Add comments to Python ECDSA implementation
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1 changed files with 56 additions and 16 deletions
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@ -1,18 +1,17 @@
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# Copyright (c) 2019 Pieter Wuille
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# Distributed under the MIT software license, see the accompanying
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# file COPYING or http://www.opensource.org/licenses/mit-license.php.
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"""Test-only secp256k1 elliptic curve implementation
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WARNING: This code is slow, uses bad randomness, does not properly protect
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keys, and is trivially vulnerable to side channel attacks. Do not use for
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anything but tests.
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"""
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anything but tests."""
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import random
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def modinv(a, n):
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"""Compute the modular inverse of a modulo n
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See https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm#Modular_integers
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See https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm#Modular_integers.
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"""
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t1, t2 = 0, 1
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r1, r2 = n, a
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@ -30,8 +29,9 @@ def jacobi_symbol(n, k):
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"""Compute the Jacobi symbol of n modulo k
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See http://en.wikipedia.org/wiki/Jacobi_symbol
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"""
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assert k > 0 and k & 1
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For our application k is always prime, so this is the same as the Legendre symbol."""
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assert k > 0 and k & 1, "jacobi symbol is only defined for positive odd k"
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n %= k
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t = 0
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while n != 0:
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@ -47,11 +47,18 @@ def jacobi_symbol(n, k):
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return 0
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def modsqrt(a, p):
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"""Compute the square root of a modulo p
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"""Compute the square root of a modulo p when p % 4 = 3.
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For p = 3 mod 4, if a square root exists, it is equal to a**((p+1)/4) mod p.
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The Tonelli-Shanks algorithm can be used. See https://en.wikipedia.org/wiki/Tonelli-Shanks_algorithm
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Limiting this function to only work for p % 4 = 3 means we don't need to
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iterate through the loop. The highest n such that p - 1 = 2^n Q with Q odd
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is n = 1. Therefore Q = (p-1)/2 and sqrt = a^((Q+1)/2) = a^((p+1)/4)
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secp256k1's is defined over field of size 2**256 - 2**32 - 977, which is 3 mod 4.
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"""
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assert(p % 4 == 3) # Only p = 3 mod 4 is implemented
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if p % 4 != 3:
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raise NotImplementedError("modsqrt only implemented for p % 4 = 3")
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sqrt = pow(a, (p + 1)//4, p)
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if pow(sqrt, 2, p) == a % p:
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return sqrt
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@ -65,7 +72,9 @@ class EllipticCurve:
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self.b = b % p
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def affine(self, p1):
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"""Convert a Jacobian point tuple p1 to affine form, or None if at infinity."""
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"""Convert a Jacobian point tuple p1 to affine form, or None if at infinity.
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An affine point is represented as the Jacobian (x, y, 1)"""
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x1, y1, z1 = p1
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if z1 == 0:
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return None
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@ -101,7 +110,9 @@ class EllipticCurve:
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return (x, y, 1)
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def double(self, p1):
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"""Double a Jacobian tuple p1"""
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"""Double a Jacobian tuple p1
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See https://en.wikibooks.org/wiki/Cryptography/Prime_Curve/Jacobian_Coordinates - Point Doubling"""
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x1, y1, z1 = p1
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if z1 == 0:
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return (0, 1, 0)
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@ -119,10 +130,13 @@ class EllipticCurve:
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return (x2, y2, z2)
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def add_mixed(self, p1, p2):
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"""Add a Jacobian tuple p1 and an affine tuple p2"""
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"""Add a Jacobian tuple p1 and an affine tuple p2
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See https://en.wikibooks.org/wiki/Cryptography/Prime_Curve/Jacobian_Coordinates - Point Addition (with affine point)"""
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x1, y1, z1 = p1
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x2, y2, z2 = p2
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assert(z2 == 1)
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# Adding to the point at infinity is a no-op
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if z1 == 0:
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return p2
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z1_2 = (z1**2) % self.p
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@ -131,7 +145,9 @@ class EllipticCurve:
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s2 = (y2 * z1_3) % self.p
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if x1 == u2:
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if (y1 != s2):
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# p1 and p2 are inverses. Return the point at infinity.
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return (0, 1, 0)
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# p1 == p2. The formulas below fail when the two points are equal.
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return self.double(p1)
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h = u2 - x1
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r = s2 - y1
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@ -144,13 +160,17 @@ class EllipticCurve:
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return (x3, y3, z3)
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def add(self, p1, p2):
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"""Add two Jacobian tuples p1 and p2"""
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"""Add two Jacobian tuples p1 and p2
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See https://en.wikibooks.org/wiki/Cryptography/Prime_Curve/Jacobian_Coordinates - Point Addition"""
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x1, y1, z1 = p1
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x2, y2, z2 = p2
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# Adding the point at infinity is a no-op
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if z1 == 0:
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return p2
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if z2 == 0:
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return p1
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# Adding an Affine to a Jacobian is more efficient since we save field multiplications and squarings when z = 1
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if z1 == 1:
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return self.add_mixed(p2, p1)
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if z2 == 1:
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@ -165,7 +185,9 @@ class EllipticCurve:
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s2 = (y2 * z1_3) % self.p
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if u1 == u2:
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if (s1 != s2):
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# p1 and p2 are inverses. Return the point at infinity.
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return (0, 1, 0)
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# p1 == p2. The formulas below fail when the two points are equal.
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return self.double(p1)
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h = u2 - u1
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r = s2 - s1
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@ -214,6 +236,8 @@ class ECPubKey():
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x = int.from_bytes(data[1:33], 'big')
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if SECP256K1.is_x_coord(x):
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p = SECP256K1.lift_x(x)
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# if the oddness of the y co-ord isn't correct, find the other
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# valid y
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if (p[1] & 1) != (data[0] & 1):
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p = SECP256K1.negate(p)
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self.p = p
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@ -243,8 +267,14 @@ class ECPubKey():
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return bytes([0x04]) + p[0].to_bytes(32, 'big') + p[1].to_bytes(32, 'big')
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def verify_ecdsa(self, sig, msg, low_s=True):
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"""Verify a strictly DER-encoded ECDSA signature against this pubkey."""
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"""Verify a strictly DER-encoded ECDSA signature against this pubkey.
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See https://en.wikipedia.org/wiki/Elliptic_Curve_Digital_Signature_Algorithm for the
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ECDSA verifier algorithm"""
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assert(self.valid)
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# Extract r and s from the DER formatted signature. Return false for
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# any DER encoding errors.
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if (sig[1] + 2 != len(sig)):
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return False
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if (len(sig) < 4):
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@ -275,11 +305,15 @@ class ECPubKey():
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if (slen > 1 and (sig[6+rlen] == 0) and not (sig[7+rlen] & 0x80)):
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return False
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s = int.from_bytes(sig[6+rlen:6+rlen+slen], 'big')
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# Verify that r and s are within the group order
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if r < 1 or s < 1 or r >= SECP256K1_ORDER or s >= SECP256K1_ORDER:
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return False
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if low_s and s >= SECP256K1_ORDER_HALF:
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return False
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z = int.from_bytes(msg, 'big')
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# Run verifier algorithm on r, s
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w = modinv(s, SECP256K1_ORDER)
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u1 = z*w % SECP256K1_ORDER
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u2 = r*w % SECP256K1_ORDER
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@ -331,7 +365,10 @@ class ECKey():
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return ret
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def sign_ecdsa(self, msg, low_s=True):
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"""Construct a DER-encoded ECDSA signature with this key."""
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"""Construct a DER-encoded ECDSA signature with this key.
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See https://en.wikipedia.org/wiki/Elliptic_Curve_Digital_Signature_Algorithm for the
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ECDSA signer algorithm."""
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assert(self.valid)
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z = int.from_bytes(msg, 'big')
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# Note: no RFC6979, but a simple random nonce (some tests rely on distinct transactions for the same operation)
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@ -341,6 +378,9 @@ class ECKey():
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s = (modinv(k, SECP256K1_ORDER) * (z + self.secret * r)) % SECP256K1_ORDER
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if low_s and s > SECP256K1_ORDER_HALF:
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s = SECP256K1_ORDER - s
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# Represent in DER format. The byte representations of r and s have
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# length rounded up (255 bits becomes 32 bytes and 256 bits becomes 33
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# bytes).
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rb = r.to_bytes((r.bit_length() + 8) // 8, 'big')
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sb = s.to_bytes((s.bit_length() + 8) // 8, 'big')
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return b'\x30' + bytes([4 + len(rb) + len(sb), 2, len(rb)]) + rb + bytes([2, len(sb)]) + sb
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