Thanks in advance!

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- Apr 6th 2008, 03:01 PMshawnMore Graph Questions
Thanks in advance!

- Apr 6th 2008, 04:21 PMPlato
You do know, I hope, that these definitions are not unique or standard.

So if we mean by a*walk*a sequence of alternating vertices and edges, beginning with a vertex and ending with a vertex, not necessarily the same.

A*path*is a walk in which all edges are distinct.

(Again these may not agree with your definitions)

So is a star-like graph with n+1 vertices, 0 to n, both “010” and “102” are walks but only “102” is a path.

Thus the number of walks of length two is $\displaystyle {\binom {n} {2}} + n$. WHY?

But the number of paths of length two is $\displaystyle {\binom {n} {2}}$. WHY? - Apr 6th 2008, 07:10 PMshawn
Your interpretation is correct.

The number of walks of length two is $\displaystyle {\binom {n} {2}} + n$ because you either just choose two edges from n (all the paths) or you can walk from 0 to any vertex and back and there are n ways to do this.

and the number of paths of length two is $\displaystyle {\binom {n} {2}}$ because you just choose two edges from n and because they are all adjacent through 0 you have your path. - Apr 7th 2008, 03:32 PMPlato
**Please do not use PM's to ask for more help!**

If n=4 then $\displaystyle Adj = \left( {\begin{array}{*{20}c} 0 & 1 & 1 & 1 & 1 \\

1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\

1 & 1 & 1 & 1 & 0 \\\end{array}} \right)$

$\displaystyle \left( {Adj} \right)^2 = \left( {\begin{array}{*{20}c}

4 & 3 & 3 & 3 & 4 \\ 3 & 4 & 3 & 3 & 4 \\ 3 & 3 & 4 & 3 & 4 \\

3 & 3 & 3 & 4 & 4 \\ 4 & 4 & 4 & 4 & 5 \\\end{array}} \right)$ - Apr 7th 2008, 03:39 PMshawn