Hamilton Paths in Complete graph $K_n$












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In complete graph $K_n$, is it true that we can have at least $2*n$ Hamilton paths?






graph-theory algorithms hamiltonian-path







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    0












    $begingroup$


    In complete graph $K_n$, is it true that we can have at least $2*n$ Hamilton paths?






    graph-theory algorithms hamiltonian-path







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      In complete graph $K_n$, is it true that we can have at least $2*n$ Hamilton paths?






      graph-theory algorithms hamiltonian-path







      share|cite|improve this question









      $endgroup$




      In complete graph $K_n$, is it true that we can have at least $2*n$ Hamilton paths?






      graph-theory algorithms hamiltonian-path




      graph-theory algorithms hamiltonian-path






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 18 '18 at 22:01









      JuneJune

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          $begingroup$

          Since in a complete graph $K_n$ each two vertices are adjacent, the set of Hamiltonian paths of $K_n$ is exactly a set $S_n$ of permutations of its vertices. But $|S_n|=n!$, which is at least $2n$ when $nge 3$, but it is less than $2n$ for $nle 2$.






          share|cite|improve this answer









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          • $begingroup$
            I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
            $endgroup$
            – Bob Krueger
            Dec 19 '18 at 8:50










          • $begingroup$
            @BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
            $endgroup$
            – Alex Ravsky
            Dec 21 '18 at 7:02






          • 1




            $begingroup$
            Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
            $endgroup$
            – Bob Krueger
            Dec 21 '18 at 9:05











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          $begingroup$

          Since in a complete graph $K_n$ each two vertices are adjacent, the set of Hamiltonian paths of $K_n$ is exactly a set $S_n$ of permutations of its vertices. But $|S_n|=n!$, which is at least $2n$ when $nge 3$, but it is less than $2n$ for $nle 2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
            $endgroup$
            – Bob Krueger
            Dec 19 '18 at 8:50










          • $begingroup$
            @BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
            $endgroup$
            – Alex Ravsky
            Dec 21 '18 at 7:02






          • 1




            $begingroup$
            Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
            $endgroup$
            – Bob Krueger
            Dec 21 '18 at 9:05
















          1












          $begingroup$

          Since in a complete graph $K_n$ each two vertices are adjacent, the set of Hamiltonian paths of $K_n$ is exactly a set $S_n$ of permutations of its vertices. But $|S_n|=n!$, which is at least $2n$ when $nge 3$, but it is less than $2n$ for $nle 2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
            $endgroup$
            – Bob Krueger
            Dec 19 '18 at 8:50










          • $begingroup$
            @BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
            $endgroup$
            – Alex Ravsky
            Dec 21 '18 at 7:02






          • 1




            $begingroup$
            Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
            $endgroup$
            – Bob Krueger
            Dec 21 '18 at 9:05














          1












          1








          1





          $begingroup$

          Since in a complete graph $K_n$ each two vertices are adjacent, the set of Hamiltonian paths of $K_n$ is exactly a set $S_n$ of permutations of its vertices. But $|S_n|=n!$, which is at least $2n$ when $nge 3$, but it is less than $2n$ for $nle 2$.






          share|cite|improve this answer









          $endgroup$



          Since in a complete graph $K_n$ each two vertices are adjacent, the set of Hamiltonian paths of $K_n$ is exactly a set $S_n$ of permutations of its vertices. But $|S_n|=n!$, which is at least $2n$ when $nge 3$, but it is less than $2n$ for $nle 2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 6:45









          Alex RavskyAlex Ravsky

          41.4k32282




          41.4k32282












          • $begingroup$
            I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
            $endgroup$
            – Bob Krueger
            Dec 19 '18 at 8:50










          • $begingroup$
            @BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
            $endgroup$
            – Alex Ravsky
            Dec 21 '18 at 7:02






          • 1




            $begingroup$
            Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
            $endgroup$
            – Bob Krueger
            Dec 21 '18 at 9:05


















          • $begingroup$
            I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
            $endgroup$
            – Bob Krueger
            Dec 19 '18 at 8:50










          • $begingroup$
            @BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
            $endgroup$
            – Alex Ravsky
            Dec 21 '18 at 7:02






          • 1




            $begingroup$
            Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
            $endgroup$
            – Bob Krueger
            Dec 21 '18 at 9:05
















          $begingroup$
          I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
          $endgroup$
          – Bob Krueger
          Dec 19 '18 at 8:50




          $begingroup$
          I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path.
          $endgroup$
          – Bob Krueger
          Dec 19 '18 at 8:50












          $begingroup$
          @BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
          $endgroup$
          – Alex Ravsky
          Dec 21 '18 at 7:02




          $begingroup$
          @BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too.
          $endgroup$
          – Alex Ravsky
          Dec 21 '18 at 7:02




          1




          1




          $begingroup$
          Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
          $endgroup$
          – Bob Krueger
          Dec 21 '18 at 9:05




          $begingroup$
          Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle.
          $endgroup$
          – Bob Krueger
          Dec 21 '18 at 9:05


















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