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I'm trying to figure out to solve this problem: Given a graph G = (V, E) prove e <= n(n-1)/2 for all n, where e is the number of edges and n is the number of vertices.

I'm thinking that I should somehow be using math induction to figure out the correct answer and use n = 1 or 0 for my hypothesis, but I'm getting a little stuck on what to do after -- if I assume n = k, then: e <= (k+1)k/2. and if n = k+1 then e <= k(k-1)/2.

As I understand it, each vertex has n-1 possible edges coming out, and there are n total vertices, which is where n(n-1) comes from and dividing by 2 gets rid of the repeats. But I am unsure how I am to prove this.

winsticknova
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1 Answers1

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The statement is false for multi-graphs. Take the graph:

 /---\
O-----O

There are two vertices (O) and two edges; therefore n=2,e=2 and substituting into n(n-1)/2 <= e gives 1 <= 2 which is false.

However, if you restrict the graph to be simple - disallowing looping edges (where both ends of the edge terminate at the same vertex), multi-edges (where two edges connect the same pair of vertices) and that the graph is undirected - then the property holds.

Consider a complete graph K_n (with n vertices): each of the n vertices is incident to the other n-1 vertices via a connecting edge therefore there are n(n-1) connections from one vertex to another; given that edges are undirected then this will count each edge twice (i.e counting from vertex A to vertex B and vice versa) then the total number of edges will be n(n-1)/2.

Any graph G_n (with n vertices) will be a sub-graph of K_n (since you cannot add any more edges to K_n without creating multi- or looping edges) then there must be equal or fewer edges in G_n than in K_n.

Thus e <= n(n-1)/2 for all simple graphs.

If you further restrict the graph to be planar then you can state that e <= 3n - 6 (when n > 2).

MT0
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