Complexity and monotonicity results for domination games

Theoretical Computer Science 628 (2016) 1–29

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Theoretical Computer Science www.elsevier.com/locate/tcs

Complexity and monotonicity results for domination games ✩ Stephan Kreutzer a , Sebastian Ordyniak b,∗ a b

School of Electrical Engineering and Computer Science, Technical University Berlin, Berlin, Germany Faculty of Informatics, TU Wien, Vienna, Austria

a r t i c l e

i n f o

Article history: Received 23 July 2014 Received in revised form 29 July 2015 Accepted 2 March 2016 Available online 4 March 2016 Communicated by D. Peleg Keywords: Graph searching games Cops and robber games Domination search games Parameterized complexity Monotonicity

a b s t r a c t In this paper we study Domination Games, a class of games introduced by Fomin, Kratsch, and Müller in [8]. Domination games are a variant of the well-known graph searching games (also called cops and robber games), where a number of cops tries to capture a robber hiding on the vertices of a graph. Variants of these games are often used to provide a game-theoretic characterization of important graph parameters such as pathwidth, treewidth, and hypertreewidth. We are primarily interested in questions concerning the complexity and monotonicity of these games. We show that dominating games are computationally much harder than standard cops and robber games and establish strong non-monotonicity results for various notions of monotonicity that arise naturally in the context of domination games. Answering a question of [8], we show that there are graphs where the shortest winning strategy for a minimal number of cops must necessarily be of exponential length. © 2016 Elsevier B.V. All rights reserved.

1. Introduction

Graph searching games are a form of two-player games played on graphs. A wide range of such games have been studied in the literature but they all share the common scheme that a number of cops tries to catch a robber who is hiding in the graph. The problem is to guide a party of as few cops as possible so that the robber is guaranteed to be captured regardless of his moves. In the model of graph searching games known as node searching, the cops and the robber occupy vertices of the graph. At each step of the play, the player controlling the cops can lift some of the cops from the graph and place them somewhere else. While they are in transit, the robber can move in the graph following any path from his current to his new position as long as this path does not go through a vertex occupied or “blocked” by a remaining cop (in which case the robber would have been captured). Variants of this game are obtained by varying the abilities of the cops, for instance, whether or not they know the current position of the robber, and by the precise definition of “blocking”. The minimal number of cops needed to catch a robber on a graph yields an interesting graph invariant related to the global connectivity of the graph. See [6] for a comprehensive survey of the subject.

✩ Parts of this paper appeared in a preliminary and shortened form in the Proceedings of WG 2009, the 35th International Workshop on Graph-Theoretic Concepts in Computer Science [12]. Kreutzer’s research was partially supported by DFG Emmy-Noether Grant KR 2898/1-3 Games. Ordyniak’s research was partially supported by the Employment of Newly Graduated Doctors of Science for Scientific Excellence (CZ.1.07/2.3.00/30.0009). Corresponding author. E-mail addresses: [email protected] (S. Kreutzer), [email protected] (S. Ordyniak).

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http://dx.doi.org/10.1016/j.tcs.2016.03.003 0304-3975/© 2016 Elsevier B.V. All rights reserved.

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S. Kreutzer, S. Ordyniak / Theoretical Computer Science 628 (2016) 1–29

Graph searching games have found a wide range of applications in Computer Science in seemingly unrelated areas: there is a strong resemblance of graph searching games to pebble games modeling sequential computation as described in [10]. In [9], graph searching games have been employed as a model for privacy in distributed systems, where the cops model eavesdroppers or intruders in networks. Furthermore, applications of graph searching games can be found in VLSI design as the game theoretical approach to important graph layout parameters providing valuable tools for the design of efficient algorithms. Of particular importance is the connection between graph searching games and well-known graph parameters such as pathwidth and treewidth (see e.g. [4,1,3]). For instance, Seymour and Thomas [14] characterized the treewidth of a graph in terms of a variant of graph searching games where the robber is visible and hides on the vertices of the graph. An important concept in the theory of graph searching games is monotonicity. Intuitively, a strategy for the cops is monotone if they can catch the robber without allowing him to revisit vertices from which he has previously been expelled. Monotonicity has featured highly in research on graph searching games for a number of reasons. For instance, monotone strategies correspond directly to graph decompositions such as tree- or path-decompositions. Also, for many game variants, winning strategies for the robber can often be characterized by simple combinatorial structures, such as brambles for the case of games corresponding to treewidth, and hence provide natural and intuitive obstructions for treewidth and similar measures. However, these structures usually provide a winning strategy even against cops following a non-monotone strategy. Hence, showing for a game variant that the number of cops needed to win against a robber is always the same as the number of cops needed for a monotone strategy brings all these concepts together and establishes a smooth theory of decompositions and games in terms of min-max or duality theorems. From an algorithmic perspective, an important property of monotone strategies is that their length is usually linearly bounded in the order of the graph, whereas non-monotone strategies can have up to exponential length, although almost no game variant actually requires such long strategies. Hence, monotone strategies often provide polynomial certificates and thereby yield NP-algorithms for deciding the number of cops needed to catch a robber. In this paper we study domination games that have first been introduced by Fomin, Kratsch and Müller [8]. These games are closely related to the cops and robber games used to characterize pathwidth and treewidth. The main difference between domination games and the games used to characterize pathwidth and treewidth is that by occupying a vertex of the graph the cop-player not only “blocks” the vertex but also its neighborhood for the robber. In [8], the authors develop the fundamental theory of domination games and establish a relationship between domination games and the size of a minimum dominating set of a graph and an interesting connection between these games and a graph parameter called domination target number introduced in [11]. The focus of [8] is on establishing bounds on the domination search number – the minimal number of cops required to win the domination game on a graph – for various classes of graphs such as k-dimensional cubes, asteroidal-triple free graphs, claw-free graphs, and graphs with certain types of spanning trees and caterpillars. They also exhibit an example showing that domination games are non-monotone. In this paper we study domination games with a focus on complexity and monotonicity. Following the initial results on monotonicity of graph searching games mentioned above, monotonicity proofs for a large number of graph searching games and also non-monotonicity proofs for some games have been obtained (see e.g. [6]). Most variants of graph searching games are either monotone or, if not, at least a bound on the difference between the number of cops needed for arbitrary or monotone strategies can be established. As it turns out, domination games exhibit a completely different behavior in this respect. Organization and results. Sections 2 and 3 introduce the necessary preliminaries of this paper related to graphs, decomposition parameters, parameterized complexity, the visible and invisible variant of the domination game as well as the concept of monotonicity for these games. We then continue by stating some preliminary results about domination games and their relation to other well-known parameters in Section 5. Because domination games turn out to be technically difficult to handle, we introduce a novel variant of graph searching games, which we call set games, in Section 4. We show that these games are for all intents and purposes of this paper equivalent to domination games, i.e. there are polynomial-time reductions between set games and domination games in both directions and more importantly these reductions preserve important properties of both games such as number of cops needed, length of strategies, as well as monotonicity properties of strategies etc. We believe that set games provide a novel perspective on domination games and apart from simplifying many of the proofs this perspective is also interesting in its own right. Set games are then used intensively in the remainder of this paper as a shortcut to prove results about domination games. The next two sections then deal with the monotonicity and complexity properties of the invisible variant of the domination game. Namely, in Section 6 we show that the invisible domination game is neither robber- nor cop-monotone and moreover there can be an arbitrary difference between the number of cops needed in a non-monotone strategy and the number of cops needed using a monotone strategy. We then argue that the traditional and well-established notions of monotonicity are ill-equipped to capture the properties of domination games and introduce a novel notion of monotonicity, which we call selective-monotonicity, for domination games. Even though we show that domination games are also not selective-monotone there is still hope that non-monotone and selective-monotone strategies are closely related to each other. In particular, we conjecture that the selective-monotone and non-monotone variants of domination games lie within a constant factor of each other implying that the selective-monotone variant could serve as an (NP-)approximation to the non-monotone variant.

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Section 7 then contains various complexity results for the invisible domination game. It is known that domination games are NP-hard, however, so far no corresponding complexity upper bound has been known. We show that it is Pspace-complete to decide whether just two cops have a winning strategy in the invisible domination game. This comes as quite a surprise, because the majority of graph searching games are NP-complete and also settles an open question of [8] by showing that there are graphs for which a shortest winning strategy for the cop-player for a minimal number of cops must necessarily be of exponential length. We then study the monotone versions of the game, which are known to be NP-complete, and show that even deciding whether two cops have a cop-monotone winning strategy is already NP-complete. This is in contrast to cops and robber games that define pathwidth, which are known to be fixed-parameter tractable parameterized by the number of cops. A problem with a distinguished parameter k given in the input is fixedparameter tractable if it can be solved in time O ( f (k)n O (1) ), where n is the input size and f is an arbitrary but computable function. Section 8 is devoted to the visible variant of the domination game. We show that also this variant is not robber-monotone and provide evidence that even the visible variant of the game is not fixed-parameter tractable parameterized by the number of cops. This is in contrast to the cops and robber game defining treewidth, which is known to be fixed-parameter tractable for the same parameterization. Finally, we conclude in Section 9. 2. Preliminaries We use standard notation from graph theory as can be found in, e.g., [4]. In particular, we write V (G ) for the vertex set of a graph G and E (G ) for its edge set. All graphs in this paper are simple and undirected and all graphs and hypergraphs are finite. Let G be a graph. The (open) neighborhood of a vertex v in G  is N G ( v ) := {u : 0{ v , u } ∈ E (G )}. The closed G G G neighborhood of v is N [ v ] := N ( v ) ∪ { v }. If X is a set, we define N [ X ] := v ∈ X N G [ v ]. We omit the index G whenever G is clear from the context. Pathwidth and treewidth The notions of treewidth and pathwidth were introduced by Robertson and Seymour as part of their work on graph minors. We refer to [1,4] for definitions and further information. A tree decomposition T of an (undirected) graph G = ( V , E ) is a pair ( T , χ ), where T is a tree and χ is a function that assigns to each tree node t a set χ (t ) ⊆ V of vertices such that the following conditions hold: (P1) For every vertex u ∈ V , there is a tree node t such that u ∈ χ (t ). (P2) For every edge {u , v } ∈ E (G ) there is a tree node t such that u , v ∈ χ (t ). (P3) For every vertex v ∈ V (G ), the set of tree nodes t with v ∈ χ (t ) forms a subtree of T . The sets χ (t ) are called bags of the decomposition T and χ (t ) is the bag associated with the tree node t. The width of a tree decomposition ( T , χ ) is the size of a largest bag minus 1. A tree decomposition of minimum width is called optimal. The treewidth of a graph G, denoted by tw(G ), is the width of an optimal tree decomposition of G. A path decomposition of a graph G is a tree decomposition ( T , χ ) such that T is a path instead of a tree. All notions and definitions introduced for tree decompositions above apply in the same way for path decompositions. The pathwidth of G, denoted by pw(G ), is the width of an optimal path decomposition of G. Parameterized complexity Here we introduce the relevant concepts of parameterized complexity theory. For more details, we refer to text books on the topic [5,7,13]. An instance of a parameterized problem is a pair ( I , k) where I is the main part of the instance, and k is the parameter. A parameterized problem is fixed-parameter tractable if instances ( I , k) can be solved in time f (k)| I | O (1) , where f is a computable function of k. FPT denotes the class of all fixed-parameter tractable problems. If a problem is not fixed-parameter tractable, it might still be solvable in polynomial-time for every fixed value of the parameter. This is capture by the class XP that contains all parameterized problems, whose instances ( I , k) can be solved in time | I | f (k) , for some computable function f . Hardness for parameterized complexity classes is based on fpt-reductions. A parameterized problem L is fpt-reducible to another parameterized problem L  if there is a mapping R from instances of L to instances of L  such that (i) ( I , k) ∈ L if and only if ( I  , k ) = R ( I , k) ∈ L  , (ii) k ≤ g (k) for a computable function g, and (iii) R can be computed in time O ( f (k)| I | O (1) ) for a computable function f . Next, we will define the classes capturing fixed-parameter intractability needed in this work. For further details we refer to the literature on parameterized complexity theory. The class W[2] contains all problems that are fpt-reducible to Dominating Set when parameterized by the size of the solution, i.e. the dominating set [5,7]. Showing W[2]-hardness for a problem rules out the existence of a fixed-parameter algorithm under the usual complexity theoretic assumption FPT = W[2]. To rule out the existence of a polynomial-time algorithm for every fixed value of the parameter (under the assumption that P = NP), i.e. to show that a parameterized problem is not in XP, we will show that the problem remains NP-hard even after fixing the parameter to a constant. The following relations between the parameterized complexity classes mentioned above hold: FPT ⊆ W[1] ⊆ W[2] ⊆ XP.

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3. Domination games In this section we introduce domination games. A domination game on a graph G is played between two players, the cop and the robber, where the goal of the cops is to capture the robber. At each step of the play, the robber occupies a vertex of the graph and the cop player controls a finite number of cops each occupying vertices. A play starts by the robber choosing an initial position. In each step of the game, the cops either place a new cop on a vertex or they remove an existing cop from the graph. Suppose X is the set of vertices currently occupied by the cops and they want to place a new cop on vertex v. They first have to announce this to the robber. The robber can then run away, but is not allowed to run through a vertex that is in the closed neighborhood of a vertex occupied by a cop, i.e. he can pick a new position u anywhere on the graph as long as there is a path from his current position to u that contains no vertex in N G [ X ]. After the robber has chosen his new position, the new cop is placed on v and the play continues. The cops win a play if they can capture the robber, i.e. if they can place a cop in the neighborhood of the vertex occupied by the robber so that the robber cannot escape. If the robber can escape forever, the robber wins. Domination games are a variant of the well-known cops and robber games used to characterize graph parameters such as treewidth or pathwidth (see e.g. [14]). The difference is that in a cops and robber game, a cop only occupies his current position but does not block the neighborhood of this position. We will distinguish between two variants of domination games, i.e. the invisible and the visible variant. Because we will mostly deal with the invisible variant of the game, we will implicitly assume the invisible variant, whenever no variant is explicitly mentioned. For instance, in the sequel we will refer to the invisible domination game simply as domination game In the invisible case, the cops do not see the robber and hence have to search the graph independent of the robber movements. In this case, we are essentially dealing with a one player game and in describing the game, we can discard the robber positions. We can therefore represent any cop strategy on a graph G in the invisible domination game by a sequence S := ( S 1 , . . . , S n ), where, for 1 ≤ i ≤ n, S i ⊆ V (G ) is the cop position after step i. With any strategy S := ( S 1 , . . . , S n ) we associate the corresponding sequence R 0 , . . . , R n of robber spaces as follows: R 0 := V (G ) and for all i > 0, R i := { v ∈ V (G ) \ N G [ S i ]: there is u ∈ R i −1 and a path from u to v in G \ N G [ S i −1 ∩ S i ]}, where we take S 0 := ∅. Hence, R i is the set of vertices available to the robber after i steps of the play. Vertices in V (G ) \ R i are called clear at stage i and all vertices outside of R i are called contaminated at stage i. Definition 3.1. Let S := ( S 1 , . . . , S n ) be a strategy and ( R 0 , . . . , R n ) be the corresponding robber spaces. 1. S is a winning strategy if R n = ∅. 2. The width w (S ) of S is defined as w (S ) := max{| S i | : 1 ≤ i ≤ n}. 3. The domination search number ds(G ) := min{ w (S ) : S is a winning strategy on G } of G is the minimal number of cops required to win the invisible domination game on G. Clearly, every graph G can be searched by | V (G )| cops (by placing one cop on every vertex of the graph). Hence ds(G ) is well-defined. An important aspect of domination games (and of cops and robber games in general) is monotonicity. There are two common notions of monotonicity for cops and robber games – cop-monotonicity and robber-monotonicity, which are defined as follows. Definition 3.2. Let S := ( S 1 , . . . , S n ) be a strategy and ( R 0 , . . . , R n ) be the corresponding robber spaces. 1. S is robber-monotone, if R i ⊇ R j for all i < j. 2. S is cop-monotone if for all i < j < l and all v ∈ V (G ), if v ∈ S i \ S j then v ∈ S l . 3. The cop-monotone domination search number is defined as c-ds(G ) := min{ w (S ) : S is a cop-monotone winning strategy on G }. The robber-monotone domination search number r-ds(G ) is defined analogously by using robber-monotone winning strategies. In a non-monotone strategy, a vertex v ∈ R j \ R i , for j > i, is called recontaminated. Note that, unlike cops and robber games, in domination games cop-monotone strategies might not be robber-monotone and vice versa. We will see in Section 6 that the above notions of monotonicity are not well suited to capture the “monotone behavior” of domination games. This is due to the property of domination games that if the cop-player occupies a vertex it indiscriminately dominates all the vertices in its neighborhood, even though, dominating the entire neighborhood might not be necessary to win the game. Hence, even though dominating the entire neighborhood is not necessary to win the game for the cop-player it still influences the monotonicity properties of a strategy. We therefore introduce the notion of selective monotonicity, where we allow the cop-player to dominate only a part of the neighborhood. More formally, selective monotonicity is defined as follows. We say that a strategy S := ( S 1 , . . . , S n ) in the invisible domination game on a

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graph G is selective-monotone if there exists a sequence ( D 1 , . . . , D n ) with D i ⊆ N G [ S i ] for every 1 ≤ i ≤ n such that the sequence R 0 , . . . , R n with: R 0 := V (G ) and for all i > 0, R i := { v ∈ V (G ) \ D i : there is u ∈ R i −1 and a path from u to v in G \ D i −1 ∩ D i ∩ N G [ S i −1 ∩ S i ]}, where we take D 0 := S 0 := ∅, is monotonously decreasing, i.e. R i ⊆ R i −1 for every i with 1 ≤ i ≤ n. We say the strategy is winning if R n = ∅. The selective-monotone domination search number is defined as s-ds(G ) := min{ w (S ) : S is a selective-monotone winning strategy on G }. We now want to briefly comment on the visible variant of domination games. In the visible domination game, the cops can see the robber and can adapt their strategy accordingly. Therefore, a strategy for the cops in the visible domination game is a function that tells the cops where to move from the current position ( X , v ) of the play, where the cops are on X ⊆ V (G ) and the robber is on the vertex v. Therefore, formally, cop strategies in the visible game are functions

f : P ( V (G )) × V (G ) → P ( V (G )) assigning to each pair ( X , v ), with X ⊆ V (G ) and v ∈ V (G ), a new set f ( X , v ) ⊆ V (G ). Such a strategy f is winning for the cop-player if for all plays of the game, where the cop-player plays according f , the game eventually reaches a position ( X , v ) such that v ∈ N G [ X ]. All other notions and definitions corning strategies and monotonicity naturally carry over from the invisible variant. We define the visible domination search number vis-ds(G ) := min{ w ( f ) : f is a winning strategy on G } of G is the minimal number of cops required to win the visible domination game on G. The cop-monotone visible domination search number is defined as c-vis-ds(G ) := min{ w ( f ) : f is a cop-monotone winning strategy on G }. The robber-monotone visible domination search number r-vis-ds(G ) and the selective-monotone visible domination search number s-vis-ds(G ) are defined analogously by using robber-monotone or selective-monotone winning strategies, respectively. 4. Set games In this section, we will introduce the set game. The main purpose of introducing this game is to simplify the proofs of our results for the domination game in the following sections. Because of the somewhat more powerful and intuitive nature of set games, we also believe these games are interesting in their own right. To do so we will show that the set game is for the intents and purposes of this paper equivalent to the domination game. We believe that the set game provides a clearer way of looking at domination games, because it can be handled without the many tedious technicalities of the domination game. Informally, in the set game the cop-player chooses vertex sets from a given family of vertex subsets of the input graph and by doing so “blocks” all vertices for the robber that are in the union of the chosen vertex sets. More formally, in the set game the input is a graph G together with a family F of vertex subsets of G and a position for the cop-player is a subset of F . All notions and definitions about strategies, search number, monotonicity for the invisible and visible variant of the set game naturally carry over from the corresponding variant of the domination game. As an illustrative example we will briefly reintroduce the most important notions for the invisible set game. In the invisible variant of the set game, a strategy for the cop-player is again a sequence S := ( S 1 , . . . , S n ) of cop-positions, where S i ⊆ F for every i with 1 ≤ i ≤ n. Given such a strategy for the cop-player S := ( S 1 , . . . , S n ) the corresponding sequence R 0 , . . . , R n of robber spaces is defined as follows: R 0 := V (G ) and  for every i with 1 ≤ i ≤ n, we set R i to be the set { v ∈ V (G ) \ s∈ S i s] : there is u ∈ R i −1 and a path from u to v in G \ ( s∈( S i−1 ∩ S i ) s)}, where we take S 0 := ∅. To avoid situations, where the cop player cannot win (even by choosing all vertex subsets of F ), we will henceforth implicitly assume that every vertex of G is contained in at least one subset of F . This ensures that | V (G )| cops can win on any graph G. Given the definition of strategies and robber-spaces all notions and definitions defined for the domination game can be defined for the set game in the obvious way. Let G be a graph and F a family of vertex subsets of G. We denote by ss(G ) the set search number of G and F , by c-ss(G ) the cop-monotone set search number, and by r-ss(G ) the robber-monotone set search number. It is easily seen that the (visible) set game is at least as general as the (visible) domination game. To see this consider an instance G and k of the (visible) domination game. Then, G, F and k with F := { N G [ v ] : v ∈ V (G )} is an equivalent instance of the (visible) set game. Moreover, because there is a one-to-one mapping between cop-positions in the domination game instance and cop-positions in the set game instance, this translation preserves both the length as well as the monotonicity properties of winning strategies for the domination game instance, i.e., there is a non-monotone, cop-monotone, robbermonotone, or selective-monotone winning strategy in the (visible) domination game on G using k cops if and only if there is such a strategy of the same length in the (visible) set game on G and F . The opposite direction from set games to domination games is however not that obvious. The following lemma shows that set games can be reduced to domination games. In particular, the reduction preserves monotonicity and therefore allows to transfer monotonicity results from set games to domination games. Lemma 4.1. Let G be a graph, F a set of vertex subsets of G and k an integer. Then, in polynomial-time (with respect to G and F ), one can construct a graph G  such that k cops have a non-monotone/cop-monotone/robber-monotone/selective-monotone winning strategy in the (visible) set game on G and F if and only if they have such a strategy of the same length in the (visible) domination game on G  .

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Proof. To facilitate the reduction, we will need the following notions and definitions. For k > 0, let K k be the k-clique, i.e. the complete graph on k vertices. Further, if X is a set, we write K [ X ] for the complete graph with vertex set X . For each k > 0 and d > 0, we define S k as the graph (up to isomorphism) obtained from K k by subdividing each edge 2-times, i.e. replacing each edge by a path of length 3. We call S k a subdivided k-clique. Note that S k contains more than k vertices but in the following the vertices in the paths replacing edges will usually not play a role. We say that S is the subdivided clique over a set X if S is obtained from K [ X ] by subdividing each edge 2 times. We write S [ X ] for this graph and call X the original vertices of S [ X ]. Observe that ds( S k ) = k. For a graph G, k > 0, and a set F of vertex subsets of G, we define the subdivided k-clique graph of G, denoted by SC (G , k, F ), to be the graph obtained from G by 1) replacing each vertex v ∈ V (G ) by a distinct copy of S k , denoted SC ( v ), and 2) replacing each edge {u , v } ∈ E (G ) by a perfect matching between the original vertices in SC (u ) and the original vertices in SC ( v ), and 3) for each f ∈ F we add a new vertex denoted v f such that { v f : v ∈ V (G )} induces a clique in SC (G , k, F ) and for each f ∈ F , SC (G , k, F ) contains all edges between v f and all vertices in every SC (u ) for u ∈ f . We are now ready to define the reduction from the (visible) set game to the (visible) domination game. Assume that G, F , and k is an instance of the (visible) set game. We claim that the graph G  := SC (G , | V (G )| + 1, F ) satisfies the claim of the lemma. We will show the correctness of the above reduction only for the invisible domination game. The proof for the visible variant is analogous. Suppose that S := ( S 1 , . . . , S n ) is a (non-monotone/cop-monotone/robber-monotone/selective-monotone) winning strategy for k cops in the set game on G and F . It is now straightforward to verify that S  := ( S 1 , . . . , S n ) with S i := { v f : f ∈ S i } for every i with 1 ≤ i ≤ n is such a winning strategy for k cops in the domination game on G  of the same length. For the reverse direction suppose that S  := ( S 1 , . . . , S n ) is a (non-monotone, cop-monotone, robber-monotone, selectivemonotone) winning strategy for k cops in the domination game on G  . Observe that because k < | V (G )| + 1, placing a cop on a vertex G  \ { v f : f ∈ F } does not help the cop-player in any way, i.e. neither to preserve monotonicity nor to capture the robber. Hence, S  := ( S 1 , . . . , S n ), where S i := S i ∩ { v f : f ∈ F } is still a winning strategy for k cops in the domination game on G  , which has the same length and the same monotonicity properties as S  . Hence, we can assume that S i ⊆ { v f : f ∈ F } for every i with 1 ≤ i ≤ n. It is now straightforward to verify that S := ( S 1 , . . . , S n ) with S i := { f : v f ∈ S i } is a winning strategy for k cops in the set search game on G and F , which has the same length and the same monotonicity properties as S  . 2 Together with the previously mentioned reduction from domination games to set games, we obtain that set games and domination games are polynomial-time equivalent. This is captured by the following theorem, which follows easily from the above discussion. Theorem 4.2. Any variant (visible/invisible) of the set game is polynomial-time equivalent to the corresponding variant of the domination game. Moreover, the reductions in both directions preserve monotonicity, number of cops, and length of strategies. Observe that the above theorem shows that any property that we show for some variant of the set game also holds for the corresponding variant of the domination game (and vice versa). Because the set game is technically easier to handle than the domination game, we will use this fact frequently in the remainder of the paper and show most properties for the set game. 5. Preliminary results and relationship to other games In this section we present some preliminary results about domination games and also say how the games are related to other graph searching games like the ones used to characterize pathwidth and treewidth. The cops and robber games that characterize pathwidth and treewidth are the invisible and the visible variant of a game that is often just called the cops and robber game. The rules of the cops and robber game are very similar to the rules of the domination game, the only difference is that instead of “blocking” the whole neighborhood of the current cop-position, the cops merely block the vertices themselves. The game characterizing pathwidth is the invisible variant of the cops and robber game and the game characterizing treewidth is its visible variant. There is a straightforward (polynomial-time) reduction from both variants of the cops and robber game to the corresponding variant of the set game and hence also the domination game. Namely, given a graph G and an integer k, then k cops have a winning strategy in the visible/invisible cops and robber game if and only if k cops have a winning strategy in the visible/invisible set game played on G and F , where F contains one set F v := { v } for every vertex v ∈ V (G ). However, the complexity and monotonicity results of the following sections show that there is no polynomial-time reduction in the other direction and hence domination games are significantly more general than cops and robber games. To provide a better understanding of domination games, we want to briefly point out some similarities and differences between these games and the cops and robber game. We hope that this will be especially useful for readers familiar with cops and robber games. Let G be a graph and k an integer. If k cops can win the visible/invisible cops and robber game on G, then the same number of cops can win in the corresponding domination game on G. On the other hand if k cops can

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win the invisible domination game, then ((G ) + 1)k cops can win the invisible cops and robber game (by occupying the closed neighborhood of the current cop-position). These relationships are captured by the following proposition. Proposition 5.1. For all graphs G, ds(G ) ≤ pw(G ) + 1 ≤ ((G ) + 1)ds(G ) and vis-ds(G ) ≤ tw(G ) + 1 ≤ ((G ) + 1)vis-ds(G ), where pw(G ) denotes the pathwidth of G, tw(G ) denotes the treewidth of G, and (G ) the maximum degree of G. Note that the +1 in the above statement comes from that fact that both the pathwidth and the treewidth are defined to be one less than the number of cops used in the corresponding cops and robber games. A classical example of a class of graphs where a high number of cops is needed to capture the robber in the cops and robber game is the class of grids. Not surprisingly, on a grid the additional power of the cops in domination games is use. For k > 0, let G k×k be the k × k-grid, i.e. the graph with vertex set {1, . . . , k} × {1, . . . , k} and edges  of limited {(i , j ), (i  , j  )} : |i − i  | + | j − j  | = 1 . Proposition 5.2. For all k ≥ 1,

  k 3

≤ ds(G k×k ) ≤

  k 3

+ 1.

Proof. Let k ≥ 1 and G := G k×k . It is easily seen that

  k 3

+ 1 cops are sufficient to catch the robber on G k×k . To see that   fewer cops are not sufficient to catch the robber, we first note that for every position of the cop-player using < 3k there are i and j with 1 ≤ i , j ≤ k such that no vertex of row i and no vertex of row j is dominated by a cop. Hence, no matter how the cop-player moves the robber can always escape to a free row and a free column. 2 It is well known that the search number of the cops and robber games are preserved under taking minors, i.e. the pathwidth of a minor of a graph is no larger than the pathwidth of the graph. This fails for domination games, in fact the domination game is not even closed under taking (induced) subgraphs. This is easily seen by taking the class of apex-grids, i.e. graphs G k obtained from G k×k by adding one apex vertex adjacent to all other vertices in the grid. Clearly, on any such graph G k , one cop has a winning strategy in the visible/invisible domination game by playing   on the apex. However, the induced subgraph obtained by removing the apex is a grid of height k for which we need

k 3

cops.

6. Monotonicity of domination games In this section we study the monotonicity properties of the invisible domination game. In particular, we establish strong non-monotonicity results for common notions of monotonicity – cop- and robber-monotonicity – in showing that in general more cops are needed to catch a robber with a monotone strategy than with an unrestricted strategy and that the ratio between the monotone and the non-monotone case is unbounded. In [8], Stefan Dobrev exhibited an example where three cops can win the domination game but four cops are needed to search the graph using a monotone strategy. We now strengthen this result considerably by showing that the ratio between the (robber- or cop-) monotone and the non-monotone search numbers is unbounded. Theorem 6.1. For every k > 2, there is a graph G k such that ds(G k ) = 2 but r-ds(G k ) = c-ds(G k ) = k. Proof. To simplify the construction, we show the result for the invisible set game. Because of Theorem 4.2, the result then also follows for the invisible domination game. For k ∈ N we define G k and Fk as follows. Let ρ be a permutation of (1, . . . , k) and let P ρ be the path of length k on the ρ ρ vertices v 1 , . . . , v k . Then, G k is defined as the disjoint union of all P ρ such that ρ is a permutation of (1, . . . , k). Furtherρ

more, Fk contains the sets F 1 , . . . , F k with F i := { v ρ (i ) : ρ is a permutation of (1, . . . , k)}. This completes the construction of G k and Fk . An illustration of the construction is given in Fig. 1. It is easily seen that two cops win the invisible set game on G k and Fk as follows. Observe that for every permutation ρ of (1, . . . , k), the strategy Sρ := ({ F ρ −1 (1) , F ρ −1 (2) }, { F ρ −1 (2) , F ρ −1 (3) }, . . . , { F ρ −1 (k−1) , F ρ −1 (k) }) completely clears the path P ρ in G k . Then, the whole graph G k can be searched by executing Sρ for every permutation ρ . It remains to show that k − 1 cops do not have a cop-monotone or a robber-monotone winning strategy on G k and Fk . Suppose for a contradiction that such a winning strategy exists. W.l.o.g. we can assume that the cops start by playing on ρ F \ F i for some i with 1 ≤ i ≤ k. Then, for each path P ρ , the vertex v ρ (i ) is still contaminated. Furthermore, in the next move the cop-player has to remove a token from some set F j for j = i. But then, there is a permutation ρ such that ρ (i ) ρ and ρ ( j ) are consecutive numbers and thus in P ρ the vertex v ρ ( j ) becomes recontaminated. This shows that the strategy ρ

is not robber-monotone. Since the vertex v ρ ( j ) can only be cleared again by playing on F j at a later step of the strategy, we obtain that the strategy is also not cop-monotone either. This concludes the proof of the theorem. 2

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Fig. 1. The graph G 3 from Theorem 6.1. The vertices are labeled by the sets in F3 containing them.

Fig. 2. The graph G from Theorem 6.2. The vertex labels consists of the vertex name as well as all sets of F that contain the vertex separated by a colon.

Hence, in general there is no hope to find a function f : N → N such that whenever k cops have a winning strategy on a graph G then f (k) cops have a robber- or cop-monotone winning strategy on G. However, there are interesting classes of graphs where such a bound can be established. This is easily seen on classes of graphs of treewidth at most k, as there k cops always have a cop-monotone winning strategy. Proposition 5.1 implies that on graph classes of maximal degree d, whenever k cops have a winning strategy in the domination game, then (d + 1) · k cops have a cop-monotone winning strategy, by following the (cop-)monotone strategy of the invisible cops and robber game. Other examples, such as the class of apex graphs, can be found using techniques from graph minor theory. A full exploration of classes where such bounds can be found is left to future research. Considering again the example above exhibiting non-monotone strategies for the cops, the main source for nonmonotonicity appears to be that while clearing some parts of the graph, the cops accidentally and unintentionally clear other parts of the graph also – which later on they have to allow to be recontaminated. For instance, in the example above,  while clearing a sub-graph P ρ they also clear parts of other sub-graphs P ρ but in the wrong order. If we gave the cops the power to choose which vertices in a set they really want to block, then they could easily search the graphs G k with a robber- and cop-monotone strategy. We call this selective monotonicity. It seems conceivable, thus, that such selective strategies are always sufficient, i.e. whenever k cops can win in any form, they can do so with a selective monotone strategy. Such a result would be extremely interesting as it would imply a linear upper bound for the length of minimal winning strategies for the cop player. This hope is dashed, though, by the following theorem. Theorem 6.2. There exists a graph such that ds(G ) = 2 but more than two cops are needed to win the invisible domination game with any selective-monotone strategy on G. Proof. To simplify the construction, we show the result for the invisible set game. Because of Theorem 4.2, the result then also follows for the invisible domination game. The graph G is shown in Fig. 2. More formally, G consists of the paths ( v 4 , v 3 , v 2 , v 1 , v 0 , v 1¯ , v 2¯ , v 3¯ , v 4¯ ) and the path ( w 4 , w 3 , w 2 , w 1 , w 1¯ , w 2¯ , w 3¯ , w 4¯ ) plus the connecting edges as shown in the figure. Moreover, the set F consists of the sets F 4 , F 3 , F 2 , F 1 , F 0 , F 1¯ , F 2¯ , F 3¯ , F 4¯ with F 4 := { v 4 , w 4 , w 3 , w 2 }, F 3 := { v 3 , w 1 }, F 2 := { v 2 , w 3 }, F 1 := { v 1 , w 2 }, F 0 := { v 0 , w 1 , w 1¯ }, F 1¯ := { v 1¯ , w 2¯ }, F 2¯ := { v 2¯ , w 3¯ }, F 3¯ := { v 3¯ , w 1¯ }, and F 4¯ := { v 4¯ , w 2¯ , w 3¯ , w 3¯ }.

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9

Now, ss(G , F ) = 2 as witnessed by the following strategy:

 S := { F 4 , F 3 }, { F 3 , F 2 }, { F 2 , F 1 }, { F 1 , F 0 },  { F 0 , F 1¯ }, { F 1¯ , F 2¯ }, { F 2¯ , F 3¯ }, { F 3¯ , F 4¯ } Note that S is not robber monotone. Namely, the vertex w 1 becomes recontaminated in the step from { F 3 , F 2 } to { F 2 , F 1 }. Further, observe that in order for S to be a winning strategy, at each step all the vertices contained in sets occupied by a cop need to be blocked. Hence, S is also not selective monotone. We claim that there is no selective monotone strategy with only two cops. It is easily seen that the path of length 9 on the vertices v 4 , v 3 , v 2 , v 1 , v 0 , v 1¯ , v 2¯ , v 3¯ , v 4¯ can be searched in only two ways by two cops using a monotone strategy, i.e., from left to right or from right to left. If follows that the only possible strategies are S or its reverse and neither of them is selective monotone. 2 Note that there is still hope that the non-monotone and selective-monotone variant of the invisible domination game lie within a constant factor of each other. This would be an interesting and important result, because the selective-monotone variant of the domination game is in NP and could hence provide a constant factor NP-approximation for the non-monotone variant of the game. We do not know yet whether this is indeed the case and leave the question for future research. As argued above, an important aspect of monotonicity for a variant of graph searching games is that in this way a bound on the maximal number of steps in a strategy is obtained. As invisible domination games are strongly non-monotone, no such bound can be achieved using this approach. In Corollary 7.6 below we show that there exist graphs such that the number of steps needed by a strategy in the domination game is exponential in the size of the graph and thus cannot be bounded by a polynomial. 7. Complexity of domination games In this section we study the complexity of deciding whether k cops have a (monotone) winning strategy in the invisible domination game on a graph G. Because there is a polynomial-time reduction from the invisible cops and robber game to the invisible domination game (see Section 5) and it is NP-hard to decide the number of cops needed in the invisible cops and robber game, we obtain that the invisible domination game is NP-hard (this has also already been observed in [8]). Theorem 7.1. (See [8].) Let G be a graph and k an integer. Then, deciding whether k cops can win the invisible domination game is NP-hard. No upper bound for the complexity of the problem was given. We settle this problem by giving precise complexity bounds for this problem. Theorem 7.2. Deciding whether two cops can win the invisible domination game is Pspace-complete. Proof. To simplify the reduction, we show the result for the invisible set game. Because of Theorem 4.2, the result then also follows for the invisible domination game. We start by showing that the problem is in NPspace and hence also in Pspace. To see this let G be a graph, F a set of vertex subsets of G, and k an integer and assume we want to decide whether k cops have a winning strategy in the invisible set game on G and F . Note that a position in the invisible set game is completely determined by the pair (C , R ), where C ⊆ F is the current position of the cop-player and R ⊆ V (G ) is the set of vertices of G that are still contaminated. In particular, any position can be described in polynomial space and furthermore, there are only exponentially many distinct positions. Hence, a non-deterministic Turing machine can guess a strategy for the cop-player by storing the current position of the game on its tape and then guessing at each computational step the next move of the cop-player. More specifically, the Turing machine will start by writing the position (∅, V (G )) on its tape. Given the current position stored on its tape, the Turing machine will then guess the next (legal) move for the cop-player, i.e. whether he adds or removes a cop from a certain set, and update the tape with the game position reached after the guessed move by the cop-player. Furthermore, the machine can keep a counter which it increases in each step. It will accept if the tape contains a game position such that the set of contaminated vertices is empty. Otherwise, it will reject once the counter has reached the maximum number of possible positions. At this point, the play would become cyclic and therefore the robber can escape forever. Because the space needed to store the current position is at most k log |F | + | V (G )| log | V (G )| and hence polynomial in the input size, this shows that Set Search is in NPspace, as required. We will show Pspace-hardness of Set Search by reducing the well-known Quantified Boolean Formula (QBF) problem to the problem whether two cops have a winning strategy on a graph. The Quantified Boolean Formula (QBF) problem is the problem to decide whether a given quantified boolean (propositional) formula (containing existential and universal

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quantifiers as well as the usual logical connectives) is satisfiable. The problem is well-known to be Pspace-complete even if the input formula is given in prenex normal form with strictly alternating quantifiers and the quantifier-free part of the formula is in conjunctive normal form with each clause having at most three literals. Let

 = ∃x1 ∀x2 ∃x3 · · · ∀xn−1 ∃xn

i ≤m

Ci

i =1

be a quantified boolean formula in prenex normal form with strictly alternating quantifiers, such that all clauses C i are the disjunctions of at most three literals. We will construct a graph G and a family F of vertex subsets of G such that two cops have a winning strategy in the invisible set game on G and F , iff  is satisfiable. Before we twelve into the details of the construction, let us provide a brief overview about the main ideas used in the construction. First note that the satisfiability of the formula  can be checked by using the following procedure. At any time during the procedure every existential quantifier, every universal quantifier, and the set of all clauses is in one of the following states: an existential quantifier is either zero, one, accepted, or initialized, an universal quantifier is either zero, one, zero-accepted, one-accepted, or initialized, and the set of all clauses is either accepted or initialized. Furthermore, the procedure can change the state of the quantifiers and the set of all clauses according to the following rules: RI At the beginning of the procedure all quantifiers as well as the set of all clauses are in the state initialized. RG In order for a quantifier (existential or universal) or the set of all clauses to change its state all its predecessors have to be either initialized, zero, or one, and all the successors of the next quantifier (including the set of all clauses) have to be initialized. RE For any existential quantifier the following additional rules apply: RE1 At any time it can change its status from initialized to either zero or one. RE2 If the next (universal) quantifier is one-accepted (or in case of the last existential quantifier, given that the set of all clauses is accepted), it can change its status (from zero, one, or initialized) to accepted. If it does so, the next (universal) quantifier is set to initialized (or in case of the last existential quantifier the set of all clauses is set to initialized). RU For any universal quantifier the following additional rules apply: RU1 At any time it can change its status from initialized to zero. RU2 If the next (existential) quantifier is accepted (or in case of the last universal quantifier, given that the set of all clauses is accepted), it can change its status from zero to zero-accepted. If it does so, the next (existential) quantifier is set to initialized (or in case of the last universal quantifier the set of all clauses is set to initialized). RU3 At any time it can change its status from zero-accepted to one. RU4 If the next (existential) quantifier is accepted (or in case of the last universal quantifier, given that the set of all clauses is accepted), it can change its status from one to one-accepted. If it does so, the next (existential) quantifier is set to initialized (or in case of the last universal quantifier the set of all clauses set to initialized). RC If the current assignment of the variables (represented by the corresponding quantifiers to be in any of the states zero or one) satisfies all the clauses of , the set of all clauses can change its status from initialized to accepted. Then, the formula  is satisfiable if and only if the first existential quantifier can become accepted using the above procedure. The main idea of our reduction is to construct a grid-like graph that can be searched by two cops only if they follow a strategy that implicitly follows the above procedure. In this graph every quantifier as well as the set of all clauses is associated to a distinct part around a specific “center” column of the grid. The parts of the grid that correspond to the quantifiers and the set of all clauses are aligned from left to right, starting with the first existential quantifier and ending with the part that corresponds to the set of all clauses. At every step of a strategy for the cop-player on the grid-like graph, the status of a quantifier (and also the status of the set of all clauses), with respect to the above procedure, is then represented by which subset of the cells that belong to the part of the grid that correspond to the quantifier or the set of all clauses, respectively, are already cleared by the cop-player. The definition of the sets (that can be used by the cop-player) ensure that, when cleaning the cells representing a particular state of the quantifiers or the set of all clauses, respectively, the cop-player has to follow the rules of the procedure given above and that the whole graph can only be searched by two cops after the first existential quantifier becomes accepted. Example 7.3. Consider the formula  := ∃x1 ∀x2 (x1 ∨ x2 ). The satisfiability of  can be shown by the application of the sequence (RI, RE1, RU1, RC , RU2, RU3, RC , RU4, RE2) of rules, i.e., a) Rule RI is used to set all quantifiers as well as the set of all clauses to initialized, b) Rule RE1 is used to set the first quantifier to one, c) Rule RU1 is used to set the second quantifier to zero, d) Rule RC is used to set the set of all clauses to accepted, e) Rule RU2 is used to set the second quantifier to zero-accepted, f) Rule RU3 is used to set the second quantifier to one, g) Rule RC is used to set the set of all clauses to accepted, h) Rule RU4 is used to set the second quantifier to one-accepted, and i) Rule RE2 is used to set the first quantifier

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Fig. 3. An illustration of a winning strategy for two cops in the invisible set game on G given the formula  := ∃x1 ∀x2 x1 ∨ x2 . The satisfiability of  can be shown by the sequence (RI, RE1, RU1, RC , RU2, RU3, RC , RU4, RE2) of rules, which as will be seen later in the proof corresponds to the strategy Q START ◦ S1∃ (1) ◦ S0∀ (2) ◦ SC ◦ S0∀A (2) ◦ S1∀ (2) ◦ SC ◦ S1∀A (2) ◦ S ∃A (1) ◦ END (the sub-strategies START, SC , END, and S S (i ) for every S ∈ {0, 1, A , 0 A , 1 A }, Q ∈ {∃, ∀}, and i ∈ {1, 2} will be defined later in the proof). For ease of presentation, the grid-like graph G is represented by a grid, i.e., the center vertices as well as the vertices resulting from the subdivision of edges are omitted. Each of the figures a)–j) illustrates the set of cells (represented by the black cells) that are clear after the part of the strategy corresponding to one of the above rules has been executed, i.e., the set of cleared cells a) after the application of Rule RI (realized by the sub-strategy START) that sets all quantifiers as well as the set of all clauses to initialized, b) after the application of Rule RE1 (realized by the sub-strategy S1∃ (1)) that sets the first quantifier to one, c) after the application of Rule RU1 (realized by the sub-strategy S0∀ (2)) that sets the second quantifier to zero, d) after the first application of Rule RC (realized by the sub-strategy SC ) that sets the set of all clauses to accepted, e) after the application of Rule RU2 (realized by the sub-strategy S0∀A (2)) that sets the second quantifier to zero-accepted, f) after the application of Rule RU3 (realized by the sub-strategy S1∀ (2)) that sets the second quantifier to one, g) after the second application of Rule RC (realized by the sub-strategy SC ) that sets the set of all clauses to accepted, h) after the application of Rule RU4 (realized by the sub-strategy S1∀A (2)) that sets the second quantifier to one-accepted, i) after the application of Rule RE2 (realized by the sub-strategy S ∃A (1)) that sets the first quantifier to accepted, j) after using the sub-strategy END to clear the remaining cells of G.

to accepted. This sequence of rules will then corresponds to an order in which the cells of the grid-like graph have to be cleared by any winning strategy for two cops in the invisible set domination game, which is illustrated in Fig. 3 a)–i). We are now ready to give a formal definition of the construction. We start by defining the grid-like graph G. An illustration of G is shown in Fig. 4. Let r = max{12, 3m + 4} and c = 27n + 4. Then G is the r × c-grid, where every edge has been subdivided exactly once and where every cell contains one additional (center) vertex that is connected to all vertices in the border of the cell. More formally, G contains one vertex xi , j for every i and j with 1 ≤ i ≤ r and 1 ≤ j ≤ c, one vertex h i , j for every i and j with 1 ≤ i ≤ r and 1 ≤ j < c, one vertex v i , j for every i and j with 1 ≤ i < r and 1 ≤ j ≤ c, and one vertex c i , j for every i and j with 1 ≤ i < r and 1 ≤ j < c. Furthermore, G has the following edges:

• one edge between xi , j and hi , j and one edge between hi , j and xi , j+1 for every i and j with 1 ≤ i ≤ r and 1 ≤ j < c, • one edge between xi , j and v i , j and one edge between v i , j and xi +1, j for every i and j with 1 ≤ i < r and 1 ≤ j ≤ c, • eight edges between c i , j and the vertices xi , j , xi , j+1 , xi +1, j , xi +1, j+1 , hi , j , hi +1, j , v i , j , and v i , j +1 for every i and j with 1 ≤ i < r and 1 ≤ j < c. Let a, b, i, j, and l be natural numbers with a ≤ b. We define the following sets of vertices of G.

• R [a,b] (l) contains all vertices of the row with index l of G, which are in between the columns with index a and b, i.e., R [a,b] (l) contains all the vertices xl, j with a ≤ j ≤ b and all vertices hl, j with a ≤ j < b.

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Fig. 4. The grid-like graph G from the proof of Theorem 7.2 for r = 4 and c = 8.

• C [a,b] (l) contains all vertices of the column with index l of G, which are in between the rows with index a and b, i.e., C [a,b] (l) contains all the vertices xi ,l with a ≤ i ≤ b and all vertices v i ,l with a ≤ i < b. • Z (i , j ) contains the vertices belonging to the cell (i , j ) of G, i.e., Z (i , j ) contains the vertices xi , j , xi +1, j , xi , j +1 , xi +1, j +1 , h i , j , h i +1, j , v i , j , v i , j +1 , and c i , j . For every i with 1 ≤ i ≤ n, we define posi := 16 + 27(i − 1) – this represents the center column of the part of G associated to quantifier i. We can start now to define a number of vertex sets of G, that will be needed for general purposes.

• This set specifies the bottom line for all quantifiers and clauses: Bottom_line := R [1,pos1 ] (2) ∪ R [pos1 +1,pos2 ] (2) ∪ · · · ∪ R [posn−1 +1,posn ] (2) ∪ R [posn +1,c −2] (2) ∪ R [c −1,c ] (2)

• This set defines (a part of) the border of the part corresponding to the initialized state of the i-th quantifier: Q_initi := C [2,4] (posi ) ∪ R [posi ,posi +1] (4) ∪ C [2,4] (posi + 1) The following sets are illustrated in Fig. 5 and are associated to the part of G that represents the i-th existential quantifier of , where i ∈ [1, 3, · · · , n − 2, n].

• The following sets will be used to clear the cells necessary for the quantifier to change into the accepted state: – Let 1 ≤ j ≤ 3. The following sets will be used, when checking that the next universal quantifier is one-accepted (or in the case of the last existential quantifier that the set of clauses is accepted): j

E_check_nexti := C [2,3+ j ] (posi ) ∪ C [2,3+ j ] (posi + 1) ∪ Z (3 + j , posi ) – The following set will be used, when resetting the state of the next universal quantifier to initialized (or in the case of the last existential quantifier when resetting the state of the set of clauses to initialized):

E_save_nexti := C [2,7] (posi ) ∪ C [2,7] (posi + 1) ∪ Z (7, posi ) – Let 1 ≤ j ≤ 3. The following sets will be used, when resetting the state of the next universal quantifier to initialized (or in the case of the last existential quantifier when resetting the state of the set of clauses to initialized): j

E_reset_nexti := C [2,7+ j ] (posi ) ∪ C [2,7+ j ] (posi + 1) ∪ Z (7 + j , posi )

• The following set will be used (by the previous quantifier) to verify that this quantifier is accepted: E_verifyi := C [2,11] (posi ) ∪ C [2,11] (posi + 1) ∪ R [posi ,posi +1] (11)

• The following set will be used (by the previous quantifier) to reset the quantifier to initialized: E_savei := E_verifyi ∪ R [posi ,posi +1] (4)

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Fig. 5. A simplified illustration of the sets defined for the existential quantifiers. The part of the graph G, containing vertices from the illustrated set, is represented by a grid (omitting the center vertices c i , j as well as the vertices h i , j and v i , j ), which is given by the dotted lines. The vertices contained in the illustrated set are represented by the bold lines as well as the black cells as follows: a bold line indicates that the corresponding vertices (i.e. vertices of the form xi , j , h i , j , and v i , j ) belong to the set and a black cell indicates that all vertices of the corresponding cell and its border (including its center vertex) belong to the set.

• Let 1 ≤ j ≤ 3. The following sets will be used to clear the cells necessary for the quantifier to change into the zero state: j

E_seti (0) := C [2,4] (posi ) ∪ C [2,3] (posi + 1) ∪ R [posi ,posi + j ] (4) ∪ R [posi +1,posi + j ] (3) ∪ Z (3, posi + j )

• Let 1 ≤ j ≤ 3. The following sets will be used to clear the cells necessary for the quantifier to change into the one state: j

E_seti (1) := C [2,4] (posi + 1) ∪ C [2,3] (posi ) ∪ R [posi +1− j ,posi +1] (4) ∪ R [posi +1− j ,posi ] (3) ∪ Z (3, posi − j )

• The following set will be used (by the set of all clauses) to verify that this quantifier is zero: E_checki (0) := C [2,4] (posi ) ∪ C [2,3] (posi + 1) ∪ R [posi ,posi +4] (4) ∪ R [posi +1,posi +4] (3) ∪ C [3,4] (posi + 4)

• The following set will be used (by the set of all clauses) to verify that this quantifier is one:

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Fig. 6. A simplified illustration of the first third of all sets defined for the universal quantifiers. The part of the graph G, containing vertices from the illustrated set, is represented by a grid (omitting the center vertices c i , j as well as the vertices h i , j and v i , j ), which is given by the dotted lines. The vertices contained in the illustrated set are represented by the bold lines as well as the black cells as follows: a bold line indicates that the corresponding vertices (i.e. vertices of the form xi , j , h i , j , and v i , j ) belong to the set and a black cell indicates that all vertices of the corresponding cell and its border (including its center vertex) belong to the set.

E_checki (1) := C [2,4] (posi + 1) ∪ C [2,3] (posi ) ∪ R [posi −3,posi +1] (4) ∪ R [posi −3,posi ] (3) ∪ C [3,4] (posi − 3)

• The following set will be used to preserve the state of the quantifier whenever it is either zero or one: E_preservei := C [2,4] (posi ) ∪ C [2,4] (posi + 1) ∪ R [posi −3,posi +4] (4) ∪ R [posi −3,posi ] (3) ∪ R [posi +1,posi +4] (3) ∪ C [3,4] (posi − 3) ∪ C [3,4] (posi + 4) The following sets are illustrated in Figs. 6, 7, and 8 and are associated to the part of G that represents the i-th universal quantifier of , where i ∈ [2, 4, . . . , n − 3, n − 1].

• Let 1 ≤ j ≤ 3. The following sets will be used to clear the cells necessary for the quantifier to change into the zero state: j

j

F_seti (0) := E_seti (0)

• The following sets will be used to clear the cells necessary for the quantifier to change into the zero-accepted state: – Let 1 ≤ j ≤ 3. The following sets will be used to verify that the next existential quantifier has accepted after this quantifier was zero: j

F_check_nexti (0) := C [2,4] (posi ) ∪ C [2,3] (posi + 1) ∪ R [posi ,posi +3+ j ] (4) ∪ R [posi +1,posi +3+ j ] (3) ∪ Z (3, posi + 3 + j )

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Fig. 7. A simplified illustration of the second third of all sets defined for the universal quantifiers. The part of the graph G, containing vertices from the illustrated set, is represented by a grid (omitting the center vertices c i , j as well as the vertices h i , j and v i , j ), which is given by the dotted lines. The vertices contained in the illustrated set are represented by the bold lines as well as the black cells as follows: a bold line indicates that the corresponding vertices (i.e. vertices of the form xi , j , h i , j , and v i , j ) belong to the set and a black cell indicates that all vertices of the corresponding cell and its border (including its center vertex) belong to the set.

– The following set will be used to reset the state of the next quantifier to initialized after this quantifier was zero:

F_save_nexti (0) := C [2,4] (posi ) ∪ C [2,3] (posi + 1) ∪ R [posi ,posi +7] (4) ∪ R [posi +1,posi +7] (3) ∪ Z (3, posi + 7) – Let 1 ≤ j ≤ 3. The following sets will be used to reset the state of the next quantifier to initialized after this quantifier was zero: j

F_reset_nexti (0) := C [2,4] (posi ) ∪ C [2,3] (posi + 1) ∪ R [posi ,posi +7+ j ] (4) ∪ R [posi +1,posi +7+ j ] (3)

∪ Z (3, posi + 7 + j ) • Let 1 ≤ j ≤ 3. The following sets will be used to verify that the quantifier was zero-accepted (before it can be set to one): j

F_check_zeroi := C [2,3] (posi ) ∪ C [2,3] (posi + 1) ∪ R [posi +1− j ,posi +11] (4) ∪ R [posi +1− j ,posi ] (3) ∪ R [posi +1,posi +11] (3) ∪ Z (3, posi − j )

• The following set will be used to change the quantifier from zero-accepted to one: F_save_zeroi := C [2,3] (posi ) ∪ C [2,4] (posi + 1) ∪ R [posi −3,posi +11] (4) ∪ R [posi −3,posi ] (3) ∪ R [posi +1,posi +11] (3) ∪ Z (3, posi − 4)

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Fig. 8. A simplified illustration of the last third of all sets defined for the universal quantifiers. The part of the graph G, containing vertices from the illustrated set, is represented by a grid (omitting the center vertices c i , j as well as the vertices h i , j and v i , j ), which is given by the dotted lines. The vertices contained in the illustrated set are represented by the bold lines as well as the black cells as follows: a bold line indicates that the corresponding vertices (i.e. vertices of the form xi , j , h i , j , and v i , j ) belong to the set and a black cell indicates that all vertices of the corresponding cell and its border (including its center vertex) belong to the set.

• Let 1 ≤ j ≤ 3. The following sets will be used to set this quantifier to one: j

F_seti (1) := C [2,3] (posi ) ∪ C [2,4] (posi + 1) ∪ R [posi −3− j ,posi +1] (4) ∪ R [posi −3− j ,posi ] (3) ∪ Z (3, posi − 4 − j )

• The following sets will be used to clear the cells necessary for the quantifier to change into the zero-accepted state: – Let 1 ≤ j ≤ 3. The following sets will be used to verify that the next quantifier has accepted after this quantifier was set to one: j

F_check_nexti (1) := C [2,3] (posi ) ∪ C [2,4] (posi + 1) ∪ R [posi −6− j ,posi +1] (4) ∪ R [posi −6− j ,posi ] (3) ∪ Z (3, posi − 7 − j ) – The following set will be used to reset the next existential quantifier to initialized after this quantifier was one:

F_save_nexti (1) := C [2,3] (posi ) ∪ C [2,4] (posi + 1) ∪ R [posi −10,posi +1] (4) ∪ R [posi −10,posi ] (3) ∪ Z (3, posi − 11) – Let 1 ≤ j ≤ 3. The following sets will be used to reset the next existential quantifier to initialized after this quantifier was one: j

F_reset_nexti (1) := C [2,3] (posi ) ∪ C [2,4] (posi + 1) ∪ R [posi −10− j ,posi +1] (4) ∪ R [posi −10− j ,posi ] (3) ∪ Z (3, posi − 11 − j )

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• The following set will be used (by the previous quantifier) to verify that this quantifier is zero-accepted: F_verifyi := C [2,3] (posi ) ∪ C [2,4] (posi + 1) ∪ R [posi −14,posi +1] (4) ∪ R [posi −14,posi ] (3) ∪ C [3,4] (posi − 14)

• The following set will be used (by the previous quantifier) to reset this quantifier to initialized: F_savei := C [2,4] (posi ) ∪ C [2,4] (posi + 1) ∪ R [posi −14,posi +1] (4) ∪ R [posi −14,posi ] (3) ∪ C [3,4] (posi − 14)

• The following set will be used (by the set of all clauses) to verify that this quantifier is zero: F_checki (0) := E_checki (0)

• The following set will be used (by the set of all clauses) to verify that this quantifier is one: F_checki (1) := C [2,4] (posi + 1) ∪ C [2,3] (posi ) ∪ R [posi −7,posi +1] (4) ∪ R [posi −7,posi ] (3) ∪ C [3,4] (posi − 7)

• The following set will be used to preserve the state of the quantifier whenever it is either zero or one: F_preservei := C [2,4] (posi + 1) ∪ C [2,4] (posi ) ∪ R [posi −7,posi +4] (4) ∪ R [posi −7,posi ] (3) ∪ R [posi +1,posi +4] (3) ∪ C [3,4] (posi − 7) ∪ C [3,4] (posi + 4) We define the set Q_checki ( j ) to be E_checki ( j ) if i is odd, i.e. the quantifier on position i is existential, and F_checki ( j ) if i is even, i.e. position i represents an universal quantifier. Similarly, let Q_preservei be E_preservei if i is odd and F_preservei if i is even. The following sets are associated to the part of G that represents the set of all clauses of .

• This set defines (a part of) the border of the part corresponding to the initialized state of the set of all clauses: C_save := R [c −2,c −1] (2). • For every clause C i with 1 ≤ i ≤ m and every j with 1 ≤ j ≤ 3 the following set will be used to clear one of the cells necessary for the set of all clauses to become accepted: j

C_seti := C [2,1+3(i −1)+ j ] (c − 2) ∪ C [2,1+3(i −1)+ j ] (c − 1) Z (1 + 3(i − 1) + j , c − 2)

• The following set will be used (by the previous existential quantifier) to verify that the set of all clauses is accepted: C_check := C [2,2+3m] (c − 2) ∪ C [2,2+3m] (c − 1) ∪ R [c −2,c −1] (2 + 3m) Finally, we will introduce the following sets, which provide shortcuts for the definition of the sets in F .

• Let 1 ≤ i < n. These set represents the bottom-line, when changing the state of the i-th quantifier: BL_quanti := Bottom_line ∪ C_save ∪

(



Q_preserve j ) ∪ (

1≤ j < i



Q_init j )

i +1< j ≤n

• This set represents the bottom-line, when the last quantifier verifies that the set of all clauses is accepted: BL_check_clauses := Bottom_line ∪ C_check ∪

(



1≤ j
Q_preserve j )

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Fig. 9. A simplified illustration of the sets in F defined for i-th existential quantifier, where i ∈ [1, 3, · · · , n − 2, n]. The figure only shows the part of G that corresponds to the i-th and i + 1-th quantifier of  and which contains vertices from the illustrated set. The part of the graph G that is shown in the figure is represented by a grid (omitting the center vertices c i , j as well as the vertices h i , j and v i , j ), which is given by the dotted lines. The vertices contained in the illustrated set are represented by the bold lines as well as the black cells as follows: a bold line indicates that the corresponding vertices (i.e. vertices of the form xi , j , h i , j , and v i , j ) belong to the set and a black cell indicates that all vertices of the corresponding cell and its border (including its center vertex) belong to the set.

• This set represents the bottom-line, before the set of all clauses will be reset to initialized: BL_save_clauses = Bottom_line ∪ C_check ∪ C_save ∪ (



Q_preserve j )

1≤ j
• Let 1 ≤ i ≤ n. These sets represent the bottom-line, when verifying that the state of the i-th quantifier (being either zero or one) satisfies one of the clauses:

BL_set_clausei = Bottom_line ∪

(



Q_preserve j ) ∪ (

1≤ j < i



Q_preserve j )

i < j ≤n

We are now ready to define the sets contained in F . For all but the last existential quantifier in , thus for every i with i ∈ [1, 3, · · · , n − 2] and for every j with 1 ≤ j ≤ 3, the set F contains the following sets (Parts of these sets are illustrated in Fig. 9): j

j

• e_seti (0) := BL_quanti ∪ E_seti (0) ∪ Q_initi +1 •

j e_seti (1)

j

:= BL_quanti ∪ E_seti (1) ∪ Q_initi +1 j

j

• e_check_nexti := BL_quanti ∪ E_check_nexti ∪ F_verifyi +1 • e_save_nexti := BL_quanti ∪ E_save_nexti ∪ F_savei +1 j j • e_reset_nexti := BL_quanti ∪ E_reset_nexti ∪ Q_initi +1 And for the last (existential) quantifier, the following sets are contained in F :

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Fig. 10. A simplified illustration of the first half of the sets in F defined for i-th universal quantifier, where i ∈ [2, 4, · · · , n − 3, n − 1]. The figure only shows the part of G that corresponds to the i-th and i + 1-th quantifier of  and which contains vertices from the illustrated set. The part of the graph G that is shown in the figure is represented by a grid (omitting the center vertices c i , j as well as the vertices h i , j and v i , j ), which is given by the dotted lines. The vertices contained in the illustrated set are represented by the bold lines as well as the black cells as follows: a bold line indicates that the corresponding vertices (i.e. vertices of the form xi , j , h i , j , and v i , j ) belong to the set and a black cell indicates that all vertices of the corresponding cell and its border (including its center vertex) belong to the set.

• • • • •

j

j

e_setn (0) := BL_quantn ∪ E_setn (0) j e_setn (1)

j

:= BL_quantn ∪ E_setn (1) j j e_check_nextn := BL_check_clauses ∪ E_check_nextn e_save_nextn := BL_save_clauses ∪ E_save_nextn j j e_reset_nextn := BL_quantn ∪ E_reset_nextn

Furthermore, for every universal quantifier of , i.e., for every i ∈ [2, 4, · · · , n − 1] and 1 ≤ j ≤ 3, F contains the following sets (Parts of these sets are illustrated in Figs. 10 and 11): j

j

• f_seti (0) := BL_quanti ∪ F_seti (0) ∪ Q_initi +1 • • • • • •

j

j

f_check_nexti (0) := BL_quanti ∪ F_check_nexti (0) ∪ E_verifyi +1 f_save_nexti (0) := BL_quanti ∪ F_save_nexti (0) ∪ E_savei +1 j

j

f_reset_nexti (0) := BL_quanti ∪ F_reset_nexti (0) ∪ Q_initi +1 j

j

f_check_zeroi := BL_quanti ∪ F_check_zeroi ∪ Q_initi +1 f_save_zeroi := BL_quanti ∪ F_save_zeroi ∪ Q_initi +1 j

j

f_seti (1) := BL_quanti ∪ F_seti (1) ∪ Q_initi +1 j

j

• f_check_nexti (1) := BL_quanti ∪ F_check_nexti (1) ∪ E_verifyi +1 • f_save_nexti (1) := BL_quanti ∪ F_save_nexti (1) ∪ E_savei +1 j j • f_reset_nexti (1) := BL_quanti ∪ F_reset_nexti (1) ∪ Q_initi +1 The following sets are used to search the structure of G corresponding to the clauses of . For every i with 1 ≤ i ≤ m and every j with 1 ≤ j ≤ 3, the set F contains the following set: If the l-th literal of C i is xk for some k with 1 ≤ k ≤ n, j j then F contains the set c_seti (l) := BL_set_clausek ∪ Q_checkk (1) ∪ C_seti . Otherwise, i.e., if the l-th literal of C i is x¯ k for j

j

some k with 1 ≤ k ≤ n, then F contains the set c_seti (l) := BL_set_clausek ∪ Q_checkk (0) ∪ C_seti . Finally, F contains the following sets, which are used at the start and at the end of any strategy.

• The sets start1 , . . . , startc+2n are defined in such a way that the sequence: START := ({start1 , start2 }, . . . , {startc +2n−1 , startc +2n })

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Fig. 11. A simplified illustration of the second half of the sets in F defined for i-th universal quantifier, where i ∈ [2, 4, · · · , n − 3, n − 1]. The figure only shows the part of G that corresponds to the i-th and i + 1-th quantifier of  and which contains vertices from the illustrated set. The part of the graph G that is shown in the figure is represented by a grid (omitting the center vertices c i , j as well as the vertices h i , j and v i , j ), which is given by the dotted lines. The vertices contained in the illustrated set are represented by the bold lines as well as the black cells as follows: a bold line indicates that the corresponding vertices (i.e. vertices of the form xi , j , h i , j , and v i , j ) belong to the set and a black cell indicates that all vertices of the corresponding cell and its border (including its center vertex) belong to the set.

clears exactly all the cells in the first row of G plus the cells of the second and third row of each column posi for every i with 1 ≤ i ≤ n. More formally, the above sequence clears the cells (1, 1), . . . , (1, c + 2n) as well as the cells (2, pos1 ), (3, pos1 ) . . . , (2, posn ), (3, posn ). Furthermore, these cells are cleared such that at most one cell is cleared at each step, i.e., each set contains at most one center vertex c i , j of some cell, and such  that the border of all these cells is contained in startc +2n , i.e., startc +2n contains all vertices in Bottom_line ∪ C_save ∪ ( 1≤i ≤n Q_initi ). • The sets end1 , . . . , endc+2n are defined in such a way that the sequence:

END := ({end1 , end2 }, . . . , {endr ·c −(c +2n)−1 , endr ·c −(c +2n) }) clears exactly all the cells of G apart from cells cleared by the sequence START and the cells (4, pos1 ), . . . , (11, pos1 ). More formally, the above sequence clears the cells in {(i , j ) : 2 ≤ i ≤ r ∧ 1 ≤ j ≤ c } minus the cells in {(i , pos j ) : 2 ≤ i ≤ 3 ∧ 1 ≤ j ≤ n} ∪ {(i , pos1 ) : 2 ≤ i ≤ 11}. Furthermore, these cells are cleared such that at most one cell is cleared at each step, i.e., each set contains at most one center vertex c i , j of some cell, and such that the border of all these cells is contained in end1 , i.e., end1 contains all vertices in Bottom_line ∪ C_save ∪ E_verify1 ∪ ( 2≤i ≤n Q_initi ). This finishes the construction of G and F . Clearly, G and F can be constructed in polynomial time for a given input formula . It hence remains to show that two cops have a winning strategy in the invisible set game on G and F if and only if  is satisfiable.

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Before doing so, let us mention how exactly the states of the quantifiers and the set of all clauses (used by the procedure given at the beginning of this proof to verify the satisfiability of ) are related to the set of cells that have already been cleared at a certain step of a strategy for the cop-player. In the following, let S := ( S 1 , . . . , S l ) be a strategy for two cops in the invisible set game on G and F together with its associated sequence R := ( R 1 , . . . , R l ) of robber-spaces and let s with 1 ≤ s ≤ l be an arbitrary step of S . We will say that a vertex v ∈ V (G ) is clear at step s if it is not contained in R s . Furthermore, we will say that a cell (i , j ) of G is clear at step s if the vertices xi , j , xi +1, j , xi , j +1 , xi +1, j +1 , h i , j , h i +1, j , v i , j , v i , j +1 , c i , j are not contained in R s . We say that the set of all clauses is accepted (at step some step s of S ) if the cells (1, c − 2), . . . , (1 + 3m, c − 2) are clear and we say that the set of all clauses is initialized if the cell (1, c − 2) is clear and it is not accepted. For an existential quantifier, i.e., for every i with i ∈ {1, 3, . . . , n − 2, n}, we say that it is:

• • • •

accepted, if the cells (1, posi ), . . . , (11, posi ) are clear, zero, if the cells (3, posi + 1), . . . , (3, posi + 4) and the cells (1, posi ), . . . , (3, posi ) are clear, one, if the cells (3, posi − 1), . . . , (3, posi − 4) and the cells (1, posi ), . . . , (3, posi ) are clear, initialized, if the cells (1, posi ), . . . , (3, posi ) are clear and none of the previous applies. Similarly, for an universal quantifier, i.e., for every i with i ∈ {2, 4, . . . , n − 2, n}, we say it is:

• • • • •

zero-accepted, if the cells (3, posi + 1), . . . , (3, posi + 11) and the cells (1, posi ), . . . , (3, posi ) are clear, one-accepted, if the cells (3, posi − 1), . . . , (3, posi − 14) and the cells (1, posi ), . . . , (3, posi ) are clear, zero, if the cells (3, posi + 1), . . . , (3, posi + 4) and the cells (1, posi ), . . . , (3, posi ) are clear, one, if the cells (3, posi − 1), . . . , (3, posi − 7) and the cells (1, posi ), . . . , (3, posi ) are clear, initialized, if the cells (1, posi ), . . . , (3, posi ) are clear and none of the above states applies.

We are now ready to show the forward direction. Suppose that we are given a winning strategy S := ( S 1 , . . . , S l ) for two cops in the invisible set game on G and F together with its associated sequence R := ( R 1 , . . . , R l ) of robber-spaces. We start by showing some simply properties of S . W.l.o.g. we can assume that | S s | = 2 for every 1 ≤ s ≤ l, | S s  S s+1 | = 2 for every 1 ≤ s < l, i.e., between every two steps of S the cop-player changes the position for exactly one token, and that S is of minimal length (given the previous properties). Furthermore, we can also assume (in almost all cases) that after the cop-player changes a certain token at some step, he will not change the same token again immediately afterwards. More formally, for every step s of S with 1 < s < l, if f ∈ S s \ S s−1 and f ∈ / {start1 , endr ·c−(c+2n) }, then f ∈ S s+1 . To see this note that start1 and endr ·c −(c +2n) are the only sets in F that can clear an entire component of G \ f  for any f  ∈ F (in particular start1 and endr ·c −(c +2n) can clear the unique component of start2 and endr ·c −(c +2n)−1 , respectively, that contains exactly one center vertex). This is because every set in F contains at most one center vertex but every component of G \ f  for every f  ∈ F \ {start2 , endr ·c −(c +2n)−1 } contains at least two center vertices. Hence, if there were such an f ∈ S s \ S s−1 with f ∈ / {start1 , endr ·c−(c+2n) }, we could obtain a shorter strategy by removing the step S s from S , contradicting the minimality of S . Note that from the above properties, we obtain that S has to be of the form ({ f 1 , f 2 }, { f 2 , f 3 }, . . . , { f l−1 , f l }, { f l , f l+1 }), where f 1 , . . . , f l+1 ∈ F and f i = f i +2 for every i with 1 ≤ i ≤ l − 1. The following claim shows that whenever the cop-player removes a token from a set f ∈ F and places it on a token f  ∈ F , then the component of G \ f  that contains the unique cell whose center vertex is contained in f is clear immediately after f is removed. Claim 7.4. Let s be a step with 1 < s < l − 1 and f ∈ S s \ S s−1 . Then, the unique cell whose center vertex is in f , remains clear after f is removed (at step s + 2). Using the properties for the strategy S shown above, we can assume that the sub-sequence S  of S starting at step s − 1 and ending at step s + 2 has the form S  := ({ f 1 , f 2 }, ( f 2 , f ), ( f , f 3 ), ( f 3 , f 4 )) for some f 1 , f 2 , f 3 , f 4 ∈ F . Now, suppose for a contradiction that the unique cell whose center vertex is in f is recontaminated after f is removed. Hence, the occurrence of f in S  does not lead to any additional center vertices to be clear after step s + 2. Hence, replacing S  in S with the sequence S  := ({ f 1 , f 2 }, { f 2 , f 3 }, { f 3 , f 4 }) leads to the same cop-position { f 3 , f 4 } and the set of (center) vertices that are cleared after the execution of ( S 1 , . . . , S s−2 ) ◦ S  is contained in the set of (center) vertices that are cleared after the execution of ( S 1 , . . . , S s−2 ) ◦ S  . Consequently, we obtain a shorter winning strategy for two cops by replacing S  in S with S  , which contradicts the minimality of S . We now show that S has a very particular global structure. Let S M be the following sequence:

S M := T (e_check_next1 ) ◦

({e_check_next31 , e_save_next1 }, {e_save_next1 , e_reset_next11 }) ◦ T (e_reset_next1 ) Furthermore, let S E be the sequence:

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S. Kreutzer, S. Ordyniak / Theoretical Computer Science 628 (2016) 1–29

S E := ({e_reset_next31 , end1 }, {end1 , end2 },

··· , {endr ·c−(c+2n)−1 , endr ·c−(c+2n) }) The next claim shows that w.l.o.g. S can be split into a prefix START ◦ S S and a suffix S M ◦ S E , such that the prefix (or more precisely S S ) has a number of important properties and will contain the sequences used to verify the formula . Claim 7.5. S is of the form START ◦ S S ◦ S M ◦ S E and: A During the whole execution of S S at least one center vertex of the cells (4, pos1 ), . . . , (11, pos1 ) is contaminated. B S S does not contain any set used by the sequence START (apart from startc +2n ). C S S does not contain any set used by the sequence END. Proof. We first show that S has to contain the sequence S M . Because the only sets in F , which contain center vertices of the cells (4, pos1 ), . . . , (11, pos1 ) are the sets e_check_next11 , . . . , e_check_next31 , e_save_next1 , e_reset_next11 , . . . , e_reset_next31 , i.e., the sets used by the sequence S M , we obtain that these cells have to be cleared using these sets. Furthermore, given the definition of these sets, it is straightforward to check, that the only way to clear these cells is to use the sequence S M or its reverse. Because of the symmetry of strategies we can assume that S M occurs at least once in S , i.e., if only the reverse of S M occurs in S , we choose the reverse of S as our winning strategy. This shows that w.l.o.g. we can assume that S contains the sequence S M . We will show next that S has to end with the sequence S M ◦ S E and everything before does not contain the sequence S M . Clearly, S M cannot occur at the end of S , because this is only possible if the whole graph G is already cleared before the execution of S M and hence we could delete the last occurrence of S M in S and obtain a shorter winning strategy. Hence, after S M or its reverse are executed in S at least one of the two components of G \ e_check_next11 or G \ e_reset_next31 , respectively, is still contaminated. Furthermore, because of Claim 7.4, we obtain that the component of G \ e_reset_next31 containing the first row of G is clear after each execution of S M and similarly the component of G \ e_check_next11 containing the last row of G is clear after each execution of the reverse of S M in S . Now consider the first occurrence of S M in S . Because after its execution the component of G \ e_reset_next31 containing the first row is clear and the other component is contaminated, it follows that we can (minimally) replace the whole part of S after S M with the sequence S E . Hence, we have shown that S has to end with the sequence S M ◦ S E , and up to that point S does not contain S M . We show next that S has to start with the sequence START and does not contain START or its reverse again. Because the component of G \ e_check_next11 containing the first row of G has to be clear before the execution of S M in S and the only way to clear parts of this component is to use the sequence START or its reverse, we obtain that START or its reverse have to be contained in S before the execution of S M . We first show that S cannot contain the reverse of START. Assume that it does, then because of Claim 7.4, the component of G \ start1 containing the last row of G has to be clear after the execution of the reverse of START. However, then the whole graph G is clear and everything coming behind the reverse of START in S could be deleted contradicting the minimality of S . This shows that S does not contain the reverse of START and hence the sequence START has to be contained in S . We show next that this does only happen at the start of S . To see this consider the last occurrence of START in S S . Because of Claim 7.4 the component of G \ startc +2n containing the first row has to be clear after the execution of START. However, because S does not end after START the other component of G \ startc +2n has to be contaminated at this point. Hence, we could obtain a shorter winning strategy by deleting everything up to (not including) the occurrence of START from S contradicting the minimality of S . We show next that S S does also not contain the reverse of S M . To see this consider the last occurrence of the reverse of S M in S S . As we showed above immediately after the execution of the reverse of S M in S S the component of G \ e_check_next11 containing the first row of G is contaminated. Since we know that this component has to be clear after the execution of S S , it follows that S S has to contain (at least) the sequence START or its reverse after the last occurrence of the reverse of S M . However, this is not possible because the sequence START or its reverse occur only at the beginning of S S as shown above. Hence, S S does not contain the reverse of S M (and also not S M due to the choice of S S ). This shows that during the whole execution of S S at least one of the cells (4, pos1 ), . . . , (11, pos1 ) is contaminated and finishes the proof of part (A) of the claim. It remains to show parts (B) and (C). Towards showing part (C), suppose for a contradiction that S S uses a set used by the sequence END. Because for every such set e, it holds that all the cells that can only be cleared by the sequence START or its reverse are in the same component of G \ e as the center vertices of the cells (4, pos1 ), . . . , (11, pos1 ), it follows from part (A) that using any of these sets in S S would recontaminate all the cells that can only be cleared by the sequence START or its reverse. However, as we have shown above that S S does not contain the sequence START or its reverse, part (C) of the claim follows. Towards showing part (C), suppose for a contradiction that S S contains a set used in START and let s be the last such set that occurs in S S . It is sufficient to show that s = startc +2n . Because of part (A) of this claim, we obtain that the component

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of G \ s containing the last row of G is contaminated and if s = startc +2n , then this component contains at least the center vertex of the unique cell contained in startc +2n . However, this center vertex has to be clear at the end of S S as argued before. Hence, the set startc +2n has to be contained in S S after the occurrence of s contradicting our choice of s. This shows that s = startc +2n and concludes the proof of the claim. 2 We show now the exact sequences that S S has to contain in order for the quantifiers and the set of all clauses to enter each of the above states. See also Fig. 3 for an illustration of the effects of these sub-strategies for a simple example formula. EA For each i ∈ [1, 3, 5, · · · , n − 2, n] if the i-th existential quantifier becomes accepted (at some step of S S ), then S S contains the sequence:

S ∃A (i ) := T (e_check_nexti ) ◦

({e_check_next3i , e_save_nexti }, {e_save_nexti , e_reset_next1i }) T (e_reset_nexti ) Proof. From of the definition of F , we obtain that the only sets containing the center vertices from the cells (4, posi ), . . . , (11, posi ) are the sets used by the sequence S A∃ (i ) as well as exactly one set contained in END for each of these center vertices. Because of Claim 7.5 (C) none of the sets contained in the sequence END can be contained in S S . Consequently, either S A∃ (i ) or its reverse has to be contained in S S in order for the i-th quantifier to become accepted. Because of Claim 7.4, if the reverse of S A∃ (i ) is used, then the component of G \ e_check_next1i containing the last row of G has to be clear after its execution. However, because S is not finished after this execution, we obtain that the other component of G \ e_check_next1i , i.e., the component containing the first row of G is still contaminated at this point. Hence, in order to clear this component S S would have to contain START or its reverse, which is not possible at this point due to Claim 7.5. 2 EZ For each i ∈ [1, 3, 5, · · · , n − 2, n], if the i-th existential quantifier becomes zero, then S S contains the following sequence:

S0∃ (i ) := T (e_seti (0)) The proof uses the same arguments as the proof of Claim EA and is not repeated here. EO For each i ∈ [1, 3, 5, · · · , n − 2, n], if the i-th existential quantifier becomes one, then S S contains the following sequence:

S1∃ (i ) := T (e_seti (1)) The proof uses the same arguments as the proof of Claim EA and is not repeated here. UZ For each i ∈ [2, 4, · · · , n − 3, n − 1], if the i-th universal quantifier becomes zero, then S S contains the following sequence:

S0∀ (i ) := T (f_set1i (0)) Again the proof uses the same arguments as the proof of Claim EA and is not repeated here. UZA For each i ∈ [2, 4, · · · , n − 3, n − 1], if the i-th universal quantifier becomes zero-accepted, then S S contains the following sequence:

S0∀A (i ) := T (f_check_nexti (0)) ◦

({f_check_next3i (0), f_save_nexti (0)}, {f_save_nexti (0), f_reset_next1i (0)}) ◦ T (f_reset_nexti (0)) Again the proof uses the same arguments as the proof of Claim EA and is not repeated here. UO For each i ∈ [2, 4, · · · , n − 3, n − 1], if the i-th universal quantifier becomes one, then S S contains the following sequence:

S1∀ (i ) := T (f_check_zeroi ) ◦

({f_check_zero3i , f_save_zeroi }, {f_save_zero3i , f_set1i (1)}) ◦ T (f_seti (1)) Again the proof uses the same arguments as the proof of Claim EA and is not repeated here.

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UOA For each i ∈ [2, 4, · · · , n − 3, n − 1], if the i-th universal quantifier becomes one-accepted, then S S contains the following sequence:

S1∀A (i ) := T (f_check_nexti (1)) ◦

({f_check_next3i (1), f_save_nexti (1)}, {f_save_nexti (1), f_reset_next1i (1)}) ◦ T (f_reset_nexti (1)) Again the proof uses the same arguments as the proof of Claim EA and is not repeated here. C If the clauses become accepted, then S S contains a sequence of the following form:

SC := ({c_set11 (l11 ), c_set21 (l21 )},

{c_set21 (l21 ), c_set31 (l31 )}, {c_set31 (l31 ), c_set12 (l12 )}, {c_set12 (l12 ), c_set22 (l22 )}, {c_set22 (l22 ), c_set32 (l32 )}, ··· 1 1 2 2 (lm ), c_setm (lm )}, ({c_setm 2 2 3 3 (lm ), c_setm (lm )}, {c_setm j

where for every i and j with 1 ≤ i ≤ m and 1 ≤ j ≤ 3, li is an index (between 1 and 3) of a literal of C i that is satisfied j

by the current states of the quantifiers, i.e., if the li -th literal of C i is xk or x¯ k for some k with 1 ≤ k ≤ n, then the k-th quantifier is one or zero, respectively, before the occurrence of SC in S S . Again the proof uses the same arguments as the proof of claim EA and is not repeated here. Using these claims and again Claims 7.4 and 7.5 it is straightforward to show that S S has to change the set of cleared cells and therefore the states of the quantifiers and the state of the set of all clauses according to the rules RI, RG, RE, RU, and RC given for the procedure to verify the satisfiability of  at the beginning of this proof. Because after the execution of START ◦ S S ◦ S M , the first existential quantifier has accepted, this implies that the formula  is satisfiable. Towards showing the reverse direction, suppose that  is satisfiable and let P := ( p 1 , . . . , pn ) be a sequence of state changes according to the rules RE1, RE2, RU1–RU4, and RC witnessing this. We will now show how to construct a winning strategy S for two cops in the invisible set game on G and F . We set S := S n , where S i for every 0 ≤ i ≤ n is inductively defined as follows: To ensure rule RI, we set S 0 := START. Furthermore, for every i with 1 ≤ i ≤ n, we set S i := S i −1 ◦ −1 i −1 i i ({ S |iS | \ S |S |−1 , S 1 ( p ) \ S 2 ( p )}) ◦ S ( p i ), where S j is the j-th entry of S , S j ( p i ) is the j-th entry of S ( p i ), and S ( p ) for every p ∈ P is defined depending on the nature of the state change p as follows:

• If p is a state change according to rule RE1, i.e., the state of the i-th existential quantifier changes from initialized to zero or one, we set S ( p ) := S0∃ (i ) or S ( p ) := S1∃ (i ), respectively. • If p is a state change according to rule RE2, i.e., the state of the i-th existential quantifier changes to accepted, we set S ( p ) := S A∃ (i ). • If p is a state change according to rule RU1, i.e., the state of the i-th universal quantifier changes from initialized to zero, we set S ( p ) := S0∀ (i ). • If p is a state change according to rule RU2, i.e., the state of the i-th universal quantifier changes from zero to zeroaccepted, we set S ( p ) := S0∀A (i ). • If p is a state change according to rule RU3, i.e., the state of the i-th universal quantifier changes from zero-accepted to one, we set S ( p ) := S1∀ (i ). • If p is a state change is according to rule RU4, i.e., the state of the i-th universal quantifier changes from one to one-accepted, we set S ( p ) := S1∀A (i ). • If p is a state change according to rule RC, i.e., the state of the set of all clauses changes from initialized to accepted, we set S ( p ) := SC . Note that a sequence of the form SC always exists due to the condition of rule RC that the current assignment of the variables (represented by the corresponding quantifiers being either zero or one) satisfies all the clauses of . This completes the definition of S (recall that S := Sn ) and it is straightforward to verify that S is a winning strategy for two cops in the invisible set game on G and F . 2

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Fig. 12. Partial illustration of the construction in the proof of Theorem 7.7. The formula ϕ := (x1 ∨ x¯ 2 ∨ x3 ) ∧ (¯x1 ∨ x2 ∨ x¯ 3 ) ∧ (¯x1 ∨ x¯ 2 ∨ x¯ 3 ). The figure contains three copies of the cycle corresponding to the first clause and the sets in F induced by the literals.

In [8, Problem 7], Fomin et al. raise the question whether for every graph G there is a winning strategy of length O (n) using ds(G ) cops in the invisible domination game. As a consequence of the proof of the previous theorem we answer this question negatively by showing that there exist graphs on which the number of steps needed by a strategy in the domination game is at least exponential in the size of the graph and thus cannot be bounded by a polynomial. Clearly, exponential length of strategies is also the worst possible. Corollary 7.6. There exists a family C of graphs such that two cops have a winning strategy in the invisible domination game on each G ∈ C but any such strategy is at least of exponential length, i.e. there is no polynomial p (n) so that the length of these strategies is bounded by p (|G |). We now consider the problem to decide for a given graph G whether k cops have a cop-monotone winning strategy in the invisible domination game. Clearly, as the length of monotone strategies is polynomially bounded in the size of the graph, the problem is necessarily in NP. We again give tight complexity bounds by showing that even deciding whether two cops have monotone winning strategies is NP-hard. Theorem 7.7. Deciding whether two cops have a cop-monotone winning strategy in the invisible domination game is NP-complete. Proof. To simplify the presentation, we show the result for the invisible set game. Because of Theorem 4.2, the result then also follows for the invisible domination game. We reduce 3-SAT to the problem whether two cops have a cop-monotone winning strategy in the invisible set game. Let  be a 3-SAT formula with variables x1 , . . . , xn and clauses C 1 , . . . , C m . We construct a graph G and a set F of vertex subsets of G such that  is satisfiable if, and only if, two cops have a cop-monotone winning strategy in the invisible set game on G and F . The graph G is the disjoint union of m cycles of length 12. More formally, for every 1 ≤ j ≤ m the graph G contains the cycle Y j := ( y 1j , a1j , . . . , y 6j , a6j ). Furthermore, for every 1 ≤ i ≤ n, the set F contains three sets F xi , F x¯ i and F i , which are defined as follows: Let C j be a clause and let L be a literal of C j involving the variable xi (positively or negatively) for some 1 ≤ i ≤ n. Hence, L = xi or L = x¯ i . If L is the k-th literal of C j then:

• The set F L contains the vertices ykj and ykj+3 .

• The set F L¯ contains all vertices of the component of Y j \ F L that contains yk+1 . Here, L¯ is the dual literal, i.e. L¯ = xi if L = x¯ i and L¯ = x¯ i otherwise. • The set F i contains the remaining vertices of Y j , i.e., all vertices in Y j \ ( F L ∪ F L¯ ). This completes the construction of G and F . See Fig. 12 for an illustration. The figure illustrates parts of G and F for the formula ϕ := (x1 ∨ x¯ 2 ∨ x3 ) ∧ (¯x1 ∨ x2 ∨ x¯ 3 ) ∧ (¯x1 ∨ x¯ 2 ∨ x¯ 3 ). Hence, the graph has 3 cycles of length 12. For clarity, the figure contains three copies of the first cycle corresponding to the clause (x1 ∨ x¯ 2 ∨ x3 ). Here, the vertices labeled by numbers i = 1, . . . , 6 are the vertices y 1i and the black dots represent the vertices a1i . Each of the three literals in the clause yields three sets as illustrated in the figure. It remains to show that  is satisfiable if, and only if, two cops have a cop-monotone winning strategy in the invisible set game on G and F . Towards this aim, note first that no cycle can be covered by only two sets in F : for two sets F 1 , F 2 ∈ F corresponding to the same literal this is obvious from the construction and for two sets corresponding to different literals this follows from the fact that the sets are “shifted” along the cycle. Now suppose that  is satisfiable and let β be a satisfying assignment. Let C j be a clause and let L be a literal occurring in C j such that β sets L to 1. If L = x¯ i , then we can clear the cycle Y j using the following strategy Si0 := ({ F x¯ i , F i }, { F x¯ i , F xi }). Otherwise, if L = xi , then we can clear the cycle Y j using the strategy Si1 := ({ F xi , F i }, { F xi , F x¯ i }).

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Hence, we can clear the entire graph by choosing a minimal set X of variables such that every clause in  is satisfied by a literal containing a variable from X . Let X := {x1 , . . . , xk }. Then, for i = 1, . . . , k, if β(xi ) = 1 we play the partial strategy Si1 and otherwise we play Si0 . By playing Si1 or Si0 , resp., we clear every cycle whose corresponding clause is satisfied by a literal containing xi . Hence, once we played the strategy for all xi ∈ X we have cleared the entire graph. It is easily seen that this is indeed a cop-monotone strategy (but it is not robber monotone). We now show the converse, i.e. whenever two cops have a cop-monotone winning strategy then  is satisfiable. For this, we essentially show that the graph can only be cleared using some of the strategies Si1 or Si0 (possibly in reverse direction) as above. As a first step, consider a single cycle Y j corresponding to a clause C j and the restriction of the sets in F to this cycle. We claim that any cop-monotone strategy ( F 1 , . . . , F k ) clearing this cycles must contain a subsequence F i , . . . , F j , for some i < j, such that for some literal L ∈ C j involving variable x v , 1. F i = { F L , F L¯ }, F j = { F L , F v } and F L ∈ F s for all i < s < j or 2. F i = { F L , F v }, F j = { F L , F L¯ } and F L ∈ F s for all i < s < j. Suppose the claim was false. Let ( F 1 , . . . , F k ) be a cop-monotone winning strategy for two cops witnessing this such that k is of minimal length among all such strategies. As Y j cannot be cleared by a single cop we may assume that the first move of the cop-player is to place cops on two different sets F 1 = F 2 ∈ F . As there are only two cops, it follows that one cop must be removed from F 1 or F 2 and be placed on some F 3 ∈ F . W.l.o.g. we may assume that the cop is moved from F 1 to F 3 . But now, if { F 1 , F 2 } = { F L , F L¯ } and { F 1 , F 2 } = { F L , F v }, for some literal L ∈ C j involving variable x v , then after removing the first cop from F 1 , all vertices of Y j not in F 2 are contaminated again. Hence, placing a cop in F 1 was not necessary and therefore ( F 2 , . . . , F k ) is a shorter winning strategy for two cops, contradicting the choice of ( F 1 , . . . , F k ). Hence we can assume that { F 1 , F 2 } = { F L , F L¯ } or { F 1 , F 2 } = { F L , F v }. Assume F 1 = F L¯ and F 2 = F L and the cop player removes F 1 to place the cop in F 3 . But then, by the same argument, in any winning strategy of minimal length, a cop must remain on F L until the second cop is eventually placed on F v to clear Y j . The other case, where { F 1 , F 2 } = { F L , F v } is analogous. This proves the claim above. Note that if any cop-monotone strategy ( F 1 , . . . , F k ) for two cops contains for some literal L involving variable x v a subsequence satisfying Condition (1) or (2) above, then due to cop-monotonicity it cannot contain a subsequence satisfying ¯ Hence, for any literal L the strategy contains at most one subsequence satisfying Condition (1) or (2) above for the literal L. Condition (1) or (2) above. Now let S := ( F 1 , . . . , F k ) be a cop-monotone winning strategy for two cops. We define an assignment β of the variables x1 , . . . , xn as follows. If S contains a subsequence satisfying Condition (1) or (2) above for the literal L = xi , then we set β(xi ) = 1. Otherwise we set β(xi ) = 0. We claim that β satisfies the formula . Let C j be a clause in . As S clears Y j it follows from above that for some literal L ∈ C j the strategy S contains a subsequence satisfying Condition (1) or (2) for the ¯ it follows that if L = xi then β(xi ) = 1 and otherwise, if L = x¯ i , then literal L. As no such sequence can exist in S for L, β(xi ) = 0. In both cases β satisfies C j and hence . This completes the proof. 2 We do not know corresponding results for robber-monotone or selective-monotone strategies and leave this as an open problem. The previous results settle the classical complexity of the domination game problem. We now study the parameterized complexity of this problem. The parameterized domination search problem p-Domination Search is defined as the problem, given a graph G and k ∈ N as input, to decide if k cops have a winning strategy in the invisible domination game on G. We take k as the parameter. The following is an immediate consequence of Theorem 7.2 and Theorem 7.7. Corollary 7.8. p-Domination Search is not in XP (unless P = NP). The same is true for the cop-monotone version of the problem. The previous results establish fixed-parameter intractability for domination games. Hence, domination games are considerably more complex than standard cops and robber games, which are NP-complete and fixed-parameter tractable. The latter follows from the parameterized tractability of treewidth and pathwidth and the monotonicity of the games. We now turn to special cases where tractability can be obtained. A natural choice of graph classes where the problem might be easier are classes of bounded tree- or pathwidth. One is tempted to think that fixed-parameter tractability of domination search on classes C of graphs of treewidth at most d could be established along the following lines: given G ∈ C and k ∈ N, we first compute a tree-decomposition of G of width d and then use dynamic programming to decide whether there is a winning strategy of width k. This is the approach taken to show that the analogous questions for cops and robber games (visible and invisible) can be solved by linear time parameterized algorithms. Typically, one proceeds bottom-up along the tree-decomposition and for each node in the decomposition tree one computes a constant size data structure containing information about the sub-graph induced by the vertices in the sub-tree rooted at this node. For domination games, however, this approach fails as a vertex in a bag can be dominated by vertices not contained in this bag. The ways

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Fig. 13. The graph G k for k = 3 in the proof of Theorem 8.1. Thick lines between two ellipses represent edges from all vertices contained in one to all vertices contained in the other ellipse. The filled vertices are all vertices contained in F r for the root r of T k .

in which this happens can be rather complex and hence a constant size data structure seems difficult to obtain. It is still possible, though, that domination search is fpt on classes of bounded treewidth and we leave this for future work. However, we believe that parameterized algorithms for classes of graphs of bounded degree can be obtained by suitably adapting the algorithm in [2] which uses dynamic programming to compute an optimal path-decomposition of a graph given a bounded width tree-decomposition (recall that the problem is already NP-hard on the class of graphs of degree at most 3). To adapt this algorithm for the cop-monotone domination game on bounded degree classes of graphs we have to extend the characteristics used in this algorithm to not only contain information about when a vertex in B t is occupied by a cop, but also to contain information about when it is dominated by some cop on a vertex in the sub-tree below t. Since every vertex in G has degree at most d and we are restricting ourselves to cop-monotone strategies, the status of a vertex in v ∈ B t , i.e. whether v is currently dominated, occupied or neither dominated nor occupied, can change at most O (d) times. However, the proof in [2] is very long and technical and we would need to adapt and reprove every part of it which would add another very long and technical part to this paper. We therefore refrain from doing so and pose the following as a conjecture. Conjecture 7.9. The problem, given G ∈ C and k ∈ N, to decide whether k cops have a cop-monotone winning strategy on G is fixed-parameter tractable with parameter d + w, where w is the tree-width of G and d is its maximum degree. 8. Visible robber In this section we consider the visible variant of the domination game. We start by showing that also the visible domination game is not robber-monotone. Theorem 8.1. For every k > 2, there is a graph G k such that vis-ds(G k ) = 2 but r-vis-ds(G k ) = k. Proof. To simplify the construction, we show the result for the visible set game. Because of Theorem 4.2, the result then also follows for the visible domination game. For k > 2, let T k be the complete k-ary tree of height k that is rooted at the unique root r balancing T k . Then, G k is obtained from T k after replacing every vertex t of T k with a clique C t of size k (and adding all edges between all vertices of C t and all vertices of C t  if and only if t and t  are adjacent in T k ). For t ∈ V ( T k ), we denote by l(t ) the level of t in T k . Then, for every vertex t ∈ V ( T k ), the set Fk contains a set F t , which contains all vertices in C t as well as the l(t )-th vertex of C l for every leaf l of T k . This completes the construction of G k and Fk . An illustration of the construction is given in Fig. 13. We start by showing that two cops have a (non-monotone) winning strategy in the visible set game on G k and Fk . The strategy is defined as follows: 1. In the beginning of the game one cop is placed on the set F r for the root r of T k . 2. Now the robber has k choices, i.e. he can go into any sub tree rooted at the k children of r in T k . Then, the cop-player responds by placing one additional cop F r  , where r  is the child of r corresponding to the choice of the robber. Moreover, the cop-player can now safely remove the cop on F r without allowing the robber to leave his current sub tree rooted at r  . By setting r to r  and repeating step (2), the cop-player will eventually capture the robber in a leaf of T k . Observe that this step is not monotone, since the robber can now revisit the l(r )-th original vertex of every clique corresponding to a leaf of his current sub tree. Note that the above strategy can be turned into a robber-monotone winning strategy for k cops by not allowing the copplayer to remove any token. Towards showing that less than k cops are not able to capture the robber in a robber-monotone way consider the following strategy for the robber: The robber stays on the vertices C r for the root r of T k until a cop is placed on F r . Observe that if the cop-player eventually places a cop on F r , then there exits at least one child r  of r in T k such that no cop has yet been placed on F s for any s in the subtree of T k rooted in r  . Hence, once the cop-player places a token on r the robber is free to move to some vertex in C r  . The robber then continues this strategy with r replaced by r  , i.e., he stays on

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its vertex in C r  until a cop is placed on F r  in which case he chooses a child r  of r  such that no cop has yet been placed on F s for any s in the subtree of T k rooted in r  . Observe that the cop-player can never remove a cop from F a for any ancestor a of the current robber position in T k without violating robber-monotonicity. Hence, when the robber is eventually captured on a vertex in C l for a leave l of T k the cop occupies F l together with F a for every of the k − 1 ancestors a of l in T k . Hence, the cop-player needs at least k tokens to capture the robber with a robber-monotone strategy. This completes the proof of the theorem. 2 Finally, we consider the complexity of the visible domination game. In terms of classical complexity, we can show the following. It is easily seen that all visible game variants except for the selective monotone variant are in XP, as the current cop and robber position completely determine the current state of the play and there are only nO(k) such positions. We show next that the problem is not in FPT unless FPT = W[2]. As observed in [8], domination search is closely related to dominating sets in graphs. A dominating set of a graph G is a set X such that for all v ∈ V (G ) either v ∈ X or there is a u ∈ X such that {u , v } ∈ E (G ). The domination number of G, denoted by γ (G ), is the minimal size of a dominating set of G. Lemma 8.2. (See [8].) Let G be a graph and H be the graph obtained from G by connecting every pair of non-adjacent vertices in G by a path of length three. Then γ (G ) ≤ ds( H ) ≤ γ (G ) + 1. We establish a similar but exact correspondence for the visible/invisible set game (and the corresponding domination game) using a slightly different construction. Theorem 8.3. For every graph G there is a graph G  and a set F  of vertex subsets of G  , both of size polynomial in |G |, such that γ (G ) + 1 is equal to the number of cops required to win the visible/invisible set game on G  and F  . Proof. Let G  be the graph obtained from G after adding | V (G )| + 1 new vertices v 1 , . . . , v | V (G )|+1 to G and adding an edge from every new vertex v i to all vertices of G. Furthermore, the set F  contains one set F v i := { v i } for every i with 1 ≤ i ≤ | V (G )| + 1 as well as one set F v := N G [ v ] for every v ∈ V (G ). This completes the construction of G  and F  and it remains to show that G has a dominating set of size k if and only if the cop-player wins the visible/invisible set game on G  and F  using at most k + 1 cops. Suppose that G has a dominating set D. Then, | D | + 1 cops have a winning strategy in the visible/invisible set game on G  and F  by first placing | D | cops on { F d : d ∈ D } and then using one additional cop playing on the sets F v 1 , . . . , F v | V (G )|+1 to clear the remaining vertices of G  . For the reverse direction, we will show that the robber has a winning strategy against γ (G ) cops in the visible/invisible set game on G  . Hence, any winning strategy for the cop-player uses at least γ (G ) + 1 cops and the reverse direction  follows. Observe that if  S ⊆ F is a position for the cop-player such that neither V (G ) ⊆ s∈ S s nor { v 1 , . . . , v | V (G )|+1 } ⊆ s∈ S s, then the graph G \ ( s∈ S s) is connected. Furthermore, with the cop-player  using at most γ (G ) cops  and γ (G ) < | V (G )| + 1, we obtain that for any position S of the cop-player either V (G ) ⊆ s∈ S s or the graph G  \ ( s∈ S s) is connected. Hence, whenever the union of the sets currently occupied by the cops is not a dominating set of G, the robber has a unique component to which he can escape. Hence, if the cop-player wants to capture the robber, the cop-player eventually has to  go on a position S such that V (G ) ⊆ ( s∈ S s). Observe that in this case the set { v ∈ V (G ) : F v ∈ S } is a dominating set of G and | S | = γ (G ). Consequently, when the cop-player plays on S, the robber can escape to any of the vertices v 1 , . . . , v | V (G )|+1 and  in its next move the cop-player has to remove at least one cop and hence again ends up in a position S  such that G \ ( s∈ S  s) is connected, which implies that the robber can escape forever. 2 The theorem immediately gives a parameterized reduction from the dominating set problem, parameterized by the size of the solution, to the visible/invisible set and hence also domination search problem (using Theorem 4.2), parameterized by the number k of cops. The following result follows from the W[2]-hardness of the dominating set problem, where W[2] is a parameterized complexity class strongly believed to be different from FPT. Theorem 8.4. The problem p-Domination Search: “given a graph G and k ∈ N, with parameter k, decide whether k cops have a winning strategy in the (in-)visible set or domination game on G” is W[2]-hard. 9. Conclusion We have studied the complexity and monotonicity properties of the domination game. Our main results show that domination games behave quite differently from related graph searching games in that they are strongly not monotone and Pspace-complete. This identifies domination games as one of the first variants of graph searching games, where strategies for the cop-player are necessarily of exponential length. On the way to these results, we have introduced set games, which we believe provide a much more elegant way to handle games such as the domination game. We also introduced a novel

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version of monotonicity for domination games, which is conceivably a better fit for the peculiarities of domination games, and provided some initial monotonicity and complexity results for the visible domination game. There remain a number of open questions, however, which are mostly related to the monotone versions of the visible and invisible variants of the domination game. For instance, can it be shown that the selective-monotone variant is within a constant factor of the non-monotone variant of the domination game? If it is, the selective-monotone variant would provide a constant factor NP-approximation for the non-monotone variant of the game. And if so, what is the exact parameterized complexity of the selective-monotone variant of the visible and invisible domination game? Acknowledgements We are grateful to Fedor Fomin for bringing domination games to our attention and thereby stimulating the research reported in this paper and to Fedor Fomin, Paul Hunter and Paul Dorbec for valuable discussions on the subject. References [1] H. Bodlaender, A partial k-arboretum of graphs with bounded tree-width, Theoret. Comput. Sci. 209 (1998) 1–45. [2] H. Bodlaender, T. Kloks, Efficient and constructive algorithms for the pathwidth and treewidth of graphs, J. Algorithms 21 (2) (1996) 358–402. [3] H.L. Bodlaender, Treewidth: algorithmic techniques and results, in: Proc. of Mathematical Foundations of Computer Science, MFCS, in: LNCS, vol. 1295, 1997, pp. 19–36. [4] R. Diestel, Graph Theory, 3rd edition, Springer-Verlag, 2005. [5] R.G. Downey, M.R. Fellows, Fundamentals of Parameterized Complexity, Springer, 2014. [6] F. Fedor, D.M. Thilikos, An annotated bibliography on guaranteed graph searching, Theoret. Comput. Sci. 399 (3) (2008) 236–245. [7] Jörg Flum, Martin Grohe, Parameterized Complexity Theory, Texts in Theoretical Computer Science. An EATCS Series, vol. XIV, Springer Verlag, Berlin, 2006. [8] F. Fomin, D. Kratsch, H. Müller, On the domination search number, Discrete Appl. Math. 127 (3) (2003) 565–580. [9] M.K. Franklin, Z. Galil, M. Yung, Eavesdropping games: a graph-theoretic approach to privacy in distributed systems, J. ACM 47 (2) (2000) 225–243. [10] L.M. Kirousis, C.H. Papadimitriou, Searching and pebbling, Theoret. Comput. Sci. 47 (3) (1986) 205–218. [11] T. Kloks, D. Kratsch, H. Müller, On the structure of graphs with bounded asteroidal number, Graphs Combin. 17 (2) (2001) 295–306. [12] Stephan Kreutzer, Sebastian Ordyniak, Distance d-domination games, in: Paul Christophe, Michel Habib (Eds.), Graph-Theoretic Concepts in Computer Science, 35th International Workshop, Revised Papers, WG 2009, Montpellier, France, June 24–26, 2009, in: Lecture Notes in Computer Science, vol. 5911, 2009, pp. 308–319. [13] Rolf Niedermeier, Invitation to Fixed-Parameter Algorithms, Oxford Lecture Series in Mathematics and Its Applications, Oxford University Press, Oxford, 2006. [14] P.D. Seymour, R. Thomas, Graph searching and a min-max theorem for tree-width, J. Combin. Theory Ser. B 58 (1) (1993) 22–33.