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src/ec/ec_prime_i15.c  view on Meta::CPAN

/*
 * We use a custom interpreter that uses a dozen registers, and
 * only six operations:
 *    MSET(d, a)       copy a into d
 *    MADD(d, a)       d = d+a (modular)
 *    MSUB(d, a)       d = d-a (modular)
 *    MMUL(d, a, b)    d = a*b (Montgomery multiplication)
 *    MINV(d, a, b)    invert d modulo p; a and b are used as scratch registers
 *    MTZ(d)           clear return value if d = 0
 * Destination of MMUL (d) must be distinct from operands (a and b).
 * There is no such constraint for MSUB and MADD.
 *
 * Registers include the operand coordinates, and temporaries.
 */
#define MSET(d, a)      (0x0000 + ((d) << 8) + ((a) << 4))
#define MADD(d, a)      (0x1000 + ((d) << 8) + ((a) << 4))
#define MSUB(d, a)      (0x2000 + ((d) << 8) + ((a) << 4))
#define MMUL(d, a, b)   (0x3000 + ((d) << 8) + ((a) << 4) + (b))
#define MINV(d, a, b)   (0x4000 + ((d) << 8) + ((a) << 4) + (b))
#define MTZ(d)          (0x5000 + ((d) << 8))
#define ENDCODE         0

/*
 * Registers for the input operands.
 */
#define P1x    0
#define P1y    1
#define P1z    2
#define P2x    3
#define P2y    4
#define P2z    5

/*
 * Alternate names for the first input operand.
 */
#define Px     0
#define Py     1
#define Pz     2

/*
 * Temporaries.
 */
#define t1     6
#define t2     7
#define t3     8
#define t4     9
#define t5    10
#define t6    11
#define t7    12

/*
 * Extra scratch registers available when there is no second operand (e.g.
 * for "double" and "affine").
 */
#define t8     3
#define t9     4
#define t10    5

/*
 * 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
 *
 * If y = 0 (P has order 2) then this yields infinity (z' = 0), as it
 * should. This case should not happen anyway, because our curves have
 * prime order, and thus do not contain any point of order 2.
 *
 * If P is infinity (z = 0), then again the formulas yield infinity,
 * which is correct. Thus, this code works for all points.
 *
 * Cost: 8 multiplications
 */
static const uint16_t code_double[] = {
	/*
	 * Compute z^2 (in t1).
	 */
	MMUL(t1, Pz, Pz),

	/*
	 * Compute x-z^2 (in t2) and then x+z^2 (in t1).
	 */
	MSET(t2, Px),
	MSUB(t2, t1),
	MADD(t1, Px),

	/*
	 * Compute m = 3*(x+z^2)*(x-z^2) (in t1).
	 */
	MMUL(t3, t1, t2),
	MSET(t1, t3),
	MADD(t1, t3),
	MADD(t1, t3),

	/*
	 * Compute s = 4*x*y^2 (in t2) and 2*y^2 (in t3).
	 */
	MMUL(t3, Py, Py),
	MADD(t3, t3),
	MMUL(t2, Px, t3),
	MADD(t2, t2),

	/*
	 * Compute x' = m^2 - 2*s.
	 */
	MMUL(Px, t1, t1),
	MSUB(Px, t2),
	MSUB(Px, t2),

	/*
	 * Compute z' = 2*y*z.
	 */
	MMUL(t4, Py, Pz),
	MSET(Pz, t4),
	MADD(Pz, t4),

	/*
	 * Compute y' = m*(s - x') - 8*y^4. Note that we already have
	 * 2*y^2 in t3.
	 */
	MSUB(t2, Px),
	MMUL(Py, t1, t2),
	MMUL(t4, t3, t3),
	MSUB(Py, t4),
	MSUB(Py, t4),

	ENDCODE
};

/*
 * 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
 *
 * If both P1 and P2 are infinity, then z1 == 0 and z2 == 0, implying that
 * z3 == 0, so the result is correct.
 * If either of P1 or P2 is infinity, but not both, then z3 == 0, which is
 * not correct.
 * h == 0 only if u1 == u2; this happens in two cases:
 * -- if s1 == s2 then P1 and/or P2 is infinity, or P1 == P2
 * -- if s1 != s2 then P1 + P2 == infinity (but neither P1 or P2 is infinity)
 *
 * Thus, the following situations are not handled correctly:
 * -- P1 = 0 and P2 != 0
 * -- P1 != 0 and P2 = 0
 * -- P1 = P2
 * All other cases are properly computed. However, even in "incorrect"
 * situations, the three coordinates still are properly formed field
 * elements.
 *
 * The returned flag is cleared if r == 0. This happens in the following
 * cases:
 * -- Both points are on the same horizontal line (same Y coordinate).
 * -- Both points are infinity.
 * -- One point is infinity and the other is on line Y = 0.
 * The third case cannot happen with our curves (there is no valid point
 * on line Y = 0 since that would be a point of order 2). If the two
 * source points are non-infinity, then remains only the case where the
 * two points are on the same horizontal line.
 *
 * This allows us to detect the "P1 == P2" case, assuming that P1 != 0 and
 * P2 != 0:
 * -- If the returned value is not the point at infinity, then it was properly
 * computed.
 * -- Otherwise, if the returned flag is 1, then P1+P2 = 0, and the result
 * is indeed the point at infinity.
 * -- Otherwise (result is infinity, flag is 0), then P1 = P2 and we should
 * use the 'double' code.
 *
 * Cost: 16 multiplications
 */
static const uint16_t code_add[] = {
	/*
	 * Compute u1 = x1*z2^2 (in t1) and s1 = y1*z2^3 (in t3).
	 */
	MMUL(t3, P2z, P2z),
	MMUL(t1, P1x, t3),
	MMUL(t4, P2z, t3),
	MMUL(t3, P1y, t4),

	/*
	 * Compute u2 = x2*z1^2 (in t2) and s2 = y2*z1^3 (in t4).
	 */

src/ec/ec_prime_i15.c  view on Meta::CPAN

	MMUL(t2, P1y, P1y),

	/* Compare y^2 with x^3 - 3*x + b; they must match. */
	MSUB(t1, t2),
	MTZ(t1),

	/* Set z to 1 (in Montgomery representation). */
	MMUL(P1z, P2x, P2z),

	ENDCODE
};

/*
 * Conversion back to affine coordinates. This code snippet assumes that
 * the z coordinate of P2 is set to 1 (not in Montgomery representation).
 */
static const uint16_t code_affine[] = {

	/* Save z*R in t1. */
	MSET(t1, P1z),

	/* Compute z^3 in t2. */
	MMUL(t2, P1z, P1z),
	MMUL(t3, P1z, t2),
	MMUL(t2, t3, P2z),

	/* Invert to (1/z^3) in t2. */
	MINV(t2, t3, t4),

	/* Compute y. */
	MSET(t3, P1y),
	MMUL(P1y, t2, t3),

	/* Compute (1/z^2) in t3. */
	MMUL(t3, t2, t1),

	/* Compute x. */
	MSET(t2, P1x),
	MMUL(P1x, t2, t3),

	ENDCODE
};

static uint32_t
run_code(jacobian *P1, const jacobian *P2,
	const curve_params *cc, const uint16_t *code)
{
	uint32_t r;
	uint16_t t[13][I15_LEN];
	size_t u;

	r = 1;

	/*
	 * Copy the two operands in the dedicated registers.
	 */
	memcpy(t[P1x], P1->c, 3 * I15_LEN * sizeof(uint16_t));
	memcpy(t[P2x], P2->c, 3 * I15_LEN * sizeof(uint16_t));

	/*
	 * Run formulas.
	 */
	for (u = 0;; u ++) {
		unsigned op, d, a, b;

		op = code[u];
		if (op == 0) {
			break;
		}
		d = (op >> 8) & 0x0F;
		a = (op >> 4) & 0x0F;
		b = op & 0x0F;
		op >>= 12;
		switch (op) {
			uint32_t ctl;
			size_t plen;
			unsigned char tp[(BR_MAX_EC_SIZE + 7) >> 3];

		case 0:
			memcpy(t[d], t[a], I15_LEN * sizeof(uint16_t));
			break;
		case 1:
			ctl = br_i15_add(t[d], t[a], 1);
			ctl |= NOT(br_i15_sub(t[d], cc->p, 0));
			br_i15_sub(t[d], cc->p, ctl);
			break;
		case 2:
			br_i15_add(t[d], cc->p, br_i15_sub(t[d], t[a], 1));
			break;
		case 3:
			br_i15_montymul(t[d], t[a], t[b], cc->p, cc->p0i);
			break;
		case 4:
			plen = (cc->p[0] - (cc->p[0] >> 4) + 7) >> 3;
			br_i15_encode(tp, plen, cc->p);
			tp[plen - 1] -= 2;
			br_i15_modpow(t[d], tp, plen,
				cc->p, cc->p0i, t[a], t[b]);
			break;
		default:
			r &= ~br_i15_iszero(t[d]);
			break;
		}
	}

	/*
	 * Copy back result.
	 */
	memcpy(P1->c, t[P1x], 3 * I15_LEN * sizeof(uint16_t));
	return r;
}

static void
set_one(uint16_t *x, const uint16_t *p)
{
	size_t plen;

	plen = (p[0] + 31) >> 4;
	memset(x, 0, plen * sizeof *x);
	x[0] = p[0];
	x[1] = 0x0001;