1. The Vertex Arboricity of Integer Distance Graph with a Special Distance Set Juan Liu* and Qinglin Yu Center for Combinatorics, LPMC Nankai University, Tianjin 300071, P. R. China

2. Outline Definitions and Notations Background and Known Results Main Theorem

3. Definitions and Notations Vertex arboricity Given a graph G, a k-coloring of G is a mapping from V(G) to [1, k]. denotes the set of all vertices of G colored with i, and denotes the subgraph induced by in G.

4. Chromatic Number VS Vertex Arboricity a proper k-coloring: each is an independent set. chromatic number = min {k|G has a proper k-coloring} a tree k-coloring: each induces a forest. vertex arboricity va(G) va(G) = min {k|G has a tree k-coloring }

5. Vertex Arboricity Vertex arboricity is the minimum number of subsets into which V(G) can be partitioned so that each subset induces an acyclic subgraph of G. Clearly, for any graph G.

6. Examples

7. Known results for va(G) (Kronk & Mitchem, 1975) For any graph G, (Catlin & Lai, 1995) If G is neither a cycle nor a clique, then

8. Known results for va(G) (Skrekovski, 1975) For a locally planar graph G, ; For a triangle-free locally planar graph G,. (Jorgensen, 2001) Every graph without a -minor has vertex arboricity at most 4.

9. Definitions and Notations Distance graph If and, then the distance graph G(S, D) is defined by the graph with vertex set S and two vertices x and y are adjacent if and only if where the set D is called the distance set.

10. Definitions and Notations Integer distance graph if and all elements of D are positive integers, then the graph G(Z, D)=G(D) is called the integer distance graph and the set D is called its integer distance set.

11. Examples of Integer Distance Graph D={2} - 6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 D={1, 3} - 6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

12. Background The distance graph was introduced by Eggleton et al in 1985. Coloring problems on distance graphs are motivated by the famous Hadwiger-Nelson coloring problem on the unit distance plane.

13. Known results Chromatic number of integer distance graph; Vertex arboricity of integer distance graph.

14. Results on (Eggleton, Erdos & Skilton, 1984) where D is an interval between 1 and for.

15. Results on (Eggleton, Erdos & Skilton, 1985) If a and b are relatively prime positive integers of opposite parity, then. (Eggleton, Erdos & Skilton, 1986) where P is the set of primes.

16. Results on (Chen, Chang & Huang, 1994) If D={a,b,a+b}, where and gcd{a,b}=1 then

17. Integer Distance Graph m=4, k=2, D={1, 3, 4} -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

18. Results on graph (Chang, Liu & Zhu, 1999) Let then

19. Vertex arboricity of (Yu, Zuo & Wu) For any integer

20. Vertex arboricity of (Yu, Zuo & Wu) Let with for a positive integer, we have

21. Vertex arboricity of (Yu, Zuo & Wu) For any with, we have

22. Main theorem Theorem for Proof (1) Upper bound (2) Lower bound

23. (1) Upper bound 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 8(l+2) 8(l+2)+6 8(l+2)+14

24. (2) Lower bound Lemma (Shifting Lemma) Let, be subgraphs of G(D) induced by vertices and vertices for any respectively. Then has a tree n- coloring if and only if has a tree n- coloring.

25. (2) Lower bound By contradiction. Assume, on the contrary, that then has a tree (2l+3)-coloring f.

26. (2) Lower bound Find a finite subgraph H and try to get a contradiction. Question: How to find such a subgraph H? How to get a contradiction?

27. How to find such a subgraph H ? By hypothesis, f is also a tree (2l+3)- coloring of H. ● We consider a subgraph H induced by the vertex subset [0, m+5].

28. How to get a contradiction in H ? Note that |V(H)|=m+6. There exist at least five vertices in H, say, which are colored by the same color. Question: Can we prove that any color except just colors four vertices? Answer: Yes!

29. How to prove? There are at most five vertices receiving the color in H. There isn't any other color, except, coloring five vertices in H.

30. There are at most five vertices receiving the color in H. By contradiction. Suppose the color colors six vertices in H. Need only to consider the case when because we can shift the interval to [-1, m+4] when, and get a contradiction similarly in H.

31. There isn't any other color, except coloring five vertices in H. By contradiction. Note that m=8l+7, use the divisibility of m.

32. Get a contradiction in H Main idea: Find a vertex whose coloring will result in a cycle in some color set. Get information about the location of the five vertices as much as possible. Make cases needed to consider as few as possible. Use vertices as few as possible.

33. Thank you!