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B.Tech - Odd Sem : Semester in Exam-I
Academic Year:2024-2025
22CS2224F - CRYPT ANALYSIS & CYBER DEFENSE
Set No: 1
Time: Max.Marks: 50
CO COI
S.NO Answer All Questions Choice Options Marks CO
BTL BTL
1. Answer all the Questions 12Marks CO2 3 2
1.A. Explain encryption equation for Hill Cipher. 2Marks CO1 3 2
1.B. Discuss the length of Key in DES 2Marks CO1 3 2
1.C. Interpret the components of CIA triangle. 2Marks CO1 3 2
1.D. Discuss length of key in AES. 2Marks CO2 3 2
1.E. Interpret True Random Number Generator. 2Marks CO2 3 2
1.F. Describe Stream Cipher ib cryptography 2Marks CO2 3 2
2. Answer all the Questions 16Marks CO2 3 2
2.A. Illustrate Symmetric Cipher Model with neat diagram. 4Marks CO1 3 2
2.B. Interpret the steps involved in key generation process of DES algorithm. 4Marks CO1 3 2
2.C. Describe the steps involved in the Encryption process of AES algorithm. 4Marks CO2 3 2
2.D. Demonstrate Double DES with a neat diagram. 4Marks CO2 3 2
choice
3. Answer all the Questions 11Marks CO1 3 3
Q-4
3.A. Apply Playfair cipher and encrypt plaintext "TRANSPORT" using the key "HUT". 5Marks CO1 3 3
Apply SDES encryption algorithm and solve the following: Plaintext 1111 0111;
Key 1 (K1): 1 1 1 1 1 1 1 1; Key 2 (K2): 1 0 0 1 1 0 0 1 Initial Permutation (IP): 3 1
2 6 4 8 5 7, E/P: 3 2 4 1 4 3 2 1, P4: 2 4 1 3
3.B. 6Marks CO1 3 3
4. Answer all the Questions 11Marks CO1 3 3
Apply Vigenere Cipher on the plaintext "COMMUNICATION" using the key
4.A. 5Marks CO1 3 3
"SKILL".
Apply columnar transposition and encrypt the plaintext "WELCOME TO KLU”
4.B. 6Marks CO1 3 3
with the keyword "MONDAY"
choice
5. Answer all the Questions 11Marks CO2 3 3
Q-6
Apply SRC4 algorithm given a 4 x 3-bit key of K = [2 1 3 8], S= [0 1 2 3 4 5 6 7]
5.A. 5Marks CO2 3 3
and T = [2 1 3 8 2 1 3 8] and perform initial permutation on S.
5.B. Apply SAES and encrypt the plaintext P: "1000 0111 1010 1111", K0 = W0W1= 6Marks CO2 3 3
1101 1101 0010 1000 K1 = W1W2=0100 1010 1111 0101 K2 = W3W4= 1010 0110
1001 0000
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6. Answer all the Questions 11Marks CO2 3 3
Apply LCG given modulus (m) = 17, multiplier (a) = 5, increment (c) = 3, and seed
6.A. 5Marks CO2 3 3
(X0) = 7, generate the next 10 pseudorandom numbers.
Apply Blum Blum Shub Generator by considering the following values p=11, q=19,
6.B. 6Marks CO2 3 3
s=7 and generate random numbers.
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