Then replace each plaintext letter with the letter that forms the other corner of the rectangle that lies on the same row as that plaintext letter (being careful to maintain the order). Otherwise, form the rectangle for which the two plaintext letters are two opposit corners.If the two letters appear in the same column in the square, then replace each letter by the letter immediately below it in the square (cycling round to the top of the square if necessary).If the two letters appear on the same row in the square, then replace each letter by the letter immediately to the right of it in the square (cycling round to the left hand side if necessary).If the digraph consists of the same letter twice (or there is only one letter left by itself at the end of the plaintext) then insert the letter "X" between the same letters (or at the end), and then continue with the rest of the steps.There are a number of steps involved: Build a table like the following with the key square. p h q g m e a y n o f d x k r c v s z w b u t i l The key square is a 5 by 5 square containing all the letters except 'j'. On each digraph we peform the following encryption steps: The 'key' for a ADFGX cipher is a 'key square' and a key word. (For example, the results of A Message and a message should be the same. The extended version is actually very simple You simply also pick a keyword and write that into. When decoding, it translates 42 to (i/j). The polybius square was originally covered in CC1 Lesson 4. We must now split the plaintext up into digraphs (that is pairs of letters). For the Polybius square (example: Polybius('message') > '23513434112251'), the tests that you write should test that the following is true: When encoding, it translates the letters i and j to 42.
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