In both cases of the Playfair and Polybius with variants, the amount of material that the cryptanalyst is able to access is extraordinarily useful. By having large amounts of enciphered messages, the cryptanalysist can combine the messages to search for common numbers and digraphs. If one were able to read a large amount of Polybius with Variants encipherments, no doubt the large amount of three different numbers in nearly every message would led to the deduction of what 'e' was being represented by. Of course, obtaining large amounts of enciphered text can be difficult but it is still the most useful ability.

NOTE: The sender of messages can limit large amounts of messages from being useful to the cryptanalyst by changing the keyword frequently. There is no keyword for the Polybius variant system, though.

To solve Polybius Variant ciphers, one can take advantage of the fact that the second number of each letter indicates that it can be one of three letters. By making special frequency tables in which all numbers with a certain number as their second digit, say '3', are placed together, the cryptanalyst will be able to more accurately see how many letters have been represented with more than one letter. Evenutally, through probably word trial and error and frequency analaysis, Variants will fall under enough cryptanalytic time and effort.

In the case of the Playfair system, the probable word method and use of common digraph charts is the most important method. To help the cryptanalyst along in this task, though, there are several rules which they can take advantage of.

For starter, when the cryptanalyst is reconstructing the cipher alphabet grid, they should recognize that twice as many digraphs are formed from rectangular intersections than those of the column/row. Any letter in the grid can form a rectangle with sixteen other letters while only eight letters are in the same column or row. In short, when recreating the table, the cryptanalyst should look for retectangular arrangements first because of the 50% better chance of being right.

Still, how do you even know how to begin reconstruction of the cipher alphabet grid? It all begins with a hypothesis. You should try to identify one of the more common digraphs or find a word. If, for example, you spot anagrammed digraphs, like 'atta', you should realize that the plaintext represented by those symbols follows the same pattern -- so that the digraph may be part of a word like 'follow,' 'clippings,' letter,' or other words. This is caused by the reversibility of Playfair system, in which opposite digraph pairs represent opposite plaintext pairs. If you spot any sequences like this, you should try to figure out likely words which would fit into place.

After you've identified one digraph and it's plaintext equivalent, you should focus on expanding on your accomplisment by exploring the other pairings. If you found that 'qk' represented 'th', for example, you should then find all other digraphs with a 'q' for the first letter. Each of these digraphs have about a 20% chance of having a plaintext equivilent in which 't' is also the first letter. Similarly, all the digraphs with 'k' have about a 20% chance of representing 'h'. The origin of the percents are clear when one see all of the possible digraphs for the ciphertext 'q' as the first letter.

Of the twenty four cases, 'q' represents plaintext 's' in eight cases and 'l,' 'm,' 'x,' and 't,' four times each. In our case of the discovering plaintext 'th' was cipher digraph 'qk', we were a bit unluckly. This would mean that the chances of other digraphs beginning with 'q' and representing 't' is only 4 in 24, or about 17%. If we had identified a cipher digraph beginning with q as representing 's', though, the chances of other digraphs beginning with 'q' and representing 's' would have been 8 in 24, or about 33%. BUT, since when you discover what a digraph represents you don't know whether you're dealing with a 17% or 33%, the 20% chance is a good estimate.

One final tip: Remember the reversibility of Playfair digraphs! If you believe that 'qk' represents 'th', make sure you check for 'kq' digraphs in the message, which would then represent 'ht'. 'ht' is a fairly rare digraph so you should find very few of these and an abundance of 'qk'. When you begin identifying digraphs, remember to note that you also identified the recipricol and search the message for those. The reversibility of the Playfair cipher is a major weakness that you can exploit! You can test your hypothesises of what a digraph represents by seeing the recipricol meaning makes sense in another area.

Solving a Playfair cipher is more difficult than those trivial monoalphabetic substitutions. You can't rely on the frequency distribution alone anymore. You'll have to test theories and spend a good deal of time erasing failed attempts the first time you solve a Playfair. But just stick with it!

Well, it's the end of the lesson again, which means it's time for another CryptAgent Challenge! Since the advanced mono alphabetic substitutions require a bit more effort, we're giving you two different challenges. One of the challenges requires you to enchiper specified text using a playfair substitution and a keyword that we give you (pretty easy), while the second challenge requires you to crack a same of enciphered text created using a playfair substitution (sorta hard). Good luck with the challenges, and click here to open the CryptAgent Challenge window, or on the button below... If you're not a CryptAgent, you can find out more and register if you like. It's free, and it's fun!