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

2. Given an infinite sequence of symbols
We say S has a repetition if two adjacent blocks are identical.
… 11 …
… 1212 …
… …
… …
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.

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

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

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

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

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

8. 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.

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

10. Nonrepetitive sequences
No two overlapping blocks are identical.
Proof:
Choose the shortest overlapping identical blocks.

11. Nonrepetitive sequences 1 1

12. 1 1

13. 10 1

14. 10 10

15. 10
101

16. 101
101

17. 1010
101

18. 1010
1010

19. 1010
10101

20. 10101
10101

21. The starting point of symbolic dynamics, or combinatorial words.
(Thue 1906) There exist infinite sequences over 3 symbols without identical adjacent blocks.

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

23. 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

24. (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.

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

26. 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.

27. 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.

28. 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

29. Question: Which classes of graphs have bounded ?

30. 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.

31. 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.

32. There are basically two methods for proving upper bounds for

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

34. 2 1 2 1 2

35. Such a path cannot be a repetition2 1 2 1 2

36. However, this is a repetition2 1 2 1 2

37. y x x

38. 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
The new sequence has no repetition and no palindrome.

39. 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.

40. 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

42. 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.

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

44. 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). 

45. 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). 

46. 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

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

48. 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 !

49. 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 !

50. 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 . ≤

53. Whatever happened here

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

55. 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.

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

57. 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

58. 謝謝