2078

CSC327 · TU past paper

Cryptography 2078 question paper

The complete TU 2078 exam paper for Cryptography (CSC327), all 12 questions with solved model answers written to the mark scheme.

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  1. 1Double DESAnswer

    CIA Triad, Double & Triple DES, and S-Box in DES

    --- The CIA Triad is the foundational model of information security consisting of three core principles: "Computer Security is the protection afforded to an automated information system in order to attain the applicable objectives of pre...

  2. 210 marksNumericalDigital SignaturesAnswer

    Explain the generic model of digital signature process. Consider the two prime numbers 7 and 19. Select 29 as public key and 41 as private key. Encrypt the plaintext 4 and decrypt the cipher text 3 using RSA.[10]

    Digital Signature and RSA

    Part 1: Generic Model of Digital Signature Process

    A digital signature uses asymmetric (public key) cryptography to provide authentication, integrity, and non-repudiation. The sender signs using their private key; the receiver verifies using the sender's public key.

    Model

    Sender side:

    1. Message: Sender has the original plaintext message $M$.
    2. Hashing: A hash function $H$ (e.g., SHA-1, MD5) is applied to $M$ to produce a fixed-length message digest $h = H(M)$.
    3. Signing: The digest is encrypted with the sender's private key to form the digital signature $S = E_{PR_{sender}}(h)$.
    4. Transmission: The message $M$ and signature $S$ are sent together.

    Receiver side: 5. Recompute hash: Receiver applies the same hash function to the received message: $h_1 = H(M)$. 6. Decrypt signature: Receiver decrypts $S$ with the sender's public key to recover $h_2 = D_{PU_{sender}}(S)$. 7. Compare: If $h_1 = h_2$, the message is authentic and unaltered; otherwise it has been tampered with or is forged.

    Sender:   M --H--> h --Encrypt(PR_sender)--> S ;  send (M, S)
    Receiver: M --H--> h1
              S --Decrypt(PU_sender)--> h2
              compare h1 == h2 ? valid : invalid
    

    Properties provided: Authentication, Integrity, Non-repudiation.


    Part 2: RSA

    Given data

    • $p = 7$, $q = 19$
    • Public key $e = 29$
    • Private key $d = 41$
    • Plaintext to encrypt: $M = 4$
    • Ciphertext to decrypt: $C = 3$

    Step 1: Parameters

    $$n = p \times q = 7 \times 19 = 133$$ $$\phi(n) = (p-1)(q-1) = 6 \times 18 = 108$$

    Check: $e \times d = 29 \times 41 = 1189$, and $1189 \mod 108 = 1189 - 11\times108 = 1189-1188 = 1$. ✓ Keys valid.

    Step 2: Encrypt $M = 4$

    $$C = M^e \bmod n = 4^{29} \bmod 133$$

    Successive squaring mod 133:

    • $4^1 = 4$
    • $4^2 = 16$
    • $4^4 = 16^2 = 256 \equiv 123$
    • $4^8 = 123^2 = 15129 \equiv 6$ (since $15129 - 113\times133 = 15129-15029=100$... recheck)

    Recheck $4^8$: $15129 \div 133 = 113.75$, $113\times133 = 15029$, remainder $= 100$. So $4^8 \equiv 100$, not 6.

    Redo carefully:

    • $4^4 \equiv 123$
    • $4^8 = 123^2 = 15129$; $133\times113 = 15029$; $15129-15029 = 100 \Rightarrow 4^8 \equiv 100$
    • $4^{16} = 100^2 = 10000$; $133\times75 = 9975$; $10000-9975 = 25 \Rightarrow 4^{16} \equiv 25$

    Binary: $29 = 16+8+4+1$, so $$4^{29} = 4^{16}\cdot 4^{8}\cdot 4^{4}\cdot 4^{1} \equiv 25 \times 100 \times 123 \times 4 \pmod{133}$$

    Step by step:

    • $25 \times 100 = 2500;\ 2500 - 18\times133 = 2500 - 2394 = 106$
    • $106 \times 123 = 13038;\ 13038 - 98\times133 = 13038 - 13034 = 4$
    • $4 \times 4 = 16$

    $$\boxed{C = 4^{29} \bmod 133 = 16}$$

    Step 3: Decrypt $C = 3$

    $$M = C^d \bmod n = 3^{41} \bmod 133$$

    Successive squaring mod 133:

    • $3^1 = 3$
    • $3^2 = 9$
    • $3^4 = 81$
    • $3^8 = 81^2 = 6561;\ 6561 - 49\times133 = 6561-6517 = 44$
    • $3^{16} = 44^2 = 1936;\ 1936 - 14\times133 = 1936-1862 = 74$
    • $3^{32} = 74^2 = 5476;\ 5476 - 41\times133 = 5476-5453 = 23$

    Binary: $41 = 32 + 8 + 1$, so $$3^{41} = 3^{32}\cdot 3^{8}\cdot 3^{1} \equiv 23 \times 44 \times 3 \pmod{133}$$

    • $23 \times 44 = 1012;\ 1012 - 7\times133 = 1012 - 931 = 81$
    • $81 \times 3 = 243;\ 243 - 133 = 110$

    $$\boxed{M = 3^{41} \bmod 133 = 110}$$

    Final Results

    • Encryption of plaintext $4$: $C = 16$
    • Decryption of ciphertext $3$: $M = 110$
  3. 310 marksNumericalFinite FieldsAnswer

    Define Galois field with an example. Explain any two modes of block cipher encryption. Determine the quadratic residues of 7.[10]

    Galois Field, Block Cipher Modes, and Quadratic Residues

    Step 1 - Extract (Given Data)

    • Modulus for quadratic residues: $n = 7$
    • Tasks: (a) define Galois field with example, (b) explain any two block cipher modes, (c) find quadratic residues of 7.

    No numeric data is missing. All parts are solvable.


    Step 2 - Solve

    Part 1: Galois Field

    Definition: A Galois Field (Finite Field), denoted $GF(p^n)$, is a field containing a finite number of elements, satisfying all field axioms (closure, associativity, commutativity, additive and multiplicative identities, additive and multiplicative inverses for non-zero elements, and distributivity).

    The order (number of elements) must be a prime power $p^n$, where $p$ is prime and $n \geq 1$.

    • $GF(p)$: elements ${0, 1, \ldots, p-1}$ with arithmetic mod $p$.
    • $GF(2^8)$ is used in AES.

    Example: $GF(7) = {0, 1, 2, 3, 4, 5, 6}$ (arithmetic mod 7)

    Addition: $5 + 4 = 9 \equiv 2 \pmod 7$

    Multiplication: $3 \times 5 = 15 \equiv 1 \pmod 7$

    Multiplicative inverse: since $3 \times 5 \equiv 1 \pmod 7$, we have $3^{-1} = 5$.

    Every non-zero element has an inverse, so $GF(7)$ is a valid Galois field.


    Part 2: Two Modes of Block Cipher Encryption

    Mode 1: Electronic Codebook (ECB)

    • Plaintext is split into fixed-size blocks ($P_1, P_2, \ldots$).
    • Each block is encrypted independently with the same key $K$: $$C_i = E(K, P_i)$$
    • Advantages: Simple, fast, fully parallelizable; random access to blocks.
    • Disadvantages: Identical plaintext blocks give identical ciphertext blocks, exposing data patterns; weak for structured data.

    Mode 2: Cipher Block Chaining (CBC)

    • Each plaintext block is XORed with the previous ciphertext block before encryption. An Initialization Vector (IV) seeds the first block: $$C_i = E(K,; P_i \oplus C_{i-1}), \qquad C_0 = IV$$
    • Decryption: $P_i = D(K, C_i) \oplus C_{i-1}$
    • Advantages: Identical plaintext blocks give different ciphertext (hides patterns); more secure than ECB.
    • Disadvantages: Encryption is sequential (not parallelizable); a transmission error in one block corrupts that block and the next during decryption.

    Part 3: Quadratic Residues of 7

    An integer $a$ is a quadratic residue mod $n$ if there exists $x$ with $x^2 \equiv a \pmod n$ and $\gcd(a,n)=1$.

    Compute $x^2 \bmod 7$ for $x = 1$ to $6$:

    $x$$x^2$$x^2 \bmod 7$
    111
    244
    392
    4162
    5254
    6361

    Distinct quadratic residues: $$QR(7) = {1, 2, 4}$$

    Non-residues: ${3, 5, 6}$

    Check with formula: For prime $p$, number of QRs $= \dfrac{p-1}{2} = \dfrac{7-1}{2} = 3$. ✓ (matches ${1,2,4}$)


    Final Answers

    • $GF(7) = {0,\ldots,6}$ under mod-7 arithmetic is the example Galois field.
    • ECB and CBC modes explained as above.
    • Quadratic residues of 7: ${1, 2, 4}$.
  4. 45 marksIntrusion Detection SystemAnswer

    What does intrusion mean? How the system detect intrusion? List any four types of firewall. [5]

    Intrusion refers to any unauthorized attempt to access, manipulate, or compromise a computer system, network, or data. It involves a set of actions that attempt to bypass the security mechanisms of a system to gain unauthorized access, s...

  5. 55 marksNumericalSubstitution TechniquesAnswer

    Decrypt the message 'GVPJ' using Hill cipher taking the key as {Row1 = 3 7, Row2= 5 12}. [5]

    • Ciphertext: GVPJ - Key matrix: $K = \begin{pmatrix} 3 & 7 \ 5 & 12 \end{pmatrix}$ - Letter encoding: A=0, B=1, ..., Z=25 - Decryption: $P = K^{-1} \cdot C \pmod{26}$ --- G V P J ------------ 6 21 15 9 Pairs (column vectors): $$C1 = \b...
  6. 65 marksPKI trust modelsAnswer

    Describe the PKI trust model. [5]

    PKI Trust Model

    Introduction

    Public Key Infrastructure (PKI) is a framework that manages the creation, distribution, storage, and revocation of digital certificates. The PKI trust model defines how trust is established and maintained among entities using public key certificates.


    Key Elements of the PKI (PKIX) Model

    The PKIX (Public Key Infrastructure X.509) model consists of the following elements:

    1. End Entity

    A generic term used to denote end users, devices, or any other entity that can be identified in the subject field of a public key certificate. End entities are the ultimate consumers of PKI services.

    2. Certification Authority (CA)

    • The trusted third party that issues and signs digital certificates.
    • The CA binds a public key to an identity by digitally signing the certificate with its own private key.
    • Trust in the PKI system ultimately rests on trust in the CA.

    3. Registration Authority (RA)

    • Acts as an intermediary between the end entity and the CA.
    • Verifies the identity of the end entity before the CA issues a certificate.

    4. Certificate Repository

    • A publicly accessible store (e.g., LDAP directory) where certificates and Certificate Revocation Lists (CRLs) are published and retrieved.

    5. Digital Certificates

    • A certificate binds a public key to an individual or entity.
    • It is signed by the CA, so any party that trusts the CA can verify the certificate.
    • Benefits include:
      • Authentication: The individual's unique private key was used to apply the signature, so recipients can be confident of the signer's identity.
      • Integrity: Even the slightest change to the original document causes signature verification to fail.
      • Confidentiality: Content encrypted with a public key can only be decrypted with the corresponding private key.

    How the Trust Model Works

    End Entity  <---->  Registration Authority (RA)
                              |
                              v
                      Certification Authority (CA)
                              |
                        Issues Certificate
                              |
                        Certificate Repository
                              |
                        Relying Party verifies
    
    1. An end entity requests a certificate, providing proof of identity to the RA.
    2. The RA validates the request and forwards it to the CA.
    3. The CA issues a signed digital certificate binding the entity's identity to its public key.
    4. The certificate is stored in a repository for public access.
    5. A relying party retrieves the certificate, verifies the CA's signature, and trusts the public key.

    Public Key Distribution in PKI

    Several techniques are used for distributing public keys:

    MethodDescriptionWeakness
    Public AnnouncementUsers broadcast their public keys (e.g., append to emails)Vulnerable to forgery
    Publicly Available DirectoryKeys stored in a trusted directoryDirectory must be trusted
    CA-based DistributionCA signs and vouches for public keysRelies on CA trustworthiness

    Summary

    The PKI trust model establishes a chain of trust rooted in a trusted CA. By using digital certificates, it ensures authentication, integrity, and confidentiality in public key cryptography, solving the key distribution problem in a scalable and verifiable manner.

  7. 75 marksAuthentication SystemAnswer

    Define authentication system. Illustrate the need of mutual authentication over one way authentication with an example. [5]

    --- An authentication system is a security mechanism that verifies the identity of a user, process, or device before granting access to a system or resource. It ensures that the entity claiming an identity is genuinely who they say they ...

  8. 85 marksNumericalNumber TheoryAnswer

    Find the value of $7^{2019} \mod 13$ using Fermat's Little theorem. Define Euler totient function with an example. [5]

    • Expression to evaluate: $7^{2019} \bmod 13$ - Base $a = 7$, modulus $p = 13$ (prime), exponent $= 2019$ - Task: define Euler totient function with an example --- Fermat's Little Theorem: If $p$ is prime and $\gcd(a,p)=1$, then $$a^{p-1...
  9. 95 marksProperties of Hash functionsAnswer

    List the properties of hash function. Discuss the first pass of MD4. [5]

    A hash function maps a message of any length into a fixed-length hash value, which serves as an authenticator. The key properties are: 1. One-way property (Pre-image resistance): It is computationally hard to find the input (message) fro...

  10. 105 marksNumericalSubstitution TechniquesAnswer

    Differentiate between Symmetric and Asymmetric cipher. Encrypt the message 'HELL' using the key 'FAIL' using Vernam cipher. [5]

    • Plaintext: HELL - Key: FAIL - Cipher: Vernam cipher - Letter mapping used: $A=0, B=1, \dots, Z=25$ --- Feature Symmetric Cipher Asymmetric Cipher --------- Keys used Single shared secret key for both encryption and decryption Two keys:...
  11. 115 marksNumericalFinite FieldsAnswer

    Divide $3x^2 + 4x + 3$ by $5x + 6$ over GF(7). [5]

    • Dividend: $3x^2 + 4x + 3$ - Divisor: $5x + 6$ - Field: $GF(7)$, all arithmetic mod 7. We need $5^{-1} \pmod 7$: since $5 \cdot 3 = 15 \equiv 1 \pmod 7$, we have $$5^{-1} = 3 \text{ in } GF(7)$$ $$\frac{3x^2}{5x} = 3 \cdot 5^{-1} x = 3 ...
  12. 125 marksSecure Socket LayerAnswer

    Define SSL protocol. Mention the services provided by PGP. [5]

    --- Definition: SSL (Secure Sockets Layer) is a protocol designed to provide secure communication over the Internet between a client and a server. It operates between the application layer and the transport layer (TCP/IP) and ensures tha...