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Although SHA-1 theoretically has collisions, HMAC-SHA-1 which is based on SHA-1 is still widely used (in TLS for example) and is considered to be secure. How is that possible?

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    $\begingroup$ cseweb.ucsd.edu/~mihir/papers/hmac-new.html $\;$ $\endgroup$ – user991 Aug 10 '14 at 23:19
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    $\begingroup$ The gist is that the HMAC construction has properties which allow its security definition to be met even if the hash used internally is not a cryptographic hash. It need only meet a weaker definition, which is why HMAC-MD5 is also still considered secure. $\endgroup$ – pg1989 Aug 10 '14 at 23:43
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As shown in the paper Ricky Demer linked in the comments, HMAC can be secure even when the underlying hash function is not collision resistant. Only PRF-ness of the hash function is required, and SHA-1 is not known to lack it. Or a couple of other conditions can suffice even if it isn't a PRF.

Intuitively, it makes sense that HMAC is secure as a MAC even with SHA-1, because a MAC does not allow a collision search. The only way to find the key would be to compromise the preimage resistance of SHA-1. HMAC in turn prevents length extension attacks and the like that would allow a forgery without knowing the key.

As an aside, even HMAC-MD5 hasn't been broken. However, for the same reason as there – attacks only get better – I would recommend not choosing SHA-1 as the HMAC hash for new applications if you can just as easily use SHA-2.

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  • $\begingroup$ You should update your answer. An attack on HMAC-MD5 has been found. $\endgroup$ – mikeazo May 11 '15 at 1:53
  • $\begingroup$ @mikeazo AFAICT that doesn't seem to qualify as an attack after all? $\endgroup$ – otus May 16 '15 at 7:43

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