# Cracking the Polybius Cipher

A Polybius square cipher is not difficult to crack, even though it initially appears so unlike normal English because so few characters are used. But this apparent strength of the cipher is actually a weakness - it means we only have a few combinations of symbols to deal with. If using a 5x5 square, each plaintext letter is turned into one of only 25 possible two-character coordinates, and each letter is always turned into exactly the same coordinate. The Polybius cipher, therefore, is just a monoalphabetic substitution cipher, with the only difference being that each letter in the plaintext has been replaced not by a single unique letter or symbol, but by a unique pair of letters or symbols.

#### Step 1: create a Polybius square

By studying the ciphertext, it is easy to determine the headings used in the Polybius square. In the example above, it is clear that it is the letters A-E for both the row and column headings.

(It is possible, but unusual, for different symbols to be used for the row headings and for the column headings. This will not make the process much more difficult: the row headings will be used for all the odd-numbered characters in the ciphertext, and the column headings for all the even-numbered ones.)

Once the headings have been extracted from the ciphertext, create a Polybius square based on those headings, and fill it randomly with 25/26 letters.

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

#### Step 2: rewrite the ciphertext using the Polybius square

AE DE EA DB BB AE AE CA EA DB EA AE DA EA EA BE AE DE DC AB EA ...

Work through each pair in the ciphertext: these pairs must not overlap, so the second pair in the example above is DE, not ED. Look up the first pair in the newly created Polybius square to find the corresponding letter. This letter becomes the first letter in our rewritten ciphertext. Repeat the process for all the remaining pairs:

• AE >>> E
• DE >>> U
• EA >>> V
• DB >>> R
• BB >>> G
• ...
If we continue this process for the ciphertext above, we get the following:

EUVRGEELVRVEQVVKEUSBVEWFTKNESCVVHEUVTWBVXWVEAVBBGNVBBVXYWVGKPEUSBVEWFTKNESTKEVWXVHEGKPYKWGMVLEUVATBAYXULTCVEUVKVMVWVKPTKNGWABWGXVRVEQVVKHWVPGESWGKPHWVFTKEUVQTLPGBEUVHWVFGKTAGLBHVXTVBVMSLMVBESPVMVLSHGRVEEVWPTBNYTBVGKPESWYKDGBEVWBSEUVHWVPGESWVMSLMVBESPVMVLSHRVEEVWVFVBTNUEGKPEUVXGHGXTEFESWYKDGBEVWDSWXTKNEUVHWVFESVMSLMVGNGTKESBYWMTMVBTATLGWLFGKVQXTHUVWTBXWVGEVPGKPAGFTKTETGLLFBYWMTMVGEEVAHEBESXWGXCTERYESKXVMYLKVWGRTLTETVBTKTEGWVPTBXSMVWVPBSEUVXSPVAGCVWBEWFESPVMVLSHASWVBSHUTBETXGEVPXTHUVWBESNYGWGKEVVEUVBVXYWTEFSDEUVTWAVBBGNVBKSQUSQVIGXELFPSQVOYTZEUSBVXUVVCFGKEB

#### Step 3: solve the simple substitution cipher

The rewritten ciphertext is basically a simple substitution cipher. These can be solved automatically here. Alternatively, ou can read more about the process of cracking a subsitution cipher here.

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