$(2,4)$-Colorability of Planar Graphs Excluding $3$-, $4$-, and $6$-Cycles

TL;DR

This paper proves that planar graphs excluding 3-, 4-, and 6-cycles are also $(2,4)$-colorable, extending previous results on defective colorings of such graphs.

Contribution

The paper introduces a new defective coloring result, showing these graphs are $(2,4)$-colorable, which was not previously established.

Findings

Graphs are $(2,4)$-colorable without 3-, 4-, and 6-cycles

Extends known defective coloring results for planar graphs

Builds on prior work showing $(0,6)$ and $(3,3)$-colorability

Abstract

A defective $k$-coloring is a coloring on the vertices of a graph using colors $1,2, \dots, k$ such that adjacent vertices may share the same color. A $(d_1,d_2)$-\emph{coloring} of a graph $G$ is a defective $2$-coloring of $G$ such that any vertex colored by color $i$ has at most $d_i$ adjacent vertices of the same color, where $i\in\{1,2\}$. A graph $G$ is said to be $(d_1,d_2)$-\emph{colorable} if it admits a $(d_1,d_2)$-coloring. Defective $2$-coloring in planar graphs without $3$-cycles, $4$-cycles, and $6$-cycles has been investigated by Dross and Ochem, as well as Sittitrai and Pimpasalee. They showed that such graphs are $(0,6)$-colorable and $(3,3)$-colorable, respectively. In this paper, we proved that these graphs are also $(2,4)$-colorable.