Talk:Security of cryptographic hash functions: Difference between revisions

Revision as of 18:18, 9 January 2019

Revision as of 06:53, 31 July 2019

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The text says that finding a collision in a particular hash function "is supposed to be hard, at least PSPACE-complete." But, this can't be true unless NP = PSPACE, since finding a collision is trivially in NP. Right?! <!-- Template:Unsigned --><small class="autosigned">—&nbsp;Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:NoahSD|NoahSD]] ([[User talk:NoahSD#top|talk]] • [[Special:Contributions/NoahSD|contribs]]) 18:17, 9 January 2019 (UTC)</small> <!--Autosigned by SineBot-->

== Parts of the article need to be rewritten ==

As a computer scientist with knowledge of complexity theory but not of cryptographic hashing I am a bit puzzled about what is being said in the article.

As far as I can understand, the article seems to say that commonly used cryptographic hash functions are "hard" only in the sense that nobody seems to have figured out to "break" them (whatever notion of "breaking" is being used.) This seems like a very weak form of "hardness".

Then, there are other cryptographic hash functions, that are "hard" in the complexity theory sense, e.g. NP-hard, so that breaking seems unlikely given that no NP-hard problem has a polynomial time solution algorithm.

However, also this latter hardness is not very strong: very large instances of NP-hard problems are solved, as not all instances of NP-hard problems are at all difficult to solve. There are large classes of instances from any NP-hard problem that can be solved in linear time: they are trivial.

To have a convincing ''proof'' of hardness also in the complexity theory sense, one would need to somehow tie the hardness to some empirically hard class of instances of an NP-hard problem.