Add comments to Python ECDSA implementation

This commit is contained in:
John Newbery 2019-04-18 13:47:24 -04:00 committed by Pieter Wuille
parent 8c7b9324ca
commit b67978529a

View File

@ -1,18 +1,17 @@
# Copyright (c) 2019 Pieter Wuille
# Distributed under the MIT software license, see the accompanying
# file COPYING or http://www.opensource.org/licenses/mit-license.php.
"""Test-only secp256k1 elliptic curve implementation
WARNING: This code is slow, uses bad randomness, does not properly protect
keys, and is trivially vulnerable to side channel attacks. Do not use for
anything but tests.
"""
anything but tests."""
import random
def modinv(a, n):
"""Compute the modular inverse of a modulo n
See https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm#Modular_integers
See https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm#Modular_integers.
"""
t1, t2 = 0, 1
r1, r2 = n, a
@ -30,8 +29,9 @@ def jacobi_symbol(n, k):
"""Compute the Jacobi symbol of n modulo k
See http://en.wikipedia.org/wiki/Jacobi_symbol
"""
assert k > 0 and k & 1
For our application k is always prime, so this is the same as the Legendre symbol."""
assert k > 0 and k & 1, "jacobi symbol is only defined for positive odd k"
n %= k
t = 0
while n != 0:
@ -47,11 +47,18 @@ def jacobi_symbol(n, k):
return 0
def modsqrt(a, p):
"""Compute the square root of a modulo p
"""Compute the square root of a modulo p when p % 4 = 3.
For p = 3 mod 4, if a square root exists, it is equal to a**((p+1)/4) mod p.
The Tonelli-Shanks algorithm can be used. See https://en.wikipedia.org/wiki/Tonelli-Shanks_algorithm
Limiting this function to only work for p % 4 = 3 means we don't need to
iterate through the loop. The highest n such that p - 1 = 2^n Q with Q odd
is n = 1. Therefore Q = (p-1)/2 and sqrt = a^((Q+1)/2) = a^((p+1)/4)
secp256k1's is defined over field of size 2**256 - 2**32 - 977, which is 3 mod 4.
"""
assert(p % 4 == 3) # Only p = 3 mod 4 is implemented
if p % 4 != 3:
raise NotImplementedError("modsqrt only implemented for p % 4 = 3")
sqrt = pow(a, (p + 1)//4, p)
if pow(sqrt, 2, p) == a % p:
return sqrt
@ -65,7 +72,9 @@ class EllipticCurve:
self.b = b % p
def affine(self, p1):
"""Convert a Jacobian point tuple p1 to affine form, or None if at infinity."""
"""Convert a Jacobian point tuple p1 to affine form, or None if at infinity.
An affine point is represented as the Jacobian (x, y, 1)"""
x1, y1, z1 = p1
if z1 == 0:
return None
@ -101,7 +110,9 @@ class EllipticCurve:
return (x, y, 1)
def double(self, p1):
"""Double a Jacobian tuple p1"""
"""Double a Jacobian tuple p1
See https://en.wikibooks.org/wiki/Cryptography/Prime_Curve/Jacobian_Coordinates - Point Doubling"""
x1, y1, z1 = p1
if z1 == 0:
return (0, 1, 0)
@ -119,10 +130,13 @@ class EllipticCurve:
return (x2, y2, z2)
def add_mixed(self, p1, p2):
"""Add a Jacobian tuple p1 and an affine tuple p2"""
"""Add a Jacobian tuple p1 and an affine tuple p2
See https://en.wikibooks.org/wiki/Cryptography/Prime_Curve/Jacobian_Coordinates - Point Addition (with affine point)"""
x1, y1, z1 = p1
x2, y2, z2 = p2
assert(z2 == 1)
# Adding to the point at infinity is a no-op
if z1 == 0:
return p2
z1_2 = (z1**2) % self.p
@ -131,7 +145,9 @@ class EllipticCurve:
s2 = (y2 * z1_3) % self.p
if x1 == u2:
if (y1 != s2):
# p1 and p2 are inverses. Return the point at infinity.
return (0, 1, 0)
# p1 == p2. The formulas below fail when the two points are equal.
return self.double(p1)
h = u2 - x1
r = s2 - y1
@ -144,13 +160,17 @@ class EllipticCurve:
return (x3, y3, z3)
def add(self, p1, p2):
"""Add two Jacobian tuples p1 and p2"""
"""Add two Jacobian tuples p1 and p2
See https://en.wikibooks.org/wiki/Cryptography/Prime_Curve/Jacobian_Coordinates - Point Addition"""
x1, y1, z1 = p1
x2, y2, z2 = p2
# Adding the point at infinity is a no-op
if z1 == 0:
return p2
if z2 == 0:
return p1
# Adding an Affine to a Jacobian is more efficient since we save field multiplications and squarings when z = 1
if z1 == 1:
return self.add_mixed(p2, p1)
if z2 == 1:
@ -165,7 +185,9 @@ class EllipticCurve:
s2 = (y2 * z1_3) % self.p
if u1 == u2:
if (s1 != s2):
# p1 and p2 are inverses. Return the point at infinity.
return (0, 1, 0)
# p1 == p2. The formulas below fail when the two points are equal.
return self.double(p1)
h = u2 - u1
r = s2 - s1
@ -214,6 +236,8 @@ class ECPubKey():
x = int.from_bytes(data[1:33], 'big')
if SECP256K1.is_x_coord(x):
p = SECP256K1.lift_x(x)
# if the oddness of the y co-ord isn't correct, find the other
# valid y
if (p[1] & 1) != (data[0] & 1):
p = SECP256K1.negate(p)
self.p = p
@ -243,8 +267,14 @@ class ECPubKey():
return bytes([0x04]) + p[0].to_bytes(32, 'big') + p[1].to_bytes(32, 'big')
def verify_ecdsa(self, sig, msg, low_s=True):
"""Verify a strictly DER-encoded ECDSA signature against this pubkey."""
"""Verify a strictly DER-encoded ECDSA signature against this pubkey.
See https://en.wikipedia.org/wiki/Elliptic_Curve_Digital_Signature_Algorithm for the
ECDSA verifier algorithm"""
assert(self.valid)
# Extract r and s from the DER formatted signature. Return false for
# any DER encoding errors.
if (sig[1] + 2 != len(sig)):
return False
if (len(sig) < 4):
@ -275,11 +305,15 @@ class ECPubKey():
if (slen > 1 and (sig[6+rlen] == 0) and not (sig[7+rlen] & 0x80)):
return False
s = int.from_bytes(sig[6+rlen:6+rlen+slen], 'big')
# Verify that r and s are within the group order
if r < 1 or s < 1 or r >= SECP256K1_ORDER or s >= SECP256K1_ORDER:
return False
if low_s and s >= SECP256K1_ORDER_HALF:
return False
z = int.from_bytes(msg, 'big')
# Run verifier algorithm on r, s
w = modinv(s, SECP256K1_ORDER)
u1 = z*w % SECP256K1_ORDER
u2 = r*w % SECP256K1_ORDER
@ -331,7 +365,10 @@ class ECKey():
return ret
def sign_ecdsa(self, msg, low_s=True):
"""Construct a DER-encoded ECDSA signature with this key."""
"""Construct a DER-encoded ECDSA signature with this key.
See https://en.wikipedia.org/wiki/Elliptic_Curve_Digital_Signature_Algorithm for the
ECDSA signer algorithm."""
assert(self.valid)
z = int.from_bytes(msg, 'big')
# Note: no RFC6979, but a simple random nonce (some tests rely on distinct transactions for the same operation)
@ -341,6 +378,9 @@ class ECKey():
s = (modinv(k, SECP256K1_ORDER) * (z + self.secret * r)) % SECP256K1_ORDER
if low_s and s > SECP256K1_ORDER_HALF:
s = SECP256K1_ORDER - s
# Represent in DER format. The byte representations of r and s have
# length rounded up (255 bits becomes 32 bytes and 256 bits becomes 33
# bytes).
rb = r.to_bytes((r.bit_length() + 8) // 8, 'big')
sb = s.to_bytes((s.bit_length() + 8) // 8, 'big')
return b'\x30' + bytes([4 + len(rb) + len(sb), 2, len(rb)]) + rb + bytes([2, len(sb)]) + sb