Take the 2-minute tour ×
Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. It's 100% free, no registration required.

I've read something to the effect that the HMAC construct is able to lessen the problem of collisions in the underlying hash.

Does that mean that something like HMAC-MD5 still might be considered safe for authenticating encrypted data?

share|improve this question
    
Collisions are irrelevant in typical MAC usage scenarios. –  CodesInChaos Jul 20 '13 at 12:24
    
Somewhat relevant crypto.stackexchange.com/a/6753/5231 –  rath Jul 20 '13 at 12:31
    
@CodesInChaos does that mean that HMAC-MD5 is secure for use in an encrypt-then-mac scheme? –  Nuoji Jul 20 '13 at 15:04

1 Answer 1

up vote 8 down vote accepted

Yes, there are currently no known attacks on HMAC-MD5.

In particular, after the first collision attacks on MD5, Mihir Bellare (one of the inventors of HMAC) came up with a new security proof for HMAC that doesn't require collision resistance:

"Abstract: HMAC was proved by Bellare, Canetti and Krawczyk (1996) to be a PRF assuming that (1) the underlying compression function is a PRF, and (2) the iterated hash function is weakly collision-resistant. However, recent attacks show that assumption (2) is false for MD5 and SHA-1, removing the proof-based support for HMAC in these cases. This paper proves that HMAC is a PRF under the sole assumption that the compression function is a PRF. This recovers a proof based guarantee since no known attacks compromise the pseudorandomness of the compression function, and it also helps explain the resistance-to-attack that HMAC has shown even when implemented with hash functions whose (weak) collision resistance is compromised. We also show that an even weaker-than-PRF condition on the compression function, namely that it is a privacy-preserving MAC, suffices to establish HMAC is a secure MAC as long as the hash function meets the very weak requirement of being computationally almost universal, where again the value lies in the fact that known attacks do not invalidate the assumptions made."

However, this does not mean you should use HMAC-MD5 in new cryptosystem designs. To paraphrase Bruce Schneier, "attacks only get better, never worse." We already have practical collision attacks for MD5, showing that it does not meet its original security goals; it's possible that, any day now, someone might figure out a way to turn those into a preimage attack, which would compromise the security of HMAC-MD5. A much better choice would be to use HMAC with a hash function having no known attacks, such as SHA-2 or SHA-3.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.