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No, if you can prove that a secure hash function exists, then you have proved that P != NP. (This is what alexbecker said.) If you can prove P = NP, then you h
by Thorrez 7y ago
No, if you can prove that a secure hash function exists, then you have proved that P != NP. (This is what alexbecker said.)
If you can prove P = NP, then you have proved that no secure hash functions exist. (This is the contrapositive.)
- repsilat 7y agoFor convenient definitions of "secure" :-). "Unscrambling" a particular function could be in P but still effectively intractable. (For a fun angle not related to complexity, the algorithm itself could be so large that it wouldn't fit in the observable universe without breaking the Bekenstein bound.) And it has been a long time since I did proof theory, but is there the possibility of a non-constructive proof but no actual examples no matter how long you keep enumerating algorithms? Or a non-constructive proof, but no constructive proof? In that case there could be a "perfect heuristic"...
- fiddlerwoaroof 7y agoOops, this is what I get for not reading carefully :) But this is interesting: it implies, if I'm not mistaken, it implies that we haven't ever proven that a function takes exponential time, just that we haven't found any algorithms for certain problems faster than exponential time. I hadn't considered P != NP this way before...
- gpm 7y agoNo, we know that P != EXP_TIME, by the time hierarchy theorem - i.e. we know some problems (and we do have concrete examples) take exponential time. What we don't know is that there exists a problem that requires exponential time (or otherwise greater than polynomial) where we can create a proof that a answer is correct which is checkable in polynomial time. https://en.wikipedia.org/wiki/Time_hierarchy_theorem https://en.wikipedia.org/wiki/Time_hierarchy_theorem