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 _________ __________ ______ __________ ____ __________ ____.
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Related Questions

 Alice wants to use the Digital Signature algorithm for signing messages. She chooses equation, equation, equation and equation. Alice wants to sign the message equation. She chooses the secret value equation. Explain the procedure that Alice will use for computing the signature. What information will she send Bob? 

 Apply runs test to the following sequence:

 1001101000010000101111011
equation 0111010010110110010011010
equation 0110011100001100100111000
equation 1100001101010111101001110
equation 0010001111000001101010010
equation 1000110100000110100101101
equation 1110001001 

) 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

 Decrypt the ciphertext 101000111001 which was encrypted with the Toy block cipher once using the key 101010010. Show all the steps. 
b) A 64 bit key for the DES is given below
equation 11000111 equation 10000101
equation 11110111 equation 11000001
equation 11111011 equation 10101011
equation 10011101 equation 10010001
equation i) Check whether the key is error free using the parity bits.

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.

 

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

 

 Alice wants to send Bob the message equation. She chooses equation. How will she compute the cipher text? What information does she send to Bob? 

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