A simple RSA implementation in Python

rsa.py

'''
620031587
Net-Centric Computing Assignment
Part A - RSA Encryption
'''
from __future__ import division
import random


'''
Euclid's algorithm for determining the greatest common divisor
Use iteration to make it faster for larger integers
'''
def gcd(a, b):
    while b != 0:
        a, b = b, a % b
    return a

'''
d can be calculated using formula : d = (phi*i + 1)/e
where i is integer starting from 1
and we find integer value for d iteratively
'''
def multiplicative_inverse(e, phi):
    d = None
    i = 1
    exit = False
    while not exit:
        temp1 = phi*i +1
        d = float(temp1/e)
        d_int = int(d)
        i += 1
        if(d_int == d):
            exit=True
    return int(d)

'''
Tests to see if a number is prime.
'''
def is_prime(num):
    if num == 2:
        return True
    if num < 2 or num % 2 == 0:
        return False
    for n in xrange(3, int(num**0.5)+2, 2):
        if num % n == 0:
            return False
    return True

def generate_keypair(p, q):
    if not (is_prime(p) and is_prime(q)):
        raise ValueError('Both numbers must be prime.')
    elif p == q:
        raise ValueError('p and q cannot be equal')
    #n = pq
    n = p * q

    #Phi is the totient of n
    phi = (p-1) * (q-1)

    #Choose an integer e such that e and phi(n) are coprime
    e = random.randrange(1, phi)

    #Use Euclid's Algorithm to verify that e and phi(n) are comprime
    g = gcd(e, phi)
    while g != 1:
        e = random.randrange(1, phi)
        g = gcd(e, phi)

    #Use Extended Euclid's Algorithm to generate the private key
    d = multiplicative_inverse(e, phi)

    #Return public and private keypair
    #Public key is (e, n) and private key is (d, n)
    return ((e, n), (d, n))

def encrypt(pk, plaintext):
    #Unpack the key into it's components
    key, n = pk
    #Convert each letter in the plaintext to numbers based on the character using a^b mod m
    cipher = [(ord(char) ** key) % n for char in plaintext]
    #Return the array of bytes
    return cipher

def decrypt(pk, ciphertext):
    #Unpack the key into its components
    key, n = pk
    #Generate the plaintext based on the ciphertext and key using a^b mod m
    plain = [chr((char ** key) % n) for char in ciphertext]
    #Return the array of bytes as a string
    return ''.join(plain)


if __name__ == '__main__':
    '''
    Detect if the script is being run directly by the user
    '''
    print "RSA Encrypter/ Decrypter"
    p = int(raw_input("Enter a prime number (17, 19, 23, etc): "))
    q = int(raw_input("Enter another prime number (Not one you entered above): "))
    print "Generating your public/private keypairs now . . ."
    public, private = generate_keypair(p, q)
    print "Your public key is ", public ," and your private key is ", private
    message = raw_input("Enter a message to encrypt with your private key: ")
    encrypted_msg = encrypt(private, message)
    print "Your encrypted message is: "
    print ''.join(map(lambda x: str(x), encrypted_msg))
    print "Decrypting message with public key ", public ," . . ."
    print "Your message is:"
    print decrypt(public, encrypted_msg)

Thank you for your code.

This is SEVERELY broken

https://www.0xkasper.com/articles/hacktm-ctf-2020-qualifiers-write-up#rsa-is-easy-1

DO NOT USE FOR ANYTHING!

can you please rewrite the program to generate the prime number automatically without enter by user thank you, and another thing how i can add the value of public key into the encrypts message?

thank you for your code

Do you still need the prime number automatic generator? @balasmeh

This is SEVERELY broken

https://www.0xkasper.com/articles/hacktm-ctf-2020-qualifiers-write-up#rsa-is-easy-1

DO NOT USE FOR ANYTHING!

This wasnt my code, i forked it from someone and fixed few issues. any way let me know what the issues are please @stackola thanks

you are welcome @Bharghav-Baddam

@mj2266 yes please I'm interested of it I would be thankful for you

Also please I'm looking for another help from you I want to make it as client server, where client send hi and server response

@balasmeh you can contact me here https://www.linkedin.com/in/manjyotsingh-nanra-6a4999152/

does this one work on python3. Thanks