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I encountered the following problem:

Let $G(V, E)$ be a finite undirected graph with $2n$ vertices, in which every connected component has an even number of vertices. Given that exactly $n$ of the vertices of $G$ have a degree at least $3$, prove that $G$ is cyclic.

My solution to the problem goes as follows:

Let $C$ denote the set of all connected components $G_i(V_i, E_i)$ of $G$ and $|C| = k$.

Let's assume that $G$ is acyclic. That means that every $G_i$ is a tree. Then, by adding exactly $k-1$ edges to $G$, we can connect all connected components in such a way (root of one to another one's leaf, ordering them in a sequence that way) that we obtain a tree $G'(V', E')$, where $V'= V$ and $|E'| = |E| + k - 1$.

But since $G'$ is a tree, that means that $$|E'| = |V'| - 1 = |V| - 1 = 2n - 1$$

By the handshake lemma we also know that $$2|E'| = \sum_{v\in V'}\deg v \ge \underbrace{3n}_{\substack{\text{atleast $n$}\\ \text{with degree $\ge$ 3}}} + \underbrace{n}_{\substack{\text{G' is connected}\\\text{$\Rightarrow$ the remaining $n$ have $\deg\ge 1$}}}$$

Thus we obtain that $$2(2n-1) \ge 4n \Leftrightarrow -1 \ge 0$$ which is a contradiction, hence $G$ can't be acyclic and contains a cycle.

Now, what bothered me about that is that I never used the given fact, that every $G_i$ had an even number of vertices, so I felt like I was missing something huge or my solution is flawed in a fundamnetal way... and I really banged my head against the wall hard enough, but was nonetheless unable to figure out what was wrong. So my question would be: am I missing something or is the mentioned statement in the problem simply redundant? And if it is redundand, how would one go about proving it? I really got stuck somewhere about there.

Many thanks in advance!

D. Petrov
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    I think you did in fact prove that the "every connected component has an even number of vertices" restriction is redundant. – Joffan Jan 18 '21 at 16:17
  • @Joffan Thank you for the confirmation, that's what I also suspected. :) – D. Petrov Jan 18 '21 at 17:59

2 Answers2

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Here's something simple we can deduce from the extra hypothesis: because every connected component has an even number of vertices, there can be no vertices of degree $0$. This leads to a version of your argument that skips the "connect all the components" step.

That is, suppose by way of contradiction that $G$ is a union of $k$ trees. It's known that a forest with $2n$ total vertices and $k$ components has $2n - k$ edges. Therefore by Handshake we have $$\sum \deg(v_i) = 2|E| = 4n - 2k.$$

Now, $n$ vertices have degree at least 3, and $n$ vertices have degree 1 or 2. Therefore $$\sum \deg(v_i) \ge 3n + n = 4n.$$

Putting this together, we get $$4n \le 4n - 2k \implies k \le 0,$$ a contradiction.

Remark: I don't see anything wrong with your argument, so maybe the inclusion of the hypothesis is just to make life a bit easier?

Sam Freedman
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    Well, I for example wasn't aware of the fact you linked to about forests, although it is quite intuitive. In that regard, I'd say it's subjective what we consider "easier" since what I described as a solution was the very frist thing that ran through my head as a concept and therefore I found it quite intuitive and easy. :D Arguably, I could agree that your solution is as well elegant, just using a slightly different set of facts. Thank you for your time and opinion! – D. Petrov Jan 18 '21 at 15:20
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    @MishaLavrov Thanks for the comment. I was using that fact to get the $3n + n$ lower bound on the sum of the degrees. If we assume there are vertices with degree 0, doesn't it weaken the lower bound to $3n + 0$? – Sam Freedman Jan 18 '21 at 15:32
  • One fact I was able to deduce from the given hypothesis was: every $G_i$ has an even number of vertices with even degree. (Actually follows from the universal fact that every graph has an even number of vertices with odd degree). I still couldn't see how that is needed though. – D. Petrov Jan 18 '21 at 15:38
  • @SamFreedman Oh, I see, I missed that. – Misha Lavrov Jan 18 '21 at 15:39
  • @SamFreedman Your argument was actually probably the initial idea of the authors, since it is in fact a bit simpler. But hey, in the end of the day, it's always good to be able to solve a problem with fewer restrictions, right? :D – D. Petrov Jan 18 '21 at 15:41
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    @D.Petrov Proof-writing is just like engineering. We try to optimize our proofs along many axes: correctness, efficiency, generality, elegance etc. The proof can't be the best in all of them, so instead we just try to make the best tradeoffs we personally can! E.g. in this example I'd trade generality for efficiency, but you might prefer to choose it the other way around. – Sam Freedman Jan 18 '21 at 15:50
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It seems even a little simpler than that, once we can show:

Lemma: Any tree has more leaf nodes than degree-3 and higher nodes.
Proof: Let a graph $G$ be a tree on $n$ vertices with $\ell$ leaf nodes and $h$ vertices of degree $3$ or more. Then $G$ has $n{-}1$ edges and by handshaking we know:
$2(n-1) = \sum d(v_i) \; \geq \; \ell + 2(n-\ell-h) + 3h = 2n-\ell+h,\quad$ so $\ell-2\geq h$ as required.

So the fact that half the nodes are degree $3$ or more shows that not all components are trees; i.e. that $G$ contains cycles. We could join components into a tree but it isn't really needed.

(thanks to jlammy for a compact statement of the lemma proof)

Joffan
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