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Let $G$ be a triangle-free simple graph whose minimum degree is $> 2n/5$. Assume that $G$ is not a $5$-cycle. Prove that $G$ is bipartite.

darij grinberg's note: This is claimed to be a result by Andrásfai, Erdős & Sós (1974) in the Wikipedia, but the reference (Andrásfai, B.; Erdős, P.; Sós, V. T. (1974), "On the connection between chromatic number, maximal clique and minimal degree of a graph", Discrete Mathematics, 8 (3): 205–218, doi:10.1016/0012-365X(74)90133-2) is not very readable.

I have been trying to prove this but don't have an intuition on how to start I think it can be proved by contradiction.

YuiTo Cheng
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Sandeep Ipk
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1 Answers1

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Assume the converse. Then $G$ contains a cycle of odd length. Take a cycle $C$ of the smallest odd length $l$. Then $l\ge 5$ and $C$ is chordless (the latter means that if $v$ and $u$ are adjacent vertices of $C$ then they are consecutive in $C$). Let $V’$ be the set of vertices of $G\setminus C$. Then, $V' \neq \varnothing$ (else, $G$ would just be a single cycle, which easily yields a contradiction). Each vertex of $C$ is adjacent to more than $2n/5-2$ vertices of $V’$, so there are more than $l(2n/5-2)\ge 2(n-l)$ edges between $C$ and $V’$. Since $|V’|=n-l$, by the pigeonhole principle there exists a vertex $v\in V’$ adjacent to at least $3$ vertices of $C$. There exist two of them joined by a path in $C$ of odd length less than $l-2$. When we connect them with $v$ we obtain a cycle of odd length less than $l$, a contradiction.

darij grinberg
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Alex Ravsky
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  • I cannot see how the pidgeonhole principle gives you a vertex $v$ with three neighbors in $C$. If all vertices of $V'$ are adjacent to exactly two vertices of $C$, then there are $2(n-l)$ vertices between $V'$ and $C$, so the inequality is satisfied. I think this is no major problem to complete the proof as this creates just one more case to check. – M. Winter Sep 11 '17 at 07:46
  • @M.Winter There are bigger than $2(n-l)$ edges between $V′$ and $C$. – Alex Ravsky Sep 11 '17 at 08:23
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    I see. I only read the $\geq$ but skipped over the "bigger" in front of it. Thank you. Amazing proof. – M. Winter Sep 11 '17 at 08:28
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    What intelligence! Amazing. – Sandeep Ipk Sep 13 '17 at 09:36
  • Brilliant solution! +1 – RFZ Jul 22 '22 at 01:40
  • I think you can say that there exist two ot them joined by a path in $C$ of odd length less than $l-3$. It is because the the distance between these 3 specified vertices in $C$ should be at least 2. Otherwise, there is a triangle in $G$. – RFZ Jul 22 '22 at 01:48