RSA Encryption/ Decryption
RootOfTheNull Aug 14th, 2018 1,483 Never
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- #!/usr/bin/env python
- # -*- coding: utf-8 -*-
- # @Author: john
- # @Date: 2016-08-25 14:33:10
- # @Last Modified by: john
- # @Last Modified time: 2016-12-03 11:16:39
- from Crypto.Util.number import getPrime, inverse
- import binascii
- This Python code tries to illustrate how RSA is done at a basic level.
- # At RSA's core, there are two PRIME factors, p and q.
- # With this code, I just generate two large random prime numbers.
- # p and q are typically NOT given to you in an RSA challenge.
- bits_size = 256
- p = getPrime( bits_size )
- q = getPrime( bits_size )
- # These two prime factors multiply to create n, which is "the modulus".
- # n is typically GIVEN to you in an RSA challenge.
- n = p * q
- # The plaintext is typically referred to as m. Since we are working
- # with numbers and math, we treat the string information as hex.
- # Obviously this is NEVER given to you in an RSA challenge.
- m = "This is the clear text message."
- m = binascii.hexlify(m)
- m = int(m, 16)
- # Another value, called e, is "the exponent".
- # Both e and n make up "the public key", which is meant to be common
- # knowledge. This is why typically n and e are BOTH GIVEN in an RSA challenge.
- # Typcially, e is 65537, which is 0x10001 in hex
- e = 0x10001
- # The ciphertext, c, is created by this ENCRYPTION formula here.
- # c = ( m ^ e ) % n
- # See how e is the exponent and n is the modulus?
- # That is how the ENCRYPTION takes place.
- # Typically you are given the c in an RSA challenge and it is your task
- # to DECRYPT it to the m value, the plaintext.
- c = pow( m, e, n )
- # Now how do we decrypt? We need "the private key"....
- # We call this d in RSA. Thanks to some handy mathematical functions,
- # you can find "Euler's Totient", or "the Phi function" of a number.
- # This is 'the number of numbers that are less than a certain number and share
- # a common denominator with that certain number'.
- # I know that is hard to wrap your mind around here, but thankfully it is MUCH
- # easier for a prime number. For a prime number, it is simply that number minus 1!
- # So, if you are given n, what you need to do is find the factors of n.
- # Like we've seen, this is typically p and q. So, can we find the phi function
- # of n? Well, n is prime, and so are its factors, p and q, so phi should just be:
- phi = ( q - 1 ) * ( p - 1 )
- # Now, we can kind of unravel that modular arithmetic that was done during the
- # ENCRYPTION formula. We can find the private key, d, with the MODULAR INVERSE,
- # of e and phi.
- # I use a module to do this that is reworked to use Trey's function
- # https://github.com/JohnHammond/primefac_fork
- d = inverse( e, phi )
- # Now, we can do a similar thing like before, but this time for DECRYPTION.
- # m = ( c ^ d ) % n
- # This time we raise to our private key as an exponent, but still take the modulus.
- # And we have successfully decrypted RSA!
- m = pow( c, d, n )
- print repr(binascii.unhexlify(hex(m)[2:-1]))
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