Category: Cryptography & Network Security Objective Questions

Cryptography Multiple Choice Questions on “Whirlpool Algorithm”.

1. What is the correct order of operation within each round of the Whirlpool algorithm?
a) Add round key, Substitute bytes, Mix rows, Shift columns
b) Substitute bytes, Add round key, Shift columns, Mix rows
c) Mix rows, Substitute bytes, Shift columns, Add round key
d) Substitute bytes, Shift columns, Mix rows, Add round key

Answer: d
Clarification: The correct order is Substitute bytes, Shift columns, Mix rows, Add round key.

2. What is the size of the s-box table/matrix in Whirlpool?
a) 8 × 8
b) 16 × 16
c) 8 × 16
d) 16 × 8

Answer: b
Clarification: The size of the matrix is 16 × 16.

3. The S-box implements linear mapping thus providing diffusion.
a) True
b) False

Answer: b
Clarification: Non-linear mapping provides diffusion. The s-box provides Non-linear mapping.

4. Another name for the Whirlpool key is
a) CState
b) LState
c) Estate
d) KState

Answer: d
Clarification: Whirlpool uses a 512 bit key called KState.

5. How many round constants (RCs) are required in Whirlpool?
a) 10
b) 11
c) 12
d) 21

Answer: a
Clarification: There are 10 RCs required for modifying 10 Keys in each step.

6. The Round Constant is dependent on the s-box values.
a) True
b) False

Answer: a
Clarification: The Round Constant is given by S[8(r-1)+j] where S is the s-box permutation.

7. The Round Constant is given by S[8(r-1)+j]. What are the limits of ‘r’?
a) 0 <= r <= 11
b) 0 <= r <= 12
c) 0 <= r <= 10
d) 0 <= r <=16

Answer: c
Clarification: r can have values from 0 to 10.

8. When the Round Constant is given by S[8(r-1)+j]. What are the limits of ‘i’ and ‘j’?
a) i = 1 ; 0 <= j <= 7
b) i = 0 ; 0 <= j <= 7
c) 0 <= i <= 7; 0 <= j <= 7
d) 0 <= i <= 7; j=0

Answer: b
Clarification: All rows except the first one are 0 i.e. RC=0 for ‘i’ not equal to 0.

9. The 7th row in the 8 × 8 matrix undergoes a ___________ shift.
a) 6 bit
b) 6 byte
c) 7 bit
d) 7 byte

Answer: b
Clarification: The nth row in the 8 × 8 matrix undergoes an (n-1) byte shift.

10. How many XOR gate operations are involved in the S-box permutation?
a) 12
b) 8
c) 16
d) 4

Answer: a
Clarification:12 XORs are involved in the S-box permutation in Whirlpool.

11. Which round provides linear diffusion in the Whirlpool Algorithm?
a) Add Key
b) Substitute Bytes
c) Mix Rows
d) Shift Rows

Answer: c
Clarification: MR function is the linear diffusion layer of the Whirlpool block cipher.

Network Security Assessment Questions on “TCP/IP and OSI Reference Model”.

Answer the following full forms in terms of TCP/IP Reference Model.

1. HTTP stands for ________
a) Hash Text Transfer Protocol
b) Hyper Text Transfer Protocol
c) Hash Transfer Text Protocol
d) none of the mentioned

Answer: b
Clarification: HTTP stands for Hyper Text Transfer Protocol.

2. SMTP stands for ________
a) Service Message Transmission Permission
b) Secure Message Transfer Protocol
c) Simple Mail Transfer Protocol
d) Simple Message Transfer Protocol

Answer: c
Clarification: SMTP stands for Simple Mail Transfer Protocol.

3. UDP stands for ________
a) User Datagram Protocol
b) Used Data Protocol
c) Unified Definition Protocol
d) Undefined Diagnostic Protocol

Answer: a
Clarification: UDP stands for User Datagram Protocol.

4. ICMP stands for ________
a) Internal Control Message Protocol
b) Internet Cipher Mail Protocol
c) Internal Cipher Mail Protocol
d) Internet Control Message Protocol

Answer: d
Clarification: ICMP stands for Internet Control Message Protocol.

5. SONET stands for ________
a) Secure Offline Network
b) Synchronous Optical Network
c) Service Offline Network
d) Secure Optical Netwrok

Answer: b
Clarification: SONET stands for Synchronous Optical Network.

6. DSL stands for ________
a) Data Storage Line
b) Digital Subscriber Line
c) Data Service Language
d) Data Secure Language

Answer: b
Clarification: DSL stands for Digital Subscriber Line.

7. Which of the following does not lie in the Application Layer of the TCP/IP Model?
a) HTTP
b) SMTP
c) RTP
d) UDP

Answer: d
Clarification: UDP lies in the Transport layer.

8. Which one of these does not lie in the Link Layer of the TCP/IP Model?
a) DSL
b) IP
c) SONET
d) 802.11

Answer: b
Clarification: IP (Internet Protocol) is a member of the Internet layer.

9. Which of these is a connection oriented service?
a) X.25
b) Frame Relay
c) ATM
d) All of the mentioned

Answer: d
Clarification: All the mentioned are connection oriented services.

Network Security Questions & Answers for entrance exams on “IEEE 802.11i WLAN Security”.

1. What was the security algorithm defined for the IEEE 802.11?
a) WEP
b) RSN
c) WPA
d) SSL

Answer: a
Clarification: Wired Equivalency Privacy was the security algorithm defined for the IEEE 802.11.

2. The final form of the 802.11i standard is referred to as –
a) Wi-Fi Protected Access
b) Robust Security Network
c) Wired Equivalency Privacy
d) None of the mentioned

Answer: b
Clarification: The final form of the 802.11i standard is the Robust Security Network (RSN).

3. EAP stands for –
a) Extended Application Protocol
b) Extensible Authentication Protocol
c) Embedded Application Protocol
d) Embedded Authentication Protocol

Answer: b
Clarification: EAP stands for Extensible Authentication Protocol.

4. TKIP is an access control protocol.
a) True
b) False

Answer: b
Clarification: TKIP stands for Temporal Key Integrity Protocol and falls under “Confidentiality, Data Origin Authentication and Integrity and Replay Protection.”

5. In which phase of operation does the STA and AS prove their identities to each other?
a) Discovery
b) Authentication
c) Key generation and distribution
d) Protected data transfer

Answer: b
Clarification: The STA and AS prove their identities to each other in the Authentication phase.

6. The specification of a protocol, along with the chosen key length (if variable) is known as –
a) cipher suite
b) system suite
c) key set
d) service set

Answer: a
Clarification: The specification of a protocol, along with the chosen key length (if variable) is known as cipher suite.

7. Which the 3rd phase of operation in the IEEE 802.11i Protocol?
a) Protected Data Transfer
b) Discovery
c) Authentication
d) Key Management

Answer: d
Clarification: Key management is the 3rd Phase of operation in the IEEE 802.11i Protocol.

8. Which phase uses the Extensible Authentication Protocol?
a) Discovery
b) Authentication
c) Key Management
d) Protected Data Transfer

Answer: b
Clarification: EAP belongs to the Authentication Phase and is defined in the IEEE 802.1X standard.

9. There are a number of possible EAP exchanges that can be used during authentication phase. Typically the message flow between the STA and AP employs the ___________ protocol.
a) RADUIS
b) EAPOL
c) TKIP
d) KSN

Answer: b
Clarification: The message flow between the STA and AP employs the EAP over LAN (EAPOL) protocol.

10. Another name for the AAA key (Authentication, Authorization and Accounting Key) is –
a) pre-shared key
b) pairwise transient key
c) master session key
d) key conformation key

Answer: c
Clarification: The AAA key (Authentication, Authorization and Accounting Key) is also known as master session key.

Cryptography Multiple Choice Questions on “Substitution and Transposition Techniques”.

1. Use Caesar’s Cipher to decipher the following
HQFUBSWHG WHAW
a) ABANDONED LOCK
b) ENCRYPTED TEXT
c) ABANDONED TEXT
d) ENCRYPTED LOCK

Answer: b
Clarification: Caesar Cipher uses C =(p+3) mod 26 to encrypt.

2. Caesar Cipher is an example of
a) Poly-alphabetic Cipher
b) Mono-alphabetic Cipher
c) Multi-alphabetic Cipher
d) Bi-alphabetic Cipher

Answer: b
Clarification: Caesar Cipher is an example of Mono-alphabetic cipher, as single alphabets are encrypted or decrypted at a time.

3. Monoalphabetic ciphers are stronger than Polyalphabetic ciphers because frequency analysis is tougher on the former.
a) True
b) False

Answer: b
Clarification: Monoalphabetic ciphers are easier to break because they reflect the frequency of the original alphabet.

4. Which are the most frequently found letters in the English language ?
a) e,a
b) e,o
c) e,t
d) e,i

Answer: c
Clarification: The relativity frequency of these letters in percent : e-12.702, a-8.167, t-9.056, i-6.996, o-7.507.

5. Choose from among the following cipher systems, from best to the worst, with respect to ease of decryption using frequency analysis.
a) Random Polyalphabetic, Plaintext, Playfair
b) Random Polyalphabetic, Playfair, Vignere
c) Random Polyalphabetic, Vignere, Playfair, Plaintext
d) Random Polyalphabetic, Plaintext, Beaufort, Playfair

Answer: c
Clarification: Random Polyalphabetic is the most resistant to frequency analysis, followed by Vignere, Playfair and then Plaintext.

6. On Encrypting “thepepsiisintherefrigerator” using Vignere Cipher System using the keyword “HUMOR” we get cipher text-
a) abqdnwewuwjphfvrrtrfznsdokvl
b) abqdvmwuwjphfvvyyrfznydokvl
c) tbqyrvmwuwjphfvvyyrfznydokvl
d) baiuvmwuwjphfoeiyrfznydokvl

Answer: b
Clarification: Cipher text:= Ci = Pi + ki mod m (mod 26).

7. On Encrypting “cryptography” using Vignere Cipher System using the keyword “LUCKY” we get cipher text
a) nlazeiibljji
b) nlazeiibljii
c) olaaeiibljki
d) mlaaeiibljki

Answer: a
Clarification: Cipher text:= Ci = Pi + ki mod m (mod 26).

8. The Index of Coincidence for English language is approximately
a) 0.068
b) 0.038
c) 0.065
d) 0.048

Answer: c
Clarification: The IC for the English language is approximately 0.065.

9. If all letters have the same chance of being chosen, the IC is approximately
a) 0.065
b) 0.035
c) 0.048
d) 0.038

Answer: d
Clarification: If all letters have the same chance of being chosen, the IC is approximately 0.038, about half of the IC for the English language.

10. Consider the cipher text message with relative frequencies:
4 0 10 25 5 32 24 15 6 11 5 5 1 2 6 6 15 19 10 0 6 28 8 2 3 2
The Index of Coincidence is
a) 0.065
b) 0.048
c) 0.067
d) 0.042

Answer: c
Clarification: Number of letters = 250. From this, IC=0.0676627. This is very strong evidence that the message came from a Monoalphabetic ciphering scheme.

11.Consider the cipher text message:
YJIHX RVHKK KSKHK IQQEV IFLRK QUZVA EVFYZ RVFBX UKGBP KYVVB QTAJK TGBQO ISGHU CWIKX QUXIH DUGIU LMWKG CHXJV WEKIH HEHGR EXXSF DMIIL UPSLW UPSLW AJKTR WTOWP IVXBW NPTGW EKBYU SBQWS

Relative Frequencies –
3 7 2 2 5 5 7 9 11 4 14 4 2 1 3 4 6 5 6 5 7 10 9 8 4 2

The Index of Coincidence is –
a) 0.065
b) 0.048
c) 0.067
d) 0.044

Answer: d
Clarification: Number of letters = 145.From this, IC=0.0438697 .This is very strong evidence that the message came from a polyalphabetic ciphering scheme.

12. A symmetric cipher system has an IC of 0.041. What is the length of the key ‘m’?
a) 1
b) 3
c) 2
d) 5

Answer: d
Clarification: Using the formula for calculating ‘m’ we get m=5, where
m≈0.027n/(I_c (n-1)-0.038n+0.065).

Cryptography Questions experienced people on “DES Modes of Operation”.

1. Which mode of operation has the worst “error propagation” among the following?
a) OFB
b) CFB
c) CBC
d) ECB

Answer: d
Clarification: The ECB or electronic code book mode of operation propagates the most errors. A single bit error is carried onto the next block and so on.

2. Which block mode limits the maximum throughput of the algorithm to the reciprocal of the time for one execution?
a) OFB
b) CTR
c) CBC
d) ECB

Answer: b
Clarification: The CTR mode of operation limits the maximum throughput of the algorithm to the reciprocal of the time for one execution.

3. Which mode requires the implementation of only the encryption algorithm?
a) ECB
b) CBC
c) CTR
d) OFB

Answer: c
Clarification: The CTR mode only requires the implementation of either the encryption or decryption phase. Both the phases are somewhat similar.

4. Which of the following modes of operation does not involve feedback?
a) ECB
b) CBC
c) CTR
d) OFB

Answer: a
Clarification: Electronic code book does not involve feedback.

5. Which of the following is a natural candidates for stream ciphers?
a) OFB
b) CFB
c) CBC
d) ECB

Answer: a
Clarification: OFB and CTR both produce outputs that are independent of both PT and CT. Thus they are ideal candidates for stream ciphers.

6. The XTS-AES mode was approved by NIST in
a) 1999
b) 2010
c) 2006
d) 2002

Answer: b
Clarification: The XTS-AES mode of operation was approved by NIST in 2010. It is the most recent mode of operation for block ciphers.

7. The XTS-AES mode is based on the concept of tweakable block cipher.
a) True
b) False

Answer: a
Clarification: The XTS-AES mode uses a tweak thus the name.

8. The purpose of a ‘tweak’ in XTS-AES mode is to
a) secure the public key
b) provide security
c) provide variability
d) all of the mentioned

Answer: c
Clarification: The purpose of the ‘tweak’ in the XTS-AES mode is to provide variability in each round.

9. A tweak is used in XTS-AES mode to provide a different output for the same input PT and same key.
a) True
b) False

Answer: a
Clarification: The statement is true for XTS-AES mode of operation.

10. XTS-AES mode of operation is a better version of
a) OFB
b) CFB
c) CTR
d) ECB

Answer: d
Clarification: XTS-AES mode overcomes the principle security weakness of ECB, which is that two encryptions of the same CT yeild the same PT.

11. What is the size of the XTS-AES key?
a) 1024 bits
b) 64 bits
c) 512 bits
d) 128 bits

Answer: c
Clarification: The key size can be either 256 bits or 512 bits. There are 2 keys of the same size K1 and K2.

12. Which of the following represent the tweak?
a) j
b) i
c) alpha
d) alpha^j

Answer: b
Clarification: ‘i’ represents the tweak value.

13. Which of the following is true for the tweak in XTS-AES mode?
a) they are non-negative integers
b) they are of size 128 bits
c) they are assigned consecutively
d) all of the mentioned

Answer: d
Clarification: All of the statements are true in relation to the tweak ‘i’ from the XTS-AES mode of operation.

14. Which of the following is the correct encryption statement representation for the XTS-AES mode?
a) E(K1,P) XOR T
b) E(K2,T) XOR P
c) E(K1,T XOR P) XOR P
d) E(K1,P XOR T) XOR T

Answer: d
Clarification: The correct encryption representation would be : C = E(K1,P XOR T) XOR T.

15. The last two blocks of the XTS-AES mode are –
a) padded as 10*
b) encrypted/ decrypted using ciphertext-stealing
c) padded as 10*1
d) padded and then swapped after encryption/ decryption

Answer: b
Clarification: The correct term used to encrypt/ decrypt the last 2 blocks is ‘cipher-text stealing’ where C(m) and C(m-1) are interchanged with each other.

Cryptography Multiple Choice Questions on “Knapsack/ Merkle – Hellman/ RSA Cryptosystem”.

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
Clarification: {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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: {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
Clarification: n = pq = 11 × 19 = 187.
C=M^e 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
Clarification: n = pq = 11 × 19 = 187.
C=M^e 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
Clarification: 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

Answer: c
Clarification: Since n=233 ´ 241=56,153, p=233 and q=241
f(n) = (p – 1)(q – 1) = 55,680
Using Extended Euclidean algorithm, we obtain
d = 23–1 mod 55680 = 19,367.