250+ TOP MCQs on DES Modes of Operation and Answers

Cryptography Multiple Choice Questions on “DES Modes of Operation”.

1. How many modes of operation are there in in DES and AES?
a) 4
b) 3
c) 2
d) 5

Answer: d
Clarification: DES has 5 modes of operation.

2. Which one of the following modes of operation in DES is used for operating short data?
a) Cipher Feedback Mode (CFB)
b) Cipher Block chaining (CBC)
c) Electronic code book (ECB)
d) Output Feedback Modes (OFB)

Answer: c
Clarification: The Electronic code book mode is used for operating on short data as the same key is used for each block. Thus repetitions in Plain Text lead to repetitions in Cipher Text.

3. Which of the following is false for ECB mode of operation
i) The Plain text is broken into blocks of size 128 bytes
ii) Blocks can be swapped, repeated, replaced without recipient noticing
iii) Good for short data
iv) Encryption of each block is done separately using a randomly generated key for each block

a) i) only
b) ii) and iii)
c) i) and iv)
d) i) ii) and iv)

Answer: c
Clarification: Block size is 64 bits. The same Key is used for each block.

4. Which of the following statements are true
i) In the CBC mode, the plaintext block is XORed with previous ciphertext block before encryption
ii) The CTR mode does not require an Initialization Vector
iii) The last block in the CBC mode uses an Initialization Vector
iv) In CBC mode repetitions in plaintext do not show up in ciphertext

a) iii)
b) ii) and iv)
c) All the Statements are true
d) i) ii) and iv)

Answer: d
Clarification: The first block in CBC mode uses an IV.

5. There is a dependency on the previous ‘s’ bits in every stage in CFB mode. Here ‘s’ can range from ___
a) 8-16 bits
b) 8-32 bits
c) 4-16 bits
d) 8-48 bits

Answer: b
Clarification: The range of the output of each stage of the cipher system is 8-32 bits for a 64 bit system.

6. The counter value in CTR modes repeats are a regular interval.
a) True
b) False

Answer: b
Clarification: The Counter value in CTR mode should never be repeated, else it leads to vulnerability of the mode. We must ensure never reuse key/counter values; otherwise it could break (OFB).

7. Which of the following can be classified under advantages and disadvantages of OFB mode?
i) Transmission errors
ii) A bit error in a ciphertext segment
iii) Cannot recover from lost ciphertext segments
iv) Ciphertext or segment loss

a) Advantages: None; Disadvantages: All
b) Advantages: All; Disadvantages: None
c) Advantages: i); Disadvantages: ii) iii) iv)
d) Advantages: i); ii) Disadvantages: iii) iv)

Answer: d
Clarification: Advantages:
More resistant to transmission errors.
A bit error in a ciphertext segment affects only the decryption of that segment.
Disadvantages:
Cannot recover from lost ciphertext segments.
If a ciphertext segment is lost, all following segments will be decrypted incorrectly (if the receiver is not aware of the segment loss).

8. In OFB Transmission errors do not propagate: only the current ciphertext is affected, since keys are generated “locally”.
a) True
b) False

Answer: a
Clarification: Yes, transmission errors do not propagate in OFB mode because of the locally generated key.

9. Which of the following modes does not implement chaining or “dependency on previous stage computations”?
a) CTR, ECB
b) CTR, CFB
c) CFB, OFB
d) ECB, OFB

Answer: a
Clarification: Only CTR and ECB do not implement chaining.

250+ TOP MCQs on Rabin/ Elgamal Algorithm and Answers

Cryptography Multiple Choice Questions on “Rabin/ Elgamal Algorithm”.

1. “Rabin Cryptosystem is a variant of the Elgamal Cryptosystem”
a) True
b) False

Answer: b
Clarification: Rabin Cryptosystem is a variant of the RSA Cryptosystem.

2. Using Rabin cryptosystem with p=23 and q=7
Encrypt P=24 to find ciphertext. The Cipher text is
a) 42
b) 93
c) 74
d) 12

Answer: b
Clarification: Calculate n = p × q = 161
Plaintext P = 24
Ciphertext = C ≡ P2 (mod n)
= 242 mod 161 = 93 mod 161
Ciphertext transmitted = 93.

3. Which Cryptographic system uses C1 = (e1r) mod p and C1 = (e2r x P) mod p at the encryption side?
a) Elgamal
b) RSA
c) Rabin
d) Whirlpool

Answer: a
Clarification: The Elgamal cryptographic system uses the above formulae to compute the CT.

4. Sender chooses p = 107, e1 = 2, d = 67, and the random integer is r=45. Find the plaintext to be transmitted if the ciphertext is (28,9).
a) 45
b) 76
c) 66
d) 13

Answer: c
Clarification: P = [C2 (C1d)-1] mod p = 66.

5. In Elgamal cryptosystem, given the prime p=31.
Choose e1= first primitive root of p and d=10, calculate e2.
a) 24
b) 36
c) 25
d) 62

Answer: c
Clarification: We choose e1=3 (a primitive root of p = 31) and d=10. Then we have e2 = 310 mod 31 = 25.

6. In Elgamal cryptosystem, given the prime p=31.
Encrypt the message “HELLO”; use 00 to 25 for encoding. The value of C2 for character ‘L’ is
a) 12
b) 07
c) 20
d) 27

Answer: d
Clarification:The Common factor for the calculation of C2 is e7 mod 31 = 257 mod 31 = 25.
P = “L” = 11; C1 = 37 mod 31 = 17; C2 = 11 x 25 mod 31 = 27; C = (17,27).

7. In Elgamal cryptosystem, given the prime p=31.
Encrypt the message “HELLO”; use 00 to 25 for encoding. The value of C2 for character ‘O’ is
a) 09
b) 07
c) 23
d) 27

Answer: a
Clarification:The Common factor for the calculation of C2 is e7 mod 31 = 257 mod 31 = 25.
P = “O” = 14; C1 = 37 mod 31 = 17; C2 = 14 x 25 mod 31 = 09; C = (17,09).

8. In Elgamal cryptosystem, given the prime p=31.
What is the respective plaintext character for C = (27, 20)?
a) H
b) L
c) O
d) M

Answer: a
Clarification: The Common factor for the calculation of C2 is e7 mod 31 = 257 mod 31 = 25.
C = 17, 20); P = 20 X (1710)-1 mod 31 = 07; “07” = “H”.

250+ TOP MCQs on Whirlpool Algorithm and Answers

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.

250+ TOP MCQs on TCP/IP and OSI Reference Model and Answers

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.

250+ TOP MCQs on IEEE 802.11i WLAN Security Answers

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.

250+ TOP MCQs on Substitution and Transposition Techniques

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).