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Cryptography

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    $\begingroup$ What is the purpose of interleaving, rather than concatenating the bits? And what properties do you believe this combination would have that others do not? $\endgroup$ Commented Jan 16, 2016 at 8:30
  • $\begingroup$ Interleaving the hash outputs to make the HMAC key means that each HMAC depends on bits from every hash algorithm, while concatenation would end up with each HMAC just being $HMAC_{H_n}(H_n(x), x)$. If the key depends on all the base hash functions, forging any single element of the final concatenation of HMACs should require breaking all the base hash functions (or at least the bits of each hash used for that key). $\endgroup$ Commented Jan 16, 2016 at 22:32
  • $\begingroup$ With your equation that uses full $K$ for each HMAC, concatenation would work just as well – I missed the part where you said one could drop some of the bits. By "what properties", I meant what you think the collision resistance, preimage resistances etc. would be here. $\endgroup$ Commented Jan 17, 2016 at 10:04
  • $\begingroup$ I would expect collision resistance at least as good as the strongest hash (as for just concating the base hashes, since HMAC is ultimately the output of a call to the base hash). I would expect similar properties for preimage resistance, since each part of the final output depends not only on the data but also a pseudo-random key derived from all the base hashes. Unlike simple hashes, finding a preimage for one of the base hashes still doesn't let you use that string to generate the other HMACs without the key, so you can't use the others to select among the possible preimages. $\endgroup$ Commented Jan 17, 2016 at 19:32