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I have just started looking into the cracking of the Vernam cipher with two ciphertexts encrypted with the same key by XORing them and then crib dragging and I was wondering if it is made easier if you have more ciphertexts like 3 or 4, and if so how that would be used to the crackers advantage?

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Yes, in practice the more ciphertexts you have with the same key, the easier it is to break the encryption:

  • With more ciphertexts, you'll have more chances of hitting a crib, the more plaintext you'll reveal when you do, and the easier it is to be sure that a particular match is correct and not just plausible-looking by chance.

  • The more ciphertexts you have, the more opportunities you'll have for extending a decrypted segment, since you only need to guess how one of the messages continues outside the segment.

This homework exercise with 11 messages XORed with the same key is a good example: you can solve the puzzle almost entirely simply by starting with the assumption that the plaintexts (mostly) consist of ASCII letters and spaces, and observing that — due to the way the ASCII code is arranged — XORing a space with a letter has a specific and distinctive effect (it just flips the case of the letter) that is easily distinguishable from the result of XORing two letters (producing numbers, punctuation and/or non-printable control characters) or two spaces (which just produces a null byte). With 11 plain English ASCII messages, most positions in the aligned ciphertexts end up containing a space in at least one of the messages, and the few remaining gaps are then easy to fill in.

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