added rsa_factorization.py (#1556)

* added RSA_factorization.py

This algorithm can effectively factor RSA large prime N given public key e and private key d.

* Rename RSA_factorization.py to rsa_factorization.py

* Add definitions for d, e, and N

ciphers/rsa_factorization.py

@@ -0,0 +1,51 @@
+"""
+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
+large number can take minutes to factor, therefore are not included in doctest.
+"""
+import math
+import random
+from typing import List
+
+
+def rsafactor(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
+    while p == 0:
+        g = random.randint(2, N - 1)
+        t = k
+        if t % 2 == 0:
+            t = t // 2
+            x = (g ** t) % N
+            y = math.gcd(x - 1, N)
+            if x > 1 and y > 1:
+                p = y
+                q = N // y
+    return sorted([p, q])
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()