Category: Cryptography & Network Security Objective Questions

Cryptography Questions & Answers for campus interviews on “Whirlpool Algorithm”.

1. Mix Row functions operate on the 8 × 8 matrix (A) to form the new matrix. The operation can be considered to be B = AC. The matrix C is formed by
a) each successive even row having left shifts
b) each successive row having a left shifts
c) each successive odd row having right shifts
d) each successive row having a right shifts

Answer: d
Clarification: The matrix C is formed by each successive row having right shifts.

2. Which of the following hexadecimal values are not present in the matrix C of the Mix Rows function?
a) 01
b) 09
c) 05
d) 03

Answer: d
Clarification: 03 is not a valid value in matrix C.

3. There’s no known successful attacks made against Whirlpool.
a) True
b) False

Answer: a
Clarification: The reason for this is because the Whirlpool algorithm is a new algorithm and hasn’t been used very much.

4. Which hash algorithm is the most hardware intensive among the following?
a) SHA-1
b) SHA-2
c) MD-5
d) Whirlpool

Answer: d
Clarification: None.

5. Which algorithm has the most execution latency?
a) SHA-1
b) SHA-2
c) MD-5
d) Whirlpool

Answer: b
Clarification: Whirlpool has the lowest latency. Latency is defined by the number of iteration rounds. SHA-2 has 80 rounds, whereas Whirlpool has 10.

6. Which among the following has not been broken yet( still collision free)?
a) MD-5
b) SHA-1
c) DES
d) SHA-2

Answer: d
Clarification: SHA-2 has not yet been broken, but since SHA-1 has been found to be not collision free (and SHA-2 has the same logic as SHA-1), SHA-3 algorithm was developed and officialised in 2012.

7. Before the first round of the Whirlpool algorithm, which is the operation that takes place?
a) Addition of Key
b) Substitution of Bytes
c) Mixing of Rows
d) Shifting of Columns

Answer: a
Clarification: Before the first round of the Whirlpool algorithm, addition of key round takes place.

8. How many rounds does the Whirlpool algorithm undergo?
a) 12
b) 14
c) 10
d) 16

Answer: c
Clarification: There are 10 rounds in the algorithm.

9. Each round uses a different key in the algorithm.
a) True
b) False

Answer: a
Clarification: Whirlpool produces keys for each round using the same algorithm itself. Keys are updated per round.

10. What is the size of the key in the algorithm?
a) 256 bits
b) 512 bits
c) 128 bits
d) 1024 bits

Answer: b
Clarification: The size of the key is 512 bits.

11. How many add round key functions are present in the Whirlpool algorithm?
a) 16
b) 18
c) 11
d) 10

Answer: c
Clarification: There are 11 round key addition functions in the Whirlpool algorithm. One key addition per round. One key addition before the 1st round.

Advanced Network Security Questions & Answers on “TCP/IP and OSI Reference Model”.

1. In terms of the size of the network the correct order (ascending) is –
a) PAN, MAN, LAN, WAN
b) LAN, MAN, WAN, PAN
c) PAN, LAN, MAN, WAN
d) LAN, PAN, MAN, WAN

Answer: c
Clarification: Personal Area Network, Local Area Network, Metropolitan Area Network, Wide Area Network is the correct order in terms of size of the network from smallest to largest.

2. How many layers are there in the OSI reference model?
a) 4
b) 5
c) 6
d) 7

Answer: d
Clarification: The 7 layers are : Physical Layer, Data Link Layer, Network Layer, Transport Layer, Session Layer, Presentation Layer, Application Layer.

3. Logical Addressing and Routing are functions of which layer?
a) Physical Layer
b) Transport Layer
c) Data Link Layer
d) Network Layer

Answer: d
Clarification: . Logical Addressing and Routing are functions of the Network Layer.

4. Flow Control and Error Control are functions of which layer?
a) Physical Layer
b) Application Layer
c) Data Link Layer
d) Network Layer

Answer: c
Clarification: Flow Control and Error Control are functions of the Data Link Layer.

5. Dialog Control and Synchronization are function of which layer?
a) Presentation Layer
b) Application Layer
c) Session Layer
d) Data Link Layer

Answer: c
Clarification: Dialog Control and Synchronization are function of the Session Layer.

6.Encryption and Compression are functions of which OSI layer?
a) Presentation Layer
b) Application Layer
c) Session Layer
d) Data Link Layer

Answer: a
Clarification: Encryption and Compression are functions of the Presentation Layer.

7. File, Transfer, Access and Management (FTAM) is a function of which layer ?
a) Presentation Layer
b) Application Layer
c) Session Layer
d) Data Link Layer

Answer: b
Clarification: FTAM is an Application Layer function.

8. How many layers are present in the TCP/IP Reference model?
a) 6
b) 7
c) 5
d) 4

Answer: d
Clarification: There are 4 layers in the TCP/IP reference model : Link, Internet, Transport and Application.

Network Security Questions & Answers for campus interviews on “IEEE 802.11i WLAN Security”.

1. What is the size of the Temporal Key for the case of TKIP?
a) 64 bits
b) 128 bits
c) 256 bits
d) 512 bits

Answer: c
Clarification: In the case of TKIP -256 bits and CCMP – 128 bits.

2. What is a valid size of the Group Temporal Key (GTK) in WEP?
a) 40 bits
b) 128 bits
c) 512 bits
d) 80 bits

Answer: a
Clarification: GTK in WEP is of 40 bits or 104 bits.

3. Which key in the PTK protects the confidentiality of keys and data during RSN associated procedure?
a) EAPOL KCK
b) EAPOL KEK
c) TK
d) All of the mentioned

Answer: b
Clarification: This is the function of the EAPOL Key Encryption Key.

4. What is the size of the Message Integrity Code Key?
a) 64 bits
b) 128 bits
c) 256 bits
d) 512 bits

Answer: a
Clarification: The MIC Key has a size of 64 bits and it is used by TCIP’s Michael MIS to provide integrity protection of messages.

5. Which one of these is not a Traffic control key?
a) GTK
b) TK
c) MIC Key
d) WEP Key

Answer: c
Clarification: The MIC Key is a message integrity key.

6. The MPDU exchange for distributing pairwise keys is known as __________
a) 2-way handshake
b) 4-way handshake
c) 5 way handshake
d) 3 way handshake

Answer: b
Clarification: The MPDU exchange for distributing pairwise keys is a 4 way handshaking process and hence the name.

7. GTK is encrypted using:
a) Either RC4 or AES
b) DES
c) Rabin
d) ECC

Answer: a
Clarification: GTK is encrypted using either the RC4 or AES cipher system.

8. Message Integrity Code (MIC) is generated by an algorithm called
a) Susan
b) Michael
c) Barbara
d) Rijndael

Answer: b
Clarification: Message Integrity Code (MIC) is generated by an algorithm called Michael.

9. The TKIP and CCMP both provide 3 services: Message Integrity, Data Confidentiality and Protected Data Transfer.
a) True
b) False

Answer: b
Clarification: The TKIP and CCMP both provide the services: Message Integrity and Data Confidentiality. Protected Data Transfer isn’t TKIP and CCMP functionality.

10. CCMP uses which block cipher mode of operation?
a) OFB
b) CFB
c) CTR
d) CBC

Answer: c
Clarification: CCMP uses the CTR mode of operation for AES encryption.

Cryptography Multiple Choice Questions on “Block Cipher Systems”.

1. In affine block cipher systems if f(m)=Am + t, what is f(m1+m2) ?
a) f(m1) + f(m2) + t
b) f(m1) + f(m2) + 2t
c) f(m1) + t
d) f(m1) + f(m2)

Answer: a
Clarification: In general f(∑(i=1 to n) m_i = ∑(i=1 to n) f(m_i) + tδ_n) where δ_n=0 if n is odd and 1 if n is even.

2. In affine block cipher systems if f(m)=Am + t, what is f(m1+m2+m3) ?
a) f(m1) + f(m2) + f(m3) + t
b) f(m1) + f(m2) + f(m3) +2t
c) f(m1) + f(m2) + f(m3)
d) 2(f(m1) + f(m2) + f(m3))

Answer: c
Clarification: In general f(∑(i=1 to n) m_i =∑(i=1 to n) f(m_i ) + tδ_n) where δ_n=0 if n is odd and 1 if n is even.

3. If the block size is ‘s’, how many affine transformations are possible ?
a) 2^s (2^s-1)(2^s-1)(2^s-1²)………(2^s-1^(s-1))
b) 2^s (2^s-1)(2^s-2)(2^s-2²)………(2^s-2^(s-2))
c) 2^ss (2^s-1)(2^s-2)(2^s-2²)………(2^s-2^(s-1))
d) 2^s (2^s-1)(2^s-2)(2^s-2²)………(2^s-2^(s-3))

Answer: c
Clarification: 2^s (2^s-1)(2^s-2)(2^s-2²)………(2^s-2^(s-1)) is the maximum number of affine transformations possible for a block size ‘s’ matrix.

4. What is the number of possible 3 x 3 affine cipher transformations ?
a) 168
b) 840
c) 1024
d) 1344

5. Super-Encipherment using two affine transformations results in another affine transformation.
a) True
b) False

Answer: a
Clarification: f(g(m))=A_1 g(m)+c_1
f(g(m))=A_1 (A_2 m+c_2)+c_1 f(g(m))=A_1 A_2 m+A_1 c_2+c_1 f(x)=A_3 m+c_3
where
A_3=A_1 A_2
c_3=A_1 c_2+c_1
This results in another affine transformation, and does not improve the security.

6. If the key is 110100001, the output of the SP network for the plaintext: 101110001 is
a) 110100011
b) 110101110
c) 010110111
d) 011111010
Answer: b

7. If the key is 110100001 where,
If ki=0, then S_i (x)=((1 1 0 | 0 1 1 | 1 0 0 ))x+((1 1 1))
and If ki=1, then S_i (x)=((0 1 1 | 1 0 1 | 1 0 0))x+((0 1 1))
then the output of the SP network for the plaintext: 101110001 is
a) 010110011
b) 111000011
c) 110110111
d) 010110110
Answer: a

8. Confusion hides the relationship between the ciphertext and the plaintext.
a) True
b) False

Answer: b
Clarification: Confusion hides the relationship between the ciphertext and the key.

9. The S-Box is used to provide confusion, as it is dependent on the unknown key.
a) True
b) False

Answer: a
Clarification: The S-Box is used to provide confusion, as it is dependent on the unknown key.
The P-Box is fixed, and there is no confusion due to it, but it provides diffusion.

10. Which of the following slows the cryptographic algorithm –
1) Increase in Number of rounds
2) Decrease in Block size
3) Decrease in Key Size
4) Increase in Sub key Generation

a) 1 and 3
b) 2 and 3
c) 3 and 4
d) 2 and 4

Answer: b
Clarification: Increase in any of the above 4 leads to slowing of the cipher algorithm i.e. more computational time will be required.

Cryptography Multiple Choice Questions on “Principles of PRNG”.

1. PRNG stands for
a) Personal Random Number Generation
b) Pseudo Random Number Generation
c) Primitive Number Generators
d) Private Number Generators

Answer: b
Clarification: PRNG stands for Pseudo Random Number Generation.

2. TRNG stands for
a) True Random Number Generation
b) Trust Random Number Generation
c) Text Ring Number Generators
d) None of the mentioned

Answer: a
Clarification: TRNG stands for True Random Number Generation.

3. A rule to check for a large number N being a prime number
a) by dividing N by every even interger less than N/2
b) by dividing N by every odd interger less than root(N)
c) by dividing N by every even interger less than root(N)
d) by dividing N by every interger less than N/2

Answer: b
Clarification: A rule for checking if a number N is a prime is by dividing N by every odd interger less than root(N).

4. PRNGs are derived through algorithms.
a) True
b) False

Answer: a
Clarification: PRNGs are derived through algorithms. These algorithms are deterministic and therefore produce sequences of numbers that are not statistically random. Thus we should choose an algorithm which provides a high degree of randomness.

5. TRNGs take in an input which is referred to as
a) random variable
b) external variable
c) entropy source
d) bit stream

Answer: c
Clarification: TRNGs use an input known as “entropy source” which is an input source that is effectively random.

6. PRNGs take in an input which is referred to as
a) bit stream
b) seed
c) entropy source
d) external variable

Answer: b
Clarification: PRNGs use an input known as “seed”. The seed is usually generated using a TRNG.

7. Which of the following produces an output of fixed length?
a) PRNG
b) TRNG
c) PRF
d) All of the mentioned

Answer: c
Clarification: Only PRFs produce an output of fixed length. TRNGs and PRNGs have open ended outputs.

8. Which of the following a is NOT a check for randomness?
a) Uniformity
b) Scalability
c) Consistency
d) All of the mentioned

Answer: d
Clarification: Uniformity, Scalability and Consistency are all checks for randomness of a PRNG.

9. The property that there should be equally likely number of 1s and 0s in a Pseudo Random Number sequence is
a) Scalability
b) Uniformity
c) Stability
d) Consistency

Answer: b
Clarification: This is the property of Uniformity.

10. The property that any extracted subsequence should pass the test for randomness is
a) Scalability
b) Uniformity
c) Stability
d) Consistency

Answer: a
Clarification: Scalability is the property where any extracted subsequence should pass the test for randomness.

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 )

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

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=M^e 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.