# Thread: Vertices of a simple graph

1. ## Vertices of a simple graph

Prove that a smiple graph with at least two vertices would have two or more vertices of the same degree.

thanks.

2. Realize that if a simple graph has n vertices then the maximum degree of any vertex is n-1. Thus suppose that we have a simple graph G with a degree sequence of <0,1,2,…,n-1>, that is all the n-vertices are of different degree. Now remove the vertex of degree 0. We now have a new graph G’ with (n-1) vertices and the same number of edges. But it has a vertex of degree (n-1). But that is impossible; so the original graph is impossible. Thus the graph G would have to have at least two vertices of the same degree.

3. ## Discrete maths solutions - option?

TTcommander

By the way what is the option of using the "piegonhole" principle of answering the question - 2nd form.

Thanks

Fortuna

4. ## Sorry for getting the name wrong my apology Plato

Plato

By the way what is the option of answering the question using the pigeonhole princple - 2nd form? Look forward to your comments.

Cheers

Fortuna

5. Originally Posted by fortuna
Pluto
Thanks heaps for your explanation on Q. 2.
Originally Posted by Jhevon
It's "Plato", lol
It’s a dog’s life is it not.

6. Originally Posted by fortuna
By the way what is the option of answering the question using the pigeonhole principle - 2nd form?
Actually the graph theory proof above does use the pigeonhole principle implicitly. Think of 15 pigeonholes numbered 0,1,2,…,14. Assign each person to the hole that matches the number of acquaintances at the meeting that he/she has. If the hole #14 is empty then the fifteen will be assigned to 14 holes so there is a hole 0-13 with at least two. If hole #14 is not empty then someone there knows each other person there. Hence hole #0 must be empty(WHY?) Thus the fifteen will be assigned to 14 holes so there is a hole 1-14 with at least two.

Plato