Chromatic Coloring with a Maximum Color Class

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. Definitions and Notations • 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.

5. Definitions and Notations • 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.

6. Definitions and Notations • 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.

7. Definitions and Notations • 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 amaximum independent set in G, then we say that the color class S is maximum.

8. Problem

9. Problem • 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 coloringin which some color class is maximumif 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 Gin which some color class is maximum.

17. Preliminaries

18. Preliminaries • 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)  1if and only ifeach 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 Gi of GS such that (Gi)  (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) (Gi). Then, by (Gi)  (G)  (G)  (Gi) andBrooks’ Theorem, either Gi is an odd cycle with (Gi)  (G)  3, or Gi is a complete graph with |V(Gi)|  (Gi)  (G)  3. This is a contradiction.

21. Definitions and Notations • An odd path-component oran odd cycle-component of a graph G is a component of G isomorphic to an odd path or an odd cycle. AKn-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.

22. Definitions and Notations Given a nonempty proper subset SofV(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 NG(S).

23. Preliminaries • Lemma 2. Let G be a connected graph with χ(G) (G)  3. The there exists a maximum independent setS 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 someodd cycle-component of GS.Consider any vertex v1 in C. Then there must exist exactly a vertex w1 of S adjacent to v1 in G and (GS) (G)  1 2. Now, let Pv1w1  v2w2 be a maximal(S, S)-chain from v1 in G.Thenvi Sandwi  Sfor alli 1.

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

27. w2 w1 wn S P v1 GS u2 v2 un vn un+1 vn+1 (1) If r does not exist, then

28. w2 w1 wn S P v1 GS u2 v2 un vn un+1 vn+1 (1) If r does not exist, then

29. w2 w1 wr1 wr S P v1 GS vr+1 u2 v2 ur1 vr1 ur vr (2) If r exists, then

30. w2 w1 wr1 wr S P v1 GS vr+1 u2 v2 ur1 vr1 ur vr (2) If r exists, then

31. wr v1 GS u2 v2 ur1 vr1 ur vr (2) If r exists, then

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 G1, G2, …, Gt, where t  1. It suffices to claim that there exists a maximum independent set Si in each component Gi such that χ(GiSi)  χ(G)  1. First, if χ(Gi)  χ(G)  1, then any maximum independent set Si in Gi has the property that χ(GiSi)  χ(Gi)  χ(G)  1. Next, if χ(Gi)  χ(G)  (Gi), then Gi is an odd cycleor a complete graph. Moreover, if χ(Gi)  χ(G)  (Gi)  2, then Gi is a path or an even cycle. In each of these two cases, it is not difficult to find a maximum independent set Si in Gi such that χ(GiSi)  χ(Gi)  1 χ(G)  1. Finally, if χ(Gi)  χ(G)  (Gi)  3, then there exists a maximum independent set Si in Gi such that χ(GiSi)  χ(Gi)  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 Gin which some color class is maximum.

36. Definitions and Notations • Let χmax(G)denotethe least ksuch thata graph G has a proper k-coloring in which some color class is maximum.

37. Main Results • Proposition 6. χ(G)  χmax(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 haveχ(GS)  χ(G). Then it is easy to obtain that χmax(G)  χ(GS) + 1  χ(G) + 1 byadding the additional color class S to a chromatic coloring of GS. Besides, it is trivial thatχmax(G)  χ(G).

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

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

41. Proof. By Brooks’ Theorem, we have χ(G) (G). If χ(G) (G)  1, then χmax(G)  χ(G) + 1  ((G)  1) + 1 (G). If χ(G) (G), then χmax(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 K3,3; 2.  (G K3,3)  |V(G K3,3) | / 3.

43. Thank you for your attention!