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ECDSA
Elliptic Curve Digital Signature Algorithm
Signature algorithm
- Use the secret scalar to compute the public point, by doing a scalar multiplication with : .
- Pick a random (secret) scalar , and perform a scalar multiplication with to get a random point.
- Use the above variables in the general equation of ECDSA is: , where, , is the 256-bit message being signed
- Simplify the resulting equation to get the component of the signature:
from random import randint
@dataclass
class PrivateKey:
secret: int
def sign(self, z: int) -> "Signature":
e = self.secret
k = randint(0, N)
R = k * G
r = R.x.value
k_inv = pow(k, -1, N) # Python 3.8+
s = ((z + r*e) * k_inv) % N
return Signature(r, s)
Apart from the fact that
e
is a secret number, the security of ECDSA also relies on the condition that k
is also very random and secret.We'll learn about the consequences of not having a random
k
in the next section.
Verification algorithm
- Given: (r, s) is the signature, z is the 256 bit message being signed, and P is the public key of the signer.
- Calculate: , .
- Calculate .
- Signature is valid is is equal to .
@dataclass
class Signature:
r: int
s: int
def verify(self, z: int, pub_key: Point) -> bool:
s_inv = pow(self.s, -1, N) # Python 3.8+
u = (z * s_inv) % N
v = (self.r * s_inv) % N
return (u*G + v*pub_key).x.value == self.r
pub = Point(
x=0x887387E452B8EACC4ACFDE10D9AAF7F6D9A0F975AABB10D006E4DA568744D06C,
y=0x61DE6D95231CD89026E286DF3B6AE4A894A3378E393E93A0F45B666329A0AE34,
curve=secp256k1
)
# Test case 1: verify authenticity
z = 0xEC208BAA0FC1C19F708A9CA96FDEFF3AC3F230BB4A7BA4AEDE4942AD003C0F60
r = 0xAC8D1C87E51D0D441BE8B3DD5B05C8795B48875DFFE00B7FFCFAC23010D3A395
s = 0x68342CEFF8935EDEDD102DD876FFD6BA72D6A427A3EDB13D26EB0781CB423C4
assert Signature(r, s).verify(z, pub)
# Test case 2: verify authenticity for different signature w/ same P
z = 0x7C076FF316692A3D7EB3C3BB0F8B1488CF72E1AFCD929E29307032997A838A3D
r = 0xEFF69EF2B1BD93A66ED5219ADD4FB51E11A840F404876325A1E8FFE0529A2C
s = 0xC7207FEE197D27C618AEA621406F6BF5EF6FCA38681D82B2F06FDDBDCE6FEAB6
assert Signature(r, s).verify(z, pub)
# Test case 3: sign and verify
e = PrivateKey(randint(0, N)) # generate a private key
pub = e.secret * G # public point corresponding to e
z = randint(0, 2 ** 256) # generate a random message for testing
signature: Signature = e.sign(z)
assert signature.verify(z, pub)
Last modified 2mo ago