Cryptography online test on “Knapsack/ Merkle – Hellman/ RSA Cryptosystem”.
1. Find the ciphertext for the message {100110101011011} using superincreasing sequence { 1, 3, 5, 11, 35 } and private keys a = 5 and m=37.
a) C = ( 33, 47, 65 )
b) C = ( 65, 33, 47 )
c) C = ( 47, 33, 65 )
d) C = ( 47, 65, 33 )
Answer: c
Clarification: {vi} = { 1, 3, 5, 11, 35 }
a = 5 and m = 37
Public key generation:
{wi} = avi mod m
wi = {5, 15, 25, 18, 27}
Break the message into k-bit tuple i.e. 5-bit tuple
10011 01010 11011
Encoding of M as follows:
M Ci
10011 47
01010 33
11011 65
Ciphertext sent will be: C = (47, 33, and 65).
2. Suppose that plaintext message units are single letters in the usual 26-letter alphabet with A-Z corresponding to 0-25. You receive the sequence of ciphertext message units 14, 25, 89. The public key is the sequence {57, 14, 3, 24, 8} and the secret key is b = 23, m = 61.
Decipher the message. The Plain text is
a) TIN
b) INT
c) KIN
d) INK
Answer: b
Clarification: Solve using Knapsack Cryptosystem.
Wi = {57, 14, 3, 24, 8}
b = 23 and m = 61
a = b-1 mod m
61 = 2 x23 + 15
23 = 1x 15 + 8 Therefore 1= 8 x 23 – 3 x 61
15 = 1x 8 + 7 b-1 = 23-1= 8
8 = 1x 7 + 1 a = 8
v_i=a^(-1) w_i mod m
=bw_i mod m
v_i={ 30, 17, 8, 3, 1}
Cipher text V = bC mod m Plaintext
14 23 x 14 mod 61 = 17 01000 = 8 = I
25 23 x 25 mod 61 = 26 01101 = 13 = N
89 23 x 89 mod 61 = 34 10011 = 19 = T.
3. RSA is also a stream cipher like Merkel-Hellman.
a) True
b) False
Answer: a
Clarification: RSA is a block cipher system.
4. In the RSA algorithm, we select 2 random large values ‘p’ and ‘q’. Which of the following is the property of ‘p’ and ‘q’?
a) p and q should be divisible by Ф(n)
b) p and q should be co-prime
c) p and q should be prime
d) p/q should give no remainder
Answer: c
Clarification: ‘p’ and ‘q’ should have large random values which are both prime numbers.
5. In RSA, Ф(n) = _______ in terms of p and q.
a) (p)/(q)
b) (p)(q)
c) (p-1)(q-1)
d) (p+1)(q+1)
Answer: c
Clarification: Ф(n) = (p-1)(q-1).
6. In RSA, we select a value ‘e’ such that it lies between 0 and Ф(n) and it is relatively prime to Ф(n).
a) True
b) False
Answer: b
Clarification: gcd(e, Ф(n))=1; and 1 < e < Ф(n).
7. For p = 11 and q = 19 and choose e=17. Apply RSA algorithm where message=5 and find the cipher text.
a) C=80
b) C=92
c) C=56
d) C=23
Answer: a
Clarification: n = pq = 11 × 19 = 209.
8. For p = 11 and q = 19 and choose d=17. Apply RSA algorithm where Cipher message=80 and thus find the plain text.
a) 54
b) 43
c) 5
d) 24
Answer: c
Clarification: n = pq = 11 × 19 = 209.
C=Me mod n ; C=517 mod 209 ; C = 80 mod 209.
9. USENET falls under which category of public key sharing?
a) Public announcement
b) Publicly available directory
c) Public-key authority
d) Public-key certificates
Answer: a
Clarification: Many users have adopted the practice of appending their public key to messages that they send to public forums, such as USENET newsgroups and Internet mailing lists.
Perform encryption on the following PT using RSA and find the CT.
10. p = 3; q = 11; M = 5
a) 28
b) 26
c) 18
d) 12
Answer: b
Clarification: n = 33; f(n) = 20; d = 3; C = 26.
11. p = 5; q = 11; M = 9
a) 43
b) 14
c) 26
d) 37
Answer: b
Clarification: n = 55; f(n) = 40; d = 27; C = 14.
12. p = 7; q = 11; M = 8
a) 19
b) 57
c) 76
d) 59
Answer: b
Clarification: n = 77; f(n) = 60; d = 53; C = 57.
13. p = 11; q = 13; M = 7
a) 84
b) 124
c) 106
d) 76
Answer: c
Clarification: n = 143; f(n) = 120; d = 11; C = 106.
14. p = 17; q = 31; M = 2
a) 254
b) 423
c) 128
d) 523
Answer: c
Clarification: n = 527; f(n) = 480; d = 343; C = 128.
15. n = 35; e = 5; C = 10. What is the plaintext (use RSA) ?
a) 3
b) 7
c) 8
d) 5
Answer: d
Clarification: Use RSA system to decrypt and get PT = 5.