[英]How to find the prime factors of a number with python
I'm writing a program that will calulate the private key for a weak RSA public key. 我正在编写一个程序,它将为较弱的RSA公钥计算私钥。 I am wondering how I would go about determining the values for
p
and q
from the value n
. 我想知道如何从值
n
确定p
和q
的值。 Here is the Python code so far: 到目前为止,这是Python代码:
from Crypto.PublicKey import RSA #PyCryptoDome
import .math as cm # My own module
with open(public_keyfile, 'rb') as key: # Public Keyfile Is in PEM format
public_key = RSA.import_key(key)
n = public_key.n # N value of the public_key
e = public_key.e # E value of the public_key
p, q = get_factors_of(n) # This I don't know how to do, though there is a question that might help [see bottom]
t = cm.lcm(p-1, q-1) # Get the lowest common multiple of q and q
d = cm.mod_inverse(e, t) # Get d, the modular inverse of e % t
private_key = RSA.construct((n, e, d, p, q) # Construct the RSA private_key
The .math
module referenced above: .math
模块:
from math import gcd
def mod_inverse(a, b):
a = a % b
for x in range(1, b):
if (a * x) % b == 1:
return x
return 1
def lcm(x, y):
return x * y // gcd(x, y)
What I need to do appears to be referenced here but this code is in Java. 我需要执行的操作似乎在这里被引用,但是此代码是Java语言。
If anyone knows how to get p
and q
from n
with python, help would be appreciated. 如果有人知道如何使用python从
n
获取p
和q
,将不胜感激。
Many thanks, Legorooj. 非常感谢,Legolooj。
Mandatory warning: if you are after performance, you will need to investigate the details of the algorithms yourself. 强制警告:如果您追求性能,则需要亲自调查算法的详细信息。 Even "weak" public keys will take forever to crack with a simplistic algorithm (eg Erathostene's sieve).
即使是“弱”的公钥,也将通过简单的算法(例如Erathostene的筛子)永远破解。
That being said, sympy.ntheory.factorint()
might be what you need: 话虽这么说,
sympy.ntheory.factorint()
可能就是您需要的:
from sympy.ntheory import factorint
print(factorint(54)) # {2: 1, 3: 3} i.e. 54 == 2**1 * 3**3
After lots of googling, and pdf reading, I found an algorithm that works. 经过大量的Google搜索和pdf阅读后,我发现了一种有效的算法。 Here is a python implementation:
这是一个python实现:
import math
def get_factors_of(num):
poss_p = math.floor(math.sqrt(num))
if poss_p % 2 == 0: # Only checks odd numbers, it reduces time by orders of magnitude
poss_p += 1
while poss_p < num:
if num % poss_p == 0:
return poss_p
poss_p += 2
This algorithm effectively finds the P/Q factors of a small RSA key. 该算法有效地找到了一个小的RSA密钥的P / Q因子。 (I have tested it against a 64-bit PEM public key)
(我已经针对64位PEM公钥对其进行了测试)
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