Easy instances of the coloring problem on graphs with degree at most 4












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Given a graph with some set of colors, the goal of the coloring problem is to color the input graph with as few number of colors as possible, such that no adjacent vertices have the same color.



In general, the problem is hard. For trees and cycles, the problem is solvable in polynomial time. I am interested in bounded degree instances or more precisely graphs of degree at most 4. Is there any non-trivial subclass of graphs of degree at most 4 for which the coloring problem is solvable in polynomial time?






complexity-theory graphs colorings







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$endgroup$

















    2












    $begingroup$


    Given a graph with some set of colors, the goal of the coloring problem is to color the input graph with as few number of colors as possible, such that no adjacent vertices have the same color.



    In general, the problem is hard. For trees and cycles, the problem is solvable in polynomial time. I am interested in bounded degree instances or more precisely graphs of degree at most 4. Is there any non-trivial subclass of graphs of degree at most 4 for which the coloring problem is solvable in polynomial time?






    complexity-theory graphs colorings







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      2



      $begingroup$


      Given a graph with some set of colors, the goal of the coloring problem is to color the input graph with as few number of colors as possible, such that no adjacent vertices have the same color.



      In general, the problem is hard. For trees and cycles, the problem is solvable in polynomial time. I am interested in bounded degree instances or more precisely graphs of degree at most 4. Is there any non-trivial subclass of graphs of degree at most 4 for which the coloring problem is solvable in polynomial time?






      complexity-theory graphs colorings







      share|cite|improve this question











      $endgroup$




      Given a graph with some set of colors, the goal of the coloring problem is to color the input graph with as few number of colors as possible, such that no adjacent vertices have the same color.



      In general, the problem is hard. For trees and cycles, the problem is solvable in polynomial time. I am interested in bounded degree instances or more precisely graphs of degree at most 4. Is there any non-trivial subclass of graphs of degree at most 4 for which the coloring problem is solvable in polynomial time?






      complexity-theory graphs colorings




      complexity-theory graphs colorings






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited 20 hours ago









      Discrete lizard♦

      4,21411535




      4,21411535










      asked 22 hours ago









      I_wil_break_wallI_wil_break_wall

      995




      995






















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          $begingroup$

          You can of course take any graph class for which coloring is easy, and additionally require that the maximum degree is at most 4. For example, every bipartite graph of maximum degree at most 4 works. Or "bipartite" could be replaced with say "outerplanar" or more generally "small cliquewidth".



          But having maximum degree at most 4 alone will not work. That is, coloring remains hard for (i) 4-regular graphs and (ii) planar graphs of maximum degree 4.






          share|cite|improve this answer









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            2












            $begingroup$

            As the other answer states, any class that is already easy to color can obviously be additionally restricted to bounded-degree graphs. But if you are looking for restrictions that only yield easily colored graphs in combination with the bounded-degree restriction, that also works.



            For example: We know that graphs with bounded degree $d$ can be greedily colored with $d+1$ colors. But that by itself isn't optimal, since there are graphs in that class that can be colored with even fewer colors.



            One way to fix that is by only considering the subclass of bounded-degree graphs that also need at least $d+1$ colors, ensuring that the greedy coloring is also the optimal one. So for example, the subclass of graphs of degree 4 that contain a 5-clique.



            (Edit: There's also something called Brooks' theorem, which says that unless the graph really does contain such a 5-clique, it will actually be 4-colorable. But I can't see if it also says that this coloring can be found in polynomial time.)






            share|cite|improve this answer










            New contributor




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






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              2 Answers
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              $begingroup$

              You can of course take any graph class for which coloring is easy, and additionally require that the maximum degree is at most 4. For example, every bipartite graph of maximum degree at most 4 works. Or "bipartite" could be replaced with say "outerplanar" or more generally "small cliquewidth".



              But having maximum degree at most 4 alone will not work. That is, coloring remains hard for (i) 4-regular graphs and (ii) planar graphs of maximum degree 4.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                You can of course take any graph class for which coloring is easy, and additionally require that the maximum degree is at most 4. For example, every bipartite graph of maximum degree at most 4 works. Or "bipartite" could be replaced with say "outerplanar" or more generally "small cliquewidth".



                But having maximum degree at most 4 alone will not work. That is, coloring remains hard for (i) 4-regular graphs and (ii) planar graphs of maximum degree 4.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  You can of course take any graph class for which coloring is easy, and additionally require that the maximum degree is at most 4. For example, every bipartite graph of maximum degree at most 4 works. Or "bipartite" could be replaced with say "outerplanar" or more generally "small cliquewidth".



                  But having maximum degree at most 4 alone will not work. That is, coloring remains hard for (i) 4-regular graphs and (ii) planar graphs of maximum degree 4.






                  share|cite|improve this answer









                  $endgroup$



                  You can of course take any graph class for which coloring is easy, and additionally require that the maximum degree is at most 4. For example, every bipartite graph of maximum degree at most 4 works. Or "bipartite" could be replaced with say "outerplanar" or more generally "small cliquewidth".



                  But having maximum degree at most 4 alone will not work. That is, coloring remains hard for (i) 4-regular graphs and (ii) planar graphs of maximum degree 4.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 21 hours ago









                  JuhoJuho

                  15.4k54191




                  15.4k54191























                      2












                      $begingroup$

                      As the other answer states, any class that is already easy to color can obviously be additionally restricted to bounded-degree graphs. But if you are looking for restrictions that only yield easily colored graphs in combination with the bounded-degree restriction, that also works.



                      For example: We know that graphs with bounded degree $d$ can be greedily colored with $d+1$ colors. But that by itself isn't optimal, since there are graphs in that class that can be colored with even fewer colors.



                      One way to fix that is by only considering the subclass of bounded-degree graphs that also need at least $d+1$ colors, ensuring that the greedy coloring is also the optimal one. So for example, the subclass of graphs of degree 4 that contain a 5-clique.



                      (Edit: There's also something called Brooks' theorem, which says that unless the graph really does contain such a 5-clique, it will actually be 4-colorable. But I can't see if it also says that this coloring can be found in polynomial time.)






                      share|cite|improve this answer










                      New contributor




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






                      $endgroup$


















                        2












                        $begingroup$

                        As the other answer states, any class that is already easy to color can obviously be additionally restricted to bounded-degree graphs. But if you are looking for restrictions that only yield easily colored graphs in combination with the bounded-degree restriction, that also works.



                        For example: We know that graphs with bounded degree $d$ can be greedily colored with $d+1$ colors. But that by itself isn't optimal, since there are graphs in that class that can be colored with even fewer colors.



                        One way to fix that is by only considering the subclass of bounded-degree graphs that also need at least $d+1$ colors, ensuring that the greedy coloring is also the optimal one. So for example, the subclass of graphs of degree 4 that contain a 5-clique.



                        (Edit: There's also something called Brooks' theorem, which says that unless the graph really does contain such a 5-clique, it will actually be 4-colorable. But I can't see if it also says that this coloring can be found in polynomial time.)






                        share|cite|improve this answer










                        New contributor




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






                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          As the other answer states, any class that is already easy to color can obviously be additionally restricted to bounded-degree graphs. But if you are looking for restrictions that only yield easily colored graphs in combination with the bounded-degree restriction, that also works.



                          For example: We know that graphs with bounded degree $d$ can be greedily colored with $d+1$ colors. But that by itself isn't optimal, since there are graphs in that class that can be colored with even fewer colors.



                          One way to fix that is by only considering the subclass of bounded-degree graphs that also need at least $d+1$ colors, ensuring that the greedy coloring is also the optimal one. So for example, the subclass of graphs of degree 4 that contain a 5-clique.



                          (Edit: There's also something called Brooks' theorem, which says that unless the graph really does contain such a 5-clique, it will actually be 4-colorable. But I can't see if it also says that this coloring can be found in polynomial time.)






                          share|cite|improve this answer










                          New contributor




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






                          $endgroup$



                          As the other answer states, any class that is already easy to color can obviously be additionally restricted to bounded-degree graphs. But if you are looking for restrictions that only yield easily colored graphs in combination with the bounded-degree restriction, that also works.



                          For example: We know that graphs with bounded degree $d$ can be greedily colored with $d+1$ colors. But that by itself isn't optimal, since there are graphs in that class that can be colored with even fewer colors.



                          One way to fix that is by only considering the subclass of bounded-degree graphs that also need at least $d+1$ colors, ensuring that the greedy coloring is also the optimal one. So for example, the subclass of graphs of degree 4 that contain a 5-clique.



                          (Edit: There's also something called Brooks' theorem, which says that unless the graph really does contain such a 5-clique, it will actually be 4-colorable. But I can't see if it also says that this coloring can be found in polynomial time.)







                          share|cite|improve this answer










                          New contributor




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









                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 14 hours ago





















                          New contributor




                          Christoph Burschka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                          answered 17 hours ago









                          Christoph BurschkaChristoph Burschka

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                          1214




                          New contributor




                          Christoph Burschka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                          New contributor





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






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






























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