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squaldig's avatar

How do I show that a simple graph of size n >= 2 always has at least two vertices of the same degree?

Asked by squaldig (5points) February 7th, 2007
Observing members: 0 Composing members: 0

20 Answers

ben's avatar
Here's one idea... just use counting:
ben's avatar
Start by picking the vertex with the highest degree, k
ben's avatar
we can assume k >= 2, (if it's 1, we're done).
ben's avatar
Now we know each connected vertex, k(i), must have a different degree (or we're done)
ben's avatar
so k actually must be > 3
ben's avatar
and then iterate through k(i) the same way... should be able to outnumber it
ben's avatar
I'm positive there's a far simpler solution, this was just my first idea... maybe it will help get the wheels turning
ben's avatar
Yeah, this should work. Just for the second step pick the vertex connected to k with the highest degree
bhs's avatar
I don't see how that's a proof... It seems like more of a test for verification that this is true?
ben's avatar
It's an loose idea for a proof
ben's avatar
you're right, it's not a proof
bhs's avatar
hmm... I don't think that will work formally... I think you need to start from the N=2 case and use induction to go up
ben's avatar
Okay, I was close, this proof is indeed much simpler:
ben's avatar
We assume there are no two vertices with the same degree
ben's avatar
so degrees must go from 0 to n-1 (we have n vertices, and an vertex can't connect to itself)
ben's avatar
but have a vertex with degree 0 and n-1 is a contradiction... the n-1 must be connected to every other vertex
ben's avatar
that should do it
finkelitis's avatar
I think this is a pigeonhole argument: What are the possible degrees? 1 through n-1 (let's assume it's connected, else we just start with the smaller pieces). How many vertices? n. So two of them have the same degree by the pigeonhole principle.
finkelitis's avatar
what you said, ben
finkelitis's avatar
essentially

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