If lower bound of a problem is exponential then is it NP?

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Assuming that we have a problem $p$ and we showed that the lower bound for solving $p$ is $mathcalOmega(2^n)$.



  • can lower bound $mathcalOmega(2^n)$ implies the problem in $NP$?





time-complexity np-complete







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    Assuming that we have a problem $p$ and we showed that the lower bound for solving $p$ is $mathcalOmega(2^n)$.



    • can lower bound $mathcalOmega(2^n)$ implies the problem in $NP$?





    time-complexity np-complete







    share|cite|improve this question







    New contributor




    kelalaka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Assuming that we have a problem $p$ and we showed that the lower bound for solving $p$ is $mathcalOmega(2^n)$.



      • can lower bound $mathcalOmega(2^n)$ implies the problem in $NP$?





      time-complexity np-complete







      share|cite|improve this question







      New contributor




      kelalaka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      Assuming that we have a problem $p$ and we showed that the lower bound for solving $p$ is $mathcalOmega(2^n)$.



      • can lower bound $mathcalOmega(2^n)$ implies the problem in $NP$?





      time-complexity np-complete




      time-complexity np-complete






      share|cite|improve this question







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      kelalaka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question







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      kelalaka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 16 mins ago









      kelalaka

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          No. For example, the halting problem has an $Omega(2^n)$ lower bound, but it is not in NP (since it is not computable).



          NP is an upper bound on the complexity of a problem.






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          • do you know a computable example, also?
            – kelalaka
            11 mins ago






          • 1




            Yes. Take any NEXPTIME-complete problem.
            – Yuval Filmus
            10 mins ago










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          1 Answer
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          up vote
          2
          down vote













          No. For example, the halting problem has an $Omega(2^n)$ lower bound, but it is not in NP (since it is not computable).



          NP is an upper bound on the complexity of a problem.






          share|cite|improve this answer




















          • do you know a computable example, also?
            – kelalaka
            11 mins ago






          • 1




            Yes. Take any NEXPTIME-complete problem.
            – Yuval Filmus
            10 mins ago














          up vote
          2
          down vote













          No. For example, the halting problem has an $Omega(2^n)$ lower bound, but it is not in NP (since it is not computable).



          NP is an upper bound on the complexity of a problem.






          share|cite|improve this answer




















          • do you know a computable example, also?
            – kelalaka
            11 mins ago






          • 1




            Yes. Take any NEXPTIME-complete problem.
            – Yuval Filmus
            10 mins ago












          up vote
          2
          down vote










          up vote
          2
          down vote









          No. For example, the halting problem has an $Omega(2^n)$ lower bound, but it is not in NP (since it is not computable).



          NP is an upper bound on the complexity of a problem.






          share|cite|improve this answer












          No. For example, the halting problem has an $Omega(2^n)$ lower bound, but it is not in NP (since it is not computable).



          NP is an upper bound on the complexity of a problem.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 12 mins ago









          Yuval Filmus

          183k12171333




          183k12171333











          • do you know a computable example, also?
            – kelalaka
            11 mins ago






          • 1




            Yes. Take any NEXPTIME-complete problem.
            – Yuval Filmus
            10 mins ago
















          • do you know a computable example, also?
            – kelalaka
            11 mins ago






          • 1




            Yes. Take any NEXPTIME-complete problem.
            – Yuval Filmus
            10 mins ago















          do you know a computable example, also?
          – kelalaka
          11 mins ago




          do you know a computable example, also?
          – kelalaka
          11 mins ago




          1




          1




          Yes. Take any NEXPTIME-complete problem.
          – Yuval Filmus
          10 mins ago




          Yes. Take any NEXPTIME-complete problem.
          – Yuval Filmus
          10 mins ago










          kelalaka is a new contributor. Be nice, and check out our Code of Conduct.









           

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          kelalaka is a new contributor. Be nice, and check out our Code of Conduct.











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