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This assignment involves multiple cryptographic scenarios, including the Diffie-Hellman key exchange, analysis of a login protocol's security flaws, and RSA key generation, encryption, and decryption. The tasks encompass detailed step-by-step calculations, explanations of protocol weaknesses, and methods to enhance security. Below is a comprehensive analysis and solution for each component.

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Diffie-Hellman Key Exchange Calculation

The Diffie-Hellman key exchange allows two parties, Alan and Bill, to generate a shared secret over an insecure channel, based on agreed public parameters. Given q=7 (a prime) and generator g=2, Alan randomly chooses a private number CA=5, which he uses to compute his public value DA. Bill randomly chooses CB=6, which he uses to compute his public value DB. The shared secret is then computed using these public values.

Part a: Calculation of DA

DA is computed as g^CA mod q. Applying the values:

DA = 2^5 mod 7

Calculating 2^5:

2^5 = 32

Then, 32 mod 7:

32 ÷ 7 = 4 with a remainder of 4

Therefore, DA = 4

Part b: Calculation of DB

Similarly, DB is calculated as g^CB mod q:

DB = 2^6 mod 7

Calculating 2^6:

2^6 = 64

Calculating 64 mod 7:

64 ÷ 7 = 9 with a remainder of 1

Thus, DB = 1

Part c: Computing the Common Secret Key

The secret key for Alan is (DB)^CA mod q:

Key_A = 1^5 mod 7 = 1

The secret key for Bill is (DA)^CB mod q:

Key_B = 4^6 mod 7

Calculating 4^6:

4^6 = (4^3)^2

First, 4^3:

4^3 = 64

64 mod 7:

64 ÷ 7 = 9 remainder 1, so 64 mod 7 = 1

Now, (64)^2:

(64)^2 = 4096

4096 mod 7:

4096 ÷ 7 = 585 with a remainder of 1, since 7 × 585=4095

Thus, 4096 mod 7 = 1

Therefore, Key_B = 1^6 mod 7 = 1

The shared secret key is 1, which both parties obtain independently, confirming the correctness of the key exchange.

Analysis of the Login Protocol

The described login protocol involves the user computing X = Hash(P) XOR Hash(R) and submitting X to the machine, which then verifies the login by recomputing Y = Hash(P) XOR Hash(R) and comparing with X. The protocol aims to authenticate a user without transmitting the password directly.

Part a: Security Flaws and Vulnerabilities

The core weakness lies in the fact that both X and Y are XORed with Hash(P) and Hash(R). Since Hash(R) is sent only once in the form of X, an attacker who intercepts X can perform a replay attack or extract information about Hash(P). Specifically, because Hash(R) is used as a one-time secret during this process, but the protocol does not establish a challenge-response mechanism, it is vulnerable to replay attacks. An attacker can capture a valid X value and replay it later, gaining access without knowing the password. Furthermore, the protocol does not generate a session-specific nonce or timestamp, increasing its vulnerability to replay and impersonation attacks.

Part b: Strengthening the Protocol

To mitigate these vulnerabilities, one effective approach is to introduce a challenge-response mechanism with nonces and timestamps. For example, before user input, the server generates a unique, random challenge R' and sends it to the user. The user then computes X' = Hash(P) XOR Hash(R') and transmits X' back. The server verifies the correctness of the response by recomputing Y' = Hash(P) XOR Hash(R') and comparing it with the received X'. Since R' is unique per session, replay attacks are thwarted. Additionally, incorporating timestamps or session tokens can prevent the reuse of stale data and further enhance security.

RSA Key Generation, Encryption, and Decryption

Using prime numbers p=13 and q=17, we can generate RSA keys, encrypt a message, and decrypt it as follows.

Part a: Private Key Generation

The modulus n is p × q = 13 × 17 = 221. The totient φ(n) = (p-1)(q-1) = 12 × 16 = 192. Given the public exponent e=5, we need to find the private exponent d such that e × d ≡ 1 mod φ(n).

• Find d such that (5 × d) mod 192 = 1.

Using the Extended Euclidean Algorithm:

To find d:

- Solve 5d ≡ 1 mod 192

- 5d - 192k = 1 for integers d, k

Applying the Extended Euclidean Algorithm:

gcd(5, 192):

192 = 5×38 + 2

5 = 2×2 + 1

2 = 1×2 + 0

Back substitution:

1 = 5 - 2×2

But, 2 = 192 - 5×38

Thus:

1 = 5 - (192 - 5×38)×2 = 5 - 192×2 + 5×76 = 5×77 - 192×2

Therefore, d ≡ 77 mod 192

Hence, the private key component d = 77.

Part b: Encryption of Message M=25

The encryption function: C = M^e mod n

Calculating C = 25^5 mod 221

Calculate 25^5:

- 25^1 = 25

- 25^2 = 625

- 25^3 = 625×25=15,625

- 25^4 = 15,625×25=390,625

- 25^5 = 390,625×25=9,765,625

Now, reduce modulo 221:

Using modular exponentiation to avoid large numbers, employ successive squaring:

• 25^1 ≡ 25 mod 221
• 25^2 ≡ 625 mod 221

  625 ÷ 221 = 2, remainder 183, so 625 mod 221 = 183

• 25^4 = (25^2)^2 ≡ 183^2 mod 221

  183^2 = 33,489

  33,489 ÷ 221 ≈ 151, remainder: 151×221=33,371; 33,489 - 33,371=118

  So 183^2 ≡118 mod 221

Now, 25^5 = 25^4 × 25^1 ≡ 118×25=2,950 mod 221

2,950 ÷ 221 ≈ 13, with remainder: 13×221=2,873

2,950 - 2,873=77

Thus, the ciphertext C=77.

Decrypting the Message

The decryption function: M = C^d mod n = 77^77 mod 221.

This computation is complex; applying successive modular exponentiation via efficient algorithms yields the original message as 25. For brevity, the process involves binary exponentiation or software tools to compute 77^77 mod 221, resulting in M=25.

Conclusion

This comprehensive review illustrates the implementation of Diffie-Hellman key exchange, highlights vulnerabilities in simple authentication protocols and suggests improvements, and demonstrates RSA key generation along with encryption and decryption procedures. Ensuring cryptographic security requires meticulous protocol design and rigorous validation of cryptographic primitives, as exemplified in these scenarios.

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