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Optimize ScalarMult with NAF
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Use Non-Adjacent Form (NAF) of large numbers to reduce ScalarMult computation times.

Preliminary results indicate around a 8-9% speed improvement according to BenchmarkScalarMult.

The algorithm used is 3.77 from Guide to Elliptical Curve Crytography by Hankerson, et al.

This closes #3
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jimmysong committed Feb 5, 2015
1 parent 95b23c2 commit 6c36218
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Showing 4 changed files with 171 additions and 36 deletions.
8 changes: 8 additions & 0 deletions bench_test.go
Expand Up @@ -78,6 +78,14 @@ func BenchmarkScalarMult(b *testing.B) {
}
}

// BenchmarkNAF benchmarks the NAF function.
func BenchmarkNAF(b *testing.B) {
k := fromHex("d74bf844b0862475103d96a611cf2d898447e288d34b360bc885cb8ce7c00575")
for i := 0; i < b.N; i++ {
btcec.NAF(k.Bytes())
}
}

// BenchmarkSigVerify benchmarks how long it takes the secp256k1 curve to
// verify signatures.
func BenchmarkSigVerify(b *testing.B) {
Expand Down
157 changes: 125 additions & 32 deletions btcec.go
Expand Up @@ -666,6 +666,78 @@ func (curve *KoblitzCurve) moduloReduce(k []byte) []byte {
return k
}

// NAF takes a positive integer k and returns the Non-Adjacent Form (NAF)
// as two byte slices. The first is where 1's should be. The second is where
// -1's should be.
// NAF is also convenient in that on average, only 1/3rd of its values are
// non-zero.
// The algorithm here is from Guide to Elliptical Cryptography 3.30 (ref above)
// Essentially, this makes it possible to minimize the number of operations
// since the resulting ints returned will be at least 50% 0's.
func NAF(k []byte) ([]byte, []byte) {

// The essence of this algorithm is that whenever we have consecutive 1s
// in the binary, we want to put a -1 in the lowest bit and get a bunch of
// 0s up to the highest bit of consecutive 1s. This is due to this identity:
// 2^n + 2^(n-1) + 2^(n-2) + ... + 2^(n-k) = 2^(n+1) - 2^(n-k)
// The algorithm thus may need to go 1 more bit than the length of the bits
// we actually have, hence bits being 1 bit longer than was necessary.
// Since we need to know whether adding will cause a carry, we go from
// right-to-left in this addition.
var carry, curIsOne, nextIsOne bool
// these default to zero
retPos := make([]byte, len(k)+1)
retNeg := make([]byte, len(k)+1)
for i := len(k) - 1; i >= 0; i-- {
curByte := k[i]
for j := uint(0); j < 8; j++ {
curIsOne = curByte&1 == 1
if j == 7 {
if i == 0 {
nextIsOne = false
} else {
nextIsOne = k[i-1]&1 == 1
}
} else {
nextIsOne = curByte&2 == 2
}
if carry {
if curIsOne {
// This bit is 1, so we continue to carry and
// don't need to do anything
} else {
// We've hit a 0 after some number of 1s.
if nextIsOne {
// We start carrying again since we're starting
// a new sequence of 1s.
retNeg[i+1] += 1 << j
} else {
// We stop carrying since 1s have stopped.
carry = false
retPos[i+1] += 1 << j
}
}
} else if curIsOne {
if nextIsOne {
// if this is the start of at least 2 consecutive 1's
// we want to set the current one to -1 and start carrying
retNeg[i+1] += 1 << j
carry = true
} else {
// this is a singleton, not consecutive 1's.
retPos[i+1] += 1 << j
}
}
curByte >>= 1
}
}
if carry {
retPos[0] = 1
}

return retPos, retNeg
}

// ScalarMult returns k*(Bx, By) where k is a big endian integer.
// Part of the elliptic.Curve interface.
func (curve *KoblitzCurve) ScalarMult(Bx, By *big.Int, k []byte) (*big.Int, *big.Int) {
Expand All @@ -676,66 +748,85 @@ func (curve *KoblitzCurve) ScalarMult(Bx, By *big.Int, k []byte) (*big.Int, *big
// see Algorithm 3.74 in Guide to Elliptical Curve Cryptography by
// Hankerson, et al.
k1, k2, signK1, signK2 := curve.splitK(curve.moduloReduce(k))
k1Len := len(k1)
k2Len := len(k2)
m := k1Len
if k2Len > m {
m = k2Len
}

// 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
p1x, p1y := curve.bigAffineToField(Bx, By)
// For NAF, we need the negative point
p1yNeg := new(fieldVal).NegateVal(p1y, 1)
p1z := new(fieldVal).SetInt(1)
// Note ϕ(x,y) = (βx,y), the Jacobian z coordinate is 1, so this math
// goes through.
p2x := new(fieldVal).Set(p1x).Mul(curve.beta)
p2x := new(fieldVal).Mul2(p1x, curve.beta)
p2y := new(fieldVal).Set(p1y)
// For NAF, we need the negative point
p2yNeg := new(fieldVal).NegateVal(p2y, 1)
p2z := new(fieldVal).SetInt(1)

// If k1 or k2 are negative, we only need to flip the y of the respective
// Jacobian point. In ECC terms, we're reflecting the point over the
// x-axis which is guaranteed to still be on the curve.
// If k1 or k2 are negative, we flip the positive/negative values
if signK1 == -1 {
p1y.Negate(1)
p1y, p1yNeg = p1yNeg, p1y
}
if signK2 == -1 {
p2y.Negate(1)
p2y, p2yNeg = p2yNeg, p2y
}

// We use the left to right binary addition method.
// At each bit of k1 and k2, we add the current part of the
// k * P = k1 * P + k2 * ϕ(P) equation (that is, P1 and P2) and double.
// A further optimization using NAF is possible here but unimplemented.
var byteVal1, byteVal2 byte
// NAF versions of k1 and k2 should have a lot more zeros
// the Pos version of the bytes contain the +1's and the Neg versions
// contain the -1's
k1PosNAF, k1NegNAF := NAF(k1)
k2PosNAF, k2NegNAF := NAF(k2)
k1Len := len(k1PosNAF)
k2Len := len(k2PosNAF)

m := k1Len
if m < k2Len {
m = k2Len
}

// We add left-to-right using the NAF optimization. This is using
// algorithm 3.77 from Guide to Elliptical Curve Cryptography.
// 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.
var k1BytePos, k1ByteNeg, k2BytePos, k2ByteNeg byte
for i := 0; i < m; i++ {
// Note that if k1 or k2 has less than the max number of bytes, we
// want to ignore the bytes at the front since we're going left to
// right.
// Since we're going left-to-right, we need to pad the front with 0's
if i < m-k1Len {
byteVal1 = 0
k1BytePos = 0
k1ByteNeg = 0
} else {
byteVal1 = k1[i-m+k1Len]
k1BytePos = k1PosNAF[i-(m-k1Len)]
k1ByteNeg = k1NegNAF[i-(m-k1Len)]
}
if i < m-k2Len {
byteVal2 = 0
k2BytePos = 0
k2ByteNeg = 0
} else {
byteVal2 = k2[i-m+k2Len]
k2BytePos = k2PosNAF[i-(m-k2Len)]
k2ByteNeg = k2NegNAF[i-(m-k2Len)]
}
for bitNum := 0; bitNum < 8; bitNum++ {
// Q = 2*Q

for j := 7; j >= 0; j-- {
// Q = 2 * Q
curve.doubleJacobian(qx, qy, qz, qx, qy, qz)
if byteVal1&0x80 == 0x80 {
// Q = Q + P1

if k1BytePos&0x80 == 0x80 {
curve.addJacobian(qx, qy, qz, p1x, p1y, p1z, qx, qy, qz)
} else if k1ByteNeg&0x80 == 0x80 {
curve.addJacobian(qx, qy, qz, p1x, p1yNeg, p1z, qx, qy, qz)
}
if byteVal2&0x80 == 0x80 {
// Q = Q + P2

if k2BytePos&0x80 == 0x80 {
curve.addJacobian(qx, qy, qz, p2x, p2y, p2z, qx, qy, qz)
} else if k2ByteNeg&0x80 == 0x80 {
curve.addJacobian(qx, qy, qz, p2x, p2yNeg, p2z, qx, qy, qz)
}
byteVal1 <<= 1
byteVal2 <<= 1
k1BytePos <<= 1
k1ByteNeg <<= 1
k2BytePos <<= 1
k2ByteNeg <<= 1
}
}

Expand Down Expand Up @@ -801,6 +892,8 @@ func initS256() {
// 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.
// These have been independently verified by Dave Collins using
// an ecc math script.
secp256k1.lambda, _ = new(big.Int).SetString("5363AD4CC05C30E0A5261C028812645A122E22EA20816678DF02967C1B23BD72", 16)
secp256k1.beta = new(fieldVal).SetHex("7AE96A2B657C07106E64479EAC3434E99CF0497512F58995C1396C28719501EE")
secp256k1.a1, _ = new(big.Int).SetString("3086D221A7D46BCDE86C90E49284EB15", 16)
Expand Down
35 changes: 35 additions & 0 deletions btcec_test.go
Expand Up @@ -656,6 +656,41 @@ func TestSignAndVerify(t *testing.T) {
testSignAndVerify(t, btcec.S256(), "S256")
}

func TestNAF(t *testing.T) {
negOne := big.NewInt(-1)
one := big.NewInt(1)
two := big.NewInt(2)
for i := 0; i < 1024; i++ {
data := make([]byte, 32)
_, err := rand.Read(data)
if err != nil {
t.Fatalf("failed to read random data at %d", i)
break
}
nafPos, nafNeg := btcec.NAF(data)
want := new(big.Int).SetBytes(data)
got := big.NewInt(0)
// Check that the NAF representation comes up with the right number
for i := 0; i < len(nafPos); i++ {
bytePos := nafPos[i]
byteNeg := nafNeg[i]
for j := 7; j >= 0; j-- {
got.Mul(got, two)
if bytePos&0x80 == 0x80 {
got.Add(got, one)
} else if byteNeg&0x80 == 0x80 {
got.Add(got, negOne)
}
bytePos <<= 1
byteNeg <<= 1
}
}
if got.Cmp(want) != 0 {
t.Errorf("%d: Failed NAF got %X want %X", i, got, want)
}
}
}

func fromHex(s string) *big.Int {
r, ok := new(big.Int).SetString(s, 16)
if !ok {
Expand Down
7 changes: 3 additions & 4 deletions gensecp256k1.go
Expand Up @@ -35,15 +35,14 @@ func (curve *KoblitzCurve) getDoublingPoints() [][3]fieldVal {
// the possible points per 8-bit window. This is used to when generating
// secp256k1.go.
func (curve *KoblitzCurve) SerializedBytePoints() []byte {
byteSize := curve.BitSize / 8
doublingPoints := curve.getDoublingPoints()

// Segregate the bits into byte-sized windows
serialized := make([]byte, byteSize*256*3*10*4)
serialized := make([]byte, curve.byteSize*256*3*10*4)
offset := 0
for byteNum := 0; byteNum < byteSize; byteNum++ {
for byteNum := 0; byteNum < curve.byteSize; byteNum++ {
// Grab the 8 bits that make up this byte from doublingPoints.
startingBit := 8 * (byteSize - byteNum - 1)
startingBit := 8 * (curve.byteSize - byteNum - 1)
computingPoints := doublingPoints[startingBit : startingBit+8]

// Compute all points in this window and serialize them.
Expand Down

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