Question

use ELGMAL algorithm.

let p = 11. find a generator number for p in case of elgmal algorithm. alice selects an integer number x = 5. calculate public and private key for alice in elgmal algorithm. alice wants to send plaintext "AGE" to Bob. assume that Alice selects random k values as 6, 4, 7 respectively for encryption. what is the ciphertext???

Answer 1

ANS :a)

CODE FOR IMPLEMENTATION :

import random

from math import pow

  

a = random.randint(2, 10)

  

def gcd(a, b):

    if a < b:

        return gcd(b, a)

    elif a % b == 0:

        return b;

    else:

        return gcd(b, a % b)

  

# Generating large random numbers

def gen_key(q):

  

    key = random.randint(pow(10, 20), q)

    while gcd(q, key) != 1:

        key = random.randint(pow(10, 20), q)

  

    return key

  

# Modular exponentiation

def power(a, b, c):

    x = 1

    y = a

  

    while b > 0:

        if b % 2 == 0:

            x = (x * y) % c;

        y = (y * y) % c

        b = int(b / 2)

  

    return x % c

  

# Asymmetric encryption

def encrypt(msg, q, h, g):

  

    en_msg = []

  

    k = gen_key(q)# Private key for sender

    s = power(h, k, q)

    p = power(g, k, q)

      

    for i in range(0, len(msg)):

        en_msg.append(msg[i])

  

    print("g^k used : ", p)

    print("g^ak used : ", s)

    for i in range(0, len(en_msg)):

        en_msg[i] = s * ord(en_msg[i])

  

    return en_msg, p

  

def decrypt(en_msg, p, key, q):

  

    dr_msg = []

    h = power(p, key, q)

    for i in range(0, len(en_msg)):

        dr_msg.append(chr(int(en_msg[i]/h)))

          

    return dr_msg

  

# Driver code

def main():

  

    msg = 'encryption'

    print("Original Message :", msg)

  

    q = random.randint(pow(10, 20), pow(10, 50))

    g = random.randint(2, q)

  

    key = gen_key(q)# Private key for receiver

    h = power(g, key, q)

    print("g used : ", g)

    print("g^a used : ", h)

  

    en_msg, p = encrypt(msg, q, h, g)

    dr_msg = decrypt(en_msg, p, key, q)

    dmsg = ''.join(dr_msg)

    print("Decrypted Message :", dmsg);

  

  

if __name__ == '__main__':

    main()

n this cryptosystem, original message M is masked by multiplying g^ak to it. To remove the mask, a clue is given in form of g^k. Unless someone knows a, he will not be able to retrieve M. This is because of finding discrete log in an cyclic group is difficult and simplying knowing g^a and g^k is not good enough to compute g^ak.

** VALUES CAN BE INPUTTED IN THE ABOVE CODE FOR OPTIMIZATION.