What will happen to NP-Hard problems if P=NP

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I was going through the lecture of prof. Erik Demaine and he said that a problem X is NP-Hard if it is at-least as hard as every problem Y that belongs to NP class.



He further said that if we can prove that there exists a problem that belongs to NP but not to P, then we can absolutely be sure that NP hard problems don't belong to P class.



Here is my question... What happens if P=NP. Thus it mean that all NP-hard problems become polynomially solvable? Will all NP hard problems reduce to polynomial order if P=NP?










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    I was going through the lecture of prof. Erik Demaine and he said that a problem X is NP-Hard if it is at-least as hard as every problem Y that belongs to NP class.



    He further said that if we can prove that there exists a problem that belongs to NP but not to P, then we can absolutely be sure that NP hard problems don't belong to P class.



    Here is my question... What happens if P=NP. Thus it mean that all NP-hard problems become polynomially solvable? Will all NP hard problems reduce to polynomial order if P=NP?










    share|cite|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I was going through the lecture of prof. Erik Demaine and he said that a problem X is NP-Hard if it is at-least as hard as every problem Y that belongs to NP class.



      He further said that if we can prove that there exists a problem that belongs to NP but not to P, then we can absolutely be sure that NP hard problems don't belong to P class.



      Here is my question... What happens if P=NP. Thus it mean that all NP-hard problems become polynomially solvable? Will all NP hard problems reduce to polynomial order if P=NP?










      share|cite|improve this question















      I was going through the lecture of prof. Erik Demaine and he said that a problem X is NP-Hard if it is at-least as hard as every problem Y that belongs to NP class.



      He further said that if we can prove that there exists a problem that belongs to NP but not to P, then we can absolutely be sure that NP hard problems don't belong to P class.



      Here is my question... What happens if P=NP. Thus it mean that all NP-hard problems become polynomially solvable? Will all NP hard problems reduce to polynomial order if P=NP?







      complexity-theory np-complete np-hard np






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      edited 4 hours ago









      Yuval Filmus

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      asked 5 hours ago









      Abhilash Mishra

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          No. A problem is Np-Hard if all NP problems are reducible to an instance of that problem in polynomial time. Some NP-Hard problems cannot be solved in nondeterministic polynomial time, and are not in NP. Then these problems will not be polynomial time solvable regardless of whether or not P=NP.






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            The halting problem is NP-hard: given any language $L$ in P accepted by some nondeterministic machine $M$, we can come up with a polytime reduction that on input $x$ constructs a Turing machine which runs $M$ on $x$ and all possible witnesses, halting if $M$ accepts any of them, and going into an infinite loop otherwise.



            We know unconditionally that the halting problem is not computable, and so in particular, isn’t in P.






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              So just as your Prof. wikipedia says (emphasis mine):




              NP-hardness [...]is the defining property of a class of problems that are, informally, "at least as hard as the hardest problems in NP"




              (Still Informally) you can think of the ‚at least‘ as:



              NP $leq$ NP-Hard



              So you have to distinguish those NP-hard problems for which the equality holds from those which are ,strictly‘ harder. I.e problems that are NP-hard and are also in NP (a.k.a NP-complete) vs problems which are NP-hard but not in NP.



              Looking at it this way, it should be realtively easy to reason about your question. Namely, if P=NP:



              • For problems that are NP-hard and are also in NP it will obviously mean they are also in P (since P=NP by assumption)

              • For problems that are NP-hard but not in NP it will mean nothing.

              This euler diagram from wikipedia might also help to clear things up






              share|cite|improve this answer




















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                3 Answers
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                3 Answers
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                No. A problem is Np-Hard if all NP problems are reducible to an instance of that problem in polynomial time. Some NP-Hard problems cannot be solved in nondeterministic polynomial time, and are not in NP. Then these problems will not be polynomial time solvable regardless of whether or not P=NP.






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                  No. A problem is Np-Hard if all NP problems are reducible to an instance of that problem in polynomial time. Some NP-Hard problems cannot be solved in nondeterministic polynomial time, and are not in NP. Then these problems will not be polynomial time solvable regardless of whether or not P=NP.






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                    No. A problem is Np-Hard if all NP problems are reducible to an instance of that problem in polynomial time. Some NP-Hard problems cannot be solved in nondeterministic polynomial time, and are not in NP. Then these problems will not be polynomial time solvable regardless of whether or not P=NP.






                    share|cite|improve this answer












                    No. A problem is Np-Hard if all NP problems are reducible to an instance of that problem in polynomial time. Some NP-Hard problems cannot be solved in nondeterministic polynomial time, and are not in NP. Then these problems will not be polynomial time solvable regardless of whether or not P=NP.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 4 hours ago









                    HackerBoss

                    2345




                    2345




















                        up vote
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                        The halting problem is NP-hard: given any language $L$ in P accepted by some nondeterministic machine $M$, we can come up with a polytime reduction that on input $x$ constructs a Turing machine which runs $M$ on $x$ and all possible witnesses, halting if $M$ accepts any of them, and going into an infinite loop otherwise.



                        We know unconditionally that the halting problem is not computable, and so in particular, isn’t in P.






                        share|cite|improve this answer
























                          up vote
                          2
                          down vote













                          The halting problem is NP-hard: given any language $L$ in P accepted by some nondeterministic machine $M$, we can come up with a polytime reduction that on input $x$ constructs a Turing machine which runs $M$ on $x$ and all possible witnesses, halting if $M$ accepts any of them, and going into an infinite loop otherwise.



                          We know unconditionally that the halting problem is not computable, and so in particular, isn’t in P.






                          share|cite|improve this answer






















                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            The halting problem is NP-hard: given any language $L$ in P accepted by some nondeterministic machine $M$, we can come up with a polytime reduction that on input $x$ constructs a Turing machine which runs $M$ on $x$ and all possible witnesses, halting if $M$ accepts any of them, and going into an infinite loop otherwise.



                            We know unconditionally that the halting problem is not computable, and so in particular, isn’t in P.






                            share|cite|improve this answer












                            The halting problem is NP-hard: given any language $L$ in P accepted by some nondeterministic machine $M$, we can come up with a polytime reduction that on input $x$ constructs a Turing machine which runs $M$ on $x$ and all possible witnesses, halting if $M$ accepts any of them, and going into an infinite loop otherwise.



                            We know unconditionally that the halting problem is not computable, and so in particular, isn’t in P.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 4 hours ago









                            Yuval Filmus

                            185k12175335




                            185k12175335




















                                up vote
                                0
                                down vote













                                So just as your Prof. wikipedia says (emphasis mine):




                                NP-hardness [...]is the defining property of a class of problems that are, informally, "at least as hard as the hardest problems in NP"




                                (Still Informally) you can think of the ‚at least‘ as:



                                NP $leq$ NP-Hard



                                So you have to distinguish those NP-hard problems for which the equality holds from those which are ,strictly‘ harder. I.e problems that are NP-hard and are also in NP (a.k.a NP-complete) vs problems which are NP-hard but not in NP.



                                Looking at it this way, it should be realtively easy to reason about your question. Namely, if P=NP:



                                • For problems that are NP-hard and are also in NP it will obviously mean they are also in P (since P=NP by assumption)

                                • For problems that are NP-hard but not in NP it will mean nothing.

                                This euler diagram from wikipedia might also help to clear things up






                                share|cite|improve this answer
























                                  up vote
                                  0
                                  down vote













                                  So just as your Prof. wikipedia says (emphasis mine):




                                  NP-hardness [...]is the defining property of a class of problems that are, informally, "at least as hard as the hardest problems in NP"




                                  (Still Informally) you can think of the ‚at least‘ as:



                                  NP $leq$ NP-Hard



                                  So you have to distinguish those NP-hard problems for which the equality holds from those which are ,strictly‘ harder. I.e problems that are NP-hard and are also in NP (a.k.a NP-complete) vs problems which are NP-hard but not in NP.



                                  Looking at it this way, it should be realtively easy to reason about your question. Namely, if P=NP:



                                  • For problems that are NP-hard and are also in NP it will obviously mean they are also in P (since P=NP by assumption)

                                  • For problems that are NP-hard but not in NP it will mean nothing.

                                  This euler diagram from wikipedia might also help to clear things up






                                  share|cite|improve this answer






















                                    up vote
                                    0
                                    down vote










                                    up vote
                                    0
                                    down vote









                                    So just as your Prof. wikipedia says (emphasis mine):




                                    NP-hardness [...]is the defining property of a class of problems that are, informally, "at least as hard as the hardest problems in NP"




                                    (Still Informally) you can think of the ‚at least‘ as:



                                    NP $leq$ NP-Hard



                                    So you have to distinguish those NP-hard problems for which the equality holds from those which are ,strictly‘ harder. I.e problems that are NP-hard and are also in NP (a.k.a NP-complete) vs problems which are NP-hard but not in NP.



                                    Looking at it this way, it should be realtively easy to reason about your question. Namely, if P=NP:



                                    • For problems that are NP-hard and are also in NP it will obviously mean they are also in P (since P=NP by assumption)

                                    • For problems that are NP-hard but not in NP it will mean nothing.

                                    This euler diagram from wikipedia might also help to clear things up






                                    share|cite|improve this answer












                                    So just as your Prof. wikipedia says (emphasis mine):




                                    NP-hardness [...]is the defining property of a class of problems that are, informally, "at least as hard as the hardest problems in NP"




                                    (Still Informally) you can think of the ‚at least‘ as:



                                    NP $leq$ NP-Hard



                                    So you have to distinguish those NP-hard problems for which the equality holds from those which are ,strictly‘ harder. I.e problems that are NP-hard and are also in NP (a.k.a NP-complete) vs problems which are NP-hard but not in NP.



                                    Looking at it this way, it should be realtively easy to reason about your question. Namely, if P=NP:



                                    • For problems that are NP-hard and are also in NP it will obviously mean they are also in P (since P=NP by assumption)

                                    • For problems that are NP-hard but not in NP it will mean nothing.

                                    This euler diagram from wikipedia might also help to clear things up







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 1 hour ago









                                    dingalapadum

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