The exercise involves a Diffie-Hellman key exchange with the following parameters:

Given:

• $\alpha = 10$
• $p = 223$
• Secret keys:
  • $x_A = 11$ for user A
  • $x_B = 13$ for user B

Part (a) Steps:

1. Compute the public keys for A and B:

   • Public key of A, $y_A$: $y_A = \alpha^{x_A} \mod p$
   • Public key of B, $y_B$: $y_B = \alpha^{x_B} \mod p$

2. Calculate the shared secret $K$:

   • The shared secret $K$ is calculated as: $K = y_B^{x_A} \mod p = y_A^{x_B} \mod p$

3. Calculate the ciphertext of the message $m = 159$:

   • A encrypts the message $m$ using $K$: $c = E_K(m) = m \cdot K \mod p$

Calculations:

1. Compute $y_A$: $y_A = 10^{11} \mod 223$

2. Compute $y_B$: $y_B = 10^{13} \mod 223$

3. Compute $K$: $K = y_B^{11} \mod 223 = y_A^{13} \mod 223$

4. Encrypt $m = 159$: $c = 159 \cdot K \mod 223$

Let me perform these calculations.The calculations are as follows:

1. Public key of A ($y_A$): $61$
2. Public key of B ($y_B$): $79$
3. Shared secret $K$: $204$
4. Ciphertext $c$ for the message $m = 159$: $101$

Would you like further details or have any questions?

Here are five related questions to expand on this topic:

1. How is the multiplicative inverse $K^{-1}$ used in the decryption process?
2. Can you explain the importance of the prime $p$ in Diffie-Hellman key exchange?
3. What are the security implications of choosing different values for $\alpha$ in Diffie-Hellman?
4. How would you decrypt the ciphertext $c = 101$ to retrieve the original message $m$?
5. Could we use a different key exchange protocol here, and how would it compare to Diffie-Hellman?

Tip: In Diffie-Hellman, selecting large prime values for $p$ and securely chosen private keys $x_A$ and $x_B$ enhances security against brute-force attacks.