We need to make some changes: separate the explanation from the code, add a subpart about the MITM and maybe to develop more the instructions

Let's say Alice and Bob want to exchange a secret over an insecure channel. In other words, anyone can read the messages they send, but the goal is to ensure that only Alice and Bob can calculate the secret key.

Diffie-Hellman key exchange provides a solution to this seemingly impossible task. Since code may be easier to understand than a detailed explanation, I'll provide it first:

import Crypto.Util.number as cun
import Crypto.Random.random as crr
class DiffieHellman:
def __init__(self, p: int):
self.p = p
self.g = 5
self.private_key = crr.randrange(2, p-1)
def public_key(self) -> int:
return pow(self.g, self.private_key, self.p)
def shared_key(self, other_public_key: int) -> int:
return pow(other_public_key, self.private_key, self.p)
p = cun.getPrime(512)
alice = DiffieHellman(p)
bob = DiffieHellman(p)
shared_key = bob.shared_key(alice.public_key())
assert shared_key == alice.shared_key(bob.public_key())

Here's a brief explanation of the code:

  • We choose a prime pp and a generator gFpg \in \mathbb{F}_p

  • Alice picks a private key aZp1a \in \mathbb{Z}_{p-1}

  • Bob picks a private key bZp1b \in \mathbb{Z}_{p-1}

  • Alice's public key is gamodpg^a \mod p

  • Bob's public key is gbmodpg^b \mod p

  • Their shared key is gab(ga)b(gb)a(modp)g^{ab} \equiv (g^a)^b \equiv (g^b)^a \pmod p

So anybody observing the messages sent between Alice and Bob would see p,g,ga,gbp, g, g^a, g^b, but they wouldn't be able to calculate the shared key gabg^{ab}.

This is because given gg and gag^a, it should be infeasible to calculate aa. If this sounds familiar, that's because it's the Discrete Log Problem.

The original paper can be found here. It uses the group of integers modulo a prime to perform the key exchange. In practice however, any group with a hard discrete log problem can be used.