A community for students.
Here's the question you clicked on:
 0 viewing
RolyPoly
 one year ago
Given an undirected graph H, let C be the adjacency matrix of H. What is the meaning of entries in \(C^n\)?
RolyPoly
 one year ago
Given an undirected graph H, let C be the adjacency matrix of H. What is the meaning of entries in \(C^n\)?

This Question is Open

KingGeorge
 one year 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
 one year 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.

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

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

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

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.0Ehm, sorry, induction.

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

Crixpack
 one year ago
Best ResponseYou've already chosen the best response.1King George is correct. Wahid yes ;)

RolyPoly
 one year 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!

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.0Would you mind explaining the intuition behind?

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

Crixpack
 one year ago
Best ResponseYou've already chosen the best response.1This 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
 one year 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.

RolyPoly
 one year 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
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0now let's look at \[ A^2 V = A \begin{bmatrix} 0\\ 1\\ 0\\ 1\\ 0\\ 0 \end{bmatrix} = A \left( \begin{bmatrix} 0\\ 1\\ 0\\ 0\\ 0\\ 0 \end{bmatrix} +\begin{bmatrix} 0\\ 1\\ 0\\ 0\\ 0\\ 0 \end{bmatrix} \right)\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.