are all cubic graphs almost Hamiltonian?












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I apologize for my ignorance in advance as I am not very familiar with graph theory so I am likely not using the correct jargon. If someone could correct me to the correct buzzword in the literature, I would be thrilled.



Call a graph $G$ $n$-almost-Hamiltonian if there is a closed walk in $G$ that visits every vertex of $G$ exactly $n$-times. So a Hamiltonian graph is $n$-almost-Hamiltonian for all $n$. Are all cubic graphs $n$-almost-Hamiltonian for some sufficiently large $n$?






graph-theory hamiltonian-graphs







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    1












    $begingroup$


    I apologize for my ignorance in advance as I am not very familiar with graph theory so I am likely not using the correct jargon. If someone could correct me to the correct buzzword in the literature, I would be thrilled.



    Call a graph $G$ $n$-almost-Hamiltonian if there is a closed walk in $G$ that visits every vertex of $G$ exactly $n$-times. So a Hamiltonian graph is $n$-almost-Hamiltonian for all $n$. Are all cubic graphs $n$-almost-Hamiltonian for some sufficiently large $n$?






    graph-theory hamiltonian-graphs







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I apologize for my ignorance in advance as I am not very familiar with graph theory so I am likely not using the correct jargon. If someone could correct me to the correct buzzword in the literature, I would be thrilled.



      Call a graph $G$ $n$-almost-Hamiltonian if there is a closed walk in $G$ that visits every vertex of $G$ exactly $n$-times. So a Hamiltonian graph is $n$-almost-Hamiltonian for all $n$. Are all cubic graphs $n$-almost-Hamiltonian for some sufficiently large $n$?






      graph-theory hamiltonian-graphs







      share|cite|improve this question









      $endgroup$




      I apologize for my ignorance in advance as I am not very familiar with graph theory so I am likely not using the correct jargon. If someone could correct me to the correct buzzword in the literature, I would be thrilled.



      Call a graph $G$ $n$-almost-Hamiltonian if there is a closed walk in $G$ that visits every vertex of $G$ exactly $n$-times. So a Hamiltonian graph is $n$-almost-Hamiltonian for all $n$. Are all cubic graphs $n$-almost-Hamiltonian for some sufficiently large $n$?






      graph-theory hamiltonian-graphs




      graph-theory hamiltonian-graphs






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      share|cite|improve this question











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      asked 1 hour ago









      user101010user101010

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          Yes, every connected cubic graph is 3-almost-Hamiltonian.
          Replace each edge by two parallel edges then follow an Eulerian circuit.



          In the case of a bridgeless cubic graph, you can add a perfect matching instead of doubling each edge, which shows they are 2-almost-Hamiltonian.






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            $begingroup$

            Yes, every connected cubic graph is 3-almost-Hamiltonian.
            Replace each edge by two parallel edges then follow an Eulerian circuit.



            In the case of a bridgeless cubic graph, you can add a perfect matching instead of doubling each edge, which shows they are 2-almost-Hamiltonian.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Yes, every connected cubic graph is 3-almost-Hamiltonian.
              Replace each edge by two parallel edges then follow an Eulerian circuit.



              In the case of a bridgeless cubic graph, you can add a perfect matching instead of doubling each edge, which shows they are 2-almost-Hamiltonian.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Yes, every connected cubic graph is 3-almost-Hamiltonian.
                Replace each edge by two parallel edges then follow an Eulerian circuit.



                In the case of a bridgeless cubic graph, you can add a perfect matching instead of doubling each edge, which shows they are 2-almost-Hamiltonian.






                share|cite|improve this answer









                $endgroup$



                Yes, every connected cubic graph is 3-almost-Hamiltonian.
                Replace each edge by two parallel edges then follow an Eulerian circuit.



                In the case of a bridgeless cubic graph, you can add a perfect matching instead of doubling each edge, which shows they are 2-almost-Hamiltonian.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 26 mins ago









                Brendan McKayBrendan McKay

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                24.8k150104






























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