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$100 reward - The American ADFGVX Cipher (updated 7/10); aiming to forgo columnar transposition
Topic Started: Jun 22 2014, 01:58 AM (3,232 Views)
a649b9e0
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I found the challenge set a little limiting, so I ran my own. Your worries about the key character showing are well founded.
Let's for a moment assume that the headers are a known. Here are test runs encrypting each line of Alice in Wonderland and War and Peace:

Code:
 

[space]1:[space]AA,[space][space][space][space]420/2472[space]=[space]0.1699
[space]2:[space]XG,[space][space][space][space]301/2472[space]=[space]0.1218
[space]3:[space]XA,[space][space][space][space]227/2472[space]=[space]0.0918
[space]4:[space]DG,[space][space][space][space]200/2472[space]=[space]0.0809
[space]5:[space]AX,[space][space][space][space]191/2472[space]=[space]0.0773
[space]6:[space]GD,[space][space][space][space]131/2472[space]=[space]0.0530
[space]7:[space]AF,[space][space][space][space]127/2472[space]=[space]0.0514
[space]8:[space]DD,[space][space][space][space]113/2472[space]=[space]0.0457
[space]9:[space]FG,[space][space][space][space]101/2472[space]=[space]0.0409
10:[space]DX,[space][space][space][space][space]90/2472[space]=[space]0.0364
11:[space]VD,[space][space][space][space][space]72/2472[space]=[space]0.0291
12:[space]FX,[space][space][space][space][space]68/2472[space]=[space]0.0275
13:[space]VA,[space][space][space][space][space]67/2472[space]=[space]0.0271
14:[space]GG,[space][space][space][space][space]67/2472[space]=[space]0.0271
15:[space]XF,[space][space][space][space][space]54/2472[space]=[space]0.0218
16:[space]XX,[space][space][space][space][space]44/2472[space]=[space]0.0178
17:[space]AG,[space][space][space][space][space]43/2472[space]=[space]0.0174
18:[space]GX,[space][space][space][space][space]40/2472[space]=[space]0.0162
19:[space]FD,[space][space][space][space][space]35/2472[space]=[space]0.0142
20:[space]VX,[space][space][space][space][space]19/2472[space]=[space]0.0077
21:[space]AD,[space][space][space][space][space]19/2472[space]=[space]0.0077
22:[space]DF,[space][space][space][space][space]15/2472[space]=[space]0.0061
23:[space]AV,[space][space][space][space][space]15/2472[space]=[space]0.0061
24:[space]VG,[space][space][space][space][space]13/2472[space]=[space]0.0053


Code:
 

[space]1:[space]AA,[space][space][space]7268/55608[space]=[space]0.1307
[space]2:[space]XG,[space][space][space]5735/55608[space]=[space]0.1031
[space]3:[space]DD,[space][space][space]4957/55608[space]=[space]0.0891
[space]4:[space]DG,[space][space][space]4003/55608[space]=[space]0.0720
[space]5:[space]GD,[space][space][space]3971/55608[space]=[space]0.0714
[space]6:[space]AX,[space][space][space]3907/55608[space]=[space]0.0703
[space]7:[space]XA,[space][space][space]2866/55608[space]=[space]0.0515
[space]8:[space]FG,[space][space][space]2648/55608[space]=[space]0.0476
[space]9:[space]AG,[space][space][space]2552/55608[space]=[space]0.0459
10:[space]AF,[space][space][space]2375/55608[space]=[space]0.0427
11:[space]GG,[space][space][space]1956/55608[space]=[space]0.0352
12:[space]FX,[space][space][space]1870/55608[space]=[space]0.0336
13:[space]XF,[space][space][space]1833/55608[space]=[space]0.0330
14:[space]VD,[space][space][space]1723/55608[space]=[space]0.0310
15:[space]XX,[space][space][space]1654/55608[space]=[space]0.0297
16:[space]FD,[space][space][space]1512/55608[space]=[space]0.0272
17:[space]VA,[space][space][space]1190/55608[space]=[space]0.0214
18:[space]GX,[space][space][space][space]966/55608[space]=[space]0.0174
19:[space]DX,[space][space][space][space]885/55608[space]=[space]0.0159
20:[space]AD,[space][space][space][space]501/55608[space]=[space]0.0090
21:[space]DF,[space][space][space][space]421/55608[space]=[space]0.0076
22:[space]VX,[space][space][space][space]390/55608[space]=[space]0.0070
23:[space]VG,[space][space][space][space]185/55608[space]=[space]0.0033
24:[space]AV,[space][space][space][space]158/55608[space]=[space]0.0028
25:[space]DA,[space][space][space][space][space]22/55608[space]=[space]0.0004
26:[space]FF,[space][space][space][space][space]22/55608[space]=[space]0.0004
27:[space]XV,[space][space][space][space][space]21/55608[space]=[space]0.0004
28:[space]FV,[space][space][space][space][space][space]9/55608[space]=[space]0.0002
29:[space]DV,[space][space][space][space][space][space]2/55608[space]=[space]0.0000
30:[space]FA,[space][space][space][space][space][space]1/55608[space]=[space]0.0000
31:[space]VF,[space][space][space][space][space][space]1/55608[space]=[space]0.0000
32:[space]GA,[space][space][space][space][space][space]1/55608[space]=[space]0.0000
33:[space]XD,[space][space][space][space][space][space]1/55608[space]=[space]0.0000
34:[space]GV,[space][space][space][space][space][space]1/55608[space]=[space]0.0000
35:[space]GF,[space][space][space][space][space][space]1/55608[space]=[space]0.0000


These are a frequency analysis of the first digraph of the cipher texts. If we know the headers, we know AA is the 0, 0 position. Any guesses as to what the first letter of my key is? Also, we can start assembling a guess at the initial state of the alphabet in relation to the 0, 0 position. If we have plain text to the right handful of messages, it starts falling away like dominoes.

Now, let's take the opposite assumption, that which the headers are unknown. We still can make a 1 to 1 association between this state's digraphs and the letters it encrypts. Also, our set of possible key letter digraphs only goes up to 6 (AA, DD, FF, GG, VV, XX) as we know that the x and y axes of the state array share the same quantity at the key letter.
Edited by a649b9e0, Sep 15 2014, 12:56 PM.
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Elijah Cross
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What was limiting about the challenge set so far? Granted, it's a been a long time since I added any text - I've had to put this on the back burner for a while, but I'm about to work with a developer on this. Basically, I'm asking how much ciphertext would you like me to produce?


Also, I'm bumping up the reward to $100.
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Elijah Cross
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This challenge is still open, though enough time has passed that I've forgotten the keys.

Also, I think a way around the issue of a digraph possibly giving away that the plaintext/key letters are the same is to have a key that is exclusively digits; encrypt any digits in plaintext as words ("1" becomes "one", etc.) or have the letters A through J represent 0 through 9. Doing so would prevent a certain digraph from appearing in the ciphertext, and therefore, gives away which of the ADFGVX letters equals 0, and when a plain/key pair are in the same row/column, but as of right now I'm not sure that helps unravel things.

Finally, if anyone can tell me how/where to use the code provided in post #15, I'd appreciate it. Forgive me for being a complete programming noob.
Edited by Elijah Cross, Dec 14 2017, 12:52 AM.
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