On ( k , kn − k 2 − 2 k − 1 )-Choosability of n-Vertex Graphs

A graph G is called k-colorable if every vertex of G can be labeled by at most k colors and every adjacent vertex receives distinct colors. The smallest number t such that G is t-colorable is called the chromatic number of G, denoted by χ(G). A k-list assignment L of a graph G is a function which assigns a set of size k to each vertex V of G. A (k, t)-list assignment of a graph is a k-list assignment with | ⋃V∈V(G) L(V)| = t. Given a list assignment L, a proper coloring f of G is an L-coloring of G if f(V) is chosen from L(V) for each vertex V of G. A graph G is L-colorable if G has an L-coloring. Particularly, if L is a (k, k)-list assignment of G, then any L-coloring of G is a k-coloring of G. A graph G is (k, t)-choosable if G is L-colorable for every (k, t)-list assignment L. If a graphG is (k, t)-choosable for each positive number t then G is called k-choosable and the smallest number k satisfying this property is called the list chromatic number of G denoted by ch(G). 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 2-choosable graphs. For k ≥ 3, there is no characterization of k-choosable graphs. There are only results for some classes of graphs. For example, all planar graphs are 5-choosable, while some planar graphs are 3-choosable. (See [3–9].) In order to simplify the problem, (k, t)-choosability is defined. It is a partial problem of kchoosability. Instead of proving that a graph can always be colored for entire k-list assignments, we prove the graph can be colored for k-list assignments that have exactly t colors. In 2011, (k, t)-choosability of graphs was explored in [10]. They proved the following theorem.


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 V of .A (, )-list assignment of a graph is a -list assignment with | ⋃ V∈() (V)| = .Given a list assignment , a proper coloring  of  is an -coloring of  if (V) is chosen from (V) for each vertex V 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 ch().
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 2-choosable graphs.For  ≥ 3, there is no characterization of -choosable graphs.There are only results for some classes of graphs.For example, all planar graphs are 5-choosable, while some planar graphs are 3-choosable.(See [3][4][5][6][7][8][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.
Moreover, they showed that the bound is best possible by proving if  ≤ − 2 , then an -vertex graph containing  +1 is not (, )-choosable.Furthermore, they keep investigating the (, )-choosability to obtain another interesting theorem.

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  = {V}, we write  − V instead of  − {V}.The subgraph induced by , denoted by [], is the graph obtained from deleting all vertices of () outside .The notation (V) stands for the degree of V in .For a subgraph  of ,   (V) stands for the degree of V in .
Let Let  ≥ 3 and  be a (2, )-list assignment of   .Thus there are two adjacent vertices V 1 , V  ∈ () such that (V 1 ) ̸ = (V  ).Let V 2 , V 3 , . . ., V −1 be remaining vertices along the cycle   , where V  is adjacent to V +1 for  = 1, 2, . . .,  − 1.First we assign V 1 a color  in (V 1 ) which is not in (V  ) and then we assign vertex V 2 a color in (V 2 ) 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.
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.
Recently, Ohba's conjecture is proved by Noel as shown in Theorem 13.
The theorem is powerful because several interesting results can be obtained; for example, Corollaries 14 and 15.
Finally, we need one more theorem and one more lemma to prove our main results.

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.Proof.Let  be an 8-vertex graph with Δ() = 5.Assume that  has no  5 and  5 ∨  2 .
If there is a vertex V such that (V) ≤ 3, then  − V is 4-colorable by Lemma 9; hence,  is 4-colorable, as well.Suppose that () ≥ 4. Let V 1 be a vertex with (V 1 ) = 5 of  and let V 2 be a vertex which is not adjacent to Then  must be a subgraph of the graph shown in Figure 3.According to the figure,  and its subgraphs are 4-colorable.
If  1 and  2 are adjacent to different vertices in ( 4 ), then  must be the left graph in Figure 4.If  1 and  2 are adjacent to the same vertex in ( 4 ), then  must be the right graph in Figure 4.According to the figure,  is 4-colorable.
If  1 and  2 are adjacent to the same vertex in ( 4 ), then  must be the left graph in Figure 5.If  1 and  2 are adjacent to different vertices in ( 4 ), then  must be the middle graph or the right graph in Figure 5.According to the figure,  is 4-colorable.Case 3 (Δ() = 6).Let V be a vertex with (V) = 6 and  is the vertex which is not adjacent to V. Since |(V)| = |()| = 4 and |(())| = 7, (V) and () have a common color, for example, .Hence, we label V and  by color .
Case 1.There is a color  that appears in exactly 1 list.Without loss of generality, suppose  ∈ ( 1 ) but  ¡ ∈ (V) for the remaining vertices V. Then we label  1 by color .Each of the remaining six vertices has 3 available colors.Since  7 −  1 does not contain  4 or  5 ∨  1 , it is 3-colorable by Lemma 9. Hence,  7 −  1 is 3-choosable by Corollary 14.That is, the remaining vertices can be labeled.
Case 2. There is a color  that appears in exactly 2 lists.Without loss of generality, we may prove only 3 subcases because   is not adjacent to   if and only if  −  ≡ ±1 mod 7.
Case 3.There is a color  that appears in exactly 4, 5, or 6 lists.According to the fact that   is not adjacent to   if and only if  −  ≡ ±1 mod 7, we may prove only the following cases.
Case 4. Each color appears in exactly 3 or 7 lists.Let  and  be the number that appears in exactly 3 or 7 lists, respectively.Then + = 5 and 3+7 = 21.There are no integer solutions for this system.Hence, this case is impossible.
⊆ ().If  is a list assignment of , we let |  denote  restricted to  and () denote ⋃ V∈ (V).For a color set , let  −  be the new list assignment obtained from  by deleting all colors in  from (V) for each V ∈ ().When  has only one color , we write  −  instead of  − {}.The cycle   is (2, )-choosable unless  is odd and  = 2.Note that a graph  is (2, 2)-choosable if and only if  is 2colorable.Hence,   is (2, 2)-choosable if and only if  is even.It remains to show that all of the cycles are (2, )-choosable for  ≥ 3.