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I'm going to agree with @fgrieu's marvelous post above in a back-handed way. My answer is: No, you don't have to use an HMAC. Do it anyway. As you noted, some hashes, sush as SHA-3 (especially in its Keccak form), Skein (which I was a team member on), and others will work just fine. In the case of Skein, there is a one-pass Skein-MAC that has a proof of ...


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Security The level of security is likely to depend on the cryptographic primitives - the actual hash function and cipher - used. It is very likely that you can construct a function that is insecure, e.g. where the cipher is used for both the hash function an encryption. So you need to prove that the hash function and the encryption primitive are not ...


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In summary: Yes, HMAC is the way to go for construction of a MAC from an arbitrary concrete iterated hash. We have no constructive argument of security of the MAC constructs in the question; we even have a concrete attack when using some otherwise apparently fine hashes. I consider a hash constructed by iterating a compression function $F$ as ...


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Your scheme is secure under the assumption that AES is a pseudo-random permutation (PRP): In fact, message authentication codes (MACs) are theoretically modelled as pseudo-random functions (PRFs), and any PRP is also a PRF. Therefore, your usage of single-block AES essentially is a MAC. In the case that only a very small subset of the possible blocks is ...


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Summary: HMAC-MD5 with a key later revealed is a completely insecure way to do commitments of messages that can be chosen malignantly, because of the ease with which MD5 collisions can now be found. There is no compelling evidence that's so insecure for messages constrained to belong in a small arbitrary set that no adversary can choose or influence; still, ...


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The construction you are proposing is called the "envelope" or "sandwich" MAC, it predates HMAC, and it is in fact secure—provided the key and message are appropriately padded. That is, $$ \text{SHA256}(k \parallel m \parallel 1 \parallel 0^{b - 1 - (|m| \bmod b)} \parallel k) $$ is secure, as long as $k$ is the underlying hash function's block length $b$ ...


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PGP [1024-bit] digital signature vs SHA256 HMAC Comparison... First, you can compare asymmetric and symmetric algorithms. A 1024-bit asymmetric key provides about 80-bits of security. A SHA256 HMAC provides about 128-bits of security. With all other things being equal, the HMAC is stronger. Second, I believe PGP (or is it GnuPG) uses Lim-Lee primes and ...


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Questions like these are hard to answer because who can predict what the future holds, right? That said, there are some things I wanted to share. Prefer symmetric cryptography over public-key cryptography. Prefer conventional discrete-log-based systems over elliptic-curve systems; the latter have constants that the NSA influences when they can. From NSA ...



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