) { // invariants: // let u = limbs us.a[0..u.len1-1] in little endian, // let s = limbs us.a[u.len..8] in big endian, // let v = limbs vr.a[0..u.len1-1] in little endian, // let r = limbs vr.a[u.len..8] in big endian, // r,s >= 0 ; u,v >= 1 // x*-r = u*2^k mod prime // x*s = v*2^k mod prime // u*s + v*r = prime // floor(log2(u)) + floor(log2(v)) + k <= 510 // max(u,v) <= 2^k (*) see comment at end of loop // gcd(u,v) = 1 // {odd,even} = {&us, &vr} // odd->a[0] and odd->a[8] are odd // even->a[0] or even->a[8] is even // // first u/v are large and r/s small // later u/v are small and r/s large assert(odd->a[0] & 1); assert(odd->a[8] & 1); // adjust length of even. while (even->a[even->len1 - 1] == 0) { even->len1--; // if input was 0, return. // This simple check prevents crashing with stack underflow // or worse undesired behaviour for illegal input. if (even->len1 < 0) return; } // reduce even->a while it is even while (even->a[0] == 0) { // shift right first part of even by a limb // and shift left second part of even by a limb. for (i = 0; i < 8; i++) { even->a[i] = even->a[i+1]; } even->a[i] = 0; even->len1--; k += 32; } // count up to 32 zero bits of even->a. j = 0; while ((even->a[0] & (1 << j)) == 0) { j++; } if (j > 0) { // shift first part of even right by j bits. for (i = 0; i + 1 < even->len1; i++) { even->a[i] = (even->a[i] >> j) | (even->a[i + 1] << (32 - j)); } even->a[i] = (even->a[i] >> j); if (even->a[i] == 0) { even->len1--; } else { i++; } // shift second part of even left by j bits. for (; i < 8; i++) { even->a[i] = (even->a[i] << j) | (even->a[i + 1] >> (32 - j)); } even->a[i] = (even->a[i] << j); // add j bits to k. k += j; } // invariant is reestablished. // now both a[0] are odd. assert(odd->a[0] & 1); assert(odd->a[8] & 1); assert(even->a[0] & 1); assert((even->a[8] & 1) == 0); // cmp > 0 if us.a[0..len1-1] > vr.a[0..len1-1], // cmp = 0 if equal, < 0 if less. cmp = us.len1 - vr.len1; if (cmp == 0) { i = us.len1 - 1; while (i >= 0 && us.a[i] == vr.a[i]) i--; // both are equal to 1 and we are done. if (i == -1) break; cmp = us.a[i] > vr.a[i] ? 1 : -1; } if (cmp > 0) { even = &us; odd = &vr; } else { even = &vr; odd = &us; } // now even > odd. // even->a[0..len1-1] = (even->a[0..len1-1] - odd->a[0..len1-1]); temp = 1; for (i = 0; i < odd->len1; i++) { temp += 0xFFFFFFFFull + even->a[i] - odd->a[i]; even->a[i] = temp & 0xFFFFFFFF; temp >>= 32; } for (; i < even->len1; i++) { temp += 0xFFFFFFFFull + even->a[i]; even->a[i] = temp & 0xFFFFFFFF; temp >>= 32; } // odd->a[len1..8] = (odd->b[len1..8] + even->b[len1..8]); temp = 0; for (i = 8; i >= even->len1; i--) { temp += (uint64_t) odd->a[i] + even->a[i]; odd->a[i] = temp & 0xFFFFFFFF; temp >>= 32; } for (; i >= odd->len1; i--) { temp += (uint64_t) odd->a[i]; odd->a[i] = temp & 0xFFFFFFFF; temp >>= 32; } // note that // if u > v: // u'2^k = (u - v) 2^k = x(-r) - xs = x(-(r+s)) = x(-r') mod prime // u's' + v'r' = (u-v)s + v(r+s) = us + vr // if u < v: // v'2^k = (v - u) 2^k = xs - x(-r) = x(s+r) = xs' mod prime // u's' + v'r' = u(s+r) + (v-u)r = us + vr // even->a[0] is difference between two odd numbers, hence even. // odd->a[8] is sum of even and odd number, hence odd. assert(odd->a[0] & 1); assert(odd->a[8] & 1); assert((even->a[0] & 1) == 0); // The invariants are (almost) reestablished. // The invariant max(u,v) <= 2^k can be invalidated at this point, // because odd->a[len1..8] was changed. We only have // // odd->a[len1..8] <= 2^{k+1} // // Since even->a[0] is even, k will be incremented at the beginning // of the next loop while odd->a[len1..8] remains unchanged. // So after that, odd->a[len1..8] <= 2^k will hold again. } // In the last iteration we had u = v and gcd(u,v) = 1. // Hence, u=1, v=1, s+r = prime, k <= 510, 2^k > max(s,r) >= prime/2 // This implies 0 <= s < prime and 255 <= k <= 510. // // The invariants also give us x*s = 2^k mod prime, // hence s = 2^k * x^-1 mod prime. // We need to compute s/2^k mod prime. // First we compute inverse = -prime^-1 mod 2^32, which we need later. // We use the Explicit Quadratic Modular inverse algorithm. // http://arxiv.org/pdf/1209.6626.pdf // a^-1 = (2-a) * PROD_i (1 + (a - 1)^(2^i)) mod 2^32 // the product will converge quickly, because (a-1)^(2^i) will be // zero mod 2^32 after at most five iterations. // We want to compute -prime^-1 so we start with (pp[0]-2). assert(pp[0] & 1); uint32_t amone = pp[0]-1; uint32_t inverse = pp[0] - 2; while (amone) { amone *= amone; inverse *= (amone + 1); } while (k >= 32) { // compute s / 2^32 modulo prime. // Idea: compute factor, such that // s + factor*prime mod 2^32 == 0 // i.e. factor = s * -1/prime mod 2^32. // Then compute s + factor*prime and shift right by 32 bits. uint32_t factor = (inverse * us.a[8]) & 0xffffffff; temp = us.a[8] + (uint64_t) pp[0] * factor; assert((temp & 0xffffffff) == 0); temp >>= 32; for (i = 0; i < 7; i++) { temp += us.a[8-(i+1)] + (uint64_t) pp[i+1] * factor; us.a[8-i] = temp & 0xffffffff; temp >>= 32; } us.a[8-i] = temp & 0xffffffff; k -= 32; } if (k > 0) { // compute s / 2^k modulo prime. // Same idea: compute factor, such that // s + factor*prime mod 2^k == 0 // i.e. factor = s * -1/prime mod 2^k. // Then compute s + factor*prime and shift right by k bits. uint32_t mask = (1 << k) - 1; uint32_t factor = (inverse * us.a[8]) & mask; temp = (us.a[8] + (uint64_t) pp[0] * factor) >> k; assert(((us.a[8] + pp[0] * factor) & mask) == 0); for (i = 0; i < 7; i++) { temp += (us.a[8-(i+1)] + (uint64_t) pp[i+1] * factor) << (32 - k); us.a[8-i] = temp & 0xffffffff; temp >>= 32; } us.a[8-i] = temp & 0xffffffff; } // convert s to bignum style temp32 = 0; for (i = 0; i < 8; i++) { x->val[i] = ((us.a[8-i] << (2 * i)) & 0x3FFFFFFFu) | temp32; temp32 = us.a[8-i] >> (30 - 2 * i); } x->val[i] = temp32; // let's wipe all temp buffers explicit_bzero(pp, sizeof(pp)); explicit_bzero(&us, sizeof(us)); explicit_bzero(&vr, sizeof(vr)); } #endif void bn_normalize(bignum256 *a) { bn_addi(a, 0); } // add two numbers a = a + b // assumes that a, b are normalized // guarantees that a is normalized void bn_add(bignum256 *a, const bignum256 *b) { int i; uint32_t tmp = 0; for (i = 0; i < 9; i++) { tmp += a->val[i] + b->val[i]; a->val[i] = tmp & 0x3FFFFFFF; tmp >>= 30; } } void bn_addmod(bignum256 *a, const bignum256 *b, const bignum256 *prime) { int i; for (i = 0; i < 9; i++) { a->val[i] += b->val[i]; } bn_fast_mod(a, prime); } void bn_addi(bignum256 *a, uint32_t b) { int i; uint32_t tmp = b; for (i = 0; i < 9; i++) { tmp += a->val[i]; a->val[i] = tmp & 0x3FFFFFFF; tmp >>= 30; } } void bn_subi(bignum256 *a, uint32_t b, const bignum256 *prime) { assert (b <= prime->val[0]); // the possible underflow will be taken care of when adding the prime a->val[0] -= b; bn_add(a, prime); } // res = a - b mod prime. More exactly res = a + (2*prime - b). // b must be a partly reduced number // result is normalized but not reduced. void bn_subtractmod(const bignum256 *a, const bignum256 *b, bignum256 *res, const bignum256 *prime) { int i; uint32_t temp = 1; for (i = 0; i < 9; i++) { temp += 0x3FFFFFFF + a->val[i] + 2u * prime->val[i] - b->val[i]; res->val[i] = temp & 0x3FFFFFFF; temp >>= 30; } } // res = a - b ; a > b void bn_subtract(const bignum256 *a, const bignum256 *b, bignum256 *res) { int i; uint32_t tmp = 1; for (i = 0; i < 9; i++) { tmp += 0x3FFFFFFF + a->val[i] - b->val[i]; res->val[i] = tmp & 0x3FFFFFFF; tmp >>= 30; } } // a / 58 = a (+r) void bn_divmod58(bignum256 *a, uint32_t *r) { int i; uint32_t rem, tmp; rem = a->val[8] % 58; a->val[8] /= 58; for (i = 7; i >= 0; i--) { // invariants: // rem = old(a) >> 30(i+1) % 58 // a[i+1..8] = old(a[i+1..8])/58 // a[0..i] = old(a[0..i]) // 2^30 == 18512790*58 + 4 tmp = rem * 4 + a->val[i]; // set a[i] = (rem * 2^30 + a[i])/58 // = rem * 18512790 + (rem * 4 + a[i])/58 a->val[i] = rem * 18512790 + (tmp / 58); // set rem = (rem * 2^30 + a[i]) mod 58 // = (rem * 4 + a[i]) mod 58 rem = tmp % 58; } *r = rem; } // a / 1000 = a (+r) void bn_divmod1000(bignum256 *a, uint32_t *r) { int i; uint32_t rem, tmp; rem = a->val[8] % 1000; a->val[8] /= 1000; for (i = 7; i >= 0; i--) { // invariants: // rem = old(a) >> 30(i+1) % 1000 // a[i+1..8] = old(a[i+1..8])/1000 // a[0..i] = old(a[0..i]) // 2^30 == 1073741*1000 + 824 tmp = rem * 824 + a->val[i]; // set a[i] = (rem * 2^30 + a[i])/1000 // = rem * 1073741 + (rem * 824 + a[i])/1000 a->val[i] = rem * 1073741 + (tmp / 1000); // set rem = (rem * 2^30 + a[i]) mod 1000 // = (rem * 824 + a[i]) mod 1000 rem = tmp % 1000; } *r = rem; } size_t bn_format(const bignum256 *amnt, const char *prefix, const char *suffix, unsigned int decimals, int exponent, bool trailing, char *out, size_t outlen) { size_t prefixlen = prefix ? strlen(prefix) : 0; size_t suffixlen = suffix ? strlen(suffix) : 0; char *start = &out[prefixlen + suffixlen], *end = &out[outlen]; char *str = end; #define BN_FORMAT_PUSH_CHECKED(c) \ do { \ if (str == start) return 0; \ *--str = (c); \ } while (0) #define BN_FORMAT_PUSH(n) \ do { \ if (exponent < 0) { \ exponent++; \ } else { \ if ((n) > 0 || trailing || str != end || decimals <= 1) { \ BN_FORMAT_PUSH_CHECKED('0' + (n)); \ } \ if (decimals > 0 && decimals-- == 1) { \ BN_FORMAT_PUSH_CHECKED('.'); \ } \ } \ } while (0) bignum256 val; memcpy(&val, amnt, sizeof(bignum256)); if (bn_is_zero(&val)) { exponent = 0; } for (; exponent > 0; exponent--) { BN_FORMAT_PUSH(0); } unsigned int digits = bn_digitcount(&val); for (unsigned int i = 0; i < digits / 3; i++) { uint32_t limb; bn_divmod1000(&val, &limb); BN_FORMAT_PUSH(limb % 10); limb /= 10; BN_FORMAT_PUSH(limb % 10); limb /= 10; BN_FORMAT_PUSH(limb % 10); } if (digits % 3 != 0) { uint32_t limb; bn_divmod1000(&val, &limb); switch (digits % 3) { case 2: BN_FORMAT_PUSH(limb % 10); limb /= 10; //-fallthrough case 1: BN_FORMAT_PUSH(limb % 10); break; } } while (decimals > 0 || str[0] == '\0' || str[0] == '.') { BN_FORMAT_PUSH(0); } size_t len = end - str; memmove(&out[prefixlen], str, len); if (prefixlen) { memcpy(out, prefix, prefixlen); } if (suffixlen) { memcpy(&out[prefixlen + len], suffix, suffixlen); } size_t length = prefixlen + len + suffixlen; out[length] = '\0'; return length; } #if USE_BN_PRINT void bn_print(const bignum256 *a) { printf("%04x", a->val[8] & 0x0000FFFF); printf("%08x", (a->val[7] << 2) | ((a->val[6] & 0x30000000) >> 28)); printf("%07x", a->val[6] & 0x0FFFFFFF); printf("%08x", (a->val[5] << 2) | ((a->val[4] & 0x30000000) >> 28)); printf("%07x", a->val[4] & 0x0FFFFFFF); printf("%08x", (a->val[3] << 2) | ((a->val[2] & 0x30000000) >> 28)); printf("%07x", a->val[2] & 0x0FFFFFFF); printf("%08x", (a->val[1] << 2) | ((a->val[0] & 0x30000000) >> 28)); printf("%07x", a->val[0] & 0x0FFFFFFF); } void bn_print_raw(const bignum256 *a) { int i; for (i = 0; i <= 8; i++) { printf("0x%08x, ", a->val[i]); } } #endif