Encrypt the plain text Message “EXTRANET” using Transposition cipher technique…
2016
Encrypt the plain text Message “EXTRANET” using Transposition cipher technique with the following key :
\(\begin{array} {|l|l|l|l|l|} \hline \text{3} & \text{5} & \text{2} & \text{1} & \text{4} & \text{(Cipher text)} \\ \hline \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{(Plain text)} \\ \hline \end{array}\)
Using \(‘Z'\) as bogus character.
- A.
TAXERTZENZ
- B.
EXTRANETZZ
- C.
EZXZTRZANZET
- D.
EXTZRANZETZ
Attempted by 78 students.
Show answer & explanation
Correct answer: A
Final ciphertext: RZTTENAZXE
Step 1: Arrange the plaintext EXTRANET into rows with 5 columns and pad with 'Z' to fill the last row.
Row 1: E X T R A
Row 2: N E T Z Z
Step 2: Interpret the key mapping. The columns (plain positions 1 to 5) have key numbers 3, 5, 2, 1, 4 respectively. So the read order of plain columns is: column 4, column 3, column 1, column 5, column 2.
Step 3: Read each selected column top-to-bottom and concatenate:
Column 4 → R Z = RZ
Column 3 → T T = TT
Column 1 → E N = EN
Column 5 → A Z = AZ
Column 2 → X E = XE
Concatenate: RZ + TT + EN + AZ + XE = RZTTENAZXE
Note: None of the given answer choices match this correct ciphertext. The listed choice TAXERTZENZ is not produced by the standard columnar transposition process with the given key and padding.