How many Hamiltonian cycles are there in a complete graph if we discount the cycle's orientation or starting...












1












$begingroup$


Consider a complete graph G with n vertices.



Each vertex is indexed by [n] = {1,2,3...n} where n >= 4.



In this case, a Hamiltonian cycle is determined only by the collection of edges it contains, and we do not need to consider its orientation or starting point.




Question:




  1. How many Hamiltonian cycles are there in G?

  2. How many Hamiltonian cycles in G contains the edge {1,2}?

  3. How many Hamiltonian cycles in G contains the edge {1,2} and {2,3}?

  4. How many Hamiltonian cycles in G contains the edge {1,2} and {3,4}?

  5. Suppose that M is a set of k <= (n/2) edges, in which no two edges in M share a vertex. How many Hamiltonian cycles contain all edges in M? Give answer in terms of k and n

  6. How many Hamiltonian cycles in G do not contain the edge {1,2}, {2,3} and {3,4}?



The question really clusters into two parts.



PART 1: How do I discount "orientation" and "starting point"?



This has to do with 1, 2, 3, 4, and 6.



I can calculate the combinations of edges there can be, but that's not what they're asking. They only want the combinations that form a Hamiltonian cycle.



Additionally, I don't see how you can just know whether a Hamiltonian cycle has crossed through a certain edge.



The more I think about it, the more I feel this is about combinatorial numbers as opposed to graph theory. Are they trying to trick me?



PART 2: How many cycles contain a set of edges that do not share a vertex.



This has to do with question 5, specifically.



My first response is "none..."?



If the graph is a complete graph, then all edges share a vertex at some point right? In that case, M seems to be an empty set and there are no Hamiltonian cycles that cover it. But that doesn't feel right at all...






graph-theory hamiltonian-path







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Consider a complete graph G with n vertices.



    Each vertex is indexed by [n] = {1,2,3...n} where n >= 4.



    In this case, a Hamiltonian cycle is determined only by the collection of edges it contains, and we do not need to consider its orientation or starting point.




    Question:




    1. How many Hamiltonian cycles are there in G?

    2. How many Hamiltonian cycles in G contains the edge {1,2}?

    3. How many Hamiltonian cycles in G contains the edge {1,2} and {2,3}?

    4. How many Hamiltonian cycles in G contains the edge {1,2} and {3,4}?

    5. Suppose that M is a set of k <= (n/2) edges, in which no two edges in M share a vertex. How many Hamiltonian cycles contain all edges in M? Give answer in terms of k and n

    6. How many Hamiltonian cycles in G do not contain the edge {1,2}, {2,3} and {3,4}?



    The question really clusters into two parts.



    PART 1: How do I discount "orientation" and "starting point"?



    This has to do with 1, 2, 3, 4, and 6.



    I can calculate the combinations of edges there can be, but that's not what they're asking. They only want the combinations that form a Hamiltonian cycle.



    Additionally, I don't see how you can just know whether a Hamiltonian cycle has crossed through a certain edge.



    The more I think about it, the more I feel this is about combinatorial numbers as opposed to graph theory. Are they trying to trick me?



    PART 2: How many cycles contain a set of edges that do not share a vertex.



    This has to do with question 5, specifically.



    My first response is "none..."?



    If the graph is a complete graph, then all edges share a vertex at some point right? In that case, M seems to be an empty set and there are no Hamiltonian cycles that cover it. But that doesn't feel right at all...






    graph-theory hamiltonian-path







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Consider a complete graph G with n vertices.



      Each vertex is indexed by [n] = {1,2,3...n} where n >= 4.



      In this case, a Hamiltonian cycle is determined only by the collection of edges it contains, and we do not need to consider its orientation or starting point.




      Question:




      1. How many Hamiltonian cycles are there in G?

      2. How many Hamiltonian cycles in G contains the edge {1,2}?

      3. How many Hamiltonian cycles in G contains the edge {1,2} and {2,3}?

      4. How many Hamiltonian cycles in G contains the edge {1,2} and {3,4}?

      5. Suppose that M is a set of k <= (n/2) edges, in which no two edges in M share a vertex. How many Hamiltonian cycles contain all edges in M? Give answer in terms of k and n

      6. How many Hamiltonian cycles in G do not contain the edge {1,2}, {2,3} and {3,4}?



      The question really clusters into two parts.



      PART 1: How do I discount "orientation" and "starting point"?



      This has to do with 1, 2, 3, 4, and 6.



      I can calculate the combinations of edges there can be, but that's not what they're asking. They only want the combinations that form a Hamiltonian cycle.



      Additionally, I don't see how you can just know whether a Hamiltonian cycle has crossed through a certain edge.



      The more I think about it, the more I feel this is about combinatorial numbers as opposed to graph theory. Are they trying to trick me?



      PART 2: How many cycles contain a set of edges that do not share a vertex.



      This has to do with question 5, specifically.



      My first response is "none..."?



      If the graph is a complete graph, then all edges share a vertex at some point right? In that case, M seems to be an empty set and there are no Hamiltonian cycles that cover it. But that doesn't feel right at all...






      graph-theory hamiltonian-path







      share|cite|improve this question









      $endgroup$




      Consider a complete graph G with n vertices.



      Each vertex is indexed by [n] = {1,2,3...n} where n >= 4.



      In this case, a Hamiltonian cycle is determined only by the collection of edges it contains, and we do not need to consider its orientation or starting point.




      Question:




      1. How many Hamiltonian cycles are there in G?

      2. How many Hamiltonian cycles in G contains the edge {1,2}?

      3. How many Hamiltonian cycles in G contains the edge {1,2} and {2,3}?

      4. How many Hamiltonian cycles in G contains the edge {1,2} and {3,4}?

      5. Suppose that M is a set of k <= (n/2) edges, in which no two edges in M share a vertex. How many Hamiltonian cycles contain all edges in M? Give answer in terms of k and n

      6. How many Hamiltonian cycles in G do not contain the edge {1,2}, {2,3} and {3,4}?



      The question really clusters into two parts.



      PART 1: How do I discount "orientation" and "starting point"?



      This has to do with 1, 2, 3, 4, and 6.



      I can calculate the combinations of edges there can be, but that's not what they're asking. They only want the combinations that form a Hamiltonian cycle.



      Additionally, I don't see how you can just know whether a Hamiltonian cycle has crossed through a certain edge.



      The more I think about it, the more I feel this is about combinatorial numbers as opposed to graph theory. Are they trying to trick me?



      PART 2: How many cycles contain a set of edges that do not share a vertex.



      This has to do with question 5, specifically.



      My first response is "none..."?



      If the graph is a complete graph, then all edges share a vertex at some point right? In that case, M seems to be an empty set and there are no Hamiltonian cycles that cover it. But that doesn't feel right at all...






      graph-theory hamiltonian-path




      graph-theory hamiltonian-path






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 29 '18 at 21:41









      potatoguypotatoguy

      525




      525






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Hint: if we do consider starting point and orientation, then the number of Hamiltonian cycles is the number of ways that we can order $[n]$, i.e. the number of permutations. If you know the order in which to visit the vertices, this tells you exactly the cycle. Each cycle is then counted $n$ times for each possible starting point, and twice for each direction around the cycle.



          Hint for part 2: A cycle can contain ${ 1,2 }$ and ${ 3,4 }$ if it (for example) also contains edge ${ 2,3 }$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the hint, here's what I wrote for questions: Q1) n! / 2n, Q2: (n-2)!, Q3: (n-3)!
            $endgroup$
            – potatoguy
            Nov 30 '18 at 2:58












          • $begingroup$
            I'm still stuck on Q4, since the graph is separated into two sections. Do you happen to have more advise on how to approach this problem? I'm testing it out on a complete graph K5 (result seems to be 4), but I'm still trying to see how I reach this solution.
            $endgroup$
            – potatoguy
            Nov 30 '18 at 3:05










          • $begingroup$
            Well, you could try treating vxs 1 and 2 as a single vx (and same for 3 and 4), then find the number of cycles. But remember both these edges can appear in two directions on a cycle.
            $endgroup$
            – Puck Rombach
            Nov 30 '18 at 3:12



















          0












          $begingroup$

          Q(a)-Q(c) is correct, and Q(d) can be seen as {1,2}{2,3}{3,4} - {2,3} which is 2(n-2)!-(n-3)! (e) can brake into (12)(34)(56)789 and applying counting,answer=(n)!(n-k-1)!/(2n) (f)Obviously Answer=1/2(n-1)!-(n-3)!






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Hint: if we do consider starting point and orientation, then the number of Hamiltonian cycles is the number of ways that we can order $[n]$, i.e. the number of permutations. If you know the order in which to visit the vertices, this tells you exactly the cycle. Each cycle is then counted $n$ times for each possible starting point, and twice for each direction around the cycle.



            Hint for part 2: A cycle can contain ${ 1,2 }$ and ${ 3,4 }$ if it (for example) also contains edge ${ 2,3 }$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for the hint, here's what I wrote for questions: Q1) n! / 2n, Q2: (n-2)!, Q3: (n-3)!
              $endgroup$
              – potatoguy
              Nov 30 '18 at 2:58












            • $begingroup$
              I'm still stuck on Q4, since the graph is separated into two sections. Do you happen to have more advise on how to approach this problem? I'm testing it out on a complete graph K5 (result seems to be 4), but I'm still trying to see how I reach this solution.
              $endgroup$
              – potatoguy
              Nov 30 '18 at 3:05










            • $begingroup$
              Well, you could try treating vxs 1 and 2 as a single vx (and same for 3 and 4), then find the number of cycles. But remember both these edges can appear in two directions on a cycle.
              $endgroup$
              – Puck Rombach
              Nov 30 '18 at 3:12
















            1












            $begingroup$

            Hint: if we do consider starting point and orientation, then the number of Hamiltonian cycles is the number of ways that we can order $[n]$, i.e. the number of permutations. If you know the order in which to visit the vertices, this tells you exactly the cycle. Each cycle is then counted $n$ times for each possible starting point, and twice for each direction around the cycle.



            Hint for part 2: A cycle can contain ${ 1,2 }$ and ${ 3,4 }$ if it (for example) also contains edge ${ 2,3 }$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for the hint, here's what I wrote for questions: Q1) n! / 2n, Q2: (n-2)!, Q3: (n-3)!
              $endgroup$
              – potatoguy
              Nov 30 '18 at 2:58












            • $begingroup$
              I'm still stuck on Q4, since the graph is separated into two sections. Do you happen to have more advise on how to approach this problem? I'm testing it out on a complete graph K5 (result seems to be 4), but I'm still trying to see how I reach this solution.
              $endgroup$
              – potatoguy
              Nov 30 '18 at 3:05










            • $begingroup$
              Well, you could try treating vxs 1 and 2 as a single vx (and same for 3 and 4), then find the number of cycles. But remember both these edges can appear in two directions on a cycle.
              $endgroup$
              – Puck Rombach
              Nov 30 '18 at 3:12














            1












            1








            1





            $begingroup$

            Hint: if we do consider starting point and orientation, then the number of Hamiltonian cycles is the number of ways that we can order $[n]$, i.e. the number of permutations. If you know the order in which to visit the vertices, this tells you exactly the cycle. Each cycle is then counted $n$ times for each possible starting point, and twice for each direction around the cycle.



            Hint for part 2: A cycle can contain ${ 1,2 }$ and ${ 3,4 }$ if it (for example) also contains edge ${ 2,3 }$.






            share|cite|improve this answer









            $endgroup$



            Hint: if we do consider starting point and orientation, then the number of Hamiltonian cycles is the number of ways that we can order $[n]$, i.e. the number of permutations. If you know the order in which to visit the vertices, this tells you exactly the cycle. Each cycle is then counted $n$ times for each possible starting point, and twice for each direction around the cycle.



            Hint for part 2: A cycle can contain ${ 1,2 }$ and ${ 3,4 }$ if it (for example) also contains edge ${ 2,3 }$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 29 '18 at 21:49









            Puck RombachPuck Rombach

            1266




            1266












            • $begingroup$
              Thanks for the hint, here's what I wrote for questions: Q1) n! / 2n, Q2: (n-2)!, Q3: (n-3)!
              $endgroup$
              – potatoguy
              Nov 30 '18 at 2:58












            • $begingroup$
              I'm still stuck on Q4, since the graph is separated into two sections. Do you happen to have more advise on how to approach this problem? I'm testing it out on a complete graph K5 (result seems to be 4), but I'm still trying to see how I reach this solution.
              $endgroup$
              – potatoguy
              Nov 30 '18 at 3:05










            • $begingroup$
              Well, you could try treating vxs 1 and 2 as a single vx (and same for 3 and 4), then find the number of cycles. But remember both these edges can appear in two directions on a cycle.
              $endgroup$
              – Puck Rombach
              Nov 30 '18 at 3:12


















            • $begingroup$
              Thanks for the hint, here's what I wrote for questions: Q1) n! / 2n, Q2: (n-2)!, Q3: (n-3)!
              $endgroup$
              – potatoguy
              Nov 30 '18 at 2:58












            • $begingroup$
              I'm still stuck on Q4, since the graph is separated into two sections. Do you happen to have more advise on how to approach this problem? I'm testing it out on a complete graph K5 (result seems to be 4), but I'm still trying to see how I reach this solution.
              $endgroup$
              – potatoguy
              Nov 30 '18 at 3:05










            • $begingroup$
              Well, you could try treating vxs 1 and 2 as a single vx (and same for 3 and 4), then find the number of cycles. But remember both these edges can appear in two directions on a cycle.
              $endgroup$
              – Puck Rombach
              Nov 30 '18 at 3:12
















            $begingroup$
            Thanks for the hint, here's what I wrote for questions: Q1) n! / 2n, Q2: (n-2)!, Q3: (n-3)!
            $endgroup$
            – potatoguy
            Nov 30 '18 at 2:58






            $begingroup$
            Thanks for the hint, here's what I wrote for questions: Q1) n! / 2n, Q2: (n-2)!, Q3: (n-3)!
            $endgroup$
            – potatoguy
            Nov 30 '18 at 2:58














            $begingroup$
            I'm still stuck on Q4, since the graph is separated into two sections. Do you happen to have more advise on how to approach this problem? I'm testing it out on a complete graph K5 (result seems to be 4), but I'm still trying to see how I reach this solution.
            $endgroup$
            – potatoguy
            Nov 30 '18 at 3:05




            $begingroup$
            I'm still stuck on Q4, since the graph is separated into two sections. Do you happen to have more advise on how to approach this problem? I'm testing it out on a complete graph K5 (result seems to be 4), but I'm still trying to see how I reach this solution.
            $endgroup$
            – potatoguy
            Nov 30 '18 at 3:05












            $begingroup$
            Well, you could try treating vxs 1 and 2 as a single vx (and same for 3 and 4), then find the number of cycles. But remember both these edges can appear in two directions on a cycle.
            $endgroup$
            – Puck Rombach
            Nov 30 '18 at 3:12




            $begingroup$
            Well, you could try treating vxs 1 and 2 as a single vx (and same for 3 and 4), then find the number of cycles. But remember both these edges can appear in two directions on a cycle.
            $endgroup$
            – Puck Rombach
            Nov 30 '18 at 3:12











            0












            $begingroup$

            Q(a)-Q(c) is correct, and Q(d) can be seen as {1,2}{2,3}{3,4} - {2,3} which is 2(n-2)!-(n-3)! (e) can brake into (12)(34)(56)789 and applying counting,answer=(n)!(n-k-1)!/(2n) (f)Obviously Answer=1/2(n-1)!-(n-3)!






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Q(a)-Q(c) is correct, and Q(d) can be seen as {1,2}{2,3}{3,4} - {2,3} which is 2(n-2)!-(n-3)! (e) can brake into (12)(34)(56)789 and applying counting,answer=(n)!(n-k-1)!/(2n) (f)Obviously Answer=1/2(n-1)!-(n-3)!






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Q(a)-Q(c) is correct, and Q(d) can be seen as {1,2}{2,3}{3,4} - {2,3} which is 2(n-2)!-(n-3)! (e) can brake into (12)(34)(56)789 and applying counting,answer=(n)!(n-k-1)!/(2n) (f)Obviously Answer=1/2(n-1)!-(n-3)!






                share|cite|improve this answer









                $endgroup$



                Q(a)-Q(c) is correct, and Q(d) can be seen as {1,2}{2,3}{3,4} - {2,3} which is 2(n-2)!-(n-3)! (e) can brake into (12)(34)(56)789 and applying counting,answer=(n)!(n-k-1)!/(2n) (f)Obviously Answer=1/2(n-1)!-(n-3)!







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 30 '18 at 9:23









                zacahryzacahry

                1




                1






























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