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The problem with the HMAC-based solution you drew up is if the shared secret $s$ has low entropy; for example, it's actually a password that could conceivably be in a dictionary. In this case, someone could listen to the exchange $r_1, \operatorname{HMAC}(C \mathbin\| r_1, s)$, and go through his dictionary of possible values of $s$, and see if any one of ...


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The general answer is "no, you can't authenticate someone over a wire unless they know some secret information." The only way to authenticate someone is to have them do something that no one else can do. If you're trying do to it by just sending messages between the parties, the only way I can do something that someone else can't is if I know something they ...


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Yes, if your encryption algorithm is reasonable secure. Given fixed length messages (in your case the ciphertext), CBC-MAC is a secure MAC scheme, meaning that an attacker, without knowing the key, cannot produce a valid message-tag pair with non-negligible probability. Furthermore, according to this paper, Encrypt-then-MAC is the best procedure, while ...


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First of all the Additional Data (AD) is not a tag. It is data that is also authenticated by the authentication tag. This authentication tag is appended to the ciphertext by libsodium. The tag doesn't consist of separate portions for AD and the ciphertext (and IV), the AD is taken into account during calculation of the tag. The AD can be any data, including ...



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