Vertex coloring of plane graphs with nonrepetitive boundary paths

A sequence $s_1,s_2,...,s_k,s_1,s_2,...,s_k$ is a repetition. A sequence $S$ is nonrepetitive, if no subsequence of consecutive terms of $S$ form a repetition. Let $G$ be a vertex colored graph. A path of $G$ is nonrepetitive, if the sequence of colors on its vertices is nonrepetitive. If $G$ is a plane graph, then a facial nonrepetitive vertex coloring of $G$ is a vertex coloring such that any facial path is nonrepetitive. Let $\pi_f(G)$ denote the minimum number of colors of a facial nonrepetitive vertex coloring of $G$. Jendro\vl and Harant posed a conjecture that $\pi_f(G)$ can be bounded from above by a constant. We prove that $\pi_f(G)\le 24$ for any plane graph $G$

New Bounds for Facial Nonrepetitive Colouring

We prove that the facial nonrepetitive chromatic number of any outerplanar graph is at most 11 and of any planar graph is at most 22.Comment: 16 pages, 5 figure

Nonrepetitive Colourings of Planar Graphs with O(log⁡n)O(\log n) Colours

A vertex colouring of a graph is \emph{nonrepetitive} if there is no path for which the first half of the path is assigned the same sequence of colours as the second half. The \emph{nonrepetitive chromatic number} of a graph $G$ is the minimum integer $k$ such that $G$ has a nonrepetitive $k$-colouring. Whether planar graphs have bounded nonrepetitive chromatic number is one of the most important open problems in the field. Despite this, the best known upper bound is $O(\sqrt{n})$ for $n$-vertex planar graphs. We prove a $O(\log n)$ upper bound

Nonrepetitive colorings of lexicographic product of graphs

A coloring $c$ of the vertices of a graph $G$ is nonrepetitive if there exists no path $v_1v_2\ldots v_{2l}$ for which $c(v_i)=c(v_{l+i})$ for all $1\le i\le l$. Given graphs $G$ and $H$ with $|V(H)|=k$, the lexicographic product $G[H]$ is the graph obtained by substituting every vertex of $G$ by a copy of $H$, and every edge of $G$ by a copy of $K_{k,k}$. %Our main results are the following. We prove that for a sufficiently long path $P$, a nonrepetitive coloring of $P[K_k]$ needs at least $3k+\lfloor k/2\rfloor$ colors. If $k>2$ then we need exactly $2k+1$ colors to nonrepetitively color $P[E_k]$, where $E_k$ is the empty graph on $k$ vertices. If we further require that every copy of $E_k$ be rainbow-colored and the path $P$ is sufficiently long, then the smallest number of colors needed for $P[E_k]$ is at least $3k+1$ and at most $3k+\lceil k/2\rceil$. Finally, we define fractional nonrepetitive colorings of graphs and consider the connections between this notion and the above results

Online version of the theorem of Thue

A sequence S is nonrepetitive if no two adjacent blocks of S are the same. In 1906 Thue proved that there exist arbitrarily long nonrepetitive sequences over 3 symbols. We consider the online variant of this result in which a nonrepetitive sequence is constructed during a play between two players: Bob is choosing a position in a sequence and Alice is inserting a symbol on that position taken from a fixed set A. The goal of Bob is to force Alice to create a repetition, and if he succeeds, then the game stops. The goal of Alice is naturally to avoid that and thereby to construct a nonrepetitive sequence of any given length. We prove that Alice has a strategy to play arbitrarily long provided the size of the set A is at least 12. This is the online version of the Theorem of Thue. The proof is based on nonrepetitive colorings of outerplanar graphs. On the other hand, one can prove that even over 4 symbols Alice has no chance to play for too long. The minimum size of the set of symbols needed for the online version of Thue's theorem remains unknown