帮我写一段带输出的NTRU数字签名的python代码
时间: 2024-05-03 12:20:13 浏览: 7
好的,以下是带输出的NTRU数字签名的Python代码:
```python
import hashlib
import random
import numpy as np
from sympy import nextprime
# NTRU parameters
N = 503
p = 3
q = 64
d = 47
t = 300
# Generate private key polynomials f and g
f_coeffs = [random.randint(-1, 1) for i in range(N)]
g_coeffs = [random.randint(-1, 1) for i in range(N)]
while True:
r_coeffs = [random.randint(-1, 1) for i in range(N)]
r_2_coeffs = [random.randint(-1, 1) for i in range(N)]
if np.poly1d(r_coeffs).gcd(np.poly1d(f_coeffs)) == np.poly1d(1) and np.poly1d(r_2_coeffs).gcd(np.poly1d(f_coeffs)) == np.poly1d(1):
break
# Compute public key polynomial h = (t * f * g + 1) mod q
f = np.poly1d(f_coeffs)
g = np.poly1d(g_coeffs)
r = np.poly1d(r_coeffs)
r_2 = np.poly1d(r_2_coeffs)
h = (t * f * g + 1) % q
# Define hashing function
def H(m):
return int(hashlib.sha256(str(m).encode('utf-8')).hexdigest(), 16)
# Generate signature for message m
def sign(m):
while True:
s_coeffs = [random.randint(-1, 1) for i in range(N)]
e_coeffs = [random.randint(-1, 1) for i in range(N)]
s = np.poly1d(s_coeffs)
e = np.poly1d(e_coeffs)
if np.poly1d(s * h - t * e - np.poly1d([H(m)])) % q == np.poly1d(0):
break
return (s_coeffs, e_coeffs)
# Verify signature (s, e) for message m
def verify(m, s, e):
return np.poly1d(s * h - t * np.poly1d(e) - np.poly1d([H(m)])) % q == np.poly1d(0)
# Test signature generation and verification
m = "Hello, world!"
print("Message:", m)
(s, e) = sign(m)
print("Signature:", (s, e))
print("Verification:", verify(m, s, e))
```