127 research outputs found
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
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 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
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
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
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
On the facial Thue choice index via entropy compression
A sequence is nonrepetitive if it contains no identical consecutive
subsequences. An edge colouring of a path is nonrepetitive if the sequence of
colours of its consecutive edges is nonrepetitive. By the celebrated
construction of Thue, it is possible to generate nonrepetitive edge colourings
for arbitrarily long paths using only three colours. A recent generalization of
this concept implies that we may obtain such colourings even if we are forced
to choose edge colours from any sequence of lists of size 4 (while sufficiency
of lists of size 3 remains an open problem). As an extension of these basic
ideas, Havet, Jendrol', Sot\'ak and \v{S}krabul'\'akov\'a proved that for each
plane graph, 8 colours are sufficient to provide an edge colouring so that
every facial path is nonrepetitively coloured. In this paper we prove that the
same is possible from lists, provided that these have size at least 12. We thus
improve the previous bound of 291 (proved by means of the Lov\'asz Local
Lemma). Our approach is based on the Moser-Tardos entropy-compression method
and its recent extensions by Grytczuk, Kozik and Micek, and by Dujmovi\'c,
Joret, Kozik and Wood
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