Finding paths in a graph with n vertices












0












$begingroup$


Let n ≥ 2 be a natural number. Consider the graph G = (V, E) where
V ={0,1,2,...,n} and E=({0,1},{0,2},...,{0,n}) ∪ ({1,2},...,{n−1,n}) ∪ ({n,1})



For paths, it's a sequence of (non-repeating) vertices.
For cycles, we only distinguish them if they form different subgraphs.



How many paths of length 2 are there in G?
How many paths of length 3 are there in G?
How many cycles are there in G?



I can obviously draw out the first couple cases and count this, but there has to be a summation formula or something I'm missing...






discrete-mathematics graph-theory







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$endgroup$












  • $begingroup$
    If $d_0,d_1,dots,d_n$ is the degree sequence, then the number of paths of length $2$ (paths with $2$ edges, $3$ vertices) is $$sum_{i=0}^nbinom{d_i}2 = binom n2+nbinom32=frac{n(n+5)}2$$
    $endgroup$
    – bof
    Feb 8 '17 at 12:57










  • $begingroup$
    The number of cycles is $$1+2binom n2=n^2-n+1$$
    $endgroup$
    – bof
    Feb 8 '17 at 13:05
















0












$begingroup$


Let n ≥ 2 be a natural number. Consider the graph G = (V, E) where
V ={0,1,2,...,n} and E=({0,1},{0,2},...,{0,n}) ∪ ({1,2},...,{n−1,n}) ∪ ({n,1})



For paths, it's a sequence of (non-repeating) vertices.
For cycles, we only distinguish them if they form different subgraphs.



How many paths of length 2 are there in G?
How many paths of length 3 are there in G?
How many cycles are there in G?



I can obviously draw out the first couple cases and count this, but there has to be a summation formula or something I'm missing...






discrete-mathematics graph-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    If $d_0,d_1,dots,d_n$ is the degree sequence, then the number of paths of length $2$ (paths with $2$ edges, $3$ vertices) is $$sum_{i=0}^nbinom{d_i}2 = binom n2+nbinom32=frac{n(n+5)}2$$
    $endgroup$
    – bof
    Feb 8 '17 at 12:57










  • $begingroup$
    The number of cycles is $$1+2binom n2=n^2-n+1$$
    $endgroup$
    – bof
    Feb 8 '17 at 13:05














0












0








0





$begingroup$


Let n ≥ 2 be a natural number. Consider the graph G = (V, E) where
V ={0,1,2,...,n} and E=({0,1},{0,2},...,{0,n}) ∪ ({1,2},...,{n−1,n}) ∪ ({n,1})



For paths, it's a sequence of (non-repeating) vertices.
For cycles, we only distinguish them if they form different subgraphs.



How many paths of length 2 are there in G?
How many paths of length 3 are there in G?
How many cycles are there in G?



I can obviously draw out the first couple cases and count this, but there has to be a summation formula or something I'm missing...






discrete-mathematics graph-theory







share|cite|improve this question









$endgroup$




Let n ≥ 2 be a natural number. Consider the graph G = (V, E) where
V ={0,1,2,...,n} and E=({0,1},{0,2},...,{0,n}) ∪ ({1,2},...,{n−1,n}) ∪ ({n,1})



For paths, it's a sequence of (non-repeating) vertices.
For cycles, we only distinguish them if they form different subgraphs.



How many paths of length 2 are there in G?
How many paths of length 3 are there in G?
How many cycles are there in G?



I can obviously draw out the first couple cases and count this, but there has to be a summation formula or something I'm missing...






discrete-mathematics graph-theory




discrete-mathematics graph-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 3 '14 at 2:41









ConfusedGraphTheoristConfusedGraphTheorist

11




11












  • $begingroup$
    If $d_0,d_1,dots,d_n$ is the degree sequence, then the number of paths of length $2$ (paths with $2$ edges, $3$ vertices) is $$sum_{i=0}^nbinom{d_i}2 = binom n2+nbinom32=frac{n(n+5)}2$$
    $endgroup$
    – bof
    Feb 8 '17 at 12:57










  • $begingroup$
    The number of cycles is $$1+2binom n2=n^2-n+1$$
    $endgroup$
    – bof
    Feb 8 '17 at 13:05


















  • $begingroup$
    If $d_0,d_1,dots,d_n$ is the degree sequence, then the number of paths of length $2$ (paths with $2$ edges, $3$ vertices) is $$sum_{i=0}^nbinom{d_i}2 = binom n2+nbinom32=frac{n(n+5)}2$$
    $endgroup$
    – bof
    Feb 8 '17 at 12:57










  • $begingroup$
    The number of cycles is $$1+2binom n2=n^2-n+1$$
    $endgroup$
    – bof
    Feb 8 '17 at 13:05
















$begingroup$
If $d_0,d_1,dots,d_n$ is the degree sequence, then the number of paths of length $2$ (paths with $2$ edges, $3$ vertices) is $$sum_{i=0}^nbinom{d_i}2 = binom n2+nbinom32=frac{n(n+5)}2$$
$endgroup$
– bof
Feb 8 '17 at 12:57




$begingroup$
If $d_0,d_1,dots,d_n$ is the degree sequence, then the number of paths of length $2$ (paths with $2$ edges, $3$ vertices) is $$sum_{i=0}^nbinom{d_i}2 = binom n2+nbinom32=frac{n(n+5)}2$$
$endgroup$
– bof
Feb 8 '17 at 12:57












$begingroup$
The number of cycles is $$1+2binom n2=n^2-n+1$$
$endgroup$
– bof
Feb 8 '17 at 13:05




$begingroup$
The number of cycles is $$1+2binom n2=n^2-n+1$$
$endgroup$
– bof
Feb 8 '17 at 13:05










2 Answers
2






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0












$begingroup$

Assume that $nge4$: smaller cases are sometimes a bit different and can be investigated individually.



Paths of length $2$ are just (directed) edges: there are $2n$ edges and allowing for the direction gives $4n$ paths.



For paths of length $3$




  • without the centre vertex, choose the first vertex and the "direction of travel": $2n$ possibilities;

  • starting with the centre vertex, choose the second vertex and one of its two neighbours: $2n$ possibilities;

  • ending with the centre vertex: same as the previous case;

  • with the centre vertex in the middle, choose the first and last vertex: they must not be the same: $n(n-1)$ possibilities.


So the total number is $n^2+5n$.



Since $nge4$, a cycle of length $3$ can only consist of two adjacent vertices on the "circumference", together with the centre: to look at it another way, one of the edges on the "circumference" together with the two edges joining it to the centre. There are $n$ possibilities.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The graph you are describing is a Wheel graph: http://en.wikipedia.org/wiki/Wheel_graph



    To get the number of $P_{3}$ (a path of length $2$, which has $3$ vertices) in the graph, you consider the paths along the exterior of the graph. There are $n$ such paths, where $n = |V|$. Then you look at the interior paths (only interior edges are used) through the center vertex, which forms an arithmetic progression $sum_{i=1}^{n-1} i$. Finally, you count paths using an exterior and an interior edge. You again get $n$ such paths.



    To count cycles, you have $C_{i}$, for $i in {3, ..., n}$, for $n geq 3$.






    share|cite|improve this answer











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      2 Answers
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      2 Answers
      2






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      0












      $begingroup$

      Assume that $nge4$: smaller cases are sometimes a bit different and can be investigated individually.



      Paths of length $2$ are just (directed) edges: there are $2n$ edges and allowing for the direction gives $4n$ paths.



      For paths of length $3$




      • without the centre vertex, choose the first vertex and the "direction of travel": $2n$ possibilities;

      • starting with the centre vertex, choose the second vertex and one of its two neighbours: $2n$ possibilities;

      • ending with the centre vertex: same as the previous case;

      • with the centre vertex in the middle, choose the first and last vertex: they must not be the same: $n(n-1)$ possibilities.


      So the total number is $n^2+5n$.



      Since $nge4$, a cycle of length $3$ can only consist of two adjacent vertices on the "circumference", together with the centre: to look at it another way, one of the edges on the "circumference" together with the two edges joining it to the centre. There are $n$ possibilities.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Assume that $nge4$: smaller cases are sometimes a bit different and can be investigated individually.



        Paths of length $2$ are just (directed) edges: there are $2n$ edges and allowing for the direction gives $4n$ paths.



        For paths of length $3$




        • without the centre vertex, choose the first vertex and the "direction of travel": $2n$ possibilities;

        • starting with the centre vertex, choose the second vertex and one of its two neighbours: $2n$ possibilities;

        • ending with the centre vertex: same as the previous case;

        • with the centre vertex in the middle, choose the first and last vertex: they must not be the same: $n(n-1)$ possibilities.


        So the total number is $n^2+5n$.



        Since $nge4$, a cycle of length $3$ can only consist of two adjacent vertices on the "circumference", together with the centre: to look at it another way, one of the edges on the "circumference" together with the two edges joining it to the centre. There are $n$ possibilities.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Assume that $nge4$: smaller cases are sometimes a bit different and can be investigated individually.



          Paths of length $2$ are just (directed) edges: there are $2n$ edges and allowing for the direction gives $4n$ paths.



          For paths of length $3$




          • without the centre vertex, choose the first vertex and the "direction of travel": $2n$ possibilities;

          • starting with the centre vertex, choose the second vertex and one of its two neighbours: $2n$ possibilities;

          • ending with the centre vertex: same as the previous case;

          • with the centre vertex in the middle, choose the first and last vertex: they must not be the same: $n(n-1)$ possibilities.


          So the total number is $n^2+5n$.



          Since $nge4$, a cycle of length $3$ can only consist of two adjacent vertices on the "circumference", together with the centre: to look at it another way, one of the edges on the "circumference" together with the two edges joining it to the centre. There are $n$ possibilities.






          share|cite|improve this answer









          $endgroup$



          Assume that $nge4$: smaller cases are sometimes a bit different and can be investigated individually.



          Paths of length $2$ are just (directed) edges: there are $2n$ edges and allowing for the direction gives $4n$ paths.



          For paths of length $3$




          • without the centre vertex, choose the first vertex and the "direction of travel": $2n$ possibilities;

          • starting with the centre vertex, choose the second vertex and one of its two neighbours: $2n$ possibilities;

          • ending with the centre vertex: same as the previous case;

          • with the centre vertex in the middle, choose the first and last vertex: they must not be the same: $n(n-1)$ possibilities.


          So the total number is $n^2+5n$.



          Since $nge4$, a cycle of length $3$ can only consist of two adjacent vertices on the "circumference", together with the centre: to look at it another way, one of the edges on the "circumference" together with the two edges joining it to the centre. There are $n$ possibilities.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 3 '14 at 4:08









          DavidDavid

          67.8k664126




          67.8k664126























              0












              $begingroup$

              The graph you are describing is a Wheel graph: http://en.wikipedia.org/wiki/Wheel_graph



              To get the number of $P_{3}$ (a path of length $2$, which has $3$ vertices) in the graph, you consider the paths along the exterior of the graph. There are $n$ such paths, where $n = |V|$. Then you look at the interior paths (only interior edges are used) through the center vertex, which forms an arithmetic progression $sum_{i=1}^{n-1} i$. Finally, you count paths using an exterior and an interior edge. You again get $n$ such paths.



              To count cycles, you have $C_{i}$, for $i in {3, ..., n}$, for $n geq 3$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                The graph you are describing is a Wheel graph: http://en.wikipedia.org/wiki/Wheel_graph



                To get the number of $P_{3}$ (a path of length $2$, which has $3$ vertices) in the graph, you consider the paths along the exterior of the graph. There are $n$ such paths, where $n = |V|$. Then you look at the interior paths (only interior edges are used) through the center vertex, which forms an arithmetic progression $sum_{i=1}^{n-1} i$. Finally, you count paths using an exterior and an interior edge. You again get $n$ such paths.



                To count cycles, you have $C_{i}$, for $i in {3, ..., n}$, for $n geq 3$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The graph you are describing is a Wheel graph: http://en.wikipedia.org/wiki/Wheel_graph



                  To get the number of $P_{3}$ (a path of length $2$, which has $3$ vertices) in the graph, you consider the paths along the exterior of the graph. There are $n$ such paths, where $n = |V|$. Then you look at the interior paths (only interior edges are used) through the center vertex, which forms an arithmetic progression $sum_{i=1}^{n-1} i$. Finally, you count paths using an exterior and an interior edge. You again get $n$ such paths.



                  To count cycles, you have $C_{i}$, for $i in {3, ..., n}$, for $n geq 3$.






                  share|cite|improve this answer











                  $endgroup$



                  The graph you are describing is a Wheel graph: http://en.wikipedia.org/wiki/Wheel_graph



                  To get the number of $P_{3}$ (a path of length $2$, which has $3$ vertices) in the graph, you consider the paths along the exterior of the graph. There are $n$ such paths, where $n = |V|$. Then you look at the interior paths (only interior edges are used) through the center vertex, which forms an arithmetic progression $sum_{i=1}^{n-1} i$. Finally, you count paths using an exterior and an interior edge. You again get $n$ such paths.



                  To count cycles, you have $C_{i}$, for $i in {3, ..., n}$, for $n geq 3$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Apr 3 '14 at 4:14

























                  answered Apr 3 '14 at 3:15









                  ml0105ml0105

                  11.4k21538




                  11.4k21538






























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