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Differential Cryptography (Posted on 2004-09-19) Difficulty: 2 of 5
Decode this if you can!


See The Solution Submitted by Juggler    
Rating: 1.7500 (4 votes)

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re(2): Interesting Comment 7 of 7 |
(In reply to re: Interesting by Larry)

"One could have full use of all 26 letters if the cryptkeeper had elected to either blur all the words together, or insert a 27th character as a delimiter.  If there was a delimiter, of course, then we'd have to ignore it when calculating the differences."

If mod 27 were used, the given message would have come out:


Of course the starting letter is arbitrary. Just to illustrate the point, if you start with Q, it comes out:


with the blank just taking the place of the at sign in the ascii sequence--or non-computerwise, after Z and before A in the cycle of 27.

  Posted by Charlie on 2004-09-20 08:59:44
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