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We know that in the black-box sense, we cannot use one-way functions to construct Collision Resistant Hash Functions.I feel that in my impression, I have never seen CRHF based on integer factorization problem or RSA assumption

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  • $\begingroup$ One word: performance. $\endgroup$
    – swineone
    Feb 28 at 3:03
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    $\begingroup$ How about en.wikipedia.org/wiki/Very_smooth_hash $\endgroup$
    – poncho
    Feb 28 at 3:12
  • $\begingroup$ Thank you so much @poncho $\endgroup$ Feb 28 at 3:15
  • $\begingroup$ Now, I wonder if there is a hard problem that CRHF cannot be constructed on it $\endgroup$ Feb 28 at 14:08
  • $\begingroup$ As a side note that wikipedia article is very funny --- when describing an asymptotic notion of security, it contains the line "This is considered a useless assumption for practice", and then describes a concrete security assumption. $\endgroup$
    – Mark
    Mar 1 at 3:22

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Damgård constructed CRHFs from claw-free permutations, which can be based on integer factorisation (or even the discrete-log problem) [D]. That's the earliest one I am aware of (but someone feel free to correct me).

[D]: Damgård, Collision Free Hash Functions and Public Key Signature Schemes, Eurocrypt'87

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  • $\begingroup$ thanks, Now, I wonder if there is a hard problem that CRHF cannot be constructed on it. $\endgroup$ Feb 28 at 14:09
  • $\begingroup$ Interesting. The concrete hardness assumptions I can think of also yield CRHFs. Will think about it. $\endgroup$
    – ckamath
    Feb 28 at 14:41
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    $\begingroup$ Feel free to ask a question about that, but yes, there are many! For example, we don't know how to build CRHFs from one-way functions (for a generic assumption), or from the learning parity with noise assumption (for a concrete assumption - except in the extremely low-noise regime). $\endgroup$ Feb 28 at 16:24
  • $\begingroup$ @ckamath, is there a CRHF based on graph isomorphism problem? I think maybe not $\endgroup$ Mar 21 at 12:03
  • $\begingroup$ Right, but then we don't know any crypto from GI (and it might as well turn out to be easy). $\endgroup$
    – ckamath
    Mar 23 at 7:59

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