Monochromatic Loose-Cycle Partitions in Hypergraphs

András Gyárfás,Gábor Sárközy

doi:10.37236/4062

Abstract

In this paper we study the monochromatic loose-cycle partition problem for non-complete hypergraphs. Our main result is that in any $r$-coloring of a $k$-uniform hypergraph with independence number $\alpha$ there is a partition of the vertex set into monochromatic loose cycles such that their number depends only on $r$, $k$ and $\alpha$. We also give an extension of the following result of Pósa to hypergraphs: the vertex set of every graph $G$ can be partitioned into at most $\alpha(G)$ cycles, edges and vertices.

Highlights

Our main result is that in any r-coloring of a k-uniform hypergraph with independence number α there is a partition of the vertex set into monochromatic loose cycles such that their number depends only on r, k and α
It was shown in [6] that the vertex set of any r-edge-colored complete graph can be partitioned into monochromatic cycles and vertices so that their number depends only on r
A subset of vertices in a hypergraph is called independent if it does not contain any edge from H and the independence number of a hypergraph H, α(H), is defined as the maximum cardinality of an independent set in H

Summary

Introduction

It was shown in [6] that the vertex set of any r-edge-colored complete graph can be partitioned into monochromatic cycles and vertices so that their number depends only on r. A well-known result of Posa [16] (see Exercise 8.3 in [12]) states that c(1, 2, α) = α, i.e. every graph G can be partitioned into at most α(G) cycles (where we accept a vertex and an edge as a cycle) Perhaps this can be extended as follows. We prove a result that is weaker than Conjecture 1 (but still extends Posa’s theorem), replacing loose cycles by weak cycles where only cyclically consecutive edges intersect but their intersection size is not restricted. A Berge cycle in S is a tight cycle (where every set of three consecutive the electronic journal of combinatorics 21(2) (2014), #P2.36 vertices forms an edge) in the complement of S. One can derive this from a result of Katona and Kierstead [11] (for n 15) but a direct inductive argument gives it for any n 4

Sketch of the proof of Theorem 1

The main tool from Ramsey Theory

Step 1

Step 2

Step 3

Step 5