Clustered variants of Hajós' conjecture

On the conjecture of Hajós

Let G=G(n) be a graph of n vertices. Let X=X(G) denote its chromatic number and a=o(G) the largest integer I so that G contains a subdivision of K, i.e . a(G)=1 is the largest integer such that G contains a subgraph homeomorphic with complete graph of l vertices . Let us put H(G)= X(G) and H(n)=max H(G(n)) o (G) G(n) Hajós [10] conjectured that H(n)=1 and Catlin, [2] recently disproved the conj...

Triangulations and the Hajós Conjecture

The Hajós Conjecture was disproved in 1979 by Catlin. Recently, Thomassen showed that there are many ways that Hajós conjecture can go wrong. On the other hand, he observed that locally planar graphs and triangulations of the projective plane and the torus satisfy Hajós Conjecture, and he conjectured that the same holds for arbitrary triangulations of closed surfaces. In this note we disprove t...

Hajós' conjecture for line graphs

We prove that, if a graph G (without multiple edges) has maximum degree d and edge-chromatic number d + 1, then G contains two distinct vertices x, y and a collection of d pairwise edge-disjoint paths between x and y. In particular, the line graph of G satisfies Hajós’ conjecture. © 2006 Elsevier Inc. All rights reserved.

On a Coloring Conjecture of Hajós

Hajós conjectured that graphs containing no subdivision of K5 are 4-colorable. It is shown in [15] that if there is a counterexample to this conjecture then any minimum such counterexample must be 4-connected. In this paper, we further show that if G is a minimum counterexample to Hajós’ conjecture and S is a 4-cut in G then G− S has exactly two components. AMS Subject Classification: 05C15, 05...

Hajós' theorem for list colorings of hypergraphs