Question

Decrypt each of the following cipher texts:
    i) Text: "CBBGYAEBBFZCFEPXYAEBB", encrypted with affine cipher with key (7,2). 
    ii) Text:"KSTYZKESLNZUV", encrypted with Vigenère cipher with key "RESULT". 
   b) Another version of the columnar transposition cipher is the cipher using a key word. In this cipher, we encrypt as follows: Given a key word, we remove all the duplicate characters in the key word. For example, if the key word is ‘SECRET’, we remove the second ‘E’ and use ‘SECRT’ as the key word. To encrypt, we form a table as follows: In the first row, we write down the key word. In the following rows, we write the plaintext. Suppose we want to encrypt the text ‘ATTACKATDAWN’. We make a table as follows:

S E C R T
A T T A C
K A T D A
W N X X X

 

Then we read off the columns in alphabetical order. We first read the column under ‘C’, followed by the columns under ‘E’, ‘R’, ‘S’ and ‘T’. We get the cipher text TTX TAN ADX AKW CAX. To decrypt, we reverse the process. Note that, since we know the length of the keyword, we can find the length of the columns by dividing the length of the message by the length of the keyword.

 

Given the ciphertext ‘HNDWUEOESSRORUTXLARFASUXTINOOGFNEGASTORX’ and the key word ‘LANCE’, find the plaintext.

09 Jan 2026
Answer :
Word Count : 975
i) For the affine cipher, the decryption formula is: [ D(y) = a^{-1} (y - b) \mod 26 ] Given key ((a, b) = (7, 2)), we first find the modular inverse of (7 \mod 26). The inverse (7^{-1} \mod 26) is the number (x) such that (7x \equiv 1 \mod 26). Checking multiples: [ 7 \times 15 = 105 \equiv 1 \mod 26 ] So (7^{-1} = 15 \mod 26). The decryption formula becomes: [ D(y) = 15 (y - 2) \mod 26 ] We convert letters to numbers: (A=0, B=1, \dots, Z=25). Ciphertext: C B B G Y A E B B F Z C F E P X Y A E B B Numbers: 2 1 1 6 24 0 4 1 1 5 25 2 5 4 15 23 24 1 4 1 1 Now decrypt each (y): 1. (y = 2) → (D(2) = 15(2-2) = 15 \cdot 0 = 0 \mod 26) → A 2. (y = 1) → (D(1) = 15(1-2) = 15(-1) = -15 \equiv 11 \mod 26) → L 3. (y = 1) → same → L 4. (y = 6) → (15(6-2)=15 \cdot 4=60 \equiv 8 \mod 26) → I 5. (y = 24) → (15(24-2)=15 \cdot 22=330 \equiv 18 \mod 26) → S 6. (y = 0) → (15(0-2)=15(-2)=-30 \equiv 22 \mod 26) → W 7. (y = 4) → (15(4-2)=15\cdot 2=30 \equiv 4 \mod 26) → E 8. (y = 1) __________ ______ ___ _____ ________ ____ ____.
_______ __________ __________ ________ ______ _______ _________ _______.
_________ __________ ______ ____ ____ __________ __________ ____ ______ __________.
_______ _______ _________ _____ ________ _________ _________ _______ _________.
____ ____ ___ ___ ____ ________ _______ __________ ___ ______.
________ ____ ______ __________ ________ _________ ____ ______ ________ _______.
_________ __________ ___ ____ _________ _______ ___ ____.
________ ________ ____ ____ ________ __________ __________ ______.
___ ______ _____ _______ _______ ________ ___ ______.
_______ _________ _____ _______ __________ ____ ________ _________ ______ _________ ___.
_______ ____ ____ __________ ________.
_________ ___ _____ ________ _______ ________ _______ ______ _________ _____ __________.
_____ ________ ____ ________ ____ ________ ___.
__________ ___ ______ _____ _________ _________ ___ ________ ____ _________.
___ ______ _________ _____ __________ ________ __________ _________.
_________ _________ _____ ___ ___ _____ _________ ________ ________.
________ ______ ___ ____ _____ ____ _________ ________ ________ ________.
_____ _____ ____ ____ ______ __________ ___ _________.
______ _____ _________ ________ ______ __________ __________ ___.
______ __________ _________ _____ ____ ________ ______ ________.
___ __________ _________ _______ _________ _______ _______ ____ _____ _________ ______ ________.
______ _______ _______ ______ _______ ______ __________ _____ ______ ____ ___ _____.
_______ __________ _______ ____ __________.
_________ ____ ______ _______ _________ ______ ____ ___ ______.
_______ ____ ___ _______ _______ __________ _________ _____ _________ ______.
_____ _______ ___ _______ __________ ________ _______ ____ ___.
____ _____ ______ ______ _________ ________ _________ _________ ________ ____ ______ _______.
__________ ______ ______ ___ __________ _______ __________ _________ ______ _______.
__________ ____ ____ ________ _____.
___ __________ ______ _______ ______ ________ __________ ______ _______.
______ _______ ____ _______ _______ ______ ___ ______ _________.
____ _____ ___ _______ ________.
______ ___ ___ __________ _____ _________ ______.
__________ _____ ________ ___ ______.
__________ ___ _____ __________ __________.
_________ ____ _________ ____ ____ _______ _______ _________ ____ _____ __________.
______ _________ __________ ____ _______ _______.
__________ ______ _______ ________ _________.
_____ __________ ____ __________ ____ _________ ______ _______ ________.
__________ ___ ___ ________ _____.
__________ _______ _________ ____ ____ __________ ___ ______ ___.
___ ___ _________ _____ _______.
__________ _______ _______ _____ ________ _______ _______.
______ _____ _______ ________ ________ ___ ___ _____ ______ __________ ________.
__________ _________ ___ ______ ___ ____ ____ ______ ______ ____.
___ ________ _________ ____ _________ ______ ________ _____.
______ ____ ________ ___ _________ ___ ___ __________ ____ __________.
________ __________ _________ ________ _____.
_______ _______ ______ ________ ___ ________.
_____ ________ ________ _________ _________ __________ ___ ____ _________ _____ ____.
__________ _____ _______ __________ ________.
_____ __________ _______ ______ _________ ________.
______ __________ ___ _________ _________ ____ _______ __________ ______ ____ _____ ___.
____ _________ _______ ____ ________ ________ _______ ________.
____ ________ __________ ______ ____ _________ ___ _______ ____ _____.
__________ __________ ____ _____ ________.
___ ____ ___ __________ ________ ______ ______ _________ _____ _______.
________ _______ _____ _______ _____ _________ __________.
____ __________ ______ _________ ______ ___ __________ ___ _____ __________.
__________ _______ ______ __________ ___ ______.
______ _____ _____ _____ _____ ______ _______.
____ ___ _______ ______ _________ ______ _________ ______ _________ __________ ______.
___ __________ ______ _____ _______ _________ ____.
_______ __________ __________ _______ _________ ___ ________.
_________ ___ _______ __________ _______ ___ _________ ___ _________ __________.
____ ____ _______ _________ _____ _____ ______.
___ _________ ______ ________ _________ ____.
________ _______ ___ ______ __________ ________ _______ __________ ____ ______.
________ _______ ______ ___ __________.
_____ _____ _________ ___ _______ __________ ___ ________ __________ ___ ______ _____.
_________ _______ _______ ________ _______ _________ ____ ___.
_________ _________ _____ ______ _________ ___ _________ _____.
_____ _____ ____ ___ _____ __________ ____ _____ _______ _________ _________ ___.
________ ______ ______ __________ _______.
_____ __________ ___ _________ _____ __________ _________ ________ _________ ________ ________.
_______ __________ ________ __________ ________ _____ ____.
_____ _________ __________ _______ __________ ____.
______ __________ ________ ____ _____ _____.
____ __________ _________ ________ _____.
__________ ________ __________ ______ __________ ______ _________ ____ ___.
_______ ________ _______ _______ ____ ___ ___ ___ ___ _____ ______.
______ ________ _____ _______ ________ _________ _____ ______ _________ _____ _________ _________.
______ ___ __________ ________ __________ _________ _____ __________ ____.
__________ _________ ______ _________ ___.
________ _________ __________ ______ _____ ______ _____ ___ ________ __________ ______ ________.
___ ________ _______ _____ ____ ____.
___ _____ __________ _______ ______ __________ _______ ______ _____ ___.
___ _________ _______ ______ ________ _______ ____ ________ _________ _____.
______ __________ _____ ___.
Get Full Answer on WhatsApp
Related Questions

Alice wants to use the ElGamal digital signature scheme with public parameters equation, equation, secret value equation and equation. She wants to sign the message equation and send it to Bob. She chooses equation as the secret value. Explain the procedure that Alice will use for computing the signature of the message. What information will she send Bob? 

 

Find the inverse of 13 (mod 51) using extended euclidean algorithm.

 Decrypt the message equation that was encrypted using RSA algorithm with equation and equation. 

 

Use Miller-Rabin test to check whether 75521 is a strong pseuodprime to the base 2.

 Find a recurrence that generates the sequence 110110110110110. 

) Let equation. We represent the field equation by equation. Let us write equation. The table of values is given below:
 

 


i
 

  


γi
 

 Vector  


i
 

  


γi
 

 Vector
0  


1
 

  


(0,0,0,1)
 

 8  


γ2+1
 

  


(0,1,0,1)
 
1  


γ
 

  


(0,0,1,0)
 

 9  


γ3+γ
 

  


(1,0,1,0)
 
2  


γ2
 

  


(0,1,0,0)
 

 10  


γ2+γ+1
 

  


(0,1,1,1)
 
3  


γ3
 

  


(1,0,0,0)
 

 11  


γ3+γ2+γ
 

  


(1,1,1,0)
 
4  


γ+1
 

  


(0,0,1,1)
 

 12  


γ3+γ2+γ+1
 

  


(1,1,1,1)
 
5  


γ2+γ
 

  


(0,1,1,0)
 

 13  


γ3+γ2+1
 

  


(1,1,0,1)
 
6  


γ3+γ2
 

  


(1,1,0,0)
 

 14  


γ3+1
 

  


(1,0,0,1)
 
7  


γ3+γ+1
 

  


(1,0,1,1)
 

      

 

 i) Prepare logarithm and antilogarithm tables as given in page 23 of block 1. 
    ii) Compute equation and equation using the logarithm and antilogarithm tables

In this exercise, we introduce you to Hill cipher. In this cipher, we convert our message to numbers, just as in affine cipher. However, instead of encrypting character by character, we encrypt pairs of characters by multiplying them with an invertible matrix with co-efficients in Z26.

 

Here is an example: Suppose we want to ENCRYPT "ALLISWELL". Since we require the plaintext to have even number of characters, we pad the message with the character ‘X’. We break up the message into pairs of characters AL, LI, SW, EL and LX. We convert each pair of characters into a pair elements in Z26 as follows:

 

equation\underline{\overline{\begin{aligned}AL&\quad(0,11)\\LI&\quad(11,8)\\SW&\quad(18,22)\\EL&\quad(4,11)\\LX&\quad(11,23)\end{aligned}}}">

 

Next, we choose an invertible equation matrix with coefficients in equation, for example, equation.
This matrix has determinant equation and equation is a unit in equation with inverse equation. We write each pair of elements in equation as a column vector and multiply it by A:
equation
We then convert each pair of numbers to a pair of characters and write them down. In this example, we get the cipher text "LSPFYGXUEN" corresponding to the plain text "ALLWELL". To decrypt, we convert pairs of characters to pairs of numbers and multiply by equation and we have
equation
Decrypt the text "TWDXHUJLUENN" which was encrypted using the Hill’s cipher with the matrix equation as the encryption matrix.

IGNOU Solved Assignments Assignments Ignou Projects Project Previous Year Solved Exam Papers Notes Notes Sample Question Papers Sample Question Papers
Login Cart
IGNOU NEWS
Assignment Submission Last Date Extended Till 30 June 2026 Click Here★★★IGNOU June 2026 TEE Date Sheet Released Click Here★★★
Top
📞
Call Support Instant phone assistance Decrypt each of the following cipher texts:    i
→
🟢
WhatsApp Chat Fast live messaging
→
Email Us Business enquiries & support
→