Frequent These Parts

Discussion of challenges you have already solved
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lukas
Posts: 34
Joined: Wed Nov 26, 2008 1:53 pm
Location: Germany

Frequent These Parts

Post by lukas »

very simple with the right tool ^^


http://www.youtube.com/watch?v=Q7AJSAxuCkE&fmt=18
PeterS
Posts: 24
Joined: Thu Mar 05, 2009 7:17 pm

Post by PeterS »

As I read the title I immediately knew this was Vigenere, because I had it in school when we were talking about cryptography.
But the decryption tool I wrote back then was not finished (it doesn't interpret the frequency analysis, so it can't find the key by itself), so i used the online version of a script called pygenere:
http://smurfoncrack.com/pygenere/pygenere.php
theStack
Posts: 72
Joined: Sun Nov 02, 2008 12:46 am

Post by theStack »

I used the same tool than PeterS, found the link in the wikipedia article to the Vigenere cipher.
It was quite impressing to me that it can be decrypted *that* fast, I knew that it was possible but always thought it involves some bruteforcing which could take a bit of a time.
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sabretooth
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Joined: Sun Jul 12, 2009 3:13 pm

Post by sabretooth »

i used http://cs.colgate.edu/faculty/nevison/C ... acker.html

and assumed the first two words were 'this cipher' therefore built it up that way
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FreeFull
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Joined: Fri Apr 22, 2011 6:26 pm

Post by FreeFull »

I knew nothing about polyalphabetic ciphers beforehand but correctly guessed that this was a polyalphabetic cipher and that it would have about equal letter frequency (which it did), researched possible ciphers on wikipedia, guessed that this was the vigenere cipher, tried manually cracking the cipher which was doomed because I assumed the wrong key length (4 instead of 8), and then found the same tool as PeterS.
lifthrasiir
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Joined: Tue Aug 16, 2011 12:35 pm

Post by lifthrasiir »

I have used Kasiski examination to guess the key length. Four out of five four-character repetitions are found to have the distance of multiples of 8 ("iudz" for 136, "rgai" for 64, "dzbh" for 152, "sdyc" for 184), so the key length is likely 8. Since the word boundary is already known, it is easy to guess the key now. (For example, "W yt" is very likely to be "I am".) The key was not in the dictionary (two words concatenated), so I was initially confused but the first part ("This cipher...") ensured me the correct direction.

The only hard part of this problem was knowing the correct cipher algorithm.
compudemon
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Joined: Sat Aug 13, 2011 2:13 pm

Post by compudemon »

i guessed how the cipher worked then dictionary brute forced ctihuolhlacat against all known English 13 letter words thankfully the key repeated mostly over the length this word
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Hippo
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