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* from __future__ import annotations
* fixup! from __future__ import annotations
* fixup! from __future__ import annotations
* fixup! Format Python code with psf/black push
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58 lines
1.7 KiB
Python
58 lines
1.7 KiB
Python
"""
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An RSA prime factor algorithm.
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The program can efficiently factor RSA prime number given the private key d and
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public key e.
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Source: on page 3 of https://crypto.stanford.edu/~dabo/papers/RSA-survey.pdf
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More readable source: https://www.di-mgt.com.au/rsa_factorize_n.html
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large number can take minutes to factor, therefore are not included in doctest.
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"""
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from __future__ import annotations
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import math
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import random
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def rsafactor(d: int, e: int, N: int) -> list[int]:
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"""
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This function returns the factors of N, where p*q=N
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Return: [p, q]
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We call N the RSA modulus, e the encryption exponent, and d the decryption exponent.
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The pair (N, e) is the public key. As its name suggests, it is public and is used to
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encrypt messages.
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The pair (N, d) is the secret key or private key and is known only to the recipient
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of encrypted messages.
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>>> rsafactor(3, 16971, 25777)
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[149, 173]
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>>> rsafactor(7331, 11, 27233)
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[113, 241]
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>>> rsafactor(4021, 13, 17711)
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[89, 199]
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"""
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k = d * e - 1
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p = 0
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q = 0
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while p == 0:
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g = random.randint(2, N - 1)
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t = k
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while True:
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if t % 2 == 0:
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t = t // 2
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x = (g ** t) % N
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y = math.gcd(x - 1, N)
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if x > 1 and y > 1:
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p = y
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q = N // y
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break # find the correct factors
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else:
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break # t is not divisible by 2, break and choose another g
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return sorted([p, q])
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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