I have to prove that in a graph $G$ with $n$ vertices and $n$ edges which has no isolated or pendant vertices, every vertex has degree 2.

Now I know that $$\sum_{v\in V} d(v)=2|E|=2n$$

but I suppose I also have to prove every vertex has same degree and I'm not sure how to do that.

  • $\begingroup$ What is a "pendant" vertex? $\endgroup$ – JMoravitz Sep 14 '17 at 19:45
  • $\begingroup$ @JMoravitz Presumably of degree 1. $\endgroup$ – Théophile Sep 14 '17 at 19:45

Hint: From the equation you gave, we can divide both sides by $n$ to get the average degree:

$$\frac1n \sum_{v\in V} d(v)=2$$

Now, if any vertex has a degree greater than the average, there must necessarily be another vertex with a degree lower than the average. What does that imply?

  • 1
    $\begingroup$ Makes sense, thanks! $\endgroup$ – Stefan Stipanovic Sep 14 '17 at 19:47
  • $\begingroup$ The logic should go the other way (if pendant then there exists a vertex with degree greater than 2), but the core idea is excellent. $\endgroup$ – David G. Stork Sep 14 '17 at 19:49
  • $\begingroup$ @DavidG.Stork Thank you! I think the logic is solid, though: we know from the outset that there are no isolated or pendant vertices, so the only other possibility is if there is a vertex of degree greater than 2, which leads to a contradiction. $\endgroup$ – Théophile Sep 14 '17 at 20:18

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