2
$\begingroup$

It seems to me that cryptanalysis of the autokey cipher relies on patterns being found in key attempts, easy when it's English you're looking for, but how can you do this if you're just getting further cipher texts?

How would someone decrypt a plaintext enciphered with an ordinary vignere followed by an autokey cipher, where the two keys are 10-20 characters long and random?

Presumably any solution would rely on a computer, but I'd also be interested if there was a method that could be done by hand.

Thanks

$\endgroup$
1
$\begingroup$

It can still be done, because some letters of the English alphabet are used more often. Analyzing a ciphertext based on the frequency of appearance of some characters is fittingly called Frequency analysis.

This method of cryptanalysis gets better, the more ciphertext you have. If the plaintext is in English, then you would know, that the most frequently used letter would be "e".

Because the Vigenère cipher is quite an old method of encoding something it can be (and has been) broken relatively easy by hand. The security relies on the keylength and the "randomness" of the key. 10-20 random letters is actually pretty good, but with a very long ciphertext you can ultimately still break the Vigenère method.

| improve this answer | |
$\endgroup$
  • $\begingroup$ How would you decrypt the autokey cipher if it's not English that you're getting, but a Vignere cipher text? The letter frequencies would all be wrong because it's just a cipher text. Would it just be brute force? $\endgroup$ – DuckDuke Jan 22 '19 at 13:48
  • $\begingroup$ No, you would do the frequency analysis on the cipher text. You would find, that the letter in the cipher text with the highest frequency was likely a substituted "e", but that only works for very long cipher texts. The easier way to do this: First use Kasiski examination to find out the length of the key. After that the frequency analysis can be done on the ciphertext. $\endgroup$ – AleksanderRas Jan 22 '19 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.