CSC327 · TU past paper

Cryptography 2082 question paper

The complete TU 2082 exam paper for Cryptography (CSC327), all 12 questions with solved model answers written to the mark scheme.

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1NumericalFiestel Cipher StructureHideAnswer
Feistel Cipher Structure and AES Key Expansion
Feistel Cipher Structure and AES Key Expansion

Part A: Feistel Cipher Structure

A Feistel Cipher is a symmetric block cipher design that splits a plaintext block into two equal halves and processes them through several rounds of substitution and permutation using round sub-keys derived from the master key.

Working

A block of $2w$ bits is divided into $L_0$ (left) and $R_0$ (right).

For each round $i$ :

$L_i = R_{i-1}$ $R_i = L_{i-1} \oplus F(R_{i-1}, K_i)$

where $K_i$ is the round sub-key and $F$ is the round function (which need not be invertible).

Block Diagram (words)
- Split plaintext into $L_0$ and $R_0$ .
- Feed $R_0$ and $K_1$ into $F$ .
- XOR $F$ output with $L_0 \Rightarrow R_1$ .
- $R_0$ becomes $L_1$ (swap).
- Repeat for $n$ rounds; recombine $L_n, R_n$ into ciphertext.
Key Properties
- Same structure for encryption and decryption (sub-keys reversed for decryption).
- $F$ need not be invertible.
- Security grows with more rounds.
- Examples: DES (16 rounds), Blowfish, Camellia.
Part B: AES Key Expansion - First 4 Bytes After First Iteration

Given Data

Key (16 bytes) grouped into 4 words:

Word Bytes
$W_0$ 2B 7E 15 16
$W_1$ 28 AE D2 A6
$W_2$ AB F7 97 66
$W_3$ 01 02 03 04

S-Box row 0 provided (sufficient, since all bytes being substituted have upper nibble 0).

Formula

$W_4 = W_0 \oplus g(W_3)$

Step 1: RotWord of $W_3 = [01,02,03,04]$ : $[02,\ 03,\ 04,\ 01]$

Step 2: SubBytes (row 0 of S-Box):

Byte S-Box
02 → col 2 77
03 → col 3 7B
04 → col 4 F2
01 → col 1 7C

$[77,\ 7B,\ F2,\ 7C]$

Step 3: XOR Rcon(1) = [01,00,00,00]: $[77\oplus01,\ 7B,\ F2,\ 7C] = [76,\ 7B,\ F2,\ 7C]$

So $g(W_3) = [76, 7B, F2, 7C]$ .

Step 4: Compute $W_4 = W_0 \oplus g(W_3)$ :

$W_0$ $g(W_3)$ XOR
2B 76 5D
7E 7B 05
15 F2 E7
16 7C 6A

Verification of first byte: $2B = 0010\,1011$ , $76 = 0111\,0110$ , XOR $= 0101\,1101 = 5D$ . ✓

Final Answer

$\boxed{W_4 = [5D,\ 05,\ E7,\ 6A]}$

The first 4 bytes of the next key after the first iteration are 5D 05 E7 6A.
Feistel Cipher Structure and AES Key Expansion

Part A: Feistel Cipher Structure

A Feistel Cipher is a symmetric block cipher design that splits a plaintext block into two equal halves and processes them through several rounds of substitution and permutation using round sub-keys derived from the master key.

Working

A block of $2w$ bits is divided into $L_0$ (left) and $R_0$ (right).

For each round $i$:

$$L_i = R_{i-1}$$ $$R_i = L_{i-1} \oplus F(R_{i-1}, K_i)$$

where $K_i$ is the round sub-key and $F$ is the round function (which need not be invertible).

Block Diagram (words)
- Split plaintext into $L_0$ and $R_0$.
- Feed $R_0$ and $K_1$ into $F$.
- XOR $F$ output with $L_0 \Rightarrow R_1$.
- $R_0$ becomes $L_1$ (swap).
- Repeat for $n$ rounds; recombine $L_n, R_n$ into ciphertext.
Key Properties
- Same structure for encryption and decryption (sub-keys reversed for decryption).
- $F$ need not be invertible.
- Security grows with more rounds.
- Examples: DES (16 rounds), Blowfish, Camellia.
Part B: AES Key Expansion - First 4 Bytes After First Iteration

Given Data

Key (16 bytes) grouped into 4 words:

Word Bytes
$W_0$ 2B 7E 15 16
$W_1$ 28 AE D2 A6
$W_2$ AB F7 97 66
$W_3$ 01 02 03 04

S-Box row 0 provided (sufficient, since all bytes being substituted have upper nibble 0).

Formula

$$W_4 = W_0 \oplus g(W_3)$$

Step 1: RotWord of $W_3 = [01,02,03,04]$: $$[02,\ 03,\ 04,\ 01]$$

Step 2: SubBytes (row 0 of S-Box):

Byte S-Box
02 → col 2 77
03 → col 3 7B
04 → col 4 F2
01 → col 1 7C

$$[77,\ 7B,\ F2,\ 7C]$$

Step 3: XOR Rcon(1) = [01,00,00,00]: $$[77\oplus01,\ 7B,\ F2,\ 7C] = [76,\ 7B,\ F2,\ 7C]$$

So $g(W_3) = [76, 7B, F2, 7C]$.

Step 4: Compute $W_4 = W_0 \oplus g(W_3)$:

$W_0$ $g(W_3)$ XOR
2B 76 5D
7E 7B 05
15 F2 E7
16 7C 6A

Verification of first byte: $2B = 0010,1011$, $76 = 0111,0110$, XOR $= 0101,1101 = 5D$. ✓

Final Answer

$$\boxed{W_4 = [5D,\ 05,\ E7,\ 6A]}$$

The first 4 bytes of the next key after the first iteration are 5D 05 E7 6A.

Word	Bytes
$W_0$	2B 7E 15 16
$W_1$	28 AE D2 A6
$W_2$	AB F7 97 66
$W_3$	01 02 03 04

Byte	S-Box
02 → col 2	77
03 → col 3	7B
04 → col 4	F2
01 → col 1	7C

$W_0$	$g(W_3)$	XOR
2B	76	5D
7E	7B	05
15	F2	E7
16	7C	6A

Word	Bytes
$W_0$	2B 7E 15 16
$W_1$	28 AE D2 A6
$W_2$	AB F7 97 66
$W_3$	01 02 03 04

Byte	S-Box
02 → col 2	77
03 → col 3	7B
04 → col 4	F2
01 → col 1	7C

$W_0$	$g(W_3)$	XOR
2B	76	5D
7E	7B	05
15	F2	E7
16	7C	6A

210 marksNumericalMan-in-the-Middle AttackAnswer

Illustrate the man in middle attack in Diffie-Hellman key exchange protocol. Assume the prime number be 19 and 10 as its primitive root. Select 5 as private key and 4 as random integer. Find the cipher text of M = 2 using Elgamal crypto system.[10]

Man-in-the-Middle Attack on Diffie-Hellman & ElGamal Cryptosystem

Given Data

Prime number: $p = 19$
Primitive root: $g = 10$
Private key: $x = 5$
Random integer: $k = 4$
Message: $M = 2$

Part 1: Man-in-the-Middle Attack on Diffie-Hellman

Normal Diffie-Hellman

Public parameters: $p = 19$ , $g = 10$ .

Alice picks private $X_A$ , sends $Y_A = g^{X_A} \bmod p$ .
Bob picks private $X_B$ , sends $Y_B = g^{X_B} \bmod p$ .
Shared key: $K = Y_B^{X_A} \bmod p = Y_A^{X_B} \bmod p$ .

The Attack

An attacker Darth (D) sits between Alice and Bob. Since plain Diffie-Hellman has no authentication, Darth impersonates each side to the other.

Darth chooses two private keys $X_{D1}, X_{D2}$ and computes: $Y_{D1} = g^{X_{D1}} \bmod p, \qquad Y_{D2} = g^{X_{D2}} \bmod p$

Alice (A)                Darth (D)                Bob (B)
   |--- Y_A ------------->|                          |
   |                      |--- Y_D1 ---------------->|
   |                      |<-- Y_B ------------------|
   |<-- Y_D2 -------------|                          |

Steps:

Alice sends $Y_A$ ; Darth intercepts it.
Darth sends $Y_{D1}$ to Bob.
Bob sends $Y_B$ ; Darth intercepts it.
Darth sends $Y_{D2}$ to Alice.

Resulting keys:

Party	Key computed
Alice	$K_1 = Y_{D2}^{X_A} \bmod p$
Darth (with Alice)	$K_1 = Y_A^{X_{D2}} \bmod p$
Darth (with Bob)	$K_2 = Y_B^{X_{D1}} \bmod p$
Bob	$K_2 = Y_{D1}^{X_B} \bmod p$

Consequence: Alice shares $K_1$ with Darth (thinking it is Bob), Bob shares $K_2$ with Darth (thinking it is Alice). Darth decrypts, reads, possibly alters, and re-encrypts every message. The attack succeeds because DH does not authenticate the parties.

Part 2: ElGamal Cryptosystem

Step 1: Key Generation

Public key: $y = g^x \bmod p = 10^5 \bmod 19$

$10^2 = 100 \equiv 5 \pmod{19}$
$10^4 \equiv 5^2 = 25 \equiv 6 \pmod{19}$
$10^5 \equiv 6 \cdot 10 = 60 \equiv 60 - 57 = 3 \pmod{19}$

$\boxed{y = 3}$

Public key = $(p=19,\ g=10,\ y=3)$ ; Private key = $5$ .

Step 2: Encryption of $M = 2$ with $k = 4$

Compute $C_1$ : $C_1 = g^k \bmod p = 10^4 \bmod 19 = 6$

Compute $C_2$ : $y^k \bmod p = 3^4 \bmod 19 = 81 \bmod 19 = 5$ $C_2 = (y^k \cdot M)\bmod p = (5 \cdot 2)\bmod 19 = 10$

Final Answer

$\textbf{Ciphertext } (C_1, C_2) = (6,\ 10)$

Verification (Decryption)

$M = C_2 \cdot (C_1^{x})^{-1} \bmod p$

$6^5 \bmod 19$ : $6^2 = 36 \equiv 17$ ; $6^4 \equiv 17^2 = 289 \equiv 4$ ; $6^5 \equiv 4 \cdot 6 = 24 \equiv 5$ .
$5^{-1} \bmod 19 = 4$ (since $5 \cdot 4 = 20 \equiv 1$ ).
$M = 10 \cdot 4 \bmod 19 = 40 \bmod 19 = 2 \checkmark$

Recovered $M = 2$ , matching the original message.

Man-in-the-Middle Attack on Diffie-Hellman & ElGamal Cryptosystem

Given Data

Prime number: $p = 19$
Primitive root: $g = 10$
Private key: $x = 5$
Random integer: $k = 4$
Message: $M = 2$

Part 1: Man-in-the-Middle Attack on Diffie-Hellman

Normal Diffie-Hellman

Public parameters: $p = 19$, $g = 10$.

Alice picks private $X_A$, sends $Y_A = g^{X_A} \bmod p$.
Bob picks private $X_B$, sends $Y_B = g^{X_B} \bmod p$.
Shared key: $K = Y_B^{X_A} \bmod p = Y_A^{X_B} \bmod p$.

The Attack

An attacker Darth (D) sits between Alice and Bob. Since plain Diffie-Hellman has no authentication, Darth impersonates each side to the other.

csc327-dh-mitm

Darth chooses two private keys $X_{D1}, X_{D2}$ and computes: $$Y_{D1} = g^{X_{D1}} \bmod p, \qquad Y_{D2} = g^{X_{D2}} \bmod p$$

Alice (A)                Darth (D)                Bob (B)
   |--- Y_A ------------->|                          |
   |                      |--- Y_D1 ---------------->|
   |                      |<-- Y_B ------------------|
   |<-- Y_D2 -------------|                          |

Steps:

Alice sends $Y_A$; Darth intercepts it.
Darth sends $Y_{D1}$ to Bob.
Bob sends $Y_B$; Darth intercepts it.
Darth sends $Y_{D2}$ to Alice.

Resulting keys:

Party	Key computed
Alice	$K_1 = Y_{D2}^{X_A} \bmod p$
Darth (with Alice)	$K_1 = Y_A^{X_{D2}} \bmod p$
Darth (with Bob)	$K_2 = Y_B^{X_{D1}} \bmod p$
Bob	$K_2 = Y_{D1}^{X_B} \bmod p$

Part 2: ElGamal Cryptosystem

Step 1: Key Generation

Public key: $$y = g^x \bmod p = 10^5 \bmod 19$$

$10^2 = 100 \equiv 5 \pmod{19}$
$10^4 \equiv 5^2 = 25 \equiv 6 \pmod{19}$
$10^5 \equiv 6 \cdot 10 = 60 \equiv 60 - 57 = 3 \pmod{19}$

$$\boxed{y = 3}$$

Public key = $(p=19,\ g=10,\ y=3)$; Private key = $5$.

Step 2: Encryption of $M = 2$ with $k = 4$

Compute $C_1$: $$C_1 = g^k \bmod p = 10^4 \bmod 19 = 6$$

Compute $C_2$: $$y^k \bmod p = 3^4 \bmod 19 = 81 \bmod 19 = 5$$ $$C_2 = (y^k \cdot M)\bmod p = (5 \cdot 2)\bmod 19 = 10$$

Final Answer

$$\textbf{Ciphertext } (C_1, C_2) = (6,\ 10)$$

Verification (Decryption)

$$M = C_2 \cdot (C_1^{x})^{-1} \bmod p$$

$6^5 \bmod 19$: $6^2 = 36 \equiv 17$; $6^4 \equiv 17^2 = 289 \equiv 4$; $6^5 \equiv 4 \cdot 6 = 24 \equiv 5$.
$5^{-1} \bmod 19 = 4$ (since $5 \cdot 4 = 20 \equiv 1$).
$M = 10 \cdot 4 \bmod 19 = 40 \bmod 19 = 2 \checkmark$

Recovered $M = 2$, matching the original message.

310 marksProperties of Hash functionsHideAnswer
Describe the properties of hash functions. Discuss how hash value is generated using SHA-1 algorithm.[10]
A hash function is a function that maps a message of any length into a fixed-length hash value, which serves as the authenticator. Cryptographic hash functions play a fundamental role in efficient and secure information processing. Prope...
45 marksTypes of Malicious LogicHideAnswer
Describe any three types of malicious logic. [5]
Malicious Logic

Malicious Logic (Malicious Code / Rogue Programs)

Malicious logic refers to a set of instructions or programs that are intentionally designed to cause harm, unauthorized access, or disruption to a computer system or network.

1. Trojan Horse

A Trojan Horse is a program that appears to perform a legitimate or useful function but secretly contains hidden code that performs unauthorized or harmful actions when executed.
- It does not replicate itself (unlike a virus).
- It tricks the user into running it by disguising itself as a legitimate program.
- Example: A user downloads what appears to be a free game, but the program secretly installs a backdoor, allowing an attacker to gain remote access to the system.
- Damage: Data theft, unauthorized access, system damage.
2. Virus

A virus is a piece of malicious code that attaches itself to a legitimate program or file and replicates when that program is executed. It requires a host program to spread.
- It can spread from one computer to another through infected files, email attachments, or removable media.
- A virus typically has two phases:
  
  Infection phase: It attaches itself to other programs.
  
  Attack phase: It performs its malicious activity (deleting files, corrupting data, etc.).
- Example: A virus attached to a .exe file that corrupts system files when executed.
- Damage: File corruption, system crashes, data loss.
3. Worm

A worm is a self-contained, self-replicating program that spreads across networks without needing to attach itself to a host program. Unlike a virus, it does not require user interaction to propagate.
- It exploits vulnerabilities in operating systems or network services to spread automatically.
- Worms consume network bandwidth and system resources, causing slowdowns or crashes.
- Example: The Morris Worm (1988) spread across the Internet by exploiting UNIX vulnerabilities, causing widespread system slowdowns.
- Damage: Network congestion, system resource exhaustion, data theft.
Summary Table

Type Replicates? Needs Host? Spreads Via
Trojan Horse No Yes (disguise) User deception
Virus Yes Yes (host file) Infected files/media
Worm Yes No Network/vulnerabilities

Other types of malicious logic include Logic Bombs (code that triggers under specific conditions) and Trapdoors/Backdoors (hidden entry points left by developers for unauthorized access).
55 marksNumericalSubstitution TechniquesHideAnswer
Given the key "HELLOWORLD", encrypt the plaintext "TURINGTEST" using Play fair cipher. [5]
Playfair Cipher: Encrypting "TURINGTEST" with Key "HELLOWORLD"

STEP 1 - EXTRACT (Given data)
- Key: HELLOWORLD
- Plaintext: TURINGTEST
- Cipher: Playfair (5x5 matrix, I/J combined)
STEP 2 - SOLVE

Build the 5x5 Matrix

Key letters, removing duplicates (keep first occurrence): H, E, L, O, W, R, D
(HELLOWORLD → H, E, L, [L dup], O, W, [O dup], R, [L dup], D)

Fill remaining alphabet (I/J together): A, B, C, F, G, I/J, K, M, N, P, Q, S, T, U, V, X, Y, Z

C1 C2 C3 C4 C5
R1 H E L O W
R2 R D A B C
R3 F G I/J K M
R4 N P Q S T
R5 U V X Y Z

Split into digrams

TURINGTEST → TU | RI | NG | TE | ST

No double letters within a pair, length is even, so no padding needed.

Apply rules

TU: T(R4,C5), U(R5,C1) → rectangle
T → (R4, C1) = N; U → (R5, C5) = Z → NZ

RI: R(R2,C1), I(R3,C3) → rectangle
R → (R2, C3) = A; I → (R3, C1) = F → AF

NG: N(R4,C1), G(R3,C2) → rectangle
N → (R4, C2) = P; G → (R3, C1) = F → PF

TE: T(R4,C5), E(R1,C2) → rectangle
T → (R4, C2) = P; E → (R1, C5) = W → PW

ST: S(R4,C4), T(R4,C5) → same row (shift right, wrap)
S → C5 = T; T → C1 (wrap) = N → TN

Final Ciphertext

Plain pair Cipher pair
TU NZ
RI AF
NG PF
TE PW
ST TN

$\boxed{\text{Ciphertext} = \text{NZAFPFPWTN}}$
Playfair Cipher: Encrypting "TURINGTEST" with Key "HELLOWORLD"

STEP 1 - EXTRACT (Given data)
- Key: HELLOWORLD
- Plaintext: TURINGTEST
- Cipher: Playfair (5x5 matrix, I/J combined)
STEP 2 - SOLVE

Build the 5x5 Matrix

Key letters, removing duplicates (keep first occurrence): H, E, L, O, W, R, D
(HELLOWORLD → H, E, L, [L dup], O, W, [O dup], R, [L dup], D)

Fill remaining alphabet (I/J together): A, B, C, F, G, I/J, K, M, N, P, Q, S, T, U, V, X, Y, Z

C1 C2 C3 C4 C5
R1 H E L O W
R2 R D A B C
R3 F G I/J K M
R4 N P Q S T
R5 U V X Y Z

Split into digrams

TURINGTEST → TU | RI | NG | TE | ST

No double letters within a pair, length is even, so no padding needed.

Apply rules

TU: T(R4,C5), U(R5,C1) → rectangle
T → (R4, C1) = N; U → (R5, C5) = Z → NZ

RI: R(R2,C1), I(R3,C3) → rectangle
R → (R2, C3) = A; I → (R3, C1) = F → AF

NG: N(R4,C1), G(R3,C2) → rectangle
N → (R4, C2) = P; G → (R3, C1) = F → PF

TE: T(R4,C5), E(R1,C2) → rectangle
T → (R4, C2) = P; E → (R1, C5) = W → PW

ST: S(R4,C4), T(R4,C5) → same row (shift right, wrap)
S → C5 = T; T → C1 (wrap) = N → TN

Final Ciphertext

Plain pair Cipher pair
TU NZ
RI AF
NG PF
TE PW
ST TN

$$\boxed{\text{Ciphertext} = \text{NZAFPFPWTN}}$$
65 marksModes of Block Cipher EncryptionsHideAnswer
Explain any two modes of block cipher encryption. [5]
A block cipher encrypts data in fixed-size blocks. When a message is longer than one block, a mode of operation defines how the cipher is applied repeatedly. Two important modes are described below. --- Definition: In ECB mode, the plain...
75 marksDigital Signature StandardHideAnswer
Describe the working mechanism of digital signature algorithm. [5]
A Digital Signature is a cryptographic mechanism where content is digitally signed with an individual's private key and verified using the individual's public key. It provides authentication, integrity, and non-repudiation. --- - Signing...
85 marksNumber TheoryHideAnswer
State Fermat's theorem with example. What is the implication of discrete logarithm? [5]
Statement: If p is a prime number and a is a positive integer such that gcd(a, p) = 1 (i.e., a is not divisible by p), then: $a^{p-1} \equiv 1 \pmod{p}$ An alternative (and equally important) form of Fermat's theorem is: $$a^p \equiv a...
Statement: If p is a prime number and a is a positive integer such that gcd(a, p) = 1 (i.e., a is not divisible by p), then: $$a^{p-1} \equiv 1 \pmod{p}$$ An alternative (and equally important) form of Fermat's theorem is: $$a^p \equiv a...

Type	Replicates?	Needs Host?	Spreads Via
Trojan Horse	No	Yes (disguise)	User deception
Virus	Yes	Yes (host file)	Infected files/media
Worm	Yes	No	Network/vulnerabilities

	C1	C2	C3	C4	C5
R1	H	E	L	O	W
R2	R	D	A	B	C
R3	F	G	I/J	K	M
R4	N	P	Q	S	T
R5	U	V	X	Y	Z

Plain pair	Cipher pair
TU	NZ
RI	AF
NG	PF
TE	PW
ST	TN

	C1	C2	C3	C4	C5
R1	H	E	L	O	W
R2	R	D	A	B	C
R3	F	G	I/J	K	M
R4	N	P	Q	S	T
R5	U	V	X	Y	Z

Plain pair	Cipher pair
TU	NZ
RI	AF
NG	PF
TE	PW
ST	TN

95 marksFirewalls and their typesAnswer

List and explain the types of firewall. [5]

Types of Firewall

Firewall

A firewall is a network security system that monitors and controls incoming and outgoing network traffic based on predetermined security rules. It establishes a barrier between a trusted internal network and untrusted external networks.

Types of Firewall

1. Packet Filtering Firewall

Operates at the Network Layer (Layer 3) of the OSI model.
Inspects each packet passing through the network and filters them based on rules such as source IP, destination IP, port numbers, and protocols.
It does not examine the content of the packet.
Advantage: Fast and efficient.
Disadvantage: Cannot detect application-level attacks; stateless (does not track connection state).

2. Stateful Inspection Firewall (Dynamic Packet Filtering)

Operates at the Network and Transport Layer.
Tracks the state of active connections and makes filtering decisions based on the context of traffic (e.g., whether a packet is part of an established connection).
Maintains a state table to monitor ongoing sessions.
Advantage: More secure than simple packet filtering.
Disadvantage: Slower than packet filtering; cannot inspect application-layer data.

3. Application Layer Firewall (Proxy Firewall)

Operates at the Application Layer (Layer 7).
Acts as an intermediary (proxy) between the client and the server; it intercepts all messages entering and leaving the network.
Can inspect the full content of network traffic including HTTP, FTP, DNS, etc.
Advantage: Provides deep packet inspection; hides internal network details.
Disadvantage: Slower performance due to deep inspection; requires more resources.

4. Circuit-Level Gateway

Operates at the Session Layer (Layer 5).
Monitors TCP handshaking and session establishment to determine whether a requested session is legitimate.
Does not inspect the actual content of the packets.
Advantage: Faster than application-layer firewalls; hides internal network information.
Disadvantage: Cannot filter individual packets after the session is established.

5. Next-Generation Firewall (NGFW)

Combines the features of traditional firewalls with advanced capabilities such as:
- Deep Packet Inspection (DPI)
- Intrusion Prevention System (IPS)
- Application awareness and control
- SSL/TLS inspection
Can identify and block sophisticated attacks including malware and application-layer threats.
Advantage: Comprehensive security; highly effective against modern threats.
Disadvantage: Expensive and complex to configure.

Summary Table

Type	OSI Layer	Key Feature
Packet Filtering	Network (L3)	Filters by IP/port rules
Stateful Inspection	Network/Transport	Tracks connection state
Application Layer (Proxy)	Application (L7)	Deep content inspection
Circuit-Level Gateway	Session (L5)	Monitors TCP handshake
Next-Generation Firewall	Multiple Layers	DPI + IPS + App control

105 marksChallenge Response SystemAnswer

Define authentication system. Discuss about challenge response system. [5]

Authentication System and Challenge Response System

Authentication System

An authentication system is a security measure put in place to secure data and systems by requiring additional input beyond username and password for users to access a system.

Components of an Authentication System

An authentication system consists of five components required for the overall authentication process:

Component	Description
Authentication Information (A)	Information that provides identity (e.g., password, PIN)
Complementary Information (C)	Information stored on the computer used to validate authentication information
Complementation Function (F)	Function that generates complementary information from authentication information
Authentication Function (I)	Function that verifies and provides identity
Selection Function (S)	Function enabling an entity to create or alter information A or C

Types of Authentication Factors

Something the individual knows: Password, PIN
Something the individual possesses: Smart cards, cryptographic keys
Something the individual is: Fingerprint, retina, face recognition
Something the individual does: Voice pattern, handwriting recognition

Challenge Response System

A challenge response system is a handshake authentication process in which the authenticator issues a challenge to the user seeking authentication, and the user must provide a correct response in order to be authenticated.

Working Process

The process works as follows:

Step 1: User requests access to the system
Step 2: Server (authenticator) issues a CHALLENGE to the user
Step 3: User responds to the challenge
Step 4: Server verifies the response
         - If CORRECT  --> User is AUTHENTICATED
         - If INCORRECT --> User is DENIED

Forms of Challenge

The challenge may take many forms depending on the system:

A password request
A set of security questions that were answered by the user during first-time registration
A random number or token that the user must process using a shared secret key

Example (One-Way Authentication using Challenge Response)

Assume Alice and Bob share a secret key K_AB:

Alice                              Bob
  |                                  |
  |--- "I am Alice" --------------->|
  |                                  |
  |<-- Challenge: R1 ---------------|
  |                                  |
  |--- Response: K_AB(R1) -------->|
  |                                  |
  |     Bob verifies K_AB(R1)        |
  |     If correct --> Authenticated |

Alice sends her identity claim: "I am Alice"
Bob issues a challenge (random number R1)
Alice encrypts R1 using the shared key K_AB and sends it back
Bob decrypts and verifies the response
If the response matches, Alice is authenticated

Advantages

Prevents replay attacks since a new challenge is issued each time
More secure than simple password-based authentication
The actual secret (key/password) is never transmitted directly over the network

Summary: An authentication system verifies the identity of a user through its five components (A, C, F, I, S). The challenge response system strengthens this by requiring the user to prove knowledge of a secret by correctly responding to a server-issued challenge, making it resistant to eavesdropping and replay attacks.

11NumericalSubstitution TechniquesHideAnswer
Decrypt the ciphertext 'HI' using Hill Cipher where the key is. Ciphertext: HI, Key: $\begin{bmatrix} 5 & 3 \\ 4 & 3 \end{bmatrix}$
Decrypt the ciphertext 'HI' using Hill Cipher where the key is. Ciphertext: HI, Key: $$\begin{bmatrix} 5 & 3 \ 4 & 3 \end{bmatrix}$$
Hill Cipher Decryption of Ciphertext 'HI'

Given Data
- Ciphertext: HI
- Key Matrix:
$K = \begin{bmatrix} 5 & 3 \\ 4 & 3 \end{bmatrix}$
- Alphabet mapping: A=0, B=1, ..., Z=25, modulus 26.
Step 1: Convert Ciphertext to Numbers
- H = 7
- I = 8
$C = \begin{bmatrix} 7 \\ 8 \end{bmatrix}$

Step 2: Decryption Formula

$P = K^{-1} C \pmod{26}$

Step 3: Determinant of K

$\det(K) = (5)(3) - (3)(4) = 15 - 12 = 3$

Step 4: Inverse of Determinant mod 26

$3 \cdot d^{-1} \equiv 1 \pmod{26}$

$3 \times 9 = 27 \equiv 1 \pmod{26} \Rightarrow d^{-1} = 9$

Step 5: Adjugate of K

$\text{adj}(K) = \begin{bmatrix} 3 & -3 \\ -4 & 5 \end{bmatrix} \equiv \begin{bmatrix} 3 & 23 \\ 22 & 5 \end{bmatrix} \pmod{26}$

Step 6: Compute K⁻¹

$K^{-1} = 9 \cdot \begin{bmatrix} 3 & 23 \\ 22 & 5 \end{bmatrix} \pmod{26} = \begin{bmatrix} 27 & 207 \\ 198 & 45 \end{bmatrix} \pmod{26}$
- $27 \bmod 26 = 1$
- $207 \bmod 26 = 207 - 182 = 25$
- $198 \bmod 26 = 198 - 182 = 16$
- $45 \bmod 26 = 19$
$K^{-1} = \begin{bmatrix} 1 & 25 \\ 16 & 19 \end{bmatrix}$

Check: $K^{-1}K \pmod{26}$ :
- $(1)(5)+(25)(4)=5+100=105 \equiv 1$
- $(1)(3)+(25)(3)=3+75=78 \equiv 0$
- $(16)(5)+(19)(4)=80+76=156 \equiv 0$
- $(16)(3)+(19)(3)=48+57=105 \equiv 1$
Gives identity: inverse is correct.

Step 7: Multiply K⁻¹ by C

$P = \begin{bmatrix} 1 & 25 \\ 16 & 19 \end{bmatrix}\begin{bmatrix} 7 \\ 8 \end{bmatrix} \pmod{26}$
- Row 1: $7 + 200 = 207 \equiv 207 - 182 = 25$
- Row 2: $112 + 152 = 264 \equiv 264 - 260 = 4$
$P = \begin{bmatrix} 25 \\ 4 \end{bmatrix}$

Step 8: Convert to Letters
- 25 = Z
- 4 = E
Result

$\boxed{\text{Plaintext} = \textbf{ZE}}$

The ciphertext 'HI' decrypts to 'ZE'.
Hill Cipher Decryption of Ciphertext 'HI'

Given Data
- Ciphertext: HI
- Key Matrix:
$$K = \begin{bmatrix} 5 & 3 \ 4 & 3 \end{bmatrix}$$
- Alphabet mapping: A=0, B=1, ..., Z=25, modulus 26.
Step 1: Convert Ciphertext to Numbers
- H = 7
- I = 8
$$C = \begin{bmatrix} 7 \ 8 \end{bmatrix}$$

Step 2: Decryption Formula

$$P = K^{-1} C \pmod{26}$$

Step 3: Determinant of K

$$\det(K) = (5)(3) - (3)(4) = 15 - 12 = 3$$

Step 4: Inverse of Determinant mod 26

$$3 \cdot d^{-1} \equiv 1 \pmod{26}$$

$$3 \times 9 = 27 \equiv 1 \pmod{26} \Rightarrow d^{-1} = 9$$

Step 5: Adjugate of K

$$\text{adj}(K) = \begin{bmatrix} 3 & -3 \ -4 & 5 \end{bmatrix} \equiv \begin{bmatrix} 3 & 23 \ 22 & 5 \end{bmatrix} \pmod{26}$$

Step 6: Compute K⁻¹

$$K^{-1} = 9 \cdot \begin{bmatrix} 3 & 23 \ 22 & 5 \end{bmatrix} \pmod{26} = \begin{bmatrix} 27 & 207 \ 198 & 45 \end{bmatrix} \pmod{26}$$
- $27 \bmod 26 = 1$
- $207 \bmod 26 = 207 - 182 = 25$
- $198 \bmod 26 = 198 - 182 = 16$
- $45 \bmod 26 = 19$
$$K^{-1} = \begin{bmatrix} 1 & 25 \ 16 & 19 \end{bmatrix}$$

Check: $K^{-1}K \pmod{26}$:
- $(1)(5)+(25)(4)=5+100=105 \equiv 1$
- $(1)(3)+(25)(3)=3+75=78 \equiv 0$
- $(16)(5)+(19)(4)=80+76=156 \equiv 0$
- $(16)(3)+(19)(3)=48+57=105 \equiv 1$
Gives identity: inverse is correct.

Step 7: Multiply K⁻¹ by C

$$P = \begin{bmatrix} 1 & 25 \ 16 & 19 \end{bmatrix}\begin{bmatrix} 7 \ 8 \end{bmatrix} \pmod{26}$$
- Row 1: $7 + 200 = 207 \equiv 207 - 182 = 25$
- Row 2: $112 + 152 = 264 \equiv 264 - 260 = 4$
$$P = \begin{bmatrix} 25 \ 4 \end{bmatrix}$$

Step 8: Convert to Letters
- 25 = Z
- 4 = E
Result

$$\boxed{\text{Plaintext} = \textbf{ZE}}$$

The ciphertext 'HI' decrypts to 'ZE'.
125 marksCertificate Life Cycle ManagementHideAnswer
What is digital certificate? Discuss the certificate life cycle. [5]
A digital certificate is a certificate issued by a Certificate Authority (CA) to verify the identity of the certificate holder. The CA issues an encrypted digital certificate containing the applicant's public key along with a variety of ...

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Hill Cipher Decryption of Ciphertext 'HI'

Given Data

Step 1: Convert Ciphertext to Numbers

Step 2: Encryption of $M = 2$ with $k = 4$

Step 2: Encryption of $M = 2$ with $k = 4$