Game chromatic number of Cartesian and corona product graphs

The game chromatic number χg is investigated for Cartesian product G H and corona product G◦H of two graphs G and H. The exact values for the game chromatic number of Cartesian product graph of S3 Sn is found, where Sn is a star graph of order n+1. This extends previous results of Bartnicki et al. [1] and Sia [5] on the game chromatic number of Cartesian product graphs. Let Pm be the path graph on m vertices and Cn be the cycle graph on n vertices. We have determined the exact values for the game chromatic number of corona product graphs Pm ◦K1 and Pm ◦ Cn. 2010 MSC: 05C15, 05C65


Introduction
We consider the following well-known graph coloring game, played on a simple graph G with a color set C of cardinality k.Two players, Alice and Bob, alternately color an uncolored vertex of G with a color from C such that no adjacent vertices receive the same color (such a coloring of a graph G is known as proper coloring).Alice has the first move and the game ends when no move is possible any more.If all the vertices are properly colored, Alice wins, otherwise Bob wins.The game chromatic number of G, denoted by χ g (G), is the least cardinality k of the set C for which Alice has a winning strategy.In other words, necessary and sufficient conditions for k to be the game chromatic number of a graph G are: i) Bob has winning strategy for k − 1 colors or less and ii) Alice has winning strategy for k colors.This parameter is well defined, since it is easy to see that Alice always wins if the number of colors is larger than the maximum degree of G. Clearly, χ g (G) is at least as large as the ordinary chromatic number χ ( G), but it can be considerably more.For example, let G be a complete bipartite graph K n,n Syed Ahtsham Ul Haq Bokhary (Corresponding Author), Tanveer Iqbal, Usman Ali; Centre for Advanced Studies in Pure and Applied Mathematics, Bahauddin Zakariya University, Multan, Pakistan (email: sihtsham@gmail.com,tanveerm8@yahoo.com,uali@bzu.edu.pk).minus a perfect matching M and consider the following strategy for Bob.If Alice colors vertex v with color c then Bob responds by coloring the vertex u matched with v in the matching M with the same color c.Note that now c cannot be used on any other vertex in the graph.Therefore, if the number of colors is less than n, Bob wins the game.This shows that there are bipartite graphs with arbitrarily large game chromatic number and thus there is no upper bound on χ g (G) as a function of χ ( G).
The obvious bounds for the game chromatic number are: where χ(G) is the chromatic number and ∆(G) is the maximum degree of the graph G.
A lot of attempts have been made to determine the game chromatic number for several classes of graphs.This work was first initiated by Faigle et al. [3].It was proved by Kierstead and Trotter [4] that the maximum of the game chromatic number of a forest is 4, also that 33 is an upper bound for the game chromatic number of planar graphs.Bodlaender [2], found that the game chromatic number of Cartesian product is bounded above by constant in the family of planar graphs.Later, Bartnicki et al. [1] determine the exact values of χ g (G H) when G and H belong to certain classes of graphs, and showed that, in general, the game chromatic number χ g (G H) is not bounded from above by a function of game chromatic numbers of the graphs G and H.After that, Zhu [6] established a bound for game coloring number and acyclic chromatic number for Cartesian product of two graphs H and S. Sia [5] determined the exact values for the Cartesian product of different families of graph like S m P n , S m C n , P 2 W n , where P n , C n , denotes the path graph and cycle graph with n vertices, respectively.While S m is a star graph with m + 1 vertices and W n is a wheel graph with n + 1 vertices.
In this work, we have extended the study of game chromatic number and found the exact values of the game chromatic number of S 3 S n , P m • K 1 and P m • C n , where • denote corona product of graphs.

Exact value of χ g (S 3 S n )
A Cartesian product of two graphs G and H, denoted by G H, is the graph with vertex set V (G) × V (H), where two vertices (u, u ) and (v, v ) are adjacent if and only if u = v and u v ∈ E(H) or u = v and uv ∈ E(G).Note that the Cartesian product operation is both commutative and associative up to isomorphism.
Before stating our results, we introduce some definitions and notational conventions.Suppose that Alice and Bob plays the coloring game with k colors.We say that there is a threat to an uncolored vertex v if there are k − 1 colors in the neighborhood of v, and it is possible to color a vertex adjacent to v with the last color, so that all k colors would then appear in the neighborhood of v.The threat to the vertex v is said to be blocked if v is subsequently assigned a color, or it is no longer possible for v to have all k colors in its neighborhood.We shall also use the convention that color numbers are only used to differentiate distinct colors, and should not be regarded as ascribed to particular colors.For example, if only colors 1 and 2 have been used so far and we introduce a new color, color 3, then color 3 can refer to any color that is not the same as color 1 or color 2. Finally, we label figures in the following manner: vertices are labeled in the form "color(player) turn ", with the information in parentheses being omitted if the same configuration can be attained in multiple ways.
In the following result, we have found the exact value of χ g (S 3 S n ).
Proof.First, we show that Bob has a winning strategy with three or fewer colors.We give Bob's winning strategy when there are exactly three colors, it will be easy to see that Bob can win using the same strategy when there are fewer than three colors.Denote the vertices of The graph shown in Figure 1 is isomorphic to S 3 S n .Bob's winning strategy with three colors: Case 1: Let D = {1, 2, 3} be set of colors.If Alice begins by playing color 1 in the vertex a 1 of highest degree.Bob should plays color 2 in any of the vertex d k of degree two (say d 2 ), see Figure 1.Alice can not block the threats to both the vertices d 1 and a 2 , so Bob wins.Case 3: If Alice begins the game by assigning color to any of the vertices a l , 2 ≤ l ≤ n.Suppose, she plays a color 1 in the vertex a 2 .In response, Bob plays color 2 in the vertex a i , where i = 2 and i > 2 say a 3 .This forces Alice to assign color 3 to the vertex a 1 , because this is the only way, she can block the threat to this vertex.After that Bob plays color 1 in the neighbor of a l s (l ≥ 4) and wins the games.Thus χ g (S 3 S n ) ≥ 4. Next, Alice's winning strategy is given for four colors, which proves that χ g (S 3 S n ) ≤ 4. Alice's winning strategy with four colors: Alice in her strategy, first assigns colors to all the vertices of degree n + 1, because in this way, she can prevent these vertices from threats.Now a l , where l ≥ 1 are the only vertices which can come under threat.Since a l ∼ b i (l = i), a l ∼ c j (l = j) and a l ∼ d k (l = k).So, whenever, Bob plays a color in a vertex adjacent to a i , say c j , Alice will play the color of vertex a 1 or color of the vertex c j in b i (adjacent to a l ).In this way, Alice can prevent these vertices from threats, since every vertex has two same colors in its neighbor.So, in order to win the game, she need only to assign colors in the vertices of degree greater than 4. Below, we have given the Alice strategy to color these vertices with four given colors.Alice makes her first move by playing color 1 in any of three vertices having degree n + 1, say b 1 .We, now consider all the cases for Bob's first move.
Case 1: If Bob plays color 2 in any of the vertex having degree four, say a 2 .Alice makes her second move by playing color 2 in the vertex c 1 .Now Bob has two possibilities for his second move.
Case 1.1:If Bob wants that the vertex a 1 be under threat in his second move and assigns color 3 to the vertex a 3 .Alice, then has to play color 4 in a 1 , in order to prevent this vertex from threat.Alice is safe to assign suitable color in the vertex d 1 in her fourth move.Now all the vertices of degree greater than four have assigned color.After that whenever Bob plays a color in a vertex b i or c j say c j (adjacent to a l ), Alice will play the color of c j in b i where l = i = j and win the game.(strategy is shown in Figure 3) In respond, Alice assigns color 2 in a 1 .In his third move Bob plays according to the following strategy.i: If Bob plays color 2 in the vertex b i s or c j s, Alice then plays color of the vertex c j in b i (i = j).Now vertex a l (i = l = j) has two same colors in its neighbors.In this way Alice protects all the vertices of degree four from threat.Alice will play only color in a 1 , whenever Bob plays color in a i s.
Case 2: In this case, Bob plays his first move in the vertices of degree two and assigns color 1 in d k or c j say d 2 .Alice then plays color 1 in c 1 .Now we consider all the cases for Bob's second move.
Case 2.1: If Bob plays again in the neighbor of d 1 say d 3 by assigning it color 2. Alice, in respond plays color 3 in d 1 .Now there are following possibilities of Bob's third move.We distinguish 3 subcases.i: If Bob wants that the vertex a 1 be under threat and plays color 2 in a l say a 2 .Alice then should play color 4 in a 1 .Now all the vertices of degree greater than four have assigned color.After that Alice follow the same strategy as in case 1.
ii: If Bob plays in the neighbor of that a l which has already two distinct colors say a 2 and assigns color 2 in b i or c j say c 2 .Alice can prevent this vertex from threat by playing color of vertex c 2 in vertex b 2 .After that Alice follows the same strategy as in case 1.
iii: If Bob wants to plays in the neighbor of that a l which has has no color assigned say a 4 and assigns color 1 in d 4 (d 4 is adjacent to a 4 ), then Alice plays color 2 or color 4 in the vertex a 1 .In his next move if, Bob again plays in the neighbor of a 4 and assigns color 3 in c 4 .Alice then should play color of c 4 in b 4 .Now a 4 has two same colors in its neighbor.Whenever, Bob plays color 1 in any one of the remaining uncolored d k s say d 5 , Alice then plays color of vertex a 1 in c 5 .Now there are two same colors in the Case 2.2: Now vertices d 1 and a 2 have one color in its neighbor.If Bob make his second move by playing the color 2 in b i or c j say c 2 , then Alice assigns color of vertex c 2 in b 2 .In this way, Alice can block the threat of vertex a 2 .After that Alice follows the same strategies as in case 1 and case 2.1.
Case 2.3: If Bob plays suitable color in c j or b i say c j (k = j), then Alice plays color of that c j in b i (i = j).After that Alice follows the same strategies as in above cases.
Case 3: If Alice assigns color 1 in b 1 in her first move and Bob plays color 2 in c 1 .Alice then plays color in d 1 .Now all the vertices of degree n + 1 have colored.After that Alice will follow the strategy of case 1 and case 2 in her next moves to win the game.
Case 4: Bob plays color 2 in the vertex a 1 in his first move.Alice then plays color 1 in c 1 .Now, irrespective of Bob's next move.Alice plays color 3 in d 1 .In this way, Alice can succeed to assign color to all the vertices of degree greater than n + 1.Now Bob will in b i or c j or d k to create threat in a l .Alice will respond by following the strategies of case 1 and case 2 and she wins the game.

Exact values of χ
Let G and H be two graphs of order n 1 and n 2 , respectively.The corona product G • H is defined as the graph obtained from G and H by taking one copy of G and n 1 copies of H and joining by an edge each vertex from the i th −copy of H with the i th −vertex of G.We will denote by V = {v 1 , v 2 , . . ., v n } the set of vertices of G and by It is important to note that the corona product operation is not commutative.In this section we prove the exact values of χ g (P m • K 1 ) and χ g (C m • P n ).
Theorem 3.1.The game chromatic number of Proof.It is obvious that χ g (P m • K 1 ) ≥ 3.In order to show that χ g (P m • K 1 ) ≤ 3, Alice winning strategy is given.Alice's winning strategy with 3 colors: Alice starts the game by playing color 1 in the vertex w 1 .Suppose, in his first move, Bob plays one of the available color in vertex v i , then Alice responds by playing the same color in w i+1 or w i−1 .In this way she can block the threat to the vertex v i , because it has two same colors in its neighbor.Similarly if Bob plays some color in vertex w i , Alice responds by playing the same color in v i+1 or v i−1 .If during the game the following two cases arises, then Alice will play according to the following strategy.
Case 1: Suppose the vertex v j is already colored and Bob plays a color in the vertex w i .In this situation, If d(v i , v m ) is odd then Alice assigns the same color in v i−1 (if i > j) or v i+1 (if i < j), otherwise she plays any available color in the vertex v i .
Case 2: If there are two vertices say v m and v n that are already colored and Bob plays a color 1 in the vertex w i where m < i < n.In response, Alice plays the same color even).In case when v i is at odd distance to both v m and v n , she will play any available color in the vertex v i .Similarly if Bob plays some color in v i then Alice plays the same color in w i−1 or w i+1 according to the above two cases.If during the game, the following situations arise then Alice modifies her strategy according to the situation.
Situation 1: A part of graph P m • K 1 is shown in Figure 5.
If Bob plays the color 3 on w i+2 .Suppose, Bob already assigned the color 3 in the vertex v i+4 .Then according to her strategy Alice can not assign the same color in the vertex v i+3 .In this case Alice plays color 2 in v i+3 , because this is the only way she can block the threat to the vertex v i+2 .
Situation 2: If Bob plays a color (say 3) in the vertex w i+2 and the vertices v i+4 and v i−1 are already colored with colors 3 and 2 respectively.In this situation Alice plays the same color as of the vertex v i−1 in w i .Case 1: Alice makes her first move by playing color 1 in any of the vertex from the inner cycle, say a 1 .In response Bob plays the color 2 in the vertex a 3 .In her second move Alice will play color 3 in the vertex a 2 because this is the only way she can block the threat to this vertex.In his next moves Bob plays suitable color in the alternative vertices a 5 , a 7 , a 9 , . . ., respectively, forcing Alice to play suitable color in the vertices a 4 , a 6 , a 8 , . . . .Continuing in this way, Bob plays his last move according to the following two situations: i: If the number of vertices in inner cycle are even then Bob plays any suitable color in the vertex a n−1 .Alice cannot block the threat to both the vertices a n−2 and a n , and Bob wins.ii: If the number of vertices in inner cycle are odd then Bob in his second last move plays a color in any of the vertex b i,n−1 , which is different to that of a 1 .In response Alice is forced to play color in the vertex a n−1 because this is the only way to block the threat to this vertex.By playing a color different to both a 1 and a n−1 in any of the vertex b i,n−1 , Bob can succeed to assign three distinct colors in the neighbor of the vertex a n .
Case 2: Alice makes her first move by playing color 1 in any of the outer vertex b i,n , say b 1,1 .In response Bob plays color 2 in the vertex b 2,1 .In her second move Alice has to play color 3 in the vertex a 1 because this is the only way she can block the threat to this vertex.In his next moves, Bob plays in the alternative vertices a 3 , a 5 , a 7 , . . ., respectively, forcing Alice to play in the vertices a 2 , a 4 , a 6 , . . .Continuing in this way, Bob plays his last move according to Case 1. Hence χ g (C m • P n ) ≥ 4. Alice's winning strategy with 4 colors: The graph C m • P n have m + mn vertices.The vertices of degree greater than four are only the vertices of the inner cycle, as shown in the above Figure .Alice, in her strategy assigns color to the vertices a 1 , a 2 , ..., a m and consequently blocks the threat to these vertices.Since Alice has to start the game, suppose she makes her first move by playing color 1 in the vertex a 1 .In respond, Bob can play the game in the following two ways.
Case 1: Bob plays in the vertices of inner cycle say a i , for some 2 ≤ i ≤ m.Alice responds by playing any color from the set of available colors in the uncolored vertices of inner cycle.
Case 2: Bob plays in the vertices b i,j , then Alice assigns a suitable color in the corresponding a i .
Bob can only win if he succeeded to assign 4 distinct colors in the neighborhood of any vertex.But whenever he plays in the outer vertices, b i,j , Alice assigns a suitable color in the corresponding a i .In this way Alice can color all the vertices in the inner cycle and she wins the game.From both these strategies we conclude that χ g (C m • P n ) = 4.

Figure 1 .
Figure 1.Bob's winning strategy in Case 1 for the graph S3 Sn a

Figure 2 .
Figure 2. Bob's winning strategy in Case 2 for the graph S3 Sn

Case 4 :
If Alice start the game by playing color 1 in vertex b 1 , c 1 or d 1 (say b 1 ).Bob should plays color 2 in any of the vertices a i , where 2 ≤ i ≤ n, for simplicity, say a 2 .There are two distinct colors in the neighbor of vertex a 1 , so Alice must block the threat to this vertex by assigning it the color 3.In the next move Bob plays color 1 in d k (k = 2) say d 3 .Now Alice can't block the threat of d 1 and a 3 and Bob wins the game.

Figure 3 .
Figure 3. Alice's winning strategy in Case 1 for the graph S3 Sn

Figure 4 .
Figure 4. Alice's winning strategy in Case 2 for the graph S3 Sn