Suppose that you wanted to encipher the following text using Playfair...

plaintext: a life without cause is a life without effect -Barbarella

The first step is to make a cipher alphabet grid. This can easily be done by creating a 5 by 5 grid from a keyword, as shown below. You could also simply inscribe the alphabet into the grid in a spiral or other unique pattern to produce a ciphertext alphabet (similar to the routes used in the transposition ciphers).

keyword: profund

Now that you have the ciphertext alphabet grid, the next step is to encipher the plaintext. This is done by the splitting the plaintext into groups of two letters, otherwise known as digraphs (the Playfair system was the first to use digraphs). It should be noted that if two letters of the plaintext happen to be the same letter, an 'x,' any other predetermined rare letter should be used as the second letter of the digraph. In the case of our plaintext, the digraph pairs are shown below...(notice the bolded x which was inserted because of the double 'ff's.)...

plaintext: a life without cause is a life without effect
plaintext (digraphs): al if ew it ho ut ca us ei sa li fe wi th ou te fx fe ct

Now we take the first digraph of our message and bring it to the ciphertext alphabet grid. The two letters that we have, 'al', are bolded in the example grid. The ciphertext equivilent for these two will be 'nq', since those are the two letters that are found at the two cross sections of 'a' and 'l' in the grid. 'n' was put first in the ciphertext digraph not because it was higher -- but because it was the letter on the row of the first letter in the plaintext digraph. If the plaintext digraph had been 'la', the resulting ciphertext would have been 'nq'.

We can't apply the same cross section system for the next digraph, 'if', though. As you can see in the grid, both of those letters are in the same column. This means that they don't have the intersections which we used for the 'al' digraph. So, in cases where the two letters appear in the same column, the playfair system uses the letters directly below each plaintext letter in the grid. So, for 'if', we'd create a ciphertext digraph of 'sb'. Simirarly, if the each of the two letters fall in the same row, the ciphertext digraph will simply by composed of the letters directly to the right of each letter.

Of course, what happens when we are dealing with a row or column event and one of the letters happens to be on the edge of the grid, and there is no 'right' or 'down'. How, for example, would you encode the 'ca' digraph from our plaintext message?

When letters are on the edges of the grid, we simply move to the opposite end -- so that the rows and columns actually 'loop'. In our case, the letter to the 'right' of 'c' is actually 'n'. So, our digraph pair would be 'nb' for 'ca'. The letter below 'v' would be 'p' in the case of the column edge. The full ciphertext of our original message is shown below...try to follow along if you don't quite understand the system yet...

plaintext: a life without cause is a life without effect
plaintext (digraphs): al if ew it ho ut ca us ei sa li fe wi th ou te fx fe ct
ciphertext: nq sb gv ks qa cz nb ft gk qb se pi yg qk fp lk oy pi kz

The Playfair cipher's major advantage over other ciphers is that the ciphertext for a letter is not always the same and depends on its digraph pair. Plaintext 'he' and 'me' produce completely different combinations, effectively dulling the frequency distribution for popular letters. However, one small downside to the Playfair is that repreated digraphs will become the same ciphertext each time. Luckily, common digraphs tables are much more difficult to interpret than common letter tables since the most common digraph (th) is abundant only 3.25%, and the most common character (e) is 12%. The low percentages associated with digraphs makes decipherment based on digraphs much more difficult.

Of course, the Playfair is by no means unbreakable, and there are a few tips for solving these advanced mono alphabetic substitutions...