planar graphs

Graph Theory » Coloring » Vertex coloring

Nonrepetitive colourings of planar graphs ★★

Author(s): Alon N.; Grytczuk J.; Hałuszczak M.; Riordan O.

Question Do planar graphs have bounded nonrepetitive chromatic number?

Keywords: nonrepetitive colouring; planar graphs

Posted by David Wood
updated March 24th, 2014

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Graph Theory » Coloring » Homomorphisms

Circular choosability of planar graphs ★

Author(s): Mohar

Let $G = (V, E)$ be a graph. If $p$ and $q$ are two integers, a $(p,q)$ -colouring of $G$ is a function $c$ from $V$ to $\{0,\dots,p-1\}$ such that $q \le |c(u)-c(v)| \le p-q$ for each edge $uv\in E$ . Given a list assignment $L$ of $G$ , i.e.~a mapping that assigns to every vertex $v$ a set of non-negative integers, an $L$ -colouring of $G$ is a mapping $c : V \to N$ such that $c(v)\in L(v)$ for every $v\in V$ . A list assignment $L$ is a $t$ - $(p,q)$ -list-assignment if $L(v) \subseteq \{0,\dots,p-1\}$ and $|L(v)| \ge tq$ for each vertex $v \in V$ . Given such a list assignment $L$ , the graph G is $(p,q)$ - $L$ -colourable if there exists a $(p,q)$ - $L$ -colouring $c$ , i.e. $c$ is both a $(p,q)$ -colouring and an $L$ -colouring. For any real number $t \ge 1$ , the graph $G$ is $t$ - $(p,q)$ -choosable if it is $(p,q)$ - $L$ -colourable for every $t$ - $(p,q)$ -list-assignment $L$ . Last, $G$ is circularly $t$ -choosable if it is $t$ - $(p,q)$ -choosable for any $p$ , $q$ . The circular choosability (or circular list chromatic number or circular choice number) of G is $cch(G) := \inf\{t \ge 1 : G \text{ is circularly $t$-choosable}\}.$