1. Chromatic Coloring with a Maximum Color Class Bor-Liang Chen Kuo-Ching Huang Chih-Hung Yen* 30 July, 2009

2.  Throughout this talk, all graphs considered are finite, undirected, loopless and without multiple edges.

3. Definitions and Notations

4.  A proper k-coloring of a graph G is a labeling f : V(G)  {1, 2,..., k} such that adjacent vertices have different labels. The labels are colors; the vertices of one color form a color class. Definitions and Notations

5.  The chromatic number of a graph G, written χ(G), is the least k such that G has a proper k-coloring.  A chromatic coloring of a graph G is a proper coloring of G using χ(G) colors. Definitions and Notations

6.  An independent set in a graph is a set of pairwise nonadjacent vertices.  The independence number of a graph G, written  (G), is the maximum size of an independent set in G. Definitions and Notations

7.  An independent set in a graph G is maximum if it has size  (G).  Any color class S in a proper coloring of a graph G is obviously an independent set. If S is also a maximum independent set in G, then we say that the color class S is maximum. Definitions and Notations

8. Problem

9.  We would like to know whether there exists a chromatic coloring of a graph G in which some color class is maximum.

10. Problem  We would like to know whether there exists a chromatic coloring of a graph G in which some color class is maximum. However, this cannot be guaranteed if χ(G)   (G).

11.  A graph G has a chromatic coloring in which some color class is maximum if and only if there exists a maximum independent set S in G such that  (G  S) =  (G)  1.

12. Example χ(G) = 2 < 3 =  (G) χ(G  S) = 2

13. Example χ(G) = 3 < 4 =  (G) χ(G  S) = 3

14. Example χ(G) = 4 < 5 =  (G) χ(G  S) = 4

15. Example χ(G) = 4 < 5 =  (G) χ(G  S) = 4

16. Main Results  Theorem. Let G be a graph with χ(G)   (G). Then there exists a chromatic coloring of G in which some color class is maximum.

17. Preliminaries

18.  Brooks ’ Theorem. (1941) If G is a connected graph other than an odd cycle or a complete graph, then χ(G)   (G).

19. Preliminaries  Lemma 1. Let G be a graph with χ(G)   (G), and also let S be a maximum independent set in G. Then  (G  S)   (G)  1 if and only if each component of G  S is not an odd cycle when  (G)  3 or a complete graph of order  (G) when  (G)  3.

20. Proof. (  ) It is trivial. (  ) Suppose that  (G  S)   (G)  1. Then  (G  S)   (G). Hence, there must exist one component G i of G  S such that  (G i )   (G). Since S is a maximum independent set in G, each vertex of V(G)  S in G must be adjacent to some vertex of S, and  (G)   (G  S)   (G i ). Then, by  (G i )   (G)   (G)   (G i ) and Brooks ’ Theorem, either G i is an odd cycle with  (G i )   (G)  3, or G i is a complete graph with |V(G i )|   (G i )   (G)  3. This is a contradiction.

21.  An odd path-component or an odd cycle-component of a graph G is a component of G isomorphic to an odd path or an odd cycle. A K n -component of G is a component of G isomorphic to a complete graph of order n.  If a path P in G is from vertex u to vertex v, then u and v are the endpoints of P. Definitions and Notations

22. Given a nonempty proper subset S of V(G).  A (S, S)-chain in G is a path that alternates between vertices in S and vertices in S, where S denotes V(G)  S.  The set consisting of the neighbors of vertices of S in G is denoted by N G (S). Definitions and Notations

23. Preliminaries  Lemma 2. Let G be a connected graph with χ(G)   (G)  3. The there exists a maximum independent set S in G such that  (G  S)  2  χ(G)  1.

24. Proof. By Lemma 1, it suffices to show that there exists a maximum independent set S in G such that G  S contains no odd cycle-components. Hence, among all maximum independent sets in G, we let S be one satisfying that G  S contains the least number of odd cycle-components, and denote such a number by t. We claim that t  0.

25. Proof. (continued) Suppose otherwise. Then t  1, and we use C to denote some odd cycle-component of G  S. Consider any vertex v 1 in C. Then there must exist exactly a vertex w 1 of S adjacent to v 1 in G and  (G  S)   (G)  1  2. Now, let P  v 1  w 1  v 2  w 2   be a maximal (S, S)- chain from v 1 in G. Then v i  S and w i  S for all i  1.

26. Proof. (continued) Furthermore, let r denote the least i such that (1) w i has less than 3 neighbors in G or (2) the two neighbors of w i other than v i in G are not exactly the two endpoints of some odd path- component of G  S.

27. (1) If r does not exist, then S G  SG  S v1v1 u2u2 v2v2 unun vnvn u n+1 v n+1 w1w1 w2w2 wnwn P

28. (1) If r does not exist, then S G  SG  S v1v1 u2u2 v2v2 unun vnvn u n+1 v n+1 w1w1 w2w2 wnwn P

29. (2) If r exists, then S G  SG  S v1v1 u2u2 v2v2 urur vrvr ur1ur1 vr1vr1 w1w1 w2w2 wr1wr1 wrwr vr+1vr+1 P

30. S G  SG  S v1v1 u2u2 v2v2 urur vrvr ur1ur1 vr1vr1 w1w1 w2w2 wr1wr1 wrwr vr+1vr+1 P

31. G  SG  S v1v1 u2u2 v2v2 urur vrvr ur1ur1 vr1vr1 wrwr

32. Preliminaries  Lemma 3. Let G be a connected graph with χ(G)   (G)  4. The there exists a maximum independent set S in G such that  (G  S)   (G)  1.

33. Main Results  Theorem 4. Let G be a graph with χ(G)   (G). Then there exists a maximum independent set S in G such that χ(G  S)  χ(G)  1.

34. Proof. Suppose that G consists of the components G 1, G 2, …, G t, where t  1. It suffices to claim that there exists a maximum independent set S i in each component G i such that χ(G i  S i )  χ(G)  1. First, if χ(G i )  χ(G)  1, then any maximum independent set S i in G i has the property that χ(G i  S i )  χ(G i )  χ(G)  1. Next, if χ(G i )  χ(G)   (G i ), then G i is an odd cycle or a complete graph. Moreover, if χ(G i )  χ(G)   (G i )  2, then G i is a path or an even cycle. In each of these two cases, it is not difficult to find a maximum independent set S i in G i such that χ(G i  S i )  χ(G i )  1  χ(G)  1. Finally, if χ(G i )  χ(G)   (G i )  3, then there exists a maximum independent set S i in G i such that χ(G i  S i )  χ(G i )  1  χ(G)  1 by Lemmas 2 and 3.

35. Main Results  Corollary 5. Let G be a graph with χ(G)   (G). Then there exists a chromatic coloring of G in which some color class is maximum.

36.  Let χ max (G) denote the least k such that a graph G has a proper k-coloring in which some color class is maximum. Definitions and Notations

37. Main Results  Proposition 6. χ(G)  χ max (G)  χ(G) + 1 for any graph G.

38. Proof. Let S be a maximum independent set in G. Since G  S is a subgraph of G, we have χ(G  S)  χ(G). Then it is easy to obtain that χ max (G)  χ(G  S) + 1  χ(G) + 1 by adding the additional color class S to a chromatic coloring of G  S. Besides, it is trivial that χ max (G)  χ(G).

39. Main Results  Corollary 7. Let G be a graph with χ(G)   (G). Then χ max (G)  χ(G).

40. Main Results  Corollary 8. If G is a connected graph other than an odd cycle or a complete graph, then χ max (G)   (G).

41. Proof. By Brooks ’ Theorem, we have χ(G)   (G). If χ(G)   (G)  1, then χ max (G)  χ(G) + 1  (  (G)  1) + 1   (G). If χ(G)   (G), then χ max (G)  χ(G)   (G) by Corollary 7. Hence, the assertion holds.

42. Main Results  Theorem 9. A graph G with  (G)  3  χ(G) is equitably 3-colorable if and only if one of the following statements holds: 1. no components of G or at least two components of G are K 3,3 ; 2.  (G  K 3,3 )  | V (G  K 3,3 ) | / 3.

43. Thank you for your attention!