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verify.py
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215 lines (190 loc) · 6.23 KB
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from ecdsa.util import string_to_number
import hashlib
import base64
import ecdsa
VERBOSE = False
_p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
_r = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
_b = 0x0000000000000000000000000000000000000000000000000000000000000007
_a = 0x0000000000000000000000000000000000000000000000000000000000000000
_Gx = 0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798
_Gy = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
curve_secp256k1 = ecdsa.ellipticcurve.CurveFp(_p, _a, _b)
generator_secp256k1 = ecdsa.ellipticcurve.Point(curve_secp256k1, _Gx, _Gy, _r)
oid_secp256k1 = (1, 3, 132, 0, 10)
SECP256k1 = ecdsa.curves.Curve("SECP256k1", curve_secp256k1, generator_secp256k1, oid_secp256k1)
addrtype = 0
__b58chars = "123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"
__b58base = len(__b58chars)
def modular_sqrt(a, p):
""" Find a quadratic residue (mod p) of 'a'. p
must be an odd prime.
Solve the congruence of the form:
x^2 = a (mod p)
And returns x. Note that p - x is also a root.
0 is returned is no square root exists for
these a and p.
The Tonelli-Shanks algorithm is used (except
for some simple cases in which the solution
is known from an identity). This algorithm
runs in polynomial time (unless the
generalized Riemann hypothesis is false).
"""
# Simple cases
#
if legendre_symbol(a, p) != 1:
return 0
elif a == 0:
return 0
elif p == 2:
return p
elif p % 4 == 3:
return pow(a, (p + 1) // 4, p)
# Partition p-1 to s * 2^e for an odd s (i.e.
# reduce all the powers of 2 from p-1)
#
s = p - 1
e = 0
while s % 2 == 0:
s //= 2
e += 1
# Find some 'n' with a legendre symbol n|p = -1.
# Shouldn't take long.
#
n = 2
while legendre_symbol(n, p) != -1:
n += 1
# Here be dragons!
# Read the paper "Square roots from 1; 24, 51,
# 10 to Dan Shanks" by Ezra Brown for more
# information
#
# x is a guess of the square root that gets better
# with each iteration.
# b is the "fudge factor" - by how much we're off
# with the guess. The invariant x^2 = ab (mod p)
# is maintained throughout the loop.
# g is used for successive powers of n to update
# both a and b
# r is the exponent - decreases with each update
#
x = pow(a, (s + 1) // 2, p)
b = pow(a, s, p)
g = pow(n, s, p)
r = e
while True:
t = b
m = 0
for m in range(r):
if t == 1:
break
t = pow(t, 2, p)
if m == 0:
return x
gs = pow(g, 2 ** (r - m - 1), p)
g = (gs * gs) % p
x = (x * gs) % p
b = (b * g) % p
r = m
def legendre_symbol(a, p):
""" Compute the Legendre symbol a|p using
Euler's criterion. p is a prime, a is
relatively prime to p (if p divides
a, then a|p = 0)
Returns 1 if a has a square root modulo
p, -1 otherwise.
"""
ls = pow(a, (p - 1) // 2, p)
return -1 if ls == p - 1 else ls
def b58encode(v):
long_value = 0
for (i, c) in enumerate(v[::-1]):
long_value += (256**i) * ord(c)
result = ""
while long_value >= __b58base:
div, mod = divmod(long_value, __b58base)
result = __b58chars[mod] + result
long_value = div
result = __b58chars[long_value] + result
# Bitcoin does a little leading-zero-compression:
# leading 0-bytes in the input become leading-1s
nPad = 0
for c in v:
if c == '\0':
nPad += 1
else:
break
return (__b58chars[0] * nPad) + result
def msg_magic(message):
return "\x18Bitcoin Signed Message:\n" + chr(len(message)) + message
def Hash(data):
return hashlib.sha256(hashlib.sha256(data).digest()).digest()
def hash_160(public_key):
md = hashlib.new("ripemd160")
md.update(hashlib.sha256(public_key).digest())
return md.digest()
def hash_160_to_bc_address(h160):
vh160 = chr(addrtype) + h160
h = Hash(vh160)
addr = vh160 + h[0:4]
return b58encode(addr)
def public_key_to_bc_address(public_key):
h160 = hash_160(public_key)
return hash_160_to_bc_address(h160)
def encode_point(pubkey, compressed=False):
order = generator_secp256k1.order()
p = pubkey.pubkey.point
x_str = ecdsa.util.number_to_string(p.x(), order)
y_str = ecdsa.util.number_to_string(p.y(), order)
if compressed:
return chr(2 + (p.y() & 1)) + x_str
else:
return chr(4) + x_str + y_str
def verify_message(address, signature, message):
from ecdsa import numbertheory, ellipticcurve, util
curve = curve_secp256k1
G = generator_secp256k1
order = G.order()
# extract r,s from signature
sig = base64.b64decode(signature)
if len(sig) != 65:
raise BaseException("Wrong encoding")
r, s = util.sigdecode_string(sig[1:], order)
nV = sig[0]
if nV < 27 or nV >= 35:
return False
if nV >= 31:
compressed = True
nV -= 4
else:
compressed = False
recid = nV - 27
# 1.1
x = r + (recid / 2) * order
# 1.3
alpha = (x * x * x + curve.a() * x + curve.b()) % curve.p()
beta = modular_sqrt(alpha, curve.p())
y = beta if (beta - recid) % 2 == 0 else curve.p() - beta
# 1.4 the constructor checks that nR is at infinity
R = ellipticcurve.Point(curve, x, y, order)
# 1.5 compute e from message:
h = Hash(msg_magic(message))
e = string_to_number(h)
minus_e = -e % order
# 1.6 compute Q = r^-1 (sR - eG)
inv_r = numbertheory.inverse_mod(r, order)
Q = inv_r * (s * R + minus_e * G)
public_key = ecdsa.VerifyingKey.from_public_point(Q, curve=SECP256k1)
# check that Q is the public key
public_key.verify_digest(sig[1:], h, sigdecode=ecdsa.util.sigdecode_string)
# check that we get the original signing address
addr = public_key_to_bc_address(encode_point(public_key, compressed))
if address == addr:
return True
else:
return False
print(verify_message(
"14dD6ygPi5WXdwwBTt1FBZK3aD8uDem1FY",
"IPn9bbEdNUp6+bneZqE2YJbq9Hv5aNILq9E5eZoMSF3/fBX4zjeIN6fpXfGSGPrZyKfHQ/c/kTSP+NIwmyTzMfk=",
"test message"
))