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libsecp256k1/src/field_impl.h  view on Meta::CPAN

/***********************************************************************
 * Copyright (c) 2013, 2014 Pieter Wuille                              *
 * Distributed under the MIT software license, see the accompanying    *
 * file COPYING or https://www.opensource.org/licenses/mit-license.php.*
 ***********************************************************************/

#ifndef SECP256K1_FIELD_IMPL_H
#define SECP256K1_FIELD_IMPL_H

#include "field.h"
#include "util.h"

#if defined(SECP256K1_WIDEMUL_INT128)
#include "field_5x52_impl.h"
#elif defined(SECP256K1_WIDEMUL_INT64)
#include "field_10x26_impl.h"
#else
#error "Please select wide multiplication implementation"
#endif

SECP256K1_INLINE static void secp256k1_fe_clear(secp256k1_fe *a) {
    secp256k1_memclear(a, sizeof(secp256k1_fe));
}

SECP256K1_INLINE static int secp256k1_fe_equal(const secp256k1_fe *a, const secp256k1_fe *b) {
    secp256k1_fe na;
    SECP256K1_FE_VERIFY(a);
    SECP256K1_FE_VERIFY(b);
    SECP256K1_FE_VERIFY_MAGNITUDE(a, 1);
    SECP256K1_FE_VERIFY_MAGNITUDE(b, 31);

    secp256k1_fe_negate(&na, a, 1);
    secp256k1_fe_add(&na, b);
    return secp256k1_fe_normalizes_to_zero(&na);
}

static int secp256k1_fe_sqrt(secp256k1_fe * SECP256K1_RESTRICT r, const secp256k1_fe * SECP256K1_RESTRICT a) {
    /** Given that p is congruent to 3 mod 4, we can compute the square root of
     *  a mod p as the (p+1)/4'th power of a.
     *
     *  As (p+1)/4 is an even number, it will have the same result for a and for
     *  (-a). Only one of these two numbers actually has a square root however,
     *  so we test at the end by squaring and comparing to the input.
     *  Also because (p+1)/4 is an even number, the computed square root is
     *  itself always a square (a ** ((p+1)/4) is the square of a ** ((p+1)/8)).
     */
    secp256k1_fe x2, x3, x6, x9, x11, x22, x44, x88, x176, x220, x223, t1;
    int j, ret;

    VERIFY_CHECK(r != a);
    SECP256K1_FE_VERIFY(a);
    SECP256K1_FE_VERIFY_MAGNITUDE(a, 8);

    /** The binary representation of (p + 1)/4 has 3 blocks of 1s, with lengths in
     *  { 2, 22, 223 }. Use an addition chain to calculate 2^n - 1 for each block:
     *  1, [2], 3, 6, 9, 11, [22], 44, 88, 176, 220, [223]
     */

    secp256k1_fe_sqr(&x2, a);
    secp256k1_fe_mul(&x2, &x2, a);

    secp256k1_fe_sqr(&x3, &x2);
    secp256k1_fe_mul(&x3, &x3, a);

    x6 = x3;
    for (j=0; j<3; j++) {
        secp256k1_fe_sqr(&x6, &x6);
    }
    secp256k1_fe_mul(&x6, &x6, &x3);

    x9 = x6;
    for (j=0; j<3; j++) {
        secp256k1_fe_sqr(&x9, &x9);
    }
    secp256k1_fe_mul(&x9, &x9, &x3);

    x11 = x9;
    for (j=0; j<2; j++) {