| Viewing Single Post From: Brute Forcing The Adfgvx Cipher | |
|---|---|
| jdege | Apr 28 2008, 09:41 PM |
|
Elite member
![]() ![]() ![]() ![]() ![]() ![]() ![]()
|
Can I get by with "ADFGVX doesn't unfractionate, do it doesn't matter"? Didn't think so. Now if we're doing Bifid, with an even key-length, it doesn't matter, either, because then the two halves of each letter map to the two halfs of another letter:
Notice how the first letter of the ciphertext is constructed from the left halves of the first and second letters of the plaintext, and how the third letter of the ciphertext is constructed from the right halves of the first and second letters of the plaintext. In other words, with a key-length of four, the first and second letters of the plaintext entirely determine the first and third letters of the ciphertext, and the second and fourth letters of the plaintext entirely determine the second and fourth letters of the ciphertext. It can be solved as an ordinary digram substitution. With an odd key-length, this doesn't work. How to solve odd-length Bifids? I've a couple of explanations around here, but I've not studied them. (And the explanations aren't as straightforward as the even-length Bifid, so my fast skims haven't yielded anything useful.) As for the generic problem, with an unknown transposition? Not a clue. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
![]() |
|
| Brute Forcing The Adfgvx Cipher · General | |




![]](http://209.85.122.85/static/1/pip_r.png)


1:59 PM Nov 25