Eric Duchêne

Arthur Dumas

Nacim Oijid

Aline Parreau

Eric Rémila

The Maker-Maker domination game in forests *

Keywords: Positional Games, Maker-Breaker Games, Maker-Maker Games, PSPACE-completeness, domination game, Dominating set, Graphs

We study the Maker-Maker version of the domination game introduced in 2018 by Duchêne et al. Given a graph, two players alternately claim vertices. The first player to claim a dominating set of the graph wins. As the Maker-Breaker version, this game is PSPACE-complete on split and bipartite graphs. Our main result is a linear time algorithm to solve this game in forests. We also give a characterization of the cycles where the first player has a winning strategy.

Introduction

Positional games have been introduced successively by Hales and Jewett in [START_REF] Hales | Regularity and positional games[END_REF] and by Erdős and Selfridge in [START_REF] Erdös | On a combinatorial game[END_REF], and then widely studied in the literature (see the two books [START_REF] Beck | Combinatorial Games: Tic-Tac-Toe Theory[END_REF][START_REF] Hefetz | Positional Games, volume 44 of Oberwolfach seminars[END_REF] for an overview). They are played on an hypergraph of vertex set X, with a finite set F ⊆ 2 X of hyperedges. The set X is often called the board of the game, and an element of F a winning set. The game involves two players that alternately claim a previously unclaimed vertex of X. The winner of the game is defined according to a convention. The original one is called Maker-Maker (or strong convention), where both players have the same objective, i.e. filling a whole winning set with their own claimed vertices. Such games may end in a draw if each winning set contains one vertex claimed by each player. When considering positional games, the main issue consists in determining which player has a winning strategy. In particular, it is well-known that in the Maker-Maker convention, the second player has no winning strategy. Thus, resolving a Maker-Maker game consists in determining whether the first player has a winning strategy or whether it ends in a draw. As mentioned in [START_REF] Hefetz | Positional Games, volume 44 of Oberwolfach seminars[END_REF], despite this result, this convention has not been widely studied in the literature. The main reason is due to the hardness to tackle it, as the first player tries at the same time to fill a winning set while considering all the threats of his opponent. As a consequence, Maker-Maker instances often satisfy the so-called extra set paradox, which claims that adding new winning sets in the hypergraph is not necessarily better for the first player.

For all these reasons, the convention that has been mainly studied in the literature is the Maker-Breaker convention. In this convention, both players, called Maker and Breaker, have opposite objectives: Maker aims at filling a winning set while Breaker prevents her to do so. Constructive results are generally more affordable in this convention. In addition, the extra set paradox does not exist when playing in the Maker-Breaker convention: adding new winning sets is always better for Maker.

In the literature, there are many graph optimization problems that have been turned into positional games. One can cite for example the clique game, the connectivity game, the H-game [START_REF] Beck | Combinatorial Games: Tic-Tac-Toe Theory[END_REF], or the domination game [START_REF] Duchêne | Maker-Breaker domination game[END_REF]. If a large part of the studies is devoted to the case where they are played on complete or random graphs, there is a more recent approach that consists in playing such games on a general graph. It generally yields to algorithmic results, both in terms of hardness proofs or the construction of polynomial time algorithms to compute the winner of the game [START_REF] Bonnet | On the complexity of connection games[END_REF][START_REF] Duchêne | Complexity of maker-breaker games on edge sets of graphs[END_REF]. One can also find results about the parameterized complexity of such games [START_REF] Bonnet | The parameterized complexity of positional games[END_REF]. In addition, general algorithmic results about Maker-Breaker games played on k-uniforms hypergraphs (i.e., all the winning sets are of size k) have been given by Rahman and Watson [START_REF] Md | 6-uniform maker-breaker game is pspace-complete[END_REF] and Galliot [START_REF] Galliot | Structural tools for the maker-breaker game. application to hypergraphs of rank 3: strategies and tractability[END_REF]. Indeed, they respectively proved that determining the winner of a Maker-Breaker positional game is PSPACE-complete when k = 6 and polynomial when k = 3.

When switching to the Maker-Maker convention, by putting together the above result of Rahman and Watson with an argument of Byskov [START_REF] Makholm | Maker-Maker and Maker-Breaker games are PSPACE-complete[END_REF], it has been derived that Maker-Maker games are PSPACE-complete on 7-uniforms hypergraphs. Beyond this result, connections between the two types of conventions are not well established. Indeed, there are very few algorithmic results, even for particular positional games derived from optimization problems. The objective of this paper is to investigate the Maker-Maker domination game, by highlighting the similarities, the implications, and the differences with the results known in the Maker-Breaker convention. It is known that the Maker-Breaker domination game is PSPACE-complete even for bipartite graphs and split graphs, and polynomial for cographs and forests [START_REF] Duchêne | Maker-Breaker domination game[END_REF]. More precisely, it is shown that Maker has a winning strategy playing second on a forest if and only if it admits a perfect matching. When switching to the Maker-Maker convention, we will see that some of these complexity results still hold. Yet, in the case of forests, the polynomial characterization of the winning positions is far more complex than simply finding a perfect matching. The major result of the current paper consists in determining this characterization. To the best of our knowledge, this is the first time in the literature that a non-trivial algorithm is given to determine the winner of a game played according to the Maker-Maker convention.

The paper is organized as follows: in Section 2, we present the main definitions that will be useful in the Maker-Maker domination game. In Section 3, we give results about the Maker-Maker convention that are derived from the Maker-Breaker one, including the PSPACE-hardness result. Then, we fully solve the case of paths and cycles in Section 4. Sections 5 and 6 are devoted to the resolution of the Maker-Maker domination game on forests in linear time. As the proof is rather complex, we decided to split it into two parts, where Section 5 corresponds to the overview of the proof with the presentation of the algorithm, and Section 6 to the proof of the most technical elements.

Preliminaries

Standard definitions of graph theory

In this paper, we will only consider finite, undirected and simple graphs. A graph G is defined by a couple (V, E) where V denotes the set of vertices and E the edges of the graph. The (closed) neighborhood of a vertex x ∈ V , denoted by N [x], is the subset of vertices containing x and all the vertices that are adjacent to x. A vertex x is universal if N [x] = V . The degree of a vertex is the number of vertices adjacent to it. A leaf is a vertex of degree 1. If x is a vertex, G \ {x} denotes the graph on the vertex set V \ {x} with all the edges of E that are not incident to x.

Let X be a subset of vertices. X is a independent set if there are no adjacent vertices in X. X is a cutset if G \ X is disconnected. The subgraph of G induced by X, denoted by G[X], is the graph with vertex set X and edge set all the edges of E whose both extremities are in X. A matching M is a subset of edges that are two by two not incident. If |M | = |V |/2 (i.e. if all the vertices appear in some edge of M ) then M is perfect.

Let G 1 = (V 1 , E 1 ) and G 2 = (V 2 , E 2 ) be two graphs on disjoint vertex sets. The union of G 1 and G 2 is the graph

G 1 ∪ G 2 = (V 1 ∪ V 2 , E 1 ∪ E 2 ). The join of G 1 and G 2 is the graph on vertex set V 1 ∪ V 2 with edge set E 1 ∪ E 2 ∪ E × where E × = V 1 × V 2 .
A cograph is a graph that is either a single vertex or the union of two cographs or the join of two cographs. A bipartite graph is a graph whose vertex set can be partitioned into two independent sets. A split graph is a graph whose vertex set can be partitioned into two sets K and I where K induces a complete graph and I an independent set. A path is a graph whose vertex set is {v 1 , ...., v n } and v i v j is an edge if |i -j| = 1. A cycle is a path with the additional edge v 1 v n . A forest is a graph without any cycle. A tree is a connected forest.

A vertex x dominates a vertex y if y ∈ N [x]. A subset of vertices S dominates a vertex y if there exists x ∈ S that dominates y. A dominating set S of G is a subset of vertices that dominates all the vertices of the graph. The smallest size of a dominating set of G is denoted by γ(G).

The Maker-Maker domination game

The Maker-Maker domination game is played on a graph G = (V, E). Two players, Alice and Bob, alternately claim an unclaimed vertex of the graph, with Alice playing first. The game ends when the vertices claimed by one of players form a dominating set (in which case the corresponding player wins) or when all the vertices have been claimed and none of the players managed to claim a dominating set (in which case the game is a draw).

As a 2-player finite perfect information game, if both players play optimally, one of the players has a winning strategy or the game ends a draw. Furthermore, since this game is a Maker-Maker positional game, it is well known that the second player, Bob, does not have a winning strategy (see for example [START_REF] Hefetz | Positional Games, volume 44 of Oberwolfach seminars[END_REF]). As a consequence, there are only two possible outcomes for the Maker-Maker domination game played on G: either Alice has a winning strategy, which will be denoted by o(G) = A, or both players can ensure a draw, which will be denoted by o

(G) = D. The problem of deciding, given a graph G, if o(G) = A or o(G) = D is named the Maker-Maker Domination Game problem. Positions. A position P of the game is a triple (G, V A , V B ) where G = (V, E) is a graph and V A , V B ⊆ V are two subsets of vertices such that V A ∩ V B = ∅. The vertices in V A (respectively V B ) correspond to vertices claimed by Alice (resp. Bob). A vertex not in V A nor V B is unclaimed. If X is a set of vertices, the subposition induced by X is the position (G[X], V A ∩ X, V B ∩ X).
If {x, y} is a pair of unclaimed vertices of P . The position P x,y is defined by

P x,y = (G, V A ∪ {x}, V B ∪ {y}). A set of unclaimed vertices S is a winning set for Player t if S ∪ V t is a dominating set of G.
A pointed position is a position where the next player is specified, it will be denoted by a couple (P, t) where P is a position and t ∈ {A, B} is the next player (A stands for Alice and B for Bob).

The outcome o(P, t) of a pointed position (P, t) is defined for positions P satisfying the property that both V A and V B do not dominate G simultaneously. We have o(P, t) = A, if, the next player being t, considering that Alice has already claimed all the vertices V A and Bob has claimed vertices V B , there is a winning strategy for Alice, i.e. a strategy which ensures Alice to dominate the graph G before Bob. We will say in this case that the position is A-win. Otherwise, we say that the position is Draw and we denote it by o(P, t) = D. In this latter case, note that it also covers the case where Bob first dominates the graph, as the pointed position P may not be balanced depending on the sets V A and V B . Such cases will be considered in the upcoming proofs when considering the different possible sequences of moves (but their outcome will still be considered as a draw, as Alice will avoid them in her optimal sequence of moves).

Note that the starting position on the graph G is the pointed position ((G, ∅, ∅), A) and we have o(G) = o((G, ∅, ∅), A).

Ordering positions. Two positions P and P are said equivalent if for any t ∈ {A, B}, o(P, t) = o(P , t). Two pointed positions (P, t) and (P , t) with the same first player are said equivalent if o(P, t) = o(P , t).

In addition, and by analogy with combinatorial game theory, it is standard to order outcomes, stating that A > D. With this convention, a pointed position (P, t) is better for Alice (respectively Bob) than a pointed position (P , t ) if o(P, t) ≥ o(P , t ) (respectively o(P, t) ≤ o(P , t )).

As an illustration of these definitions, the two following observations ensure that:

it is always better for any player to start.

if two positions have exactly the same winning sets, there are equivalent.

Observation 2.1. For any position P , we have o(P, A) ≥ o(P, B).

The proof of the above result derives from a standard stealing strategy.

Observation 2.2. Let P = (G, V A , V B ) and P = (G , V A , V B ) be two positions such that there exists a bijection

f : V (G) \ V A ∪ V B → V (G ) \ V A ∪ V B
between the sets of unclaimed vertices that satisfies that S is a winning set for Alice (respectively Bob) if and only if f (S) is a winning set for Alice (respectively Bob). Then P and P are equivalent.

We say that a vertex x is forced for Alice (resp. Bob) in a position (P, A) (resp. (P, B), if whenever Alice (resp. Bob) claims a vertex y = x in P , the resulting position is Draw (resp. A-win). Sometimes, a move is always better than another. This is for example the case when the neighborhood of one vertex contained another neighborhood.

The following lemma gives a general formal framework which allows us to eliminate uninteresting moves. Note that this lemma or its derivatives are often used when studying positional games, sometimes not explicitly. For the sake of completeness, we give a proof here. Lemma 2.3. Let P = (G, V A , V B ) be a position. Let x, y be two unclaimed vertices. Assume that for both t ∈ {A, B}, N

[x] \ N [V t ] ⊆ N [y] \ N [V t ], where N [V t ] = ∪ v∈Vt N [v]
. Then there exists an optimal strategy in which y is claimed before x.

Roughly speaking, the condition says that y dominates a superset of the vertices dominated by x for both players. The conclusion says that if there exists an optimal strategy (for any player) starting by claiming x, one can assume that there also exists an optimal strategy starting by claiming y.

Proof. Let P = (G, V A , V B ) be a position. Suppose that Alice has a winning strategy S on P in which x is claimed before y. Up to consider more moves, consider a position P where the next vertex that will be claimed is x or y. If it is y, there is nothing to do, so suppose it is x.

If x is claimed by Bob, by definition any strategy is loosing for Bob, thus they are all equivalent, and he can claim y first instead without changing the outcome of the game. Therefore, we can suppose that Alice is going to claim x. Consider the following strategy S for Alice.

Instead of claiming x, claim y.

While Bob claims a vertex v = x, she claims the vertex w she would have claimed according to S if Bob has claimed v when she had claimed x.

If Bob claims x, she claims the vertex w she would have claimed according to S if Bob had claimed y.

As S was a winning strategy for Alice, at a certain moment of the game, she would obtain a set of vertices S A ⊃ V A that is a dominating set of G, while S B ⊃ V B , the vertices claimed by Bob, is not. If both x and y are in S A , then S will make both Alice and Bob claim exactly the same vertices, so Alice will win. Otherwise, by denoting by S A and S B the set of vertices claimed by Alice and Bob respectively after S , we have S A = S A \ {x} ∪ {y} and S B = S B \ {y} ∪ {x}. We prove that S A is still a dominating set and S B is not, which proves that S is a winning strategy for Alice. Indeed, for S A , if u is dominated only by x in S A , then it means that u /

∈ N [V A ] and thus u ∈ N [x] \ N [V A ]. Then by definition, u ∈ N [y] \ N [V A ]
and thus is dominated by y. For S B , assume that it is a dominating set. Since S B is not dominating, it means that there exists u

∈ N [x] not dominated in S B . Thus, u / ∈ N [y] but we also have u / ∈ N [V B ] as V B ⊂ S B . Hence, u is in N [x] \ N [V B ] but not in N [y] \ N [V B ], a contradiction.
The proof that if Bob has a drawing strategy is similar, by adding the fact that if Bob does not dominate the graph, then S A is not a dominating set, and thus neither is S A \ {y} ∪ {x}.

Union and decomposition of positions. Let P = (G, V A , V B ) and P = (G , V A , V B ) be two positions on disjoint sets of vertices1 . The union of P and P , denoted by P ∪ P , is the position

(G ∪ G , V A ∪ V A , V B ∪ V B ).
Note that one can remove a position where both players have a dominating set. It is a simple consequence of Observation 2.2 since the winning sets are the same.

Observation 2.4. Let P = (G, V A , V B ) and P = (G , V A , V B ) be two positions on disjoint graphs. Assume that V A and V B are dominating sets of G . Then P ∪ P and P are equivalent.

One cannot in general determine the outcome of a position P ∪ P knowing the outcome of P and P . However, when both positions are A-win when Bob starts, the union is still A-win: Proof. Alice follows her strategies as second player in both P and P until she dominates one of the component, say P . Then she just claims in P following her strategy. She might claim several times in a row in P (if Bob goes on claiming on P ), but by Observation 2.1, it is always better for Alice to play first than second. Thus, if she has to claim twice in a row in P , she will be able to dominate P , before Bob does. When considering a position, it could be useful to decompose it into several disjoint games. To do such a decomposition, the winning sets for both players should be the same. This is the case when a cut set is completely claimed and dominated by both players.

Lemma 2.6. Let P = (G, V A , V B ) be a position. Assume V (G) can be partitioned into three sets V 1 , V 2 , X such that:

There are no edges between V 1 and V 2 ;

All the vertices of X have been claimed:

X ⊆ V A ∪ V B .
The vertices in X are already dominated by V A ∩ X and V B ∩ X.

Let P 1 and P 2 be the subpositions of P induced by V 1 ∪X and V 2 ∪X respectively (vertices of P 1 and P 2 are disjoint). Then the position P and the position P 1 ∪ P 2 are equivalent.

Proof. We consider the trivial bijective map f between the unclaimed vertices of P and P 1 ∪ P 2 . Let S be a set of unclaimed vertices of P and t ∈ {A, B}. Assume first that S is a winning set of P for Player t Let S 1 = f (S) ∩ V 1 . We prove that S 1 is a winning set of P 1 for Player t. Let u be a vertex of P 1 and u = f -1 (u). Either u ∈ X and thus is dominated by a vertex of V t ∩ X in P and thus u is still dominated in P 1 . Or u / ∈ X, and then u is dominated by some vertex s of S ∪ V t in P . Since u / ∈ X and X is a cutset, s should be in V 1 ∪ X and f (s) is still in S 1 ∪ V t . Similarly, we can prove that S 2 = f (S) ∩ V 2 is a winning set of P 2 for Player t, and thus f (S) is a winning set of P 1 ∪ P 2 for Player t. The reverse is easier: the union of two winning sets in P 1 and P 2 is clearly a winning set of P . Therefore, Observation 2.2 applies and the two positions are equivalent.

Traps

We will frequently use the notion of trap that is defined in this section. Roughly speaking, a A-trap (respectively B-trap) is a vertex of a game position such that, if it is not claimed by Alice (resp. Bob) by the end of the game, means that Alice (resp. Bob) will never build a dominating set. Formally, traps can be defined as follows: Definition 2.7. Let P = (G, V A , V B ) be a position of the game. A A-trap (respectively B-trap) is an unclaimed vertex v such that there exists a vertex w with

N [w] ∩ V \V B = {v} (resp. N [w]\V A = {v})
In this definition, v corresponds to the vertex that must be claimed, and w to the vertex that will not be dominated if v is not claimed. Figure 1 illustrates the notion of traps.

The next lemma shows that if there is a A-trap in a position, one can consider that the next player will claim it immediately. Lemma 2.8. Let P be a position of the game and v be a A-trap of P . Claiming v is an optimal move for both players. Moreover, o(P, B) = D.

Proof. Let w such that N [w] ∩ V \V B = {v}. Suppose it is Bob's turn. By claiming v, Bob isolates the vertex w as now N [w] ⊆ V B . Therefore, Alice cannot dominate w any longer and the outcome is D.

If it is Alice's turn, if she does not claim v, Bob claims it and once again isolates w. Therefore, v is forced for Alice. Thus, v is an optimal move. Corollary 2.9. Let P be a position of the game. If there exist two A-traps v 1 and v 2 of P such that v 1 = v 2 , then o(P, A) = o(P, B) = D.

Proof. Since there are two distinct A-traps, the two vertices w 1 and w 2 that might be isolated are necessarily distinct (otherwise v 1 and v 2 would both be in their neighborhood, contradicting the definition of a trap). Hence, even if Bob is not the first player, he will be able to claim in one of the two traps and hence isolate either w 1 or w 2 .

The next lemma proposes another example of a forced move for Alice when there exists an unclaimed P 5 in a position as a subgraph.

Lemma 2.10. Let (P, B) be a pointed position of the game with

P = (G, V A , V B ). If there exists a path G = (v 1 , v 2 , v 3 , v 4 , v 5 ) such that G is a subgraph of G with V (G ) ∩ (V A ∪ V B ) = ∅ and v 2 , v 3 , v 4 of degree 2, then if Bob claims v 3 , Alice is then forced to answer on v 2 or v 4 .
Proof. If Bob claims v 3 , then if Alice answers elsewhere than on v 1 , v 2 or v 4 , Bob claims his second move on v 2 and creates two A-traps in v 1 and v 4 . Indeed, since the vertices v 2 and v 3 are of degree 2, there remains only one way for Alice to dominate v 2 (i.e. by claiming v 1 ) and v 3 (i.e. by claiming v 4 ). By corollary 2.9, the resulting position is Draw. If Alice claims v 1 , then Bob claims v 4 and by symmetry creates two A-traps equivalent to the previous case.

General results derived from the Maker-Breaker convention

Whereas the Maker-Maker domination game has not been studied yet, the Maker-Breaker version has been defined in [START_REF] Duchêne | Maker-Breaker domination game[END_REF]. In this section, we recall some results in this convention that have consequences for the Maker-Maker convention.

Basics

In Maker-Breaker games, only the winning condition differs from the Maker-Maker games. For more convenience and according to the terminology of [START_REF] Duchêne | Maker-Breaker domination game[END_REF], we will call the players Dominator and Staller. Dominator wins if she claims a dominating set, otherwise Staller wins. In particular, there is no draw. In this convention, since the two players have different roles, one need to precise who the first player is. Note that both players have interest to start since one can prove that if a player has a winning strategy playing second, he also has a strategy playing first.

When playing on the same graph, a winning strategy for Staller going second is a drawing strategy for Bob. Indeed, Bob will prevent Alice to claim a dominating set and thus Alice has no winning strategy. The contrary is not true since Bob can threaten Alice to create a dominating set in the Maker-Maker convention whereas Staller cannot. As a counter-example, we will prove that Bob has a drawing strategy on the cycle on ten vertices (see Theorem 4.7) whereas Dominator has a winning strategy for this graph [START_REF] Duchêne | Maker-Breaker domination game[END_REF]. However, strategies for Dominator can sometimes be used by Alice (see for example Lemma 3.4). Furthermore, if Alice manages to prevent Bob from claiming a dominating set, then she can play as Dominator in the Maker-Breaker convention in the rest of the game. Using this fact, one can prove that deciding the outcome in the Maker-Maker convention is as difficult as deciding the outcome in the Maker-Breaker convention. Theorem 3.1. Maker-Maker Domination Game is PSPACE-complete even if G is bipartite or split.

Proof. First note Maker-Maker Domination Game is in PSPACE, by application of Lemma 2.2 from Schaefer [START_REF] Thomas | On the Complexity of Some Two-Person Perfect-Information Games[END_REF], since there are most |V | turns and during each turn, at most |V | moves are available.

We prove that it is PSPACE-hard by a reduction from Maker-Breaker Domination Game: given a graph G and a first player (Dominator or Staller), who has a winning strategy ? Maker-Breaker Domination Game has been proven PSPACE-complete in [START_REF] Duchêne | Maker-Breaker domination game[END_REF], even if Staller starts, and even if the graph is split of bipartite.

We do the reduction as follows. Let G = (V, E) be a graph. Consider G = (V ∪ {v 0 }, E) with v 0 a new isolated vertex. Note that the property of being split or bipartite is maintained by this operation. We prove that Dominator wins the Maker-Breaker Domination Game on G going second, if and only if Alice wins the Maker-Maker Domination Game on G going first.

Suppose first that Dominator has a winning strategy S on G going second. We define the following strategy for Alice on G : first claim v 0 , then apply S. By hypothesis, this strategy is a winning strategy for Dominator, thus, the set of vertices claimed by Alice at the end of the game will dominate the graph. As Bob cannot dominate v 0 , he cannot dominate before her, thus Alice wins.

Reciprocally, suppose that Staller has a winning strategy S on G going first. We define the following strategy for Bob on G . If Alice does not claim first v 0 , claims it. Alice cannot dominate v 0 any longer, so the outcome is at least a Draw. Otherwise, apply S. By hypothesis, this strategy is a winning strategy for Staller, thus, the set of vertices claimed by Alice at the end of the game will not dominate G, and the outcome is Draw.

Deciding the outcome of the union of graphs when the outcome of each graph is known is trivial in Maker-Breaker convention (see [START_REF] Duchêne | Maker-Breaker domination game[END_REF]). Thus, to study a class of graphs, one just need to consider the connected case. In Maker-Maker, the situation is slightly different, one cannot in general say anything about the union.

Pairing strategies

A standard strategy in positional games is the pairing strategy. Let S be a set of disjoint pair of vertices. A pairing strategy using S consist in, whenever the opponent claims a vertex of a pair, answering by claiming the other one. This strategy ensures that the player who follows it takes at least one vertex in each pair, i.e. takes a transversal of S. For the Maker-Breaker domination game, this strategy can be used by Dominator if any transversal of S is a dominating set. This corresponds to the definition of a pairing dominating set.

Definition 3.2 ([6]). Let G = (V, E) be a graph. A subset of pair of vertices S = {(u 1 , v 1 ), ..., (u k , v k )} of V is a pairing dominating set if
all the vertices are distinct and if the intersections of the closed neighborhoods of each pair cover all the vertices of the graph:

V = ∪ k i=1 N [u i ] ∩ N [v i ].
In other words, any transversal of S is a dominating set.

Lemma 3.3 ([6]

). If G has a pairing dominating set, then in the Maker-Breaker convention, Dominator wins playing first or second.

A particular case of pairing dominating sets are perfect matchings. If G is a forest, both notions are equivalent. In general, the reverse of Lemma 3.3 is not true. However, for some classes of graphs like forests or cographs, the reverse holds: if Dominator has a winning strategy playing second (and thus first), then G admits a pairing dominating set [START_REF] Duchêne | Maker-Breaker domination game[END_REF]. In the Maker-Maker convention, Alice can also use pairing strategies. A first possibility for Alice is when there exists a pairing dominating set of size γ(G), the minimum size of a dominating set. Proof. Let G be a graph. Suppose G has a pairing dominating set of size γ(G). By claiming her γ first moves in it, and claiming in the same pair as Bob if Bob claims a first vertex in a pair, Alice can dominate in γ moves. Bob cannot dominate before by definition of γ(G). Therefore, Alice has a winning strategy in G.

This lemma can be applied in particular for connected cographs. Indeed, if G is a connected cograph with a universal vertex, then Alice wins at her first move. Otherwise, G has a pairing dominating set of size 2 and γ(G) = 2, thus by Lemma 3.4, Alice has a winning strategy. As mentioned previously, dealing with disconnected graphs is not easy in the Maker-Maker convention. Actually, we did not manage to determine the outcome of a general cograph and let it as an open problem. Note that determining the outcome for cographs in Maker-Breaker convention is polynomial (see [START_REF] Duchêne | Maker-Breaker domination game[END_REF]), but finding the minimum number of moves needed by Dominator to win is surprisingly open [START_REF] Gledel | Fast winning strategies for the maker-breaker domination game[END_REF].

Another application of pairing strategies will be when there is some B-trap in a position. If it is Alice's turn, she can claim the trap and then play any strategy of Dominator, like a pairing strategy. For that, we will use a pairing dominating set that are taking into consideration the vertices already claimed. We call such a pairing dominating set a A-pairing. Definition 3.5. Let G = (V, E) be a graph and If Alice can prevent Bob to dominate the graph, A-pairings are then enough for Alice. This can be used in particular where there are some B-traps in the game. Lemma 3.6. Let P be a position with a B-trap and a A-pairing. Then, we have o(P, A) = A.

P = (G, V A , V B ) a position. A set of disjoint pairs of unclaimed vertices S = {(u 1 , v 1 ), ..., (u k , v k )} of V is a A-pairing of position P if for each transversal T of S, the set V A ∪ T dominates V .
Proof. Let P be a position containing a B-trap v and a A-pairing S. Alice playing first can claim v. Now, Bob cannot dominate G anymore. Therefore, by following a pairing strategy using S, Alice will claim a transversal of S. Thus, she will dominate the whole graph and win. Lemma 3.7. Let P be a position with two B-traps v and v and a A-pairing that contains nor v nor v in P . Then, we have o(P, B) = A.

Proof. Alice follows a pairing strategy using S ∪ {v, v }. Bob cannot dominate since she will claim either v or v and thus, there will be a vertex not dominated by Bob. She will dominate the graph since she will claim a transversal of S.

Removing leaves

The key ingredient to solve trees in the Maker-Breaker convention is to remove leaves using the following lemma: Lemma 3.8 ( [START_REF] Duchêne | Maker-Breaker domination game[END_REF]). Let G be a graph, be a vertex of degree 1 and u be its unique neighbor. If u has degree 2, then G and G \ { , u} have the same outcome in Maker-Breaker convention: Dominator has a winning strategy in G if and only if she has a winning strategy in G \ { , u} (whoever is first).

Let T be a tree. One can apply successively Lemma 3.8 until obtaining a reduced tree T where the unique neighbor of any leaf has degree at least 3. Then there are only three possibilities for T . Either T is empty, which means that T has a perfect matching. In this case, Dominator wins playing first or second. The second possibility is that T is a single vertex or a star. Then the first player has a winning strategy. The last possibility is that T contains two disjoint cherries. A cherry is a vertex u connected to two leaves. In this last case, Staller wins playing first or second. This operation of removing leaves of a tree until obtaining a reduced tree correspond to the algorithm given in [START_REF] Duchêne | Maker-Breaker domination game[END_REF] to solve trees, and thus forests using unions of trees.

The situation is a quite different in the Maker-Maker convention. Indeed, Lemma 3.8 is not true anymore and one cannot reduce trees as in Maker-Breaker convention. Indeed, leaves are still playing an important role as shown by Figure 2. Actually, one can prove that it is always optimal for Bob to claim the unique neighbor of a leave: Lemma 3.9. Let P = (G, V A , V B ) be a position with an unclaimed leaf for which its unique neighbor u is also unclaimed. Then o(P, B) = o(P ,u , B)2 .

Proof. Suppose first that (P ,u , B) is Draw. Bob can, in (P, B), claims u first. Then is a A-trap and by Lemma 2.8, Alice must claim it. Thus, the game is now (P ,u , B) that is Draw by hypothesis. Thus, (P, B) is Draw.

We prove the other implication by contraposition. Suppose (P ,u , B) is A-win. Let S be a winning strategy for Alice in (P ,u , B). We consider a strategy S for Alice in (P, B) defined as follows:

If Bob claims a vertex in { , u}, Alice claims the other one.

Otherwise, Alice claims according to S (ignoring the moves on and u) until she dominates all the vertices of V \{ , u}. If u, have been claimed at this moment, either Bob has claimed u and Alice , and she wins since the strategy played is the same as S , or Bob has claimed and Alice u, which dominates more vertices than . Therefore, in both cases, her winning strategy in S ensures her that she dominates first. Otherwise, when she dominates all the vertices of V \{ , u}, Bob does not dominate in the game played on (P ,u , B). Thus, claiming u does not make him dominate G. Therefore, by claiming at her next move a vertex in { , u}, Alice wins.

Thus, Alice has a winning strategy in (P, B), finishing the proof.

In other words, after the first move of Alice, one can always assume that Bob has claimed all the vertices that are adjacent to leaves of G, and that Alice has answered by claiming all the leaves. This will be particularly important when dealing with trees.

Paths and cycles

In this section, we consider paths and cycles. The structure of the positions obtained after some moves will be basically union of paths where the extremities are claimed. Thus, we first deal with these paths and derive some general results on them that will help us to solve paths and cycles (and also forests in the next sections).

Bounded Paths

A bounded path is a path on at least four vertices where the four vertices at its extremities (the two leftmost and the two rightmost) are already claimed by Alice and Bob in such a way that the four vertices are already dominated by both players.

Definition 4.1. A bounded path of length n is a position (G, V A , V B ) such that: G is a path (v -1 , v 0 , v 1 , ...., v n , v n+1 , v n+2 ); the unclaimed vertices are exactly vertices v 1 to v n ; exactly one vertex among {v -1 , v 0 } (respectively {v n+1 , v n+2 }) is in V A , the other being in V B .
According to this definition, the knowledge of the label of v 0 and v n+1 is sufficient to deduce the label of v -1 and v n+2 . Therefore, for t, t in {A, B}, we will denote by [to n t ] the bounded path of size n such that v 0 ∈ V t and v n+1 ∈ V t . See Figure 3 for an illustration of [Ao 5 A] and [Bo 3 In some situations, bounded paths can be considered as a neutral structure that preserves the outcome when adjoined to another position. The next lemma illustrates a first case in which this may occur. It will also lead to a natural resolution of paths. Proof. Let P = (G, V A , V B ) be a position. We do the proof by induction on the number of unclaimed vertices in

A]. B A A B [Ao 5 A] A B A B [Bo 3 A]
P ∪ [Ao n B]. First note that if n ≤ 1,
P ∪[AoA]∪[Ao n-2 B], B), which itself is better than ([AoA]∪[Ao n-2 B], B
), as Alice already dominates P . Thus, by induction hypothesis applied with P = [AoA], which is not dominated by Bob, the position is A-win.

If v B is a vertex of [Ao n B], then v B corresponds to some vertex v k , with k ∈ {1, . . . , n}. If k = n, Alice answers v n-1 .
Otherwise, she claims v k+1 . Note that if k = n, the position obtained is better for Alice than the one obtained by k = n -1 (since she dominates a superset of vertices). Therefore, and without loss of generality, we will assume that k < n. By Lemma 2.6, the resulting position is then equivalent to

(P ∪ [Ao k-1 B] ∪ [Ao n-1-k B], B
) which has less unclaimed vertices than the original one. By induction hypothesis applied twice, it has the same outcome as (P ∪ [Ao k-1 B], B), which also has the same outcome as (P, B), which is A-win by hypothesis. This analysis ensures that o(P ∪ [Ao n B], B) = A, since for each claim of Bob, there exists an answer of Alice leading to an A-win position.

The next lemma presents another situation where the adjunction of some bounded paths with particular constraints does not change a winning outcome for Alice. We finish this subsection by proving that bounded paths [Bo n B], where n is odd, and [Ao n A], when n ≡ 0 mod 3, are not good for Alice when Bob starts. The first one is natural and give a condition for a Draw whatever the rest of the position is. The second one is more surprising. Here Bob obtains a Draw mostly by threatening Alice to dominate before her. Thus, one cannot add any position and maintain a Draw. If she claims v 6 , in the same way, the position is equivalent to ([Ao 4 B] ∪ [Ao 3(k-2) A], B), which is equivalent to ([Ao 3(k-2) A], B) which is Draw by induction.

Paths

Lemma 4.2 can be used to prove that Alice always wins on paths. Theorem 4.6. Let P n be the path of length n. Then o(P n ) = A.

Proof. Let v 1 , ..., v n be the vertices of the path. In n ≤ 3, then Alice wins at her first turn, so we can assume that n ≥ 4. Alice starts by claiming v 2 . By Lemma 3.9, we can assume that Bob claims v n-1 and Alice answers by claiming v n . Let (P, B) be the actual pointed position of the game. Using Observation 2.2, and since v 1 should be in any winning set of Bob, this position is equivalent to the position (

[AoA] ∪ [Ao n-4 B], B). By Lemma 4.2, o([AoA] ∪ [Ao n-4 B], B) = o([AoA], B) = A, which ensures that o(G) = A.
When playing in the Maker-Breaker convention, note that this result implies that Dominator also has a winning strategy on any path playing first. This result was already known since [START_REF] Duchêne | Maker-Breaker domination game[END_REF], but the strategy developed here is different from the other one.

Cycles

The case of cycles in the Maker-Maker convention is more subtle than for the Maker-Breaker convention, where Dominator always wins. More precisely, we will show that there are infinitely many A-win and Draw outcomes that depend on the size of the cycle modulo 3.

From now, we will denote by C n the cycle of order n. The vertices of C n will be denoted by v 0 to v n-1 . In all cases, the game either ends in a draw or Bob dominates the graph, thus o(C n ) = D.

We now prove the "only if" part. First consider that n ≡ 1 mod 3. Without loss of generality, one can assume that Alice will claim v 0 at her first turn, and thus we consider the pointed position (P, B) = ((C n , {v 0 }, ∅), B). Consider now the position P obtained from P by adding a vertex claimed by Bob adjacent only to v 0 . This position is better for Bob since he dominates more vertices than in P . Thus, it is enough to prove that (P , B) is A-win to ensure that (P, B) is A-win. Using Observation 2. It remains to prove that Alice wins on C 4 and C 7 . On C 4 , Alice wins by claiming any two vertices and as it is not possible to dominate in one move, she will dominate first. On C 7 , Alice can claim v 0 , and then use a pairing strategy with pairs (v 2 , v 3 ) and (v 4 , v 5 ). This way, she dominates the cycle in three moves, and Bob cannot dominate before since at least three vertices are required to dominate C 7 .

Forests

In this section, we give the essential elements to solve the case of forests, given by Theorem 5.1 below. In particular, we will first reduce the problem to any standard forest, meaning that non-standard forests correspond to cases that can be solved easily. Then we will give necessary conditions about the first move of Alice, yielding to the introduction of the skeleton of a forest. From that definition, we will be able to present the general algorithm that computes the outcome of any forest, as depicted by the decision tree of Figure 8. In the next section, we will give the full proof of the validity of the algorithm. Theorem 5.1. Deciding the outcome of a forest can be done in linear time.

Removing small components

If there is an isolated vertex in the forest, this is like playing in the Maker-Breaker convention. Lemma 5.2. Let F be a forest with an isolated vertex v 0 . Then o(F ) = A if and only if F \ {v 0 } contains a perfect matching. Proof. Alice has to claim first v 0 . Then Bob is playing first in F \ {v 0 }. Since he cannot dominate anymore, he has the same role as Staller in the Maker-Breaker Domination Game. In [START_REF] Duchêne | Maker-Breaker domination game[END_REF], it is proved that Dominator playing second in a forest in the Maker-Breaker Domination Game wins if and only if there is a perfect matching, which gives the result.

Isolated edges can be removed without changing the outcome. Proof. If Alice (respectively Bob) has a winning (resp. draw) strategy in F \ {u, v}, then she can apply her strategy in F by pairing u with v. The resulting strategy will still be winning (resp. leading to a draw) in F .

By applying Lemma 5.2 and Lemma 5.3, one can consider in what follows that all the connected components have at least three vertices.

Bottom-to-top strategies for Bob

In this subsection, we describe a strategy for Bob that will often be considered to obtain draws or to reduce trees. Let T be a tree rooted on a vertex r, vertices of T except r are labeled inductively with values 0 and 1, starting from the leaves as follows:

If all children of v are labeled 1, then v is labeled 0 (hence, all the leaves are labeled 0); If at least a child of v is labeled by 0, then v is labeled by 1. Figure 4 gives an example of such a labeling. Let T be a tree rooted in r and consider the pointed position (P, B) with P = (T, V A , ∅) and V A ⊆ {r}. A bottom-to-top strategy for Bob on (P, B) consists, at each step, in claiming a vertex v labeled by 1 such that all the successors of v labeled by 1 are already claimed by Bob. The following property is maintained during this process: Alice is forced to claim only vertices labeled by 0, and any vertex claimed by Alice (except r) has his parent claimed by Bob. Indeed, it is true before the first claim of Bob. Assume it is true before Bob's turn. Let v be a vertex labeled by 1 such that all the successors of v labeled by 1 are already claimed by Bob. By definition of the labeling and the assumption, v has a child u labeled by 0 that is unclaimed. All the children of u are by definition labeled by 1 and thus already claimed by Bob. Thus, u is a A-trap and Alice is forced to claim it, maintaining the property true. Such a strategy can also be applied when all the leaves of T are adjacent to vertices already claimed by Bob. In this strategy, Bob can claim all the vertices labeled by 1. A particular interesting case for Bob is when there exists a vertex v labeled by 1 with two children labeled by 0: Lemma 5.4. Let T be a tree and consider the labeling of T rooted in v 0 . If there exists a vertex labeled by 1 with two children labeled by 0, then the position (P, B) with P = (T, {v 0 }, ∅) is Draw.

Proof. Bob uses a bottom-to-top strategy on T . When claiming v, he will create two A-traps on the two children of v labeled by 0, which concludes the proof by Corollary 2.9.

Bob can also use bottom-to-top strategies to reduce the forest to a smaller one where he has a Draw strategy. Lemma 5.5. Let F be a forest and v 0 a vertex. Consider the labeling of F rooted in v 0 (for the components not containing v 0 , root on any vertex). Let v = v 0 be a vertex labeled by 1 and S v be the set of successors of v in the rooted tree. Let F v be the tree obtained by removing all vertices of S v and adding a leaf v connected to v.

Then, if o(F v , {v 0 , v }, {v}, B) = D, then o(F, {v 0 }, ∅, B) = D.
Proof. Bob follows by a bottom-to-top strategy on S v and can claim all the vertices of S v labeled by 1 until claiming v. Alice is always forced to claim a child of the claimed vertex. Afterward, Bob plays his strategy to obtain a Draw in (F v , {v 0 , v }, {v}, B), leading to a Draw for (T, {v 0 }, ∅, B).

Cherries

A particular case where the bottom-to-top strategy will be useful is when there are some cherries in the forest. Recall that a cherry is a vertex c connected to two leaves 1 and 2 . It will be denoted by the triple C = (c, l 1 , l 2 ). If F contains some cherries, the outcome of F can be easily computed. Suppose now that F has no such matching M . We define a strategy for Bob as follows. If Alice's first claim is not element of the cherry, then Bob claims c and creates two A-traps, leading to a Draw position. Thus, we can assume that Alice's first claim r is and element of {c, 1 , 2 }. Let T be the connected component of F containing r. If there exists another connected component T of F that has no perfect matching, then Bob can apply the strategy of Staller playing first in T that prevents Dominator to dominates T (see [START_REF] Duchêne | Maker-Breaker domination game[END_REF]). Thus, one can assume that there is a perfect matching in all the components of F distinct from T . Now label the vertices of T rooted in r as defined in Section 5.2. We want to prove that there exists a vertex labeled 1 with two children labeled 0, which will ensure a Draw strategy for Bob by applying the bottom to top strategy. If it is not the case, then consider the matching M where all the vertices labeled 1 (except c) are paired with their unique child labeled 0. We claim that M covers

V (T ) \ N [c]. Indeed, assume x ∈ V (T ) \ N [c
] is not covered by M . Then x must be labeled 0 and its parent should be r since it is the only vertex not labeled. But then x ∈ N [c]. Thus, there exists a matching in F \ {c} that cover V (F ) \ N [c]: take the union of M and the perfect matchings of all the other components.

Definition of the skeleton and easy cases

Considering Lemmas 5.6 and 5.7, we will assume now that there is no cherry in F , no isolated vertex and no isolated P 2 . From Lemma 3.9, one can assume that after the first turn of Alice, Bob will claim all the unclaimed vertices of F with a leaf as a neighbor. Alice is then forced to answer on each leaf. This motivates us to define the skeleton of F , denoted by S F , as the vertices that are not a leaf nor a parent of a leaf.

More formally, we denote by L F the leaves of F and by M F their parents. Remark that, with the hypothesis that there is neither cherry nor isolated vertex, we have L F ∩ M F = ∅ and the mapping : L F → M F , which associates to each vertex v of L F its parent, is bijective. Then, let S F be defined by S F = V (F ) \ (L F ∪ M F ). Figure 5 is an illustration of a forest with its skeleton.

In some simple cases, we can directly give the outcome of F .

S F

Figure 5: A forest F and its skeleton S F . Note that F is connected but S F is not. M F is the set of vertices in triangles, and L F is the set of vertices in squares. Proof. Alice claims the center c as a first move. Then as explained above, Bob claims all the vertices of M F and Alice answers all the leaves of L F . After that, Alice dominates the whole graph and Bob does not, since he does not dominate c.

We say that that F is standard if all its connected components have at least four vertices and a non-empty skeleton, if F has no cherries, and if S F does not induce a star of center c where c has no neighbor in M F . We now focus on the standard forests.

First move of Alice

If F is standard, next lemmas say that it can be assumed that Alice must claim outside S F and must connect it. The main idea behind this result is that, if Alice claims in S F , she claims too far from the leaves and Bob can win with a bottom-to-top strategy.

Lemma 5.11. Let F be a standard forest and v 0 ∈ S F . We have o((F, {v 0 }, ∅), B) = D.

Proof. Assume first that the graph induced by S F is either a unique vertex, a unique edge, or a star (whose center has necessarily a neighbor in M F ). Let L F = { 1 , . . . , k } and M F = {m 1 , . . . , m k } such that for 1 ≤ i ≤ k, i and m i are neighbors. Starting from (F, {v 0 }, ∅), Bob successively claims m 1 , . . . , m k , which forces Alice to reply 1 , . . . , k . When Bob has just claimed m k , he dominates the whole graph while Alice does not yet dominate k . This ensures that o((F, {v 0 }, ∅), B) = D.

It can now be assumed that the graph induced by S F is neither a unique vertex, a unique edge, nor a star. Thus, we can consider the position P = (F, {v 0 } ∪ L F , M F ), and from successive applications of Lemma 3.9, we just need to prove that o(P , B) = D.

First focus on components of S F which do not contain v 0 . Let C such a component, if each vertex of C is dominated by a vertex of M F , then Bob dominates C while Alice does not. Otherwise, let v 1 be a leaf of the subtree T C of F induced by C. Bob plays the bottom-to-top strategy in T C rooted in v 1 . It can be done since the leaves of T C are only connected to vertices of M F already claimed by Bob. If two vertices labeled by 0 have the same parent, then the strategy creates a double trap, which ensures a Draw. If v 1 has all its children labeled by 1, when Bob claims the last vertex labeled by 1, a double trap is created, since v 1 will be a trap. This also ensures the Draw. The only non directly conclusive case is when v 1 has one child labeled by 0, the reached position after Bob has followed the bottom-to-top strategy is such that Bob dominates C while Alice does not. Indeed, Bob dominates v 1 since it is a leaf of C and thus should be connected to M F but Alice does not dominate v 1 .

Bob follows this strategy on each component of S F not containing v 0 . All the answers from Alice are forced. If a double trap appears, we are done. Otherwise, each component is dominated by Bob and not by Alice. Moreover, it is Bob's turn. In such a case , we now need to focus on the component C 0 of S F which contains v 0 .

If the diameter of C 0 at most 1, (which implies that the graph induced by S F is not connected), Bob already dominates C 0 , and therefore F , but Alice does not.

If the diameter of C 0 is 2, the graph induced by C 0 is a star centered in a vertex c. Since it is assumed that the graph induced by S F is not a star, C 0 is not the only component of the graph induced by S F . Thus, if Bob does not dominate C 0 yet, he claims c or any of its neighbor and dominates the whole graph while Alice does not.

If the diameter of C 0 is at least 3, there exists a vertex v 2 at distance exactly 2 from v 0 in the subgraph induced by C 0 . Let (v 0 , v 1 , v 2 ) be the path from v 0 to v 2 . Bob can then use a bottom-to-top strategy in the tree induced by C 0 rooted in v 0 .

-If v 1 is labeled by 0, then v 2 is labeled by 1. Bob can play a bottom-to-top strategy claiming all the vertices labeled by 1 except v 2 . Then he claims v 1 . At this moment, Bob dominates F but Alice does not dominate v 2 .

-If v 1 is labeled by 1, Bob plays a bottom-to-top strategy in all branches, finishing by claiming v 1 . At this moment, he dominates F but, by construction, Alice does not dominate v 2 .

Thus, in any case, Bob has a strategy which leads to a Draw position, and thus o((F, {v 0 }, ∅), B) = D.

Lemma 5.12. If F is standard and o(F ) = A, then there exists a vertex v 0 ∈ M F such that o((F, {v 0 }, ∅), B) = A. Moreover, v 0 satisfies:

1. the subgraph of F induced by S F ∪ {v 0 } is a tree ;

2. in the labeling of F rooted in v 0 , each vertex v labeled by 1 has a unique child.

Proof. As it is supposed that o(F, A) = A, there exists a vertex v 0 ∈ V (F ) such that o((F, {v 0 }, ∅), B) = A. From Lemma 5.11, we have

v 0 ∈ M F ∪ L F . From Lemma 2.3, it can be assumed that v 0 / ∈ L F . Indeed, if l ∈ L F and m ∈ M F is its private neighbor, N [l] ⊂ N [m].
Thus, we can assume that v 0 ∈ M F . We now prove the two other properties.

1. Since F has no cherry, there exists a unique vertex v -1 ∈ L F which is a neighbor of v 0 . Assume that the graph induced by S F ∪ {v 0 } is not connected. We prove that, under this hypothesis, o((F, {v 0 }, ∅), B) = D, which gives the result by contraposition. First, using Lemma 3.9, we have o((F,

{v 0 }, ∅), B) = o((F, {v 0 } ∪ L F \ {v -1 }, M F \ {v 0 }), B).
With the same arguments as in the previous lemma, the implementation of a bottom-to-top strategy on each component of the graph induced by S F ∪ {v 0 } not containing v 0 leads to either a double trap (which gives the result), or a position where each of these components is dominated by Bob but not by Alice. Now, focus on the component C containing v 0 . Bob then plays a bottom-to-top strategy on the tree induced by C rooted in v 0 . Then Bob claims v -1 after all the vertices labeled by 1 have been claimed. By this way, if no trap has appeared before, Bob dominates the whole forest F , while Alice does not totally dominate F . This gives the result.

2. For the second item, consider the labeling of F (that is actually a tree) rooted in v 0 . First note that all the leaves of S F are labeled by 0. Thus, each vertex labeled by 1 has at least one child. Assume that there exists a vertex v labeled by 1 with has at least two children.

If two children v and v" of v are labeled by 0, Bob can then use a bottom-to-top strategy until v is claimed. When he claims v, he creates a double A-trap in v and v" and thus obtains a Draw.

Otherwise, there exists one child v of v labeled by 0 and the other one, v labeled by 1. Then Bob uses a bottom-to-top strategy but without claiming v", until all the vertices labeled by 1 (except v" are claimed). All replies of Alice remain forced. After this is done, Bob claims v -1 , the neighbor of v 0 which belongs to L F , and then dominates the whole graph while Alice does not dominate v", which ensures that o((T,

{v 0 }, ∅), B) = D.
As a consequence of this result, if F is standard and winning for Alice, then F is necessarily a tree.

Splitting the graph

By Lemma 5.12, it can be assumed that Alice first claims a vertex v 0 ∈ M F that is connected to all the components of S F (note that if S F is not connected, there is at most one such vertex). It can also be assumed that each vertex labeled by 1 in F rooted in v 0 has degree exactly 2. In all the remaining, v 0 will denote this first claim of Alice. Let v -1 be the leaf connected to v 0 . After this first move, it can be assumed, using Lemma 3.9, that Bob will claim all the other vertices of M F one by one. At each time, Alice must answer to the corresponding leaf in L F . After this step, the free vertices are the vertices in S F with the vertex v -1 . Formally, the obtained position is P = (F, L F \ {v -1 } ∪ {v 0 }, M F \ {v 0 }) and we have o(F ) = o(P, B). In that follows, we will split the graph into several components defined from the connected components of the skeleton.

Definition 5.13. For a connected component C of S F , let T be the connected component of F \ {v 0 } that contains C. The position P C is defined as the position induced by T ∪ {v 0 , v 0 } in the position P = (F,

L F \ {v -1 } ∪ {v 0 }, M F ∪ {v 0 } \ {v 0 })
, where v 0 is an additional leaf connected to v 0 and claimed by Bob.

Figure 6 illustrates the two positions P C and P C derived from the forest F of Figure 5 when played on v 0 . Lemma 5.14 shows that this splitting (with an additional [AoA] position) yields to an equivalent position.

Lemma 5.14. The position

P = (F, L F \ {v -1 } ∪ {v 0 }, M F \ {v 0 }) is equivalent to the position   C∈CC(S F ) P C   ∪ [AoA]
where CC(S F ) denotes the set of connected components of S F .

Proof. Let P be the position ( C∈CC(S F ) P C ) ∪ [AoA]. Note that the unclaimed vertices are in a one-to-one correspondence in the two positions (v -1 is corresponding to the unclaimed vertex of [AoA]). By Observation 2.2, we just need to prove that P and P have the same winning sets for both players.

A set S of unclaimed vertices is winning for Alice in P and in P if and only if it is dominating all the vertices of S F except the ones connected to (a copy of) v 0 , which corresponds to the same condition in both positions. A

v 0 A F C C v 0 B v 0 A B A B A B A P C A B A B v 0 A v 0 B P C B A A B
[AoA] =⇒ Figure 6: On the left a forest F . On the right, the equivalent position obtained by splitting F in Lemma 5.14. The component C is a fork whereas C is a P 2 . Lemma 5.17 gives us the possible structures of the connected components of S F to have a position A-win. But actually, this condition is not sufficient: there are for example trees where S F is a path, but Bob can obtain a Draw (see for example Figure 7). We need to consider which vertices of S F are already dominated by Bob. Definition 5.18. Let X, Y, ∈ {A, B}, n be a positive integer, and a subset U ⊆ {1, 2, . . . , n}. We denote by [Xo n Y ] U the position obtained from the bounded path [Xo n Y ] where for each i ∈ U a pendant edge x i y i is added to the vertex v i , with x i is linked to v i and claimed by Bob, and y i claimed by Alice.

v -1 v 0 A v 1 v 2 v 3 v 4 v 5 v 6 v 7 v 8 v 9 v 10 B A B A S F
Informally, an [Xo n Y ] U is a bounded path, where some vertices are already dominated by Bob. As an illustration, if we set that v -1 is claimed by Bob on Figure 7, then we obtain the position [Ao

8 B] {4} . When U is empty, [Xo n Y ] U = [Xo n Y ]. If Y = B (respectively X = B), one can assume that n / ∈ U (resp. 1 / ∈ U ). Indeed, v n (resp. v 1 ) is already dominated by Bob.
Finally, if C is a connected component of S F that is a path connected to v 0 by a leaf, then P C is equivalent to a position [Ao n B] U , for a fixed n and a fixed U . Indeed, set for U all the integers i such that v i is connected to a vertex of M F \ {v 0 }.

The following three observations give natural properties about bounded paths. The first one is about the existence of a pairing. that S F is connected. Let C be the tree induced by S F . Alice should claim in M F and connected to a leaf of C (Lemma 5.12). If there exists v 0 ∈ M F that is adjacent to a leaf of C such that P C is favorable to Alice, then o(F ) = A, otherwise o(F ) = D (Theorem 5.23).

In terms of complexity, almost all the operations described in the algorithm are elementary (finding a matching in a forest, identifying the isolated vertices, edges and cherries, computing L F , M F and S F , deciding if S F is connected or a star) and can be done in linear time by examining the tree from the leaves. When S F is not connected, there exists at most one v 0 that can connect them and this is easy to check. If S F is connected and thus reduce to a single component, one have to check all the possible v 0 that could make P S F favorable. If S F induces a path, there are at most two possibilities for v 0 since it should be connected to an extremity of the path. If S F induces a fork (easy to check) of center c, to be favorable, there must be at most one neighbor of c that is adjacent to M F . If there is exactly one neighbor a 1 of c adjacent to M F , then the leaf of S F adjacent to a 1 must have only one neighbor in M F and this neighbor will be the vertex v 0 . If all the neighbors at distance 1 of c in S F are not adjacent to vertices of M F , then there must be at least one leaf of S F adjacent to exactly one vertex of M F . This vertex of M F would be v 0 . In all the other cases, P S F cannot be favorable. Thus, finding v 0 can be done in linear time. Finally, once v 0 is fixed, deciding whether a component is favorable or not can also be done in linear time (because it is a fork or a path with pending edges).

6 Proof of Theorem 5.23

We here prove Theorem 5.23 by considering successively both directions of the equivalence. The next subsection proves that the favorable components are necessary conditions. The sufficient condition is then proved in Subsection 6.2.

The direct part

In this subsection, we assume that o((F, {v 0 }, ∅), B) = A and we prove that all the components are favorable and at most one is weakly favorable. We already know by Lemma 5.17 that all the connected components of S F are paths or forks. In Lemma 6.1, we prove that if a component is a fork, it should satisfy the conditions of Point 2 in Definition 5.22. In Lemma 6.2, we prove that if a component is a path, it should satisfy the conditions of Point 1 in Definition 5.22. Finally, in Lemma 6.6, we prove that if two components are weakly favorable, then the outcome is D. Proof. Proceeding by contraposition, we assume that C is not favorable. We will prove that o([AoA] ∪ P C , B) = D which gives the result using Lemma 5.15.

Let c denotes the center of the fork. Let a 1 and a 2 be the two vertices on the path between c and v 0 with a 2 adjacent to v 0 , and v -1 be the unclaimed vertex of [AoA]. Since C is not favorable, it means that either a 2 or a neighbor of c but not a 1 is dominated by Bob in P C . We consider the two cases.

If a 2 is already dominated by Bob, then Bob claims c.

-If Alice replies in v -1 then Bob can claim all the neighbors of c distinct from a 1 and Alice is forced to answer the leaf of C adjacent to the vertex claimed by Bob. Afterward, Bob claims a 1 , isolating the vertex c, which ensures that Alice will not dominate the graph.

otherwise, Bob claims v -1 and, therefore, dominates the graph before Alice.

If Bob already dominates a neighbor b 1 of c (with b 1 = a 1 ), then Bob claims all the other neighbors of c distinct from a 1 and b 1 , forcing Alice to claim the leaves adjacent to them. Then Bob claims the unclaimed vertex of [AoA]. At this point, he needs only one move to dominate the graph that can be either a 1 or a 2 , whereas Alice still need to dominate a 1 and b 2 , which cannot be dominated in a single move. Thus, Bob will dominate before Alice. Lemma 6.2. Assume that o((F, {v 0 }, ∅, B)) = A, and let C be a connected component of S F which induces a path. Then C is favorable.

If Alice replies in c , then Bob successively claims b 1 and c 1 (the replies of Alice are forced). Now Bob claims c , then Bob needs two more moves to dominate (one in (a 1 , a 2 ) and one in (a 1 , a 2 )) whereas Alice needs three moves.

If Alice replies in a 1 , then Bob claims a 2 .

-If Alice replies in a 1 , then Bob claims a 2 . At this step, Alice needs at least four moves to dominate while Bob can dominate in three moves in a lot of manners, with a center of a fork, and two vertices of the other fork. Alice cannot avoid Bob to dominate using three moves, and therefore, before Alice.

-If Alice replies in c, then Bob successively claims b 1 and c 1 . After the forced replies of Alice, Bob claims in c and dominates before Alice with one last move in (a 1 , a 2 ) while Alice cannot dominate in one move.

The case where Alice replies in c can be treated symmetrically.

- Assume now that both C and C induce paths. Since they satisfy the conditions 1.e of Definition 5.22, their length is 9 or 11. Bob first reduce the paths to length 9 if needed so that both paths have length 9. Let (v 1 , v 2 , ...., v 9 ) (respectively (v 1 , v 2 , ...., v 9 )) the path induced by C (resp. C ), with v 1 and v 1 connected to v 0 . Since the paths are weakly favorable, Bob already dominates v 3 , v 5 , v 3 and v 5 .

If
First, Bob successively claims v 7 and v 7 . By Lemma 2.10, Alice should answer first v 6 or v 8 and then v 6 or v 8 . We have three cases according to the replies of Alice.

If the replies are v 8 and v 8 , then Bob continues by claiming v 3 and v 3 which enforces Alice to reply in v 4 and v 4 (if, for instance, Alice does not reply v 4 , then Bob claims v 5 and forbids Alice to dominate at the following claim). Afterward, Bob claims v -1 , and achieves to dominate before Alice as he only needs one move in (v 1 , v 2 ) and one in (v 1 , v 2 ) while Alice needs to dominate v 2 , v 2 , v 6 and v 6 , each of them requiring a different move.

If the replies are v 6 and v 8 (note that v 8 and v 6 is symmetric), then Bob continues by claiming v 3 which enforces Alice to reply in v 4 , Afterward, Bob claims v -1 . Now Bob has a paired dominating set of size 3 {(v 1 , v 2 ), (v 1 , v 2 ), (v 3 , v 4 )}, but Alice needs to dominate v 2 , v 8 , v 2 and v 6 , each of them requiring a different move. Thus, Bob can dominate first.

If the replies are v 6 and v 6 , Bob claims v -1 . At this time Bob and Alice need four moves to dominate, but each set of four moves for Alice contains the pair {v 3 , v 3 }. Thus, after the reply of Alice, Bob can claim one element of the pair {v 3 , v 3 } and, by this way, dominate before Alice.

Assume now that C induces a path and C induces a fork. As before, C can be assumed to have exactly three branches and C nine vertices. We denote the vertices of the fork and the path as in the previous cases.

First -If Alice replies v 3 , then Bob claims c. Now (a 1 , a 2 ), (v 1 , v 2 ) and (v 4 , v 5 ) is a paired dominating set for Bob of size three, while Alice needs to dominated a 2 , b 1 , c 1 and v 8 , each of these vertices requiring a different move.

- -If Alice replies elsewhere, then Bob claims c. At this time Bob needs three moves to dominate,one in (v 1 , v 2 ), one in (a 1 , a 2 ) and one in (v 3 , v 4 , v 5 ). By hypothesis, at least one will be available in each of these sets at any moment. Alice needs at least four moves to dominate, as she does not dominate at least four of the five vertices a 1 , b 1 , c 1 , v 3 and v 8 , each of them requiring a different move.

If Alice replies v 1 or v 2 , by Lemma 2.3, as N [v 1 ] \ N [V t ] ⊂ N [v 2 ] \ N [V t ]
[v 1 ] \ N [V t ] ⊂ N [v 2 ] \ N [V t ]

The converse part

In this final subsection, we prove the reverse part of Theorem 5.23: if all the connected components of S F are favorable and at most one is weakly favorable, then Alice has a winning strategy. This part is naturally harder than the other one, as Alice cannot force Bob to answer where she would like to. Hence, all the possible answers of Bob must be considered, which was not the case previously, as Alice was often forced to answer locally to a move of Bob. We first prove in Subsection 6.2.1, that we can remove the strongly favorable components (i.e. corresponding to the cases 1.a to 1.c of Definition 5.22). In Subsection 6.2.2, we consider only one weakly favorable component and give a strategy for Alice in this case. Finally, in Subsection 6.2.3, we define a class of positions C 1 that contains the starting positions (without strongly favorable components) and for which Alice can ensure either to win or to stay in this class after a move of Bob. By induction, this will imply that Alice has a winning strategy in this class. Note that one can always consider U to be maximal in Definition 5.22. Indeed, if Alice have a strategy for U maximal, she will have a strategy with any subset of U .

Removing strongly favorable components

Lemma 6.7. Let Q be a position where Bob is not dominating. Consider a strongly favorable component

[Ao n B] U . We have o(Q, B) ≤ o(Q ∪ [Ao n B] U , B).
In other words, adding a strongly favorable component can only be favorable to Alice.

Proof. First note that since [Ao n B] U is strongly favorable, we have U ⊆ {2, 3, n -2}. Assume first that C is a fork. Let c be the center of C, a 1 the neighbor of c on the path to v 0 and a 2 the vertex between a 1 and v 0 . Let b 1 be another neighbor of c and b 2 the other neighbor of y. We prove the result by induction on the number of branches in the fork. Note that, by Lemma 2.3, as

N [a 2 ] \ N [V t ] ⊂ N [a 1 ] \ N [V t ] for t ∈ {A, B}, a 1
is always a better move than a 2 , thus we can suppose that it will not be played as next move.

if Bob claims c, then Alice replies in a 1 . Since there is a double B-trap and a A-pairing, disjoint from the traps, Alice wins by Lemma 3.7. Moreover, by following the pairing strategy, she will keep the fact there is always a A-pairing.

If Bob claims b 1 , then Alice replies in b 2 . If there are strictly more than three branches, the position is still a fork where the center is dominated by Bob (which is still weakly favorable). The result is true by induction. If there were exactly three branches, then the actual position is (( If there are strictly more than three branches, then Alice follows the strategy with one less branch. Since she dominates one more vertex, it can only be better for her. If there are exactly three branches, we have as before:

P C ∪ [AoA]) b2,b1 , B) = ([Ao 5 B] {2,3} ∪ [AoA], B). By Lemma 6.7, o((P C ∪ [AoA]) b2,b1 , B) ≥ o([AoA], B) = A.
o((P C ∪ [AoA]) b1b2 , B) ≥ o([Ao 5 B] {2,3} ∪ [AoA], B) ≥ o([AoA], B) = A.
If Bob claims a 1 , then Alice replies by claiming c. We have

o((P C ∪ [AoA]) c,a1 , B) = o([Ao 2 B] ∪ ... ∪ [Ao 2 B] ∪ [AoA], B) ≥ o([AoA], B) = A.
There is a A-pairing by pairing together the two vertices belonging to the same paths.

If Bob claims v -1 , then Alice replies by claiming c. We have

o((P C ∪ [AoA]) c,v-1 , B) ≥ o([Ao 2 A] {2} ∪ [Ao 2 B] ∪ ... ∪ [Ao 2 B], B) ≥ o([Ao 2 A] {2} , B) = A.
Again, there is a A-pairing by pairing together the two vertices belonging to the same paths.

Assume now that P C is a bounded path. Alice will never let components of the form [Bo 2 B] to Bob and thus there will always be a A-pairing. Assume first that P C has length 9. We just need to prove that Alice has a strategy in the worst case, that is for U = {2, 3, 5, 7}. Thus, let P C = [Ao 9 B] {2,3,5,7} . Note that, by Lemma 2.3, as

N [v 1 ] \ N [V t ] ⊂ N [v 2 ] \ N [V t ]
for t ∈ {A, B}, v 2 is always a better move than v 1 , thus we can suppose that it will not be played as next move. where the last inequality comes from Lemma 6.7.

A stable class of positions

In order to define our stable class C 1 , we first define the class C 0 , that informally corresponds to the positions derived from a weakly favorable position when Alice follows a winning strategy. Definition 6.9 (C 0 ). The class C 0 is defined recursively as follows. A position P is an element of C 0 if: Proof. We denote by v -1 the unclaimed element of [AoA] in Q. We assume that Bob claims y. We have several cases, according to the component that contains y.

P = P C
Assume first that y is an element of the component P of P that belongs to C 0 . If there exists a free vertex x of P such that P x,y ∈ C 0 , then we are done, since P x,y ∈ C -If y = w 2 then take x = w 3 . By Lemma 2.6, Observation 2.4 and Lemma 6.7,

o(Q x,y , B) ≥ o(Q ∪ [Ao n-3 B] {2,n-5} , B) ≥ o(Q , B).
-If y = w 3 , then take x = w 2 . There is double B-trap in Q w2,w3 . Thus, by Lemmas 3.7 and 6.13, we have o(Q w2,w3 , B) = A.

-If y = w i i ≥ 4 with i even, then take x = w i+1 We have

o(Q x,y , B) = o(Q ∪ [Ao i-1 B] {2,5} ∪ [Ao n-i B] {n-i-2} , B) ≥ o(Q ∪ [Ao i-1 B] {2,5} , B)
by application of Lemma 6.7. Note that P ∪ [Ao i-1 B] {2,5} is an element of C 1 .

if y = w i with i odd and i ≥ 5 then take x = w i-1 . We have

o(Q x,y , B) = o(Q ∪ [Ao i-2 A] {2,5} ∪ [Bo n-i B] {n-i-2} , B).
Note that P ∪ [Ao i-2 A] {2,5} ∪ [Bo n-i B] {n-i-2} is an element of C 1 , since n -i is even and since we add a position [Ao i-2 A] {2,5} .

If y is an unclaimed vertex of a component of the form [Bo 2t B] {2t-2} , then the position can be reduced by Claim 6.14.

Assume now that y is an unclaimed vertex of [Ao n A] {2,5} . As before, we denote by (w 1 , w 2 , ..., w n ) the vertices of [Ao n A] {2,5} , by P the position P without the component [Ao n A] {2,5} and by Q the position P ∪ [AoA].

If n ≤ 3, then take x = v -1 . We have o(Q v-1,y , B) = A since the position Q v-1,y admits a A-pairing by Corollary 6.13 and Bob. Thus, it can be assumed that n ≥ 4.

-If y ∈ {w 1 , w 2 }, then take x = w 3 . We have

o(Q w3,y , B) ≥ o(Q ∪ [Ao n-3 A] {2} , B).
Note that P ∪ [Ao n-3 A] {2} is in C 1 .

-If y = w 3 , then take x = w 2 . The position Q w2,w3 admits an A-pairing, by Corollary 6.13. Thus, by lemma 3.7, we have o(Q w2,w3 ) = A.

Conclusion

Putting together all the previous results, we can prove the reverse part of Theorem 5.23: Corollary 6.19. If all the components of S F are favorable to Alice and at most one of them is weakly favorable, then ((F, {v 0 }, ∅), B) is A-win.

Proof. Let Q = (F, {v 0 }, ∅) be a position such that all the components of S F are favorable to Alice and at most one of them is weakly favorable. By Lemma 6.7, one can assume that there is no strongly favorable component in S F . Let P such that Q = P ∪ [AoA]. By Remark 6.12, P is an element of C 1 . Corollary 6.18 leads to the desired result.

Observation 2 . 5 .

 25 Let P and P be two positions such that o(P, B) = o(P , B) = A. Then we have o(P ∪P , B) = A.

Figure 1 :

 1 Figure 1: On the left, v is a B-trap and w may be isolated; on the right v is a A-trap

Lemma 3 . 4 .

 34 Let G be a graph. If G has a pairing dominating set of size γ(G), then Alice has a winning strategy in G.

Figure 2 :

 2 Figure 2: Example of a graph where Bob wins in Maker-Maker, but if we remove any leaf with its neighbor, Alice wins.

Figure 3 :

 3 Figure 3: The bounded paths [Ao 5 A] and [Bo 3 A].

Lemma 4 . 2 .

 42 For any position P and any integer n, o(P, B) = o(P ∪ [Ao n B], B).

Lemma 4 . 3 .

 43 Let P = (G, V A , V B ) be a position, let n be an integer such that n ≡ 0 mod 3 and let n be an integer such that n ≡ 0 mod 2. Then if o(P, B) = A, then o(P ∪ [Ao n A] ∪ [Bo n B], B) = A. Proof. We prove this result by induction on the number m of unclaimed vertices of P ∪ [Ao n A] ∪ [Bo n B]. For initialization, if m = 1, then P has no unclaimed vertices, n = 1 and n = 0. Thus, Alice dominates P ∪ [Ao n A] ∪ [Bo n B] and the result is true, by definition. Assume now that m ≥ 2, If Alice dominates (P ∪ [Ao n A] ∪ [Bo n B]) (which implies that n ≤ 2 and n = 0) then the result is true, by definition. Otherwise, we have to prove that, for each vertex x claimed by Bob, there exists an answer y for Alice such that o(P ∪ [Ao n A] ∪ [Bo n B], B) = A, the pointed position obtained after the two claim is A-win. Denote by (u 1 , . . . , u n ) the unclaimed vertices of [Ao n A] and by (v 1 , . . . , v n ) the unclaimed vertices of [Bo n B]. We consider all the possible moves for Bob. If Bob claims u 1 or u 2 in [Ao n A] and n ≤ 2, we consider two subcases. If n = 0, then Alice dominates [Ao n A] ∪ [Bo n B], which obviously gives the result. If n ≥ 2, Alice claims v 1 . By Observation 2.4, the resulting game is equivalent to P ∪ [Ao n -1 B], which is equivalent to P by Lemma 4.2. If Bob claims u 1 or u 2 in [Ao n A] and n ≥ 4, by Lemma 2.3, we can suppose he claims u 2 . then Alice claims u 3 . Thus, the resulting position is equivalent to P ∪ [Ao n-3 A] ∪ [Bo n B], which gives the result using the induction hypothesis. The case where Bob claims u n or u n-1 is symmetric. If Bob claims a vertex u k in [Ao n A] with k / ∈ {1, 2, n -1, n}, Alice answers by claiming a vertex u i with i ∈ {k-1, k+1} such that the resulting component is [Ao j A]∪[Ao j B] with j ≡ 0[3]. Note that, since n ≡ 0[3], this vertex always exists. By Lemmas 2.6 and 4.2, the resulting position is equivalent to P ∪ [Ao j A] ∪ [Bo n B], which gives the result using the induction hypothesis.If Bob claims some vertex v k in [Bo n B], Alice answers by claiming a vertex v i with i ∈ {k -1, k + 1} such that the resulting component is [Bo j B] ∪ [Ao j B] with j ≡ 0[START_REF] Bonnet | The parameterized complexity of positional games[END_REF]. Note that as n ≡ 0[START_REF] Bonnet | The parameterized complexity of positional games[END_REF], this vertex always exists. By Lemma 4.2 and Lemma 2.6, the resulting position is equivalent to P ∪ [Ao n A] ∪ [Bo j B], which gives the result using the induction hypothesis.If Bob claims a vertex x in P , we consider two subcases. If Alice does not dominate G yet, she claims the vertex she would have claimed as an answer to x in her winning strategy in P . The resulting position has two vertices less, which gives the result using the induction hypothesis.If Alice already dominates G, and n ≥ 2, then Alice claims v 1 in [Bo n B], turning it into [Ao n -1 B]. By Lemma 4.2, the game is now equivalent to P ∪ [Ao n A], and Alice already dominates G, so by induction hypothesis, it is a winning position for her. For n = 0 and n ≥ 4 (cases where n ≤ 2 are trivial), then Alice claims u 2 . The resulting position is better for Alice than the position P ∪ [AoA] ∪ [Ao n-2 B], which, by Lemma 4.2, is equivalent to P ∪ [AoA] and therefore is a winning position for Alice.

Lemma 4 . 4 .Lemma 4 . 5 .

 4445 For any position P and any odd integer n, o(P ∪ [Bo n B], B) = D. Proof. We prove the result by induction on n. If n = 1, there is a A-trap: by Lemma 2.8, o(P ∪ [BoB], B) = D. Now, let n ≥ 3. Bob claims v 2 which forces Alice to claim v 1 . The position is then equivalent to (P ∪ [Bo n-2 B], B) which is Draw by induction. For any positive integer k, o([Ao 3k A], B) = D. Proof. We prove the result by induction on k. If k = 1, Bob claims v 2 and directly dominates [Ao 3 A]. Thus, o([Ao 3 A], B) = D. If k = 2, Bob claims v 2 . If Alice does not answer in {v 1 , v 3 , v 4 }, Bob claims v 3 at his second turn and create two A-traps in v 1 and v 4 which ensures a Draw. Thus, Alice should claim a vertex among {v 1 , v 3 , v 4 } and does not dominate the graph at her first claim. Then Bob can claim v 5 and dominates the graph. Hence o([Ao 6 A], B) = D. Assume now that k ≥ 3 and that the result is true for any positive k < k. Consider the position ([Ao 3k A], B). Bob claims v 5 at his first turn. By Lemma 2.10 applied on (v 3 , v 4 , v 5 , v 6 , v 7 ), Alice should answer on v 4 or v 6 . If she claims v 4 , then by Lemma 5.5, the position is equivalent to the position ([Ao 3 A] ∪ [Bo 3k-5 A], B), which is equivalent, by Lemma 4.2 to ([Ao 3 A], B) which is Draw.

Theorem 4 . 7 .

 47 Let n be an integer. We have o(C n ) = D if and only if n ≥ 10 and n ≡ 1 mod 3.Proof. We first prove the "if" part. Let n = 3k + 1, with k ≥ 3, and let C n be a cycle of order n By symmetry, we can assume that Alice first claims v 0 and thus o(C n ) = o(C n , {v 0 }, ∅, B). We give a strategy for Bob to obtain at least a draw. Bob first claims v 5 . By Lemma 2.10 with G = (v 3 , v 4 , v 5 , v 6 , v 7 ), Alice has to answer v 4 or v 6 .If Alice claims v 6 . Then Bob can claim v 3 which forces Alice to claim the A-trap in v 4 and then Bob can claim v 1 which forces Alice to claim the A-trap in v 2 . At this point, the position is equivalent to ([Ao

  2, (P , B) is equivalent to ([Ao n-1 A], B). By Lemma 4.3, since n -1 ≡ 0 mod 3, o([Ao n-1 A], B) = A, and thus o(C n ) = A

Figure 4 :

 4 Figure 4: Example of labelling

Lemma 5 . 3 .

 53 Let F be a forest with an isolated edge e = uv. Then o(F ) = o(F \ {u, v}).

Lemma 5 . 6 .

 56 Let F be a forest. If F has two cherries C = (c, 1 , 2 ) and C = (c , 1 , 2 ), with c = c , then o(F ) = D. Proof. Let F be such a forest. After Alice has claimed her first vertex, she cannot have claimed both c and c . Suppose without loss of generality that Alice has claimed c . Then Bob claims c. The resulting pointed position contains two A-traps and thus is Draw by Corollary 2.9. Lemma 5.7. Let F be a forest with exactly one cherry C = (c, 1 , 2 ). Then o(F ) = A if and only if there is a matching in F \ {c} that covers V (F ) \ N [c]. Proof. Suppose first that F has a matching M in F \{c} that covers V (F )\N [c]. Then Alice claims c, which creates a double B-trap in 1 and 2 . The matching M is actually a A-pairing, and, we can suppose it contains neither 1 nor 2 since these two vertices have their neighborhood included in N [c]. Then by Lemma 3.7, Thus o(F ) = A.

Lemma 5 . 8 .Lemma 5 . 9 .

 5859 If S F is empty, then o(F ) = A.Proof. Set L F = { 1 , . . . , k }. By hypothesis, F has no cherry, therefore, no vertex can be adjacent to two of these leaves. Thus, we can denote M F = {m 1 , . . . , m k } with m i adjacent to i for 1 ≤ i ≤ k. Note that k vertices (one on each set { k , m k }) are necessary and sufficient to dominate F . Thus γ(F ) = k and there is a pairing dominating set of size k (the set of pairs {( i , m i ), i ∈ {1, ..., k}). By Lemma 3.4, Alice has a winning strategy in F . Let T be a connected component of F such that S T is empty. Then o(F ) = o(F \ T ).Proof. Assume first that o(F \ T ) = A and let x be the first claim of Alice in a winning strategy. Then Alice claims x in F . By Lemma 3.9, Bob will claim all the vertices of M T and Alice will answer all the vertices of L T . After these moves, T is completely claimed and dominated by both players. By Observation 2.4, we can remove this component. The game is then equivalent to (F \ T, {x}, ∅) which is A-win when Bob starts, leading to o(F ) = A.Assume now that o(F \ T ) = D. Consider the game played in F and a first claim x of Alice. If x ∈ V (F \ T ), then as before, the position (F, {x}, ∅) is equivalent to the position (F \ T, {x}, ∅) which is Draw when Bob starts. If x ∈ V (T ), let x be the unique neighbor of x if x is a leaf or the leaf connected to x if x ∈ M T . Then we can assume by Observation 2.4 that Bob claims all the vertices in M T (except x or x ) and that Alice answers by claiming all the vertices in L T (except x or x ). At this point, all the vertices of T have been claimed except x . Then Bob claims x . Then both players dominate T and the position is equivalent to (F \ T, A) which is a Draw position. In conclusion, whatever Alice claims, Bob can ensure a draw in F . Thus o(F ) = D Lemma 5.10. If S F induces a star of center c such that c has no neighbor in M F , then o(F ) = A.

Figure 7 :

 7 Figure 7: Example of a tree F such that o(F ) = D but S F is a path. Alice should claim first v 0 or v 9 . If Alice claims v 0 , Bob claims the other vertices of M F . Then Bob can win by claiming v 2 (Alice should answer v 1 , v 3 or v 4 ), v 6 (Alice should answer v 5 ,v 7 or v 8 ) and v -1 . The case where Alice claims first v 9 is similar.

Observation 5 .

 5 19. A position [Xo n Y ] U contains a A-pairing, except if X = Y = B and n is odd.

Lemma 6 . 1 .

 61 Assume that o((F, {v 0 }, ∅), B)) = A. Let C be a connected component of S F which induces a fork. Then C is favorable.

  Alice replies in b 1 , then Bob claims successively c 1 and c and dominates before Alice by claiming one vertex in (c, b 2 ). -If Alice replies in b 1 , then Bob claims c . At this step, Alice needs at least four moves to dominate while Bob can dominate in three moves by a pairing strategy with the pairs (a 1 , a 2 ), (b 1 , b 2 ) and (c 1 , c 2 ).If Alice replies in a 2 , then Bob claims a 1 , and afterward, all is similar to the previous case.If Alice replies in b 1 or b 2 , then Bob claims c . Bob can now dominate in three moves by pairing (a 1 , a 2 ), (a 1 , a 2 ) and (c, c 1 ) while Alice needs at least four moves.

  Furthermore, there is A-pairing by Observation 5.19. If Bob claims b 2 , then Alice replies in b 1 .

If Bob claims v 1 , 7 --

 17 then Alice replies in v 3 . We haveo(([Ao 9 B] {2,3,5,7} ) v3,v1 ∪ [AoA]), B) ≥ o([Ao 6 B] {2,4} ∪ [AoA], B) ≥ o([AoA], B) = A,where the last inequality comes from Lemma 6.7.If Bob claims v 3 , then Alice replies in v 2 . Then o((P C ∪ [AoA]) w2,w3 , B) = A since there are two B-traps and a A-pairing.If Bob claims v i , 4 ≤ i ≤ 8, then Alice replies in v i+1 . We have o(([Ao 9 B] {2,3,5,7} ) vi+1,vi ∪[AoA], B) = o([Ao i-1 B] {2,3,5,7}∩{1,...i-2} ∪[Ao 9-i-1 B] 7-i-1 ∪[AoA], B) ≥ o([AoA], B) = A,where the last inequality comes from two applications of Lemma 6.7 (since both created paths satisfy hypotheses of Lemma 6.7 ).If Bob claims v 9 , then Alice replies in v 8 . We haveo(([Ao 9 B] {2,3,5,7} ) v8,v9 ∪[AoA], B) ≥ o([Ao 7 A] {2,3,5,7} ∪[AoA], B) ≥ o([Ao 7 B] {2,3,5} ∪[AoA], B) ≥ o([AoA], B) = A,where the last inequality comes from two applications of Lemma 6.7.If Bob claims v -1 , then Alice replies in v 3 . We haveo(([Ao 9 B] {2,3,5,7} ∪ [AoA]) v3,v-1 , B) ≥ o([Ao 6 B] {2,4} ∪ [Ao 2 A], B) ≥ o([Ao 2 A], B) = A,where the last inequality comes from Lemma 6.7.Assume now that P C has length 11. As before, we can assume thatP C = [Ao 11 B] {2,3,5,9} . If Bob claims v i , 1 ≤ i ≤ 8,then it can be done like for a path of length 9. If Bob claims v 9 , then Alice replies in v 8 , -If now Bob claims v 3 , then Alice replies in v 2 , creating two B-traps. Thus o(Q v2,v3 , B) = A, by Lemma 3.7, -If now Bob claims v 5 , then Alice replies in v 6 , creating two B-traps. Thus o(Q v6,v5 , B) = A, by Lemma 3.If now Bob claims v 10 (or v 11 ) , then Alice replies in v 11 (or v 11 ). The resulting position is equivalent to [Ao 7 A] {2,3,5} ∪ [AoA], and, from Lemma 6.7, o([Ao 7 A] {2,3,5} ∪ [AoA], B) ≥ o([AoA], B) = A. . If now Bob claims v i , with i ∈ {-1, 1, 2, 4}, then Alice replies in v 3 . Afterward Alice can dominate with two more claims, one in {v 5 , v 6 }, one in {v 10 , v 11 }, whatever the strategy of Bob; therefore Alice dominates before Bob -If now Bob claims v i , with i ∈ {6, 7}, then Alice replies in v 5 . Afterward Alice can dominate with two more claims, one in {v 2 , v 3 }, one in {v 10 , v 11 }, whatever the strategy of Bob; therefore Alice dominates before Bob. if Bob claims in v 10 , then Alice replies in v 11 , which reduces the problem to the previous case with 9 vertices. If Bob claims v 11 , then Alice replies in v 10 . The resulting position is equivalent to to [Ao9 A] {2,3,5,9} ∪ [AoA], and by Observation 5.20, we have o([Ao 9 A] {2,3,5,9} ∪ [AoA], B) ≥ o([Ao 9 B] {2,3,5} ∪ [AoA], B) = A, as seen before. If Bob claims v -1 , then Alice replies in v 3 . We have o(([Ao 11 B] {2,3,5,9} ∪ [AoA]) v3,v-1 , B) ≥ o([Ao 8 B] {2,6} ∪ [Ao 2 A], B) ≥ o([Ao 2 A], B) = A,

  where C is a weakly favorable component; there exists P ∈ C 0 , with two unclaimed vertices x and y, such that P = P x,y and o(P ∪ [AoA], B) = A. Lemma 6.15 (Bob does not claim v -1 ). Let P ∈ C 1 , such that P is not dominated by Alice. Let Q = P ∪ [AoA] and y be an unclaimed vertex of P . There exists an unclaimed vertex x = y such that at least one of the following alternatives holds:o(Q x,y , B) = A;or there exists P ∈ C 1 with at least two unclaimed vertices less than P , such that o(P ∪[AoA], B) ≤ o(Q x,y , B).

1 .

 1 Otherwise, by definition of C 0 , (P ∪ [AoA], B) is A-win. Since there is no winning answer in P for Alice to the claim y of Bob, we necessarily have o((P ∪ [AoA]) v-1,y , B) = A. From the position (P ∪ [AoA]) v-1,y , Bob cannot dominate the whole graph. Alice can follow her strategy on (P ∪ [AoA]) v-1,y and a pairing strategy on the rest of the graph (that exists by Corollary 6.13). This ensures that o(Q v-1,y , B) = A Assume now that y is an unclaimed vertex of [Ao n B] {2,5,n-2} , with n odd, n ≥ 9. Let (w 1 , w 2 , ..., w n ) be the sequence of free vertices of [Ao n B] {2,5,n-2} . By Lemma 2.3, asN [w 1 ] \ N [V t ] ⊂ N [w 2 ] \ N [V t ]for t ∈ {A, B}, one can assume that y = w 1 . Let P be the position such that P = P ∪[Ao n B] {2,5,n-2} and let Q = P ∪[AoA]. Note that P ∈ C 1 .

  the result is true since [Ao n B] is already dominated by both players. Hence, by Observation 2.4, the result is true when the number of unclaimed vertices is at most 1. Now assume there are at least two unclaimed vertices and n ≥ 2. Suppose first that (P, B) = D. In this case, Bob starts by claiming v n-1 in P ∪ [Ao n B] creating a A-trap in v n . By Lemma 2.8, Alice has to answer v n . By induction hypothesis o(P ∪ [Ao n-2 B], B) = D, which ensures that o(P ∪ [Ao n B], B) = D. Now assume that (P, B) is A-win and let S be a winning strategy for Alice in this pointed position. We give a strategy for Alice in (P ∪ [Ao n B], B). Let v B be the vertex claimed by Bob in (P ∪ [Ao n B], B). Alice answers as follows: If v B ∈ V (G) and if Alice does not dominate P , she claims the same vertex v A she would have answered following S if Bob had claimed v B in (P, B). The resulting position has two vertices less, and is A-win by induction hypothesis. If v B ∈ V (G) and Alice already dominates P , we necessarily have n ≥ 2, otherwise, she would already have won. She claims v 2 . The resulting position is better than (

  3(k-2) A], B) which is Draw by Lemma 4.5 (remember that k -2 ≥ 1). If Alice claims v 4 . Bob can claim successively all the v 2i+1 starting from v 7 , creating a A-trap in v 2i that Alice is forced to claim. If n is even, Bob follows this strategy until claiming v n-1 . Then Alice has to claim v n-2 and does not dominate the cycle (she does not dominate v 2 ). Then Bob can dominate the cycle by claiming v 2 . If n is odd, Bob follows this strategy until claiming v n-4 . After Alice has claimed v n-5 , Bob claims v 2 . Alice is forced to claim v n-1 , otherwise Bob wins by claiming it. Then Bob claims v n-2 , creating a trap in v n-3 , forcing Alice to claim it. At this point, Alice still does not dominate v 2 . Then Bob can dominate the cycle by claiming v 1 .

  , Bob claim v 7 . Once again, Alice has to claim either v 6 or v 8 , otherwise Bob ensure a draw by Lemma 2.10 There are two cases according to the reply of Alice. If Alice replies v 8 , then Bob successively claims b 1 , c 1 and a 1 (replies of Alice being forced in b 2 , c 2 and c). Afterward, Bob claims v 3 . Alice has to claim either v 2 or v 5 by Lemma 2.10. If she claims v 2 , Bob claims v 5 and creates two traps. If she claims v 4 , Bob claims v -1 . Now Bob only needs one move in either v 1 or v 2 to dominate while Alice needs at least two, as she does not dominate v 2 nor v 6 . If Alice replies v 6 , then Bob claims v -1 -If Alice replies c, then Bob successively claims b 1 , c 1 (replies of Alice are again forced in b 2 and c 2 ) and then v 3 . At this time, Bob needs two moves to dominate (one in v 1 , v 2 and one in (a 1 , a 2 )), while Alice needs to dominate v 2 , v 4 and v 8 , and each of them requires a different move. Therefore, Bob dominates before Alice by this way.

  for t ∈ {A, B}, we can suppose, she plays v 2 . Then Bob successively claims b 1 , c 1 , a 1 (replies are forced in b 2 , c 2 , c respectively) and then v 1 . At this time Bob needs one move to dominate in (v 3 , v 4 ), while Alice needs to dominate v 4 and v 8 which she cannot dominate in a single move.-If Alice replies a 1 or a 2 , by Lemma 2.3, asN [a 2 ] \ N [V t ] ⊂ N [a 1 ] \ N [V t ]for t ∈ {A, B}, we can suppose she claims a 1 . Bob claims a 2 . If Alice replies c, then Bob successively claims b 1 , c 1 , (replies are forced) and v 3 . At this time Bob can dominate in two moves, one in (v 1 , v 2 ) and one in (v 4 , v 5 ) while Alice cannot, as she still has to dominate v 2 , v 4 and v 8 . * If Alice replies v 1 or v 2 , by Lemma 2.3, as N

*

  for t ∈ {A, B}, we can suppose she replies v 2 . Then Bob successively claims b 1 , c 1 , (replies are forced for Alice) and v 1 . At this time Bob needs one move in (v 3 , v 4 ) to dominate, while Alice needs two moves to dominate v 4 and v 8 . * If Alice replies elsewhere, then Bob claims c. At this time Bob needs two moves to dominate,one in (v 1 , v 2 ) and one in (v 3 , v 4 , v 5 ). By hypothesis, at least one will be available. Alice needs at least three moves to dominate, as she does not dominate at least three of the four vertices v 3 , v 8 , b 1 and c 1 , each of them requiring a different move.

  6.2.2 Dealing with the weakly favorable component Lemma 6.8. Let C be a connected component of S F that is weakly favorable, then o(P C ∪[AoA], B) = A. Moreover, Alice can play to ensure that after each of her move, there is a A-pairing. Proof. Recall that by Observation 5.19, there is A-pairing in any bounded path [Xo n Y ] except if X = Y = B and n is odd.

In this paper, the union of two positions will always be done on disjoint sets of vertices. It could happen, for simplicity reasons, that two vertices in different components have the same name. They must be considered as distinct vertices.

In this equality, it is assumed that P ,u is really a position, i.e. V B ∪ {u} and V A ∪ { } do not both dominate the graph. In this latter particular case, we obviously have o(P, B) = D .

* This work has been supported by the National Research Agency (ANR) projects P-GASE (ANR-21-CE48-0001-01).

1

set S of unclaimed vertices is winning for Bob in P if and only if v -1 ∈ S and S is dominating all the vertices of S F that are not connected to a vertex of M F \ {v 0 }. In P , S is winning for Bob if it contains the unclaimed vertex of [AoA] and if it is dominating all the unclaimed vertices not already dominated by Bob, that are exactly all the vertices of S F not connected to M F \ {v 0 }. Thus, the winning sets are in bijection and by Observation 2.2, the positions are equivalent.

Using this decomposition, we now prove that Bob can just focus on a subset of components where he has a Draw strategy. Proof. Using Lemma 5.14, it suffices to prove that o(( C∈CC(S F ) P C ) ∪ [AoA], B) = D. For each C / ∈ S, Bob plays a bottom-to-top strategy on P C rooted on v 0 . At each time Alice is forced to answer in the same component C, and, at the end, Bob dominates the component C. This is successively done for all such components.

Afterward, the remaining position is equivalent to (( C∈S P C ) ∪ [AoA], B) which is Draw by hypothesis.

Favorable skeletons for Alice

In this subsection, we give some necessary and sufficient conditions for a component C to be winning for Alice.

Definition 5.16. A fork is a star with at least three branches where each is subdivided exactly once.

On Figure 6, the component C is an example of a fork with four branches.

Lemma 5.17. If ((F, {v 0 }, ∅), B) is A-win, then all the connected components of S F must induce a path or a fork that is connected to v 0 by a leaf.

Proof. Proceeding by contraposition, assume that there exists a component C of S F , which is neither a fork nor a path connected to v 0 by a leaf . We will prove o(P C ∪ [AoA], B) = D, which gives the result using Lemma 5.15. Since C is not a path connected by a leaf to v 0 , there exists a vertex c ∈ C of degree at least 3 in the tree induced by C ∪ {v 0 }. Let P a = (c, a 1 , a 2 , ...a p ) be the path linking c to v 0 (a p is adjacent to v 0 ), P b = (c, b 1 , b 2 , ..., b q ) be a path of C of maximal length such that b 1 = a 1 and P c = (c, c 1 , c 2 , ..., c r ) be another maximal path of S F starting in c. Note that possibly p = 0, but that q ≥ r ≥ 1. In the labeling of F rooted in v 0 , the vertex c is necessarily labeled by 0 since it has degree 3. This implies that both q and r are even since b q and c r , as leaves of C, are labeled by 0. Moreover, all the vertices labeled by 1 have degree 2. This implies that all the vertices of the three paths P a , P b and P c that are connected to other vertices of F must be labeled by 0. In particular, using Lemma 5.5, it is enough to prove that there is a Draw strategy when C is reduced to these three paths.

Assume first that q ≥ 4. By Lemma 5.5, it is enough to give a Draw strategy for Bob for q = 4 and r = 2 with C reduced to the union of the three paths Assume now that r = q = 2. As before, it is enough to give a strategy for C restricted to the three paths.

If p ≥ 5, then Bob claims a 3 , which enforces Alice to reply either a 2 or a 4 by Lemma 2.10.

-If Alice claims a 4 , then Bob successively claims b 1 , c 1 (with forced Alice to claim b 2 and c 2 ), and then a 1 , which creates two A-traps in a 2 and c.

Roughly speaking, the next two observations say that it is always better for Alice to play on a bounded path with the extremities claimed by her, and with fewer vertices dominated by Bob in U .

Observation 5.20. For any integer n and any set U ∈ {1, ..., n}, and any X, ∈ {A, B}, we have

Observation 5.21. For any integer n and any sets U ⊆ U ⊆ {1, ..., n}, and each X, Y, Z ∈ {A, B}, we have

In the next definition, we define the components C that are favorable for Alice, and among them, the ones that are strongly and weakly favorable. Theorem 5.23 will justify this terminology. Definition 5.22. Let v 0 be the first move of Alice, satisfying Lemma 5.12. Let C be a connected component of S F . We say that C is favorable for Alice if it satisfies one of the following case: (e) n ∈ {9, 11} and {3, 5} ⊆ U ⊆ {2, 3, 5, n -2};

2. C induces a fork and the only vertices of C that can be connected to M F are the center c, the leaves except the one connected to v 0 , and eventually the neighbor of c between c and v 0 .

Moreover, if C belongs to the cases (1.a), (1.b) or (1.c), we say that it is strongly favorable. On the opposite, if C belongs to the cases (1.e) or (2), we say that it is weakly favorable. 3 We can now state the final theorem that ends the characterization of trees. This theorem is actually technical to prove, and thus the proof is postponed to Section 6.

Theorem 5.23. Let F be a tree and v 0 ∈ M F be a first move of Alice that satisfies the condition of Lemma 5.12. The position ((F, {v 0 }, ∅), B) is A-win if and only if all the components of S F are favorable to Alice and at most one of them is weakly favorable.

Proof of Theorem 5.1

We now have all the ingredients to complete the study of forests and prove Theorem 5.1. We summarize the algorithm into the diagram of Figure 8.

Proof of Theorem 5.1. Let F be a forest. If F has an isolated vertex v 0 , then by Lemma 5.2, o(F ) = A if and only if F \ {v 0 } has a perfect matching. If F has an isolated edge e = uv, then by Lemma 5.3, F has the same outcome as F \ {u, v}. Thus, we can assume that all the connected components of F have at least three vertices. If F has two cherries, then o(F ) = D by Lemma 5.6. If it has one cherry (c, l, l ), then by Lemma 5.7, o(F ) = A if and only if there is a matching in F \ {c} that covers V (F ) \ N [c]. Thus, we can assume that F has no cherry. If S F = ∅, then o(F ) = A (Lemma 5.8). Otherwise, one can remove components of F that have an empty skeleton (Lemma 5.9). If S F induces a star of center c such that c is not adjacent to

Then Alice wins if and only if all the components of S F are favorable to her and at most one is weakly favorable (Theorem 5.23). Assume now Proof. Let P C = [Ao n B] U . We proceed by contraposition. Assume that C is not favorable. We will prove that o([AoA] ∪ P C , B) = D, which gives the result by using Lemma 5.15. We split the proof into a few claims. We denote by v -1 the unclaimed vertex of [AoA] and by v 1 , ..., v n the unclaimed vertices of P C . We consider for the labeling that

Proof of the claim. When n = 6, Bob claims v 3 , which enforces Alice to reply in v 4 (if Alice replies in v 2 , then Bob claims v 5 and create a double trap). Then Bob successively claims v 1 and v -1 and, therefore dominates before Alice.

If n ≥ 8, v 7 is labeled by 1. By Lemma 5.5, the position can be reduced to the case n = 6, and thus is also Draw.

Proof of the claim. Assume first that i = 1.

If n ∈ {4, 5}, then Bob first claims v 3 , (which forces Alice to answer v 4 or v 2 ). Then Bob claims v -1 and dominates before Alice (who does not dominate v 2 or v 4 ).

If n ≥ 6, either v 5 or v 6 is labelled by 1. Then as before, we can apply Lemma 5.5 to reduce to the case n ∈ {4, 5}.

We can now assume that i ≥ 4. There are two cases, according to the parity of the value n -i. In each case, Lemma 5.5 is used.

, for some set U . We thus have

The first equality comes from Lemma 4.2 and the second one is obvious since Bob can dominate in one move. This gives the desired result in this case. * For n = 5, Alice can also reply in v n-1 . In such a situation, Bob claims v -1 and directly dominates before Alice. If Alice claims v n-2 we are in the same situation as the previous case.

-If n -i ≥ 3, then v i+2 is labeled by 1 and Lemma 5.5 applies to reduce the instance to the case where

, then v i+5 is labeled by 1, and Lemma 5.5 applies to reduce the instance to the case where n -i = 4. Thus, we can assume that i = n -4. Furthermore, since i / ∈ {1, 2, 3, 5}, we can assume that n ≥ 8 and n = 9.

- Then, Bob claims v 7 , which enforces the reply v 9 from Alice. Finally, Bob claims either v 3 or v 4 (one of these vertices is free) and dominates before Alice.

We can now finish the proof of the lemma. Since C is not favorable, we can split the cases according to n as follows:

If n = 1, then the result is obvious, since there is no unfavorable component.

If n = 2 and 1 ∈ U , then Bob claims v -1 and then dominates the graph before Alice, which is a contradiction.

If n = 3 and 1, 2 ∈ U , then Bob claims v -1 and then dominates the graph before Alice, which is a contradiction If n ∈ {4, 5, 6, 7} then Claim 6.3 and Claim 6.4 give the result.

If n ≥ 8 and n is even, the combination of Claims 6.3 and 6.4 gives that U ⊆ {2, 3, n -2}.

If n is odd and n ≥ 13, then the combination of Claims 6.4 and 6.5 gives that either U ⊆ {2, 3, n -2} or U ⊆ {2, 5, n -2}, which gives the result .

If n ∈ {9, 11}, then Claim 6.4 allows to conclude. Lemma 6.6. Assume there are two connected components C, C of S F that are weakly favorable.

Then o(F, {v 0 }, ∅, B)) = D.

Proof. Assume first that both C and C both induce forks. We prove the result for forks with exactly three branches. Indeed, if Bob has a strategy in this case, he will have a strategy for forks with more branches using Lemma We prove by induction of the number p of unclaimed vertices of Q ∪

if G is already dominated by Alice:

if 2 ≤ n ≤ 4, then Alice claims v 3 and dominates the whole graph before Bob; if n = 5, the Alice claims v 3 . We have:

. where the first inequality comes from the fact that Q is dominated by Alice, the second inequality comes from Lemma 2.6 and from the fact that U ⊆ {2, 3}, which is strongly favorable according to Definition 5.22.1.a, and the final equality is obtained by induction hypothesis, using

where the first inequality comes from the fact that Q is dominated by Alice, the second inequality comes from Lemma 2.6 and from the fact that U ⊆ {2, 3, n -2}, which is strongly favorable according to Definition 5.22.1.b, and the final equality is obtained by induction hypothesis, using

if n ≥ 7, then Alice claims v 3 . We have:

, where the first inequality comes from the fact that Q is dominated by Alice, the second inequality comes from Lemma 2.6 and from the fact that U ⊆ {2, 3, n -2}, which is strongly favorable according to Definition 5.22.1.c, and the final equality is obtained by induction hypothesis, using

where the first equality comes from Lemma 2.6 and the inequality comes from two applications of the induction hypothesis.

If i = 1, and n ≥ 3 then Alice claims v 3 . We have

Next corollary is a direct application of Lemma 6.8. Corollary 6.10. For each P ∈ C 0 , we have o(P ∪ [AoA], B) = A.

We can now define the stable class of positions C 1 Definition 6.11 (C 1 ). A position P belongs to the class C 1 if P is the union of the following positions:

1. at most one position in C 0 .

2. a union of positions of the type [Ao n B] {2,5,n-2} n ≥ 9, n odd. Next corollary is a direct application of Lemma 6.8 and Observation 5.19. Corollary 6.13. For each P ∈ C 1 , P admits an A-pairing.

We will now prove that the class C 1 (with a trap [AoA] adjoined) is either stable after Alice's answer to Bob's claim, or directly winning for Alice. We will consider the two cases according to whether Bob claims the unclaimed vertex v -1 of [AoA] or not. The first case where Bob does not claim v -1 is proved by Lemma 6.15 that requires Claim 6.14. The second case is when Bob claims v -1 and is solved by Lemma 6.17 that requires Claim 6.16 as a particular case. Claim 6.14. Let Q be any position, and t be a positive integer. Let (w 1 , w 2 , ..., w 2t ) be the unclaimed vertices of [Bo 2t B] {2t-2} . For each w i , there exists w j = w i and 0 ≤ t < t such that

(with the convention that [Bo 0 B] {-2} is empty).

The idea behind this claim is that if a position contains a bounded path [Bo 2t B] {2t-2} and Bob claims a vertex in it, then Alice can reply in the same bounded path, preserving the global structure of the position.

Proof. We have three cases.

If i = 2t then Alice claims w 2t-1 , using Observations 5.20 and 5.21, we have

Lemma 6.7 applies, and we get

.

Lemma 6.7 applies, and we get

if y = w i , with i ≥ 4, then take x = w i-1 . We have

according to Lemma 6.7. We are done since P ∪ [Ao i-2 A] {2,5} is in C 1 .

Claim 6.16. Let P be any position, and t be a nonnegative integer. We have

where [AooAooA] {2,5} is the position obtained from [Ao In the position ((P ∪ [AoA]) v-1,y , B), Alice has a strategy which allows her to dominate P . Thus, Alice can win in R v2,y playing component by component, until, the game ends as follows:

-If Bob claims in P , Alice also claims in P according to the strategy in ((P ∪ [AoA]) v-1,y , B).

-If Bob claims in [AooAooA] {2,5} , then Alice isolates one vertex using one of the B-trap.

- -If Bob claims a vertex in P x , Alice answers in P x according to her strategy in (P x , B).

-If Bob claims a vertex in [Bo 2t B] {2t-2} , then Claim 6.14] can be used.

-

There exists an unclaimed vertex x of P such that at least one of the following alternatives holds:

there exists P ∈ C 1 with at least one unclaimed vertex less than P , such that o(P ∪

Proof. We have different cases according to the structure of P .

If P contains a component [Ao n B] {2,5,n-2} , n ≥ 9, n odd, let v 1 , v 2 , ..., v n denote the sequence of free vertices of this component. Take x = v 2 . Let P be the position such that P = P ∪ [Ao n B] {2,5,n-2} and let

from Lemma 6.7.

If P contains a component [Ao n A] {2,5} , let w 1 , w 2 , ..., w n denote the sequence of vertices of [Ao n A] {2,5} . Let P be the position such that P = P ∪ [Ao n A] {2,5} and let Q = P ∪ [AoA].

-If n ≥ 6, then take x = w n-1 . We have

and note that P ∪ [Ao n-2 A] {2,5} is an element of C 1 .

-If 3 ≤ n ≤ 4, then take x = w 2 . We have

and we conclude as previously.

-If n = 1, then take x = w 1 . The position Q w1,v-1 admits an A-pairing from Corollary 6.13. Thus, from 3.6, we have o(Q w1,v-1 , B) = A.

-If p = 5 and k ≥ 1, take x = w 3 , which creates an [AooAooA]. We define P such that