All the graphs we consider in this paper are undirected, simple, and have no loops. For every such graph , we denote by and its vertex set and edge set, respectively, by its maximum degree, and by the set of neighbors of the vertex in .
The graph coloring game is a two-player game introduced by Steven J. Brams (reported by Martin Gardner in his column Mathematical Games in Scientific American in 1981 ) and rediscovered ten years later by Bodlaender . Given a graph and a set of colors, the two players, Alice and Bob, take turns coloring properly an uncolored vertex of , Alice having the first move. Alice wins the game if and only if all the vertices of are eventually colored. In other words, Bob wins the game if and only if, at some step of the game, all the colors appear in the neighborhood of some uncolored vertex.
The game chromatic number of is then defined as the smallest integer for which Alice has a winning strategy when playing the graph coloring game on with colors. The problem of determining the game chromatic number of several graph classes has attracted much interest in recent years (see  for a comprehensive survey of this problem), with a particular focus on planar graphs (see e.g. [6, 13, 14, 19, 20, 21]) for which the best known upper bound up to now is 17 .
Most of the known upper bounds on the game chromatic number of classes of graphs are derived from upper bounds on the game coloring number of these classes, a parameter defined through the so-called graph marking game, formally introduced by Zhu in . This game is somehow similar to the graph coloring game, except that the players mark the vertices instead of coloring them, with no restriction. The game coloring number of is then defined as the smallest integer for which Alice has a strategy such that, when playing the graph marking game on , every unmarked vertex has at most marked neighbors. It is worth noting here that the game coloring number is monotonic, which means that for every subgraph of , while this property does not hold for the game chromatic number .
Let be a graph with and consider the winning strategy of Alice for the marking game on . Applying the same strategy for the coloring game on , Alice ensures that each uncolored vertex has at most colored neighbors, so that we get . Hence, the following inequalities clearly hold for every graph .
For every graph , .
In this paper, we introduce and study a new version of the graph coloring game (resp. of the graph marking game), by requiring that, after each player’s turn, the subgraph induced by the set of colored (resp. marked) vertices is connected. In other words, on their turn, each player must color an uncolored vertex (resp. mark an unmarked vertex) having at least one colored (resp. marked) neighbor, except for Alice on her first move.
We call this new game the connected graph coloring game (resp. the connected graph marking game). We will denote by the connected game chromatic number of a graph , that is, the smallest integer for which Alice has a winning strategy when playing the connected graph coloring game on with colors, and by the connected game coloring number of , that is, the smallest integer for which Alice has a strategy such that, when playing the connected graph marking game on , every unmarked vertex has at most marked neighbors. It is not difficult to observe that, similarly to the ordinary case, the following inequalities hold for every graph .
For every graph , .
It is proved in  that for every positive integer , , where denotes the complete bipartite graph with vertices in each part, minus a perfect matching. We prove in Section 2 that for every nonempty bipartite graph , which shows, since the graph is bipartite, that the difference can be arbitrarily large.
One of the main open, and rather intriguing, question concerning the graph coloring game is the following: assuming that Alice has a winning strategy for the graph coloring game on a graph with colors, is it true that she has also a winning strategy with colors? We will prove in Section 2 that the answer is “no” for the connected version of the coloring game. More precisely, we will prove that for every integer , there exist bipartite graphs on which Bob wins the connected coloring game with colors, while Alice wins the connected coloring game with two colors on every bipartite graph.
The “connected variant” of other types of games on graphs have been considered in the literature. This is the case for instance for the domination game [5, 12], the surveillance game [7, 10], the graph searching game [1, 3, 8], or Hajnal’s triangle-free game [16, 17]. However, to our knowledge, the connected variant of the graph coloring game has not been considered yet.
2 Bipartite graphs
We consider the case of bipartite graphs in this section. We will prove that for every integer , there exist bipartite graphs on which Bob wins the connected coloring game with colors, while Alice wins the connected coloring game with two colors on every bipartite graph.
It is easy to see that Alice always wins when playing the connected coloring game on a bipartite graph with two colors: thanks to the connectivity constraint, the first move of Alice forces all the next moves to be consistent with a proper 2-coloring of .
For every bipartite graph , .
Proof. Let be any bipartite graph. The strategy of Alice is as follows. On her first move, she picks any vertex of and gives it color 1. From now on, each play will color some vertex having at least one of its neighbors already colored, so that, since is bipartite, this eventually leads to a proper 2-coloring of .
However, for every integer , there are bipartite graphs on which Bob wins the connected coloring game with colors.
For every integer , there exists a bipartite graph on which Bob wins the connected coloring game with colors.
Proof. Let be any -free bipartite graph with minimum degree and let and denote the partite sets of . Let now be the bipartite graph obtained from by adding, for each -subset of , a new vertex adjacent to all vertices of .
We now define the strategy of Bob for playing the connected coloring game on as follows. In his first moves (at most three, depending on the moves of Alice), Bob colors two vertices of , say and , with two different colors. In his next two moves, Bob colors a neighbor of in with the same color as , and a neighbor of in with the same color as . Since the minimum degree of is and is -free, Alice cannot prevent Bob from doing so.
Now, Bob colors a -subset containing and with distinct colors. Again, Alice cannot prevent Bob from doing so since each move of Alice “forbids” at most uncoloured vertices in (each vertex of has at most neighbors in this set).
After that, the vertex cannot be colored and Bob wins the game.
3 Outerplanar graphs
We consider in this section the case of outerplanar graphs. An outerplanar graph is a graph that can be embedded on the plane in such a way that there are no edge crossings and all its vertices lie on the outer face.
Concerning the ordinary coloring game, Kierstead and Trotter proved in  that there exist outerplanar graphs with game chromatic number at least 6, and Guan and Zhu proved in  that the game chromatic number of every outerplanar graph is at most 7. This bound has then been proven to be tight by Kierstead and Yang in . We will prove that the connected game chromatic number of every outerplanar graph is at most 5 and that there exist outerplanar graphs with connected game chromatic number 4.
Recall that an outerplanar graph is maximal if adding any edge makes it non outerplanar. It is not difficult to see that an outerplanar graph is maximal if and only if, in all its outerplanar embeddings, all faces are triangles, except possibly the outer face. Our goal in this section is to prove that the connected coloring number of every outerplanar graph is at most 5.
When playing the connected coloring game on a graph , we will say that an uncolored vertex in is saturated if each of the available colors appears in its neighborhood. Observe that Bob wins the connected coloring game on if and only if he has a strategy such that, at some point in the game, an uncolored vertex in becomes saturated. Similarly, when trying to prove that the connected game coloring number of some graph is at most , we will say that an unmarked vertex in is saturated if it has at least marked neighbors. Again, the connected game coloring number of is at least if and only if Bob has a strategy such that, at some point in the game, an unmarked vertex in becomes saturated.
Finally, we will say that a vertex in is playable if it is uncolored (resp. unmarked) and has at least one colored (resp. marked) neighbor. Moreover, when considering the connected marking game, we will say that a vertex is threatened if it is unmarked, has differently marked neighbors and at least one playable neighbor. In that case, note that if Bob plays on a playable neighbor of any threatened vertex, then Alice loses the game. A winning strategy of Alice for the connected marking game thus consists in ensuring that, after each of her moves, the considered graph has no threatened vertex.
We first describe more precisely the structure of maximal outerplanar graphs, which will be used for defining the strategy of Alice. Let be a maximal (embedded) outerplanar graph. An edge belonging to the outer face of is an outer edge of . Let us choose and fix any outer edge of . The distance from any vertex to the edge is defined as . For every integer , let denote the set of vertices at distance from . It is not difficult to observe that the subgraph of induced by each set is a linear forest, that is, a disjoint union of paths. In particular, is the edge and is a single path. Therefore, can be viewed as a “tree of trapezoids”, as illustrated in Figure 1(a).
Each of these trapezoids has the structure depicted in Figure 1(b). Both and belong to some , , while , , belong to . The vertices and are the parents of the children vertices , the edge is the root edge of the trapezoid, the unique vertex , , which is joined by an edge to both and is the pivot of the trapezoid. We will denote by the trapezoid whose root edge is , and by the pivot of . Note that each vertex , , is a neighbor of at least one of its parents, and that only the pivot is a neighbor of both its parents. Moreover, we will say that , , is the left neighbor of , while , , is the right neighbor of .
Observe that if there is no trapezoid of the form or in for some vertex , then the degree of is at most 4 (it is 4 only if is the pivot of some trapezoid), so that cannot be a threatened vertex. Note also that every vertex belongs to at most two root edges.
Based on the drawing of the outerplanar graph depicted in Figure 1(a), we can define a total ordering of the vertices of , obtained by listing the vertices of from left to right, then the vertices of from left to right, and so on. Finally, we will say that a vertex , belonging to a trapezoid , lies above a vertex , belonging to a trapezoid , if every shortest path from to goes through or .
We now describe the strategy of Alice when playing the connected marking game on an outerplanar graph . Let be any outer edge of , and be any maximal outerplanar graph containing as a subgraph, and such that is also an outer edge of . In the following, we assume that we are given a trapezoidal representation of , starting from the edge , as described above. Moreover, we can also assume that for every trapezoid of , at least one of the edges , belongs to . This will allow us to speak about children or parent vertices (with respect to ) even if the corresponding edges do not belong to , and to use the ordering of the vertices of .
Let us denote by , , the vertex marked by Alice on her -th move, and by , , the vertex marked by Bob on his -th move, so that the sequence of moves (that is, marked vertices) is Hence, is the vertex marked by Alice on her first move and, for every , is the “response” of Alice to the move of Bob.
The strategy of Alice will then consist in applying the first of the following rules that can be applied (see Figure 2 for an illustration of Rules R1, R2 and R3) for each of her moves.
If is a playable unmarked parent of , then .
If belongs to a root edge or , is marked and is playable, then .
If belongs to a root edge or , is unmarked, is a pivot and is playable, then .
If none of the above rules can be applied, and there are still unmarked vertices in , then we let , where is the smallest (with respect to the ordering ) playable vertex.
Note that on his first move, Bob must mark either the vertex , in which case Alice will apply Rule R2 on her second move, or some neighbor of , in which case Alice will apply rule R1 and mark on her second move (recall that the edge belongs to ). Moreover, if Bob marks a child vertex of some trapezoid , then at least one of , must be marked (by the connectivity constraint), and Alice will immediately apply Rule R1 if one of them is unmarked and is an edge in . These remarks are summarized in the two following observations.
After the second move of Alice, both vertices and are marked.
After each move of Alice, if is a marked child vertex of a trapezoid and is an edge in , then both and are marked.
We are now able to prove the main result of this section.
If is an outerplanar graph, then .
Proof. We assume that we are given an outerplanar embedding of and its trapezoidal representation, as previously discussed. Clearly, it suffices to prove that if Alice applies the above described strategy, then, after each move of Alice, contains no threatened vertex. This is clearly the case after the first and second move of Alice since, at that point, only one or three vertices have been marked, respectively.
Suppose to the contrary that, after Bob has marked the vertex and Alice has marked the vertex , , is a threatened vertex in , and that is the smallest index with this property, which implies that or is a marked neighbor of . Thanks to Observation 5, we know that both and have been marked. Therefore, is necessarily a child vertex of some trapezoid (we may have ). Let and denote the left and right neighbors of (in ), if they exist. Note that at least one of them must exist, since otherwise would have at most two marked neighbors, and thus could not be a threatened vertex. Since has four marked neighbors, at least one of , must be marked, since otherwise no vertex lying below could have been marked, due to the connectivity constraint, so that, again, would have at most two marked neighbors. Thanks to Observation 6, we thus get that both and are marked if is an edge in .
We now claim that neither nor contains a marked child vertex which is a neighbor of . Indeed, such a vertex, say , cannot have been marked by Bob since, by Rule R1, Alice would have marked just after Bob had marked the first such child vertex of the corresponding trapezoid. The vertex has thus been marked by Alice which implies, since is unmarked, that none of the edges , , or belong to (otherwise would have been marked in priority by Alice), and that is the only marked neighbor of , so that cannot be a threatened vertex.
Therefore, the four marked neighbors of are necessarily , , and . Hence, is the pivot of , which implies, since is unmarked, that has been marked after , and that has been marked after , so that . (Note here that we cannot have , since this would imply , contradicting the priority of rule R2.) But in each case, that is, or , would have been marked by Alice, thanks to Rule R3.
We thus get a contradiction in each case, which concludes the proof of Theorem 7.
Concerning the connected game chromatic number of outerplanar graphs, we can now prove the following.
If is an outerplanar graph, then . Moreover, there exist outerplanar graphs with .
Proof. From Observation 2 and Theorem 7, we get . For the second part of the statement, consider the outerplanar graph depicted in Figure 3. We will prove that Bob has a winning strategy when playing the connected coloring game on with three colors. Thanks to the symmetries in , and up to permutation of colors, Alice has three possible first moves, that we consider separately.
If Alice colors with color , then Bob colors with color . Now, if Alice colors with color then Bob colors with color so that is saturated, while if Alice colors or with color (resp. with color ), then Bob colors with color (resp. with color ), so that or is saturated.
If Alice colors with color , then Bob colors with color , and the so-obtained configuration is similar to that of the previous case.
If Alice colors with color , then Bob colors with color . Now, if Alice colors with color then Bob colors with color so that is saturated, while if Alice colors with color (note that using color would saturate ), then Bob colors with color , so that is saturated.
This concludes the proof of Theorem 8.
There exist outerplanar graphs with .
We have introduced in this paper a connected version of the graph coloring and graph marking games. We have proved in particular that the connected game coloring number of every outerplanar graph is at most 5, and that there exist infinitely many bipartite graphs on which Alice wins the connected coloring game with two colors but loses the game if the number of colors is at least three.
We conclude this paper by listing some open questions that should be considered for future work.
What is the optimal upper bound on the connected game coloring number and on the connected game chromatic number of outerplanar graphs? We know that both these values are either 4 or 5.
What is the optimal upper bound on the connected game coloring number and on the connected game chromatic number of planar graphs?
Does there exist, for every integers and , a graph on which Alice wins the connected coloring game with colors, while Bob wins the game with colors?
Is the connected game coloring number a monotonic parameter, that is, is it true that for every subgraph of , the inequality holds?
Does there exist a graph for which ? or ? (That is, is it possible that the connectivity constraint is in favour of Bob?)
Acknowledgments. The first and third authors have been supported by the ANR-14-CE25-0006 project of the French National Research Agency, and the fourth author by Grant number NSFC 11571319. The work presented in this paper has been initiated while the first author was visiting LaBRI, whose hospitality was greatly appreciated. It has been pursued during the 9th Slovenian International Conference on Graph Theory (Bled’19), attended by the third and fourth authors, who warmly acknowledge the organizers for having provided a very pleasant and inspiring atmosphere.
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