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

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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.

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.

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

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