How you solved 5B?

[Anonymous]

Would anyone like to discuss how they solved 5B and what the encryption key was because I didn’t do it myself and it would be good to know how to solve it (even though I can’t now anyways!)

[Anonymous]

5B is a Beaufort cipher with a key of Arcana Imperii.

[Anonymous]

Can’t say how we solved it since there are hints in that for this challenge, but there are lots of hints in 5A, and also the official help thread. Check it out!

[Anonymous]

Well, our team decrypted part A, by removing the massive sequence of double letters, then using frequency analysis and decoding it monalphabetically. We then used the key to decrypt the sequence of U’s and E’s. Apparently the sequence was a stegonography, so our team found the factors of the number of characters in the picture to find the picture’s dimensions and made up the picture, which had the encryption key for Part B in it. We then attempted different polyalphabetic ciphers before solving it as a Beaufort.

[Anonymous]

We solved 5B by reversing all the letters (A—>Z, B—>X etc.) and then running it through a Vigenere Cracker I have.

[Anonymous]

Thank you.

[Anonymous]

5B was a Beaufort cipher, and a quick Google tells you that a Beaufort cipher can be solved just like a Vigenère once you turn all the As into Zs, all the Bs into Ys, etc. This meant that it was very easy to just change my Vigenère hill-climbing solver to work for Beaufort ciphers as well.

[Anonymous]

5B used the Beaufort cipher, and the key was given in part A (“ARCANAIMPERII”).

If you didn’t find the key in part A, you could have solved it by enciphering 5B’s ciphertext using an Affine cipher with a=25 and b=0 (ax+b mod26), and then deciphering the result as you would with a standard Vigenere-enciphered text. The key you find from solving it with this method is the same as the original key being enciphered with Affine (a=25 and b=0), which would be “AJYANASOLWJSS”.
This is a shortcut to solving any Beaufort-enciphered text, by encrypting the ciphertext using Affine a=25 and b=0, then solving the result like you would with a standard Vigenere-enciphered text.

[Anonymous]

Really, same key as the previous challenge’s Vigenere? Mutter… grumble…

[Anonymous]

The algorithms for Vigenère, Beaufort-variant and Beaufort are similar:

Where C=Ciphertext(position i); p=plaintext(i); k=key(i mod keyword length)

Vig: C = p + k (Decode with p = C – k)
BF variant: C = p – k (Decode with p = C + k)
Beafort: C = k – p (Decode with symmetric p = k – C)

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