I know in great (great great great) detail about how they cracked enigma during the second world war. Looking into it, I decided to have a crack at breaking it. I've come apon a single problem with my (theoretical) attack(s) - the most efficient way to crack plugboard settings. So, is there an attack that is more efficient than bruteforcing the plugboard, when the plaintext could be anything (Cribbing being useless)

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    $\begingroup$ When the plaintext "could be anything", there is no way to break the system. Proof: for whatever setting of the machine, we can decipher the ciphertext per this setting, and we get a plaintext which enciphers to the ciphertext and does not contradict the (absence of hypothesis) on plaintext. We need to know SOMETHING about the plaintext. $\endgroup$
    – fgrieu
    Jan 6, 2020 at 18:43
  • $\begingroup$ @fgrieu there is the fact brute-forcing enigma is hard (though someone used digital ocean and did it in 13 minutes with GPU's (about 1000 high), the question was more something to do with possible side-channel attacks I was unaware of. I'll decide a plaintext though and edit with it. With information of course... $\endgroup$
    – Legorooj
    Jan 6, 2020 at 21:55
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    $\begingroup$ @fgrieu: I do not think that proof is correct. Enigma messages contained a Spruchkopf which was meant to allow the receiver to verify that their machine was set up correctly to decrypt an incoming message. To a brute-force adversary, this clearly leaks a lot of information about the key even if the messages themselves are uniformly distributed. For the same reason, information-theoretic attackers can always learn something about the key (or at least about the authentication key) e.g. in authenticated encryption schemes, even if nothing is known about plaintexts. $\endgroup$
    – Polytropos
    Jan 7, 2020 at 21:56
  • $\begingroup$ @Polytropos : agreed, my proof considers the Spruchkopf as part of what's known about the plaintext. IIRC, there was one, but it has evolved, and the amount of information it reveals is low compared to Enigma's key entropy. $\endgroup$
    – fgrieu
    Jan 8, 2020 at 4:15


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