Hindawi Journal of Applied Mathematics Volume 2020, Article ID 6475427, 4 pages https://doi.org/10.1155/2020/6475427

Research Article Game Chromatic Number of Generalized Petersen Graphs and Jahangir Graphs

Ramy Shaheen , Ziad Kanaya, and Khaled Alshehada

Department of Mathematics, Faculty of Science, Tishreen University, Lattakia, Syria

Correspondence should be addressed to Ramy Shaheen; [email protected]

Received 20 December 2019; Accepted 10 March 2020; Published 14 April 2020

Academic Editor: Kannan Krithivasan

Copyright © 2020 Ramy Shaheen et al. %is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Let G � (V, E) be a graph, and two players Alice and Bob alternate turns coloring the vertices of the graph G 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 χg(G), while Bob’s goal is to prevent Alice’s goal. In this paper, we investigate the game chromatic number χg(G) of Generalized Petersen Graphs GP(n, k) for k ≥ 3 and arbitrary n, n-Crossed Graph, and Jahangir Graph Jn,m.

1. Introduction In [8], Shaheen et al. investigated the game chromatic number for some special circulant graphs and Generalized n 1981, Brams invented the game as a tool to provide a Petersen Graphs when k � 1, 2, 3. theoretical proof of the Four Color theorem without e Coloring Number was defined in [9] as the following: using computers [1]. But, it soon became apparent that Let Π(G) be the set of linear orderings on the vertex set there were many planar graphs which had game chro- of a graph G, and let L ∈ Π(G). For a vertex v ∈ V(G), the matic numbers greater than 4. In 1991, Bodlaender back degree of v with respect to L is the number of its + reinvented the game as a two-person game, Alice and Bob neighbors that precede it in L, i.e., b(v)tn � q|hNL(v)| + + [2]. Players take turns coloring the vertices of a given ⟶ b(v) � |NL(v)| where NL(v) � �u ∈ N(v) and u < Lv�. finite graph G � (V, E) and a color set X � {}1, 2, ... , k . %e back degree of L is the maximum back degree of the Alternately, players take turns over the elements of the set vertices in L. %e coloring number of G denoted by col(G) V; Alice starts the game, and her goal is to color the is 1+ b where b is the minimum back degree of a linear or- vertices with only k � |X| colors, while Bob’s goal is to dering L over all L ∈ Π(G), i.e., col(G) � 1 + minL∈Π(G) prevent her. Bob wins the game if at any stage there are maxv∈V(G)(b(v)). Clearly, χ(G) ≤ col(G). some vertices without available colors. %e game chro- e Game Coloring Number colg(G) is the competitive matic number of a graph G, denoted by χg(G), is defined version of the coloring number, which was introduced for as the smallest integer k such that Alice has a winning the first time by Zhu in [10] to give an improved upper strategy on only k colors. In [3], Faigle et al. proved that bound of the game chromatic number of planar graphs. %e χg (T) ≤ 4 for a tree graph. In [4], Guan and Zhu introduced game coloring number is a two-person game, which is an upper bound to the game chromatic number of out- played by Alice and Bob, who alternate turns and with Alice erplanar graphs, which is χg(OP) ≤ 7. In [5], Bartnicki et al. starting first. After every move, a vertex is selected (among studied the game chromatic number for the Cartesian product the remaining vertices) and is added to the end of the linear of two graphs. In [6], Sia studied the game for some families of ordering which was formed by previous moves. After Cartesian product graphs. Raspaud and Wu proved that selecting all the vertices, a linear ordering L of V(G) is χg(C2mCn) � 5 ⟶ χg(C2m□Cn) � 5 [7]. obtained, let b be its back degree and then colg(G) � 1 + b. 2 Journal of Applied Mathematics

Alice’s goal is to minimize the back degree b, while Bob’s 2. Main Results goal is to maximize it. Both players use their optimal strategies to form L. In this section, we investigate the game chromatic number of A combinatorial game is a finite two-player game with generalized Petersen graph, n-crossed prism graph, and perfect information and no chance. %e game coloring Jahangir graph. number and the game chromatic number are combinatorial games [11]. Lemma 1. For n ≥ 9 and k ≥ 2, every Generalized Petersen In 2009, Kierstead and Kostochka were the first to graph GP(n, k) has S8 as a subgraph like the graph in Figure 1. present an application of coloring games to a nongame problem, which was the graph packing problem [12]. %e proof comes clearly from the definition of the Several studies were presented in 2019 and discussed the Generalized Petersen graph. game chromatic number of some specific graphs [11, 13, 14]. In [14], C. Chamberlin et al. investigated the game chromatic Theorem 2. For n ≥ 9 and k ≥ 4, the game chromatic number number of G which is a graph produced by subdividing each of the Generalized Petersen Graph GP(n, k) is edge of GP(n, k) with n large enough and gcd(n, k) � 1. χg(GP(n, k)) � 4. %e uncolored vertex v is called critical vertex if there is just one available color c and there are some uncolored Proof. Generalized Petersen Graphs are 3-regular graphs; neighbors of v which can be signed with c. In any stage of the therefore, χg(GP(n, k)) ≤ Δ + 1 � 4. For k ∈ {}1, 2, 3 , game, if there are some critical vertices in Bob’s turn, then he χg(GP(n, k)) � 4 is proven in %eorem 1. wins the game. Also, if there are critical vertices in Alice’s Now, by contradiction, we will prove that Player A does turn, then she must defend them by her turn [7]. not have any winning strategy for any integer k ≥ 4. th We denote the i turn of Alice by Ai and denote the Suppose that we have a color set C � {}1, 2, 3 ; by using process of assigning the vertex v with the color c by v ⟵ c. Lemma 2.1, we notice that, in GP(n, k), there is the subgraph Tow vertices are adjacent if there is an edge joining them as the graph in Figure 1. together. For a vertex v in the graph G(V, E), the open Whatever the vertex is chosen by player A to start with, neighbor set is defined as N(v) � {u ∈ V ; ∃uv ∈ E} and the we will get the same state, so without loss of generality v (v) � |N(v)| (v) � |N(v)| G � degree of is ρ ⟶ ρ . A graph assume that player A starts the game with v1, i.e., (V, E) r (v) � r v V is -regular if ρ for every ∈ . A1: v1 ⟵ 1 then B1: uk+1 ⟵ 2; now, u1 is a critical vertex so player A will defend it by coloring it with the third color Definition 1. Let the integer k, n satisfy k < n/2; Generalized or coloring its unique uncolored neighbor with a legal color, Petersen Graph GP(n, k) is the graph whose vertex set is and whatever his choice for A2 is, then B2: vk ⟵ 1 which means that the vertex v will become a critical vertex and V∪U where V � �v1, ..., vn � and U � �u1, ..., un � and its edge k+1 also player A will have to defend it so whatever his choice is set is E � �vivi+1, viui, uiui+k�, where i � 1, 2, ..., n and sub- scripts are reduced modulo n. for A3 then B3: un ⟵ 2 and now there are two critical vertices �uk, vn � and player A cannot defend them together so three colors are not enough for the first player to have a Definition 2. For a positive even number n ≥ 4, n-crossed winning strategy. %en, 3 < χ (GP(n, k)) � 4. □ prism graph G � (V, E) is a graph obtained by taking two g 1 1 distinct cycles, outer cycle Cn where V(Cn) � �u1, ... , un �, 2 2 Theorem 3. For n ≥ 6, let G be a n-crossed prism graph, and and the inner one Cn where V(Cn) � �v1, ... , vn � and adding the game chromatic number of G is χg(G) � 4. edges between them as �uivi+1, ujvj−1�; i ∈ {1, 3, ... , n − 1} , j ∈ {}2, 4, ... , n . Proof. χg(G) ≤ 4 since Δ(G) � 3; therefore, we are going to contradict that three colors are not sufficient. Let Definition 3. Jahangir graph, Jn,m for m ≥ 3, is a cycle Cnm X � {}1, 2, 3 and A1: v1 ⟵ 1; then, B1: v3 ⟵ 2, and now, with adding one vertex v0 adjacent to m vertices of Cnm v2 is a critical vertex so Alice is forced to defend it in only two which are at distance n from each other in Cnm. ways. We will discuss Alice’s second turn A2 as follows: A : v B : u u Definition 4. %e n-sunlet graph S is a cycle C where Case 1. 2 2 ⟵ 3 then 2 2 ⟶ 3, that gives two n n u , v V(C ) � �u , u , ... , u � with n additional vertex �v , v , ... , critical vertices 2 4 and Alice cannot defend them at n 1 2 n 1 2 the same time vn}, and viui ∈ E(Sn) for i � 1, ..., n. Case 2. A2: u1 ⟵ c where c ∈ {}1, 2

Proposition 1 (see [10]). χg(G) ≤ colg(G). Subcase 2.1. c � 1 then B2: vn−1 ⟵ 2 then un, vn will be critical vertices and Alice cannot defend them at Proposition 2 (see [15]). e game chromatic number of the the same time Subcase 2.2. c � 2 then u will become a critical vertex Wn is χg(Wn) � 4. 2 during Bob’s turn who will color u3 with the color 3, and Alice loses the game Theorem 1 (see [8]). For n ≥ 6 and k ∈ {}1, 2, 3 , we have χg(GPt(n, k)) � 4 ⟶ χg(GP(n, k)) � 4. %at means χg(G) > 3, so χg(G) � 4. □ Journal of Applied Mathematics 3

v1

v1 u 1 v6 vn v2

u un k+1 v0

uk vk+1

vk v5 v3

Figure 1: S8 as subgraph of GP(n, k). v4 Figure 2: Jahangir graph J2,3. Theorem 4. For m ≥ 3, let G � Jn,m be the Jahangir graph, and the game chromatic number

4, if n � 1 or (n � 2 and m ≠ 3), Let D2 � �v ∈ V| ρ(v) � 2� and D3 � �v ∈ V| ρ(v) � 3�. χg(G) �� %e chosen vertex set is C � �v ∈ D | v is a added to L�. 3, if (n � 2 and m � 3) or (n ≥ 3 and m ≥ 3). 3 At the beginning, Alice chooses the central vertex v0. Bob (1) chooses a vertex v ∉ L. Alice follows the next strategy to respond. (1) If v ∈ D and ∃u ∈ N(v)∩​ D and u ∉ L then Alice Proof. First, because the Jahangir graph can be divided into 2 3 adds u to L cycles, then it is easy to show that χg(Jn,m) > 2. Now, we will discuss the following cases. (2) Else, if v ∈ D2 and ∃u ∈ D3C ⟶ ∃u ∈ D3\C then Alice chooses u and adds it to L Case 1. n � 1 or (n � 2 and m ≠ 3) (3) Else, Alice chooses any unchosen vertex u Subcase 1.1. n � 1 then G � W , and using Proposi- m Using this strategy on J in case n ≥ 3 produces an tion 1, χ (G) � χ ( W ) � 4. n,m g g m optimal linear order L with back degree b � Subcase 1.2. n � 2 and m ≠ 3 then three colors are not max |N+(v)| � 2. So, col (J ) � b + 1 � 3, and then, sufficient because if we suppose the color set X � v∈L L g n,m χg(Jn,m) ≤ 3 by using Proposition 1. %erefore, χg(Jn,m) � 3 {}1, 2, 3 and A1: v0 ⟵ 1 then B1: v2 ⟵ 2, then there since χg(Jn,m) > 2. □ will be two critical vertices v1, v3, and Alice cannot defend them together on her next turn, so she loses the game. Conflicts of Interest Case 2. (n � 2 and m � 3) or( n ≥ 3 and misarbitrary) %e authors declare that they have no conflicts of interest. Subcase 2.1. When n � 2 and m � 3, as it is shown in Figure 2. Acknowledgments n � m � If 2 and 3, Alice can defend any critical %is work was supported by Faculty of Science, Tishreen vertices with three colors even with Bob’s strategy University, Syria. which was mentioned in Subcase 1.2. We can explain Alice’s winning strategy as follows: let the color set be References X � {}1, 2, 3 and A1: v0 ⟵ 1 then B1: v2 ⟵ 2, so there will be two critical vertices v1, v3, and then when [1] T. Bartnicki, J. Grytczuk, H. A. Kierstead, and X. Zhu, “%e Alice chooses the next move to be A2: v5 ⟵ 3; map-coloring game,” e American Mathematical Monthly, therefore, Bob cannot use the color 3 on the vertices vol. 114, no. 9, pp. 793–803, 2007. v4, v6 to win the game and Alice wins with only three [2] H. L. Bodlaender, “On the complexity of some coloring colors. games,” International Journal of Foundations of Computer Subcase 2.2. When n ≥ 3 and m ≥ 3. Science, vol. 2, no. 2, pp. 133–147, 1991. [3] U. Faigle, U. Kem, H. Kierstead, and W. T. Trotter, “On the We will use the game coloring number as a tool to game chromatic number of some classes of graphs,” Ars investigate the upper bound of χg(Jn,m). Combinatorica, vol. 35, pp. 143–150, 1993. 4 Journal of Applied Mathematics

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