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Am a bit confused about the relationship between undecidable problems and NP hard problems. Whether NP hard problems are a subset of undecidable problems, or are they just the same and equal, or is it that they are not comparable?

For me, I have been arguing with my friends that undecidable problems are a superset to the NP hard problems. There would exist some problems that are not in NP hard but are undecidable. But i am finding this argument to be weak and am confused a bit. Are there NP-complete problems that are undecidable.? is there any problem in NP hard which is decidable.??

Some discussion would be of great help! Thanks!

svick
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akaHuman
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3 Answers3

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Undecidable = unsolvable for some inputs. No matter how much (finite) time you give your algorithm, it will always be wrong on some input.

NP-hard ~= super-polynomial running time (assuming P != NP). That's hand-wavy, but basically NP-hard means it is at least as hard as the hardest problem in NP.

There are certainly problems that are NP-hard which are not undecidable (= are decidable). Any NP-complete problem would be one of them, say SAT.

Are there undecidable problems which are not NP-hard? I don't think so, but it isn't easy to rule it out - I don't see an obvious argument that there must be a reduction from SAT to all possible undecidable problems. There could be some weird undecidable problems which aren't very useful. But the standard undecidable problems (the halting problem, say) are NP-hard.

Keith Randall
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  • Well, i seem to have come out to a conclusion..that undecidable problems are a subset of NP hard problems..this is based on the following scenario- All problems in NP are decidable. There are some problems in NP hard which are not undecidable (=decidable, and which i guess are NP complete). Therefore, undecidable problems comprise of a subset of NP hard. Am i right? – akaHuman May 08 '12 at 17:34
  • You've lost me at the "therefore". Certainly the containment doesn't go the other way, but the two sets may be incomparable. You need to prove that an arbitrary undecidable problem is in NP-hard (i.e. can be used as an oracle to solve SAT in poly time). – Keith Randall May 08 '12 at 19:46
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    Okay! I seem to almost get it. – akaHuman May 09 '12 at 10:48
  • @akaHuman There may be problems that are np complete and undecidable. – christopher clark Mar 12 '19 at 17:23
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    @christopherclark: No, all NP complete problems are decidable. – Keith Randall Apr 09 '19 at 03:19
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An NP-hard is a problem that is at least as hard as any NP-complete problem.

Therefore an undecidable problem can be NP-hard. A problem is NP-hard if an oracle for it would make solving NP-complete problems easy (i.e. solvable in polynomial time). We can imagine an undecidable problem such that, given an oracle for it, NP-complete problems would be easy to solve. For example, obviously every oracle that solves the halting problem can also solve an NP-complete problem, so every Turing-complete problem is also NP-hard in the sense that a (fast) oracle for it would make solving NP-complete problems a breeze.

Therefore Turing-complete undecidable problems are a subset of NP-hard problems.

Antti Huima
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Undecidable problem e.g. Turing Halting Problem is NP-Hard only.

                   <---------NP Hard------>
|------------|-------------||-------------|------------|--------> Computational Difficulty

|<----P--->|

|<----------NP---------->|

|<-----------Exponential----------->|

|<---------------R (Finite Time)---------------->|

In this diagram, that small pipe shows overlapping of NP and NP-Hard and which shows NP-Completeness, i.e. set of those problems which are NP as well as NP-Hard.

Undecidable problems are NP Hard problems which do not have solution and which are not in NP.

John Kugelman
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Utkarsh
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