Computer Science: Graph proof

Rabbit_Alex

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Joined
Jan 23, 2007
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Hi there, this is my first post. I signed up because I need help with my Formal Models of Computation class, which is a continuation of Discrete Math/Structures that I struggled through last semester.

Here is my problem.
Let the graph G = (V, E), where V is the set of all vertices in the graph, and E is the set of all edges in the graph. If there is a walk between vertex v1 which exists in V, and vertex v2 which exists in V, then there must be a path whose length is no larger than |V| - 1 (the number of vertices minus 1), between the two vertices. I underlined walk and path because I think they are important for solving the proof.

What I have so far.
I am trying to argue that any graph is composed of "sub-graphs" which fit into categories like wheel graph, complete bipartite graph, circle graph, etc. The path length between any two vertices in these types of graph is always less than |V| - 1. Since any graph will be made of subgraphs which have this property, it must be true for all graphs. So far all I have are some pictures of different types of graphs showing how this property is true for all of them. What I need is a way to combine them into a general proof for all graphs.

I am assuming that my reasoning is completely wrong.
Sorry about the long post. Thank you for any help. :D
 
This is more of an informal approach... you may or may not need a lemma or two for the things in quotes.

"Obviously" the shortest path from one vertex to another will be one in which each edge walked across is only walked across once, otherwise "you are visiting nodes you need not visit", or "you are backtracking". So, each node in this path is visited at most once. By the Pigeonhole Principle, since there are |V| nodes, if you take |V| or more walks, you will be revisiting a node you've already seen.

-Daon
 
daon said:
This is more of an informal approach... you may or may not need a lemma or two for the things in quotes.

-Daon

I think a proof of Djikstra's algorithm will come in useful here. Thanks for the help.
 
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