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When you are e.g. sending TLS encrypted data over a SSH tunnel, there are two things in particular that should be noted: The TLS handshake will only commence, once the SSH connection has been established. The bulk encryption keys of TLS will be completely independent of the SSH encryption keys. Since the handshakes and keys are completely independent, ...


1

To answer the question "why does a block cipher use a Mode of Operation", we need to first examine the question "what is a block cipher?" A block cipher is a keyed operation that converts a string of N bits to a string of N bits (where N is usually fixed by the block cipher; for AES, N=128), in a way that, without the key, looks like a random permutation, ...


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Such a weak key schedule was chosen since the key schedule "theory" was not well developed by that time. Designers just modified the key schedule of DES a bit. Remember that these key schedules had to be optimized for hardware, and any extra operation would cost something in terms of area.


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AES was designed to behave like an ideal cipher. An ideal cipher has no weaknesses when used with a non-uniformly distributed key (beyond that inherent in the non-uniform distribution of the key). Therefore, if AES does indeed meet its design goals, there are no shortcut attacks on AES that exploit special properties of AES, when using non-uniformly ...


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As far as I know, NO, there has not been any cryptanalysis of AES under a non-uniformly distributed key. That holds even if we let the adversary decide what the non-uniform distribution is. Of course we should adjust the expected difficulty of attack according to the entropy remaining per the distribution; but typically, hashing a low-entropy password looses ...


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I assume you are asking how the password a user enters to unlock their saved user data is translated into the keys used by 3DES? According to this link DES-EDE-CBC, which is only two-key 3DES. The first and third keys are set equal to each other, resulting in two keys of 56 bits or a total size of 112 bits. To get that key material from a password, a ...


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If you use a key for close to $2^{n/2}$ blocks in CBC mode, then the chance of getting a collision in the ciphertext is getting rather high because of the birthday paradox. As the ciphertext is used as a vector for the next calculation, and since that vector should be unpredictable, you would likely lose confidentiality. Note that the author seems to have ...



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