Isn't the structure of a potential plaintext of a ciphertext generated by a one-time pad dispositive, cryptanalysis-wise?

That is, if I generate every possible result, and analyze the structure against, for example, a dictionary, won't that identify which of the solutions in the domain is most likely the correct one?

I realize the numbers involved, but this does ultimately reduce the problem to a hardware one. If this is not true, then I'm not understanding something correctly about one-time pads.

TIA.

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1 Answer

No that is not correct, here is the thing, given a ciphertext say ezcle, there exists a key such that this would decrypt to hello, another key such that this decrypts to harry, another key which will decrypt to frank, another key which will result in world. And every other 5 letter word in the dictionary, and every other 5 letter combination of letters (I'm restricting to letters to make things simple).

Sure there are a lot of outputs that are garbage, but there are a lot of outputs that are not garbage, but every 5 letter word in the dictionary will also appear. Thus, before doing this cryptanalysis, you had a pretty good idea that the ciphertext encrypted a 5 letter english word say. After your cryptanalysis, you have not eliminated a single 5 letter word from the possible plaintexts. Thus you have gained no extra information by doing the cryptanalysis.

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Short and sweet. Excellent answer. –  pg1989 May 25 '13 at 7:20