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A path is simple if no vertices are repeated. How many simple paths exist between two vertices in a complete graph?

One way is listing all simple paths using depth-first search. but I think it should be more simple to find the number of all simple paths between 2 nodes in a complete graph.

Here is the same problem but mine is for a complete graph: Algorithm that finds the number of simple paths from $s$ to $t$ in $G$

Ali
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  • $countOfAllpathInClique(n)=\sum_{k=2}^n \binom{n}{k} \frac{k!}{2} = \Omega(2^n)$

    $countPathBetween2vertices=countOfAllpathInClique(n-2)+2*(n-1)-1$

    because we can assume we have a (n-2)clique + 2 other vertices. so after we computed the all path of (n-2)clique we should add 2*(n-1) new edge and one of them is not new.

    I'm not sure about the correctness of this.

     – Ali Sep 01 '14 at 06:35
  • Your question already contains what you claim to be an answer so I'm not sure what you're asking. – David Richerby Sep 01 '14 at 07:28
  • I'm Asking : How many simple paths between two vertices in n-clique? – Ali Sep 01 '14 at 08:58
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    How is this different from "how many s-t-paths are there in the complete graph"? Also, what have you tried and where did you get stuck? – Raphael Sep 01 '14 at 10:59
  • I won't to compute the complexity of my algorithm,so i need the size of simple paths between two vertices. – Ali Sep 01 '14 at 13:51
  • i have an error in the title: i want the number of path from 2 vertexes in a Complement graph – Ali Sep 02 '14 at 06:53
  • Neither the body of the question nor your answer fits the modified title. – FrankW Sep 02 '14 at 08:03
  • Thanks for your comment :) I edited the body of my question :) – Ali Sep 03 '14 at 05:21
  • Sorry for my mistakes, They was because of my bad English. I mean Complete Graph and i fixed it. Sorry. – Ali Sep 03 '14 at 06:57

1 Answers1

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$c(1)=0,$

$c(n)=(n-2)*c(n-1)+1$

suppose we have c(n-1) path in (n-1)-Complete Graph. when we add one more vertices, we add $(n-2)*c(n-1)$ new path and one for directed path.

the result is same as real:

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  5. 16

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    ===============================EDITED

    $c(n)=\sum_{i=0}^{n-2} \frac{(n-2)!}{i!}=(n-2)!*\sum_{i=0}^{n-2}\frac{1}{i!}$

    .

    This shows that's about $(2\text{≈}3)*(n-2)!$

Ali
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