Abstract
Let be a graph, and two players Alice and Bob alternate turns coloring the vertices of the graph a proper coloring where no two adjacent vertices are signed with the same color. Alice's goal is to color the set of vertices using the minimum number of colors, which is called game chromatic number and is denoted by , while Bob's goal is to prevent Alice's goal. In this paper, we investigate the game chromatic number of Generalized Petersen Graphs for and arbitrary , -Crossed Prism Graph, and Jahangir Graph .
1. Introduction
n 1981, Brams invented the game as a tool to provide a theoretical proof of the Four Color theorem without using computers [1]. But, it soon became apparent that there were many planar graphs which had game chromatic numbers greater than 4. In 1991, Bodlaender reinvented the game as a two-person game, Alice and Bob [2]. Players take turns coloring the vertices of a given finite graph and a color set . Alternately, players take turns over the elements of the set ; Alice starts the game, and her goal is to color the vertices with only colors, while Bob's goal is to prevent her. Bob wins the game if at any stage there are some vertices without available colors. The game chromatic number of a graph , denoted by , is defined as the smallest integer such that Alice has a winning strategy on only colors. In [3], Faigle et al. proved that for a tree graph. In [4], Guan and Zhu introduced an upper bound to the game chromatic number of outerplanar graphs, which is . In [5], Bartnicki et al. studied the game chromatic number for the Cartesian product of two graphs. In [6], Sia studied the game for some families of Cartesian product graphs. Raspaud and Wu proved that [7].
In [8], Shaheen et al. investigated the game chromatic number for some special circulant graphs and Generalized Petersen Graphs when .
The Coloring Number was defined in [9] as the following:
Let be the set of linear orderings on the vertex set of a graph G, and let . For a vertex , the back degree of with respect to is the number of its neighbors that precede it in , i.e., where . The back degree of is the maximum back degree of the vertices in . The coloring number of denoted by is 1+ where is the minimum back degree of a linear ordering over all , i.e., . Clearly, .
The Game Coloring Number is the competitive version of the coloring number, which was introduced for the first time by Zhu in [10] to give an improved upper bound of the game chromatic number of planar graphs. The game coloring number is a two-person game, which is played by Alice and Bob, who alternate turns and with Alice starting first. After every move, a vertex is selected (among the remaining vertices) and is added to the end of the linear ordering which was formed by previous moves. After selecting all the vertices, a linear ordering of is obtained, let be its back degree and then . Alice's goal is to minimize the back degree , while Bob's goal is to maximize it. Both players use their optimal strategies to form .
A combinatorial game is a finite two-player game with perfect information and no chance. The game coloring number and the game chromatic number are combinatorial games [11].
In 2009, Kierstead and Kostochka were the first to present an application of coloring games to a nongame problem, which was the graph packing problem [12].
Several studies were presented in 2019 and discussed the game chromatic number of some specific graphs [11, 13, 14]. In [14], C. Chamberlin et al. investigated the game chromatic number of G which is a graph produced by subdividing each edge of with large enough and .
The uncolored vertex is called critical vertex if there is just one available color and there are some uncolored neighbors of which can be signed with . In any stage of the game, if there are some critical vertices in Bob's turn, then he wins the game. Also, if there are critical vertices in Alice's turn, then she must defend them by her turn [7].
We denote the turn of Alice by and denote the process of assigning the vertex with the color by .
Tow vertices are adjacent if there is an edge joining them together. For a vertex in the graph , the open neighbor set is defined as and the degree of is . A graph is -regular if for every .
Definition 1. Let the integer satisfy ; Generalized Petersen Graph is the graph whose vertex set is where and and its edge set is , where and subscripts are reduced modulo n.
Definition 2. For a positive even number , -crossed prism graph is a graph obtained by taking two distinct cycles, outer cycle where , and the inner one where and adding edges between them as .
Definition 3. Jahangir graph, for , is a cycle with adding one vertex adjacent to vertices of which are at distance from each other in .
Definition 4. The -sunlet graph is a cycle where with additional vertex , and for .
Proposition 1 (see [10]). .
Proposition 2 (see [15]). The game chromatic number of the wheel graph is .
Theorem 1 (see [8]). For and , we have .
2. Main Results
In this section, we investigate the game chromatic number of generalized Petersen graph, -crossed prism graph, and Jahangir graph.
Lemma 1. For and , every Generalized Petersen graph has as a subgraph like the graph in Figure 1.
The proof comes clearly from the definition of the Generalized Petersen graph.
Theorem 2. For , the game chromatic number of the Generalized Petersen Graph is .
Proof. Generalized Petersen Graphs are 3-regular graphs; therefore, . For , is proven in Theorem 1.
Now, by contradiction, we will prove that Player A does not have any winning strategy for any integer .
Suppose that we have a color set ; by using Lemma 2.1, we notice that, in , there is the subgraph as the graph in Figure 1.
Whatever the vertex is chosen by player A to start with, we will get the same state, so without loss of generality assume that player A starts the game with , i.e., then ; now, is a critical vertex so player A will defend it by coloring it with the third color or coloring its unique uncolored neighbor with a legal color, and whatever his choice for is, then which means that the vertex will become a critical vertex and also player A will have to defend it so whatever his choice is for then and now there are two critical vertices and player A cannot defend them together so three colors are not enough for the first player to have a winning strategy. Then, .
Theorem 3. , let be a -crossed prism graph, and the game chromatic number of is .
Proof. since ; therefore, we are going to contradict that three colors are not sufficient. Let and ; then, , and now, is a critical vertex so Alice is forced to defend it in only two ways. We will discuss Alice's second turn as follows: Case 1. then , that gives two critical vertices and Alice cannot defend them at the same time Case 2. where Subcase 2.1. then then will be critical vertices and Alice cannot defend them at the same time Subcase 2.2. then will become a critical vertex during Bob's turn who will color with the color 3, and Alice loses the gameThat means , so
Theorem 4. For , let be the Jahangir graph, and the game chromatic number
Proof. First, because the Jahangir graph can be divided into cycles, then it is easy to show that . Now, we will discuss the following cases. Case 1. Subcase 1.1. then , and using Proposition 1, . Subcase 1.2. and then three colors are not sufficient because if we suppose the color set and then , then there will be two critical vertices , and Alice cannot defend them together on her next turn, so she loses the game. Case 2. Subcase 2.1. When , as it is shown in Figure 2. If and , Alice can defend any critical vertices with three colors even with Bob's strategy which was mentioned in Subcase 1.2. We can explain Alice's winning strategy as follows: let the color set be and then , so there will be two critical vertices , and then when Alice chooses the next move to be ; therefore, Bob cannot use the color 3 on the vertices to win the game and Alice wins with only three colors. Subcase 2.2. When .We will use the game coloring number as a tool to investigate the upper bound of .
Let and . The chosen vertex set is .
At the beginning, Alice chooses the central vertex . Bob chooses a vertex . Alice follows the next strategy to respond.(1)If then Alice adds to (2)Else, if and then Alice chooses and adds it to (3)Else, Alice chooses any unchosen vertex Using this strategy on in case produces an optimal linear order with back degree . So, , and then, by using Proposition 1. Therefore, since .
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This work was supported by Faculty of Science, Tishreen University, Syria.