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1. Imagine you had a set of weights {62, 93, 26, 52, 166, 48, 91, and 141}. Find subset that sums to V = 302.
a) {62, 48, 166, 52}
b) {141, 26, 52, 48}
c) {93, 26, 91, 48}
d) {62, 26, 166, 48}

Answer: d [Reason:] {62, 26, 166, 48} =302.

2. For the Knapsack: {1 6 8 15 24}, Find the cipher text value for the plain text 10011.
a) 40
b) 22
c) 31
d) 47

Answer: a [Reason:] 1+15+24 = 40.

3. For the Knapsack: {1 6 8 15 24}, find the plain text code if the ciphertext is 38.
a) 10010
b) 01101
c) 01001
d) 01110

Answer: b [Reason:] If someone sends you the code 38 this can only have come from the plain text 01101.

4. Set {1, 2, 3, 9, 10, and 24} is superincreasing.
a) True
b) False

Answer: b [Reason:] It is not because 10 < 1+2+3+9.

5. A superincreasing knapsack problem is ____ to solve than a jumbled knapsack.
a) Easier
b) Tougher
c) Shorter
d) Lengthier

Answer: a [Reason:] A superincreasing knapsack is chosen to make computations easier while manual calculations of knapsack problems.

6. Consider knapsack that weighs 23 that has been made from the weights of the superincreasing series {1, 2, 4, 9, 20, and 38}. Find the ‘n’.
a) 011111
b) 010011
c) 010111
d) 010010

Answer: b [Reason:] v0=1, v1=2, v2=4, v3=9, v4=20, v5=38 K=6, V=23 Starting from largest number: v5 > V then ϵ_5=0 v4 < V then V = V – v4 = 23 – 20 = 3 ϵ_4=1 v3 > V then ϵ_3=0 v2> V then ϵ_2=0 v1 < V then V = V – v1= 3 – 2 = 1 ϵ_1=1 v0 =1 then V = V – v0= 1 – 1 = 0 ϵ_0=1 n= ϵ_5 ϵ_4 ϵ_3 ϵ_2 ϵ_1 ϵ_0 = 010011.

7. Another name for Merkle-Hellman Cryptosystem is
a) RC4
b) Knapsack
c) Rijndael
d) Diffie-Hellman

Answer: b [Reason:] Knapsack is another name for Merkel-Hellman Cryptosystem.

8. In Merkle-Hellman Cryptosystem, the hard knapsack becomes the private key and the easy knapsack becomes the public key.
a) True
b) False

Answer: b [Reason:] The hard knapsack becomes the public key and the easy knapsack becomes the private key.

9. In Merkle-Hellman Cryptosystem, the public key can be used to decrypt messages, but cannot be used to decrypt messages. The private key encrypts the messages.
a) True
b) False

Answer: b [Reason:] The public key can be used to encrypt messages, but cannot be used to decrypt messages. The private key decrypts the messages.

10. The plaintext message consist of single letters with 5-bit numerical equivalents from (00000)2 to (11001)2. The secret deciphering key is the superincreasing 5-tuple (2, 3, 7, 15, 31), m = 61 and a = 17. Find the ciphertext for the message “WHY”.
a) C= (148, 143, 50)
b) C= (148, 143, 56)
c) C= (143, 148, 92)
d) C= (148, 132,92)

Answer: a [Reason:] {wi }= {a vi mod m} {wi} = { 17×2 mod 61, 17×3 mod 61, 17×7 mod 61, 17×15 mod 61, 17×31 mod 61} {wi} = {34, 51, 58, 11, and 39} PlainText In binary Ci W- 22 10110 148 H – 7 00111 143 Y – 24 11000 50 So that the ciphertext sent will be C= (148, 143, 50).

11. For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where PT message=88 and thus find the CT.
a) 23
b) 64
c) 11
d) 54

Answer: c [Reason:] n = pq = 11 × 19 = 187. C=Me mod n ; C=887 mod 187 ; C = 11 mod 187.

12. For p = 11 and q = 17 and choose e=7. Apply RSA algorithm where Cipher message=11 and thus find the plain text.
a) 88
b) 122
c) 143
d) 111

Answer: a [Reason:] n = pq = 11 × 19 = 187. C=Me mod n ; C=1123 mod 187 ; C = 88 mod 187.

13. In an RSA system the public key of a given user is e = 31, n = 3599. What is the private key of this user?
a) 3031
b) 2412
c) 2432
d) 1023

Answer: a [Reason:] By trail and error, we determine that p = 59 and q = 61. Hence f(n) = 58 x 60 = 3480. Then, using the extended Euclidean algorithm, we find that the multiplicative inverse of 31 modulo f(n) is 3031.

14. Compute private key (d, p, q) given public key (e=23, n=233 ´ 241=56,153).
a) 35212
b) 12543
c) 19367
d) 32432