Every planar graph is $1$-defective $(9,2)$-paintable

TL;DR

This paper proves that every planar graph can be colored on-line with a defective coloring scheme where each vertex receives two colors, using at most nine colors, with a maximum degree of one in the induced subgraph for each color.

Contribution

It introduces and proves the first result that all planar graphs are 1-defective (9,2)-paintable in an on-line coloring setting.

Findings

Every planar graph is 1-defective (9,2)-paintable.

The paper establishes a winning strategy for Painter in the on-line game.

It advances understanding of defective coloring in on-line graph coloring contexts.

Abstract

Assume $L$ is a $k$-list assignment of a graph $G$. A $d$-defective $m$-fold $L$-colouring $\phi$ of $G$ assigns to each vertex $v$ a set $\phi(v)$ of $m$ colours, so that $\phi(v) \subseteq L(v)$ for each vertex $v$, and for each colour $i$, the set $\{v: i \in \phi(v)\}$ induces a subgraph of maximum degree at most $d$. In this paper, we consider on-line list $d$-defective $m$-fold colouring of graphs, where the list assignment $L$ is given on-line, and the colouring is constructed on-line. To be precise, the $d$-defective $(k,m)$-painting game on a graph $G$ is played by two players: Lister and Painter. Initially, each vertex has $k$ tokens and is uncoloured. In each round, Lister chooses a set $M$ of vertices and removes one token from each chosen vertex. Painter colours a subset $X$ of $M$ which induces a subgraph $G[X]$ of maximum degree at most $d$. A vertex $v$ is fully coloured if $v$ has received $m$ colours. Lister wins if at the end of some round, there is a vertex with no more tokens left and is not fully coloured. Otherwise, at some round, all vertices are fully coloured and Painter wins. We say $G$ is $d$-defective $(k,m)$-paintable if Painter has a winning strategy in this game. This paper proves that every planar graph is $1$-defective $(9,2)$-paintable.