Constructing all non-isomorphic Hamiltonian graphs on n vertices











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I've recently been taking a combinatorics class and have become interested in Hamiltonian graphs.



This OEIS entry lists the number of Hamiltonian graphs on $n$ vertices up to $n =12$. However, I am having difficulty finding how these numbers were found. Is it naive to expect an algorithm which actually constructs all the hamiltonian graphs (at least for $n$ not too large) to exist?



Thanks.






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    up vote
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    favorite












    I've recently been taking a combinatorics class and have become interested in Hamiltonian graphs.



    This OEIS entry lists the number of Hamiltonian graphs on $n$ vertices up to $n =12$. However, I am having difficulty finding how these numbers were found. Is it naive to expect an algorithm which actually constructs all the hamiltonian graphs (at least for $n$ not too large) to exist?



    Thanks.






    graph-theory







    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I've recently been taking a combinatorics class and have become interested in Hamiltonian graphs.



      This OEIS entry lists the number of Hamiltonian graphs on $n$ vertices up to $n =12$. However, I am having difficulty finding how these numbers were found. Is it naive to expect an algorithm which actually constructs all the hamiltonian graphs (at least for $n$ not too large) to exist?



      Thanks.






      graph-theory







      share|cite|improve this question















      I've recently been taking a combinatorics class and have become interested in Hamiltonian graphs.



      This OEIS entry lists the number of Hamiltonian graphs on $n$ vertices up to $n =12$. However, I am having difficulty finding how these numbers were found. Is it naive to expect an algorithm which actually constructs all the hamiltonian graphs (at least for $n$ not too large) to exist?



      Thanks.






      graph-theory




      graph-theory






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      edited 22 hours ago

























      asked 22 hours ago









      Tom Holt

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      8711714






















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          I expect by constructing all graphs, probably using Brendan McKay’s program “geng” and extracting the Hamiltonian ones. Needs quite a few computer cores to do n=12, and I wouldn’t try 13.






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            The 'filter' approach of generating all non-isomorphic graphs on $n$ vertices is the simplest possibility.



            The OEIS entry references McKay (1996), but I can't see a paper in this list : https://users.cecs.anu.edu.au/~bdm/publications.html for Hamiltonian graphs (only _hypo_hamiltonian graphs).



            Presumably a more efficient approach than filtering would be to extend a Hamiltonian graph on $n$ vertices to one on $n + 1$ vertices, but maybe that is not possible.






            share|cite|improve this answer





















            • There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
              – Gordon Royle
              7 hours ago











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            I expect by constructing all graphs, probably using Brendan McKay’s program “geng” and extracting the Hamiltonian ones. Needs quite a few computer cores to do n=12, and I wouldn’t try 13.






            share|cite|improve this answer

























              up vote
              1
              down vote













              I expect by constructing all graphs, probably using Brendan McKay’s program “geng” and extracting the Hamiltonian ones. Needs quite a few computer cores to do n=12, and I wouldn’t try 13.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                I expect by constructing all graphs, probably using Brendan McKay’s program “geng” and extracting the Hamiltonian ones. Needs quite a few computer cores to do n=12, and I wouldn’t try 13.






                share|cite|improve this answer












                I expect by constructing all graphs, probably using Brendan McKay’s program “geng” and extracting the Hamiltonian ones. Needs quite a few computer cores to do n=12, and I wouldn’t try 13.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 22 hours ago









                Gordon Royle

                459310




                459310






















                    up vote
                    0
                    down vote













                    The 'filter' approach of generating all non-isomorphic graphs on $n$ vertices is the simplest possibility.



                    The OEIS entry references McKay (1996), but I can't see a paper in this list : https://users.cecs.anu.edu.au/~bdm/publications.html for Hamiltonian graphs (only _hypo_hamiltonian graphs).



                    Presumably a more efficient approach than filtering would be to extend a Hamiltonian graph on $n$ vertices to one on $n + 1$ vertices, but maybe that is not possible.






                    share|cite|improve this answer





















                    • There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
                      – Gordon Royle
                      7 hours ago















                    up vote
                    0
                    down vote













                    The 'filter' approach of generating all non-isomorphic graphs on $n$ vertices is the simplest possibility.



                    The OEIS entry references McKay (1996), but I can't see a paper in this list : https://users.cecs.anu.edu.au/~bdm/publications.html for Hamiltonian graphs (only _hypo_hamiltonian graphs).



                    Presumably a more efficient approach than filtering would be to extend a Hamiltonian graph on $n$ vertices to one on $n + 1$ vertices, but maybe that is not possible.






                    share|cite|improve this answer





















                    • There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
                      – Gordon Royle
                      7 hours ago













                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    The 'filter' approach of generating all non-isomorphic graphs on $n$ vertices is the simplest possibility.



                    The OEIS entry references McKay (1996), but I can't see a paper in this list : https://users.cecs.anu.edu.au/~bdm/publications.html for Hamiltonian graphs (only _hypo_hamiltonian graphs).



                    Presumably a more efficient approach than filtering would be to extend a Hamiltonian graph on $n$ vertices to one on $n + 1$ vertices, but maybe that is not possible.






                    share|cite|improve this answer












                    The 'filter' approach of generating all non-isomorphic graphs on $n$ vertices is the simplest possibility.



                    The OEIS entry references McKay (1996), but I can't see a paper in this list : https://users.cecs.anu.edu.au/~bdm/publications.html for Hamiltonian graphs (only _hypo_hamiltonian graphs).



                    Presumably a more efficient approach than filtering would be to extend a Hamiltonian graph on $n$ vertices to one on $n + 1$ vertices, but maybe that is not possible.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 21 hours ago









                    gilleain

                    8741713




                    8741713












                    • There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
                      – Gordon Royle
                      7 hours ago


















                    • There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
                      – Gordon Royle
                      7 hours ago
















                    There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
                    – Gordon Royle
                    7 hours ago




                    There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
                    – Gordon Royle
                    7 hours ago


















                     

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