A Connected Version of the Graph Coloring Game
Eric Sopena, Clément Charpentier, Hervé Hocquard, Xuding Zhu
aa r X i v : . [ c s . D M ] M a r A Connected Version of the Graph Coloring Game
Cl´ement Charpentier Herv´e Hocquard ´Eric Sopena Xuding Zhu March 17, 2020
Abstract
The graph coloring game is a two-player game in which, given a graph G and a set of k colors, thetwo players, Alice and Bob, take turns coloring properly an uncolored vertex of G , Alice having thefirst move. Alice wins the game if and only if all the vertices of G are colored. The game chromaticnumber of a graph G is then defined as the smallest integer k for which Alice has a winning strategywhen playing the graph coloring game on G with k colors.In this paper, we introduce and study a new version of the graph coloring game by requiringthat the starting graph is connected and, after each player’s turn, the subgraph induced by the setof colored vertices is connected. The connected game chromatic number of a connected graph G isthen the smallest integer k for which Alice has a winning strategy when playing the connected graphcoloring game on G with k colors. We prove that the connected game chromatic number of everyconnected outerplanar graph is at most 5 and that there exist connected outerplanar graphs withconnected game chromatic number 4.Moreover, we prove that for every integer k ≥
3, there exist connected bipartite graphs on whichBob wins the connected coloring game with k colors, while Alice wins the connected coloring gamewith two colors on every connected bipartite graph. Keywords:
Coloring game; Marking game; Game coloring number; Game chromatic number.
MSC 2010:
All the graphs we consider in this paper are undirected, simple, and have no loops. For every suchgraph G , we denote by V ( G ) and E ( G ) its vertex set and edge set, respectively, by ∆( G ) its maximumdegree, and by N G ( v ) the set of neighbors of the vertex v in G .The graph coloring game is a two-player game introduced by Steven J. Brams (reported by MartinGardner in his column Mathematical Games in Scientific American in 1981 [11]) and rediscovered tenyears later by Bodlaender [5]. Given a graph G and a set C of k colors, the two players, Alice andBob, take turns coloring properly an uncolored vertex of G , Alice having the first move. Alice winsthe game if and only if all the vertices of G are colored. In other words, Bob wins the game if andonly if, at some step of the game, all the colors appear in the neighborhood of some uncolored vertex.The game chromatic number χ g ( G ) of G is then defined as the smallest integer k for which Alicehas a winning strategy when playing the graph coloring game on G with k colors. The problem ofdetermining the game chromatic number of several graph classes has attracted much interest in recentyears (see [3] for a comprehensive survey of this problem), with a particular focus on planar graphs [email protected] Univ. Bordeaux, CNRS, Bordeaux INP, LaBRI, UMR 5800, F-33400, Talence, France. (Herve.Hocquard,Eric.Sopena)@u-bordeaux.fr Department of Mathematics, Zhejiang Normal University, China. [email protected] see e.g. [8, 16, 17, 24–26]) for which the best known upper bound up to now is 17 [26]. Very recently,Costa, Pessoa, Sampaio and Soares [7] proved that given a graph G and an integer k , deciding whether χ g ( G ) ≤ k is a PSPACE-Complete problem, thus answering a longstanding open question.Most of the known upper bounds on the game chromatic number of classes of graphs are derivedfrom upper bounds on the game coloring number of these classes, a parameter defined through theso-called graph marking game , formally introduced by Zhu in [25]. This game is somehow similar tothe graph coloring game, except that the players mark the vertices instead of coloring them, withno restriction. The game coloring number col g ( G ) of G is then defined as the smallest integer k forwhich Alice has a strategy such that, when playing the graph marking game on G , every unmarkedvertex has at most k − monotonic , which means that col g ( H ) ≤ col g ( G ) for every subgraph H of G , while this property doesnot hold for the game chromatic number [23].Let G be a graph with col g ( G ) = k and consider the winning strategy of Alice for the markinggame on G . Applying the same strategy for the coloring game on G , Alice ensures that each uncoloredvertex has at most k − χ g ( G ) ≤ k . Hence, the following inequalitiesclearly hold for every graph G . Observation 1
For every graph G , χ ( G ) ≤ χ g ( G ) ≤ col g ( G ) ≤ ∆( G ) + 1 . In this paper, we introduce and study a new version of the graph coloring game (resp. of the graphmarking game), played on connected graphs, by requiring that, after each player’s turn, the subgraphinduced by the set of colored (resp. marked) vertices is connected. In other words, on their turn, eachplayer 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 markinggame ). We will denote by χ cg ( G ) the connected game chromatic number of a connected graph G , thatis, the smallest integer k for which Alice has a winning strategy when playing the connected graphcoloring game on G with k colors, and by col cg ( G ) the connected game coloring number of G , thatis, the smallest integer k for which Alice has a strategy such that, when playing the connected graphmarking game on G , every unmarked vertex has at most k − G . Observation 2
For every connected graph G , χ ( G ) ≤ χ cg ( G ) ≤ col cg ( G ) ≤ ∆( G ) + 1 . It is proved in [23] that for every positive integer n , χ g ( K n,n − M ) = n , where K n,n − M denotesthe complete bipartite graph with n vertices in each part, minus a perfect matching. We prove inSection 2 that χ cg ( G ) = 2 for every nonempty connected bipartite graph G , which shows, since thegraph K n,n − M is bipartite, that the difference χ g ( G ) − χ cg ( G ) can be arbitrarily large.One of the main open, and rather intriguing, question concerning the graph coloring game is thefollowing: assuming that Alice has a winning strategy for the graph coloring game on a graph G with k colors, is it true that she has also a winning strategy with k + 1 colors? We will prove in Section 2that the answer is “no” for the connected version of the coloring game. More precisely, we will provethat for every integer k ≥
3, there exist connected bipartite graphs on which Bob wins the connectedcoloring game with k colors, while Alice wins the connected coloring game with two colors on everyconnected 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 [6, 15], the surveillance game [9, 12], the graphsearching game [2, 4, 10], or Hajnal’s triangle-free game [20, 22]. However, to our knowledge, theconnected variant of the graph coloring game has not been considered yet.Our paper is organized as follows. We consider bipartite graphs in Section 2, and outerplanargraphs in Section 3. We finally propose some directions for future research in Section 4. Bipartite graphs
We consider the case of bipartite graphs in this section. We will prove that for every integer k ≥ k 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 connectedbipartite graph G with two colors: thanks to the connectivity constraint, the first move of Alice forcesall the next moves to be consistent with a proper 2-coloring of G . Theorem 3
For every connected bipartite graph G , χ cg ( G ) ≤ . Proof.
Let G be any connected bipartite graph. The strategy of Alice is as follows. On her firstmove, she picks any vertex v of G and gives it color 1. From now on, each play will color some vertexhaving at least one of its neighbors already colored, so that, since G is bipartite, this eventually leadsto a proper 2-coloring of G . (cid:3) However, for every integer k ≥
3, there are connected bipartite graphs on which Bob wins theconnected coloring game with k colors. Theorem 4
For every integer k ≥ , there exists a connected bipartite graph G k on which Bob winsthe connected coloring game with k colors. Proof.
Let H k be any C -free bipartite graph with minimum degree at least k and let A and B denote the partite sets of H k . (Consider for instance the incidence graph of a projective plane ofdimension d ≥ k ; such a graph is a ( d + 1)-regular bipartite graph with girth 6.) Let now G k be the(connected) bipartite graph obtained from H k by adding, for each k -subset S of B , a new vertex v S adjacent to all vertices of S .We now define the strategy of Bob for playing the connected coloring game on G k as follows. Inhis first moves (at most three, depending on the moves of Alice), Bob colors two vertices of A , say u and v , with two different colors. In his next two moves, Bob colors a neighbor u ′ of u in B with thesame color as v , and a neighbor v ′ of v in B with the same color as u . Since the minimum degree of H k is at least k and H k is C -free, Alice cannot prevent Bob from doing so.Now, Bob colors a k -subset X ⊆ N H k ( u ) ∪ N H k ( v ) containing u ′ and v ′ with k distinct colors.Again, Alice cannot prevent Bob from doing so since each move of Alice “forbids” at most k uncolouredvertices in N H k ( u ) ∪ N H k ( v ) (each vertex of G k has at most k neighbors in this set).After that, the vertex v X cannot be colored and Bob wins the game. (cid:3) We consider in this section the case of outerplanar graphs. An outerplanar graph is a graph that canbe embedded on the plane in such a way that there are no edge crossings and all its vertices lie onthe outer face. Recall that a graph is outerplanar if and only if it does not contain K or K , as aminor.Concerning the ordinary coloring game, Kierstead and Trotter proved in [17] that there existouterplanar graphs with game chromatic number at least 6, and Guan and Zhu proved in [13] thatthe game chromatic number of every outerplanar graph is at most 7. In [18] Kierstead and Yang haveproved that the bound is tight for the game coloring number of outerplanar graphs. We will provethat the connected game chromatic number of every outerplanar graph is at most 5 and that thereexist outerplanar graphs with connected game chromatic number 4.Recall that an outerplanar graph is maximal if adding any edge makes it non outerplanar. There-fore, an outerplanar graph is maximal if and only if, in all its outerplanar embeddings, all faces aretriangles, except possibly the outer face (an outerplanar graph is thus a triangulation of a plane cycle). V V u u ′ Etc. (a) v v ′ v v v p v k − v k (b)Figure 1: The structure of a maximal outerplanar graph. In particular, every maximal outerplanar graph is connected. Our goal in this section is to prove thatthe connected coloring number of every connected outerplanar graph is at most 5.When playing the connected coloring game on a connected graph G , we will say that an uncoloredvertex in G is saturated if each of the available colors appears in its neighborhood. Observe thatBob wins the connected coloring game on G if and only if he has a strategy such that, at some pointin the game, an uncolored vertex in G becomes saturated. Similarly, when trying to prove that theconnected game coloring number of some graph G is at most k , we will say that an unmarked vertexin G is saturated if it has at least k marked neighbors. Again, the connected game coloring numberof G is at least k + 1 if and only if Bob has a strategy such that, at some point in the game, anunmarked vertex in G becomes saturated.Finally, we will say that a vertex in G is playable if it is uncolored (resp. unmarked) and has atleast 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 k − triangle graph T ( G ) of amaximal (embedded) outerplanar graph G , whose vertices are the triangle faces of G , incident facesbeing linked by an edge, is necessarily a tree.Let G be a maximal (embedded) outerplanar graph. An edge belonging to the outer face of G isan outer edge of G . Let us choose and fix any outer edge e = uu ′ of G . The distance from any vertex v to the edge e is defined as d G ( v, e ) = min { d G ( v, u ) , d G ( v, u ′ ) } . For every integer i ≥
0, let V i denotethe set of vertices at distance i from e . Observe that the subgraph G [ V i ] of G induced by each set V i is a linear forest, that is, a disjoint union of paths, since otherwise G would contain a K , as aminor. In particular, G [ V ] is the edge uu ′ and G [ V ] is a path. Therefore, G can be viewed as a “treeof trapezoids”, as illustrated in Figure 1(a).Each of these trapezoids has the structure depicted in Figure 1(b). Both v and v ′ belong to some V i , i ≥ vv ′ being an edge-cut of G , while v , . . . , v k , k ≥
2, belong to V i +1 . The vertices v and v ′ arethe parents of the children vertices v , . . . , v k , the edge vv ′ is the root edge of the trapezoid, the uniquevertex v p , 1 ≤ p ≤ k , which is joined by an edge to both v and v ′ is the pivot of the trapezoid (its i a i b i b i a i b i Rule R1 (the vertex a i must be playable) b i b i a i b i b i a i Rule R2 ( a i is a pivot and at least one of the dashed edges incident with a i must exist) b i a i b i b i b i a i Rule R3 (the vertex a i must be playable)Figure 2: The strategy of Alice on outerplanar graphs (Rules R1, R2 and R3). uniqueness comes from the fact that G does not contain K as a minor). We will denote by T vv ′ thetrapezoid whose root edge is vv ′ , and by p ( vv ′ ) the pivot of T vv ′ . Note that each vertex v i , 1 ≤ i ≤ k ,is a neighbor of at least one of its parents, and that only the pivot v p is a neighbor of both its parents.Moreover, we will say that v i − , 2 ≤ i ≤ k , is the left neighbor of v i , while v i +1 , 1 ≤ i ≤ k −
1, is the right neighbor of v i (this ordering is well defined since the embedding of G is given).Observe that if there is no trapezoid of the form T vv ′ or T v ′ v in G for some vertex v , then the degreeof v is at most 4 (it is 4 only if v is the pivot of some trapezoid), so that v cannot be a threatenedvertex. 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 totalordering ≤ G of the vertices of G , obtained by listing the vertices of V from left to right, then thevertices of V from left to right, and so on. Finally, we will say that a vertex w , belonging to atrapezoid T v v ′ , lies above a vertex w , belonging to a trapezoid T v v ′ , if every shortest path from { v , v ′ } to { u, u ′ } goes through v or v ′ .We now describe the strategy of Alice when playing the connected marking game on a connectedouterplanar graph G . Let uu ′ be any outer edge of G , and G m be any maximal outerplanar graphcontaining G as a subgraph, and such that uu ′ is also an outer edge of G m . In the following, weassume that we are given a trapezoidal representation of G m , starting from the edge uu ′ , as describedabove. Moreover, we can also assume that for every trapezoid T vv ′ of G m , the pivot p ( vv ′ ) has beenchosen in such a way that it is linked by an edge in G to at least one vertex from { v, v ′ } . This willallow us to speak about children or parent vertices (with respect to G m ) even if the correspondingedges do not belong to G , and to use the ordering ≤ G m of the vertices of G .Let us denote by a i , i ≥
0, the vertex marked by Alice on her ( i + 1)-th move, and by b i , i ≥
1, the ertex marked by Bob on his i -th move, so that the sequence of moves (that is, marked vertices) is a , b , a , . . . , b i , a i , . . . Hence, a is the vertex marked by Alice on her first move and, for every i ≥ a i is the “response” of Alice to the move b i of Bob.The strategy of Alice will then consist in applying the first of the following rules that can beapplied (see Figure 2 for an illustration of Rules R1, R2 and R3) for each of her moves.R0: a := u .R1: If v is a playable unmarked parent of b i , then a i := v .R2: If b i belongs to a root edge vb i or b i v , v is marked and p ( vb i ) is playable, then a i := p ( vb i ).R3: If b i belongs to a root edge vb i or b i v , v is unmarked, v is a pivot and v is playable, then a i := v .R4: If none of the above rules can be applied, and there are still unmarked vertices in G , then welet a i := w , where w is the smallest (with respect to the ordering ≤ G m ) playable vertex.Note that on his first move, Bob must mark either the vertex u ′ , in which case Alice will applyRule R2 on her second move, or some neighbor v = u ′ of u , in which case Alice will apply rule R1and mark u ′ on her second move (recall that the edge uu ′ belongs to G ). Moreover, if Bob marks achild vertex w of some trapezoid T vv ′ , then at least one of v , v ′ must be marked (by the connectivityconstraint), and Alice will immediately apply Rule R1 if one of them is unmarked and vv ′ is an edgein G . These remarks are summarized in the two following observations. Observation 5
After the second move of Alice, both vertices u and u ′ are marked. Observation 6
After each move of Alice, if w is a marked child vertex of a trapezoid T vv ′ and vv ′ is an edge in G , then both v and v ′ are marked. We are now able to prove the main result of this section.
Theorem 7 If G is a connected outerplanar graph, then col cg ( G ) ≤ . Proof.
We assume that we are given an outerplanar embedding of G m and its trapezoidal represen-tation, as previously discussed. Clearly, it suffices to prove that if Alice applies the above describedstrategy, then, after each move of Alice, G contains no threatened vertex. This is clearly the case afterthe 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 b i and Alice has marked the vertex a i , i ≥ t is a threatened vertex in G , and that i is the smallest index with this property, whichimplies that a i or b i is a marked neighbor of t . Thanks to Observation 5, we know that both u and u ′ have been marked. Therefore, t is necessarily a child vertex of some trapezoid T vv ′ (we may have vv ′ = uu ′ ). Let t ℓ and t r denote the left and right neighbors of t (in G m ), if they exist. Note thatat least one of them must exist, since otherwise t would have at most two marked neighbors, andthus could not be a threatened vertex. Since t has four marked neighbors, at least one of t ℓ , t r mustbe marked, since otherwise no vertex lying below t could have been marked, due to the connectivityconstraint, so that, again, t would have at most two marked neighbors. Thanks to Observation 6, wethus get that both v and v ′ are marked if vv ′ is an edge in G .We now claim that neither T t ℓ t nor T tt r contains a marked child vertex which is a neighbor of t .Indeed, such a vertex, say w , cannot have been marked by Bob since, by Rule R1, Alice would havemarked t just after Bob had marked the first such child vertex of the corresponding trapezoid. Thevertex w has thus been marked by Alice which implies, since t is unmarked, that none of the edges tv , tv ′ , tt ℓ or tt r belong to G (otherwise t would have been marked in priority by Alice), and that w is the only marked neighbor of t , so that t cannot be a threatened vertex.Therefore, the four marked neighbors of w are necessarily v , v ′ , t ℓ and t r . Hence, t is the pivotof T vv ′ , which implies, since t is unmarked, that t ℓ has been marked after v , and that t r has been v v v v v Figure 3: An outerplanar graph with connected game chromatic number 4. marked after v ′ , so that b i ∈ { t ℓ , t r } . (Note here that we cannot have b i ∈ { v, v ′ } , since this wouldimply a i ∈ { t ℓ , t r } , contradicting the priority of rule R2.) But in each case, that is, b i = t ℓ or b i = t r , t 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. (cid:3) Concerning the connected game chromatic number of connected outerplanar graphs, we can nowprove the following.
Theorem 8 If G is a connected outerplanar graph, then χ cg ( G ) ≤ . Moreover, there exist connectedouterplanar graphs with χ cg ( G ) = 4 . Proof.
From Observation 2 and Theorem 7, we get χ cg ( G ) ≤ col cg ( G ) ≤
5. For the second partof the statement, consider the outerplanar graph G depicted in Figure 3. We first prove that Bobhas a winning strategy when playing the connected coloring game on G with three colors. Thanks tothe symmetries in G , and up to permutation of colors, Alice has three possible first moves, that weconsider separately.1. If Alice colors v with color 1, then Bob colors v with color 2. Now, if Alice colors v withcolor 3 then Bob colors v with color 1 so that v is saturated, while if Alice colors v or v withcolor 1 (resp. with color 3), then Bob colors v with color 3 (resp. with color 1), so that v or v is saturated.2. If Alice colors v with color 1, then Bob colors v with color 2, and the so-obtained configurationis similar to that of the previous case.3. If Alice colors v with color 1, then Bob colors v with color 2. Now, if Alice colors v withcolor 3 then Bob colors v with color 2 so that v is saturated, while if Alice colors v withcolor 2 (note that using color 3 would saturate v ), then Bob colors v with color 3, so that v is saturated.We thus get χ cg ( G ) ≥
4. To finish the proof, we need to show that Alice has a winning strategywhen playing the connected coloring game on G with four colors. Since v and v are the only verticeswith degree at least 4 in G , and since they are connected by an edge, Alice can color these two verticesin her first two moves. The remaining uncolored vertices can then always be colored since their degreeis less than the number of available colors. (cid:3) By Observation 2, the second part of the statement of Theorem 8 directly implies the following.
Corollary 9
There exist connected outerplanar graphs with col cg ( G ) ≥ . Discussion
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 connected outerplanargraph is at most 5, and that there exist infinitely many connected bipartite graphs on which Alicewins the connected coloring game with two colors but loses the game if the number of colors is atleast three.We conclude this paper by listing some open questions that should be considered for future work.1. What is the optimal upper bound on the connected game coloring number and on the connectedgame chromatic number of connected outerplanar graphs? We know that both these values areeither 4 or 5.2. What is the optimal upper bound on the connected game coloring number and on the connectedgame chromatic number of connected planar graphs?3. Does there exist, for every two integers k ≥ p ≥
1, a connected graph G k,p on which Alicewins the connected coloring game with k colors, while Bob wins the game with k + p colors?4. Is the connected game coloring number a monotonic parameter, that is, is it true that for everyconnected subgraph H of a connected graph G , the inequality col cg ( H ) ≤ col cg ( G ) holds?5. Does there exist a connected graph G for which χ g ( G ) < χ cg ( G )? or col g ( G ) < col cg ( G )? (Thatis, 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-CE25-0006project of the French National Research Agency, and the fourth author by Grant mumbers NSFC11971438, ZJNSF LD19A010001 and 111 project of the Ministry of Education of China.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 InternationalConference on Graph Theory (Bled’19), attended by the third and fourth authors, who warmlyacknowledge the organizers for having provided a very pleasant and inspiring atmosphere.
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