[BearSSL] / src / ec / ec_p256_i15.c

/*
 * Copyright (c) 2017 Thomas Pornin <pornin@bolet.org>
 *
 * Permission is hereby granted, free of charge, to any person obtaining 
 * a copy of this software and associated documentation files (the
 * "Software"), to deal in the Software without restriction, including
 * without limitation the rights to use, copy, modify, merge, publish,
 * distribute, sublicense, and/or sell copies of the Software, and to
 * permit persons to whom the Software is furnished to do so, subject to
 * the following conditions:
 *
 * The above copyright notice and this permission notice shall be 
 * included in all copies or substantial portions of the Software.
 *
 * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, 
 * EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
 * MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND 
 * NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS
 * BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN
 * ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
 * CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
 * SOFTWARE.
 */

#include "inner.h"

/*
 * If BR_NO_ARITH_SHIFT is undefined, or defined to 0, then we _assume_
 * that right-shifting a signed negative integer copies the sign bit
 * (arithmetic right-shift). This is "implementation-defined behaviour",
 * i.e. it is not undefined, but it may differ between compilers. Each
 * compiler is supposed to document its behaviour in that respect. GCC
 * explicitly defines that an arithmetic right shift is used. We expect
 * all other compilers to do the same, because underlying CPU offer an
 * arithmetic right shift opcode that could not be used otherwise.
 */
#if BR_NO_ARITH_SHIFT
#define ARSH(x, n)   (((uint32_t)(x) >> (n)) \
                    | ((-((uint32_t)(x) >> 31)) << (32 - (n))))
#else
#define ARSH(x, n)   ((*(int32_t *)&(x)) >> (n))
#endif

/*
 * Convert an integer from unsigned big-endian encoding to a sequence of
 * 13-bit words in little-endian order. The final "partial" word is
 * returned.
 */
static uint32_t
be8_to_le13(uint32_t *dst, const unsigned char *src, size_t len)
{
        uint32_t acc;
        int acc_len;

        acc = 0;
        acc_len = 0;
        while (len -- > 0) {
                acc |= (uint32_t)src[len] << acc_len;
                acc_len += 8;
                if (acc_len >= 13) {
                        *dst ++ = acc & 0x1FFF;
                        acc >>= 13;
                        acc_len -= 13;
                }
        }
        return acc;
}

/*
 * Convert an integer (13-bit words, little-endian) to unsigned
 * big-endian encoding. The total encoding length is provided; all
 * the destination bytes will be filled.
 */
static void
le13_to_be8(unsigned char *dst, size_t len, const uint32_t *src)
{
        uint32_t acc;
        int acc_len;

        acc = 0;
        acc_len = 0;
        while (len -- > 0) {
                if (acc_len < 8) {
                        acc |= (*src ++) << acc_len;
                        acc_len += 13;
                }
                dst[len] = (unsigned char)acc;
                acc >>= 8;
                acc_len -= 8;
        }
}

/*
 * Multiply two 4-word integers. Basis is 2^13 = 8192. Individual words
 * may have value up to 32766. Result integer consists in eight words;
 * each word is in the 0..8191 range, except the last one (d[7]) which
 * gathers the excess bits.
 *
 * Max value for d[7], depending on max word value:
 *
 *   32764: 131071
 *   32765: 131080
 *   32766: 131088
 */
static inline void
mul4(uint32_t *d, const uint32_t *a, const uint32_t *b)
{
        uint32_t t0, t1, t2, t3, t4, t5, t6;

        t0 = MUL15(a[0], b[0]);
        t1 = MUL15(a[0], b[1]) + MUL15(a[1], b[0]);
        t2 = MUL15(a[0], b[2]) + MUL15(a[1], b[1]) + MUL15(a[2], b[0]);
        t3 = MUL15(a[0], b[3]) + MUL15(a[1], b[2])
                + MUL15(a[2], b[1]) + MUL15(a[3], b[0]);
        t4 = MUL15(a[1], b[3]) + MUL15(a[2], b[2]) + MUL15(a[3], b[1]);
        t5 = MUL15(a[2], b[3]) + MUL15(a[3], b[2]);
        t6 = MUL15(a[3], b[3]);

        /*
         * Maximum value is obtained when adding the carry to t3, when all
         * input words have maximum value. When the maximum is 32766,
         * the addition of t3 with the carry (t2 >> 13) yields 4294836223,
         * which is still below 2^32 = 4294967296.
         */

        d[0] = t0 & 0x1FFF;
        t1 += t0 >> 13;
        d[1] = t1 & 0x1FFF;
        t2 += t1 >> 13;
        d[2] = t2 & 0x1FFF;
        t3 += t2 >> 13;
        d[3] = t3 & 0x1FFF;
        t4 += t3 >> 13;
        d[4] = t4 & 0x1FFF;
        t5 += t4 >> 13;
        d[5] = t5 & 0x1FFF;
        t6 += t5 >> 13;
        d[6] = t6 & 0x1FFF;
        d[7] = t6 >> 13;
}

/*
 * Normalise an array of words to a strict 13 bits per word. Returned
 * value is the resulting carry. The source (w) and destination (d)
 * arrays may be identical, but shall not overlap partially.
 */
static uint32_t
norm13(uint32_t *d, const uint32_t *w, size_t len)
{
        size_t u;
        uint32_t cc;

        cc = 0;
        for (u = 0; u < len; u ++) {
                int32_t z;

                z = w[u] + cc;
                d[u] = z & 0x1FFF;
                cc = ARSH(z, 13);
        }
        return cc;
}

/*
 * Multiply two 20-word values together, result over 40 words. Each word
 * contains 13 bits of data; the top words (a[19] and b[19]) may contain
 * up to 15 bits each. Therefore this routine handles integer up to 2^262-1.
 * All words in the output have length 13 bits, except possibly the top
 * one, in case the sources had excess bits.
 */
static void
mul20(uint32_t *d, const uint32_t *a, const uint32_t *b)
{
        /*
         * We first split the 20-word values into a 16-word value and
         * a 4-word value. The 16x16 product uses two levels of Karatsuba
         * decomposition. The preparatory additions are done word-wise
         * without carry propagation; since the input words have size
         * at most 13 bits, the sums may be up to 4*8191 = 32764, which
         * is in the range supported by mul4().
         */
        uint32_t u[72], v[72], w[144];
        uint32_t cc;
        int i;

        /*
         * Karatsuba decomposition, two levels.
         *
         * off   src
         *       0..7 * 0..7
         *   0      0..3 * 0..3
         *   4      4..7 * 4..7
         *  16      0..3+4..7 * 0..3+4..7
         *
         *       8..15 * 8..15
         *   8      8..11 * 8..11
         *  12      12..15 * 12..15
         *  20      8..11+12..15 * 8..11+12..15
         *
         *       0..7+8..15 * 0..7+8..15
         *  24      0..3+8..11 * 0..3+8..11
         *  28      4..7+12..15 * 4..7+12..15
         *  32      0..3+4..7+8..11+12..15 * 0..3+4..7+8..11+12..15
         */

#define WADD(z, x, y)    do { \
                u[(z) + 0] = u[(x) + 0] + u[y + 0]; \
                u[(z) + 1] = u[(x) + 1] + u[y + 1]; \
                u[(z) + 2] = u[(x) + 2] + u[y + 2]; \
                u[(z) + 3] = u[(x) + 3] + u[y + 3]; \
                v[(z) + 0] = v[(x) + 0] + v[y + 0]; \
                v[(z) + 1] = v[(x) + 1] + v[y + 1]; \
                v[(z) + 2] = v[(x) + 2] + v[y + 2]; \
                v[(z) + 3] = v[(x) + 3] + v[y + 3]; \
        } while (0)

        memcpy(u, a, 16 * sizeof *a);
        memcpy(v, b, 16 * sizeof *b);
        WADD(16, 0, 4);
        WADD(20, 8, 12);
        WADD(24, 0, 8);
        WADD(28, 4, 12);
        WADD(32, 16, 20);
        memcpy(u + 36, a, 16 * sizeof *a);
        memcpy(v + 52, b, 16 * sizeof *b);
        for (i = 0; i < 4; i ++) {
                memcpy(u + 52 + (i << 2), a + 16, 4 * sizeof *a);
                memcpy(v + 36 + (i << 2), b + 16, 4 * sizeof *b);
        }
        memcpy(u + 68, a + 16, 4 * sizeof *a);
        memcpy(v + 68, b + 16, 4 * sizeof *b);

#undef WADD

        /*
         * Perform the elementary 4x4 multiplications:
         *   16x16: 9 multiplications (Karatsuba)
         *   16x4 (1): 4 multiplications
         *   16x4 (2): 4 multiplications
         *   4x4: 1 multiplication
         */
        for (i = 0; i < 18; i ++) {
                mul4(w + (i << 3), u + (i << 2), v + (i << 2));
        }

        /*
         * mul4 cross-product adjustments:
         *   subtract 0 and 8 from 32 (8 words)
         *   subtract 16 and 24 from 40 (8 words)
         *   subtract 48 and 56 from 64 (8 words)
         */
        for (i = 0; i < 8; i ++) {
                w[i + 32] -= w[i +  0] + w[i +  8];
                w[i + 40] -= w[i + 16] + w[i + 24];
                w[i + 64] -= w[i + 48] + w[i + 56];
        }

        /*
         * complete the three 8x8 products:
         *   add 32 to 4 (8 words)
         *   add 40 to 20 (8 words)
         *   add 64 to 52 (8 words)
         */
        for (i = 0; i < 8; i ++) {
                w[i +  4] += w[i + 32];
                w[i + 20] += w[i + 40];
                w[i + 52] += w[i + 64];
        }

        /*
         * Adjust the 16x16 product:
         *   subtract 0 from 48 (16 words)
         *   subtract 16 from 48 (16 words)
         *   add 48 to 8 (16 words)
         */
        for (i = 0; i < 16; i ++) {
                w[i + 48] -= w[i +  0];
                w[i + 48] -= w[i + 16];
        }
        for (i = 0; i < 16; i ++) {
                w[i + 8] += w[i + 48];
        }

        /*
         * At that point, the product of the low chunks (0..15 * 0..15)
         * is in words 0..31. We must add the three other partial products,
         * which begin at word 72 in w[]. Words 32 to 39 are first set to
         * the product of the high chunks (16..19 * 16..19), then the
         * low-high cross products are added in.
         */
        memcpy(w + 32, w + 136, 8 * sizeof w[0]);
        for (i = 0; i < 8; i ++) {
                w[i + 16] += w[i + 72] + w[i + 104];
                w[i + 20] += w[i + 80] + w[i + 112];
                w[i + 24] += w[i + 88] + w[i + 120];
                w[i + 28] += w[i + 96] + w[i + 128];
        }

        /*
         * We did all the additions and subtractions in a word-wise way,
         * which is fine since we have plenty of extra bits for carries.
         * We must now do the carry propagation.
         */
        cc = norm13(d, w, 40);
        d[39] += (cc << 13);
}

/*
 * Modulus for field F256 (field for point coordinates in curve P-256).
 */
static const uint32_t F256[] = {
        0x1FFF, 0x1FFF, 0x1FFF, 0x1FFF, 0x1FFF, 0x1FFF, 0x1FFF, 0x001F,
        0x0000, 0x0000, 0x0000, 0x0000, 0x0000, 0x0000, 0x0400, 0x0000,
        0x0000, 0x1FF8, 0x1FFF, 0x01FF
};

/*
 * The 'b' curve equation coefficient for P-256.
 */
static const uint32_t P256_B[] = {
        0x004B, 0x1E93, 0x0F89, 0x1C78, 0x03BC, 0x187B, 0x114E, 0x1619,
        0x1D06, 0x0328, 0x01AF, 0x0D31, 0x1557, 0x15DE, 0x1ECF, 0x127C,
        0x0A3A, 0x0EC5, 0x118D, 0x00B5
};

/*
 * Perform a "short reduction" in field F256 (field for curve P-256).
 * The source value should be less than 262 bits; on output, it will
 * be at most 257 bits, and less than twice the modulus.
 */
static void
reduce_f256(uint32_t *d)
{
        uint32_t x;

        x = d[19] >> 9;
        d[19] &= 0x01FF;
        d[17] += x << 3;
        d[14] -= x << 10;
        d[7] -= x << 5;
        d[0] += x;
        norm13(d, d, 20);
}

/*
 * Perform a "final reduction" in field F256 (field for curve P-256).
 * The source value must be less than twice the modulus. If the value
 * is not lower than the modulus, then the modulus is subtracted and
 * this function returns 1; otherwise, it leaves it untouched and it
 * returns 0.
 */
static uint32_t
reduce_final_f256(uint32_t *d)
{
        uint32_t t[20];
        uint32_t cc;
        int i;

        memcpy(t, d, sizeof t);
        cc = 0;
        for (i = 0; i < 20; i ++) {
                uint32_t w;

                w = t[i] - F256[i] - cc;
                cc = w >> 31;
                t[i] = w & 0x1FFF;
        }
        cc ^= 1;
        CCOPY(cc, d, t, sizeof t);
        return cc;
}

/*
 * Perform a multiplication of two integers modulo
 * 2^256-2^224+2^192+2^96-1 (for NIST curve P-256). Operands are arrays
 * of 20 words, each containing 13 bits of data, in little-endian order.
 * On input, upper word may be up to 15 bits (hence value up to 2^262-1);
 * on output, value fits on 257 bits and is lower than twice the modulus.
 */
static void
mul_f256(uint32_t *d, const uint32_t *a, const uint32_t *b)
{
        uint32_t t[40], cc;
        int i;

        /*
         * Compute raw multiplication. All result words fit in 13 bits
         * each.
         */
        mul20(t, a, b);

        /*
         * Modular reduction: each high word in added/subtracted where
         * necessary.
         *
         * The modulus is:
         *    p = 2^256 - 2^224 + 2^192 + 2^96 - 1
         * Therefore:
         *    2^256 = 2^224 - 2^192 - 2^96 + 1 mod p
         *
         * For a word x at bit offset n (n >= 256), we have:
         *    x*2^n = x*2^(n-32) - x*2^(n-64)
         *            - x*2^(n - 160) + x*2^(n-256) mod p
         *
         * Thus, we can nullify the high word if we reinject it at some
         * proper emplacements.
         */
        for (i = 39; i >= 20; i --) {
                uint32_t x;

                x = t[i];
                t[i - 2] += ARSH(x, 6);
                t[i - 3] += (x << 7) & 0x1FFF;
                t[i - 4] -= ARSH(x, 12);
                t[i - 5] -= (x << 1) & 0x1FFF;
                t[i - 12] -= ARSH(x, 4);
                t[i - 13] -= (x << 9) & 0x1FFF;
                t[i - 19] += ARSH(x, 9);
                t[i - 20] += (x << 4) & 0x1FFF;
        }

        /*
         * Propagate carries. Since the operation above really is a
         * truncature, followed by the addition of nonnegative values,
         * the result will be positive. Moreover, the carry cannot
         * exceed 5 bits (we performed 20 additions with values smaller
         * than 256 bits).
         */
        cc = norm13(t, t, 20);

        /*
         * Perform modular reduction again for the bits beyond 256 (the carry
         * and the bits 256..259). This time, we can simply inject full
         * word values.
         */
        cc = (cc << 4) | (t[19] >> 9);
        t[19] &= 0x01FF;
        t[17] += cc << 3;
        t[14] -= cc << 10;
        t[7] -= cc << 5;
        t[0] += cc;
        norm13(d, t, 20);
}

/*
 * Jacobian coordinates for a point in P-256: affine coordinates (X,Y)
 * are such that:
 *   X = x / z^2
 *   Y = y / z^3
 * For the point at infinity, z = 0.
 * Each point thus admits many possible representations.
 *
 * Coordinates are represented in arrays of 32-bit integers, each holding
 * 13 bits of data. Values may also be slightly greater than the modulus,
 * but they will always be lower than twice the modulus.
 */
typedef struct {
        uint32_t x[20];
        uint32_t y[20];
        uint32_t z[20];
} p256_jacobian;

/*
 * Convert a point to affine coordinates:
 *  - If the point is the point at infinity, then all three coordinates
 *    are set to 0.
 *  - Otherwise, the 'z' coordinate is set to 1, and the 'x' and 'y'
 *    coordinates are the 'X' and 'Y' affine coordinates.
 * The coordinates are guaranteed to be lower than the modulus.
 */
static void
p256_to_affine(p256_jacobian *P)
{
        uint32_t t1[20], t2[20];
        int i;

        /*
         * Invert z with a modular exponentiation: the modulus is
         * p = 2^256 - 2^224 + 2^192 + 2^96 - 1, and the exponent is
         * p-2. Exponent bit pattern (from high to low) is:
         *  - 32 bits of value 1
         *  - 31 bits of value 0
         *  - 1 bit of value 1
         *  - 96 bits of value 0
         *  - 94 bits of value 1
         *  - 1 bit of value 0
         *  - 1 bit of value 1
         * Thus, we precompute z^(2^31-1) to speed things up.
         *
         * If z = 0 (point at infinity) then the modular exponentiation
         * will yield 0, which leads to the expected result (all three
         * coordinates set to 0).
         */

        /*
         * A simple square-and-multiply for z^(2^31-1). We could save about
         * two dozen multiplications here with an addition chain, but
         * this would require a bit more code, and extra stack buffers.
         */
        memcpy(t1, P->z, sizeof P->z);
        for (i = 0; i < 30; i ++) {
                mul_f256(t1, t1, t1);
                mul_f256(t1, t1, P->z);
        }

        /*
         * Square-and-multiply. Apart from the squarings, we have a few
         * multiplications to set bits to 1; we multiply by the original z
         * for setting 1 bit, and by t1 for setting 31 bits.
         */
        memcpy(t2, P->z, sizeof P->z);
        for (i = 1; i < 256; i ++) {
                mul_f256(t2, t2, t2);
                switch (i) {
                case 31:
                case 190:
                case 221:
                case 252:
                        mul_f256(t2, t2, t1);
                        break;
                case 63:
                case 253:
                case 255:
                        mul_f256(t2, t2, P->z);
                        break;
                }
        }

        /*
         * Now that we have 1/z, multiply x by 1/z^2 and y by 1/z^3.
         */
        mul_f256(t1, t2, t2);
        mul_f256(P->x, t1, P->x);
        mul_f256(t1, t1, t2);
        mul_f256(P->y, t1, P->y);
        reduce_final_f256(P->x);
        reduce_final_f256(P->y);

        /*
         * Multiply z by 1/z. If z = 0, then this will yield 0, otherwise
         * this will set z to 1.
         */
        mul_f256(P->z, P->z, t2);
        reduce_final_f256(P->z);
}

/*
 * Double a point in P-256. This function works for all valid points,
 * including the point at infinity.
 */
static void
p256_double(p256_jacobian *Q)
{
        /*
         * Doubling formulas are:
         *
         *   s = 4*x*y^2
         *   m = 3*(x + z^2)*(x - z^2)
         *   x' = m^2 - 2*s
         *   y' = m*(s - x') - 8*y^4
         *   z' = 2*y*z
         *
         * These formulas work for all points, including points of order 2
         * and points at infinity:
         *   - If y = 0 then z' = 0. But there is no such point in P-256
         *     anyway.
         *   - If z = 0 then z' = 0.
         */
        uint32_t t1[20], t2[20], t3[20], t4[20];
        int i;

        /*
         * Compute z^2 in t1.
         */
        mul_f256(t1, Q->z, Q->z);

        /*
         * Compute x-z^2 in t2 and x+z^2 in t1.
         */
        for (i = 0; i < 20; i ++) {
                t2[i] = (F256[i] << 1) + Q->x[i] - t1[i];
                t1[i] += Q->x[i];
        }
        norm13(t1, t1, 20);
        norm13(t2, t2, 20);

        /*
         * Compute 3*(x+z^2)*(x-z^2) in t1.
         */
        mul_f256(t3, t1, t2);
        for (i = 0; i < 20; i ++) {
                t1[i] = MUL15(3, t3[i]);
        }
        norm13(t1, t1, 20);

        /*
         * Compute 4*x*y^2 (in t2) and 2*y^2 (in t3).
         */
        mul_f256(t3, Q->y, Q->y);
        for (i = 0; i < 20; i ++) {
                t3[i] <<= 1;
        }
        norm13(t3, t3, 20);
        mul_f256(t2, Q->x, t3);
        for (i = 0; i < 20; i ++) {
                t2[i] <<= 1;
        }
        norm13(t2, t2, 20);
        reduce_f256(t2);

        /*
         * Compute x' = m^2 - 2*s.
         */
        mul_f256(Q->x, t1, t1);
        for (i = 0; i < 20; i ++) {
                Q->x[i] += (F256[i] << 2) - (t2[i] << 1);
        }
        norm13(Q->x, Q->x, 20);
        reduce_f256(Q->x);

        /*
         * Compute z' = 2*y*z.
         */
        mul_f256(t4, Q->y, Q->z);
        for (i = 0; i < 20; i ++) {
                Q->z[i] = t4[i] << 1;
        }
        norm13(Q->z, Q->z, 20);
        reduce_f256(Q->z);

        /*
         * Compute y' = m*(s - x') - 8*y^4. Note that we already have
         * 2*y^2 in t3.
         */
        for (i = 0; i < 20; i ++) {
                t2[i] += (F256[i] << 1) - Q->x[i];
        }
        norm13(t2, t2, 20);
        mul_f256(Q->y, t1, t2);
        mul_f256(t4, t3, t3);
        for (i = 0; i < 20; i ++) {
                Q->y[i] += (F256[i] << 2) - (t4[i] << 1);
        }
        norm13(Q->y, Q->y, 20);
        reduce_f256(Q->y);
}

/*
 * Add point P2 to point P1.
 *
 * This function computes the wrong result in the following cases:
 *
 *   - If P1 == 0 but P2 != 0
 *   - If P1 != 0 but P2 == 0
 *   - If P1 == P2
 *
 * In all three cases, P1 is set to the point at infinity.
 *
 * Returned value is 0 if one of the following occurs:
 *
 *   - P1 and P2 have the same Y coordinate
 *   - P1 == 0 and P2 == 0
 *   - The Y coordinate of one of the points is 0 and the other point is
 *     the point at infinity.
 *
 * The third case cannot actually happen with valid points, since a point
 * with Y == 0 is a point of order 2, and there is no point of order 2 on
 * curve P-256.
 *
 * Therefore, assuming that P1 != 0 and P2 != 0 on input, then the caller
 * can apply the following:
 *
 *   - If the result is not the point at infinity, then it is correct.
 *   - Otherwise, if the returned value is 1, then this is a case of
 *     P1+P2 == 0, so the result is indeed the point at infinity.
 *   - Otherwise, P1 == P2, so a "double" operation should have been
 *     performed.
 */
static uint32_t
p256_add(p256_jacobian *P1, const p256_jacobian *P2)
{
        /*
         * Addtions formulas are:
         *
         *   u1 = x1 * z2^2
         *   u2 = x2 * z1^2
         *   s1 = y1 * z2^3
         *   s2 = y2 * z1^3
         *   h = u2 - u1
         *   r = s2 - s1
         *   x3 = r^2 - h^3 - 2 * u1 * h^2
         *   y3 = r * (u1 * h^2 - x3) - s1 * h^3
         *   z3 = h * z1 * z2
         */
        uint32_t t1[20], t2[20], t3[20], t4[20], t5[20], t6[20], t7[20];
        uint32_t ret;
        int i;

        /*
         * Compute u1 = x1*z2^2 (in t1) and s1 = y1*z2^3 (in t3).
         */
        mul_f256(t3, P2->z, P2->z);
        mul_f256(t1, P1->x, t3);
        mul_f256(t4, P2->z, t3);
        mul_f256(t3, P1->y, t4);

        /*
         * Compute u2 = x2*z1^2 (in t2) and s2 = y2*z1^3 (in t4).
         */
        mul_f256(t4, P1->z, P1->z);
        mul_f256(t2, P2->x, t4);
        mul_f256(t5, P1->z, t4);
        mul_f256(t4, P2->y, t5);

        /*
         * Compute h = h2 - u1 (in t2) and r = s2 - s1 (in t4).
         * We need to test whether r is zero, so we will do some extra
         * reduce.
         */
        for (i = 0; i < 20; i ++) {
                t2[i] += (F256[i] << 1) - t1[i];
                t4[i] += (F256[i] << 1) - t3[i];
        }
        norm13(t2, t2, 20);
        norm13(t4, t4, 20);
        reduce_f256(t4);
        reduce_final_f256(t4);
        ret = 0;
        for (i = 0; i < 20; i ++) {
                ret |= t4[i];
        }
        ret = (ret | -ret) >> 31;

        /*
         * Compute u1*h^2 (in t6) and h^3 (in t5);
         */
        mul_f256(t7, t2, t2);
        mul_f256(t6, t1, t7);
        mul_f256(t5, t7, t2);

        /*
         * Compute x3 = r^2 - h^3 - 2*u1*h^2.
         */
        mul_f256(P1->x, t4, t4);
        for (i = 0; i < 20; i ++) {
                P1->x[i] += (F256[i] << 3) - t5[i] - (t6[i] << 1);
        }
        norm13(P1->x, P1->x, 20);
        reduce_f256(P1->x);

        /*
         * Compute y3 = r*(u1*h^2 - x3) - s1*h^3.
         */
        for (i = 0; i < 20; i ++) {
                t6[i] += (F256[i] << 1) - P1->x[i];
        }
        norm13(t6, t6, 20);
        mul_f256(P1->y, t4, t6);
        mul_f256(t1, t5, t3);
        for (i = 0; i < 20; i ++) {
                P1->y[i] += (F256[i] << 1) - t1[i];
        }
        norm13(P1->y, P1->y, 20);
        reduce_f256(P1->y);

        /*
         * Compute z3 = h*z1*z2.
         */
        mul_f256(t1, P1->z, P2->z);
        mul_f256(P1->z, t1, t2);

        return ret;
}

/*
 * Decode a P-256 point. This function does not support the point at
 * infinity. Returned value is 0 if the point is invalid, 1 otherwise.
 */
static uint32_t
p256_decode(p256_jacobian *P, const void *src, size_t len)
{
        const unsigned char *buf;
        uint32_t tx[20], ty[20], t1[20], t2[20];
        uint32_t bad;
        int i;

        if (len != 65) {
                return 0;
        }
        buf = src;

        /*
         * First byte must be 0x04 (uncompressed format). We could support
         * "hybrid format" (first byte is 0x06 or 0x07, and encodes the
         * least significant bit of the Y coordinate), but it is explicitly
         * forbidden by RFC 5480 (section 2.2).
         */
        bad = NEQ(buf[0], 0x04);

        /*
         * Decode the coordinates, and check that they are both lower
         * than the modulus.
         */
        tx[19] = be8_to_le13(tx, buf + 1, 32);
        ty[19] = be8_to_le13(ty, buf + 33, 32);
        bad |= reduce_final_f256(tx);
        bad |= reduce_final_f256(ty);

        /*
         * Check curve equation.
         */
        mul_f256(t1, tx, tx);
        mul_f256(t1, tx, t1);
        mul_f256(t2, ty, ty);
        for (i = 0; i < 20; i ++) {
                t1[i] += (F256[i] << 3) - MUL15(3, tx[i]) + P256_B[i] - t2[i];
        }
        norm13(t1, t1, 20);
        reduce_f256(t1);
        reduce_final_f256(t1);
        for (i = 0; i < 20; i ++) {
                bad |= t1[i];
        }

        /*
         * Copy coordinates to the point structure.
         */
        memcpy(P->x, tx, sizeof tx);
        memcpy(P->y, ty, sizeof ty);
        memset(P->z, 0, sizeof P->z);
        P->z[0] = 1;
        return NEQ(bad, 0) ^ 1;
}

/*
 * Encode a point into a buffer. This function assumes that the point is
 * valid, in affine coordinates, and not the point at infinity.
 */
static void
p256_encode(void *dst, const p256_jacobian *P)
{
        unsigned char *buf;

        buf = dst;
        buf[0] = 0x04;
        le13_to_be8(buf + 1, 32, P->x);
        le13_to_be8(buf + 33, 32, P->y);
}

/*
 * Multiply a curve point by an integer. The integer is assumed to be
 * lower than the curve order, and the base point must not be the point
 * at infinity.
 */
static void
p256_mul(p256_jacobian *P, const unsigned char *x, size_t xlen)
{
        /*
         * qz is a flag that is initially 1, and remains equal to 1
         * as long as the point is the point at infinity.
         *
         * We use a 2-bit window to handle multiplier bits by pairs.
         * The precomputed window really is the points P2 and P3.
         */
        uint32_t qz;
        p256_jacobian P2, P3, Q, T, U;

        /*
         * Compute window values.
         */
        P2 = *P;
        p256_double(&P2);
        P3 = *P;
        p256_add(&P3, &P2);

        /*
         * We start with Q = 0. We process multiplier bits 2 by 2.
         */
        memset(&Q, 0, sizeof Q);
        qz = 1;
        while (xlen -- > 0) {
                int k;

                for (k = 6; k >= 0; k -= 2) {
                        uint32_t bits;
                        uint32_t bnz;

                        p256_double(&Q);
                        p256_double(&Q);
                        T = *P;
                        U = Q;
                        bits = (*x >> k) & (uint32_t)3;
                        bnz = NEQ(bits, 0);
                        CCOPY(EQ(bits, 2), &T, &P2, sizeof T);
                        CCOPY(EQ(bits, 3), &T, &P3, sizeof T);
                        p256_add(&U, &T);
                        CCOPY(bnz & qz, &Q, &T, sizeof Q);
                        CCOPY(bnz & ~qz, &Q, &U, sizeof Q);
                        qz &= ~bnz;
                }
                x ++;
        }
        *P = Q;
}

static const unsigned char P256_G[] = {
        0x04, 0x6B, 0x17, 0xD1, 0xF2, 0xE1, 0x2C, 0x42, 0x47, 0xF8,
        0xBC, 0xE6, 0xE5, 0x63, 0xA4, 0x40, 0xF2, 0x77, 0x03, 0x7D,
        0x81, 0x2D, 0xEB, 0x33, 0xA0, 0xF4, 0xA1, 0x39, 0x45, 0xD8,
        0x98, 0xC2, 0x96, 0x4F, 0xE3, 0x42, 0xE2, 0xFE, 0x1A, 0x7F,
        0x9B, 0x8E, 0xE7, 0xEB, 0x4A, 0x7C, 0x0F, 0x9E, 0x16, 0x2B,
        0xCE, 0x33, 0x57, 0x6B, 0x31, 0x5E, 0xCE, 0xCB, 0xB6, 0x40,
        0x68, 0x37, 0xBF, 0x51, 0xF5
};

static const unsigned char P256_N[] = {
        0xFF, 0xFF, 0xFF, 0xFF, 0x00, 0x00, 0x00, 0x00, 0xFF, 0xFF,
        0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xBC, 0xE6, 0xFA, 0xAD,
        0xA7, 0x17, 0x9E, 0x84, 0xF3, 0xB9, 0xCA, 0xC2, 0xFC, 0x63,
        0x25, 0x51
};

static const unsigned char *
api_generator(int curve, size_t *len)
{
        (void)curve;
        *len = sizeof P256_G;
        return P256_G;
}

static const unsigned char *
api_order(int curve, size_t *len)
{
        (void)curve;
        *len = sizeof P256_N;
        return P256_N;
}

static uint32_t
api_mul(unsigned char *G, size_t Glen,
        const unsigned char *x, size_t xlen, int curve)
{
        uint32_t r;
        p256_jacobian P;

        (void)curve;
        r = p256_decode(&P, G, Glen);
        p256_mul(&P, x, xlen);
        if (Glen >= 65) {
                p256_to_affine(&P);
                p256_encode(G, &P);
        }
        return r;
}

static uint32_t
api_muladd(unsigned char *A, const unsigned char *B, size_t len,
        const unsigned char *x, size_t xlen,
        const unsigned char *y, size_t ylen, int curve)
{
        p256_jacobian P, Q;
        uint32_t r, t, z;
        int i;

        (void)curve;
        r = p256_decode(&P, A, len);
        r &= p256_decode(&Q, B, len);
        p256_mul(&P, x, xlen);
        p256_mul(&Q, y, ylen);

        /*
         * The final addition may fail in case both points are equal.
         */
        t = p256_add(&P, &Q);
        reduce_final_f256(P.z);
        z = 0;
        for (i = 0; i < 20; i ++) {
                z |= P.z[i];
        }
        z = EQ(z, 0);
        p256_double(&Q);

        /*
         * If z is 1 then either P+Q = 0 (t = 1) or P = Q (t = 0). So we
         * have the following:
         *
         *   z = 0, t = 0   return P (normal addition)
         *   z = 0, t = 1   return P (normal addition)
         *   z = 1, t = 0   return Q (a 'double' case)
         *   z = 1, t = 1   report an error (P+Q = 0)
         */
        CCOPY(z & ~t, &P, &Q, sizeof Q);
        p256_to_affine(&P);
        p256_encode(A, &P);
        r &= ~(z & t);
        return r;
}

/* see bearssl_ec.h */
const br_ec_impl br_ec_p256_i15 = {
        (uint32_t)0x00800000,
        &api_generator,
        &api_order,
        &api_mul,
        &api_muladd
};