BICHROMATIC COLORING GAME ON TRIANGULATIONS

. In this paper, we study bichromatic coloring game on a disk triangulation, which is introduced by Aichholzer et al. in 2005. They proved that if a disk triangulation has at most two inner vertices, then the second player can force a tie in the bichromatic coloring game on the disk triangulation. We prove that the same statement holds for any disk triangulation with at most four inner vertices, and that the bound of the number of inner vertices is the best possible. Furthermore, we consider the game on topological triangulations


Introduction
As we know, there are a lot of two-player combinatorial games on graphs; for example, Generalized geography [9,12], Domination game [5], Distinguishing game [10] and so on.In particular, games on graphs related to graph colorings are well studied since the corresponding game invariants give us a good knowledge or evaluation for its original invariants; e.g. , Ramsey game [6], Graph coloring game [7] and On-line list coloring [13].On the other hand, there are not so many combinatorial games on geometric graphs.By this circumstance, in 2005, Aichholzer et al. [1] introduced many interesting combinatorial games on geometric planar triangulations, which can be classified into three types; the constructing, transforming and marking: In constructing type, starting from no edge of a finite planar point set, two players alternately add one or more edges to make a desired configuration.In transforming type, starting from a triangulation with all edges originally colored by black, two players alternately flip a black edge and color edges surrounding the flipped edge and itself for their purpose of the game.In marking type, similarly to the transforming type, two players alternately color by own colors one or more uncolored edges or points of a triangulation to make a desired colored configuration.
In this paper, we investigate one of games with marking type, called the bichromatic coloring game.In particular, this game does not require the geometric setting, and so we consider the game on disk triangulation which will be precisely defined in the next subsection.Then we improve the result in [1] for the bichromatic coloring game and it is best possible in terms of the number of inner points.In what follows, we introduce several definitions and our results.

Definitions and known results
A disk triangulation  is a simple graph embedded on the plane such that each finite face is triangular.Note that the (unique) outer face of  might not be triangular.A vertex  of a disk triangulation is inner if  does not lie on the boundary of the outer face, denoted by  , and let Ex( ) be the set of vertices lying on  .
The bichromatic coloring game on a disk triangulation is defined as follows.
Definition 1.1 (Bichromatic coloring game).There are two players ℛ and ℬ, starting with ℛ, and a disk triangulation  is given.At the beginning of the game, all edges are uncolored.Both players ℛ and ℬ alternately color an uncolored edge of  by red and blue, respectively.The first player who makes a monochromatic finite face by his own color wins the game.If all edges of  are colored without a monochromatic finite face, then the game ends in a tie.
The bichromatic coloring game can be regarded as a kind of the Ramsey game: In the Ramsey game, two players alternately color an edge by their own color to make/avoid a fixed monochromatic subgraph.In fact, considering the dual of the bichromatic coloring game (see Sect. 2 for the precise definition), the bichromatic coloring game of dual form can be regarded as the Ramsey game to make a monochromatic complete bipartite graph  1,3 .
In the bichromatic coloring game, ℬ can never win the game since otherwise ℛ can win the game by using the ℬ's winning strategy (by the strategy-stealing argument [11]).In more details, if ℬ can win the game for any strategy S of ℛ, that is, there is a sequence of edges according to S which leads ℬ to win by coloring those edges in order, then ℛ can win by coloring those edges in order according to S ′ where S ′ is a strategy (of ℬ) from the second move of S. (In this case, we can regard that ℬ passes the first move in S.) Moreover, every disk triangulation allows a 2-coloring of edges without a monochromatic finite face, where both colors are equally used (up to one edge for an odd number of edges) Lemma 4 of [1].Therefore, it is the most interesting problem on the bichromatic coloring game to clarify the sufficient condition for ℬ to force a tie in the game.For this problem, Aichholzer et al. proved the following theorem.
Theorem 1.2 (Theorem 6 in [1]).Player ℬ can force a tie in the bichromatic coloring game on every disk triangulation with at most two inner vertices.

Our contributions
In [1], they do not mention whether the number of inner vertices in Theorem 1.2 is the best possible or not.We improve the theorem with respect to the number of inner vertices, as follows.
Theorem 1.3.Player ℬ can force a tie in the bichromatic coloring game on every disk triangulation with at most four inner vertices.
The number of inner vertices in Theorem 1.3 is the best possible, as follows.
Theorem 1.4.For any integer  ≥ 16, there is a disk triangulation with  vertices and (at least) five inner vertices on which ℛ can win the bichromatic coloring game.
The remaining of the paper is organized as follows.In the next section, we prove Theorem 1.3, and then we prove Theorem 1.4 in Section 3. In final section, Section 4, we remark on the bichromatic coloring game in topological setting, that is, every face is triangular and each player can win the game by coloring all three edges of any face with own color.Figure 1.A disk triangulation  (left), the dual graph   (center), where  * denotes the edge corresponding to , and the resulting dual graph obtained from   after  is colored by ℬ (right).

Proof of Theorem 1.3
To handle the game easily, we define the dual form of the bichromatic coloring game.(The dual form of the bichromatic coloring game is also introduced in [1], Subsect.5.5.)Here we introduce some terms and notations in graph theory.For a graph , let  () and () be the sets of vertices and edges of , respectively.For an edge , each of  and  is called an end vertex.Two edges  and  of a graph are adjacent if they have a common end vertex.A cycle is a connected graph with all vertices of degree 2. A forest is an acyclic graph; a tree is an acyclic connected graph.A claw  1,3 is a tree of four vertices such that one vertex is adjacent to all other three vertices.If a graph  is drawn on the plane without crossing edges, then a facial cycle is a cycle of  bounding a 2-cell region that contains no vertex in its interior.
Let  be a disk triangulation and the vertices in Ex() = { 0 ,  1 , . . .,  −1 } are labeled in clockwise order.The dual graph   of  is defined as follows: - (  ) is the union of the set  1 of finite faces of  and the set  2 = { 0 ,  1 , . . .,  −1 } of  vertices which are the edges of the outer cycle.-Two vertices in  1 are adjacent if the two corresponding faces share an edge in  , and  ∈  1 and   ∈  2 are adjacent if the face corresponding to  has    +1 on its boundary, where subscripts are modulo .
Note that each vertex of   is of degree 1 or 3 and that an inner vertex in  corresponds to a finite face of   .The dual form of the bichromatic coloring game is defined as follows.
Definition 2.1 (Dual form of the bichromatic coloring game [1]).Let   be the dual graph of a given disk triangulation  .Two players ℛ and ℬ, starting with ℛ, alternately choose an uncolored edge  ∈ (  ) and then; ℛ colors  by red, and ℬ deletes  and then splits each end of  of degree 3 into two vertices of degree 1 as shown in the right of Figure 1; in this case, if an end of  of degree 1, then we delete it instead of splitting it.(In the game on  , if ℬ colors an edge  ∈ ( ), then two triangular faces sharing  are not used for ℛ to win the game.In addition to this, to apply component-wise arguments to the dual graph, we adopt the above splitting operation instead of coloring by blue or using vertices of degree 2 in the dual graph.)The game stops if no more legal moves are possible.If ℛ makes a claw with all edges red, then ℛ wins the game.Otherwise, the game ends in a tie.
In the remaining of this section, we consider the dual form of the bichromatic coloring game in the dual graph of a disk triangulation.For the sake of simplicity, we say that ℛ (ℬ) colors (deletes) an edge  ∈   if ℛ (ℬ) colors the edge corresponding to  in the original game of  .

Observation 2.2 (Dual graph during the game).
Let  be a disk triangulation and suppose that several edges of  are already colored by ℛ and ℬ in the game.Let  be the sequence of moves of both players.Then, by Definition 2.1, the dual graph   of  is the graph obtained from the dual graph of  at beginning of the game (i.e., all edges of  are not colored) by a sequence of coloring/deleting edges corresponding to .Definition 2.3 (Goodness).A connected component  of a dual graph is good if ℬ can force a tie in the game on .In particular, if every component of a dual graph is good, then the dual graph is good.
We first prepare the following two lemmas.
Lemma 2.4.Let  be a disk triangulation with at most one inner vertex, and let   be the dual graph.Suppose that there is at most one red edge  ∈ (  ) which does not lie on a cycle and that all edges of   except  are uncolored.Then   is good.
Proof.Let  and   be as in the statement of the lemma.First suppose that  has no inner vertex, i.e.,   is a forest.In this case, regardless of ℛ's move, there is at most one component having two red edges after ℛ's move.Thus, by suitably deleting an edge of such a component, ℬ can transform the dual graph into a forest such that each connected component is a tree with at most one red edge.Therefore ℛ can never win the game.
Next we suppose that  has exactly one inner vertex, i.e.,   has exactly one cycle .Moreover, we may assume that exactly one edge  ∈ (  ) ∖ () is already colored by red.Let  ∈ (  ) be an edge colored by ℛ.Let  =  0  1 . . . ℓ be a shortest path between  and  , where   is an edge for each  ∈ {0, 1, . . ., ℓ},  0 =  and  ℓ =  .If ℓ = 1, then ℬ deletes the third edge sharing a vertex with  and  .Otherwise, ℬ colors an edge Case (iii).If  ̸ =   , then we can prove the case similarly to Case (i).(Note that if  ∈ (  ) ∖ {  }, then ℬ can leave only good components by deleting   .)Thus, we may assume that  =   .
In this case, ℬ first deletes an edge in   incident to  which lies on the path between   and   passing no edge of   ; see Figure 3.If the resulting dual graph  has only one cycle (  ), then we are done.Thus, we may suppose that two facial cycles   and   remain in .
Let  be the path in   between   and   , and let  be the edge of  colored by ℛ at his next move.There are four possibilities: (1)  is   or   .
It is easy to see that in the cases ( 1) and (4) (case (2)), ℬ can leave only two good components by deleting   (an edge of   adjacent to   ).In the case (3), ℬ deletes an edge in ( ) ∪ (  ) ∪ ( , ) ∖ {  } adjacent to  so that all red edges belong to distinct components each.After that, exactly one of the resulting connected components might not be yet good, i.e., it has two facial cycles and exactly one colored edge which is not lying on any cycle.Since the order of such a component is smaller than the original one and the graph is finite, one of the cases (1), ( 2) and ( 4) eventually occurs.This completes the proof of the case.Case 3.  induces a claw  1,3 .
Without loss of generality,  is the center vertex of   component) after his first move.Therefore, we can prove this case similarly to Cases 2 and 3, which completes the proof of the theorem.

Proof of Theorem 1.4
To prove Theorem 1.4, we show that there exists a disk triangulation with five inner vertices such that ℛ can win the game, as follows.
Lemma 3.1.Player ℛ can win the bichromatic coloring game on the disk triangulation shown in Figure 4.
We first introduce definitions used to prove the theorem.A ℛ  -face (ℬ  -face)  is a finite face of a disk triangulation such that exactly  ≥ 1 edges of its boundary cycle are colored by red (blue).and all other edges are not colored.A finite face is full-colored if all edges of its boundary are colored.Note that ℛ (ℬ) can win the game if there exists an ℛ 2 -face (ℬ 2 -face) in the graph at his turn.Conversely, ℬ (ℛ) must color the uncolored edge of an ℛ 2 -face (ℬ 2 -face) if it exists at his turn.Thus, when no ℬ 2 -face exists, if ℛ can make two ℛ 2 -faces such that their uncolored edges do not coincide, then ℛ can win the game.In this sense, we call a pair of such two ℛ 2 -faces a winning pair.
Proof of Lemma 3.1.Let  be the disk triangulation shown in Figure 4. We show that there is a winning strategy of ℛ for the game on  .First, ℛ colors the edge .If ℬ does not color any uncolored edge lying on the boundaries of two faces sharing , then one of two sequences  13 ,  12 ,  11 ,  10 and  ′ ,  ′ 2 ,  ′ 1 ,  is a winning sequence.Thus, by symmetry, we may suppose that ℬ first colors  or  ′ .
Suppose the former, i.e., ℬ colors  .(We can similarly prove the case when ℬ colors  ′ .)Then ℛ colors  ′ at his next turn.If ℬ colors  5 ( 7 ), then the sequence  13 ,  12 ,  11 ,  9 ,  8 ,  7 ,  6 ,  3 ( 13 ,  12 ,  11 ,  9 ,  8 ,  4 ,  5 ) is a winning sequence.Otherwise, ℬ necessarily colors  1 or  ′ 1 to make a ℬ 2 -face as discussion in the first paragraph.It is important to note that after ℛ colors the uncolored edge ( 1 or  ′ 1 ) of a ℬ 2 -face, there are always at least two edge-disjoint winning sequences.For example, if ℬ colors  1 and ℛ colors  ′ 1 , then two sequences  13 ,  12 ,  11 ,  9 ,  8 ,  4 ,  5 and  ′ ,  ′ 2 are winning sequences.Thus, even if ℬ continues to color an edge with producing a ℬ 2 -face, he can never break two winning sequences simultaneously.Moreover, such ℬ's coloring will be not applicable eventually; in fact, he colors an edge of a face containing  or  ′ at last in the above coloring (with continuing to make a ℬ 2 -face).Therefore, ℛ can win the game using one of winning sequences.
By the proof of Lemma 3.1, we see that for any integer  ≥ 16, there is a disk triangulation with  vertices and five inner vertices since ℛ can win the game on the disk triangulation regardless of the degree of the leftmost inner vertex in Figure 4 if the degree is at least 5. Similarly, by adding inner vertices into the leftmost finite face  in Figure 4 and joining the added inner vertices to vertices on the boundary of  , we can increase the number of inner vertices.Therefore, Lemma 3.1 implies Theorem 1.4.

Remark on topological setting
It is natural to extend the bichromatic coloring game on a disk triangulation to that on a topological triangulation, i.e., a simple graph which is 2-cell embedded on a closed surface with each face triangular.In the topological version of the game, for every face  of a topological triangulation, ℛ wins the game when  becomes monochromatic by his own color.Similarly to the original version, it seems to be hard to determine whether ℛ wins the game or not.However, using the topological condition, we could find a good structure for ℛ, as follows.
Let  be a triangulation on a surface, and suppose that several edges of  are already colored by ℛ and ℬ in the game on .An edge  of  is critical if  is shared by two ℛ 1 -faces, i.e., if ℛ colors , then he can make a winning pair.A ℬ-chain  1 = ,  2 , . . .,   =  ′ from an edge  to another  ′ is a sequence of blue edges in which for each  ∈ {1, 2, . . .,  − 1},   and  +1 are shared by a full-colored ℬ 2 -face.In other words, a ℬ-chain can be obtained from a ℬ 1 -face with two non-colored edges (where  is one of them) by successively coloring by ℬ to make ℬ 2 -faces in each move (until he colors  ′ ).Proof.Let ( ) be the 4-region consisting of two faces of  sharing an edge  , that is, ( ) is a graph obtained from the complete graph of order 4, denoted by  4 , on the plane by removing an edge.Suppose that ℛ first colors .Then ℬ colors an uncolored edge of (), say , after that, ℛ colors  ′ .(The case when ℬ colors  can be similarly proved.)Note that ℬ needs to color (at least) an uncolored edge of ( ′ ) to avoid that ℛ wins the game by coloring successively edges incident to  ′ or  ′ .Moreover, if ℬ colors an uncolored edge of ( ′ ) in his second turn, then ℛ can win the game by the similar way in the proof of Lemma 3.1.Thus, ℬ has just one chance of making a ℬ-chain starting from  to some edge on ( ′ ) and he needs to make it for coloring an uncolored edge of ( ′ ).
By the above facts, it suffices to consider the following two cases.
Case 1. ℬ colors an edge in ( ′ ) via making a ℬ-chain starting from  which passes only edges of the sub-map.We first suppose that ℬ colors  ′ (in his second turn).In this case,  3 ,  is a winning sequence after ℛ naturally colors  ′ .Thus, ℬ has to color both an edge in ( ′ ) and an edge around  3 and  to obstruct that ℛ wins.Observe that the ℬ-chain from  passing  ′ can reach either  ′  ′ or  ′  ′ without making a ℛ 2 -face.So, after ℬ extends ℬ-chain to either edge, he colors an edge around  3 and  to break the winning sequence.However, there remains another winning sequence around  ′ ; for example, if the ℬ-chain is ,  ′ ,  ′  ′ ,  ′  ′ , then  ′  2 ,  ′  ′ is a winning sequence.
Next suppose that ℬ colors  ′ (in his second turn).If ℬ colors  3 ,  and one edge of ( ′ ) in this order, then there is a winning sequence from  3 to  ′ .Otherwise, he colors ) and then colors an edge around  (ℬ and ℛ alternately color  ′  ′ ,  ′  ′ ,  ′ ).In each case, there remains a winning sequence from  ′  2 to  ′  ′ (from  3 to ).Case 2. ℬ colors an edge in ( ′ ) via making a ℬ-chain starting from  which passes a face of the outside of the sub-map.
There are two possibilities of ℬ-chains going to the outside without making ℛ 2 -faces: ( In the case (1), since there are two winning sequences  ′  ′ ,  ′  ′ and  ′  ′ ,  ′  ′ , and a critical edge , ℬ has to extend the ℬ-chain back to the inside of the sub-map (without making ℛ 2 -faces) so that either (1-a) one of edges of ( ′ ) is colored as the end of the ℬ-chain, or (1-b)  3 or  4 is colored as the end of the ℬ-chain.
In the case (1-a), the ℬ-chain is extended to the desired edge passing  1  2 or  1  4 .After this extending, ℬ colors  (or an edge around it).In this case, ℛ can win the game since at least one of the two winning sequences and the critical edge remains (it is possible that an edge of such winning sequences becomes a critical edge).In the case (1-b), the ℬ-chain must pass  3  4 .Similarly to the case (1-a), at least one of the two winning sequences and the critical edge remains or a critical edge  remains.Thus, ℛ wins the game in either case.
In the latter case (2), after ℬ colors  1  4 (and ℛ colors  1 ), there remain three critical edges  ′  ′ ,  ′  ′ and .It is clear that ℬ cannot simultaneously break all of them, and hence, ℛ can win the game.This completes the proof of the theorem.
Furthermore, we found two triangulations of small orders containing no sub-map shown in Figure 5; see Figures 6 and 7, where ℬ (resp.ℛ) can force a tie (resp.win) in the game on the graph shown in Figure 6 (resp.7).Note that ℬ can force a tie in the game on the complete graph  4 on the plane.In general, few triangulations are known where ℬ can force a tie in the game.So the following is a challenging open problem:   is added.For a graph  on a surface and an integer  ≥ 0, let   () denote the graph obtained from  by applying  face subdivisions, where  0 () = .We conclude the paper with proposing the following problem which is more reasonable than Problem 4.2.

Figure 4 .
Figure 4.A disk triangulation with five inner vertices.

Theorem 4 . 1 .
Let  be a triangulation on a surface.If  contains the sub-map shown in Figure5, then ℛ wins the bichromatic coloring game on .

Problem 4 . 2 .
Characterize all triangulations on surfaces where ℬ can force a tie the bichromatic coloring game.The graph shown in Figure6is obtained from  4 by a single face subdivision, where a face subdivision inserts a vertex into each face and joins the added vertex  and all vertices of the boundary of the face to where
Let  be as in the statement of the lemma and let   be the dual graph of  .By Theorem 1.2, we may suppose that  has exactly three inner vertices, and then let ,  and  be the three inner vertices.We denote by   ,   and   facial cycles of   corresponding to ,  and , respectively.We show that ℬ can leave only good components after several moves.By symmetry, it suffices to prove the following four cases.Case 1.  has a face , i.e.,   and   share an edge   for each pair (, ) ∈ {(, ), (, ), (, )}, where   denotes the edge of   corresponding to an edge  of  .Regardless of ℛ's first move, ℬ deletes one of   and   .It is clear that the resulting dual graph has no cycle, and hence, it is good by Corollary 2.5.Case 2.  forms a path of length 2, i.e.,   and   share an edge   for each pair (, ) ∈ {(, ), (, )} but   and   do not share any edge.Case 3.  and  are adjacent but  is adjacent to neither, i.e.,   and   share   but   does not share any edge with   and   .By symmetry, we let   =  0  1 . . . ℓ be the path in   joining two cycles   and   , where   is an edge and exactly one of ends of  0 ( ℓ ) is contained in   (  ); see Figure 2. If  ∈ (  ), then ℬ deletes  ℓ ; hence, by Corollary 2.5 (1) and (2),   is good.If  / ∈ (  ), then ℬ deletes an edge adjacent with   ; hence, by Corollary 2.5 (2),   is good.Case 4. ,  and  are not adjacent to each other, i.e.,   ,   and   do not share any edge each other.Based on the above cases, it is easy to see that ℬ can leave two good components for any ℛ's first move.As in the proof of Lemma 2.6, it suffices by symmetry to consider the following four cases.Throughout the proof, let  denote the edge of   colored by ℛ at his first move.Case 1.  has three faces ,  and , i.e.,   and   share an edge   for each pair (, ) with ,  ∈ {, , , }, where   denotes the edge of   corresponding to an edge  of  .Let  = {  ,   ,   ,   ,   ,   } ⊂ (  ).If  ∈  , then ℬ deletes one of   ,   ,   .Regardless of ℛ's move, the resulting dual graph has only one cycle  with  / ∈ (), and hence, it is good by Corollary 2.5.If  / ∈  , then ℬ can leave only good components by deleting   ,   or   according to ℛ's first move; for example, ℬ deletes   if  ∈ (  ) ∖  .Case 2.  has a face , i.e.,   and   share an edge   for each pair (, ) with ,  ∈ {, , }.By Case 1, it suffices by symmetry to consider three subcases; (i)  is not adjacent to any vertex in {, , }, i.e.,   does not share any edge with   ,   and   .(ii)  is a face of  , i.e.,   shares   and   with   and   , respectively, but does not share any edge with   .(iii)  is adjacent to only , i.e.,   shares   with   , but does not share any edge with   and   .Case (i).By symmetry, we let   =  0  1 . . . ℓ be the path in   joining two cycles   and   , where exactly one of ends of  0 ( Figure 3.The ℬ's first move in Subcase (iii).Case (ii).Let  ′ = {  ,   ,   ,   ,   } ⊂ (  ).If  ∈  ′ with  ̸ =   ( =   ), then by deleting   (  ), ℬ can leave only one good component with a cycle not containing .If  / ∈  ′ , then ℬ can leave only one good component by deleting  or  according to ℛ's first move; for example, ℬ deletes   (resp. ℓ − 1} :  +1 ∈ ()} otherwise.In each case,  and  belong to distinct components in the dual graph and they do not lie on a cycle since  / ∈ (), that is, each component is good.Therefore ℛ can never win the game.By Theorem 1.2 and Lemma 2.4, we have the following.Corollary 2.5.A component  in a dual graph is good if (1)  has no colored edge and at most two facial cycles, or (2)  has at most one facial cycle  and at most edge  ∈ () ∖ () is red.Lemma 2.6.Let  be a disk triangulation with at most three inner vertices.Then the dual graph   is good.Proof.If ℛ colors an edge  which is neither   nor   , then ℬ can delete   or   so that  is not contained in any cycle of the resulting dual graph.Thus, without loss of generality, we may suppose that ℛ first colors   .In this case, ℬ deletes an edge adjacent to   .Since the resulting dual graph has at most one cycle  and   is not contained in , it is good by Corollary 2.5.Now we shall prove Theorem 1.3.Proof of Theorem 1.3.Let  be a disk triangulation with at most four inner vertices and let   be the dual graph of  .By Lemma 2.6, we may suppose that  has exactly four inner vertices, and let  = {, , , } be the set of four inner vertices.Let   ,   ,   and   be facial cycles in   corresponding to , ,  and , respectively, and let  = (  ) ∪ (  ) ∪ (  ) ∪ (  ).ℓ ) is contained in   (  ).If  ∈ (  ) ∪ (  ), then ℬ can transform   into a union of two good components by deleting  0 or an edge of   adjacent to  ℓ .If  / ∈ (  ) ∪ (  ), then ℬ can transform   into a graph with exactly one cycle  and  / ∈ () by deleting   or   .Therefore ℬ can leave only good components in each case.
1,3 .If  / ∈  , then ℬ can leave only good components by deleting one of   ,   and   to separate  and components with cycles.Moreover, if  ∈  but  / ∈ (  ), then we are done similarly to the above.Thus we may assume  ∈ (  ).If  / ∈ {  ,   ,   } ( ∈ {  ,   ,   }), then by deleting one in {  , ,   } (an edge of   adjacent to ), ℬ can leave a component with two facial cycles and exactly one colored edge not on any cycle (and a good component).Note that the remaining two facial cycles do not share an edge in this case, but we can prove the case similarly to Subcase (iii) in Case 2. Case 4.  induces a forest in which each component is a path of order at most 4.
It is easy to check that regardless of ℛ's first move, ℬ can leave either only good components or a component with at most two facial cycles and exactly one colored edge not on any cycle (and a good