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Explain Playfair cipher technique in detail. Find the cipher text for theplain text ‘GTUINSEXAM’ using ‘STUDY’ as a key.

Question

Explain Playfair cipher technique in detail.

Find the cipher text for the plain text ‘GTUINSEXAM’ using ‘STUDY’ as a key.

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Solution

The Playfair cipher is a manual symmetric encryption technique and was the first literal digraph substitution cipher. The scheme was invented in 1854 by Charles Wheatstone but was named after Lord Playfair who promoted the use of the cipher.

The technique encrypts pairs of letters (digraphs), instead of single letters as in the simple substitution cipher and rather more complex Vigenère cipher systems then in use. The Playfair is thus significantly harder to break since the frequency analysis used for simple substitution ciphers does not work with it.

The Playfair cipher uses a 5 by 5 grid of letters, filled with the letters of the keyword (omitting duplicates), followed by the remaining letters of the alphabet (usually omitting "J" to reduce the alphabet to fit). The keyword together with the conventions for filling in the 5 by 5 table constitute the cipher key.

To encrypt a message, you break your message into digraphs (groups of 2 letters) such that, for example, "HELLO" becomes "HE" "LL" "O". If a pair is a repeated letter, insert a filler like 'X', ex: "Communication" becomes "CO" "MM" "UN" "IC" "AT" "IO" "NX". If the last letter is a single, also add a filler at the end.

The rules for transforming the plaintext into ciphertext are as follows:

  1. If both letters are the same (or only one letter is left), add an "X" after the first letter. Encrypt the new pair and continue. Some variants of Playfair use "Q" instead of "X", but any letter, itself uncommon as a repeated pair, will do.

  2. If the letters appear on the same row of your table, replace them with the letters to their immediate right respectively (wrapping around to the left side of the row if a letter in the original pair was on the right side of the row).

  3. If the letters appear on the same column of your table, replace them with the letters immediately below respectively (wrapping around to the top side of the column if a letter in the original pair was on the bottom side of the column).

  4. If the letters are not on the same row or column, replace them with the letters on the same row respectively but at the other pair of corners of the rectangle defined by the original pair. The order is important – the first letter of the encrypted pair is the one that lies on the same row as the first letter of the plaintext pair.

To illustrate this, let's use your example. The key is 'STUDY' and the plain text is 'GTUINSEXAM'.

First, we create a 5x5 matrix using the key. The matrix looks like this:

S T U D Y
A B C E F
G H I K L
M N O P Q
R V W X Z

Now, we break the plaintext into digraphs: GT UI NS EX AM.

Applying the rules:

  • GT: These letters are not in the same row or column. So, we form a rectangle with these letters and take the letters on the same row but at the other corners. This gives us 'YH'.
  • UI: These letters are not in the same row or column. So, we form a rectangle with these letters and take the letters on the same row but at the other corners. This gives us 'DY'.
  • NS: These letters are not in the same row or column. So, we form a rectangle with these letters and take the letters on the same row but at the other corners. This gives us 'CM'.
  • EX: These letters are not in the same row or column. So, we form a rectangle with these letters and take the letters on the same row but at the other corners. This gives us 'KC'.
  • AM: These letters are in the same column. So, we take the letters immediately below respectively. This gives us 'MR'.

So, the cipher text for the plain text ‘GTUINSEXAM’ using ‘STUDY’ as a key is 'YHDYCMKCMR'.

This problem has been solved

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