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anonymous
 2 years ago
Given an undirected graph H, let C be the adjacency matrix of H. What is the meaning of entries in \(C^n\)?
anonymous
 2 years ago
Given an undirected graph H, let C be the adjacency matrix of H. What is the meaning of entries in \(C^n\)?

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KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1If I'm remembering correctly...and I may not be, the \(i,j\)th entry in \(C^n\) is the number of paths of length \(n\) between the \(i\)the vertex and the \(j\)th vertex.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1I think I saw this in a combinatorics class I had last year. I could probably dig out the proof if you want.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Ah! I would love to know the proof, at least to understand why. :)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0By the way, if the proof involves MI, please, please do not post it...

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1What do you mean by MI?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Ehm, sorry, induction.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1Ah. Well my proof does do it by induction...

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0King George is correct. Wahid yes ;)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Haha, because the hint of the proof is to prove by induction, and I don't want to copy your work, lol!

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Would you mind explaining the intuition behind?

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1Well, it's actually a pretty short proof. The main part that uses induction, is where you write\[C^n=C^{n1}C,\]and for a given entry \(C_{ij}\), you can write\[(C^{n1}C)_{ij}=\sum_{k}(C^{n1})_{ik}C_{kj}\]

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1After that, it's a counting argument.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0This is really rough but you can think of it like this: the i,jth entry of C is 1 only if i and j are connected. Now if we multiply C by itself, to get the i,kth entry (of C^2), we would be multiply the ith row of C by the kth column of C now the ith row would be 1's in places where i is adjacent to some other node and 0's elsewhere and the kth column would be the same, 1's only where k was adjacent to some node, and 0's elsewhere so to get the i,kth entry of C^2, it would only be nonzero if i and k were adjacent to some other node, which would imply that there was a path of length 2 from i to k, and everytime they shared a common node, it would add 1 to the entry, so the entry counts the number of patsh of length 2 and then u go up from there, each power basically extends the path by 1 edge if it is possible, (if it not possible the product would just be 0)

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1So I guess the underlying idea of the proof, is to count the number of paths of length \(n1\) from a vertex \(i\) to a vertex \(k\), and then the number of paths of length \(1\) from the vertex \(k\) to the vertex \(j\). And then let \(k\) vary over the whole graph.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Not sure if it looks good. Base case: When n=1, \(A^1 = A\), which shows the number of path to vertex j, \(v_j\), to vertex i, \(v_i\), with the path length being 1. So, it is true for n=1. Induction step: Assume it is true for n=m1. Consider the i,jth entry of \(A^m\), \[(A^m)_{ij}\]\[ = (AA^{m1})_{ij} \]\[= A_{i1}(A^{m1})_{1j} + A_{i2}(A^{m1})_{2j} +...+A_{in}(A^{m1})_{nj}\]\[=\sum_{k}A_{ik}(A^{m1})_{kj}\] Consider \( A_{i1}(A^{m1})_{1j}\), \( A_{i1}(A^{m1})_{1j}\) is the number of paths of length 1 from \(v_i\) to \(v_1\) times the number of paths of length m1 from \(v_1\) to \(v_j\), which is equal to the number of paths of length m from \(v_i\) to \(v_j\) with \(v_1\) being the second vertex visited. This argument follows for each term \( A_{ik}(A^{m1})_{kj}\), where \( A_{ik}(A^{m1})_{kj}\) is the number of paths of length m from \(v_i\) to \(v_j\) with \(v_k\) being the second vertex visited. Hence, \(\sum_{k}A_{ik}(A^{m1})_{kj}\) is the total number of all possible paths with length m from \(v_i\) to \(v_j\).

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0let's take this graph for example http://en.wikipedia.org/wiki/Adjacency_matrix#Examples also let's take that we are at vertex 3 of that graph.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0from 3, you can go to vertex 2 or vertex 4. now let us represent the vertex 3 by \[ V = \begin{bmatrix} 0\\ 0\\ 1\\ 0\\ 0\\ 0 \end{bmatrix} \] Let the adjacency matrix be A. \[ A V = \begin{bmatrix} 0\\ 1\\ 0\\ 1\\ 0\\ 0 \end{bmatrix} \]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0it means that ... after one step, either you end up in vertex 2 or vertex 4. therefore it makes sense to construct adjacency matrix this way, and represent state by vectors.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0now let's look at \[ A^2 V = A \begin{bmatrix} 0\\ 1\\ 0\\ 1\\ 0\\ 0 \end{bmatrix} = A \left( \begin{bmatrix} 0\\ 1\\ 0\\ 0\\ 0\\ 0 \end{bmatrix} +\begin{bmatrix} 0\\ 1\\ 0\\ 0\\ 0\\ 0 \end{bmatrix} \right)\]
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