Abstract
A -list assignment of a graph is a mapping which assigns a set of size to each vertex of and . A graph is -choosable if has a proper coloring such that for each -list assignment . In 2011, Charoenpanitseri et al. gave a characterization of -choosability of -vertex graphs when and left open problems when . Recently, Ruksasakchai and Nakprasit obtain the results when . In this paper, we extend the results to case .
1. Introduction
A graph is called -colorable if every vertex of can be labeled by at most colors and every adjacent vertex receives distinct colors. The smallest number such that is -colorable is called the chromatic number of , denoted by . A -list assignment of a graph is a function which assigns a set of size to each vertex of . A -list assignment of a graph is a -list assignment with . Given a list assignment , a proper coloring of is an -coloring of if is chosen from for each vertex of . A graph is -colorable if has an -coloring. Particularly, if is a -list assignment of , then any -coloring of is a -coloring of . A graph is -choosable if is -colorable for every -list assignment . If a graph is -choosable for each positive number then is called -choosable and the smallest number satisfying this property is called the list chromatic number of denoted by .
List coloring is a well-known problem in the field of graph theory. It was first studied by Vizing [1] and by Erdös et al. [2]. They gave a characterization of -choosable graphs. For , there is no characterization of -choosable graphs. There are only results for some classes of graphs. For example, all planar graphs are -choosable, while some planar graphs are -choosable. (See [3–9].) In order to simplify the problem, -choosability is defined. It is a partial problem of -choosability. Instead of proving that a graph can always be colored for entire -list assignments, we prove the graph can be colored for -list assignments that have exactly colors. In 2011, -choosability of graphs was explored in [10]. They proved the following theorem.
Theorem 1 (see [10]). For an -vertex graph , if , then is -choosable.
Moreover, they showed that the bound is best possible by proving if , then an -vertex graph containing is not -choosable. Furthermore, they keep investigating the -choosability to obtain another interesting theorem.
Theorem 2 (see [10]). Let . A -free graph with vertices is -choosable for .
Again, they showed that the bound is best possible for -free graphs with vertices by proving an -vertex graph containing is not -choosable for . In conclusion, they gave a characterization of -choosability of -vertex graphs when .
In 2013, Ruksasakchai and Nakprasit gave a characterization of -choosability of -vertex graphs as shown in the following theorem.
Theorem 3 (see [11]). Let be a graph with vertices and . If does not contain and , then is -choosable for .
Results on -choosability of -vertex graphs are almost completed by [12, 13]. Here, we focus on . We will prove that if an -vertex graph does not contain and , then is -choosable for .
2. Preliminaries
Throughout the paper, denotes a simple, undirected, finite, connected graph; and are the vertex set and the edge set of . For , is the graph obtained from deleting all vertices of from . In case , we write instead of . The subgraph induced by , denoted by , is the graph obtained from deleting all vertices of outside . The notation stands for the degree of in . For a subgraph of , stands for the degree of in .
Let . If is a list assignment of , we let denote restricted to and denote . For a color set , let be the new list assignment obtained from by deleting all colors in from for each . When has only one color , we write instead of .
Example 4. The cycle is -choosable unless is odd and .
Note that a graph is -choosable if and only if is -colorable. Hence, is -choosable if and only if is even. It remains to show that all of the cycles are -choosable for .
Let and be a -list assignment of . Thus there are two adjacent vertices such that . Let be remaining vertices along the cycle , where is adjacent to for . First we assign a color in which is not in and then we assign vertex a color in different from and so on. This algorithm guarantees that each pair of adjacent vertices receives distinct colors.
Theorem 5 is a powerful tool when combined to Lemma 6. Everyone uses both of them to obtain a result on -choosability of graphs.
Theorem 5 (see [14]). Let be a list assignment of a graph and let be a maximal nonempty subset such that . If is -colorable, then is -colorable.
Lemma 6 (see [10]). Let be -sets and . If , then .
The following statements appear in [10, 11]. They use the tools to obtain characterizations of -choosability of -vertex graphs. We also need the tools, as well.
Lemma 7 (see [10]). Let be an -vertex graph. If and is -free, then is -choosable for any positive integer .
Lemma 8 (see [10]). If a -vertex graph is -free, then it is -choosable for .
Lemma 9 (see [11]). Let be an -vertex graph, where . If does not contain and , then is -colorable.
Lemma 10 (see [11]). Let be a graph in Figure 1. If each list has size exactly two lists of size and or , then is -colorable.
Lemma 11 (see [11]). Let be a graph with vertices and . If does not contain and , then and .
Corollary 12 (see [11]). If is a -vertex graph having no and , then is -choosable.
Recently, Ohba’s conjecture is proved by Noel as shown in Theorem 13.
Theorem 13 (see [15]). If , then .
The theorem is powerful because several interesting results can be obtained; for example, Corollaries 14 and 15.
Corollary 14. Let be an -vertex graph with . If is -colorable, then is -choosable.
Proof. Let be an -vertex graph with . Assume that is -colorable. Then . Then we add some edges to to obtain a graph such that . Since , we obtain that by Theorem 13. Hence is -choosable because is a subgraph of .
Corollary 15. Let be an -vertex graph with . If is -colorable, then is -choosable.
Proof. Let be an -vertex graph with . Assume that is -colorable. Then . Then we add some edges to to obtain a graph such that . Since , we obtain that by Theorem 13. Hence is -choosable because is a subgraph of .
Finally, we need one more theorem and one more lemma to prove our main results.
Theorem 16 (see [16]). If is a graph other than odd cycle and complete graph, then .
Lemma 17 (see [11, 17]). Let be a graph with and . If has no and , then must be one of the graphs in Figure 2.
3. Main Results
In this section, the main result is in Theorem 24. In order to prove the main result, we need to prove Theorems 19 and 23. Lemma 18 is established for Theorem 19 while Lemmas 20, 21, and 22 are established for Theorem 23.
Lemma 18. Let be an -vertex graph with . If has no and , then is -colorable.
Proof. Let be an -vertex graph with . Assume that has no and .
If there is a vertex such that , then is -colorable by Lemma 9; hence, is -colorable, as well. Suppose that . Let be a vertex with of and let be a vertex which is not adjacent to . Notice that is or .
Case 1 ( contains ). Under the condition , must be . Then must be a subgraph of the graph shown in Figure 3. According to the figure, and its subgraphs are -colorable.
Case 2 ( contains ). Let , , and .
Case 2.1 (). Notice that has at most edges because has vertices and are not adjacent. Then . However, and . Then . Hence, there are at least edges between and . Since and for , there are at most edges between and .
Hence, there are exactly edges between and . Moreover, we obtain that has exactly edges, , and . Furthermore, each of is adjacent to exactly one vertex in .
If and are adjacent to different vertices in , then must be the left graph in Figure 4. If and are adjacent to the same vertex in , then must be the right graph in Figure 4. According to the figure, is -colorable.
Case 2.2 (). Notice that has at most edges because has vertices and are not adjacent. Then . However, , , and . Then . Hence, there are at least edges between and .
Since does not contain , is not adjacent to at least one vertex from , for example, . If , then the proof is done by Case 2.1 because and are nonadjacent vertices with degree . Suppose that . Since and for and there are at most edges between and .
Hence, there are exactly edges between and . Moreover, we obtain that has exactly edges, and , and . Furthermore, each of is adjacent to exactly one vertex in .
If and are adjacent to the same vertex in , then must be the left graph in Figure 5. If and are adjacent to different vertices in , then must be the middle graph or the right graph in Figure 5. According to the figure, is -colorable.
Case 3 ( contains neither nor ). By Lemma 9, is -colorable. Hence, all vertices from are labeled by colors and are labeled by the fourth color. Therefore, is -colorable.
Theorem 19. If is an -vertex graph having no and , then is -choosable.
Proof. Assume that is an -vertex graph having no and . Then for every vertex , have no and .
Case 1 (). Then is -colorable by Theorem 16. Hence, is -choosable by Corollary 15.
Case 2 (). Then is -colorable by Lemma 18. Hence, is -choosable by Corollary 15.
Case 3 (). Let be a vertex with and is the vertex which is not adjacent to . Since and , and have a common color, for example, . Hence, we label and by color .
By Lemma 9, is -colorable because it has no and . By Corollary 14, is -choosable. That is, we can label the remaining vertices.
Case 4 (). Let be a vertex with .
Case 4.1 (). Then we can label in order to obtain that each of the remaining vertices still has available colors. Since has no and , it is -colorable by Lemma 9. Then, it is -choosable by Corollary 15. That is, we can label the remaining vertices.
Case 4.2 (). Then we label in order to obtain that each of the remaining seven vertices has at least available colors and the number of the remaining colors is . According to Corollary 12, is -choosable. That is, we can label the remaining vertices.
Lemma 20. is -colorable if and each list has size except for one list of size .
Proof. Let , , and be partite sets of . Let be a list assignment of such that and each list has size except for one list of size . Without loss of generality, suppose that and for the remaining vertices. Since , there is a color . Then we label and by color . Hence, each of the remaining vertices has available colors. Since is -choosable, the remaining vertices can be labeled.
Lemma 21. is -choosable.
Proof. Let be a -list assignment of . Since , we obtain that or . Without loss of generality, suppose that .
Since but , there is a color . Then and are labeled by color . Let be the list assignment of . Since for and , is -colorable by Example 4.
Lemma 22. is -choosable.
Proof. Notice that is not adjacent to if and only if mod 7. Let be a -list assignment of .
Case 1. There is a color that appears in exactly list. Without loss of generality, suppose but for the remaining vertices . Then we label by color . Each of the remaining six vertices has available colors. Since does not contain or , it is -colorable by Lemma 9. Hence, is -choosable by Corollary 14. That is, the remaining vertices can be labeled.
Case 2. There is a color that appears in exactly lists. Without loss of generality, we may prove only subcases because is not adjacent to if and only if mod 7.
Case 2.1 (). Then we label and by color . Hence, is -colorable by Lemma 9. Consequently, it is -choosable by Corollary 14.
Case 2.2 (). Then we label by color . Lemma 20 confirms that the remaining vertices can be labeled.
Case 2.3 (). Then we label by color . Again, Lemma 20 confirms that the remaining six vertices can be labeled.
Case 3. There is a color that appears in exactly , , or lists. According to the fact that is not adjacent to if and only if mod 7, we may prove only the following cases.
If appears in exactly lists, we suppose that or or or . Then we label and by color and Lemma 10 confirms that the remaining five vertices can be labeled.
If appears in exactly lists, we suppose that or or . Again, we label and by color and Lemma 10 confirms that the remaining five vertices can be labeled.
If appears in exactly lists, we suppose that . Similarly, we label and by color and Lemma 10 confirms that the remaining five vertices can be labeled.
Case 4. Each color appears in exactly or lists. Let and be the number that appears in exactly or lists, respectively. Then and . There are no integer solutions for this system. Hence, this case is impossible.
Theorem 23. If is a -vertex graph having no and , then is -choosable.
Proof. Assume that is a -vertex graph having no and . Let be a -list assignment of . According to Lemma 9, is -colorable. Then . If , we apply Corollary 14. Hence, is -choosable. Otherwise, suppose that . By Lemma 11, we obtain that and . By Lemma 17, must be one of the seven graphs in Figure 2. By Lemma 21, , , , and are -choosable. By Lemma 22, , , and are -choosable.
Theorem 24. Let . If an -vertex graph does not contain and , then is -choosable for .
Proof. Let be a graph with vertices and . Assume that does not contain and .
Let be a -list assignment such that . To apply Theorem 5, let such that . We need to prove that is -colorable.
By Lemma 6, . Then . It follows that .
Case 1 (). Then is -colorable by Lemma 7.
Case 2 (). Then is -colorable by Lemmas 8 and 9.
Case 3 (). Then . Hence, or .
Case 3.1 (). Then . Consequently, . Since , we obtain that . If , then is -colorable by Corollary 12. If , then is -colorable by Theorem 23.
Case 3.2 (). Then . Consequently, . Since , we obtain that . By Theorem 19, is -colorable.
4. Open Problem
Thanks to [10, 11], a characterization of -choosability of -vertex graphs when and is obtained. According to [13], a -choosability of -vertex graphs when is obtained. In this paper, we give a characterization of -choosability of -vertex graphs when and .
Conjecture 25. Let . If an -vertex graph has no and , then it is -choosable of .
Conflict of Interests
The author declares that there is no conflict of interests regarding the publication of this paper.
Acknowledgment
The author would like to thank Rangsit University for all support.