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Simple interepretation problem regarding Polynomial Hierarchy?


Why isn't this undecidable problem in NP?Do any decision problems exist outside NP and NP-Hard?P is contained in NP ∩ Co-NP?Constructing a promise problem equivalent to XSAT from subset sumIs a problem in NP if it is decided by some non-deterministic, polynomial time turing machine?Is there a task that is solvable in polynomial time but not verifiable in polynomial time?A clarification on $NP=coNP$?NP-complete problem with a polynomial number of yes-instances?Is This A Solution for does P=NP problem?How does a co-NP problem differ from an NP (its complement) one?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








6












$begingroup$


So $NP$ stands for problems where we have small verifiable witnesses for $YES$ instances and $coNP$ for small verifiable witnesses for $NO$ instances. How does this work for



  1. $P^NP$


  2. $NP^NP$


  3. $coNP^NP$


  4. and so on?










share|cite|improve this question









$endgroup$


















    6












    $begingroup$


    So $NP$ stands for problems where we have small verifiable witnesses for $YES$ instances and $coNP$ for small verifiable witnesses for $NO$ instances. How does this work for



    1. $P^NP$


    2. $NP^NP$


    3. $coNP^NP$


    4. and so on?










    share|cite|improve this question









    $endgroup$














      6












      6








      6





      $begingroup$


      So $NP$ stands for problems where we have small verifiable witnesses for $YES$ instances and $coNP$ for small verifiable witnesses for $NO$ instances. How does this work for



      1. $P^NP$


      2. $NP^NP$


      3. $coNP^NP$


      4. and so on?










      share|cite|improve this question









      $endgroup$




      So $NP$ stands for problems where we have small verifiable witnesses for $YES$ instances and $coNP$ for small verifiable witnesses for $NO$ instances. How does this work for



      1. $P^NP$


      2. $NP^NP$


      3. $coNP^NP$


      4. and so on?







      complexity-theory np






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 9 hours ago









      TurboTurbo

      1,2726 silver badges18 bronze badges




      1,2726 silver badges18 bronze badges




















          1 Answer
          1






          active

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          5












          $begingroup$

          There is a logical interpretation of the various levels of the polynomial hierarchy, which extends the witness characterization of $mathsfNP$ and $mathsfcoNP$.



          A language $L$ is in $Sigma_k^P$ if there is a polytime predicate $f$ and a polynomial $ell$ such that
          $$
          x in L Leftrightarrow exists |y_1| le ell(|x|) ; forall |y_2| le ell(|x|) ; cdots Q |y_k| le ell(|x|) ; f(x,y_1,ldots,y_k).
          $$

          Here:




          • $exists |y| le ell(|x|)$ means that there exists a number $y$ whose length is at most $ell(|x|)$ such that ...


          • $forall |y| le ell(|x|)$ mean that for all $y$ whose length is at most $ell(|x|)$, the following holds ...


          • $Q$ is $exists$ if $k$ is odd and $forall$ if $k$ is even.

          Similarly, $L$ is in $Pi_k^P$ if it can be written in a similar way, only starting with $forall$.



          As an example, $mathsfNP^mathsfNP$ is $Sigma_2^P$, and consists of all languages such that
          $$
          x in L Leftrightarrow exists |y_1| leq ell(|x|) ; forall |y_2| leq ell(|x|) ; f(x,y_1,ldots,y_k).
          $$

          As another example, $mathsfcoNP^mathsfNP$ is $Pi_2^P$.



          Your third example is $mathsfP^mathsfNP$, which is $Delta_2^P$. I'm not sure what is the logical characterization.






          share|cite|improve this answer









          $endgroup$















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            1 Answer
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            active

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            active

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            5












            $begingroup$

            There is a logical interpretation of the various levels of the polynomial hierarchy, which extends the witness characterization of $mathsfNP$ and $mathsfcoNP$.



            A language $L$ is in $Sigma_k^P$ if there is a polytime predicate $f$ and a polynomial $ell$ such that
            $$
            x in L Leftrightarrow exists |y_1| le ell(|x|) ; forall |y_2| le ell(|x|) ; cdots Q |y_k| le ell(|x|) ; f(x,y_1,ldots,y_k).
            $$

            Here:




            • $exists |y| le ell(|x|)$ means that there exists a number $y$ whose length is at most $ell(|x|)$ such that ...


            • $forall |y| le ell(|x|)$ mean that for all $y$ whose length is at most $ell(|x|)$, the following holds ...


            • $Q$ is $exists$ if $k$ is odd and $forall$ if $k$ is even.

            Similarly, $L$ is in $Pi_k^P$ if it can be written in a similar way, only starting with $forall$.



            As an example, $mathsfNP^mathsfNP$ is $Sigma_2^P$, and consists of all languages such that
            $$
            x in L Leftrightarrow exists |y_1| leq ell(|x|) ; forall |y_2| leq ell(|x|) ; f(x,y_1,ldots,y_k).
            $$

            As another example, $mathsfcoNP^mathsfNP$ is $Pi_2^P$.



            Your third example is $mathsfP^mathsfNP$, which is $Delta_2^P$. I'm not sure what is the logical characterization.






            share|cite|improve this answer









            $endgroup$

















              5












              $begingroup$

              There is a logical interpretation of the various levels of the polynomial hierarchy, which extends the witness characterization of $mathsfNP$ and $mathsfcoNP$.



              A language $L$ is in $Sigma_k^P$ if there is a polytime predicate $f$ and a polynomial $ell$ such that
              $$
              x in L Leftrightarrow exists |y_1| le ell(|x|) ; forall |y_2| le ell(|x|) ; cdots Q |y_k| le ell(|x|) ; f(x,y_1,ldots,y_k).
              $$

              Here:




              • $exists |y| le ell(|x|)$ means that there exists a number $y$ whose length is at most $ell(|x|)$ such that ...


              • $forall |y| le ell(|x|)$ mean that for all $y$ whose length is at most $ell(|x|)$, the following holds ...


              • $Q$ is $exists$ if $k$ is odd and $forall$ if $k$ is even.

              Similarly, $L$ is in $Pi_k^P$ if it can be written in a similar way, only starting with $forall$.



              As an example, $mathsfNP^mathsfNP$ is $Sigma_2^P$, and consists of all languages such that
              $$
              x in L Leftrightarrow exists |y_1| leq ell(|x|) ; forall |y_2| leq ell(|x|) ; f(x,y_1,ldots,y_k).
              $$

              As another example, $mathsfcoNP^mathsfNP$ is $Pi_2^P$.



              Your third example is $mathsfP^mathsfNP$, which is $Delta_2^P$. I'm not sure what is the logical characterization.






              share|cite|improve this answer









              $endgroup$















                5












                5








                5





                $begingroup$

                There is a logical interpretation of the various levels of the polynomial hierarchy, which extends the witness characterization of $mathsfNP$ and $mathsfcoNP$.



                A language $L$ is in $Sigma_k^P$ if there is a polytime predicate $f$ and a polynomial $ell$ such that
                $$
                x in L Leftrightarrow exists |y_1| le ell(|x|) ; forall |y_2| le ell(|x|) ; cdots Q |y_k| le ell(|x|) ; f(x,y_1,ldots,y_k).
                $$

                Here:




                • $exists |y| le ell(|x|)$ means that there exists a number $y$ whose length is at most $ell(|x|)$ such that ...


                • $forall |y| le ell(|x|)$ mean that for all $y$ whose length is at most $ell(|x|)$, the following holds ...


                • $Q$ is $exists$ if $k$ is odd and $forall$ if $k$ is even.

                Similarly, $L$ is in $Pi_k^P$ if it can be written in a similar way, only starting with $forall$.



                As an example, $mathsfNP^mathsfNP$ is $Sigma_2^P$, and consists of all languages such that
                $$
                x in L Leftrightarrow exists |y_1| leq ell(|x|) ; forall |y_2| leq ell(|x|) ; f(x,y_1,ldots,y_k).
                $$

                As another example, $mathsfcoNP^mathsfNP$ is $Pi_2^P$.



                Your third example is $mathsfP^mathsfNP$, which is $Delta_2^P$. I'm not sure what is the logical characterization.






                share|cite|improve this answer









                $endgroup$



                There is a logical interpretation of the various levels of the polynomial hierarchy, which extends the witness characterization of $mathsfNP$ and $mathsfcoNP$.



                A language $L$ is in $Sigma_k^P$ if there is a polytime predicate $f$ and a polynomial $ell$ such that
                $$
                x in L Leftrightarrow exists |y_1| le ell(|x|) ; forall |y_2| le ell(|x|) ; cdots Q |y_k| le ell(|x|) ; f(x,y_1,ldots,y_k).
                $$

                Here:




                • $exists |y| le ell(|x|)$ means that there exists a number $y$ whose length is at most $ell(|x|)$ such that ...


                • $forall |y| le ell(|x|)$ mean that for all $y$ whose length is at most $ell(|x|)$, the following holds ...


                • $Q$ is $exists$ if $k$ is odd and $forall$ if $k$ is even.

                Similarly, $L$ is in $Pi_k^P$ if it can be written in a similar way, only starting with $forall$.



                As an example, $mathsfNP^mathsfNP$ is $Sigma_2^P$, and consists of all languages such that
                $$
                x in L Leftrightarrow exists |y_1| leq ell(|x|) ; forall |y_2| leq ell(|x|) ; f(x,y_1,ldots,y_k).
                $$

                As another example, $mathsfcoNP^mathsfNP$ is $Pi_2^P$.



                Your third example is $mathsfP^mathsfNP$, which is $Delta_2^P$. I'm not sure what is the logical characterization.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 8 hours ago









                Yuval FilmusYuval Filmus

                204k15 gold badges198 silver badges360 bronze badges




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