Can Romeo and Juliet Meet? Or Rendezvous Games with Adversaries on Graphs
CCan Romeo and Juliet Meet?Or Rendezvous Games with Adversaries on Graphs ∗ Fedor V. Fomin † Petr A. Golovach † Dimitrios M. Thilikos ‡ For never was a story of more woe than this ofJuliet and her Romeo. — William Shakespeare, Romeo and Juliet
Abstract
We introduce the rendezvous game with adversaries. In this game, two players,
Facilitator and
Disruptor , play against each other on a graph. Facilitator has two agents, and Disruptorhas a team of k agents located in some vertices of the graph. They take turns in movingtheir agents to adjacent vertices (or staying). Facilitator wins if his agents meet in somevertex of the graph. The goal of Disruptor is to prevent the rendezvous of Facilitator’sagents. Our interest is to decide whether Facilitator can win. It appears that, in general,the problem is PSPACE -hard and, when parameterized by k , co - W [2]-hard. Moreover, eventhe game’s variant where we ask whether Facilitator can ensure the meeting of his agentswithin τ steps is co - NP -complete already for τ = 2. On the other hand, for chordal and P -free graphs, we prove that the problem is solvable in polynomial time. These algorithmsexploit an interesting relation of the game and minimum vertex cuts in certain graph classes.Finally, we show that the problem is fixed-parameter tractable parameterized by both thegraph’s neighborhood diversity and τ . We introduce the
Rendezvous Game with Adversaries on graphs. In our game, a team ofdisruptors tries to prevent two passionate lovers, say Romeo and Juliet, from meeting eachother. We are interested in the minimum size of the team of disruptors sufficient to “separate”Romeo from Juliet. In the static setting, when disruptors do not move, this is just the problemof computing the minimum vertex cut between the pair of vertices occupied by Romeo andJuliet. But in the dynamic variant, when disruptors are allowed to change their position, theteam’s size can be significantly smaller than the size of the minimum cut. In fact, this givesrise to a new interactive form of connectivity that is much more challenging both from thecombinatorial and the algorithmic point of view.Our rendezvous game rules are very similar to the rules of the classical
Cops-and-Robber game of Nowakowski-Winkler and Quillioit [28, 29], see also the book of Bonato and Nowakowski[4]. The difference is that in the
Cops-and-Robber game, a team of k cops tries to capture ∗ This research was supported by the French Ministry of Europe and Foreign Affairs, via the Franco-Norwegianproject PHC AURORA. The two first authors have been supported by the Research Council of Norway via theproject “MULTIVAL” (grant no. 263317). The last author was supported by the ANR projects DEMOGRAPH(ANR-16-CE40-0028), ESIGMA (ANR-17-CE23-0010), ELIT (ANR-20-CE48-0008), and the French-German Col-laboration ANR/DFG Project UTMA (ANR-20-CE92-0027). † Department of Informatics, University of Bergen, Norway. ‡ LIRMM, Univ Montpellier, CNRS, Montpellier, France. a r X i v : . [ c s . D M ] F e b robber in a graph, while in our game, the group of k disruptors tries to keep the two loversseparated.A bit more formally. The game is played on a finite undirected connected graph G by twoplayers: Facilitator and
Disruptor . Facilitator has two agents R and J that are initially placedin designated vertices s and t of G . Disruptor has a team of k ≥ D , . . . , D k thatare initially placed in some vertices of V ( G ) \ { s, t } selected by Disruptor. Several disruptoragents can occupy the same vertex. Then the players make their moves by turn, starting withFacilitator. At every move, each player moves some of his agents to adjacent vertices or keepsthem in their old positions. No agent can be moved to a vertex that is currently occupied byadversary agents. Both players have complete information about G and the positions of all theagents. Facilitator aims to ensure that R and J meet; that is, they are in the same vertex.The task of Disruptor is to prevent the rendezvous of R and J by maintaining D , . . . , D k inpositions that block the possibility to meet. Facilitator wins if R and J meet, and Disruptorwins if he succeed in preventing the meeting of R and J forever.We define the following problem: Input:
A graph G with two given vertices s and t , and a positive integer k . Task:
Decide whether Facilitator can win on G starting from s and t againstDisruptor with k agents. Rendezvous
Another variant of the game is when the number of moves of the players is at most someparameter τ . Then Facilitator wins if R and J meet within the first τ moves, and Disruptorwins otherwise. Thus the problem is. Input:
A graph G with two given vertices s and t , and positive integers k and τ . Task:
Decide whether Facilitator can win on G starting from s and t in atmost τ steps against Disruptor with k agents. Rendezvous in Time
Our results.
We start with combinatorial results. If s = t or if s and t are adjacent, thenFacilitator wins by a trivial strategy. However, if s and t are distinct nonadjacent vertices, thenDisruptor wins provided that he has sufficiently many agents. For example, the agents can beplaced in the vertices of an ( s, t )-separator and stay there. Then R and J never meet. Wecall the minimum number k of the agents of Disruptor that is sufficient for his winning, the( s, t ) -disruption number and denote it by d G ( s, t ). We put d G ( s, t ) = + ∞ for s = t or when s and t are adjacent. The disruption number can be seen as a dynamic analog of the minimumsize λ G ( s, t ) of a vertex ( s, t )-separator in G . (The minimum number of vertices whose removalleaves s and t in different connected components.) Then Rendezvous can be restated as theproblem of deciding whether d G ( s, t ) > k .The first natural question is: What is the relation between d G ( s, t ) and λ G ( s, t )? Clearly, d G ( s, t ) ≤ λ G ( s, t ). We show that d G ( s, t ) = 1 if and only if λ G ( s, t ) = 1. If d G ( s, t ) ≥
2, then weconstruct examples demonstrating that the difference λ G ( s, t ) − d G ( s, t ) can be arbitrary evenfor sparse graphs. Interestingly, there are graph classes where both parameters are equal. Inparticular, we show that λ G ( s, t ) = d G ( s, t ) holds for P -free graphs and chordal graphs. Thisalso yields a polynomial time algorithm computing d G ( s, t ) on these classes of graphs.Then we turn to the computational complexity of Rendezvous and
Rendezvous in Time on general graphs. Both problems can be solved it n O ( k ) time by the standard backtrack-ing technique. We show that this running time is asymptotically tight by proving that they2re co - W [2]-hard when parameterized by k (we prove that it is W [2]-hard to decide whether d G ( s, t ) ≤ k ) and cannot be solved in f ( k ) · n o ( k ) time for any function f of k , unless ETH fails.Moreover,
Rendezvous in Time is hard even if τ = 2. If τ is a constant, then Rendezvousin Time is in co - NP and our co - W [2]-hardness proof implies that Rendezvous in Time is co - NP -complete in this case. However, for the general case, the problems are harder and weprove that Rendezvous and
Rendezvous in Time are
PSPACE -hard.Finally, we initiate the study of the complexity of the problems under structural parameter-ization of the input graphs. We show that
Rendezvous in Time is FPT when parameterizedby the neighborhood diversity of the input graph and τ . Related work.
The classical rendezvous game introduced by Alpern [2] is played by two agentsthat are placed in some unfamiliar area and whose task is to develop strategies that maximizethe probability that they meet. We refer to the book of Alpern and Gal [3] for detailed studyof the subject. Deterministic rendezvous problem was studied by Ta-Shma and Zwock [31].
Rendezvous is closely related to the
Cops-and-Robber game. The game was defined (forone cop) by Winkler and Nowakowski [28] and Quilliot [29] who also characterized graphs forwhich one cop can catch the robber. Aigner and Fromme [1] initiated the study of the problemwith several cops. The minimum number of cops that are required to capture the robber is calledthe cop number of a graph. This problem was studied intensively and we refer to the book ofBonato and Nowakowski [4] for further references. Kinnersley [24] established that the problemis
EXPTIME -complete. The
Cops-and-Robber game can be seen as a special case of searchgames played on graphs, surveys [5, 13] provide further references on search and pursuit-evasiongames on graphs. A related variant of
Cops-and-Robber game is the guarding game studiedin [11, 12, 27, 30]. Here the set of cops is trying to prevent the robber from entering a specifiedsubgraph in a graph.
Organization of the paper.
In Section 2, we give the basic definitions and introduce notationused throughout the paper. We also show that
Rendezvous and
Rendezvous in Time canbe solved in n O ( k ) time. In Section 3, we investigate relations between d G ( s, t ) and λ G ( s, t ).In Section 4, we give algorithmic lower bounds for Rendezvous and
Rendezvous in Time .In Section 5, we show that
Rendezvous in Time is fixed-parameter tractable (
FPT ) whenparameterized by τ and the neighborhood diversity of the input graph. We conclude in Section 6by stating some open problems. Graphs.
All graphs considered in this paper are finite undirected graphs without loops ormultiple edges, unless it is said explicitly that we consider directed graphs. We follow the stan-dard graph theoretic notation and terminology (see, e.g., [9]). For each of the graph problemsconsidered in this paper, we let n = | V ( G ) | and m = | E ( G ) | denote the number of verticesand edges, respectively, of the input graph G if it does not create confusion. For a graph G and a subset X ⊆ V ( G ) of vertices, we write G [ X ] to denote the subgraph of G induced by X .For a set of vertices S , G − S denotes the graph obtained by deleting the vertices of S , that is, G − S = G [ V ( G ) \ S ]; for a vertex v , we write G − v instead of G − { v } . For a vertex v , we denoteby N G ( v ) the (open) neighborhood of v , i.e., the set of vertices that are adjacent to v in G . Weuse N G [ v ] to denote the closed neighborhood , that is N G ( v ) ∪ { v } . For two nonadjacent vertices s and t , a set of vertices S ⊆ V ( G ) \ { s, t } is an ( s, t ) -separator if s and t are in distinct connectedcomponents of G − S . We use λ G ( s, t ) to denote the minimum size of an ( s, t )-separator of G ; λ G ( s, t ) = + ∞ if s = t or s and t are adjacent. A path is a connected graph with at leat one andmost two vertices (called end-vertices ) of degree at most one whose remaining vertices (called internal ) have degrees two. We say that a path with end-vertices u and v is an (u,v)-path .3he length of a path P , denoted by (cid:96) ( P ) , is the number of its edges. The distance dist G ( u, v )between two vertices u and v of G in the length of a shortest ( u, v )-path. We use v · · · v k todenote the path with the vertices v , . . . , v k and the edges v i − v i for i ∈ { , . . . , k } . A cycle is aconnected graph with all the vertices of degree two. The length (cid:96) ( C ) of a cycle C is the numberof edges of C .Let X and Y be multisets of vertices of a graph G (i.e., X and Y can contain several copiesof the same vertex). We say that X and Y of the same size are adjacent if there is a bijectivemapping α : X → Y such that for x ∈ X , either x = α ( x ) or x and α ( x ) are adjacent in G . Itis useful to observe the following. Observation 1.
For multisets X and Y of vertices of G , it can be decided in polynomial timewhether X and Y are adjacent.Proof. It is trivial to check whether X and Y have the same size. If this holds, we constructthe bipartite graph H with the vertex set V ∪ V , where | V | = | V | = | X | = | Y | , and the nodesof V correspond to the elements of X and the nodes of V correspond to the elements of Y . Anode of V is adjacent to a node of V if and only if the corresponding vertices of G are eitherthe same or adjacent. Then X and Y are adjacent if and only if H has a perfect matching. Theexistence of a perfect matching in a bipartite graph can be verified in polynomial time (see, e.g.,[26]) and the claim follows. Parameterized Complexity.
We obtain a number of results about the parameterized com-plexity of
Rendezvous and
Rendezvous in Time . We refer to the resent book of Cygan etal. [8] for the introduction to the area. Here we just remind that an instanse of the parame-terized version Π p of a decision problem Π is a pair ( I, k ), where I is an instance of Π and k is an integer parameter associated with I . It is said that Π p is fixed-parameter tractable ( FPT )if it can be solved in time f ( k ) | I | O (1) for a computable function f ( k ) of the parameter k . TheParameterized Complexity theory also provides tools that allow to show that a parameterizedproblem cannot be solved in FPT time (up to some reasonable complexity assumptions). Forthis, Downey and Fellows (see [10]) introduced a hierarchy of parameterized complexity classes,namely
FPT ⊆ W [1] ⊆ W [2] ⊆ · · · ⊆ X P, and the basic conjecture is that all inclusions in thehierarchy are proper. The usual way to show that it is unlikely that a parameterized problemadmit an
FPT algorithm is to show that it is W [1] or W [2]-hard using a parameterized reduction from a known hard problem in the corresponding class. The most common tool for establish-ing fine-grained complexity lower bound for parameterized problems is the Exponential TimeHypothesis ( ETH ) proposed by Impagliazzo, Paturi, and Zane [20, 21]. This is the conjecturestating that there is ε > -Satisfiability cannot be solved in O ∗ (2 εn ) time onformulas with n variables. Rendezvous Games with Adversaries.
Suppose that the game is played on a connectedgraph G , and s and t are initial positions of the agents of Facilitator. Let also k be the numberof agents of Disruptor.Notice that a placements of the agents of Facilitator is defined by a multiset of two vertices,as R and J can occupy the same vertex. We denote by F G the family of all multisets of twovertices. Similarly, a placement of k agents of Disruptor is defined by a multiset of k vertices,because several agents can occupy the same vertex. Let D kG be the family of all multisets of k vertices. We say that F ∈ F G and D ∈ D kG are compatible if F ∩ D = ∅ . Notice that the numberof pairs of compatible F ∈ F G and D ∈ D kG is n (cid:0) n + k − k (cid:1) + (cid:0) n (cid:1)(cid:0) n + k − k (cid:1) . We denote by P kG = { ( F, D ) | F ∈ F G , D ∈ D kG s.t. F and D are compatible } the set of positions in the game. 4ormally, a strategy of Facilitator for Rendezvous is a function f : P kG → F G that maps( F, D ) ∈ P kG to F (cid:48) ∈ F G such that F and F (cid:48) are adjacent and F (cid:48) is compatible with D . Inwords, given a position ( F, D ), Facilitator moves R and J from F to F (cid:48) if this is his turn tomove. Similarly, a strategy of Disruptor is a function d : P kG → D kG that maps ( F, D ) ∈ P kG to D (cid:48) ∈ F kG such that D and D (cid:48) are adjacent and D (cid:48) is compatible with F , that is, Disruptor moveshis agents from D to D (cid:48) if this is his turn to move. To accommodate the initial placement, weextend the definition of d for the pair ( { s, t } , ∅ ) and let d (( { s, t } , ∅ ) = D (cid:48) , where D (cid:48) ∈ D kG iscompatible with { s, t } .The definitions of strategies for Rendezvous in Time are more complicated, because thedecisions of the players also depend on the number of the current step. A strategy of Facilitatorfor
Rendezvous is a family of functions f i : P kG → F G for i ∈ { , . . . , τ } such that f i maps( F, D ) ∈ P kG to F (cid:48) ∈ F G , where F and F (cid:48) are adjacent and F (cid:48) is compatible with D . Facilitatoruses f i for the move in the i -th step of the game. A strategy of Disruptor is a family of functions d i : P kG → D kG for i ∈ { , . . . , τ − } such that for i ∈ { , . . . , τ − } , d i maps ( F, D ) ∈ P kG to D (cid:48) ∈ F kG , where D and D (cid:48) are adjacent and D (cid:48) is compatible with F , and d maps ( { s, t } , ∅ ) to D (cid:48) ∈ D kG compatible with { s, t } (slightly abusing notation we do not define d for the elementsof P kG ).In the majority of the proofs in our paper, we rather explain the strategies of the playersin an informal way, to avoid defining functions for all elements of P kG , because the majority ofpositions never occur in the game. However, the above notation is useful in some cases.As it is common for various games on graphs (see, e.g., the book of Bonato and Nowakowski [4]about Cops-and-Robber games), our Rendezvous Game with Adversaries can be resolved bybacktracking. As the approach is standard, we only briefly sketch the proof of the followingtheorem.
Theorem 1.
Rendezvous and
Rendezvous in Time can be solved in n O ( k ) time.Proof. Let G be a connected graph on which the game is played and let s, t ∈ V ( G ). Let also k be a positive integer denoting the number of agents of Disruptor.We define the game graph G (also the name arena could be found in the literature) as thedirected graph, whose nodes correspond to positions and turns to move. We denote the nodesof G by v ( h ) F,D for (
F, D ) ∈ P kG , and h ∈ { , } ; if h = 1, then Facilitator makes a move, and if h = 2, then this is Disruptor’s turn. For h ∈ { , } , we set V h = { v ( h ) F,D | ( F, D ) ∈ P kG } . Forevery two nodes v (1) F,D ∈ V and v (2) F (cid:48) ,D (cid:48) ∈ V , we construct arcs as follows. We construct the arc( v (1) F,D , v (2) F (cid:48) ,D (cid:48) ) if D = D (cid:48) , and F and F (cid:48) are adjacent. Symmetrically, we construct ( v (2) F (cid:48) ,D (cid:48) , v (1) F,D )if F = F (cid:48) , and D (cid:48) and D are adjacent. We denote by A the set of arcs of G .Observe, that G can be constructed in n O ( k ) time. The number of nodes is 2 · ( n (cid:0) n + k − k (cid:1) + (cid:0) n (cid:1)(cid:0) n + k − k (cid:1) ) = n O (1) . Given two nodes v (1) F,D and v (2) F (cid:48) ,D (cid:48) , the arcs between these nodes can beconstructed in polynomial time by Observation 1. Hence, the construction of the arc set can bedone in time k O ( k ) · n O ( k ) · n O ( k ) = n O ( k ) .Let (cid:96) ≥ W (cid:96) ⊆ V of winning positions for Facilitator inat most (cid:96) moves. A node v (1) F,D is in W (cid:96) if Facilitator can win on G in at most (cid:96) moves providedthat R and J are placed in F and the agents of Disruptor are occupying D . We explain how toconstruct W (cid:96) for (cid:96) = 0 , , . . . by dynamic programming.It is straightforward to verify that v (1) F,D ∈ W if and only if F = { x, x } for x ∈ V ( G ).For (cid:96) ≥ W (cid:96) = W (cid:96) − ∪ U , (1)where U = { v (1) F,D | there is ( v (1) F,D , v (2) F (cid:48) ,D ) ∈ A s.t. for every ( v (2) F (cid:48) ,D , v (1) F (cid:48) ,D (cid:48) ) ∈ A , v (1) F (cid:48) ,D (cid:48) ∈ W (cid:96) − } . (2)5nformally, v (1) F,D ∈ U if there is a move of Facilitator such that for every response of Disruptor,the obtained position is in W (cid:96) − , that is, Facilitator can win in at most (cid:96) − W (cid:96) using (1) and (2) is proved by completely standard argu-ments and we leave this to the reader. Notice that given W (cid:96) − , (1) and (2) allow to compute W (cid:96) in n O (1) time.We compute the sets W (cid:96) consecutively starting from (cid:96) = 0 until we obtain W (cid:96) = W (cid:96) − forsome (cid:96) ≥
1. Observe that if W (cid:96) = W (cid:96) − , then W (cid:96) (cid:48) = W (cid:96) for every (cid:96) (cid:48) ≥ (cid:96) . Notice also that westop after at most |V | iterations, because W (cid:96) ⊆ V . This implies that all the sets W (cid:96) can beconstructed in n O (1) time. Let W (cid:96) ∗ be the last constructed set.To solve Rendezvous for an instance (
G, s, t, k ), it is sufficient to observe that (
G, s, t, k )is a yes-instance if and only if v (1) F,D ∈ W (cid:96) ∗ for F = { s, t } and every D ∈ D kG that is compatiblewith F , that is, Facilitator can win starting from s and t for every choice the initial placementof the k agents of Disruptor.Solving Rendezvous in Time is slightly more complicated, because the parameter τ isexpected to be encoded in binary. Let ( G, s, t, k, τ ) be an instance of
Rendezvous in Time .If τ ≥ (cid:96) ∗ , we observe that ( G, s, t, k, τ ) is a yes-instance of
Rendezvous in Time if and only if(
G, s, t, k ) is a yes-instance of
Rendezvous . If τ < (cid:96) ∗ , then recall that we already constructedthe set W τ . Then ( G, s, t, k, τ ) is a yes-instance if and only if v (1) F,D ∈ W τ for F = { s, t } andevery D ∈ D kG that is compatible with F .Summarizing the running time of all the steps, we obtain that Rendezvous and
Ren-dezvous in Time can be solved in n O ( k ) time.We conclude this section by the observation that a strategy of Disruptor in RendezvousGames with Adversaries for τ steps, that is, for Rendezvous in Time , can be representedas a rooted tree of height τ . Suppose that the game is played on a graph G , and s and t areinitial positions of the agents of Facilitator. Let also k be the number of agents of Disruptor.Suppose that Disruptor has a strategy defined by the family of functions d i : P kG → D kG for i ∈ { , . . . , τ − } . We define the tree T kG ( τ ) such that every node v of T kG ( τ ) is associated witha position P v ∈ P kG inductively starting from the root: • P r = ( { s, t } , d ( { s, t } , ∅ )) is associated with the root r of T kG ( τ ). • for every node v ∈ V ( T kG ( τ )) with P v = ( F, D ) at distance i ≤ τ − u of v for every ( F (cid:48) , D (cid:48) ) ∈ P kG such that (i) F (cid:48) is adjacent to F andcompatible with D , and (ii) D (cid:48) = d i ( F (cid:48) , D ), and associate u with P u = ( F (cid:48) , D (cid:48) ).In words, the children of every node correspond to all possible moves of Facilitator from theposition ( D, F ) and are the positions obtained by the responses of Disruptor. Observe thatevery node has at most |F | = (cid:0) n +12 (cid:1) children. Therefore, the total number of nodes of T kG ( τ ) isat most (cid:0) n +12 (cid:1) τ +1 . It is straightforward to make the following observation. Observation 2.
A strategy { d i | ≤ i ≤ τ − } is a winning strategy for Disruptor with k agents in Rendezvous Games with Adversaries for τ steps on G against Facilitator starting from s and t if and only if F is a set of two distinct vertices for every P v = ( F, D ) for v ∈ V ( T kG ( τ )) . In particular, this allows to observe the following.
Observation 3.
Rendezvous in Time is in co - NP if τ is a fixed constant.Proof. If (
G, s, t, k, τ ) is a no-instance of
Rendezvous in Time , then Disruptor has a winningstrategy that allows to prevent R and J from meeting in at most τ steps. Then the tree T kG ( τ )can be used as a certificate. Since the tree has n O ( τ ) nodes, we can check whether a given treeencodes a winning strategy in polynomial time using Observations 2 and 1.6 Disruption number and separators
In this section we investigate relations between d G ( s, t ) and λ G ( s, t ). Given a connected graph G and two vertices s and t , it is straightforward to see that d G ( s, t ) ≤ λ G ( s, t ). Indeed, if S ⊆ V ( G ) \ { s, t } is an ( s, t )-separator of size k = λ G ( s, t ), then Disruptor with k agents canput then in the vertices of S in the beginning of the game. Then he can use the trivial strategythat keeps the agents D , . . . , D k in their positions. However, d G ( s, t ) and λ G ( s, t ) can be farapart. Still, d G ( s, t ) = 1 if and only if λ G ( s, t ) = 1, and this is the first result of the section. Theorem 2.
Let G be a connected graph and let s, t ∈ V ( G ) . Then d G ( s, t ) = 1 if and only if λ G ( s, t ) = 1 .Proof. As we already observed, d G ( s, t ) ≤ λ G ( s, t ). Hence, if λ G ( s, t ) = 1, then d G ( s, t ) = 1.This means that it is sufficient to show that if d G ( s, t ) = 1, then λ G ( s, t ) = 1. We prove thisby contradiction. Assume that λ G ( s, t ) ≥
2. We show that Facilitator has a winning strategywhen starting from s and t on G against Disruptor with one agent.Let C be a cycle in G . For every two distinct vertices u and v of C , C has two internallyvertex disjoint ( u, v )-paths P and P in C . We say that C has a ( u, v ) -shortcut if there is a( u, v )-path P in G − ( V ( C ) \ { u, v } ) that is shorter than P and P . That is, (cid:96) ( P ) < (cid:96) ( P )and (cid:96) ( P ) < (cid:96) ( P ). We say that C has a shortcut if there are distinct u, v ∈ V ( C ) that have a( u, v )-shortcut.We claim the following. Claim 1. If R and J occupy vertices of a cycle C of G that has a shortcut, then Facilitator hasa strategy such that in at most (cid:96) ( C ) steps R and J are moved into vertices of a cycle C (cid:48) with (cid:96) ( C (cid:48) ) < (cid:96) ( C ) .Proof of Claim 1. Suppose that R and J occupy vertices x and y of C , respectively. Assume thata path P is a ( u, v )-shortcut for some distinct u, v ∈ V ( G ). Denote by P and P , respectively,the internally vertex disjoint ( u, v )-paths in C . Let C be the cycle of G composed by P and P , and let C be the cycle composed by P and P . Because P is a shortcut for C , we have that (cid:96) ( C ) < (cid:96) ( C ) and (cid:96) ( C ) < (cid:96) ( C ). If x, y ∈ V ( P ), then x, y ∈ V ( C ) and the claim holds trivially,since R and J are already on cycle C with (cid:96) ( C ) < (cid:96) ( C ). Symmetrically, if x, y ∈ V ( P ), thenthe claim holds. Assume that this is not the case. Then x and y are internal vertices of P and P belonging to distinct paths. We assume without loss of generality that x is an internalvertex of P and y is an internal vertex of P .Facilitator uses the following strategy. In each step, R is moved along P toward u , unlessthe next vertex is occupied by D . In the last case, R stays in the current position. Similarly, J moves toward v in P whenever this is possible and stays in the current position if the way isblocked. Notice that, since the unique agent D of Disruptor occupies a unique vertex in eachstep, at least one of the agents R or J moves to an adjacent vertex. Therefore, either R reaches u or J reaches v in at most (cid:96) ( C ) steps. If R is in u , then R and J are in the vertices of C and (cid:96) ( C ) < (cid:96) ( C ). Symmetrically, if J reaches v , then R and J reach C with (cid:96) ( C ) < (cid:96) ( C ).Next, we show that Facilitator can win if R and J are in a cycle without shortcuts and D is in the same cycle. Claim 2. If R and J occupy vertices of a cycle C of G without a shortcut, and the unique agent D of Disruptor is in a vertex of C as well, then Facilitator has a winning strategy with at most (cid:96) ( C ) / steps.Proof of Claim 2. Suppose that R and J occupy vertices x and y of C , respectively, and that D occupies z ∈ V ( C ). Denote by P the unique ( x, y )-path in C − z . Facilitator uses the following7trategy. In every step, R and J move towards each other along P except if they appear tooccupy adjacent vertices. In the last case, R stays and J moves to the vertex occupied by R .We show that this strategy is a feasible winning strategy.The proof is by induction on the length of P . The claim is trivial when (cid:96) ( P ) ≤
2. Assumethat (cid:96) ( P ) ≥ x (cid:48) , y (cid:48) and z (cid:48) of R , J and D , respectively, ifthe length of the ( x (cid:48) , y (cid:48) )-path in C − z (cid:48) is at most (cid:96) ( P ) − R and J move to the neighbors x (cid:48) and y (cid:48) of x and y , respectively, in P . If D moves to a vertex z (cid:48) ∈ V ( C ), then we apply the inductive assumption and, since the lengthof the ( x (cid:48) , y (cid:48) )-subpath P (cid:48) is (cid:96) ( P ) − z (cid:48) / ∈ V ( P (cid:48) ), obtain that the strategy of Facilitator iswinning. Assume that by the first move Disruptor removes D from C . If D does not return toa vertex of C in (cid:96) ( P ) / h ≤ (cid:96) ( P ) /
2, at the h -th move, D steps back on a vertex z (cid:48) ∈ V ( C ). ≤ h zz (cid:48) x yx (cid:48)(cid:48) y (cid:48)(cid:48) hh Figure 1: The position placement after h steps (up to symmetry).By the assumption, cycle C has no shortcuts. In particular, there is no ( z, z (cid:48) )-shortcut. Thisimplies, that the length of one of the two ( z, z (cid:48) )-paths in C is at most h . Observe that in h steps, R and J reach vertices x (cid:48)(cid:48) and y (cid:48)(cid:48) that are at the distance h in P from x and y , respectively.Therefore (see Figure 1), the ( x (cid:48)(cid:48) , y (cid:48)(cid:48) )-subpath P (cid:48)(cid:48) of P does not contain z (cid:48) . Since (cid:96) ( P (cid:48)(cid:48) ) < (cid:96) ( P ),we can apply the inductive assumption. This proves that the Facilitator’s strategy is a feasiblewinning strategy and the claim holds.Notice that the total number of steps is (cid:100) (cid:96) ( P ) / (cid:101) ≤ (cid:96) ( C ) /
2. This completes the proof.Now we are ready to complete the proof of the theorem. If s = t or s and t are adjacent,then Facilitator has a straightforward winning strategy. Assume that s and t are distinct andnonadjacent. Since λ G ( s, t ) ≥
2, by Menger’s theorem (see, e.g., [9]), there are two internallyvertex disjoint ( s, t )-paths P and P . The union of these two paths forms cycle C . If the agent D of Disruptor occupies a vertex of C (cid:48) , then Facilitator has a winning strategy by Claim 2. If D is outside C (cid:48) , then Facilitator moves R and J along C (cid:48) towards each other. Then either R and J meet or D steps on C (cid:48) at some moment. In this case, Facilitator switches to the strategyfrom Claim 2 that guarantees him to win.We observed that d G ( s, t ) ≤ λ G ( s, t ) and, by Theorem 2, d G ( s, t ) = 1 if and only if λ G ( s, t ) =1. However, if d G ( s, t ) ≥
2, then the difference betweem λ G ( s, t ) and d G ( s, t ) may be arbitrary.To see this, consider the following example.Let p ≥ • Construct a set U = { u , . . . , u p } of pairwise adjacent vertices. • Add a vertex s and join s with each vertex u i ∈ U by a path sx i u i . • Add a vertex t and join t with each vertex u i ∈ U by a path ty i u i .8enote the obtained graph by G (see the left part of Figure 2). Observe that λ G ( s, t ) = p .We show that d G ( s, t ) = 2 by demonstrating a winning strategy for Disruptor with two agents D and D . Initially, D and D are placed in arbitrary vertices of the clique U . Then D “shadows” R and D “shadows” J in U in the following sense. If R moves to x i for some i ∈ { , . . . , p } , Disruptor responds by moving D to u i . Symmetrically, if R moves to y j forsome j ∈ { , . . . , p } , then D is moved to u j . It is easy to verify that if Disruptor follows thisstrategy, then neither R no J can enter U . Therefore, Disruptor wins. Since p can be arbitrary,we have that λ G ( s, t ) − d G ( s, t ) = p − P (cid:48) s tu u x y y x u u s tu = wG HP P P (cid:48) Figure 2: The construction of G and H for p = 4.The family of graphs G for p ≥ G contains a clique with p vertices. However, exploiting the same idea as for G , we can showthat there are sparse graphs with the same property. For this, we considered the following morecomplicated example.Let p ≥ • Construct a path P = u · · · u p on p vertices. • Add a vertex s and join s with each vertex u i ∈ V ( P ) by an ( s, u i )-path P i of length h = (cid:98) p/ (cid:99) + 1. • Add a vertex t and join t with each vertex u i ∈ V ( P ) by an ( t, u i )-path P (cid:48) j of length h = (cid:98) p/ (cid:99) + 1.Denote the obtained graph by H (see the right part of Figure 2). Clearly, λ H ( s, t ) = p . Weclaim that d H ( s, t ) = p . The idea behind the winning strategy for Disruptor with two agents D and D is similar to the one from the first example: D “shadows” R and D “shadows” J on P . Let w = u (cid:98) p/ (cid:99) . Initially, D and D are placed in w . Then D is moved as follows. If R moves to/stays in s , then D moves to/stays in w . If R is moved into an internal vertex x of P i for some i ∈ { , . . . , p } , then Disruptor responds my moving D toward u i or keeping D in thecurrent position maintaining the following condition: D is in a vertex u j at minimum distancefrom w such that the distance between x and u i in P i is more than the distance between u j and u i in P . The construction of the strategy for D is symmetric. It is easy to see that thedescribed strategy for Disruptor is feasible, and the strategy allows neither R no J to enter avertex of P . Therefore, d H ( s, t ) = 2.Notice that the graph H for each p ≥ H is at most 3 (we refer to [8, 9] for the formal treewidth definition), that is, graphs H are,indeed, sparse.Our examples indicate that λ G ( s, t ) may differ from d G ( s, t ) if G has sufficiently long inducedpaths and cycles. We observe that λ G ( s, t ) = d G ( s, t ) if G belongs to graph classes that haveno graphs of this type.A graph G is P -free if G has no induced subgraph isomorphic to the path with 5 vertices.9 roposition 1. If G is a connected P -free graph, then for every s, t ∈ V ( G ) , d G ( s, t ) = λ G ( s, t ) .Proof. Let G be a P -free graph and let s, t ∈ V ( G ). The statement is trivial if s = t or s and t are adjacent. Assume that s and t are distinct nonadjacent vertices. Since d G ( s, t ) ≤ λ G ( s, t ),it is sufficient to show the opposite inequality. We prove that if Disruptor with k agents has awinning strategy on G against Facilitator starting from s and t , then λ G ( s, t ) ≥ k . Let P bean induced ( s, t )-path. Since G is P -free, (cid:96) ( P ) ≤
3. Suppose that (cid:96) ( P ) = 2, that is P = sxt for some x ∈ V ( G ). Then Disruptor has to place an agent in x in the beginning of the game.Otherwise, R and J move to x in the first step and Facilitator wins. Suppose that (cid:96) ( P ) = 3,that is, P = sxyt for some x, y ∈ V ( G ). Observe that Disruptor has to place an agent in x or y in the beginning of the game. Otherwise, R moved to x , J moved to y , and Facilitator winsin the next step, as x and y are adjacent. Let S be the set of vertices containing the agents ofDisruptor in the beginning of the game. Clearly, k ≥ | S | . We have that every induced ( s, t )-path P contains a vertex of S . This means that S is an ( s, t )-separator. Therefore, k ≥ | S | ≥ λ G ( s, t )and the claim follows.A graph G is chordal if G does not contain induced cycles on at least 4 vertices, that is, if C is a cycle in G of length at least 4, then there is a chord , i.e., an edge of G with end-vertices intwo nonconsecutive vertices of C . We need some properties of chordal graphs (we refer to [6, 19]for the detailed introduction).It follows from the results of Gavril [17] that a graph G is chordal if and only of it has atree decomposition with every bag being a clique. Formally, G is a chordal graph if and only ifthere is a pair T = ( T, { X i } i ∈ V ( T ) ), where T is a tree whose every node i is assigned a vertexsubset X i ⊆ V ( G ), called a bag , such that X i is a clique and the following holds:(i) (cid:83) i ∈ V ( T ) X i = V ( G ),(ii) for every uv ∈ E ( G ), there exists a node i of T such that u, v ∈ X i , and(iii) for every u ∈ V ( G ), the set T u = { i ∈ V ( T ) | u ∈ X i } , i.e., the set of nodes whosecorresponding bags contain u , induces a connected subtree of T .We use the following well-known property of tree decomposition (see, e.g., [8, 9]). Assume that G is connected. Let x, y ∈ V ( T ) be adjacent nodes of T with S = X x ∩ X y , and let T and T be the connected components of T − xy . Then for every u ∈ (cid:0) (cid:83) i ∈ V ( T ) X i (cid:1) \ S and every v ∈ (cid:0) (cid:83) i ∈ V ( T ) X i (cid:1) \ S , S is a ( u, v )-separator. Proposition 2. If G is a connected chordal graph, then for every s, t ∈ V ( G ) , d G ( s, t ) = λ G ( s, t ) .Proof. Let G be a chordal graph and let s, t ∈ V ( G ). As before, let us notice that the propositionis trivial if s = t or s and t are adjacent, and we assume that s and t are distinct nonadjacentvertices. Recall also that it is sufficient to show that λ G ( k ) ≤ d G ( s, t ). We prove that Facilitatorhas a winning strategy against Disruptor with k agents if k < λ G ( s, t ). For this, we show that R can reach t occupied by J .Since G is a chordal graph, there is a tree decomposition T = ( T, { X i } t ∈ V ( T ) ) of G suchthat every bag X i is a clique. Let i, j ∈ V ( T ) be nodes of T such that s ∈ X i , t ∈ X j and the( i, j )-path P in T has minimum length. Since s and t are nonadjacent, i (cid:54) = j . Let P = i · · · i r ,where i = i and j = s r , and let S h = X i h − ∩ X i h for h ∈ { , . . . , r } . By the choice of P , s ∈ X i \ X i h for h ∈ { , . . . , r } and t / ∈ X j \ X i h for h ∈ { , . . . , r − } . By the properties oftree decompositions, we obtain that S , . . . , S r are ( s, t )-separators. Since λ G ( s, t ) > k , we havethat | S h | > k for every h ∈ { , . . . , r } . 10e describe the strategy of Facilitator, where R is moved from s to t via vertices of S , . . . , S r .Since | S | > k , there is a vertex v ∈ S that is not occupied by the agents of Disruptor. By thefirst move, Facilitator moves R from s to v . Assume now that R is in a vertex v ∈ S h for some h ∈ { , . . . , r } . If h = r , then R is moved to t . Otherwise, if h < r , then since | S h +1 | > k , thereis v (cid:48) ∈ S h +1 that is not occupied by the agents of Disruptor. Then Facilitator either keeps R in v if v (cid:48) = v or moves R from v to v (cid:48) otherwise. Note that v and v (cid:48) that are adjacent in thelast case, because S h , S h +1 ⊆ X i h +1 . Then we proceed from v (cid:48) . It follows that R reaches t in r steps. This completes the proof.The chordality of a graph is biggest smallest size of a induced cycle in it. Clearly, chordalgraphs are the graphs of chordality three. It is natural to ask whether for graphs of biggerchordality the difference betweem λ G ( s, t ) and d G ( s, t ) may be arbitrary. In the example wegave after the Proof of Claim 2, we have seen that, for the graph G (depicted in the left partof Figure 2), it holds that λ G ( s, t ) − d G ( s, t ) = p − G haschordality five. Notice that this graph G can be further enhanced so to obtain chordality four:just add a clique between the vertices in { x , . . . , x p } and a a clique between the vertices in { y , . . . , y p } . This indicates a sharp transition of d G away from λ G when graphs are not chordalany more.Since λ G ( s, t ) can be computed in polynomial time by the standard maximum flow algorithms(see, e.g., the recent textbook [32]), we obtain the following corollary. Corollary 1.
Rendezvous can be solved in polynomial time on the classes of P -free andchordal graphs. In this section, we discuss algorithmic lower bounds for
Rendezvous and
Rendezvous inTime .We proved in Theorem 1 that
Rendezvous and
Rendezvous in Time can be solved in n O ( k ) . We show that it is unlikely that the dependance on k can be improved. For this, weshow that both problems are co - W [2]-hard (i.e, it is W [2]-had to decide whether the input is ano-instance; in fact, we show that it is W [2]-hard to decide whether d G ( s, t ) ≤ k ) and, therefore,cannot be solved in time f ( k ) · n O (1) for any computable function f ( k ), unless FPT = W [2]; theresult for Rendezvous in Time holds even if τ is a constant. Our proof also implies that bothproblems cannot be solved in time f ( k ) · n o ( k ) unless ETH fails.Observe that
Rendezvous in Time can be solved in polynomial time if τ = 1, because ofthe following straightforward observation. Observation 4.
Facilitator can with in Rendezvous Game with Adversaries game in one stepon G starting from s and t against Disruptor with k agents if and only if one of the followingholds: (i) s = t , (ii) s and t are adjacent, or (iii) | N G ( s ) ∩ N G ( t ) | > k . However, if τ ≥ Rendezvous in Time becomes hard.
Theorem 3.
Rendezvous and
Rendezvous in Time for any constant τ ≥ are co - W [2] -hard when parameterized by k . Moreover, the problems cannot be solved in time f ( k ) · n o ( k ) unless ETH fails.Proof.
We show theorem by reducing the
Set Cover problem. Given a universe U , a family S of subsets of U , and a positive integer k , the task of Set Cover is to decide whether thereis a subfamily S (cid:48) ⊆ S of size at most k that covers U , that is, every element of U is in at leastone set of S (cid:48) . This problem is well-known to be W [2]-complete when parameterized by k (see,e.g., [8]). 11 (cid:48) k s z tUS (1) S ( k ) w w k x x n x (cid:48) x (cid:48) n y y k y (cid:48) Figure 3: The construction of G .Let ( U, S , k ) be an instance of Set Cover . Let U = { u , . . . , u n } and S = { S , . . . , S m } .We construct the graph G as follows (see Figure 3). • Construct a set of n vertices U = { u , . . . , u n } corresponding to the universe. • For every i ∈ { , . . . , k } , construct a set of m vertices S ( i ) = { s ( i )1 , . . . , s ( i ) m } ; each S ( i ) corresponds to a copy of S . • For every i ∈ { , . . . , k } , h ∈ { , . . . , m } and h ∈ { , . . . , n } , make s ( i ) j and u h adjacent ifthe element of the universe u h is in S j ∈ S . • For every i ∈ { , . . . , k } , construct a vertex w i and make it adjacent to s ( i )1 , . . . , s ( i ) m . • Construct two vertices s and t . • For every h ∈ { , . . . , n } , join s and u h by a path sx h u h and joint u h and t by a path u h x (cid:48) h t . • For every i ∈ { , . . . , k } , join s and w i by a path sy i w i and join w i and t by a path w i y (cid:48) i t . • Construct a vertex z and make it adjacent to s and t .We show that if ( U, S , k ) is a yes-instance of Set Cover , then Disruptor with k + 1 agentscan win in Rendezvous Game with Adversaries. Let S (cid:48) = { S i , . . . , S i k } be a set cover; weassume without loss of generality that S (cid:48) has size exactly k . We describe a winning strategyfor Disruptor with the agents D , . . . , D k +1 . Initially, Disruptor puts D j in the vertex s ( j ) i j foreach j ∈ { , . . . , k } , and D k +1 is placed in z . Then the following strategy is used. The agents D , . . . , D k +1 are keeping their position until either R or J are moved from s or t , respectively.Assume by symmetry that R is moved by Facilitator from s ( J can either move or stay in t ). If R is moved from s to y j for some j ∈ { , . . . , k } , then Disruptor moves D j from s ( j ) i j to w j and D k +1 is moved from z to s . Notice that R is in the vertex y j of degree two and both neighborsof y j are occupied by the agents of Disruptor. Hence, R cannot move and J cannot reach y j .This implies that Disruptor wins by keeping the agents in their current positions. Assume that12 is moved from s to x h for some h ∈ { , . . . , n } . Since S (cid:48) is a set cover, there is j ∈ { , . . . , k } such that the element of the universe u h ∈ S i j . Then D j is in the vertex s ( j ) i j that is adjacent tothe vertex u h . Disruptor responds by moving D j from s ( j ) i j to u h and D k +1 is moved from z to s . Now we have that R is blocked in x h by the agents in s and u h . This means that Disruptorwins.Next, we claim that if ( U, S , k ) is a no-instance of Set Cover , then Facilitator wins in atmost two steps against Disruptor with k +1 agents. Assume that Disruptor completed the initialplacement of the agents. If z is not occupied, then Facilitator moves R and J to z and wins inone step. Assume that z is occupied by D k +1 . If there is i ∈ { , . . . , k } such that there is noagent of Disruptor in a vertex of N G [ w i ], then Facilitator moves R to y i and J to y (cid:48) i by the firstmove. Since Disruptor has no agents in N G [ w i ], for any his move, w i remains unoccupied by hisagents. Therefore, Facilitator can move R and J to w i and win in two steps. Suppose from nowthat for every i ∈ { , . . . , k } , D i is in N G [ w i ]. Because Disruptor has k + 1 agents, this meansthat for every h ∈ { , . . . , h } , x h , u h and x (cid:48) h are not occupied by the agents of Disruptor, andfor every i ∈ { , . . . , k } , at most one agent is in S ( i ) . Let X be the set of vertices of (cid:83) ki =1 S ( i ) occupied by the agents of Disruptor in the beginning of the game. Since | X | ≤ k and ( U, S , k )is a no-instance of Set Cover , there is h ∈ { , (cid:48) . . . , n } such that the vertices N H [ u h ] are notoccupied by the agents if Disruptor. Therefore, Facilitator can move R from s to x h and thento u h and, symmetrically, move J from t to x (cid:48) h and then to u h . We obtain that R and J meetin u h in two steps, that is, Facilitator wins in two steps.These arguments imply that ( U, S , k ) is a yes-instance of Set Cover if and only if (
G, s, t, k +1) is a no-instance of
Rendezvous . This mens that
Rendezvous is co - W [2]-hard. For Ren-dezvous in Time for τ ≥
2, notice that if ( U, S , k ) is a no-instance of Set Cover , thenFacilitator can win in at most two steps against Disruptor with k + 1 agents, and if ( U, S , k )is a yes-instance, then Disruptor has a strategy that prevents Facilitator from winning in anynumber of steps. It follows that Rendezvous in Time is co - W [2]-hard for any fixed τ ≥ Set Cover cannot be solved in time f ( k ) · ( n + m ) o ( k ) unless ETH fails. To show co - W [2]-hardness of Rendezvous and
Rendezvousin Time , we constructed a polynomial reduction and the obtained parameter for
Rendezvous and
Rendezvous in Time is k + 1, i.e., is linear in the input parameter for Set Cover . Thus,any algorithm for
Rendezvous or Rendezvous in Time with running time f ( k ) · n o ( k ) wouldimply the algorithm for Set Cover with running time f ( k ) · ( n + m ) o ( k ) .We proved Theorem 3 by giving a polynomial reduction from Set Cover . Since
Set Cover is NP -complete (see [16]), we obtain the following corollary using Observation 3. Corollary 2.
Rendezvous in Time is co - NP -complete for every fixed constant τ ≥ . Using the reduction from the proof of Theorem 3, we can conclude that
Rendezvous and
Rendezvous in Time are co - NP -hard. However, the problem is harder. Theorem 4.
Rendezvous and
Rendezvous in Time are
PSPACE -hard.Proof.
We show that
Rendezvous in Time is PSPACE -hard and then explain how to modifyour reduction for
Rendezvous . We prove that it is
PSPACE -hard to decide whether Disruptorcan win in at most τ steps in Rendezvous Game with Adversaries.The reduction is from the Quantified Boolean Formula in Conjunctive NormalForm (QBF) problem with alternating quantifiers that is well-known to be
PSPACE -complete(see, e.g., [16]). The task of QBF is, given 2 n Boolean variables x , . . . , x n and m clauses13 , . . . , C m , where every C i is a disjunction of literals over the variables, to decide whether theformula ϕ = ∀ x ∃ x . . . ∀ x n − ∃ x n [ C ∧ . . . ∧ C m ]evaluates true . w (cid:48) s tx x (cid:48) x (cid:48)(cid:48) z z (cid:48) x (cid:48)(cid:48) x (cid:48)(cid:48) x x x (cid:48) x (cid:48) x x x x x x (cid:48)(cid:48) x (cid:48)(cid:48) x (cid:48)(cid:48) x (cid:48)(cid:48) y y y y y (cid:48) y (cid:48) y (cid:48) y (cid:48) c c x (cid:48) P P P P P P x (cid:48)(cid:48) P P Q Q Q Q R R R R u u u v v v v v w w w (cid:48) Figure 4: The construction of G for ϕ = ∀ x ∃ x ∀ x ∃ x [( x ∨ x ∨ x ) ∧ ( x ∨ x ∨ x )].Given a formula ϕ = ∀ x ∃ x . . . ∀ x n − ∃ x n [ C ∧ . . . ∧ C m ], we construct the graph G asfollows (see Figure 4). • Construct m vertices c , . . . , c m corresponding to the clauses of ϕ . • Construct vertices s , u , . . . , u n , and x i − , x i − for i ∈ { , . . . , n } , and for each i ∈{ , . . . , n } , make x i − , x i − adjacent to u i − and u i . • Make s and u adjacent, and join u n with c , . . . , c m by paths u n w j c j for j ∈ { , . . . , m } . • Construct 2 n + 1 vertices v , . . . , v n and construct the path tv · · · v n . Then join v n with c , . . . , c m by paths v n w (cid:48) j c j for j ∈ { , . . . , m } . • Construct two vertices z and z (cid:48) , and make them adjacent to s and t . • For every i ∈ { , . . . , n } , construct vertices x i , x i , x (cid:48)(cid:48) i , x (cid:48)(cid:48) i and y i , y (cid:48) i , and then make y i adjacent to x i , x i and make y (cid:48) i adjacent to s and t . • For every i ∈ { , . . . , n } , – construct ( x i − , x (cid:48)(cid:48) i − ) and ( x i − , x (cid:48)(cid:48) i − )-paths P i − and P i − , respectively, oflength 2( n − i ) + 1, 14 construct ( x i − , v i − ) and ( x i − , v i − )-paths Q i − and Q i − , respectively, oflength 4( n − i ) + 5, – construct a ( y i − , y (cid:48) i − )-path R i − of length 2 i − • For every i ∈ { , . . . , n } , – construct ( x i , x (cid:48)(cid:48) i ) and ( x i , x (cid:48)(cid:48) i )-paths P i and P i , respectively, of length 2( n − i )+1, – construct a ( y i , y (cid:48) i )-path R i of length 2 i − G . We define τ = 2 n + 3 and k = 2 n + 2.We claim that ϕ evaluates true if and only if Disruptor with k agents has a winning strategythat prevents R and J from meeting in at most τ steps.Assume that ϕ = true . We describe a winning strategy for Disruptor.Assume that after the i -th move of Facilitator R and J are occupying some vertices a and b and the agents of Disruptor are in the vertices of a set X . If dist G − X ( a, b ) > τ − i , thenDisruptor wins by keeping the agents in their current positions, because R and J are unable tomeet in the remaining τ − i moves. In this case, we say that Disruptor has a trivial winningstrategy .Disruptor has k = 2 n + 2 agents. We place D i in y (cid:48) i for i ∈ { , . . . , n } . The remainingtwo agents D n +1 and D n +2 are placed in z and z (cid:48) , respectively. For i ≥
1, we use X i todenote the set of vertices occupied by the agents of Disruptor after the i -th step of the game; X = { y (cid:48) , . . . , y (cid:48) n } ∪ { z, z (cid:48) } .Observe that dist G − X ( s, t ) = 2 τ . This means that if either R or J is not moved, thenDisruptor wins by the trivial winning strategy. We assume that this is not the case and R ismoved to u and J is moved to v . Disruptor responds by moving D n +1 and D n from z and z (cid:48) to s and t , respectively; these agents remain in s and t until the end of the game and theonly role of them is to prevent R and J from entering these vertices. The agents D , . . . , D n are moved to the neighbors of y (cid:48) , . . . , y (cid:48) n in the paths R , . . . , R n , respectively.The general idea of the reduction is that by the further 2 n steps, the players define thevalues of the Boolean variables x , . . . , x n , and the values of the variables x , x , . . . , x n − arechosen by Facilitator, and Disruptor chooses the values of x , x , . . . , x n . Disruptor chooses thevalue of his variables to achieve ψ = C ∧ . . . ∧ C m = true . To describe this process, we showinductively for i = 0 , . . . , n that after the 2 i + 1-th step of the game, either Disruptor wins bythe trivial strategy or the following configuration is maintained. • The values of the variables x j for j ≤ i are chosen and the values of the variables x j for j > i are unassigned. Moreover, the values of x , . . . , x i are chosen in such a way that ϕ evaluates true if the values of x , . . . , x i are constrained by the choice. • R is in u i and J is in v i . • For j ∈ { i + 1 , . . . , n } , D j − and D j are on the paths R j − and R j , respectively, atdistance (cid:96) ( R j − ) − i − (cid:96) ( R j ) − i − j − i ) − y j − and y j , respectively;in particular, D i +1 and D i +2 are in y i +1 and y i +2 , respectively, if i < n . • For j ∈ { , . . . , i } , – if the variable x j − = true , then D j − is on the path P j − at distance (cid:96) ( P j − ) + (cid:96) ( R j − ) − i = 2( n − i ) from x (cid:48)(cid:48) j − , – if the variable x j − = false , then D j − is on the path P j − at distance (cid:96) ( P j − ) + (cid:96) ( R j − ) − i = 2( n − i ) from x (cid:48)(cid:48) j − ,15 if the variable x j = true , then D j is on the path P j at distance (cid:96) ( P j )+ (cid:96) ( R j ) − i =2( n − i ) from x (cid:48)(cid:48) j , – if the variable x j = false , then D j is on the path P j at distance (cid:96) ( P j ) + (cid:96) ( R j ) − i = 2( n − i ) from x (cid:48)(cid:48) j .It is straightforward to verify that the claim holds for i = 0. Assume inductively that theclaim holds for 0 ≤ i < n . We show that either Disruptor wins by the trivial strategy appliedfrom the steps 2 i + 2 or 2 i + 3, or the configuration is maintained for i (cid:48) = i + 1.Observe that dist G − X i +1 ( u i , v i ) = 2( τ − i − R and J has to move. Moreover, they have to move along a shortest ( u i , v i )-path in G − X i +1 . Hence, Facilitator moves J from v i to v v +1 and R is moved either to x (cid:48) i +1 orto x (cid:48) i +1 . If R is moved to x (cid:48) i +1 , then we assign the variable x i +1 = true , and x i +1 = false otherwise. Disruptor responds by moving D i +1 from y i +1 to x i +1 if R is in x (cid:48) i +1 , and D i +1 ismoved to x i +1 if R is in x (cid:48) i +1 . For D i +2 , Disruptor chooses one of the vertices x i +2 and x i +2 and moves the agent there. By this move, Disruptor selects the value of the Boolean variable x i +2 , and x i +2 = true if D i +2 in x i +2 , and x i +2 = false otherwise. Note that Disruptorknows the values of x , . . . , x i +1 and selects the move for D i +1 to ensure that the final valueof ψ = true . The agents D h for h ∈ { , . . . , n } such that h (cid:54) = 2 i + 1 , i + 2 are moved toadjacent vertices in the corresponding paths. For j ∈ { i + 2 , . . . , n } , D j − and D j are movedalong R j − and R j toward y j − and y j , respectively. For j ∈ { , . . . , i } , D j − and D j aremoved along P j − ( P j − ) and P j ( P j ) toward x (cid:48)(cid:48) j − ( x (cid:48)(cid:48) j − ) and x (cid:48)(cid:48) j ( x (cid:48)(cid:48) j ), respectively.Assume without loss of generality that R occupies x (cid:48) i +1 , because the case when R is in x i +1 is symmetric. We have that dist G − X i +2 ( x (cid:48) i +1 , v i +1 ) = 2( τ − i − R of J are notmoved toward each other by the next step, Disruptor wins by the trivial strategy. Assume thatthis is not the case. Recall that D i +1 is in x . We conclude that R is moved to u i +1 and J ismoved to v i +2 . Disruptor responds as follows. For j ∈ { i + 2 , . . . , n } , D j − and D j are movedalong R j − and R j toward y j − and y j , respectively. For j ∈ { , . . . , i + 1 } , D j − and D j are moved along P j − ( P j − ) and P j ( P j ) toward x (cid:48)(cid:48) j − ( x (cid:48)(cid:48) j − ) and x (cid:48)(cid:48) j ( x (cid:48)(cid:48) j ), respectively.We obtain that for i (cid:48) = i + 1, the players are in the required configuration.By the above claim, we have that after 2 n + 1 steps of the game either Disruptor wins bythe trivial strategy applied from some step or the following configuration is achieved: • The values of the Boolean variables x , . . . , x n are chosen and ψ = true for them. • R is in u n and J is in v n . • For i ∈ { , . . . , n } , D i is in x (cid:48)(cid:48) i if x i = true and D i is in x (cid:48)(cid:48) i otherwise.Since dist G − X n +1 ( u n , v n ) = 4, the only possibility for Facilitator to win in two steps isto move R and J toward each other along the path u n w h c h w (cid:48) h v n for some h ∈ { , . . . , m } .Otherwise, Disruptor wins by the trivial strategy. Assume that R is moved to w h for some h ∈ { , . . . , m } in the next step. Recall that ψ = true . Therefore, the clause C j contains aliteral x i or x i for some i ∈ { , . . . , n } with the value true . Assume that C h contains x i as theother case is symmetric. We have that the vertex x (cid:48)(cid:48) i is occupied by D i and x (cid:48)(cid:48) i is adjacent to c h in G . Disruptor responds to the moving R to w h by moving D i to c h . This prevents R and J from meeting in the next step. Therefore, Disruptor wins. This concludes the proof of theclaim that if ϕ evaluates true , then Disruptor with k agents has a winning strategy in the gamewith τ steps.Our next aim is to show that if ϕ evaluates false , then Facilitator can win in τ steps.It is convenient to define a special strategy for Facilitator that can be applied after somestep. Assume that after the i -th move of Facilitator R and J are occupying some vertices a and16 and G has an ( a, b )-path L whose length is at most 2( τ − i ) and the internal vertices of L havedegree two in G . If there are no agent of Disruptor occupying a vertex of L , then Facilitatorwins in at most τ − i remaining steps by moving R and J along L toward each other (except if R and J are in adjacent vertices; then R moves to the vertex occupied by J ). If Facilitator canwin this way, we say that Facilitator has a trivial winning strategy .As above, for i ∈ { , . . . , n } , we use X i to denote the set of vertices occupied by the agentsof Disruptor after i -ith step of the game and X is the set of vertices occupied in the beginningof the game.Initially, R is in s and J is in t . If there is a vertex a ∈ N G ( s ) ∩ N G ( t ) such that a / ∈ X , thenFacilitator wins in one step by moving R and J to a . Hence, we assume that N G ( s ) ∩ N G ( t ) ⊆ X .Since | N G ( s ) ∩ N G ( t ) | , X = N G ( s ) ∩ N G ( t ) = { y (cid:48) , . . . , y (cid:48) n } ∪ { z, z (cid:48) } and each vertex of X isoccupied by exactly one agent of Disruptor. Let D i be in y (cid:48) i for i ∈ { , . . . , n } and let theremaining two agents D n +1 and D n +2 be in z and z (cid:48) , respectively.The idea behind the strategy of Facilitator is that R and J are moved towards each otheralong the paths containing the vertices u , . . . , u n and v , . . . , v n with the aim to meet in somevertex c h . The trajectory of R goes trough vertices x (cid:48) i and x (cid:48) i and the choice between thesevertices defines the value of the variable x i . On the way, Facilitator forces Disruptor to behavein a certain way as, otherwise, Facilitator can win by the trivial strategy using paths Q i or Q i .It is convenient to sort-out the agents of Disruptor whose movements are irrelevant for thestrategy of Facilitator. We say that an agent D j is out of game if D j cannot block any shortestpath between the vertices occupied by R and J in G − X . Formally, assume that after the i -th step of the game R is in a vertex a and J is in b , and let d be the vertex occupied by D j .We say that D j is out of game after the i -step of the game if for every shortest ( a, b )-path L in G − X and every e ∈ V ( L ), it holds that • dist L ( a, e ) ≤ dist G ( d, e ) if dist L ( a, e ) ≤ dist L ( b, e ), • dist L ( b, e ) ≤ dist G ( d, e ), otherwise.Note that if D j is out of game after the i -th step, then D j is out of game for all subsequentsteps, because R and J are moving toward each other along some shortest path between thevertices occupied by them. In particular, D n +1 and D n are out of game from the beginning.By the first step, Facilitator moves R to u and J is moved from t to v . Further, R and J move towards each other. The moves of the players define the values of the Boolean variables x , . . . , x n . By his moves, Facilitator consecutively chooses the values of x , x , . . . , x n − andDisruptor selects the values of x , x , . . . , x n . Facilitator aim to achieve ψ = C ∧ . . . ∧ C m = false .To describe the strategy, we show inductively for i = 0 , . . . , n that after the 2 i + 1-th step of thegame, either Facilitator wins by the trivial strategy or the following configuration is maintained. • The values of the variables x j for j ≤ i are chosen and the values of the variables x j for j > i are unassigned. Moreover, the values of x , . . . , x i are chosen in such a way that ϕ evaluates false if the values of x , . . . , x i are constrained by the choice. • R is in u i and J is in v i . • For j ∈ { i + 1 , . . . , n } , – either D j − is out of game or D j − in on the path R j − at distance (cid:96) ( R j − ) − i − j − i ) − y j − , – either D j is out of game or D j is on the path R j at distance (cid:96) ( R j ) − i − j − i ) − y j . • For j ∈ { , . . . , i } , 17 if the variable x j − = true , then either D j − is out of game or D j − is on the path P j − at distance (cid:96) ( P j − ) + (cid:96) ( R j − ) − i = 2( n − i ) from x (cid:48)(cid:48) j − , – if the variable x j − = false , then either D j − is out of game or D j − is on the path P j − at distance (cid:96) ( P j − ) + (cid:96) ( R j − ) − i = 2( n − i ) from x (cid:48)(cid:48) j − , – if the variable x j = true , then either D j is out of game or D j is on the path P j at distance (cid:96) ( P j ) + (cid:96) ( R j ) − i = 2( n − i ) from x (cid:48)(cid:48) j , – if the variable x j = false , then either D j is out of game or D j is on the path P j at distance (cid:96) ( P j ) + (cid:96) ( R j ) − i = 2( n − i ) from x (cid:48)(cid:48) j .The construction of G immediately implies that the claim holds for i = 0. Assume induc-tively that the claim holds for 0 ≤ i < n . We show that either Facilitator wins by the trivialstrategy applied from the steps 2 i + 2 or 2 i + 3, or the configuration is maintained for i (cid:48) = i + 1.By the 2 i + 2-th move, Facilitator moves J to v i +1 and R is moved either to x (cid:48) i +1 or to x (cid:48) i +1 . Note that Facilitator cannot prevent these moves, because of our assumption about theconfiguration of the positions of the players and the observation that D n +1 and D n are outof game. If R is moved to x (cid:48) i +1 , then we assign the variable x i +1 = true , and x i +1 = false otherwise. Assume that R is moved to x (cid:48) i +1 (the other case is symmetric). If no agent ofDisruptor is moved to x i +1 , then by the next moves R is moved to x i +1 and J is moved alongthe path Q i +1 toward R . Because the vertices of Q i +1 are not occupied by the agents ofDisruptor, Facilitator wins by the trivial strategy. Since only D i +1 can move into x i +1 , weassume that D i +1 is moved to this vertex. Symmetrically, we assume that if R is moved to x (cid:48) i +1 , then D i +1 is moved to x i +1 .Observe that if D i +2 is out of game, then no agent of Disruptor can be moved to either x i +2 or x i +2 . Otherwise, D i +2 is in y i +2 . If the agent is not moved to neither x i +2 nor x i +2 , D i +2 is out of game. In all these cases, the value of the Boolean variable x i +2 is definedarbitrarily. Otherwise, if D i +2 is moved to x i +2 , then we set x i +2 = true , and if D i +2 ismoved to x i +2 , then we set x i +2 = false . Consider the agents D h for h ∈ { , . . . , n } suchthat h (cid:54) = 2 i + 1 , i + 2 are not out of game. If such an agent D j − ( D j , respectively) for j ∈ { i + 2 , . . . , n } is not moved along R j − toward y j − (along R j toward y j , trespectively), D j − ( D j , respectively) is out of game. Similarly, if such an agent D j − ( D j , respectively)for j ∈ { , . . . , i } is not moved along his current path P j − or P j − ( P j or P j , respectively)toward x (cid:48)(cid:48) j − or x (cid:48)(cid:48) j − ( x (cid:48)(cid:48) j or x (cid:48)(cid:48) j ), respectively, this agent is placed out of game.Now we consider the step 2 i + 3. By symmetry, we assume without loss of generality that R is in x (cid:48) i +1 (the case when R is in x i +1 is symmetric). Then Facilitator moves R to u i +1 and J to v i +2 . For each agent D h that is not out of game, observe that D h is placed on some path: for j ∈ { i + 2 , . . . , n } , D j − and D j are in R j − and R j , respectively, and for j ∈ { , . . . , i + 1 } , D j − and D j are in P j − ( P j − ) and P j ( P j ), respectively. If they do not move alongthese paths toward y i +1) , . . . , y n and the vertices x (cid:48)(cid:48) h or x (cid:48)(cid:48) h for h ≤ i , then they are out ofgame. We obtain that for i (cid:48) = i + 1, the players are in the required configuration.By the above claim, we have that after 2 n + 1 steps of the game either Facilitator wins bythe trivial strategy applied from some step or the following configuration is achieved: • The values of the Boolean variables x , . . . , x n are chosen and ψ = false for them. • R is in u n and J is in v n . • For each j ∈ { , . . . , m } , the vertices w j , c j and w (cid:48) j are not occupied by the agents ofDisruptor. • For i ∈ { , . . . , n } , if x (cid:48)(cid:48) i is occupied by an agent of Disruptor, then the vertex is occupiedby D i and the value of the variable x i = true , and if x (cid:48)(cid:48) i is occupied by an agent ofDisruptor, then the vertex is occupied by D i and the value of the variable x i = f alse .18ince ψ = false , there is j ∈ { , . . . , m } such that C j = false . Then for c j , we have that thevertices of N G [ c j ] are not occupied by the agents of Disruptor. This mean that Facilitator winsby the next two moves: R is moved to w j and then to c j , and R is moved to w (cid:48) j and then to c j .It follows that Facilitator wins on G in at most τ steps.This concludes the proof of PSPACE -hardness for
Rendezvous in Time .The proof for
Rendezvous is similar but more complicated. Observe that in the winningstrategy for Disruptor for the case ϕ = true , it is crucial that R and J are forced to move towardeach other along a shortest path between their positions, because of the limit of the number ofsteps. In Rendezvous , we have no such a limitation and Facilitator can use other strategies.However, we can modify the construction of the graph to make Facilitator behave exactly inthe same way as in the above proof or loose immediately. x (cid:48)(cid:48) x x (cid:48)(cid:48) x x x (cid:48) x c c P P P Q Q Q Q s tG u u u v v v x (cid:48) x (cid:48) x (cid:48) P u (cid:48) u (cid:48) c (cid:48) c (cid:48) v (cid:48) v (cid:48) v (cid:48) u (cid:48) v v v (cid:48) v (cid:48) a a a (cid:48) a (cid:48) a a a (cid:48) a (cid:48) L L L L (cid:48) L (cid:48) L (cid:48) L (cid:48) L (cid:48) F F S S (cid:48) S S S S (cid:48) S (cid:48) S (cid:48) x (cid:48)(cid:48) x (cid:48)(cid:48) Figure 5: The construction of G (cid:48) for G shown in Figure 4.We construct the graph G (cid:48) starting from G as follows (see Figure 5). • Construct a copy of G . • For i ∈ { , . . . , n } , construct a vertex u i , make it adjacent to s and t , and join it with u i by a path L i of length 2 i + 1. • For i ∈ { , . . . , n } , – construct vertices a i , a i , a (cid:48) i , a (cid:48) i and make them adjacent to s and t , – join a i with x i − by a path S i and join a i with x i − by a path S i of length 2 i + 1, – join a (cid:48) i with x (cid:48) i − by a path S (cid:48) i and join a (cid:48) i with x (cid:48) i − by a path S (cid:48) i of length 2 i . • For i ∈ { , . . . , n } , construct a vertex v (cid:48) i , make it adjacent to s and t , and join it with v i by a path L (cid:48) i of length i + 1. • For j ∈ { , . . . , m } , construct a vertex c (cid:48) j , make it adjacent to s and t , and join it with c j by a path F j of length 2 n + 3. 19et Y = { u , . . . , u n } ∪ (cid:0) (cid:83) ni =1 { a i , a i , a (cid:48) i , a (cid:48) i } (cid:1) ∪ { v (cid:48) , . . . , v (cid:48) n } ∪ { c (cid:48) , . . . , c (cid:48) m } . Then we define k (cid:48) = k + | Y | = 9 n + m + 4.We claim that ϕ evaluates true if and only if Disruptor with k (cid:48) agents has a winning strategyin Rendezvous Game with Adversaries.Assume that ϕ = true . We describe a winning strategy for Disruptor. The k = 2 n + 2 agents D , . . . , D k are initially placed exactly as in the proof for Rendezvous in Time . The remaining | Y | agents are placed in the vertices of the set Y ; we call these agents auxiliary . The agents D , . . . , D k are using essentially the same strategy as in the proof for Rendezvous in Time (we call this strategy old ). We use the same notation X i to denote the set of vertices occupiedby these agents after the i -th step of the game. The auxiliary agents force Facilitator to move R and J in the same way as in the previous proof. For i ≥
1, we denote by X (cid:48) i the set of verticesoccupied by the agents of Disruptor after the i -th step of the game; X (cid:48) = X ∪ Y .We can assume that Facilitator moves ether R or J to an adjacent vertex by the first move.Suppose that R is moved to u and J keeps the old position in t . Then Disruptor moves D n +1 from z to s and the agent from v (cid:48) is moved to v . Observe that R and J are now in distinctconnected components of G − X and Disruptor wins by the trivial strategy, that is, by keepingall the agents in their current position. Similarly, if J is moved to v and R remains in s , thenDisruptor moves D n +1 to t and the agent from u (cid:48) is moved to u . Again, X separates R and J ,that is, Disruptor wins. Assume that both R and J are moved in the first step of the game. ThenDisruptor responds by moving D , . . . , D n +2 using the old strategy. The auxiliary agents aremoved to adjacent vertices along the paths L i for i ∈ { , . . . , n } , S i , S i , S (cid:48) i , S (cid:48) i for i ∈ { , . . . , n } , L (cid:48) i for u ∈ { , . . . , n } and F j for j ∈ { , . . . , m } . By the subsequent moves, these agents aremoved further along these paths until they reach the end-vertices. If an auxiliary agent is unableto enter a vertex, because it is occupied by an agent of Facilitator, Disruptor waits until thevertex get vacated and then moves the agent there.Assume inductively for i = 0 , . . . , n that after the 2 i + 1-th step of the game, R is in u i , J is in v i and the agents D , . . . , D k are occupying the positions according to the old strategy.Notice if i = 0, then s and t are occupied by D n +1 and D n +2 . If i ≥
1, then the vertices x (cid:48) i − , x (cid:48) i − and the vertex v i − are occupied by auxiliary agents of Disruptor. Moreover, thevertices that are adjacent to u i in L i and to v i in L (cid:48) i are also occupied by auxiliary agents.This means that neither R or J can move “backward” or use L i or L (cid:48) i .Suppose that i < n . Notice that the vertices of S (cid:48) i +1 and S (cid:48) i +1 that are adjacent to x (cid:48) i +1 and x (cid:48) i +1 are occupied by auxiliary agents. If Facilitator does not move R to an adjacent vertex,i.e., either to x (cid:48) i +1 or x (cid:48) i +1 , then Disruptor moves the agents to x (cid:48) i +1 and x (cid:48) i +1 and wins bythe trivial strategy. Similarly, the vertex adjacent to v i +1 in L (cid:48) i +1 is occupied by an auxiliaryagent. Hence, if J is not moved, this agents enters v i +1 and J gets separated from R .Assume that R and J are moved to adjacent vertices. Disruptor responds using the oldstrategy. Assume that R is moved to x (cid:48) i +1 as the other case is symmetric. Recall that accordingto the old strategy D i +1 is moved to x i +1 . Notice also that u i and v i are occupied by auxiliaryagents. Moreover, the neighbors of x i +1 , x i +1 , x (cid:48) i +1 , x (cid:48) i +1 , u i +1 , v i +1 and v i +2 in S i +1 , S i +1 , S (cid:48) i +1 , S (cid:48) i +1 , L i +1 and L i +1 , respectively, are occupied by auxiliary agents. If R is not moved,then an agent enters u i +2 and R gets separated from J . If J is not moved, then agents enter x i +1 , x i +1 and v i +2 . Again, R and J are in distinct components of G − X i +2 . Suppose that J is moved to one of the neighbors of v i +1 in Q i +1 or Q i +1 . Disruptor responds by movingagents to v i +1 , x i +1 and x i +1 and wins. We conclude that both R and J should be moved“forward” to u i +1 and v i . Then Disruptor responds using the old strategy.Using these arguments, we obtain after 2 n + 1 steps of the game either Disruptor alreadyseparated R and J and won or the following configuration is achieved: • R is in u n and J is in v n and the vertices of the sets N G (cid:48) ( u n ) \{ w , . . . , w m } and N G (cid:48) ( v n ) \ w (cid:48) , . . . , w (cid:48) m } are occupied by auxiliary agents. • For every j ∈ { , . . . , m } , the vertex at distance two from c i in F i is occupied by auxiliaryagents. • For i ∈ { , . . . , n } , D i is either in x (cid:48)(cid:48) i or in x (cid:48)(cid:48) i , and for the corresponding choice of thevalues of the Boolean variables x , . . . , x n , ψ = C ∧ . . . ∧ C m = true .If Facilitator moves neither R nor J to adjacent vertices, Disruptor moves auxiliary agents intwo steps to c , . . . , c m and separates R and J . Assume that R is moved from u n to w j . Thenthe auxiliary agent that is in the vertex adjacent to u n in the path L n is moved to u n and oneof the agents D , . . . , D n is moved to c j . Recall that such an agent exists, because C j = true .Then R is separated from J . Similarly, if J is moved to w (cid:48) j , then an auxiliary agent is movedto v n and one of the agents D , . . . , D n is moved to c j . Then Disruptor wins.This completes the proof that if ϕ evaluates true , then Disruptor with k (cid:48) agents has a winningstrategy.To show that if ϕ evaluates false , then Facilitator has a winning strategy, we use the samearguments as in the analogous proof for Rendezvous in Time . If there is a vertex of N G ( s ) ∩ N G ( t ) that is not occupied by the agents of Disruptor, Facilitator wins in one step by moving R and J to this vertex. Assume that all the vertices of N G ( s ) ∩ N G ( t ) are occupied by the agentsof Disruptor in the beginning of the game. In particular, we have that every vertex of Y isoccupied by one agent. Now Facilitator uses the same strategy as in the proof for Rendezvousin Time . To see that this is a winning strategy, it is sufficient to observe that the agents ofDisruptor that are placed in the vertices of Y are out of game and we can ignore them in theanalysis of the strategy of Facilitator.We obtain that ϕ evaluates true if and only if Disruptor with k (cid:48) agents has a winning strategyin Rendezvous Game with Adversaries. Therefore, Rendezvous is PSPACE -hard.
In this section, we show that
Rendezvous in Time is FPT when parameterized by τ and theneighborhood diversity of the input graph.The notion of neighborhood diversity was introduced by Lampis in [25]. It is convenient forus to define this notion in terms of modules. Let G be a graph. A set of vertices U ⊆ V ( G ) is a module if for every v ∈ V ( G ) \ U , either N G ( v ) ∩ U = ∅ or U ⊆ N G ( v ). It is said that is a module U is a clique module if U is a clique, and U is an independent module if U is an independentset. We say that a partition { U , . . . , U (cid:96) } of V ( G ) into clique and independent modules is a neighborhood decomposition . The neighborhood diversity of a graph G is the minimum (cid:96) such that G has a neighborhood decomposition with (cid:96) modules; we use nd ( G ) to denote the neighborhooddiversity of G . The value of nd ( G ) and the corresponding partition of V ( G ) into clique andindependent modules can be computed in polynomial (linear) time [25]. Given a neighborhooddecomposition U = { U , . . . , U (cid:96) } , we define the quotient graph G as the graph with the vertexset { , . . . , (cid:96) } such that i is adjacent to j for distinct i, j ∈ { , . . . , (cid:96) } if and only if u ∈ U i isadjacent to v ∈ U j in G . For a vertex v ∈ V ( G ), id ( v ) = i if v ∈ U i . For a multiset of vertices X = { x , . . . , x r } , id ( X ) denotes the multiset of indices { id ( x ) , . . . , id ( x r ) } .Let U = { U , . . . , U (cid:96) } be a neighborhood decomposition of G . Notice that every bijectivemapping ϕ : V ( G ) → V ( G ) such that ϕ ( U i ) = U i for i ∈ { , . . . , (cid:96) } is an automorphism of G .We say that ϕ is an automorphism that agrees with U .21or an automorphism ϕ , we extend it on multisets of vertices in the natural way. Namely,if X = { x , . . . , x r } is a multiset of vertices of G , ϕ ( X ) = { ϕ ( x ) , . . . , ϕ ( x r ) } . Similarly, for apair ( X, Y ) of multisets without common elements, ϕ ( X, Y ) = ( ϕ ( X ) , ϕ ( Y )).Let G be a connected graph and let U = { U , . . . , U (cid:96) } be a neighborhood decomposition of G such that (cid:96) = nd ( G ). Suppose that s, t ∈ V ( G ) such that s and t are distinct modules of U .We consider our Rendezvous Game with Adversaries on G in τ steps.Consider a strategy of Disruptor with k agents, that is, a family of functions d i : P kG → D kG for i ∈ { , . . . , τ − } , where d i : P kG → D kG for i ∈ { , . . . , τ − } such that for i ∈ { , . . . , τ − } .Recall that d i maps ( F, D ) ∈ P kG to D (cid:48) ∈ F kG , where D and D (cid:48) are adjacent and D (cid:48) is compatiblewith F , and d maps ( { s, t } , ∅ ) to D (cid:48) ∈ D kG compatible with { s, t } .Recall that a strategy of Disruptor can be represented as a rooted tree T kG ( τ ) of height τ .Each node v ∈ V ( T kG ( τ )) is associated with a position P v ∈ P kG , and • P r = ( { s, t } , d ( { s, t } , ∅ )) is associated with the root r of T kG ( τ ), • for every node v ∈ V ( T kG ( τ )) with P v = ( F, D ) at at distance i ≤ τ − u of v for every ( F (cid:48) , D (cid:48) ) ∈ P kG such that (i) F (cid:48) is adjacent to F and compatiblewith D , and (ii) D (cid:48) = d i ( F (cid:48) , D ), and u is associated with P u = ( F (cid:48) , D (cid:48) ).From now, we consider such a representation.By Observation 2, T kG ( τ ) is a winning strategy for Disruptor if and only if F is a set of twodistinct vertices for every P v = ( F, D ) for v ∈ V ( T kG ( τ )). By our assumption that s and t arein distinct modules, we can refine the claim. Observation 5.
The tree T kG ( τ ) is a winning strategy for Disruptor if and only if for every v ∈ V ( T kG ( τ )) with P v = ( F, D ) for v ∈ V ( T kG ( τ )) , F contains at most one vertex of everymodule U i for i ∈ { , . . . , (cid:96) } .Proof. To see the observation, it is sufficient to note that if both agents of Disruptor are movedto the same module or if one of the agents is in a module U i and the other is moved to U i , thenFacilitator can move the agents into the same vertex.Let P = ( F, D ) and P (cid:48) = ( F (cid:48) , D (cid:48) ) be positions in Rendezvous Game with Adversaries on G .We say that P and P (cid:48) are isomorphic, if there an automorphism ϕ of G such that P (cid:48) = ϕ ( P ).We also say that P and P (cid:48) are isomorphic with respect to ϕ for such an automorphism ϕ . Weuse the following straightforward observation about positions in the game. Observation 6.
Let P and P (cid:48) be isomorphic positions in Rendezvous Game with Adversarieson G . Then Disruptor can win in at most r steps if the game starts from P if and only ifDisruptor can win in r steps if the game starts from P (cid:48) . We say that T kG ( τ ) is a uniform strategy if for every node v with P v = ( F, D ) and eachof its two children u and u with P u = ( F , D ) and P u = ( F , D ), the following holds: if( F , D ) and ( F , D ) are isomorphic with respect to some automorphism ϕ of G that agrees with U , then P u and P u are isomorphic with resect to some automorphism ψ of G that agrees with U . Informally, if possible moves of the agents of Facilitator to F and F are the same withrespect to moving them to the same modules, then the response of Disruptor is also the same(up to an automorphism that agrees with U ). Observation 6 immediately implies the following. Observation 7.
If Disruptor has a winning strategy in Rendezvous Game with Adversaries on G , then Disruptor has a uniform winning strategy. T kG ( τ ) is uniform.Let u and u be distinct children of a node v of T kG ( τ ). We say that u and u are equivalent if P u and P u are isomorphic with respect to some automorphism ϕ of G that agrees with U .We also say that two subtrees T and T rooted in u and u are equivalent if u and u areequivalent. It is straightforward to see that the introduced relation is indeed an equivalencerelation. Observe that, because the strategy is uniform, for P u = ( F , D ) and P u = ( F , D ), u and u are equivalent if and only if | F ∩ U i | = | F ∩ U i | for all i ∈ { , . . . , (cid:96) } .Since T kG ( τ ) is uniform, to represent the strategy, it is sufficient to keep one representativefrom each class of equivalent children. Given T kG ( τ ), we construct the reduced strategy ˆ T kG ( τ )obtained by the following operation applied top-down staring from the root: for a node v anda class of equivalent subtrees rooted in the children of v , delete all the elements of the classexcept one. Observe that given a reduced strategy ˆ T kG ( τ ), we can reconstruct T kG ( τ ). Noticealso that by Observations 5 and 6, the strategy is a winning strategy for Disruptor if and onlyif for every v ∈ V ( ˆ T kG ( τ )) with P v = ( F, D ) for v ∈ V ( T kG ( τ )), F contains at most one vertex ofevery module U i for i ∈ { , . . . , (cid:96) } .Now we construct the tree that represents all possible moves of Facilitator in τ steps betweenthe modules without the agents of Disruptor. We define the rooted tree T ∗ G ( τ ) of height τ witheach node v associated with a pair X v = { p, q } of not necessarily distinct elements of { , . . . , (cid:96) } such that • X r = { s, t } is associated with the root r of T ∗ G ( τ ), • for every node v ∈ V ( T ∗ G ( τ )) with X v = { p, q } at distance at most τ − u of v with X v = { p (cid:48) , q (cid:48) } for every { p (cid:48) , q (cid:48) } adjacent to { p, q } in the quotientgraph G .Observe that T ∗ G ( τ ) has at most (cid:0) (cid:96) +12 (cid:1) τ +1 nodes.The tree ˆ T kG ( τ ) can be seen as a subtree of T ∗ G ( τ ). Formally, we define an injective mapping α : V ( ˆ T kG ( τ )) → V ( ˆ T kG ( τ )) inductively top-down: • for the root r of ˆ T kG ( τ ), α ( r ) is the root of T ∗ G ( τ ), • if α ( v ) = u for v ∈ V ( ˆ T kG ( τ )), then every child v (cid:48) of v in ˆ T kG ( τ ) with P v (cid:48) = ( F, D ) ismapped to the child u (cid:48) of u in T ∗ G ( τ ) with X u (cid:48) = id ( F ).In particularly, for every v ∈ V ( ˆ T kG ( τ )) with P v = ( F, D ), X α ( v ) = id ( F ). We say that thesubtree of T ∗ G ( τ ) induced by α ( V ( ˆ T kG ( τ ))) is a projection of ˆ T kG ( τ ) to T ∗ G ( τ ). We use the followingproperty of projections that immediately follows from the definition. Observation 8.
Let u be a non-leaf node of T ∗ G ( τ ) with X u = { p, q } . Let also u = α ( v ) forsome v ∈ V ( ˆ T kG ( τ )) with P v = ( F, D ) and let I v = { i | i ∈ { , . . . , (cid:96) } and U i ⊆ D } . Then achild u (cid:48) of u in T ∗ G ( τ ) with X u (cid:48) = { p, q } is a child of u in the projection of ˆ T kG ( τ ) if and only if { p (cid:48) , q (cid:48) } ∈ {{ i, j } | i ∈ N G [ p ] \ I v and j ∈ N G [ q ] \ I v } . Note, in particular, that each leaf of the projection of ˆ T kG ( τ ) is a leaf of T ∗ G ( τ ).In our algorithm for Rendezvous in Time , we check whether Disruptor with k agents has awinning strategy on G . For this, we consider T ∗ G ( τ ) and guess the projection T of a hypotheticalreduced winning strategy tree by trying all subtrees of T ∗ G ( τ ) using brute force. For each T ,we verify whether Disruptor indeed has a strategy corresponding to T by checking whetherDisruptor can respond to the moves of Facilitator in such a way that Disruptor is able to ensurethat T has the required structure, according to Observation 8.Checking whether Disruptor has a strategy corresponding to T is based on the results ofLenstra [22] (see also [23, 14] for further improvements) about parameterized complexity of23nteger Linear Programming. The task of the Integer Linear Programming Feasibility problem is, given a q × p matrix A over Z and a vector b ∈ Z q , decide whether there is avector x ∈ Z p such that Ax ≤ b ; we write Ax ≤ b to denote that for every i ∈ { , . . . , q } ,the i -th element of the vector Ax is at most the i -th element of b . Lenstra [22] proved that Integer Linear Programming Feasibility is FPT when parameterized by p and later thisresult was improved by Kannan [23]. Further, Frank and Tardos [14] proved that IntegerLinear Programming Feasibility can be solved in polynomial space. These results can besummarized in the following statement.
Proposition 3 ([22, 23, 14]) . Integer Linear Programming Feasibility can be solved in O ( p . p + o ( p ) · L ) time and polynomial in L space, where L is the number of bits in the input. Theorem 5.
Rendezvous in Time can be solved in (cid:96) O ( τ ) · n O (1) time on graphs of neighborhooddiversity (cid:96) .Proof. Let (
G, s, t, k, τ ) be an instance of
Rendezvous in Time . If s = t or s and t areadjacent, then the problem is trivial. Assume that s and t are distinct nonadjacent vertices of G . We compute a neighborhood decomposition U = { U , . . . , U (cid:96) } of G with (cid:96) = nd ( G ). Recallthat this can be done in polynomial time [25]. Denote by n i = | U i | for i ∈ { , . . . , (cid:96) } .Suppose that s and t are in the same module U i . Since s and t are distinct and notadjacent, U i is an independent module. We have that N G ( s ) = N G ( t ) and, therefore, λ G ( s, t ) = | N G ( s ) ∩ N G ( t ) | . Notice that Facilitator wins in one step if k < | N G ( s ) ∩ N G ( t ) | by moving R and J into a vertex of N G ( s ) ∩ N G ( t ) that is not occupied by an agent of Disruptor. Weconclude that d G ( s, t ) = | N G ( s ) ∩ N G ( t ) | and, therefore, ( G, s, t, k, τ ) is a yes-instance if andonly if k < | N G ( s ) ∩ N G ( t ) | . From now, we assume that s and t are in distinct modules of U .We construct the tree T ∗ G ( τ ) by brute force with the corresponding pairs X v = { p, q } for v ∈ V ( T ∗ G ( τ )). Since T ∗ G ( τ ) has at most (cid:0) (cid:96) +12 (cid:1) τ +1 nodes, the construction can be done in (cid:96) O ( τ ) time. Denote by r the root of T ∗ G ( τ ).We consider all subtrees T of T ∗ G ( τ ) containing r and rooted in this vertex, whose leavesare leaves of T ∗ G ( τ ). Observe that the total number of such trees is at most 2 |T ∗ G ( τ ) | = 2 (cid:96) O ( τ ) .For each T , we check whether Disruptor has a winning strategy such that the projection of thecorresponding reduced strategy is T . If we find such a tree T , we conclude that Disruptor winsin the game. Otherwise, we conclude that Facilitator wins.Assume that T is given. If for some v ∈ V ( T ), X v = { p, p } for some p ∈ { , . . . , (cid:96) } , wediscard the choice of T , because T cannot be the projection of a winning strategy of Disruptorby Observation 5. Suppose that for every v ∈ V ( T ), X v = { p, q } with p (cid:54) = q .The running time of our algorithm is going to be dominated by checking all the trees T andsolving Integer Linear Programming Feasibility . Therefore, to simplify the arguments,for each node v of T we guess the set I v ⊆ { , . . . , (cid:96) } such that the agents of Disruptor occupyall the vertices of the modules U i with i ∈ I v in the position of the game corresponding to v .As standard, we do it by brute force checking of all possible assignments of sets to the nodes of T . Since the number of the assignments is at most (2 (cid:96) ) | V ( T ) | , this can be done in 2 (cid:96) O ( τ ) time.For each selection of I v for v ∈ V ( T ), we check feasibility using Observation 8. Namely, foreach non-leaf vertex v ∈ V ( T ), we consider its set X v = { p, q } and check whether the children u of v in T ∗ G ( τ ) with X u ∈ {{ i, j } | i ∈ N G [ p ] \ I v and j ∈ N G [ q ] \ I v } are exactly the children of v in T . We discard the assignment if this is not the case, and we discard the current choice of T if we fail to find a feasible assignment of sets I v .From now on, we assume that the assignment of sets I v for v ∈ V ( T ) is given.Our general idea is to express the question about existence of a winning strategy of Disruptorin terms of Integer Linear Programming Feasibility . We start with introducing twofamilies of variables x v , . . . , x v(cid:96) and y v , . . . , y v(cid:96) for each node v of T . The intuition behind these24ariables is following. For every i ∈ { , . . . , (cid:96) } , x vi is the number of vertices of U i occupiedby agents of Disruptor in the position of the game corresponding to the node v . It is moreconvenient for us to consider x vi as the number of the agents of Disruptor that occupy distinctvertices of U i ; we call these agents blockers . Disruptor may also have other agents in U i and y vi is the number of these agents and we call these agents dwellers . It is also convenient to assumethat blockers are active in the current step of the game, and dwellers are inactive and do notprevent R or J from entering the vertices occupied by them. By this convenience, we can allow,say, the situation x vi = 0 and y vi >
0, as we do not care where the dwellers are placed in thecorresponding module.We impose the following constraints on these variables for every v ∈ V ( T ): (cid:96) (cid:88) i =1 ( x vi + x vi ) = k, x vi ≥ y vi ≥ i ∈ { , . . . , (cid:96) } , (3) x vi ≤ n i for every i ∈ { , . . . , (cid:96) } \ X v , (4) x vi ≤ n i − i ∈ X v , (5) y vi = 0 for i ∈ { , . . . , (cid:96) } if n i = 1 and i ∈ X v . (6)The necessity of constraints (3) and (4) is straightforward. To see the reason behind (5) and(6), notice that if a vertex of U i is occupied by an agent of Facilitator, then at most n i blockerscan be in U i and, moreover, if n i = 1, then no agent of Disruptor can be in U i .Next, we state the constraints coming from the choice of sets I v . For every v ∈ V ( T ), x vi = n i for every i ∈ I v . (7)The variables x vi and y vi are used to express the positions of the players. However, we alsohave to express transitions between these positions, that is, the players should be able to makemoves from the position corresponding to a node of T to the positions corresponding to itschildren. For this, we need additional variables. For every v ∈ V ( T ) and every child u of v in T , and every ordered pair ( i, j ) of adjacent vertices of the quotient graph G , we introducefour variables a v,ui,j , b v,ui,j , c v,ui,j , d v,ui,j . The meaning of the variables is following. For the move ofFacilitator from the position corresponding to v to the position corresponding to u , Disruptorresponds by moving a v,ui,j blockers from X i to make them blockers in X j , b v,ui,j blockers from X i become dwellers in X j , c v,ui,j dwellers from X i become blockers in X j , and d v,ui,j dwellers from X i become dwellers in X j . Notice that if U i is a clique module, then some dwellers can move toadjacent vertices to become blockers (it has no sense for Disruptor to make a blocker a dweller).For this, we introduce a variable z v,ui for i ∈ { , . . . , (cid:96) } . The constraints for these variables arefollowing.For every non-leaf v ∈ V ( T ) and every child u of v in T , a v,ui,j ≥ , b v,ui,j ≥ , c v,ui,j ≥ , d v,ui,j ≥ i, j ) of adjacent vertices of G , (8) z v,ui ≥ z v,ui = 0 if U i is an independent module for every i ∈ { , . . . , (cid:96) } , (9) (cid:88) j ∈ N G ( i ) ( a v,ui,j + b v,ui,j ) ≤ x vi and (cid:88) j ∈ N G ( i ) ( c v,ui,j + d v,ui,j ) + z v,ui ≤ y vi for every i ∈ { , . . . , (cid:96) } , (10) x ui = x vi − (cid:88) j ∈ N G ( i ) ( a v,ui,j + b v,ui,j ) + (cid:88) j ∈ N G ( i ) ( a v,uj,i + c v,uj,i ) + z v,ui for every i ∈ { , . . . , (cid:96) } , (11) y ui = y vi − (cid:88) j ∈ N G ( i ) ( c v,ui,j + d v,ui,j ) − z v,ui + (cid:88) j ∈ N G ( i ) ( b v,uj,i + d v,uj,i ) for every i ∈ { , . . . , (cid:96) } . (12)Constraints (8) and (9) are straightforward. Constraint (10) encodes that the number of blockersthat leave a module U i is upper bounded by the number of blockers in U i and, symmetrically,25he number of dwellers that leave a module U i or become blockers in the block is at mostthe number of dwellers in U i . Finally, (11) and (12) express that the movements of agents ofDisruptor from the position associated with v lead to the position corresponding to u .We have 2 | V ( T ) | (cid:96) variables x vi , y vj , at most 8 | E ( T ) | (cid:0) (cid:96) (cid:1) variables a v,ui,j , b v,ui,j , c v,ui,j , d v,ui,j , and | E ( T ) | (cid:96) variables z u,vi , that is, (cid:96) O ( τ ) variables. We defined 5 | V ( T ) | (cid:96) constraints (3)–(6), at most | V ( T ) | (cid:96) constraints (7), and | E ( T ) | (8 (cid:0) (cid:96) (cid:1) + 5 (cid:96) ) constraints (8)–(12). Hence, in total, we have (cid:96) O ( τ ) constraints. Denote the obtained system of integer linear inequalities by ( ∗ ). Observe thatthe coefficients in ( ∗ ) are upper bounded by n . Therefore, the bit-size of ( ∗ ) is (cid:96) O ( τ ) · log n . Wesolve ( ∗ ) in 2 (cid:96) O ( τ ) · log n time by Proposition 3.We claim that ( ∗ ) is feasible, that is, has an integer solution if and only if Disruptor has awinning strategy such that the projection of the reduced strategy on T ∗ G ( τ ) is T .Suppose that Disruptor with k agents has a uniform winning strategy T kG ( τ ) such that theprojection of the reduced strategy ˆ T on T ∗ G ( τ ) is T . For every two distinct modules U i and U j , ether every vertex of U i is adjacent to every vertex of U j or the vertices of the modules arenonadjacent. This allows us to make some additional assumptions about T kG ( τ ). Namely, wecan assume that on each step the agents of Disruptor are divided into blockers and dwellers, andthen we can assume that a blocker (dweller, respectively) agent can become a dweller (blocker,respectively) only if the agent is moved to an adjacent vertex. Also we can assume that a blockeris never moved to an adjacent vertex of the same clique module to become a dweller. Thenwe define the values of all the variables according to the description given in the constructionof ( ∗ ) following the reduced strategy ˆ T kG ( τ ). Then the construction of the constraints of ( ∗ )immediately imply that these values of the variables provide a solution of ( ∗ ).For the opposite direction, given a solution of ( ∗ ), we construct the strategy T kG ( τ ). Initially,we place the agents of Disruptor on G according to the values of x r , . . . , x r(cid:96) and y r , . . . , y r(cid:96) forthe root r . For each i ∈ { , . . . , (cid:96) } , we place x ri agents (blockers) into distinct vertices of U i unoccupied by the agents of Facilitator. Then we put y ri dwellers into U i ; as we pointed above,it is convenient to assume that these agents are inactive and we can place them arbitrarily.Assume inductively that we constructed a node v of the future T kG ( τ ) with P v = ( F, D ) thatcorresponds to the node v (cid:48) of T , that is, id ( F ) = X v (cid:48) and for each i ∈ { , . . . , (cid:96) } , Disruptor hasexactly x v (cid:48) i blockers in U i that occupy distinct vertices, and also y v (cid:48) i dwellers are in U i . Assumethat F (cid:48) ∈ F is compatible with D and adjacent to F . Then because of constraints (7), thereis a child u (cid:48) of v (cid:48) in T with id ( F (cid:48) ) = X u (cid:48) . Then Disruptor responds to moving the agents ofFacilitator from F to F (cid:48) by moving his agents according to the values a v (cid:48) ,u (cid:48) i,j , b v (cid:48) ,u (cid:48) i,j , c v (cid:48) ,u (cid:48) i,j , d v (cid:48) ,u (cid:48) i,j for the ordered pairs ( i, j ) of adjacent vertices of G and according to the values of z v (cid:48) ,u (cid:48) i for i ∈ { , . . . , (cid:96) } . For the obtained node u of T kG ( τ ) with P u = ( F (cid:48) , D (cid:48) ), we have that the positioncorresponds to the configuration defined by the variables x u (cid:48) , . . . , x u (cid:48) (cid:96) and y u (cid:48) , . . . , y u (cid:48) (cid:96) . Theseinductive arguments imply that the constructed strategy is a uniform winning strategy forDisruptor and the projection of the reduced strategy is T .This completes the construction of the algorithm. To evaluate the running time, observethat we consider 2 (cid:96) O ( τ ) trees T , and for each T , we consider 2 (cid:96) O ( τ ) assignments of sets I v forthe nodes. Then for each tree T given together with the assignments of sets I v for v ∈ V ( T ),we construct a solve the system ( ∗ ) in time 2 (cid:96) O ( τ ) · log n . Taking into account the preliminarysteps where we consider special cases of s and t and construct the neighborhood decomposition U , the total running time is 2 (cid:96) O ( τ ) · n O (1) . We introduced and initiated the systematic study of Rendezvous Game with Adversaries ongraphs. We introduced the disruption number d G ( s, t ), as the minimum number of agents26eeded for Disruptor to win against Facilitator, and investigated relations between d G ( s, t ) andthe minimus size λ G ( s, t ) of an ( s, t )-separator in G . We proved that d G ( s, t ) = 1 if and only if λ G ( s, t ) = 1. While it can be easily seen that d G ( s, t ) ≤ λ G ( s, t ), the difference λ G ( s, t ) − d G ( s, t )can be arbitrary. However, we proved that for some hereditary graphs classes, namely for P -free and chordal graphs, d G ( s, t ) = λ G ( s, t ). Are there other natural graph classes with thisproperty? Is it is possible to characterize hereditatary graph classes for which the equalityholds?Further, we investigated the computational complexity of Rendezvous and
Rendezvousin Time . Both problems can be solved it n O ( k ) time. However, they are co - W [2]-hard whenparameterized by k and cannot be solved in n o ( k ) time unless FPT = W [1]. Moreover, Ren-dezvous in Time is hard even if τ = 2. We also proved that Rendezvous and
Rendezvousin Time are
PSPACE -hard. We conjecture that the problems are even harder and, similarly tovarious related games on graphs (see, e.g., [18, 24]), the problems are
EXPTIME -complete. Weleave this as an open question.Finally, we initiated the study of the complexity of
Rendezvous and
Rendezvous inTime under structural parameterization of the input graphs. We proved that
Rendezvousin Time is FPT when paramererized by the neighborhood diversity of the input graph and τ .Can this result be generalized for the parameterization by modular width (see, e.g., [15] forthe definition and the discussion of this parameterization) and τ ? Is Rendezvous in Time
FPT when parameterized by the neighborhood diversity only? The same question is openfor
Rendezvous . We believe that this problem is interesting even for the more restrictiveparameterization by the vertex cover number. Also, what can be said about
Rendezvous and
Rendezvous in Time parameterized by treewidth? The upper bound for the disruptionnumber by λ G ( s, t ) together with Theorem 1 imply that the problems can be solved it time n O ( w ) , where w is the treewidth of the input graph, that is, the problems are in X P for thisparameterization. Are the problems
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