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fix implementation errors. (#1568)

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I revised my implementation and found out that I have miss a inner loop for t.  x and y should be recalculated everytime when t is divisble by 2. I have also included a more readble source for this algorithm.
This commit is contained in:
Zizhou Zhang 2019-11-15 06:29:54 +11:00 committed by Christian Clauss
parent ea9bf0a90c
commit e3aa0f65e8
1 changed files with 13 additions and 8 deletions
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21
ciphers/rsa_factorization.py
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@ -4,6 +4,7 @@ An RSA prime factor algorithm.
The program can efficiently factor RSA prime number given the private key d and The program can efficiently factor RSA prime number given the private key d and
public key e. public key e.
Source: on page 3 of https://crypto.stanford.edu/~dabo/papers/RSA-survey.pdf 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. large number can take minutes to factor, therefore are not included in doctest.
""" """
import math import math
@ -15,7 +16,7 @@ def rsafactor(d: int, e: int, N: int) -> List[int]:
""" """
This function returns the factors of N, where p*q=N This function returns the factors of N, where p*q=N
Return: [p, q] Return: [p, q]
We call N the RSA modulus, e the encryption exponent, and d the decryption exponent. 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 The pair (N, e) is the public key. As its name suggests, it is public and is used to
encrypt messages. encrypt messages.
@ -35,13 +36,17 @@ def rsafactor(d: int, e: int, N: int) -> List[int]:
while p == 0: while p == 0:
g = random.randint(2, N - 1) g = random.randint(2, N - 1)
t = k t = k
if t % 2 == 0: while True:
t = t // 2 if t % 2 == 0:
x = (g ** t) % N t = t // 2
y = math.gcd(x - 1, N) x = (g ** t) % N
if x > 1 and y > 1: y = math.gcd(x - 1, N)
p = y if x > 1 and y > 1:
q = N // y p = y
q = N // y
break # find the correct factors
else:
break # t is not divisible by 2, break and choose another g
return sorted([p, q]) return sorted([p, q])