Thue choice number of paths Xuding Zhu Department of Applied Mathematics National Sun Yat-sen University Joint work with Jaroslaw Grytczuk and Jakub Przybylo

Given an infinite sequence of symbols We say S has a repetition if two adjacent blocks are identical. … 11 … … 1212 … … 123123 … … 120102120102 … Is there an infinite sequence of symbols with no repetitions? Trivially Yes. Is there an infinite sequence consisting of finite many symbols which has no repetitions? 3 Yes, not so trivial.

There is no infinite sequence consisting of 2 symbols which has no repetitions.

There is no infinite sequence consisting of 2 symbols which has no repetitions. 00

There is no infinite sequence consisting of 2 symbols which has no repetitions. 01

There is no infinite sequence consisting of 2 symbols which has no repetitions. 010

There is no infinite sequence consisting of 2 symbols which has no repetitions. 0101

Is there an infinite sequence consisting of 3 symbols which has no repetitions? Thue Theorem [1906]: Yes, there is an infinite sequence consisting of 3 symbols which has no repetitions.

Thue-Morse sequence 011010011001011010010110011... 0110100110010110100101100110... 0110100110010110100101100110... 0  01  0110  01101001  0110100110010110 ... Axel Thue The odd entries form the original. Each even entry is the complement of its preceding entry. Marston Morse

Nonrepetitive sequences 01101001100101101001011001101001... No two overlapping blocks are identical. Proof: Choose the shortest overlapping identical blocks. 01101001100101101001011001101001... 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 ...

Nonrepetitive sequences 01101001100101101001011001101001... 1 1

1 1

10 1

10 10

10 101

101 101

1010 101

1010 1010

1010 10101

10101 10101

The starting point of symbolic dynamics, or combinatorial words. 011010-0110-01011010-010110-011010-0... 011010-0110-01011010-010110-011010-0... 0 20 1 0020 0010 20 1 0 10 20 0 2 1000... (Thue 1906) There exist infinite sequences over 3 symbols without identical adjacent blocks.

There are many generalizations of Thue Theorem One direction of generalization is to find ``more restrictive’’ or ``less restrictive” infinite sequences.

Evdokimov (1968) – 25 colors Pleasants (1970) – 5 colors (Erdös, 1961) Is there a 4-coloring of the integers such that no two adjacent segments have the same multisets of colors? Evdokimov (1968) – 25 colors Pleasants (1970) – 5 colors Keränen (1992) – 4 colors Paul Erdös

(Lasoń, Michałek, 2007) There is a measurable 4-coloring of the real line such that no two adjacent intervals contain the same measure of every color. (Lasoń, Michałek, 2007) There is a measurable 5-coloring of the real line such that no two segments contain the same measure of every color.

(Alon, West, 1981) Every k-colored interval has a splitting of size at most k. splitting of size 3

Every k-colored necklace has a splitting of size at most k. (Alon, Grytczuk, Lasoń, Michałek, 2008) For every k > 0 there is a measurable (k+3)-coloring of the real line such that no interval has a splitting of size at most k. (Goldberg, West, 1978) Every k-colored necklace has a splitting of size at most k.

Another direction of generalization is to view the non-repetitive infinite sequence as a coloring of the infinite path. Then we can consider non-repetitive coloring of an arbitrary graph.

Such a sequence is called a repetition A k-non-repetitive-colouring of a graph G colours its vertices with k colours, so that the colour sequence of any path P in G is not a repetition. (G) = the Thue chromatic number of a graph G = min k for which G has a k-non-repetitive-colouring

Question: Which classes of graphs have bounded ?

Every graph of maximum degree at most  satisfies (G)  162. (Alon, Grytczuk, Hałuszczak, Riordan, 2001) Every graph of maximum degree at most  satisfies (G)  162. (Alon, Grytczuk, Hałuszczak, Riordan, 2001) There is a constant c > 0 such that for each  there is a graph G with maximum degree  satisfying (G)  c2/ln . (Kündgen, Pelsmajer, 2003) Every graph G of treewidth at most t satisfies (G)  4t.

The strong product of t paths satisfies (G)  4t. (Kündgen, Pelsmajer, 2003) The strong product of t paths satisfies (G)  4t. (Kündgen, Pelsmajer, 2003) Every outerplanar graph G satisfies (G)  12. (Barat,Wood, 2008) Every graph G has a subdivision S satisfying (G)  4. (Pezarski, Zmarz, 2008) Every graph G has a subdivision S satisfying (G)  3.

There are basically two methods for proving upper bounds for

Method 1: Use modification of Thue sequence. Example: every tree T has Thue chromatic number at most 4.

2 1 2 1 2

Such a path cannot be a repetition 2 1 2 1 2

However, this is a repetition 2 1 2 1 2

y x x

Such a repetition exists, because in the Thue colouring of the path, there exists palindrome. y x x x y Theorem: Every tree has Thue chromatic number at most 4 x 3 3 3 3 3 3 3 0 20 1 0020 0010 20 1 0 10 20 0 2 1000... The new sequence has no repetition and no palindrome.

Method 2: Probabilistic method Randomly pick a colour for each vertex. Prove with positive probability, the resulting colouring is nonrepetitive. Result proved by this method applies to Thue choice number.

A k-assignment L of a graph G assigns to each vertex v A set L(v) of k permissible colours. Given a k-assignment L of G, a non-repetitive L-colouring of G is a non-repetitive colouring c of G such that for each vertex v, c(v)  L(v) G is non-repetitive k-choosable For every k-assignment L of G, there is a non-repetitive L-colouring

Proof sketch: For each i, randomly pick a symbol from Li Let T be the event that the resulting sequence is non-repetitive. If there is no non-repetitive L-sequence, then Pr[T]=0 We shall prove Pr[T] > 0. This implies that there exists a non-repetitive L-sequence.

Suppose x1,x2,…,x(i-1) have been chosen. Choose xi uniform randomly from Li subject to the following: 1: xi ≠ x(i-1). 

Suppose x1,x2,…,x(i-1) have been chosen. Choose xi uniform randomly from Li subject to the following: 1: xi ≠ x(i-1). 2: If x(i-3) = x(i-1), then xi ≠ x(i-2). 

Suppose x1,x2,…,x(i-1) have been chosen. Choose xi uniform randomly from Li subject to the following: 1: xi ≠ x(i-1). 2: If x(i-3) = x(i-1), then xi ≠ x(i-2). 3: If x(i-5) = x(i-2),x(i-4)=x(i-1) then xi ≠ x(i-3). 

Suppose x1,x2,…,x(i-1) have been chosen. Choose xi uniform randomly from Li subject to the following: 1: xi ≠ x(i-1). 2: If x(i-3) = x(i-1), then xi ≠ x(i-2). 3: If x(i-5) = x(i-2),x(i-4)=x(i-1) then xi ≠ x(i-3). Conditions of (2) and (3) cannot be satisfied at the same time. When choosing xi, there are at least 2 choices, with each choice chosen with the same probability

x[i,i+s] =(xi, x(i+1), …, x(i+s)) Lemma: Pr[ none of A_{i,s} happens ] > 0.

Lovasz Local Lemma : a set of events G: a digraph with V=  For A  , (A) is the set of out-neighbors of A. A is mutually independent to event in {B: A B} By our random process, any two events are dependent !

For A  , (A) is the set of out-neighbors of A. Lovasz Local Lemma (one sided [Pegden])  : a set of events G: a digraph with V=  For A  , (A) is the set of out-neighbors of A. ≤: a quasi-order on  such that A is mutually independent to event in {B: A B} B  (A), C   (A) implies C ≤ B or C > A. By our random process, any two events are dependent !

For A  , (A) is the set of out-neighbors of A. Lovasz Local Lemma  : a set of events G: a digraph with V=  For A  , (A) is the set of out-neighbors of A. ≤: a quasi-order on  such that B  (A), C   (A) implies C ≤ B or C > A . ≤

Whatever happened here

ti = 0 if Rule 2 or Rule 3 applies to xi, and ti = 1 otherwise

Lemma Assume r  4. With probability 2/3, at least of the xi’s in x[i+s+1, i+2s] has at least 3 choices, and the xi’s has at least 2 choices. With probability 1/3, at least of the xi’s in x[i+s+1, i+2s] has at least 3 choices, and the xi’s has at least 2 choices.

Theorem: If P is a path on at least 4 vertices, then Question:

Question: Is there a constant C, such that every planar graph G satisfies ? Question: Is there a constant C, such that every tree T satisfies

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