Are there any computational problems in groups that are harder than P?












17












$begingroup$


There are several well known classes of groups for which the word problem, conjugacy etc. are solvable in polynomial time (hyperbolic, automatic).



Then there are several classes of groups like asynchronously automatic groups for which it is known that there is an exponential time algorithm to solve the word problem (and whether this can be improved to polynomial is open and conjectured as far as I'm aware).



Going several steps further, there is an algorithm to solve the word problem in one-relator groups in time not bounded by any finite tower of exponentials (and again it is open and conjectured whether this can be improved to P).



On the other, there are algorithms to solve word problems in pathological groups like the Baumslag-Gersten group:



$$G_{(1,2)} = langle a, b | a^{a^b}= a^2 rangle$$



in polynomial time. So even though general algorithms can be very bad, it is unknown whether there are group-specific algorithms for every group in a given class that solve the word problem quickly.



So my question is, are there any groups in which it is known that the word problem (or any other computational problem) is at least exponential, but still solvable? Maybe exp-complete? What are the approaches to proving such a thing?










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    17












    $begingroup$


    There are several well known classes of groups for which the word problem, conjugacy etc. are solvable in polynomial time (hyperbolic, automatic).



    Then there are several classes of groups like asynchronously automatic groups for which it is known that there is an exponential time algorithm to solve the word problem (and whether this can be improved to polynomial is open and conjectured as far as I'm aware).



    Going several steps further, there is an algorithm to solve the word problem in one-relator groups in time not bounded by any finite tower of exponentials (and again it is open and conjectured whether this can be improved to P).



    On the other, there are algorithms to solve word problems in pathological groups like the Baumslag-Gersten group:



    $$G_{(1,2)} = langle a, b | a^{a^b}= a^2 rangle$$



    in polynomial time. So even though general algorithms can be very bad, it is unknown whether there are group-specific algorithms for every group in a given class that solve the word problem quickly.



    So my question is, are there any groups in which it is known that the word problem (or any other computational problem) is at least exponential, but still solvable? Maybe exp-complete? What are the approaches to proving such a thing?










    share|cite|improve this question









    New contributor




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







    $endgroup$















      17












      17








      17


      4



      $begingroup$


      There are several well known classes of groups for which the word problem, conjugacy etc. are solvable in polynomial time (hyperbolic, automatic).



      Then there are several classes of groups like asynchronously automatic groups for which it is known that there is an exponential time algorithm to solve the word problem (and whether this can be improved to polynomial is open and conjectured as far as I'm aware).



      Going several steps further, there is an algorithm to solve the word problem in one-relator groups in time not bounded by any finite tower of exponentials (and again it is open and conjectured whether this can be improved to P).



      On the other, there are algorithms to solve word problems in pathological groups like the Baumslag-Gersten group:



      $$G_{(1,2)} = langle a, b | a^{a^b}= a^2 rangle$$



      in polynomial time. So even though general algorithms can be very bad, it is unknown whether there are group-specific algorithms for every group in a given class that solve the word problem quickly.



      So my question is, are there any groups in which it is known that the word problem (or any other computational problem) is at least exponential, but still solvable? Maybe exp-complete? What are the approaches to proving such a thing?










      share|cite|improve this question









      New contributor




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







      $endgroup$




      There are several well known classes of groups for which the word problem, conjugacy etc. are solvable in polynomial time (hyperbolic, automatic).



      Then there are several classes of groups like asynchronously automatic groups for which it is known that there is an exponential time algorithm to solve the word problem (and whether this can be improved to polynomial is open and conjectured as far as I'm aware).



      Going several steps further, there is an algorithm to solve the word problem in one-relator groups in time not bounded by any finite tower of exponentials (and again it is open and conjectured whether this can be improved to P).



      On the other, there are algorithms to solve word problems in pathological groups like the Baumslag-Gersten group:



      $$G_{(1,2)} = langle a, b | a^{a^b}= a^2 rangle$$



      in polynomial time. So even though general algorithms can be very bad, it is unknown whether there are group-specific algorithms for every group in a given class that solve the word problem quickly.



      So my question is, are there any groups in which it is known that the word problem (or any other computational problem) is at least exponential, but still solvable? Maybe exp-complete? What are the approaches to proving such a thing?







      gr.group-theory computational-complexity geometric-group-theory computational-group-theory word-problem






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      edited 2 days ago







      MSL













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      asked 2 days ago









      MSLMSL

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          4 Answers
          4






          active

          oldest

          votes


















          18












          $begingroup$

          An earlier reference for groups with this property is



          J. Avenhaus and K. Madlener. Subrekursive Komplexität der Gruppen. I. Gruppen mit vorgeschriebenen Komplexität. Acta Infomat., 9 (1): 87-104, 1977/78.



          There is a hierarchy of the recursive functions known as the (difficult to pronounce) Grzegorczyk Hierarchy $E_0 subset E_1 subset E_2 subset cdots$, where (roughly) $E_1$ contains the linearly bounded functions, $E_2$ polynomially bounded functions, and $E_3$ those functions that are bounded by iterated exponentials.



          The above paper describes constructions of finitely presented groups $G_n$ for $n ge 3$, in which solving the word problem has time complexity bounded by a function $E_n$ but not by any function in $E_{n-1}$,






          share|cite|improve this answer









          $endgroup$





















            16












            $begingroup$

            As Andreas says (in his answer and his comment to it), there are groups whose word problem is undecidable and one could similarly set up a group that encodes the halting problem of a class of Turing machines where this is decidable but difficult. However, one must be careful in the encoding. In



            Isoperimetric and Isodiametric Functions of Groups,
            Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
            Annals of Mathematics
            Second Series, Vol. 156, No. 2 (Sep., 2002), pp. 345-466



            and



            Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
            Isoperimetric functions of groups and computational complexity of the word problem.
            Ann. of Math. (2) 156 (2002), no. 2, 467–518.



            groups with NP complete word problem are constructed and other similar results. See in particular, Corollary 1.1 of the first paper listed above.



            To add more information, Corollary 1.1 says:




            There exists a finitely presented group with NP-complete word problem. Moreover for every language $Lsubseteq A^*$ for some finite alphabet $A$ there exists a finitely presented group $G$ such that the nondeterministic complexity of $G$ is polynomially equivalent to the nondeterministic complexity of $L$.




            So, for instance, if $L$ is an EXP-time complete problem, then the word problem of $G$ is in NEXP-time and not in NP (hence not in P). Of course you can replace EXP by your favorite time complexity class strictly above NP.






            share|cite|improve this answer











            $endgroup$





















              10












              $begingroup$

              There are finitely presented groups whose word problem is undecidable. See, for example, https://en.wikipedia.org/wiki/Word_problem_for_groups .






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
                $endgroup$
                – MSL
                2 days ago






              • 7




                $begingroup$
                @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
                $endgroup$
                – Andreas Blass
                2 days ago



















              6












              $begingroup$

              Classical problem which is believed not to be in P is number factoring, which can be cast as computing a decomposition of a cyclic group into simple ones.



              Several problems in permutation groups are known to be as hard as graph isomorphism, thus not believed to be in P, too.



              There are also NP-hard problems known for permutation groups, see e.g. https://www.sciencedirect.com/science/article/pii/S0012365X09001289






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                Checking if a cyclic group is simple would seem to be primality testing which is in P.
                $endgroup$
                – Benjamin Steinberg
                2 days ago










              • $begingroup$
                You are right; I have corrected my answer to reflect this.
                $endgroup$
                – Dima Pasechnik
                2 days ago






              • 1




                $begingroup$
                As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
                $endgroup$
                – Yakk
                yesterday










              • $begingroup$
                “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
                $endgroup$
                – Dima Pasechnik
                yesterday










              • $begingroup$
                Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
                $endgroup$
                – Pål GD
                yesterday











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              4 Answers
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              4 Answers
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              active

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              18












              $begingroup$

              An earlier reference for groups with this property is



              J. Avenhaus and K. Madlener. Subrekursive Komplexität der Gruppen. I. Gruppen mit vorgeschriebenen Komplexität. Acta Infomat., 9 (1): 87-104, 1977/78.



              There is a hierarchy of the recursive functions known as the (difficult to pronounce) Grzegorczyk Hierarchy $E_0 subset E_1 subset E_2 subset cdots$, where (roughly) $E_1$ contains the linearly bounded functions, $E_2$ polynomially bounded functions, and $E_3$ those functions that are bounded by iterated exponentials.



              The above paper describes constructions of finitely presented groups $G_n$ for $n ge 3$, in which solving the word problem has time complexity bounded by a function $E_n$ but not by any function in $E_{n-1}$,






              share|cite|improve this answer









              $endgroup$


















                18












                $begingroup$

                An earlier reference for groups with this property is



                J. Avenhaus and K. Madlener. Subrekursive Komplexität der Gruppen. I. Gruppen mit vorgeschriebenen Komplexität. Acta Infomat., 9 (1): 87-104, 1977/78.



                There is a hierarchy of the recursive functions known as the (difficult to pronounce) Grzegorczyk Hierarchy $E_0 subset E_1 subset E_2 subset cdots$, where (roughly) $E_1$ contains the linearly bounded functions, $E_2$ polynomially bounded functions, and $E_3$ those functions that are bounded by iterated exponentials.



                The above paper describes constructions of finitely presented groups $G_n$ for $n ge 3$, in which solving the word problem has time complexity bounded by a function $E_n$ but not by any function in $E_{n-1}$,






                share|cite|improve this answer









                $endgroup$
















                  18












                  18








                  18





                  $begingroup$

                  An earlier reference for groups with this property is



                  J. Avenhaus and K. Madlener. Subrekursive Komplexität der Gruppen. I. Gruppen mit vorgeschriebenen Komplexität. Acta Infomat., 9 (1): 87-104, 1977/78.



                  There is a hierarchy of the recursive functions known as the (difficult to pronounce) Grzegorczyk Hierarchy $E_0 subset E_1 subset E_2 subset cdots$, where (roughly) $E_1$ contains the linearly bounded functions, $E_2$ polynomially bounded functions, and $E_3$ those functions that are bounded by iterated exponentials.



                  The above paper describes constructions of finitely presented groups $G_n$ for $n ge 3$, in which solving the word problem has time complexity bounded by a function $E_n$ but not by any function in $E_{n-1}$,






                  share|cite|improve this answer









                  $endgroup$



                  An earlier reference for groups with this property is



                  J. Avenhaus and K. Madlener. Subrekursive Komplexität der Gruppen. I. Gruppen mit vorgeschriebenen Komplexität. Acta Infomat., 9 (1): 87-104, 1977/78.



                  There is a hierarchy of the recursive functions known as the (difficult to pronounce) Grzegorczyk Hierarchy $E_0 subset E_1 subset E_2 subset cdots$, where (roughly) $E_1$ contains the linearly bounded functions, $E_2$ polynomially bounded functions, and $E_3$ those functions that are bounded by iterated exponentials.



                  The above paper describes constructions of finitely presented groups $G_n$ for $n ge 3$, in which solving the word problem has time complexity bounded by a function $E_n$ but not by any function in $E_{n-1}$,







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  Derek HoltDerek Holt

                  26.7k462109




                  26.7k462109























                      16












                      $begingroup$

                      As Andreas says (in his answer and his comment to it), there are groups whose word problem is undecidable and one could similarly set up a group that encodes the halting problem of a class of Turing machines where this is decidable but difficult. However, one must be careful in the encoding. In



                      Isoperimetric and Isodiametric Functions of Groups,
                      Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                      Annals of Mathematics
                      Second Series, Vol. 156, No. 2 (Sep., 2002), pp. 345-466



                      and



                      Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                      Isoperimetric functions of groups and computational complexity of the word problem.
                      Ann. of Math. (2) 156 (2002), no. 2, 467–518.



                      groups with NP complete word problem are constructed and other similar results. See in particular, Corollary 1.1 of the first paper listed above.



                      To add more information, Corollary 1.1 says:




                      There exists a finitely presented group with NP-complete word problem. Moreover for every language $Lsubseteq A^*$ for some finite alphabet $A$ there exists a finitely presented group $G$ such that the nondeterministic complexity of $G$ is polynomially equivalent to the nondeterministic complexity of $L$.




                      So, for instance, if $L$ is an EXP-time complete problem, then the word problem of $G$ is in NEXP-time and not in NP (hence not in P). Of course you can replace EXP by your favorite time complexity class strictly above NP.






                      share|cite|improve this answer











                      $endgroup$


















                        16












                        $begingroup$

                        As Andreas says (in his answer and his comment to it), there are groups whose word problem is undecidable and one could similarly set up a group that encodes the halting problem of a class of Turing machines where this is decidable but difficult. However, one must be careful in the encoding. In



                        Isoperimetric and Isodiametric Functions of Groups,
                        Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                        Annals of Mathematics
                        Second Series, Vol. 156, No. 2 (Sep., 2002), pp. 345-466



                        and



                        Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                        Isoperimetric functions of groups and computational complexity of the word problem.
                        Ann. of Math. (2) 156 (2002), no. 2, 467–518.



                        groups with NP complete word problem are constructed and other similar results. See in particular, Corollary 1.1 of the first paper listed above.



                        To add more information, Corollary 1.1 says:




                        There exists a finitely presented group with NP-complete word problem. Moreover for every language $Lsubseteq A^*$ for some finite alphabet $A$ there exists a finitely presented group $G$ such that the nondeterministic complexity of $G$ is polynomially equivalent to the nondeterministic complexity of $L$.




                        So, for instance, if $L$ is an EXP-time complete problem, then the word problem of $G$ is in NEXP-time and not in NP (hence not in P). Of course you can replace EXP by your favorite time complexity class strictly above NP.






                        share|cite|improve this answer











                        $endgroup$
















                          16












                          16








                          16





                          $begingroup$

                          As Andreas says (in his answer and his comment to it), there are groups whose word problem is undecidable and one could similarly set up a group that encodes the halting problem of a class of Turing machines where this is decidable but difficult. However, one must be careful in the encoding. In



                          Isoperimetric and Isodiametric Functions of Groups,
                          Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                          Annals of Mathematics
                          Second Series, Vol. 156, No. 2 (Sep., 2002), pp. 345-466



                          and



                          Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                          Isoperimetric functions of groups and computational complexity of the word problem.
                          Ann. of Math. (2) 156 (2002), no. 2, 467–518.



                          groups with NP complete word problem are constructed and other similar results. See in particular, Corollary 1.1 of the first paper listed above.



                          To add more information, Corollary 1.1 says:




                          There exists a finitely presented group with NP-complete word problem. Moreover for every language $Lsubseteq A^*$ for some finite alphabet $A$ there exists a finitely presented group $G$ such that the nondeterministic complexity of $G$ is polynomially equivalent to the nondeterministic complexity of $L$.




                          So, for instance, if $L$ is an EXP-time complete problem, then the word problem of $G$ is in NEXP-time and not in NP (hence not in P). Of course you can replace EXP by your favorite time complexity class strictly above NP.






                          share|cite|improve this answer











                          $endgroup$



                          As Andreas says (in his answer and his comment to it), there are groups whose word problem is undecidable and one could similarly set up a group that encodes the halting problem of a class of Turing machines where this is decidable but difficult. However, one must be careful in the encoding. In



                          Isoperimetric and Isodiametric Functions of Groups,
                          Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                          Annals of Mathematics
                          Second Series, Vol. 156, No. 2 (Sep., 2002), pp. 345-466



                          and



                          Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                          Isoperimetric functions of groups and computational complexity of the word problem.
                          Ann. of Math. (2) 156 (2002), no. 2, 467–518.



                          groups with NP complete word problem are constructed and other similar results. See in particular, Corollary 1.1 of the first paper listed above.



                          To add more information, Corollary 1.1 says:




                          There exists a finitely presented group with NP-complete word problem. Moreover for every language $Lsubseteq A^*$ for some finite alphabet $A$ there exists a finitely presented group $G$ such that the nondeterministic complexity of $G$ is polynomially equivalent to the nondeterministic complexity of $L$.




                          So, for instance, if $L$ is an EXP-time complete problem, then the word problem of $G$ is in NEXP-time and not in NP (hence not in P). Of course you can replace EXP by your favorite time complexity class strictly above NP.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited yesterday

























                          answered 2 days ago









                          Benjamin SteinbergBenjamin Steinberg

                          23.2k265125




                          23.2k265125























                              10












                              $begingroup$

                              There are finitely presented groups whose word problem is undecidable. See, for example, https://en.wikipedia.org/wiki/Word_problem_for_groups .






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
                                $endgroup$
                                – MSL
                                2 days ago






                              • 7




                                $begingroup$
                                @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
                                $endgroup$
                                – Andreas Blass
                                2 days ago
















                              10












                              $begingroup$

                              There are finitely presented groups whose word problem is undecidable. See, for example, https://en.wikipedia.org/wiki/Word_problem_for_groups .






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
                                $endgroup$
                                – MSL
                                2 days ago






                              • 7




                                $begingroup$
                                @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
                                $endgroup$
                                – Andreas Blass
                                2 days ago














                              10












                              10








                              10





                              $begingroup$

                              There are finitely presented groups whose word problem is undecidable. See, for example, https://en.wikipedia.org/wiki/Word_problem_for_groups .






                              share|cite|improve this answer









                              $endgroup$



                              There are finitely presented groups whose word problem is undecidable. See, for example, https://en.wikipedia.org/wiki/Word_problem_for_groups .







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 2 days ago









                              Andreas BlassAndreas Blass

                              57k7135219




                              57k7135219












                              • $begingroup$
                                True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
                                $endgroup$
                                – MSL
                                2 days ago






                              • 7




                                $begingroup$
                                @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
                                $endgroup$
                                – Andreas Blass
                                2 days ago


















                              • $begingroup$
                                True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
                                $endgroup$
                                – MSL
                                2 days ago






                              • 7




                                $begingroup$
                                @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
                                $endgroup$
                                – Andreas Blass
                                2 days ago
















                              $begingroup$
                              True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
                              $endgroup$
                              – MSL
                              2 days ago




                              $begingroup$
                              True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
                              $endgroup$
                              – MSL
                              2 days ago




                              7




                              7




                              $begingroup$
                              @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
                              $endgroup$
                              – Andreas Blass
                              2 days ago




                              $begingroup$
                              @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
                              $endgroup$
                              – Andreas Blass
                              2 days ago











                              6












                              $begingroup$

                              Classical problem which is believed not to be in P is number factoring, which can be cast as computing a decomposition of a cyclic group into simple ones.



                              Several problems in permutation groups are known to be as hard as graph isomorphism, thus not believed to be in P, too.



                              There are also NP-hard problems known for permutation groups, see e.g. https://www.sciencedirect.com/science/article/pii/S0012365X09001289






                              share|cite|improve this answer











                              $endgroup$









                              • 1




                                $begingroup$
                                Checking if a cyclic group is simple would seem to be primality testing which is in P.
                                $endgroup$
                                – Benjamin Steinberg
                                2 days ago










                              • $begingroup$
                                You are right; I have corrected my answer to reflect this.
                                $endgroup$
                                – Dima Pasechnik
                                2 days ago






                              • 1




                                $begingroup$
                                As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
                                $endgroup$
                                – Yakk
                                yesterday










                              • $begingroup$
                                “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
                                $endgroup$
                                – Dima Pasechnik
                                yesterday










                              • $begingroup$
                                Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
                                $endgroup$
                                – Pål GD
                                yesterday
















                              6












                              $begingroup$

                              Classical problem which is believed not to be in P is number factoring, which can be cast as computing a decomposition of a cyclic group into simple ones.



                              Several problems in permutation groups are known to be as hard as graph isomorphism, thus not believed to be in P, too.



                              There are also NP-hard problems known for permutation groups, see e.g. https://www.sciencedirect.com/science/article/pii/S0012365X09001289






                              share|cite|improve this answer











                              $endgroup$









                              • 1




                                $begingroup$
                                Checking if a cyclic group is simple would seem to be primality testing which is in P.
                                $endgroup$
                                – Benjamin Steinberg
                                2 days ago










                              • $begingroup$
                                You are right; I have corrected my answer to reflect this.
                                $endgroup$
                                – Dima Pasechnik
                                2 days ago






                              • 1




                                $begingroup$
                                As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
                                $endgroup$
                                – Yakk
                                yesterday










                              • $begingroup$
                                “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
                                $endgroup$
                                – Dima Pasechnik
                                yesterday










                              • $begingroup$
                                Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
                                $endgroup$
                                – Pål GD
                                yesterday














                              6












                              6








                              6





                              $begingroup$

                              Classical problem which is believed not to be in P is number factoring, which can be cast as computing a decomposition of a cyclic group into simple ones.



                              Several problems in permutation groups are known to be as hard as graph isomorphism, thus not believed to be in P, too.



                              There are also NP-hard problems known for permutation groups, see e.g. https://www.sciencedirect.com/science/article/pii/S0012365X09001289






                              share|cite|improve this answer











                              $endgroup$



                              Classical problem which is believed not to be in P is number factoring, which can be cast as computing a decomposition of a cyclic group into simple ones.



                              Several problems in permutation groups are known to be as hard as graph isomorphism, thus not believed to be in P, too.



                              There are also NP-hard problems known for permutation groups, see e.g. https://www.sciencedirect.com/science/article/pii/S0012365X09001289







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 2 days ago

























                              answered 2 days ago









                              Dima PasechnikDima Pasechnik

                              9,03311851




                              9,03311851








                              • 1




                                $begingroup$
                                Checking if a cyclic group is simple would seem to be primality testing which is in P.
                                $endgroup$
                                – Benjamin Steinberg
                                2 days ago










                              • $begingroup$
                                You are right; I have corrected my answer to reflect this.
                                $endgroup$
                                – Dima Pasechnik
                                2 days ago






                              • 1




                                $begingroup$
                                As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
                                $endgroup$
                                – Yakk
                                yesterday










                              • $begingroup$
                                “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
                                $endgroup$
                                – Dima Pasechnik
                                yesterday










                              • $begingroup$
                                Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
                                $endgroup$
                                – Pål GD
                                yesterday














                              • 1




                                $begingroup$
                                Checking if a cyclic group is simple would seem to be primality testing which is in P.
                                $endgroup$
                                – Benjamin Steinberg
                                2 days ago










                              • $begingroup$
                                You are right; I have corrected my answer to reflect this.
                                $endgroup$
                                – Dima Pasechnik
                                2 days ago






                              • 1




                                $begingroup$
                                As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
                                $endgroup$
                                – Yakk
                                yesterday










                              • $begingroup$
                                “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
                                $endgroup$
                                – Dima Pasechnik
                                yesterday










                              • $begingroup$
                                Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
                                $endgroup$
                                – Pål GD
                                yesterday








                              1




                              1




                              $begingroup$
                              Checking if a cyclic group is simple would seem to be primality testing which is in P.
                              $endgroup$
                              – Benjamin Steinberg
                              2 days ago




                              $begingroup$
                              Checking if a cyclic group is simple would seem to be primality testing which is in P.
                              $endgroup$
                              – Benjamin Steinberg
                              2 days ago












                              $begingroup$
                              You are right; I have corrected my answer to reflect this.
                              $endgroup$
                              – Dima Pasechnik
                              2 days ago




                              $begingroup$
                              You are right; I have corrected my answer to reflect this.
                              $endgroup$
                              – Dima Pasechnik
                              2 days ago




                              1




                              1




                              $begingroup$
                              As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
                              $endgroup$
                              – Yakk
                              yesterday




                              $begingroup$
                              As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
                              $endgroup$
                              – Yakk
                              yesterday












                              $begingroup$
                              “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
                              $endgroup$
                              – Dima Pasechnik
                              yesterday




                              $begingroup$
                              “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
                              $endgroup$
                              – Dima Pasechnik
                              yesterday












                              $begingroup$
                              Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
                              $endgroup$
                              – Pål GD
                              yesterday




                              $begingroup$
                              Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
                              $endgroup$
                              – Pål GD
                              yesterday










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