Number of Hamiltonian cycles in complete graph Kn with constraints












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I am currently working on a exercice which aims to count the number of hamiltonian cycles in a complete graph. Since it is a completely new topic to me, I struggle to think about how to solve the problem.
I understand how to count the number of Hamiltonian cycles in a complete graph Kn but my main problem is: what if we have constraints on some particular edges that the Hamiltonian must contain?
Here is the exercice with which I'm struggling with





Suppose Kn is a complete graph whose vertices are indexed by
[n] = {1,2,3,...,n} where n >= 4. In this question, a cycle is identied solely by the collection of edges it contains; there is no particular orientation or starting point
associated with a cycle. (Give your answers in terms of n for the following questions.



(a) How many Hamiltonian cycles are there in Kn?



-> My answer : (n-1)!/2. In fact, there is no particular orientation or stationg point, so we can avoid counting the starting point (thus we have n-1)



(b) How many Hamiltonian cycles in Kn contain the edge {1,2}?



-> My answer : since I consider the edge {1,2} as a vertex, the number of HC should be (n-2)!/2



(c) How many Hamiltonian cycles in Kn contain both the edges{1,2} and {2,3}?



-> My Answer : Same reflexion as above, there are (n-3)!/2 hamiltonian cycles



(d) How many Hamiltonian cycles in Kn contain both the edges {1,2} and {3,4}?



-> My question : should I consider {1,2} and {3,4} as proper vertices also?



(e) Suppose that M is a set of k <= n
2 edges in Kn with the property that no two
edges in M share a vertex. How many Hamiltonian cycles in Kn contain all the
edges in M? Give your answer in terms of n and k.



-> Note : Do you have any hint?



(f) How many Hamiltonian cycles in Kn do not contain any edge from {1,2}, {2,3} and {3,4}?



-> Note : Do you have any hint?





Thanks by advance for your help










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    I am currently working on a exercice which aims to count the number of hamiltonian cycles in a complete graph. Since it is a completely new topic to me, I struggle to think about how to solve the problem.
    I understand how to count the number of Hamiltonian cycles in a complete graph Kn but my main problem is: what if we have constraints on some particular edges that the Hamiltonian must contain?
    Here is the exercice with which I'm struggling with





    Suppose Kn is a complete graph whose vertices are indexed by
    [n] = {1,2,3,...,n} where n >= 4. In this question, a cycle is identied solely by the collection of edges it contains; there is no particular orientation or starting point
    associated with a cycle. (Give your answers in terms of n for the following questions.



    (a) How many Hamiltonian cycles are there in Kn?



    -> My answer : (n-1)!/2. In fact, there is no particular orientation or stationg point, so we can avoid counting the starting point (thus we have n-1)



    (b) How many Hamiltonian cycles in Kn contain the edge {1,2}?



    -> My answer : since I consider the edge {1,2} as a vertex, the number of HC should be (n-2)!/2



    (c) How many Hamiltonian cycles in Kn contain both the edges{1,2} and {2,3}?



    -> My Answer : Same reflexion as above, there are (n-3)!/2 hamiltonian cycles



    (d) How many Hamiltonian cycles in Kn contain both the edges {1,2} and {3,4}?



    -> My question : should I consider {1,2} and {3,4} as proper vertices also?



    (e) Suppose that M is a set of k <= n
    2 edges in Kn with the property that no two
    edges in M share a vertex. How many Hamiltonian cycles in Kn contain all the
    edges in M? Give your answer in terms of n and k.



    -> Note : Do you have any hint?



    (f) How many Hamiltonian cycles in Kn do not contain any edge from {1,2}, {2,3} and {3,4}?



    -> Note : Do you have any hint?





    Thanks by advance for your help










    share|cite|improve this question

























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      I am currently working on a exercice which aims to count the number of hamiltonian cycles in a complete graph. Since it is a completely new topic to me, I struggle to think about how to solve the problem.
      I understand how to count the number of Hamiltonian cycles in a complete graph Kn but my main problem is: what if we have constraints on some particular edges that the Hamiltonian must contain?
      Here is the exercice with which I'm struggling with





      Suppose Kn is a complete graph whose vertices are indexed by
      [n] = {1,2,3,...,n} where n >= 4. In this question, a cycle is identied solely by the collection of edges it contains; there is no particular orientation or starting point
      associated with a cycle. (Give your answers in terms of n for the following questions.



      (a) How many Hamiltonian cycles are there in Kn?



      -> My answer : (n-1)!/2. In fact, there is no particular orientation or stationg point, so we can avoid counting the starting point (thus we have n-1)



      (b) How many Hamiltonian cycles in Kn contain the edge {1,2}?



      -> My answer : since I consider the edge {1,2} as a vertex, the number of HC should be (n-2)!/2



      (c) How many Hamiltonian cycles in Kn contain both the edges{1,2} and {2,3}?



      -> My Answer : Same reflexion as above, there are (n-3)!/2 hamiltonian cycles



      (d) How many Hamiltonian cycles in Kn contain both the edges {1,2} and {3,4}?



      -> My question : should I consider {1,2} and {3,4} as proper vertices also?



      (e) Suppose that M is a set of k <= n
      2 edges in Kn with the property that no two
      edges in M share a vertex. How many Hamiltonian cycles in Kn contain all the
      edges in M? Give your answer in terms of n and k.



      -> Note : Do you have any hint?



      (f) How many Hamiltonian cycles in Kn do not contain any edge from {1,2}, {2,3} and {3,4}?



      -> Note : Do you have any hint?





      Thanks by advance for your help










      share|cite|improve this question













      I am currently working on a exercice which aims to count the number of hamiltonian cycles in a complete graph. Since it is a completely new topic to me, I struggle to think about how to solve the problem.
      I understand how to count the number of Hamiltonian cycles in a complete graph Kn but my main problem is: what if we have constraints on some particular edges that the Hamiltonian must contain?
      Here is the exercice with which I'm struggling with





      Suppose Kn is a complete graph whose vertices are indexed by
      [n] = {1,2,3,...,n} where n >= 4. In this question, a cycle is identied solely by the collection of edges it contains; there is no particular orientation or starting point
      associated with a cycle. (Give your answers in terms of n for the following questions.



      (a) How many Hamiltonian cycles are there in Kn?



      -> My answer : (n-1)!/2. In fact, there is no particular orientation or stationg point, so we can avoid counting the starting point (thus we have n-1)



      (b) How many Hamiltonian cycles in Kn contain the edge {1,2}?



      -> My answer : since I consider the edge {1,2} as a vertex, the number of HC should be (n-2)!/2



      (c) How many Hamiltonian cycles in Kn contain both the edges{1,2} and {2,3}?



      -> My Answer : Same reflexion as above, there are (n-3)!/2 hamiltonian cycles



      (d) How many Hamiltonian cycles in Kn contain both the edges {1,2} and {3,4}?



      -> My question : should I consider {1,2} and {3,4} as proper vertices also?



      (e) Suppose that M is a set of k <= n
      2 edges in Kn with the property that no two
      edges in M share a vertex. How many Hamiltonian cycles in Kn contain all the
      edges in M? Give your answer in terms of n and k.



      -> Note : Do you have any hint?



      (f) How many Hamiltonian cycles in Kn do not contain any edge from {1,2}, {2,3} and {3,4}?



      -> Note : Do you have any hint?





      Thanks by advance for your help







      discrete-mathematics graph-theory hamiltonian-path






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      asked Nov 24 at 10:21









      Cleola

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          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





















          • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
            – José Carlos Santos
            Nov 30 at 9:22











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          1 Answer
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          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





















          • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
            – José Carlos Santos
            Nov 30 at 9:22
















          0














          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





















          • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
            – José Carlos Santos
            Nov 30 at 9:22














          0












          0








          0






          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












          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 at 9:19









          zacahry

          1




          1












          • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
            – José Carlos Santos
            Nov 30 at 9:22


















          • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
            – José Carlos Santos
            Nov 30 at 9:22
















          Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
          – José Carlos Santos
          Nov 30 at 9:22




          Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
          – José Carlos Santos
          Nov 30 at 9:22


















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