347 research outputs found
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
Pathwidth and nonrepetitive list coloring
A vertex coloring of a graph is nonrepetitive if there is no path in the
graph whose first half receives the same sequence of colors as the second half.
While every tree can be nonrepetitively colored with a bounded number of colors
(4 colors is enough), Fiorenzi, Ochem, Ossona de Mendez, and Zhu recently
showed that this does not extend to the list version of the problem, that is,
for every there is a tree that is not nonrepetitively
-choosable. In this paper we prove the following positive result, which
complements the result of Fiorenzi et al.: There exists a function such
that every tree of pathwidth is nonrepetitively -choosable. We also
show that such a property is specific to trees by constructing a family of
pathwidth-2 graphs that are not nonrepetitively -choosable for any fixed
.Comment: v2: Minor changes made following helpful comments by the referee
Nonrepetitive colorings of lexicographic product of graphs
A coloring of the vertices of a graph is nonrepetitive if there
exists no path for which for all
. Given graphs and with , the lexicographic
product is the graph obtained by substituting every vertex of by a
copy of , and every edge of by a copy of . %Our main results
are the following. We prove that for a sufficiently long path , a
nonrepetitive coloring of needs at least
colors. If then we need exactly colors to nonrepetitively color
, where is the empty graph on vertices. If we further require
that every copy of be rainbow-colored and the path is sufficiently
long, then the smallest number of colors needed for is at least
and at most . Finally, we define fractional nonrepetitive
colorings of graphs and consider the connections between this notion and the
above results
Vertex coloring of plane graphs with nonrepetitive boundary paths
A sequence is a repetition. A sequence
is nonrepetitive, if no subsequence of consecutive terms of form a
repetition. Let be a vertex colored graph. A path of is nonrepetitive,
if the sequence of colors on its vertices is nonrepetitive. If is a plane
graph, then a facial nonrepetitive vertex coloring of is a vertex coloring
such that any facial path is nonrepetitive. Let denote the minimum
number of colors of a facial nonrepetitive vertex coloring of . Jendro\vl
and Harant posed a conjecture that can be bounded from above by a
constant. We prove that for any plane graph
Anagram-free Graph Colouring
An anagram is a word of the form where is a non-empty word and
is a permutation of . We study anagram-free graph colouring and give bounds
on the chromatic number. Alon et al. (2002) asked whether anagram-free
chromatic number is bounded by a function of the maximum degree. We answer this
question in the negative by constructing graphs with maximum degree 3 and
unbounded anagram-free chromatic number. We also prove upper and lower bounds
on the anagram-free chromatic number of trees in terms of their radius and
pathwidth. Finally, we explore extensions to edge colouring and
-anagram-free colouring.Comment: Version 2: Changed 'abelian square' to 'anagram' for consistency with
'Anagram-free colourings of graphs' by Kam\v{c}ev, {\L}uczak, and Sudakov.
Minor changes based on referee feedbac
Nonrepetitive Colourings of Planar Graphs with 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 is
the minimum integer such that has a nonrepetitive -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 for -vertex planar graphs. We prove a
upper bound
A new approach to nonrepetitive sequences
A sequence is nonrepetitive if it does not contain two adjacent identical
blocks. The remarkable construction of Thue asserts that 3 symbols are enough
to build an arbitrarily long nonrepetitive sequence. It is still not settled
whether the following extension holds: for every sequence of 3-element sets
there exists a nonrepetitive sequence with
. Applying the probabilistic method one can prove that this is true
for sufficiently large sets . We present an elementary proof that sets of
size 4 suffice (confirming the best known bound). The argument is a simple
counting with Catalan numbers involved. Our approach is inspired by a new
algorithmic proof of the Lov\'{a}sz Local Lemma due to Moser and Tardos and its
interpretations by Fortnow and Tao. The presented method has further
applications to nonrepetitive games and nonrepetitive colorings of graphs.Comment: 5 pages, no figures.arXiv admin note: substantial text overlap with
arXiv:1103.381
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