CAP Cryptographic Analysis Program

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CAP Cryptographic Analysis Program

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CAP Cryptographic Analysis Program

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Presentation transcript:

CAP Cryptographic Analysis Program Breaking Autokey-Plaintext Help Presentation Press Enter or click on your mouse button to continue

Breaking the Autokey-Plaintext This is not as easy to break as the autokey-ciphertext but . . . A Kasiski-like method exists to determine the “group length” The group-length is the actual keyword length but since the keyword is not repeated, the group-length describes the length of segments of the running key These segments define the distance between a plaintext letter being enciphered and becoming a key letter

Group Length For example, if the keyword is “alice” a possible cipher structure could be given by 5 a l i c e t h i s i s a n e x t t is a key letter t t is enciphered t h i s i s a n e x a m p l e

Finding the Group Length It is possible to discover the group length from the ciphertext Assume the group length is 5 then every plaintext letter will be enciphered by the letter 5 positions to its left and in turn will encipher the letter 5 positions to its right. Since plaintext has a lot of repeated characters there will be an unusually large number of identical letters separated by the group length Example R There is a good chance that this kind of event will occur Key: . . . R . . . S Plaintext: . . . S . . . R Ciphertext: . . . J . . . J S Group length

Procedure Find the distances between all letter repetitions The most common distances are strong candidates for the group-length CAP will automatically calculate these distances:

Using the Group Length a l i c e t h i s i s a n e x t Knowledge of the group length reveals how the plaintext is divided into segments. If the group length is 5, then the first segment consists of every 5th letter beginning with the first letter We know the encipher chain for this segment, for example: a l i c e t h i s i s a n e x t s t h i s i s a n e x a m p l e a t s q u m l h v w f s m c p b

Brute Force Attack We can use a brute force attack because each segment is defined by its beginning character (which is from the key word) For example: a Guess Ciphertext (known) Key character (unknown) plaintext (unknown) t ? ? t t Try all 26 guesses l ? ? t s s a s s ? ? Use a low frequency character test