go-sample-webpage/vendor/github.com/decred/dcrd/dcrec/secp256k1/v4/curve.go
2021-11-04 02:14:51 +01:00

943 lines
38 KiB
Go
Raw Permalink Blame History

This file contains ambiguous Unicode characters

This file contains Unicode characters that might be confused with other characters. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.

// Copyright (c) 2015-2021 The Decred developers
// Copyright 2013-2014 The btcsuite developers
// Use of this source code is governed by an ISC
// license that can be found in the LICENSE file.
package secp256k1
import (
"encoding/hex"
"math/big"
)
// References:
// [SECG]: Recommended Elliptic Curve Domain Parameters
// https://www.secg.org/sec2-v2.pdf
//
// [GECC]: Guide to Elliptic Curve Cryptography (Hankerson, Menezes, Vanstone)
//
// [BRID]: On Binary Representations of Integers with Digits -1, 0, 1
// (Prodinger, Helmut)
// All group operations are performed using Jacobian coordinates. For a given
// (x, y) position on the curve, the Jacobian coordinates are (x1, y1, z1)
// where x = x1/z1^2 and y = y1/z1^3.
// hexToFieldVal converts the passed hex string into a FieldVal and will panic
// if there is an error. This is only provided for the hard-coded constants so
// errors in the source code can be detected. It will only (and must only) be
// called with hard-coded values.
func hexToFieldVal(s string) *FieldVal {
b, err := hex.DecodeString(s)
if err != nil {
panic("invalid hex in source file: " + s)
}
var f FieldVal
if overflow := f.SetByteSlice(b); overflow {
panic("hex in source file overflows mod P: " + s)
}
return &f
}
var (
// Next 6 constants are from Hal Finney's bitcointalk.org post:
// https://bitcointalk.org/index.php?topic=3238.msg45565#msg45565
// May he rest in peace.
//
// They have also been independently derived from the code in the
// EndomorphismVectors function in genstatics.go.
endomorphismLambda = fromHex("5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72")
endomorphismBeta = hexToFieldVal("7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee")
endomorphismA1 = fromHex("3086d221a7d46bcde86c90e49284eb15")
endomorphismB1 = fromHex("-e4437ed6010e88286f547fa90abfe4c3")
endomorphismA2 = fromHex("114ca50f7a8e2f3f657c1108d9d44cfd8")
endomorphismB2 = fromHex("3086d221a7d46bcde86c90e49284eb15")
// Alternatively, the following parameters are valid as well, however, they
// seem to be about 8% slower in practice.
//
// endomorphismLambda = fromHex("AC9C52B33FA3CF1F5AD9E3FD77ED9BA4A880B9FC8EC739C2E0CFC810B51283CE")
// endomorphismBeta = hexToFieldVal("851695D49A83F8EF919BB86153CBCB16630FB68AED0A766A3EC693D68E6AFA40")
// endomorphismA1 = fromHex("E4437ED6010E88286F547FA90ABFE4C3")
// endomorphismB1 = fromHex("-3086D221A7D46BCDE86C90E49284EB15")
// endomorphismA2 = fromHex("3086D221A7D46BCDE86C90E49284EB15")
// endomorphismB2 = fromHex("114CA50F7A8E2F3F657C1108D9D44CFD8")
)
// JacobianPoint is an element of the group formed by the secp256k1 curve in
// Jacobian projective coordinates and thus represents a point on the curve.
type JacobianPoint struct {
// The X coordinate in Jacobian projective coordinates. The affine point is
// X/z^2.
X FieldVal
// The Y coordinate in Jacobian projective coordinates. The affine point is
// Y/z^3.
Y FieldVal
// The Z coordinate in Jacobian projective coordinates.
Z FieldVal
}
// MakeJacobianPoint returns a Jacobian point with the provided X, Y, and Z
// coordinates.
func MakeJacobianPoint(x, y, z *FieldVal) JacobianPoint {
var p JacobianPoint
p.X.Set(x)
p.Y.Set(y)
p.Z.Set(z)
return p
}
// Set sets the Jacobian point to the provided point.
func (p *JacobianPoint) Set(other *JacobianPoint) {
p.X.Set(&other.X)
p.Y.Set(&other.Y)
p.Z.Set(&other.Z)
}
// ToAffine reduces the Z value of the existing point to 1 effectively
// making it an affine coordinate in constant time. The point will be
// normalized.
func (p *JacobianPoint) ToAffine() {
// Inversions are expensive and both point addition and point doubling
// are faster when working with points that have a z value of one. So,
// if the point needs to be converted to affine, go ahead and normalize
// the point itself at the same time as the calculation is the same.
var zInv, tempZ FieldVal
zInv.Set(&p.Z).Inverse() // zInv = Z^-1
tempZ.SquareVal(&zInv) // tempZ = Z^-2
p.X.Mul(&tempZ) // X = X/Z^2 (mag: 1)
p.Y.Mul(tempZ.Mul(&zInv)) // Y = Y/Z^3 (mag: 1)
p.Z.SetInt(1) // Z = 1 (mag: 1)
// Normalize the x and y values.
p.X.Normalize()
p.Y.Normalize()
}
// addZ1AndZ2EqualsOne adds two Jacobian points that are already known to have
// z values of 1 and stores the result in the provided result param. That is to
// say result = p1 + p2. It performs faster addition than the generic add
// routine since less arithmetic is needed due to the ability to avoid the z
// value multiplications.
//
// NOTE: The points must be normalized for this function to return the correct
// result. The resulting point will be normalized.
func addZ1AndZ2EqualsOne(p1, p2, result *JacobianPoint) {
// To compute the point addition efficiently, this implementation splits
// the equation into intermediate elements which are used to minimize
// the number of field multiplications using the method shown at:
// https://hyperelliptic.org/EFD/g1p/auto-shortw-jacobian-0.html#addition-mmadd-2007-bl
//
// In particular it performs the calculations using the following:
// H = X2-X1, HH = H^2, I = 4*HH, J = H*I, r = 2*(Y2-Y1), V = X1*I
// X3 = r^2-J-2*V, Y3 = r*(V-X3)-2*Y1*J, Z3 = 2*H
//
// This results in a cost of 4 field multiplications, 2 field squarings,
// 6 field additions, and 5 integer multiplications.
x1, y1 := &p1.X, &p1.Y
x2, y2 := &p2.X, &p2.Y
x3, y3, z3 := &result.X, &result.Y, &result.Z
// When the x coordinates are the same for two points on the curve, the
// y coordinates either must be the same, in which case it is point
// doubling, or they are opposite and the result is the point at
// infinity per the group law for elliptic curve cryptography.
if x1.Equals(x2) {
if y1.Equals(y2) {
// Since x1 == x2 and y1 == y2, point doubling must be
// done, otherwise the addition would end up dividing
// by zero.
DoubleNonConst(p1, result)
return
}
// Since x1 == x2 and y1 == -y2, the sum is the point at
// infinity per the group law.
x3.SetInt(0)
y3.SetInt(0)
z3.SetInt(0)
return
}
// Calculate X3, Y3, and Z3 according to the intermediate elements
// breakdown above.
var h, i, j, r, v FieldVal
var negJ, neg2V, negX3 FieldVal
h.Set(x1).Negate(1).Add(x2) // H = X2-X1 (mag: 3)
i.SquareVal(&h).MulInt(4) // I = 4*H^2 (mag: 4)
j.Mul2(&h, &i) // J = H*I (mag: 1)
r.Set(y1).Negate(1).Add(y2).MulInt(2) // r = 2*(Y2-Y1) (mag: 6)
v.Mul2(x1, &i) // V = X1*I (mag: 1)
negJ.Set(&j).Negate(1) // negJ = -J (mag: 2)
neg2V.Set(&v).MulInt(2).Negate(2) // neg2V = -(2*V) (mag: 3)
x3.Set(&r).Square().Add(&negJ).Add(&neg2V) // X3 = r^2-J-2*V (mag: 6)
negX3.Set(x3).Negate(6) // negX3 = -X3 (mag: 7)
j.Mul(y1).MulInt(2).Negate(2) // J = -(2*Y1*J) (mag: 3)
y3.Set(&v).Add(&negX3).Mul(&r).Add(&j) // Y3 = r*(V-X3)-2*Y1*J (mag: 4)
z3.Set(&h).MulInt(2) // Z3 = 2*H (mag: 6)
// Normalize the resulting field values to a magnitude of 1 as needed.
x3.Normalize()
y3.Normalize()
z3.Normalize()
}
// addZ1EqualsZ2 adds two Jacobian points that are already known to have the
// same z value and stores the result in the provided result param. That is to
// say result = p1 + p2. It performs faster addition than the generic add
// routine since less arithmetic is needed due to the known equivalence.
//
// NOTE: The points must be normalized for this function to return the correct
// result. The resulting point will be normalized.
func addZ1EqualsZ2(p1, p2, result *JacobianPoint) {
// To compute the point addition efficiently, this implementation splits
// the equation into intermediate elements which are used to minimize
// the number of field multiplications using a slightly modified version
// of the method shown at:
// https://hyperelliptic.org/EFD/g1p/auto-shortw-jacobian-0.html#addition-mmadd-2007-bl
//
// In particular it performs the calculations using the following:
// A = X2-X1, B = A^2, C=Y2-Y1, D = C^2, E = X1*B, F = X2*B
// X3 = D-E-F, Y3 = C*(E-X3)-Y1*(F-E), Z3 = Z1*A
//
// This results in a cost of 5 field multiplications, 2 field squarings,
// 9 field additions, and 0 integer multiplications.
x1, y1, z1 := &p1.X, &p1.Y, &p1.Z
x2, y2 := &p2.X, &p2.Y
x3, y3, z3 := &result.X, &result.Y, &result.Z
// When the x coordinates are the same for two points on the curve, the
// y coordinates either must be the same, in which case it is point
// doubling, or they are opposite and the result is the point at
// infinity per the group law for elliptic curve cryptography.
if x1.Equals(x2) {
if y1.Equals(y2) {
// Since x1 == x2 and y1 == y2, point doubling must be
// done, otherwise the addition would end up dividing
// by zero.
DoubleNonConst(p1, result)
return
}
// Since x1 == x2 and y1 == -y2, the sum is the point at
// infinity per the group law.
x3.SetInt(0)
y3.SetInt(0)
z3.SetInt(0)
return
}
// Calculate X3, Y3, and Z3 according to the intermediate elements
// breakdown above.
var a, b, c, d, e, f FieldVal
var negX1, negY1, negE, negX3 FieldVal
negX1.Set(x1).Negate(1) // negX1 = -X1 (mag: 2)
negY1.Set(y1).Negate(1) // negY1 = -Y1 (mag: 2)
a.Set(&negX1).Add(x2) // A = X2-X1 (mag: 3)
b.SquareVal(&a) // B = A^2 (mag: 1)
c.Set(&negY1).Add(y2) // C = Y2-Y1 (mag: 3)
d.SquareVal(&c) // D = C^2 (mag: 1)
e.Mul2(x1, &b) // E = X1*B (mag: 1)
negE.Set(&e).Negate(1) // negE = -E (mag: 2)
f.Mul2(x2, &b) // F = X2*B (mag: 1)
x3.Add2(&e, &f).Negate(3).Add(&d) // X3 = D-E-F (mag: 5)
negX3.Set(x3).Negate(5).Normalize() // negX3 = -X3 (mag: 1)
y3.Set(y1).Mul(f.Add(&negE)).Negate(3) // Y3 = -(Y1*(F-E)) (mag: 4)
y3.Add(e.Add(&negX3).Mul(&c)) // Y3 = C*(E-X3)+Y3 (mag: 5)
z3.Mul2(z1, &a) // Z3 = Z1*A (mag: 1)
// Normalize the resulting field values to a magnitude of 1 as needed.
x3.Normalize()
y3.Normalize()
z3.Normalize()
}
// addZ2EqualsOne adds two Jacobian points when the second point is already
// known to have a z value of 1 (and the z value for the first point is not 1)
// and stores the result in the provided result param. That is to say result =
// p1 + p2. It performs faster addition than the generic add routine since
// less arithmetic is needed due to the ability to avoid multiplications by the
// second point's z value.
//
// NOTE: The points must be normalized for this function to return the correct
// result. The resulting point will be normalized.
func addZ2EqualsOne(p1, p2, result *JacobianPoint) {
// To compute the point addition efficiently, this implementation splits
// the equation into intermediate elements which are used to minimize
// the number of field multiplications using the method shown at:
// https://hyperelliptic.org/EFD/g1p/auto-shortw-jacobian-0.html#addition-madd-2007-bl
//
// In particular it performs the calculations using the following:
// Z1Z1 = Z1^2, U2 = X2*Z1Z1, S2 = Y2*Z1*Z1Z1, H = U2-X1, HH = H^2,
// I = 4*HH, J = H*I, r = 2*(S2-Y1), V = X1*I
// X3 = r^2-J-2*V, Y3 = r*(V-X3)-2*Y1*J, Z3 = (Z1+H)^2-Z1Z1-HH
//
// This results in a cost of 7 field multiplications, 4 field squarings,
// 9 field additions, and 4 integer multiplications.
x1, y1, z1 := &p1.X, &p1.Y, &p1.Z
x2, y2 := &p2.X, &p2.Y
x3, y3, z3 := &result.X, &result.Y, &result.Z
// When the x coordinates are the same for two points on the curve, the
// y coordinates either must be the same, in which case it is point
// doubling, or they are opposite and the result is the point at
// infinity per the group law for elliptic curve cryptography. Since
// any number of Jacobian coordinates can represent the same affine
// point, the x and y values need to be converted to like terms. Due to
// the assumption made for this function that the second point has a z
// value of 1 (z2=1), the first point is already "converted".
var z1z1, u2, s2 FieldVal
z1z1.SquareVal(z1) // Z1Z1 = Z1^2 (mag: 1)
u2.Set(x2).Mul(&z1z1).Normalize() // U2 = X2*Z1Z1 (mag: 1)
s2.Set(y2).Mul(&z1z1).Mul(z1).Normalize() // S2 = Y2*Z1*Z1Z1 (mag: 1)
if x1.Equals(&u2) {
if y1.Equals(&s2) {
// Since x1 == x2 and y1 == y2, point doubling must be
// done, otherwise the addition would end up dividing
// by zero.
DoubleNonConst(p1, result)
return
}
// Since x1 == x2 and y1 == -y2, the sum is the point at
// infinity per the group law.
x3.SetInt(0)
y3.SetInt(0)
z3.SetInt(0)
return
}
// Calculate X3, Y3, and Z3 according to the intermediate elements
// breakdown above.
var h, hh, i, j, r, rr, v FieldVal
var negX1, negY1, negX3 FieldVal
negX1.Set(x1).Negate(1) // negX1 = -X1 (mag: 2)
h.Add2(&u2, &negX1) // H = U2-X1 (mag: 3)
hh.SquareVal(&h) // HH = H^2 (mag: 1)
i.Set(&hh).MulInt(4) // I = 4 * HH (mag: 4)
j.Mul2(&h, &i) // J = H*I (mag: 1)
negY1.Set(y1).Negate(1) // negY1 = -Y1 (mag: 2)
r.Set(&s2).Add(&negY1).MulInt(2) // r = 2*(S2-Y1) (mag: 6)
rr.SquareVal(&r) // rr = r^2 (mag: 1)
v.Mul2(x1, &i) // V = X1*I (mag: 1)
x3.Set(&v).MulInt(2).Add(&j).Negate(3) // X3 = -(J+2*V) (mag: 4)
x3.Add(&rr) // X3 = r^2+X3 (mag: 5)
negX3.Set(x3).Negate(5) // negX3 = -X3 (mag: 6)
y3.Set(y1).Mul(&j).MulInt(2).Negate(2) // Y3 = -(2*Y1*J) (mag: 3)
y3.Add(v.Add(&negX3).Mul(&r)) // Y3 = r*(V-X3)+Y3 (mag: 4)
z3.Add2(z1, &h).Square() // Z3 = (Z1+H)^2 (mag: 1)
z3.Add(z1z1.Add(&hh).Negate(2)) // Z3 = Z3-(Z1Z1+HH) (mag: 4)
// Normalize the resulting field values to a magnitude of 1 as needed.
x3.Normalize()
y3.Normalize()
z3.Normalize()
}
// addGeneric adds two Jacobian points without any assumptions about the z
// values of the two points and stores the result in the provided result param.
// That is to say result = p1 + p2. It is the slowest of the add routines due
// to requiring the most arithmetic.
//
// NOTE: The points must be normalized for this function to return the correct
// result. The resulting point will be normalized.
func addGeneric(p1, p2, result *JacobianPoint) {
// To compute the point addition efficiently, this implementation splits
// the equation into intermediate elements which are used to minimize
// the number of field multiplications using the method shown at:
// https://hyperelliptic.org/EFD/g1p/auto-shortw-jacobian-0.html#addition-add-2007-bl
//
// In particular it performs the calculations using the following:
// Z1Z1 = Z1^2, Z2Z2 = Z2^2, U1 = X1*Z2Z2, U2 = X2*Z1Z1, S1 = Y1*Z2*Z2Z2
// S2 = Y2*Z1*Z1Z1, H = U2-U1, I = (2*H)^2, J = H*I, r = 2*(S2-S1)
// V = U1*I
// X3 = r^2-J-2*V, Y3 = r*(V-X3)-2*S1*J, Z3 = ((Z1+Z2)^2-Z1Z1-Z2Z2)*H
//
// This results in a cost of 11 field multiplications, 5 field squarings,
// 9 field additions, and 4 integer multiplications.
x1, y1, z1 := &p1.X, &p1.Y, &p1.Z
x2, y2, z2 := &p2.X, &p2.Y, &p2.Z
x3, y3, z3 := &result.X, &result.Y, &result.Z
// When the x coordinates are the same for two points on the curve, the
// y coordinates either must be the same, in which case it is point
// doubling, or they are opposite and the result is the point at
// infinity. Since any number of Jacobian coordinates can represent the
// same affine point, the x and y values need to be converted to like
// terms.
var z1z1, z2z2, u1, u2, s1, s2 FieldVal
z1z1.SquareVal(z1) // Z1Z1 = Z1^2 (mag: 1)
z2z2.SquareVal(z2) // Z2Z2 = Z2^2 (mag: 1)
u1.Set(x1).Mul(&z2z2).Normalize() // U1 = X1*Z2Z2 (mag: 1)
u2.Set(x2).Mul(&z1z1).Normalize() // U2 = X2*Z1Z1 (mag: 1)
s1.Set(y1).Mul(&z2z2).Mul(z2).Normalize() // S1 = Y1*Z2*Z2Z2 (mag: 1)
s2.Set(y2).Mul(&z1z1).Mul(z1).Normalize() // S2 = Y2*Z1*Z1Z1 (mag: 1)
if u1.Equals(&u2) {
if s1.Equals(&s2) {
// Since x1 == x2 and y1 == y2, point doubling must be
// done, otherwise the addition would end up dividing
// by zero.
DoubleNonConst(p1, result)
return
}
// Since x1 == x2 and y1 == -y2, the sum is the point at
// infinity per the group law.
x3.SetInt(0)
y3.SetInt(0)
z3.SetInt(0)
return
}
// Calculate X3, Y3, and Z3 according to the intermediate elements
// breakdown above.
var h, i, j, r, rr, v FieldVal
var negU1, negS1, negX3 FieldVal
negU1.Set(&u1).Negate(1) // negU1 = -U1 (mag: 2)
h.Add2(&u2, &negU1) // H = U2-U1 (mag: 3)
i.Set(&h).MulInt(2).Square() // I = (2*H)^2 (mag: 2)
j.Mul2(&h, &i) // J = H*I (mag: 1)
negS1.Set(&s1).Negate(1) // negS1 = -S1 (mag: 2)
r.Set(&s2).Add(&negS1).MulInt(2) // r = 2*(S2-S1) (mag: 6)
rr.SquareVal(&r) // rr = r^2 (mag: 1)
v.Mul2(&u1, &i) // V = U1*I (mag: 1)
x3.Set(&v).MulInt(2).Add(&j).Negate(3) // X3 = -(J+2*V) (mag: 4)
x3.Add(&rr) // X3 = r^2+X3 (mag: 5)
negX3.Set(x3).Negate(5) // negX3 = -X3 (mag: 6)
y3.Mul2(&s1, &j).MulInt(2).Negate(2) // Y3 = -(2*S1*J) (mag: 3)
y3.Add(v.Add(&negX3).Mul(&r)) // Y3 = r*(V-X3)+Y3 (mag: 4)
z3.Add2(z1, z2).Square() // Z3 = (Z1+Z2)^2 (mag: 1)
z3.Add(z1z1.Add(&z2z2).Negate(2)) // Z3 = Z3-(Z1Z1+Z2Z2) (mag: 4)
z3.Mul(&h) // Z3 = Z3*H (mag: 1)
// Normalize the resulting field values to a magnitude of 1 as needed.
x3.Normalize()
y3.Normalize()
z3.Normalize()
}
// AddNonConst adds the passed Jacobian points together and stores the result in
// the provided result param in *non-constant* time.
//
// NOTE: The points must be normalized for this function to return the correct
// result. The resulting point will be normalized.
func AddNonConst(p1, p2, result *JacobianPoint) {
// A point at infinity is the identity according to the group law for
// elliptic curve cryptography. Thus, ∞ + P = P and P + ∞ = P.
if (p1.X.IsZero() && p1.Y.IsZero()) || p1.Z.IsZero() {
result.Set(p2)
return
}
if (p2.X.IsZero() && p2.Y.IsZero()) || p2.Z.IsZero() {
result.Set(p1)
return
}
// Faster point addition can be achieved when certain assumptions are
// met. For example, when both points have the same z value, arithmetic
// on the z values can be avoided. This section thus checks for these
// conditions and calls an appropriate add function which is accelerated
// by using those assumptions.
isZ1One := p1.Z.IsOne()
isZ2One := p2.Z.IsOne()
switch {
case isZ1One && isZ2One:
addZ1AndZ2EqualsOne(p1, p2, result)
return
case p1.Z.Equals(&p2.Z):
addZ1EqualsZ2(p1, p2, result)
return
case isZ2One:
addZ2EqualsOne(p1, p2, result)
return
}
// None of the above assumptions are true, so fall back to generic
// point addition.
addGeneric(p1, p2, result)
}
// doubleZ1EqualsOne performs point doubling on the passed Jacobian point when
// the point is already known to have a z value of 1 and stores the result in
// the provided result param. That is to say result = 2*p. It performs faster
// point doubling than the generic routine since less arithmetic is needed due
// to the ability to avoid multiplication by the z value.
//
// NOTE: The resulting point will be normalized.
func doubleZ1EqualsOne(p, result *JacobianPoint) {
// This function uses the assumptions that z1 is 1, thus the point
// doubling formulas reduce to:
//
// X3 = (3*X1^2)^2 - 8*X1*Y1^2
// Y3 = (3*X1^2)*(4*X1*Y1^2 - X3) - 8*Y1^4
// Z3 = 2*Y1
//
// To compute the above efficiently, this implementation splits the
// equation into intermediate elements which are used to minimize the
// number of field multiplications in favor of field squarings which
// are roughly 35% faster than field multiplications with the current
// implementation at the time this was written.
//
// This uses a slightly modified version of the method shown at:
// https://hyperelliptic.org/EFD/g1p/auto-shortw-jacobian-0.html#doubling-mdbl-2007-bl
//
// In particular it performs the calculations using the following:
// A = X1^2, B = Y1^2, C = B^2, D = 2*((X1+B)^2-A-C)
// E = 3*A, F = E^2, X3 = F-2*D, Y3 = E*(D-X3)-8*C
// Z3 = 2*Y1
//
// This results in a cost of 1 field multiplication, 5 field squarings,
// 6 field additions, and 5 integer multiplications.
x1, y1 := &p.X, &p.Y
x3, y3, z3 := &result.X, &result.Y, &result.Z
var a, b, c, d, e, f FieldVal
z3.Set(y1).MulInt(2) // Z3 = 2*Y1 (mag: 2)
a.SquareVal(x1) // A = X1^2 (mag: 1)
b.SquareVal(y1) // B = Y1^2 (mag: 1)
c.SquareVal(&b) // C = B^2 (mag: 1)
b.Add(x1).Square() // B = (X1+B)^2 (mag: 1)
d.Set(&a).Add(&c).Negate(2) // D = -(A+C) (mag: 3)
d.Add(&b).MulInt(2) // D = 2*(B+D)(mag: 8)
e.Set(&a).MulInt(3) // E = 3*A (mag: 3)
f.SquareVal(&e) // F = E^2 (mag: 1)
x3.Set(&d).MulInt(2).Negate(16) // X3 = -(2*D) (mag: 17)
x3.Add(&f) // X3 = F+X3 (mag: 18)
f.Set(x3).Negate(18).Add(&d).Normalize() // F = D-X3 (mag: 1)
y3.Set(&c).MulInt(8).Negate(8) // Y3 = -(8*C) (mag: 9)
y3.Add(f.Mul(&e)) // Y3 = E*F+Y3 (mag: 10)
// Normalize the field values back to a magnitude of 1.
x3.Normalize()
y3.Normalize()
z3.Normalize()
}
// doubleGeneric performs point doubling on the passed Jacobian point without
// any assumptions about the z value and stores the result in the provided
// result param. That is to say result = 2*p. It is the slowest of the point
// doubling routines due to requiring the most arithmetic.
//
// NOTE: The resulting point will be normalized.
func doubleGeneric(p, result *JacobianPoint) {
// Point doubling formula for Jacobian coordinates for the secp256k1
// curve:
//
// X3 = (3*X1^2)^2 - 8*X1*Y1^2
// Y3 = (3*X1^2)*(4*X1*Y1^2 - X3) - 8*Y1^4
// Z3 = 2*Y1*Z1
//
// To compute the above efficiently, this implementation splits the
// equation into intermediate elements which are used to minimize the
// number of field multiplications in favor of field squarings which
// are roughly 35% faster than field multiplications with the current
// implementation at the time this was written.
//
// This uses a slightly modified version of the method shown at:
// https://hyperelliptic.org/EFD/g1p/auto-shortw-jacobian-0.html#doubling-dbl-2009-l
//
// In particular it performs the calculations using the following:
// A = X1^2, B = Y1^2, C = B^2, D = 2*((X1+B)^2-A-C)
// E = 3*A, F = E^2, X3 = F-2*D, Y3 = E*(D-X3)-8*C
// Z3 = 2*Y1*Z1
//
// This results in a cost of 1 field multiplication, 5 field squarings,
// 6 field additions, and 5 integer multiplications.
x1, y1, z1 := &p.X, &p.Y, &p.Z
x3, y3, z3 := &result.X, &result.Y, &result.Z
var a, b, c, d, e, f FieldVal
z3.Mul2(y1, z1).MulInt(2) // Z3 = 2*Y1*Z1 (mag: 2)
a.SquareVal(x1) // A = X1^2 (mag: 1)
b.SquareVal(y1) // B = Y1^2 (mag: 1)
c.SquareVal(&b) // C = B^2 (mag: 1)
b.Add(x1).Square() // B = (X1+B)^2 (mag: 1)
d.Set(&a).Add(&c).Negate(2) // D = -(A+C) (mag: 3)
d.Add(&b).MulInt(2) // D = 2*(B+D)(mag: 8)
e.Set(&a).MulInt(3) // E = 3*A (mag: 3)
f.SquareVal(&e) // F = E^2 (mag: 1)
x3.Set(&d).MulInt(2).Negate(16) // X3 = -(2*D) (mag: 17)
x3.Add(&f) // X3 = F+X3 (mag: 18)
f.Set(x3).Negate(18).Add(&d).Normalize() // F = D-X3 (mag: 1)
y3.Set(&c).MulInt(8).Negate(8) // Y3 = -(8*C) (mag: 9)
y3.Add(f.Mul(&e)) // Y3 = E*F+Y3 (mag: 10)
// Normalize the field values back to a magnitude of 1.
x3.Normalize()
y3.Normalize()
z3.Normalize()
}
// DoubleNonConst doubles the passed Jacobian point and stores the result in the
// provided result parameter in *non-constant* time.
//
// NOTE: The point must be normalized for this function to return the correct
// result. The resulting point will be normalized.
func DoubleNonConst(p, result *JacobianPoint) {
// Doubling a point at infinity is still infinity.
if p.Y.IsZero() || p.Z.IsZero() {
result.X.SetInt(0)
result.Y.SetInt(0)
result.Z.SetInt(0)
return
}
// Slightly faster point doubling can be achieved when the z value is 1
// by avoiding the multiplication on the z value. This section calls
// a point doubling function which is accelerated by using that
// assumption when possible.
if p.Z.IsOne() {
doubleZ1EqualsOne(p, result)
return
}
// Fall back to generic point doubling which works with arbitrary z
// values.
doubleGeneric(p, result)
}
// splitK returns a balanced length-two representation of k and their signs.
// This is algorithm 3.74 from [GECC].
//
// One thing of note about this algorithm is that no matter what c1 and c2 are,
// the final equation of k = k1 + k2 * lambda (mod n) will hold. This is
// provable mathematically due to how a1/b1/a2/b2 are computed.
//
// c1 and c2 are chosen to minimize the max(k1,k2).
func splitK(k []byte) ([]byte, []byte, int, int) {
// All math here is done with big.Int, which is slow.
// At some point, it might be useful to write something similar to
// FieldVal but for N instead of P as the prime field if this ends up
// being a bottleneck.
bigIntK := new(big.Int)
c1, c2 := new(big.Int), new(big.Int)
tmp1, tmp2 := new(big.Int), new(big.Int)
k1, k2 := new(big.Int), new(big.Int)
bigIntK.SetBytes(k)
// c1 = round(b2 * k / n) from step 4.
// Rounding isn't really necessary and costs too much, hence skipped
c1.Mul(endomorphismB2, bigIntK)
c1.Div(c1, curveParams.N)
// c2 = round(b1 * k / n) from step 4 (sign reversed to optimize one step)
// Rounding isn't really necessary and costs too much, hence skipped
c2.Mul(endomorphismB1, bigIntK)
c2.Div(c2, curveParams.N)
// k1 = k - c1 * a1 - c2 * a2 from step 5 (note c2's sign is reversed)
tmp1.Mul(c1, endomorphismA1)
tmp2.Mul(c2, endomorphismA2)
k1.Sub(bigIntK, tmp1)
k1.Add(k1, tmp2)
// k2 = - c1 * b1 - c2 * b2 from step 5 (note c2's sign is reversed)
tmp1.Mul(c1, endomorphismB1)
tmp2.Mul(c2, endomorphismB2)
k2.Sub(tmp2, tmp1)
// Note Bytes() throws out the sign of k1 and k2. This matters
// since k1 and/or k2 can be negative. Hence, we pass that
// back separately.
return k1.Bytes(), k2.Bytes(), k1.Sign(), k2.Sign()
}
// nafScalar represents a positive integer up to a maximum value of 2^256 - 1
// encoded in non-adjacent form.
//
// NAF is a signed-digit representation where each digit can be +1, 0, or -1.
//
// In order to efficiently encode that information, this type uses two arrays, a
// "positive" array where set bits represent the +1 signed digits and a
// "negative" array where set bits represent the -1 signed digits. 0 is
// represented by neither array having a bit set in that position.
//
// The Pos and Neg methods return the aforementioned positive and negative
// arrays, respectively.
type nafScalar struct {
// pos houses the positive portion of the representation. An additional
// byte is required for the positive portion because the NAF encoding can be
// up to 1 bit longer than the normal binary encoding of the value.
//
// neg houses the negative portion of the representation. Even though the
// additional byte is not required for the negative portion, since it can
// never exceed the length of the normal binary encoding of the value,
// keeping the same length for positive and negative portions simplifies
// working with the representation and allows extra conditional branches to
// be avoided.
//
// start and end specify the starting and ending index to use within the pos
// and neg arrays, respectively. This allows fixed size arrays to be used
// versus needing to dynamically allocate space on the heap.
//
// NOTE: The fields are defined in the order that they are to minimize the
// padding on 32-bit and 64-bit platforms.
pos [33]byte
start, end uint8
neg [33]byte
}
// Pos returns the bytes of the encoded value with bits set in the positions
// that represent a signed digit of +1.
func (s *nafScalar) Pos() []byte {
return s.pos[s.start:s.end]
}
// Neg returns the bytes of the encoded value with bits set in the positions
// that represent a signed digit of -1.
func (s *nafScalar) Neg() []byte {
return s.neg[s.start:s.end]
}
// naf takes a positive integer up to a maximum value of 2^256 - 1 and returns
// its non-adjacent form (NAF), which is a unique signed-digit representation
// such that no two consecutive digits are nonzero. See the documentation for
// the returned type for details on how the representation is encoded
// efficiently and how to interpret it
//
// NAF is useful in that it has the fewest nonzero digits of any signed digit
// representation, only 1/3rd of its digits are nonzero on average, and at least
// half of the digits will be 0.
//
// The aforementioned properties are particularly beneficial for optimizing
// elliptic curve point multiplication because they effectively minimize the
// number of required point additions in exchange for needing to perform a mix
// of fewer point additions and subtractions and possibly one additional point
// doubling. This is an excellent tradeoff because subtraction of points has
// the same computational complexity as addition of points and point doubling is
// faster than both.
func naf(k []byte) nafScalar {
// Strip leading zero bytes.
for len(k) > 0 && k[0] == 0x00 {
k = k[1:]
}
// The non-adjacent form (NAF) of a positive integer k is an expression
// k = ∑_(i=0, l-1) k_i * 2^i where k_i ∈ {0,±1}, k_(l-1) != 0, and no two
// consecutive digits k_i are nonzero.
//
// The traditional method of computing the NAF of a positive integer is
// given by algorithm 3.30 in [GECC]. It consists of repeatedly dividing k
// by 2 and choosing the remainder so that the quotient (kr)/2 is even
// which ensures the next NAF digit is 0. This requires log_2(k) steps.
//
// However, in [BRID], Prodinger notes that a closed form expression for the
// NAF representation is the bitwise difference 3k/2 - k/2. This is more
// efficient as it can be computed in O(1) versus the O(log(n)) of the
// traditional approach.
//
// The following code makes use of that formula to compute the NAF more
// efficiently.
//
// To understand the logic here, observe that the only way the NAF has a
// nonzero digit at a given bit is when either 3k/2 or k/2 has a bit set in
// that position, but not both. In other words, the result of a bitwise
// xor. This can be seen simply by considering that when the bits are the
// same, the subtraction is either 0-0 or 1-1, both of which are 0.
//
// Further, observe that the "+1" digits in the result are contributed by
// 3k/2 while the "-1" digits are from k/2. So, they can be determined by
// taking the bitwise and of each respective value with the result of the
// xor which identifies which bits are nonzero.
//
// Using that information, this loops backwards from the least significant
// byte to the most significant byte while performing the aforementioned
// calculations by propagating the potential carry and high order bit from
// the next word during the right shift.
kLen := len(k)
var result nafScalar
var carry uint8
for byteNum := kLen - 1; byteNum >= 0; byteNum-- {
// Calculate k/2. Notice the carry from the previous word is added and
// the low order bit from the next word is shifted in accordingly.
kc := uint16(k[byteNum]) + uint16(carry)
var nextWord uint8
if byteNum > 0 {
nextWord = k[byteNum-1]
}
halfK := kc>>1 | uint16(nextWord<<7)
// Calculate 3k/2 and determine the non-zero digits in the result.
threeHalfK := kc + halfK
nonZeroResultDigits := threeHalfK ^ halfK
// Determine the signed digits {0, ±1}.
result.pos[byteNum+1] = uint8(threeHalfK & nonZeroResultDigits)
result.neg[byteNum+1] = uint8(halfK & nonZeroResultDigits)
// Propagate the potential carry from the 3k/2 calculation.
carry = uint8(threeHalfK >> 8)
}
result.pos[0] = carry
// Set the starting and ending positions within the fixed size arrays to
// identify the bytes that are actually used. This is important since the
// encoding is big endian and thus trailing zero bytes changes its value.
result.start = 1 - carry
result.end = uint8(kLen + 1)
return result
}
// ScalarMultNonConst multiplies k*P where k is a big endian integer modulo the
// curve order and P is a point in Jacobian projective coordinates and stores
// the result in the provided Jacobian point.
//
// NOTE: The point must be normalized for this function to return the correct
// result. The resulting point will be normalized.
func ScalarMultNonConst(k *ModNScalar, point, result *JacobianPoint) {
// Decompose K into k1 and k2 in order to halve the number of EC ops.
// See Algorithm 3.74 in [GECC].
kBytes := k.Bytes()
k1, k2, signK1, signK2 := splitK(kBytes[:])
zeroArray32(&kBytes)
// The main equation here to remember is:
// k * P = k1 * P + k2 * ϕ(P)
//
// P1 below is P in the equation, P2 below is ϕ(P) in the equation
p1, p1Neg := new(JacobianPoint), new(JacobianPoint)
p1.Set(point)
p1Neg.Set(p1)
p1Neg.Y.Negate(1).Normalize()
// NOTE: ϕ(x,y) = (βx,y). The Jacobian z coordinates are the same, so this
// math goes through.
p2, p2Neg := new(JacobianPoint), new(JacobianPoint)
p2.Set(p1)
p2.X.Mul(endomorphismBeta).Normalize()
p2Neg.Set(p2)
p2Neg.Y.Negate(1).Normalize()
// Flip the positive and negative values of the points as needed
// depending on the signs of k1 and k2. As mentioned in the equation
// above, each of k1 and k2 are multiplied by the respective point.
// Since -k * P is the same thing as k * -P, and the group law for
// elliptic curves states that P(x, y) = -P(x, -y), it's faster and
// simplifies the code to just make the point negative.
if signK1 == -1 {
p1, p1Neg = p1Neg, p1
}
if signK2 == -1 {
p2, p2Neg = p2Neg, p2
}
// NAF versions of k1 and k2 should have a lot more zeros.
//
// The Pos version of the bytes contain the +1s and the Neg versions
// contain the -1s.
k1NAF, k2NAF := naf(k1), naf(k2)
k1PosNAF, k1NegNAF := k1NAF.Pos(), k1NAF.Neg()
k2PosNAF, k2NegNAF := k2NAF.Pos(), k2NAF.Neg()
k1Len, k2Len := len(k1PosNAF), len(k2PosNAF)
m := k1Len
if m < k2Len {
m = k2Len
}
// Point Q = ∞ (point at infinity).
var q JacobianPoint
// Add left-to-right using the NAF optimization. See algorithm 3.77
// from [GECC]. This should be faster overall since there will be a lot
// more instances of 0, hence reducing the number of Jacobian additions
// at the cost of 1 possible extra doubling.
for i := 0; i < m; i++ {
// Since k1 and k2 are potentially different lengths and the calculation
// is being done left to right, pad the front of the shorter one with
// 0s.
var k1BytePos, k1ByteNeg, k2BytePos, k2ByteNeg byte
if i >= m-k1Len {
k1BytePos, k1ByteNeg = k1PosNAF[i-(m-k1Len)], k1NegNAF[i-(m-k1Len)]
}
if i >= m-k2Len {
k2BytePos, k2ByteNeg = k2PosNAF[i-(m-k2Len)], k2NegNAF[i-(m-k2Len)]
}
for bit, mask := 7, uint8(1<<7); bit >= 0; bit, mask = bit-1, mask>>1 {
// Q = 2 * Q
DoubleNonConst(&q, &q)
// Add or subtract the first point based on the signed digit of the
// NAF representation of k1 at this bit position.
//
// +1: Q = Q + p1
// -1: Q = Q - p1
// 0: Q = Q (no change)
if k1BytePos&mask == mask {
AddNonConst(&q, p1, &q)
} else if k1ByteNeg&mask == mask {
AddNonConst(&q, p1Neg, &q)
}
// Add or subtract the second point based on the signed digit of the
// NAF representation of k2 at this bit position.
//
// +1: Q = Q + p2
// -1: Q = Q - p2
// 0: Q = Q (no change)
if k2BytePos&mask == mask {
AddNonConst(&q, p2, &q)
} else if k2ByteNeg&mask == mask {
AddNonConst(&q, p2Neg, &q)
}
}
}
result.Set(&q)
}
// ScalarBaseMultNonConst multiplies k*G where G is the base point of the group
// and k is a big endian integer. The result is stored in Jacobian coordinates
// (x1, y1, z1).
//
// NOTE: The resulting point will be normalized.
func ScalarBaseMultNonConst(k *ModNScalar, result *JacobianPoint) {
bytePoints := s256BytePoints()
// Point Q = ∞ (point at infinity).
var q JacobianPoint
// curve.bytePoints has all 256 byte points for each 8-bit window. The
// strategy is to add up the byte points. This is best understood by
// expressing k in base-256 which it already sort of is. Each "digit" in
// the 8-bit window can be looked up using bytePoints and added together.
var pt JacobianPoint
for i, byteVal := range k.Bytes() {
p := bytePoints[i][byteVal]
pt.X.Set(&p[0])
pt.Y.Set(&p[1])
pt.Z.SetInt(1)
AddNonConst(&q, &pt, &q)
}
result.Set(&q)
}
// isOnCurve returns whether or not the affine point (x,y) is on the curve.
func isOnCurve(fx, fy *FieldVal) bool {
// Elliptic curve equation for secp256k1 is: y^2 = x^3 + 7
y2 := new(FieldVal).SquareVal(fy).Normalize()
result := new(FieldVal).SquareVal(fx).Mul(fx).AddInt(7).Normalize()
return y2.Equals(result)
}
// DecompressY attempts to calculate the Y coordinate for the given X coordinate
// such that the result pair is a point on the secp256k1 curve. It adjusts Y
// based on the desired oddness and returns whether or not it was successful
// since not all X coordinates are valid.
//
// The magnitude of the provided X coordinate field val must be a max of 8 for a
// correct result. The resulting Y field val will have a max magnitude of 2.
func DecompressY(x *FieldVal, odd bool, resultY *FieldVal) bool {
// The curve equation for secp256k1 is: y^2 = x^3 + 7. Thus
// y = +-sqrt(x^3 + 7).
//
// The x coordinate must be invalid if there is no square root for the
// calculated rhs because it means the X coordinate is not for a point on
// the curve.
x3PlusB := new(FieldVal).SquareVal(x).Mul(x).AddInt(7)
if hasSqrt := resultY.SquareRootVal(x3PlusB); !hasSqrt {
return false
}
if resultY.Normalize().IsOdd() != odd {
resultY.Negate(1)
}
return true
}