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• #45154
Macss
Member

Okay, so I am just completely confused at this point. I feel like I’m most of the way there, according to all the hints going around, but I don’t know where I’ve gone wrong, so I’m just going to post what I’ve done and hope that someone can help me/ tell me where I’ve gone wrong.

So, first I converted the ciphertext into the 28 columns and got the 2’s in columns 15-21 as everyone else. After this, I swapped the columns with 2’s in (15-21) with other columns . I eventually settled on positions 4,8,12,16,20,24 and 28, for the 2’s to be in. After this is, I put it into my frequency analysis tool, and found that there were 24 quadgrams (groups of 4), possibly each one representing a single character, as Harry said that there were 24 characters in the plaintext. I had all of this done by Jan 2nd, and since then I’ve been stuck. I’ve tried replacing each of the quadgrams with a random digit of the alphabet and then done frequency anlysis and IoC on this, but I haven’t gotten any results. I seem to have tried everything I can think of, and now I’ve just been trying to solve the cipher as a Vigenere with the quadgrams represented as random letters, but that doesn’t work either.
We’ve only got 2 days left, and at this point I just want to solve the cipher for the satisfaction, and as a challenge. Hope someone can help me finish this off, as its getting ridiculous now.

Thank you.

#45162
Harry
Keymaster

You are nearly there. This is a variant on the Trifid cipher, where instead of using 3 ternary numbers to represent one of 27 characters, we have used three binary and one ternary to represent 24 letters (leaving out the usual culprits in the text). The plain text has 4620 characters in it (without spaces and punctuation. You can work out the pattern (2,2,2,3 or 2,2,3,2 or 2,3,2,2 or 3,2,2,2) by studying the patterns of twos in the cipher text, then you need to work out the chunking. Good luck!

Harry

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