![]() ![]() Here also, the last element of the column circularly follows the first element of the same column. If the alphabets in the pair occur in the same column, then replace them with the alphabet immediate below them.the last element of the row in the matrix circularly follows the first element of the same row. If an alphabet of the pair occurs at extreme right then replace it with the first element of that row, i.e. If both the alphabets of the pair occur in the same row replace them with the alphabet to their immediate right.To find cipher alphabets follow the rules below. For that, take the first pair of plain text and check for cipher alphabets for the corresponding in the matrix. Step 3: In this step, we will convert plain text into ciphertext. If a pair has ‘xx’ then we break it and add ‘z’ to the first ‘x’, i.e.In case while making pair, the last pair has only one alphabet left then we add ‘z’ to that alphabet to form a pair as in our above example, we have added ‘z’ to ‘w’ because ‘w’ was left alone at last.Like in our example letter ‘rr’ occurs in pair so, we have broken that pair and added ‘x’ to the first ‘r’. In case, pair has the same letter then break it and add ‘x’ to the previous letter. Pair of alphabets must not contain the same letter.Step 2: Now, you have to break the plain text into a pair of alphabets. Note: If a key has duplicate alphabets, then fill those alphabets only once in the matrix, and I & J should be kept together in the matrix even though they occur in the given key. Then put the remaining alphabet in the blank space. ![]() Step 1: Create a 5X5 matrix and place the key in that matrix row-wise from left to right. We will discuss the further process in steps. Now, we have to convert this plain text to ciphertext using the given key. Let us discuss the technique of this Playfair cipher with the help of an example: ![]() Playfair cipher is a substitution cipher which involves a 5X5 matrix. Then in the entire plain text wherever alphabet ‘p’ is used, it will be replaced by the alphabet ‘d’ to form the ciphertext. In simple words, if the alphabet ‘p’ in the plain text is replaced by the cipher alphabet ‘d’. Monoalphabetic cipher is a substitution cipher, where the cipher alphabet for each plain text alphabet is fixed, for the entire encryption. The hacker is also aware of the encryption and decryption algorithm. If the hacker knows that the Caesar cipher is used then to perform brute force cryptanalysis, he has only to try 25 possible keys to decrypt the plain text. There are also some drawbacks of this simple substitution technique. So, while counting further three alphabets if ‘z’ occurs it circularly follows ‘a’. For an example of how this is done, see en.Wikipedia.Note: If we have to replace the letter ‘z’ then the next three alphabets counted after ‘z’ will be ‘a’ ‘b’ ‘c’. en./w/index.php?title=File:English_letter_frequency_(alphabetic).svg&page=1 PD There are 35 choices for what A maps to, then 34 choices for what B maps to, and so on, so the total number of possibilities is 35*34*33*…*2*1 = 35! = about 10 40 en./w/index.php?title=File:Caesar3.svg&page=1. In addition to looking at individual letters, certain pairs of letters show up more frequently, such as the pair “th.” By analyzing how often different letters and letter pairs show up an encrypted message, the substitution mapping used can be deduced. ![]() Because C, A, D, and J show up rarely in the encrypted text, it is likely they are mapped to from J, Q, X, and Z. Since W is the second most frequent character, it likely that T or A maps to W. Because of this ease of trying all possible encryption keys, the Caesar cipher is not a very secure encryption method.īecause of the high frequency of the letter S in the encrypted text, it is very likely that the substitution maps E to S. In this case, a shift of 12 (A mapping to M) decrypts EQZP to SEND. ![]()
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