在过去的一个半星期里，我一直在尝试使用 Python 生成用于 RSA 加密的大素数，但没有成功。 Fermat 素性测试在 512 位尺度上是不可行的，而且我无法完全理解 Miller-Rabin。 （我今年 13 岁）所有在线脚本似乎都可以使用低于我正在使用的 Python 版本。我应该怎么做才能产生大量素数？ （是的，概率素数很好。）
这是我的米勒-拉宾质数检查器：
def isPrime(n, k=5): # miller-rabin
from random import randint
if n < 2: return False
for p in [2,3,5,7,11,13,17,19,23,29]:
if n % p == 0: return n == p
s, d = 0, n-1
while d % 2 == 0:
s, d = s+1, d/2
for i in range(k):
x = pow(randint(2, n-1), d, n)
if x == 1 or x == n-1: continue
for r in range(1, s):
x = (x * x) % n
if x == 1: return False
if x == n-1: break
else: return False
return True
如果你想要一个有保证的素数（而不是可能的素数），那么安排起来也不是很难。请参阅我的博客了解 Pocklington 提供的方法。
我也需要这个，所以我想出了这个：
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""Calculate prime candidates for cryptography uses
License: This work is marked with CC0 1.0 Universal
<https://creativecommons.org/publicdomain/zero/1.0/>"""
import math
import secrets
import sys
from typing import Set # For older version compat
from warnings import filterwarnings as filter_warnings
primes: Set[int] = {
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,
61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127,
131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193,
197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269,
271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349,
353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431,
433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503,
509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599,
601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673,
677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761,
769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857,
859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947,
953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031,
1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097,
1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187,
1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277,
1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327,
1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439,
1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499,
1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583,
1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663,
1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747,
1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847,
1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931,
1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011,
2017, 2027, 2029, 2039, 2053, 2063, 2069, 2081, 2083, 2087, 2089,
}
def test_prime(n: int, k: int = 23) -> bool:
"""Test if a number is prime
Uses Fermat's Little Theorem & Miller-Rabin primality test
Also does simple checks such as if n<0 or n is a perfect square
then `n` is not prime"""
if n < 2 or math.isqrt(n)**2 == n:
return False
for p in primes:
if n < p * p:
return True
if n % p == 0:
return False
r, s = 0, n - 1
while s % 2 == 0:
r += 1
s //= 2
for _ in range(k):
a: int = secrets.randbelow(n - 1) + 1 # [0; n)
x: int = pow(a, s, n)
if x == 1 or x == n - 1:
continue
for _ in range(r - 1):
x = pow(x, 2, n)
if x == n - 1:
break
else:
return False
return True
def generate_large_prime(n: int) -> int:
"""Generate a large n-bit prime number"""
found_prime: bool = False
p: int = 0
while not found_prime:
p = secrets.randbits(n) # Generates a random n-bit number
p |= (
1 << n - 1
) | 1 # Ensures the n-bit length and assures it is an odd number
found_prime = test_prime(p)
return p
def main() -> int:
"""Entry/main function"""
if len(sys.argv) != 2:
print(
"Please supply n bits to generate a prime candidate for",
file=sys.stderr,
)
return 1
print(generate_large_prime(int(sys.argv[1])))
return 0
if __name__ == "__main__":
assert main.__annotations__.get("return") is int, "main() should return an integer"
filter_warnings("error", category=Warning)
raise SystemExit(main())
请注意，您可以将
primes