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The letters could reference a page of randomly generated letters two recipients shared and the "key" is simply needing that page. A decipher would require that page, without it, it is completely impossible to crack...

For example lets assume something simple like A=1, B=2, C=3, etc. was a translation, and to translate that we counted along a string of randomly assigned letters to find the result. If we then continued from one position to the next and so on, and the string was completely random then nothing will solve it.

eg.

"1.1.4.6" written on pigeon leg and "telwpsmysqpt" is a random string both parties have

"1.1.4.6" applied to "telwpsmysqpt" results in "telwpsmysqpt" otherwise written as "test"

The above has no system, if you simply had 1146 there is no telling what it could be without having "telwpsmysqpt", whilst a very secure method its very slow to translate even if you have the string which would need to be fairly long to accommodate a long message. Of course it could be reused for long passages of text - but of course then it starts to have repeats and patterns and might even be crackable.

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It's funny how people point to one-click online utilities thinking it'll work when even GCHQ gave up.

then you have an appallingly boring dull view of the world and a really bad sense of humour. As it was my first post in this topic was

Thing is, if this code requires a one time pad to decipher it we're kinda screwed. The first and last block being identical could mean several things. It could be part of a key to decipher the code or it could simply indicate the beginning and end of a message.

i have no disillusions about the chances of someone on Neowin here cracking such a code, but people are having fun. You however, are just being a nob.

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Unless the key you come up with for a WW-II dead pigeon's leg letter reads: "Attack London - Signed Adolf" - I would say you could be pretty sure you cracked it then ;)

Coming up with a one-time pad to decrypt the message to say that is trivially easy, though.

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