Viewing Single Post From: Brute Forcing The Adfgvx Cipher

Viewing Single Post From: Brute Forcing The Adfgvx Cipher
Donald	Apr 28 2008, 06:58 PM
Elite member Posts: 454 Group: Trusted Member Member #8 Joined: September 2, 2005	"Jdedge" converting it back to single characters using any arbitrary Polybius square would yield a result that was only a simple substitution away from the plaintext. AH! Ok, it took me a bit, but I think I see your point. If the sequence was the standard step 1: fractionate, step 2: transpose Then we just try various transpositions until our digram frequency looks like a mono substitution frequency. And that undoes step 2. From there, any polybius square will convert the remaining crypt text into a mono alphabetic substitution, which is easy to solve. Brilliant! And NICE!!!! BUT, If the sequence is: step 1: fractionate, step 2: transpose step 3: unfractionate THIS still baffles me. How do you undo step 3? All of your results should look more or less like random data. If you don't use the correct square, you get a digraph substitution of the original step 2 data, and the original step 2 data is scrambled to start with. I don't see any way to detect that you have gotten closer to (or hit) the correct step 2 text, nor do I see any way to use an incorrect digraph substitution to your advantage. Because the connection between the step 2 data and the original plain text is involved in the relationships between the letters positions in the fractionating step that happened in step 1. It's the relative positions. Right? an incorrect digraph substitution of step 2's output doesn't give us the same cipher, but with a different square for step 1. It gives us the wrong relationships between the letters, and therefore a completely invalid and useless result. Of course, as long as the relationships between the letters used in the squares in step 1 and step 3 are the same, they do not need to be the exact same squares. But that should still eliminate all but a tiny fraction of the keyspace as useless during a brute force attack. With a playfair you can use shotgun hill climbing. Just decrypt the cipher text with a random square and then make changes to that square keep the relationships that make the decrypted text look more and more like ordinary plain text. I only understand this approach on the most theoretical level, but I can see how it works sort of. BUT, with the 3 step ADFGVX I don't see how shotgun hill climbing would work because the intermediate text does not show digram or mono text frequencies. You have to undo the transposition to get something with normal digram frequencies. Which makes a brute force attack very impractical.


Brute Forcing The Adfgvx Cipher · General