Maker-Breaker domination game on trees when Staller wins

In the Maker-Breaker domination game played on a graph G , Dominator’s goal is to select a dominating set and Staller’s goal is to claim a closed neighborhood of some vertex. We study the cases when Staller can win the game. If Dominator (resp., Staller) starts the game, then γ SMB ( G ) (resp., γ ′ SMB ( G ) ) denotes the minimum number of moves Staller needs to win. For every positive integer k , trees T with γ ′ SMB ( T ) = k are characterized and a general upper bound on γ ′ SMB is proved. Let S = S ( n 1 , . . . , n ℓ ) be the subdivided star obtained from the star with ℓ edges by subdividing its edges n 1 − 1 , . . . , n ℓ − 1 times, respectively. Then γ ′ SMB ( S ) is determined in all the cases except when ℓ ≥ 4 and each n i is even. The simplest formula is obtained when there are at least two odd n i s. If n 1 and n 2 are the two smallest such numbers, then γ ′ SMB ( S ( n 1 , . . . , n ℓ )) = ⌈ log 2 ( n 1 + n 2 + 1) ⌉ . For caterpillars, exact formulas for γ SMB and for γ ′ SMB are established.


Introduction
The Maker-Breaker game was introduced in Erdős and Selfridge (1973).The game is played on an arbitrary hypergraph by two players who alternately select a non-played vertex of the hypergraph.One player, named Maker, wants to occupy all the vertices of some hyperedge, while the other player, named Breaker, tries to prevent him from doing it.If the first situation happens, then Maker is declared as the winner of the game, otherwise Breaker wins.The game, either in its general form, or in different special cases, was investigated a lot by now, see the book Hefetz et al. (2014).For related recent developments see Clemens et al. (2021); Day and Falgas-Ravry (2021); Nicholas Day and Falgas-Ravry (2021); Glazik and Srivastav (2022); Kang et al. (2021); Stojaković and Trkulja (2021) and references therein.The Maker-Breaker games have been recently studied also on digraphs in Frieze and Pegden (2021).

2
Csilla Bujtás et al.The Maker-Breaker domination game, MBD game for short, was introduced by Duchêne, Gledel, Parreau, and Renault in Duchêne et al. (2020).Among other results they proved that deciding the winner of the MBD game is PSPACE-complete in general, and showed that the problem can be solved efficiently on trees.Following the tradition of the theory of the domination game, the two players in this version of the Maker-Breaker game are called Dominator and Staller.This game was introduced in Brešar et al. (2010), its state of the art till 2021 summarized in Brešar et al. (2021), and is still being investigated, see e.g.Bujtás et al. (2022).We also add that the Maker-Breaker total domination game was introduced in Gledel et al. (2020) and further investigated in Forcan and Mikalački (2022).
The MBD game played on a graph G = (V (G), E(G)) can be considered as the Maker-Breaker game played on the hypergraph D G whose hyperedges are the minimal dominating sets of G.In this case, Dominator is Maker, and Staller is Breaker.Moreover, this game can also be considered as the Maker-Breaker game played on the closed neighborhood hypergraph H G of G, where the hyperedges are the closed neighborhoods of the vertices of G. Now Dominator is Breaker and Staller is Maker.
A MBD game is called D-game (resp., S-game) if Dominator (resp., Staller) is the first to play a vertex.Suppose that Dominator has a winning strategy in the D-game.Then the Maker-Breaker domination number, γ MB (G), of G is the (minimum) number of moves of Dominator to win the game when both players play optimally.The corresponding invariant for the S-game is denoted by γ ′ MB (G).These concepts were introduced in Gledel et al. (2019) and further studied on prisms in Forcan and Qi (2023).Clearly, the problem of determining the Maker-Breaker domination number is interesting when Dominator wins the game.For the situations in which Staller is the winner, the Staller-Maker-Breaker domination number (SMBD-number for short), γ SMB (G), of G, is the (minimum) number of Staller's moves she needs to win the D-game if both players play optimally.If Staller has no winning strategy in the D-game, we set γ SMB (G) = ∞.For the S-game, the corresponding invariant is defined analogously and denoted by γ ′ SMB (G).These two invariants are from Bujtás and Dokyeesun (2022).In this paper we proceed the investigation of the SMBD-numbers.In Gledel et al. (2019), exact formulas for γ MB (T ) and γ ′ MB (T ) were proved if T is a tree.To determine γ SMB (T ) and γ ′ SMB (T ) turns out to be much more involved.In the main result of Section 3 we characterize, for every positive integer k, the class of trees T with γ ′ SMB (T ) = k.In the subsequent section we study subdivided stars.Since γ SMB (T ) = ∞ holds for each subdivided star T , we focus on γ ′ SMB (T ) and determine the exact value in all the cases except when the star has at least four edges, and each edge is subdivided an odd number of times.For the latter case we prove a sharp upper bound on γ ′ SMB .In Section 5 we determine γ SMB (T ) and γ ′ SMB (T ) for an arbitrary caterpillar T .

Definitions
A dominating set S is minimal if there is no dominating set among the proper subsets of S.
The residual graph R(G) of a graph G is obtained from G by iteratively removing a pendant path P 2 until no such path is present.By a pendant P 2 we mean a P 2 -component or a P 2 attached to the graph by an edge.In the latter case, when a pendant P 2 is removed, exactly two vertices and two edges are deleted, whilst in the former case we obtain the empty graph.If T is a tree, then R(T ) is either the empty graph, or R(T ) = K 1 , or each support vertex of R(T ) is of degree at least 3.It was proved in Duchêne et al. (2020) that the winner is the same in T and in R(T ).
A hypergraph H = (V (H), E(H)) is a set system over the vertex set V (H).The (hyper)edge set E(H) might contain subsets of V (H) of any cardinality that is, E(H) ⊆ 2 V (H) .If |e| = 2 holds for every hyperedge e ∈ E(H), the hypergraph H corresponds to a graph.Thus, most of the basic definitions related to graphs can be generalized to hypergraphs (see Berge (1989)).For example, if H 1 and H 2 are two hypergraphs, we say that While studying Maker-Breaker games, we will refer to two operators on hypergraphs.Given a subset X of the vertex set V (H), the hypergraph H − X is obtained from H by removing the vertices in X and all incident edges.That is, The second operator, named shrinking, creates a hypergraph denoted by H | X.Here, the vertices in X are deleted again from H but the incident hyperedges are just 'shrinked' instead of being deleted.
When a Maker-Breaker game is played on a hypergraph H, we say that the hyperedges of H are the winning sets and Maker wins the game with his i th move if he claims a winning set with this move.A winning strategy of Maker is a strategy which ensures that he wins on H no matter what strategy is applied by Breaker.Assuming that Maker has a winning strategy, we further assume that his goal is to win the game as soon as possible and that Breaker's goal is the opposite.We say that the players play optimally, if they play complying with their goals.The winning number of Maker, denoted by w M M (H) (resp., w B M (H)), is the minimum number of his moves he needs to win the game if both players play optimally and Maker (resp., Breaker) starts the game Bujtás and Dokyeesun (2022).If Maker has no winning strategy as a first (resp., second) player, we set w M M (H) = ∞ (resp., w B M (H) = ∞).These general winning numbers were introduced in Bujtás and Dokyeesun (2022), but the problem of how fast Maker can win was studied earlier in several papers including Beck (1981); Clemens et al. (2012); Clemens and Mikalački (2018); Hefetz et al. (2009)

Known results
In this subsection we collect known results that will be used later.
Proposition 2.1 (Bujtás and Dokyeesun (2022)) Suppose that a Maker-Breaker game is played on a hypergraph H and v ∈ V (H).
(i) If Breaker is the first player and she plays v, the continuation of the game corresponds to a Makerstart game on H − v.In particular, if v is an optimal first move of Breaker, then We remark that Proposition 2.3 and its proof in Bujtás and Dokyeesun (2022) directly imply that one of the optimal strategies of Maker is to play (optimally) on the component H 1 .
Proposition 2.4 (Bujtás and Dokyeesun (2022) , then the following statements hold: Proposition 2.5 (Bujtás and Dokyeesun (2022)) (i) Let G ′ be a graph obtained from G by removing a weak support vertex and the adjacent leaf.Then, the following inequalities hold: (ii) Let G ′ be a graph obtained from G by removing a weak support vertex of degree 2 and the adjacent leaf.Then, γ SMB (G) − 1 ≤ γ SMB (G ′ ) holds.
Theorem 2.6 (Duchêne et al. (2020); Gledel et al. (2019)) For a tree T , (i) γ SMB (T ) < ∞ if and only if R(T ) contains at least two strong support vertices; (ii) γ ′ SMB (T ) < ∞ if and only if T does not admit a perfect matching.
Note that, by Theorem 2.6(ii), γ ′ SMB (P n ) = ∞ if n is even.Applying the "pairing strategy" from Hefetz et al. (2014), Dominator has a winning strategy in both the D-game and S-game if the graph has a perfect matching.This fact can be extended as follows.
Lemma 2.7 (Bujtás and Dokyeesun (2022)) Consider an MBD game on G and let X and Y be the sets of vertices played by Dominator and Staller, respectively, until a moment during the game.If there exists then Dominator has a strategy to win the continuation of the game, no matter who plays the next vertex.
Theorem 2.8 (Bujtás and Dokyeesun (2022)) If n is an odd positive integer, then Moreover, Staller has an optimal strategy in the S-game on P n such that she wins on the closed neighborhood of a vertex v that is at an even distance from the ends of the path.
If n ≥ 3 and n is odd, we may use the formula ⌈log 2 n⌉ instead of ⌊log 2 n⌋ + 1.In later sections we also frequently write ⌈log 2 n⌉ instead of ⌊log 2 (n − 1)⌋ + 1 if n ≥ 2 is even.
3 Characterizing trees with γ ′ SMB = k In Gledel et al. (2019), exact formulas for γ MB (T ) and γ ′ MB (T ) were proved if T is a tree.These values and the corresponding optimal strategies of Dominator are not difficult to determine by iteratively removing a pendant path P 2 and constructing the residual graph R(T ).On the other hand, as Theorem 2.8 and its proof given in Bujtás and Dokyeesun (2022) indicate, determining γ SMB (T ) and γ ′ SMB (T ) for an arbitrary tree T turns out to be a much more difficult problem.
Thus, the aim of this section is, for a given integer k, to characterize the trees T with γ ′ SMB (T ) = k.Combining this result with properties of closed neighborhood hypergraphs, we also establish a sufficient condition under which γ ′ SMB (G) ≤ k holds for an arbitrary graph G. First, we state a lemma that will be used in the proof of the main theorem.
Lemma 3.1 Let T be a tree with 2 ≤ γ ′ SMB (T ) < ∞ and suppose that s 1 is an optimal first move of Staller in the S-game on T .
(i) If Dominator's response is playing a vertex d 1 which is a neighbor of s 1 , then Staller's optimal next move s 2 and d 1 belong to different components of T − s 1 .
(ii) After Staller's optimal move s 1 , there exists an optimal response d ′ 1 of Dominator such that d ′ 1 ∈ N T (s 1 ).
Proof: Let T 1 , . . ., T ℓ be the components of T − s 1 .(i) Suppose for a contradiction that d 1 and s 2 belong to the same component, say T 1 .Proposition 2.1 implies that, after the moves s 1 and d 1 , the game continues as a Maker-start game with the winning sets in Further, since by our assumption, s 1 and d 1 are adjacent vertices in T , the hypergraph H ′ contains the components H T2 , . . ., H T ℓ and all components of , and one the optimal strategies of Staller (Maker) is to play in the same (appropriate) component of H ′ starting with her move s 2 .As s 2 ∈ V (H 1 ) is supposed to be an optimal move, w M M (H ′ ) = w M M (H 1 ).On the other hand, H 1 is a subhypergraph of H T and Proposition 2.2 thus implies w M M (H 1 ) ≥ w M M (H T ).This gives the following contradiction: (ii) Suppose d 1 is an optimal response of Dominator to the move s 1 and d 1 / ∈ N T (s 1 ).We may also suppose, without loss of generality, that d 1 ∈ V (T 1 ).Let d ′ 1 be the neighbor of s 1 from the subtree T 1 .We will show, by using imagination strategy, that d ′ 1 is also an optimal response of Dominator.Let Game 1 be an S-game on T and let Staller start by playing s 1 and Dominator's response is d ′ 1 .After that, Staller plays optimally as follows: 1 with the possible smallest winning number and plays optimally all of her moves in this component.
Note that by Proposition 2.3 the strategy (*) is indeed an optimal one for Staller from the second move.
Let Game 2 be an S-game on T in which Dominator plays optimally and his response to the move s 1 is d 1 .In the continuation, Staller always selects an optimal move s i in Game 1 and copies her move to Game 2 if possible, while Dominator selects an optimal move d i in Game 2 and copies it to Game 1 if possible.Let t k be the number of moves Staller needs to win in Game k for k ∈ [2].
By (i), the optimal second move s 2 of Staller in Game 1 cannot belong to T 1 .Let us assume s 2 ∈ V (T j ) with j = 1.Then according to (*), all the remaining moves of Staller will be in T j .In particular, Staller never plays the vertex d 1 .On the other hand, if Dominator plays 1 in Game 2, then we replace this move by d 1 in Game 1. Otherwise Dominator's move can always be copied into Game 1.
At the end, Staller wins Game 1 with her (t 1 ) st move by claiming a winning set from H Tj .As this winning set is also claimed in Game 2 (or Staller has already won in Game 2), Staller wins Game 2 in t 2 ≤ t 1 moves.Since Staller plays optimally throughout Game 1, we infer γ ′ SMB (T ) ≥ t 1 .Similarly, t 2 ≥ γ ′ SMB (T ) holds because Dominator plays optimally in Game 2. Therefore, ) is also an optimal move for Dominator after s 1 .✷ Definition 3.2 We define a family S k of graphs for each positive integer k, and for each S ∈ S k a subset of vertices X(S), as follows.
• Let S 1 = {P 1 } and let X(P 1 ) contain the only vertex of P 1 .
• For k ≥ 2, let S * k be the set of graphs that can be obtained in the following way.Take two vertex disjoint graphs k is obtained by making adjacent the origin vertex z k with x 1 and x 2 .We define X(S) = X(S 1 ) ∪ X(S 2 ).
• The family S k is obtained as We also introduce the notation S = ∞ k=1 S k and say that S ∈ S is a substructure in the graph G if S is a subgraph of G and deg G (x) = deg S (x) holds for each x ∈ X(S).We further say that the vertices from X(S) have fixed degrees.
In Fig. 1, families S 1 , S 2 , S 3 , and S 4 , are presented, where the black vertices are the vertices with fixed degrees.Note that the vertices in V (S) \ X(S) may be incident with any number of additional edges in G. Due to the recursive definition of S k , each of the following statements can be easily verified by proceeding with induction on k, cf.Fig. 1   (i) S is a tree in which all leaves belong to the same partite class.Moreover, this partite class corresponds to X(S).
In particular, deg S (u) = 2 holds for every support vertex u of S.
(iii) If a tree T contains a substructure S, then every leaf of S is a leaf in T and every support vertex of S is a support vertex in T .
Observation 3.3(i) shows that, for each k, the family S k can be considered simply as a family of graphs, without associating each S ∈ S k with a set X(S) ⊆ V (S).The fixed vertex degrees are easy to identify by considering deg S (v) for every vertex v which is in the same partite class as the leaves of S.
In the following proposition, we state a simple characterization for the family S.For it recall that the subdivision graph, S(G), of a graph G is the graph obtained from G by subdividing each edge exactly once, cf.Burzio and Ferrarese (1998).
Proof: Set S ′ = {S(G) : G is a tree}.We are going to prove that S = S ′ .
Clearly, the elements of S ′ are trees.Moreover, if T is an arbitrary tree with partite classes A 1 and A 2 such that A 1 contains at least one leaf or an isolated vertex, then it is straightforward to observe that T ∈ S ′ if and only if If T ∈ S, Observation 3.3(i) and (ii) yield the desired property ( * ) with A 1 = X(T ) and A 2 = V (T ) \ X(T ).Hence S ⊆ S ′ .
To prove that S ′ ⊆ S also holds, suppose that T ∈ S ′ .Then T fulfils ( * ).We proceed by induction to conclude T ∈ S.This clearly holds if the order of T is at most 3. Let |V (T )| > 3.Then, A 2 = ∅, and we fix an arbitrarily chosen vertex z from A 2 .By our condition, deg T (z) = 2 and hence T − z consists of two components, say T 1 and T 2 .Observe that deg T1 (v) = deg T (v) holds for every v ∈ V (T 1 ) except for the neighbor of z from T 1 .Since z ∈ A 2 , this neighbor is from A 1 , and therefore, we infer that Applying the induction hypothesis to T 1 , we conclude T 1 ∈ S. In the same way, we may obtain T 2 ∈ S. Definition 3.2 then shows that T can be constructed from T 1 , T 2 , and the origin vertex z, and therefore T ∈ S k for an appropriate integer k. ✷ Theorem 3.5 Let T be a tree and k a positive integer.Then, γ ′ SMB (T ) = k if and only if T contains a substructure S from S k , and no substructure from ∪ k−1 i=1 S i .
Proof: The statement holds for k = 1 as Staller can win the S-game with her first move if and only if there exists a winning set N T [v] of cardinality one.It clearly means that T is just an isolated vertex P 1 or, equivalently, there is a substructure P 1 ∈ S 1 in T .From now on, we may proceed by induction on k.
First, assume that T contains a substructure S from S k , where k ≥ 2, but contains no substructure from S j if j < k.The latter condition implies, by the induction hypothesis, that γ ′ SMB (T ) ≥ k.Now consider the substructure S ∈ S k in T and its origin vertex z k .By the construction of the graphs belonging to S k , the forest T − z k contains one component T 1 with a substructure S 1 ∈ S k−1 and another component, say follows that completes the proof of one direction of the statement.Now, assume that γ ′ SMB (T ) = k.By the hypothesis, T contains no substructure from ∪ k−1 p=1 S p .Let s 1 be an optimal first move of Staller in the S-game and d 1 an optimal response of Dominator.By Lemma 3.1(ii), we may suppose that d 1 is a neighbor of s 1 .Let T 1 , . . ., T ℓ be the components of T − s 1 such that d 1 ∈ V (T 1 ).After the move d 1 , by Proposition 2.1, the game continues on H ′ = (H T | s 1 )− d 1 with Staller's next optimal move s 2 .By Lemma 3.1(i), s 2 / ∈ V (T 1 ).By our supposition, the moves s 1 and d 1 were optimal and therefore, w M M (H ′ ) = k − 1. Observe that H ′ consists of several components and ℓ − 1 of them exactly correspond to the closed neighborhood hypergraphs H T2 , . . ., It also follows that Staller's (Maker's) optimal strategy is to play in the same (appropriate) component of H ′ starting with her move s 2 .As s 2 / ∈ V (H 1 ), there exists a tree T j , 2 ≤ j ≤ ℓ, such that γ ′ SMB (T j ) = k − 1.By our hypothesis, T j contains a substructure S 1 from S k−1 .As S 1 is a subgraph but not a substructure in T , the vertex s 1 must be adjacent to a vertex t j from X(S 1 ).Note that, as T is a tree, s 1 is adjacent to exactly one vertex t i from each T i , i ∈ [ℓ].Consequently, no component T i can contain two vertex disjoint substructures from ∪ k−1 p=1 S p and, if T i contains a substructure S from ∪ k−1 p=1 S p , then t i , the neighbor of s 1 in T i , belongs to X(S).
We will show that T − s 1 contains a substructure S 2 from ∪ k−1 p=1 S p that is vertex disjoint to S 1 .Assume for a contradiction that it is not true.Thus, by the hypothesis, γ ′ SMB (T i ) ≥ k holds for every i = j.In this situation, suppose that Dominator plays d ′ 1 = t j instead of d 1 = t 1 .By Lemma 3.1(i), each remaining move of Staller belongs to a a subtree T i , i = j, and she needs at least k further moves to win the game.This contradicts the condition γ ′ SMB (T ) = k and the assumption that Staller plays optimally.To conclude the proof, we infer that T − s 1 contains two different components, T j and T s , j = s, such that an S 1 ∈ S k−1 is a substructure in T j and an S 2 ∈ ∪ k−1 p=1 S p is a substructure in T s .Further, s 1 is adjacent to t j and t s which belong to X(S 1 ) and X(S 2 ) respectively.By the definition of S k , the subgraph induced by V (S 1 ) ∪ V (S 2 ) ∪ {s 1 } is included in S k and, as the subgraph complies with the fixed vertex degrees, it is a substructure from S k in T .
✷ Remark 3.6 Observation 3.3(v) shows that P n ∈ S ⌊log 2 n⌋+1 for every odd integer n.Then, by Theorem 3.5, we may conclude γ ′ SMB (P n ) = ⌊log 2 n⌋ + 1.This provides an alternative shorter proof for the formula in Theorem 2.8.

For any graph G and a subset A ⊆ V (G), the hypergraph H −A
G is obtained from the closed neighhborhood hypergraph H G by deleting the hyperedges corresponding to the closed neighborhoods of vertices in A.
To state an upper bound on the winning numbers of hypergraphs, we define a family F k for every positive integer k as follows: Alternatively, F k can be defined recursively starting with F 1 = {B} where B is the hypergraph that consists of one vertex x and one edge {x}.Then, for every k ≥ 2, the family F * k contains a hypergraph A if A can be obtained in the following way.Choose two hypergraphs A 1 ∈ F k−1 and A 2 ∈ ∪ k−1 i=1 F i and select two edges e 1 ∈ E(A 1 ) and e 2 ∈ E(A 2 ).To finish the construction, we take a new (origin) vertex u, define e ′ i = e i ∪ {u} for i ∈ [2] and set After having the family F * k in hand, we set The following consequence of Proposition 3.7 gives an upper bound for the SMBD-numbers of graphs in general.Note that the corollary can also be obtained by analyzing the proof of Theorem 3.5.
M (H G ), this proves the statement.✷ In view of Theorem 3.5, we pose: Open Problem 3.9 Determine the computational complexity of the decision problem whether γ ′ SMB (G) ≤ k holds for a graph G and a positive integer k, where k is part of the input.In particular, what about the same question restricted to the class of trees?

Subdivided stars
In this section we consider the SMBD-numbers of subdivided stars.Let S(n 1 , . . ., n ℓ ) be the tree obtained from the paths P n1 , . . ., P n ℓ and a central vertex x by making x adjacent to one end of each path.The paths P n1 , . . ., P n ℓ obtained after the deletion of the central vertex are called branches.Throughout, we assume that ℓ ≥ 2 and n 1 , . . ., n ℓ are positive integers.
For a subdivided star T = S(n 1 , . . ., n ℓ ), the residual graph R(T ), which is obtained by iteratively removing pendant edges, clearly contains at most one strong support vertex.Therefore, by Theorem 2.6(i), γ SMB (T ) = ∞ always holds.Concerning the S-game on T , we consider three cases according to the parities of n 1 , . . ., n ℓ .If there is exactly one odd number among n 1 , . . ., n ℓ , then T has a perfect matching and γ ′ SMB (T ) = ∞ follows from Theorem 2.6(ii).If there are at least two odd numbers among n 1 , . . ., n ℓ , Theorem 4.1 will establish the explicit formula.The last case is when T is an all-even subdivided star that is, n i is even for each i ∈ [ℓ].If ℓ = 2, T is a path and Theorem 2.8 establishes the formula for γ ′ SMB (T ).If ℓ = 3, our Theorem 4.3 will answer the question.For larger values of ℓ, we provide a sharp upper bound, but the exact formula for SMBD-numbers of all-even subdivided stars with at least four branches remains an open problem.

Not-all-even subdivided stars
As Staller cannot win the S-game on T = S(n 1 , . . ., n ℓ ) if T contains exactly one odd branch, we concentrate on the case when there exist at least two odd branches.
Let T = S(n 1 , . . ., n ℓ ) and k = ⌈log 2 (n 1 + n 2 + 1)⌉.By Theorem 3.5, it suffices to show that T contains a substructure from S k and contains no substructure from ∪ k−1 i=1 S i .The first part of the statement clearly holds because the paths P n1 and P n2 together with the central vertex x form a subgraph P n1+n2+1 in T .By Observation 3.3(v), P n1+n2+1 ∈ S k .Further, by Observation 3.3(i), since n 1 and n 2 are odd numbers, x does not belong to X(P n1+n2+1 ).We infer that P n1+n2+1 is a substructure in T .
Assume that T contains a substructure S ∈ S i .By Observation 3.3(i) and (ii), S is a tree where X(S) corresponds to the partite class containing all leaves of S, while all vertices in the partite class V (S) \ X(S) are of degree 2 in S. As at least two leaves of T together with the path between them are present in S, the central vertex x also belongs to S.
• Suppose that x ∈ X(S).Then N T [x] ⊆ V (S) and hence S ∼ = T .However, as some odd numbers are present among n 1 , . . ., n ℓ , Observation 3.3(i) implies x / ∈ X(S) that contradicts the supposition x ∈ S.
• Suppose that x / ∈ X(S).By Observation 3.3(i) and (ii), deg S (x) = 2 and x is at an odd distance from each leaf of S. We infer that the substructure S is a path induced by {x} ∪ V (P na ) ∪ V (P n b ), where n a and n b are odd integers.By our condition, n a + n b + 1 ≥ n 1 + n 2 + 1 and, consequently, i ≥ k whenever S ∈ S i holds.
is a tree and the set X(S) contains all leaves of S and the vertices being at an even distance from a leaf.By Observation 3.3(iii), every leaf of S is a leaf in T and therefore, x ∈ X(S).Hence, N S (x) = N T (x) and all leaves of T are present in S. Consequently, S = T and T ∈ S hold.
Second, we prove that γ ′ SMB (T ) equals either ⌈log 2 n⌉ or ⌈log 2 n⌉ + 1.As T ∈ S, Observation 3.3(iv) and Theorem 3.5 immediately give n ≤ 2 γ ′ SMB (T ) − 1.This proves ⌈log 2 n⌉ ≤ γ ′ SMB (T ).For the upper bound, we consider the cut vertex t 3 that is the neighbor of x from the branch P 3 .The graph T − t 3 contains two components one of which is a path of order n 3 − 1 and the other one is a path of order n 1 + n 2 + 1. Recall that if m is a positive even integer, then ⌈log m⌉ = ⌊log(m − 1)⌋ + 1.Using this fact, Proposition 2.5(iii), Theorem 2.8, and inequalities n 3 < n and n 1 + n 2 + 1 < n, we can estimate γ ′ SMB (T ) as follows: We have just proved that γ ′ SMB (T ) ∈ {⌈log 2 n⌉, ⌈log 2 n⌉ + 1}.To decide which of the two possibilities holds and to prove the theorem, we first consider all non-reducible graphs and then proceed by induction on the number of vertices when dealing with reducible graphs.Let k = ⌈log 2 n⌉.
Case 1: T is a non-reducible graph.By the definition of non-reducibility, ⌈log 2 (n 3 )⌉ < k.We consider two cases.
Case 1.1: T is non-reducible and T ∈ T 0 .By (3), Considering the cut vertex t 3 ∈ N T (x) ∩ V (P 3 ) again, Proposition 2.5(iii) and Theorem 2.8 imply γ ′ SMB (T ) ≤ M + 1 ≤ k for this case.Together with the lower bound k ≤ γ ′ SMB (T ), this proves γ ′ SMB (T ) = k for every non-reducible T from T 0 .Case 1.2: T is non-reducible and T ∈ T 1 .Since Staller can win the S-game on T , there is an optimal first move s 1 for her.Note first that s 1 = x.Indeed, if s 1 = x, then Dominator plays any neighbor of x and the undominated vertices can be covered by a matching, hence Dominator will win the game according to Lemma 2.7.We may claim the same if all components of T − s 1 are even.To see it, observe that Dominator can always play a neighbor of s 1 such that the undominated vertices can be covered by a matching.We have thus demonstrated that s 1 = x and that not all components of T − s 1 are even.
If s 1 is from P 3 , let T 1 and T 2 be the two components of T −s 1 such that V (T 1 ) ⊆ V (P 3 ).Dominator may play the neighbor of s 1 that belongs to T 1 .Then, by Lemma 3.1, Staller's optimal response is a vertex from T 2 and, under optimal strategies, the game continues on T 2 and finishes with the (γ ′ SMB (T 2 ) + 1) st move of Staller.As T 2 is an all-even subdivided star or an odd path, T 2 ∈ S.Then, by (2), Observation 3.3(iv), and by our condition, we get This proves that after Staller's move s 1 ∈ V (P 3 ), Dominator has a strategy which ensures that Staller cannot win before her (k + 1) st move.The proof is similar, if s 1 is from P 1 or P 2 .We conclude γ ′ SMB (T ) ≥ k + 1, which in turn results in the desired equality γ From now on, we proceed by induction.
Case 2.1: T is reducible and T ∈ T 0 .Definition 4.2(ii) allows two possibilities for T belonging to T 0 .We first show that γ From this inequality, we may infer Let w be the (unique) vertex of P 3 such that T − w consists of a component isomorphic to T ′ and a path component of order 2 k−1 − 1.We have already seen that γ ′ SMB (T ′ ) ≤ k − 1 and the path component clearly satisfies γ ′ SMB = k − 1. Applying Proposition 2.5(iii), we get Since γ ′ SMB (T ) ≥ k also holds, we conclude γ ′ SMB (T ) = k as stated in the theorem.Case 2.2: T is reducible and T ∈ T 1 .We are going to prove that γ ′ SMB (T ) = k + 1.As γ ′ SMB (T ) ≤ k + 1 always holds, it suffices to show that there is no optimal first move s 1 for Staller that allows her to win within k (or less) moves.
• If T − s 1 contains no component of odd order, then Staller cannot win in the continuation of the game.
• If Staller plays a vertex s 1 ∈ V (P 1 ) ∪ V (P 2 ), then Dominator responds by choosing the neighbor of s 1 which is not incident to the (s 1 , x)-path.Then, by Lemma 3.1, Staller's optimal strategy is playing on the component This shows that if Staller chooses her first vertex from P 1 or P 2 , she cannot win with k moves.
• If s 1 ∈ V (P 3 ) such that T − s 1 contains a path component of order at least 2 k−1 + 1, then we can handle it as in the previous paragraph.
• Let w be the vertex from P 3 such that T − w contains a component isomorphic to T ′ and a path of order 2 k−1 − 1.If s 1 = w, Dominator may reply by playing the neighbor of w from the path and then, by Lemma 3.1, Staller's optimal strategy is to play (optimally) on T ′ in the continuation.
Hence, she needs at least γ ′ SMB (T ′ ) + 1 moves to win the game.By Definition 4.2, T ∈ T 1 only if and hence ⌈log 2 n ′ ⌉ ≥ k − 1.By hypothesis, T ′ ∈ T 1 implies γ ′ SMB (T ′ ) = ⌈log 2 n ′ ⌉ + 1 and therefore, we have γ ′ SMB (T ′ ) ≥ k by the above inequality.We may conclude again that Staller needs at least k + 1 moves to win the game.
Open Problem 4.5 Determine the value of γ ′ SMB (S(n 1 , . . ., n ℓ )) for the cases when ℓ ≥ 4 and n i is a positive even number for each i ∈ [ℓ].
Proof: Let x be the center of T , and P 1 , . . ., P ℓ be the branches of order n 1 , . . ., n ℓ respectively.The neighbor of x from P i will be denoted by t i for i ∈ [ℓ].
To prove the upper bound, we describe a strategy for Staller.In her first move, Staller plays t 1 .We have two cases.
• If Dominator replies with a move d 1 from the component of T − t 1 which contains x, Staller continues the game by playing (optimally) on the path P 1 − t 1 .After the moves t 1 , d 1 , the Maker-Breaker game continues on the hypergraph H ′ = (H T − t 1 ) | d 1 with Staller's (Maker's) move.As In this case, Staller can win the S-game on T in 1 + ⌈log 2 n 1 ⌉ moves.
• If Dominator replies with a move d 1 ∈ V (P 1 − t 1 ), then Staller plays t 2 as her next move.
In general, Staller's strategy is the following.For i ∈ [ℓ − 2], if Staller plays t i ∈ N T (x) ∩ V (P i ) as her i th move and Dominator responds by playing a vertex d i / ∈ V (P i ), then Staller continues playing (optimally) on P i − t i .It allows her to win the game in i + ⌈log 2 n i ⌉ moves.If Dominator responds by choosing a vertex d i ∈ V (P i ) and i ≤ ℓ − 3, Staller's next move is t i+1 .If Dominator's reply is d i ∈ V (P i ) and i = ℓ − 2, Staller plays on the path P ′ induced by V (P ℓ−1 ) ∪ {x} ∪ V (P ℓ ) and can win the game in ℓ − 2 + ⌈log 2 (n ℓ−1 + n ℓ + 1)⌉ moves.
Consequently, either there exists an i ∈ [ℓ − 2] such that Staller plays t i and Dominator replies with d i / ∈ V (P i ) or Staller, after playing t 1 , . . ., t ℓ−2 , can win by playing on P ′ .In the former case, Staller plays at most i + ⌈log 2 n i ⌉ vertices; in the latter case she can win in at most ℓ − 2 + ⌈log 2 (n ℓ−1 + n ℓ + 1)⌉ moves.It verifies the upper bound in the statement.

Caterpillars
A caterpillar is a tree T on at least three vertices in which a single path is incident to (or contains) every edge.There is more than one path with this property, but we select the shortest one and call it the spine of the caterpillar, see West (1996).The vertices of the spine will be denoted by v 1 , . . ., v ℓ according to their natural order.Before stating the main result of the section, some preparation is needed.
Let T be a caterpillar.Then we set where maximal is meant maximal w.r.t.inclusion.We say that a path Let further P co (T ) = {P : P ∈ P o (T ) and P is clean} .
We refer to the members of P co (T ) as P co -paths.Note that each such path contains at most two support vertices from the spine of T .If P is a collection of paths of T , then the smallest carnality of a set of vertices U such that each path from P has a vertex in U is denoted by t p (P).If T is a caterpillar containing at least one clean path, then let p(T ) be the minimum order of such a path.If t p (P co (T )) ≥ 2, then let p * (T ) be the smallest integer s such that not all clean paths of order at most s share a vertex, that is, p * (T ) = min{s : t p ({P : For the illustration of the above concepts and notation see Fig. 2. For the caterpillar T from the figure we have Clearly, p(T ) = 3. Next, t p (P co (T )) = 2 because each path from P co (T ) has a vertex in {v 3 , v 4 } and there is no such set with a single vertex.Finally, p * (T ) = 7 because each clean path of order smaller than 7 contains the vertex v 3 .Proof: Let T be a caterpillar and set p = p(T ) and p * = p * (T ).
Suppose first that P co (T ) = ∅, which also implies that there are no strong support vertices in T .We can construct a perfect matching in T as follows.First, delete the support vertices of T from the subgraph induced by the spine of T .As P co (T ) = ∅, each component of the obtained linear forest T − is an even path.Therefore, T − admits a perfect matching.For the remaining vertices of T , we may match every leaf with its support vertex.As there are no strong support vertices in T we obtain a perfect matching in T .Therefore, we may conclude by Theorem 2.6 that γ ′ SMB (T ) = γ SMB (T ) = ∞.In the rest of the proof, let P co (T ) = ∅.
We now prove the statement for γ ′ SMB (T ).Since P 1 is not a caterpillar, γ ′ SMB (T ) ≥ 2 holds.Let k = ⌈log 2 p⌉.By Theorem 3.5, γ ′ SMB (T ) = 2 if and only if T contains a strong support vertex, that is, p = 3.Hence, our statement holds for p = 3.From now on, we may suppose that p ≥ 5 which means that there is no strong support vertex in T .
Assume that P p : xv i . . .v i+p−3 y is a shortest path in P co (T ).It follows from Theorem 3.5 and Observation 3.3(v) that P p ∈ S k is a substructure in T , as the vertices v i and v i+p−3 , which may have higher degree in T than in P p , do not belong to X(P p ).By Theorem 3.5, we may infer γ ′ SMB (T ) ≤ k.Now, suppose for a contradiction that T contains a substructure S ∈ S ℓ so that ℓ ≤ k − 1. Observe first that S contains no vertex u with deg S (u) ≥ 3. Indeed, as T is a caterpillar, such a vertex u would be a support vertex in both T and S by Observation 3.3(iii).Then, deg S (u) > 2 contradicts Observation 3.3(ii).Therefore, S is an odd path.We next show that S cannot be a clean path in T .Indeed, S ∈ P co (T ) implies |V (S)| ≥ p and ℓ ≥ k that is a contradiction.Therefore, S : av q . . .v q+r b is an odd path between the leaves a and b of T and, for an integer t ∈ [r − 1], the vertex v q+t is adjacent to a leaf c in T .By choosing the smallest positive t with this property, we may further assume that S ′ : av q . . .v q+t c is a clean path in T .If t is odd then, by Observation 3.3(i), v q+t belongs to X(S) that contradicts deg S (v q+t ) = 2 < deg T (v q+t ) = 3.If t is even, then S ′ is an odd clean path and hence, |V (S ′ )| = t + 3 ≥ n min (P co (T )) = p that implies |V (S)| ≥ p + 2 and contradicts ℓ ≤ k − 1.This contradiction, together with Theorem 3.5, imply γ ′ SMB (T ) ≥ k and in turn, γ ′ SMB (T ) = k follows.From now on, suppose that a D-game is played on T and P co (T ) = ∅.If t p (P co (T )) = 1, a vertex contained in all P co -paths of T can be chosen as a support vertex v i .(Indeed, a set of intersecting clean paths must have a common endvertex and a common support vertex.)Let U be the set of leaves adjacent to v i .If Dominator plays v i as his first move, we may construct a matching satisfying the conditions of Lemma 2.7, where X = {v i } and Y = ∅.First, delete the support vertices of T from the subgraph induced by the spine of T to obtain the linear forest T − .Since every P co -path is incident to v i , each component of T − that contains neither v i−1 nor v i+1 is an even path and admits a perfect matching.If there is an odd path component which contains v i−1 , then v i−1 is the end vertex, and we can take a matching that covers all vertices of the component except v i−1 .The situation is similar if an odd path contains v i+1 .To finish path appears in T i , for i ∈ [2].Using Propositions 2.2 and 2.4, and the formula already proved for the S-game on a caterpillar, we obtain the following relation: This completes the proof for the second formula.✷

Proof:
The statement is clearly true for k = 1 as only a presence of a one-element winning set ensures that Maker can win with his first move in a Maker-start game.If k ≥ 2, we refer to the notation used in the recursive definition of F k and show that Maker has a strategy to win in at most k moves.Consider the subhypergraph H ′ ∈ F k and suppose that Maker first plays the origin vertex u.That leaves the hypergraph H | u with a subhypergraph H ′ | u for the continuation of the game.H | u therefore contains two vertex disjoint subhypergraphs A 1 ∈ F k−1 and A 2 ∈ ∪ k−1 i=1 F i .After Breaker's move v, at least one of these subhypergraphs remains untouched and (H | u) − v contains a subhypergraph from ∪ k−1 i=1 F i .By the induction hypothesis, Maker can win the game in at most k − 1 further moves.As follows, w M M (H) ≤ k that concludes the proof.✷ The discussed cases show that γ ′ SMB (T )≥ k + 1, if T is reducible and contained in T 1 .It settles the last inductive case.✷ Corollary 4.4 If k ≥ 1 is an integer, then γ ′ SMB (S(2k, 2k, 2k)) = ⌈log 2 (4k + 1)⌉ + 1 .