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).

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