rsa_factorization.py

"""
An RSA prime factor algorithm.

The program can efficiently factor RSA prime number given the private key d and
public key e.

| Source: on page ``3`` of https://crypto.stanford.edu/~dabo/papers/RSA-survey.pdf
| More readable source: https://www.di-mgt.com.au/rsa_factorize_n.html

large number can take minutes to factor, therefore are not included in doctest.
"""

from __future__ importannotations

importmath
importrandom


defrsafactor(d: int, e: int, n: int) ->list[int]:
"""
 This function returns the factors of N, where p*q=N

 Return: [p, q]

 We call N the RSA modulus, e the encryption exponent, and d the decryption exponent.
 The pair (N, e) is the public key. As its name suggests, it is public and is used to
 encrypt messages.
 The pair (N, d) is the secret key or private key and is known only to the recipient
 of encrypted messages.

 >>> rsafactor(3, 16971, 25777)
 [149, 173]
 >>> rsafactor(7331, 11, 27233)
 [113, 241]
 >>> rsafactor(4021, 13, 17711)
 [89, 199]
 """
k=d*e-1
p=0
q=0
whilep==0:
g=random.randint(2, n-1)
t=k
whileTrue:
ift%2==0:
t=t//2
x= (g**t) %n
y=math.gcd(x-1, n)
ifx>1andy>1:
p=y
q=n//y
break# find the correct factors
else:
break# t is not divisible by 2, break and choose another g
returnsorted([p, q])


if__name__=="__main__":
importdoctest

doctest.testmod()