Longest odd cycles in non-bipartite $C_{2k+1}$-free graphs

TL;DR

This paper extends classical results on odd cycles in $C_{2k+1}$-free graphs by replacing minimum degree conditions with edge count conditions, establishing bounds on the length of the longest odd cycle based on edge density.

Contribution

It introduces a new edge-based condition for the absence of long odd cycles in $C_{2k+1}$-free graphs, generalizing and strengthening previous degree-based results.

Findings

If $e(G)$ exceeds a specific quadratic bound, then $G$ contains no odd cycle longer than $r$.

The bounds are shown to be tight through constructions.

The results unify and extend prior theorems by multiple researchers.

Abstract

In strengthening a result of Andr\'asfai, Erd\H{o}s and S\'os in 1974, H\"{a}ggkvist proved that if $G$ is an $n$-vertex $C_{2k+1}$-free graph with minimum degree $\delta(G)>\frac{2n}{2k+3}$ and $n>\binom{k+2}{2}(2k+3)(3k+2)$, then $G$ contains no odd cycle of length greater than $\frac{k+1}{2}$. This result has many applications.In this paper, we consider a similar problem by replacing minimum degree condition with edge number condition. We prove that for integers $n,k,r$ with $k\geq 2,3\leq r\leq 2k$ and $n \geq 2\left(r+2\right)\left(r+1\right)\left(r+2k\right)$, if $G$ is an $n$-vertex $C_{2k+1}$-free graph with $e(G) \geq \left\lfloor\frac{(n-r+1)^2}{4}\right\rfloor+\binom{r}{2}$, then $G$ contains no odd cycle of length greater than $r$. The construction shows that the result is best possible. This extends a result of Brandt [Discrete Applied Mathematics 79 (1997)], and a result of Bollob\'as and Thomason [Journal of Combinatorial Theory, Series B. 77 (1999)], and a result of Caccetta and Jia [Graphs Combin. 18 (2002)] and independently proving by Lin, Ning and Wu [Combin. Probab. Comput. 30 (2021)]. Recently, Ren, Wang, Yang, and the second author [SIAM J. Discrete Math. 38 (2024)] show that for $3\leq r\leq 2k$ and $n\geq 318(r-2)^2k$, every $n$-vertex $C_{2k+1}$-free graph with $e(G) \geq \left\lfloor\frac{(n-r+1)^2}{4}\right\rfloor+\binom{r}{2}$ can be made bipartite by deleting at most $r-2$ vertices or deleting at most $\binom{\lfloor\frac{r}{2}\rfloor}{2}+\binom{\lceil\frac{r}{2}\rceil}{2}$ edges. As an application, we derive this result and provide a simple proof.