Abstract
A locally irregular graph is a graph in which the end vertices of every edge have distinct degrees. A locally irregular edge coloring of a graph G is any edge coloring of G such that each of the colors induces a locally irregular subgraph of G. A graph G is colorable if it allows a locally irregular edge coloring. The locally irregular chromatic index of a colorable graph G, denoted by χirr′(G), is the smallest number of colors used by a locally irregular edge coloring of G. The local irregularity conjecture claims that all graphs, except odd-length paths, odd-length cycles and a certain class of cacti are colorable by three colors. As the conjecture is valid for graphs with a large minimum degree and all non-colorable graphs are vertex disjoint cacti, we study rather sparse graphs. In this paper, we give a cactus graph B which contradicts this conjecture, i.e., χirr′(B)=4. Nevertheless, we show that the conjecture holds for unicyclic graphs and cacti with vertex disjoint cycles.
Highlights
All graphs mentioned in this paper are considered to be simple and finite
We focus our attention on locally irregular edge colorings exclusively, and we say a graph is colorable if it admits such a coloring
The consideration of the bow-tie graph gives rise to the following questions: are there any other graphs for which Conjecture 2 does not hold, do all colorable cacti admit a 4-liec, what is the thight upper bound on χirr(G) of general graphs? We believe the following conjecture holds, which is a weaker form of the Local Irregularity Conjecture
Summary
All graphs mentioned in this paper are considered to be simple and finite. An edge coloring of a graph is neighbor sum distinguishing if any two neighboring vertices differ in the sum of the colors of the edges incident to them. Assume first that a tree K is not colorable, i.e., K is an odd-length path This implies there exists in K an edge ev incident with v, such that K − ev is a collection of even paths which, admits 2-liec φa1,b. The shrub-based coloring φa,b = ∑ik=1 φai ,b would be a 2-liec of K, again a contradiction. The shrub-based coloring φa,b = ∑ik=1 φai ,b is not a 2-liec only if the a-degree of w3 or w4 by φa,b is 4. The only cases when K does not admit a 2-liec are: i) dK(v) = 3 and the a-sequence of v by the shrub-based coloring φa,b = ∑ik=1 φai ,b is 3, 2, 2, or ii) dK(v) = 4 and the a-sequence of v by φa,b is 4, 3, 3, 2.
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