The Firebreak Problem
Kathleen D. Barnetson, Andrea C. Burgess, Jessica Enright, Jared Howell, David A. Pike, Brady Ryan
aa r X i v : . [ m a t h . C O ] M a y The Firebreak Problem
Kathleen D. Barnetson ∗ , Andrea C. Burgess † , Jessica Enright ‡ ,Jared Howell § , David A. Pike ¶ , Brady Ryan k May 13, 2020
Abstract
Suppose we have a network that is represented by a graph G . Potentially a fire(or other type of contagion) might erupt at some vertex of G . We are able to respondto this outbreak by establishing a firebreak at k other vertices of G , so that the firecannot pass through these fortified vertices. The question that now arises is which k vertices will result in the greatest number of vertices being saved from the fire, assumingthat the fire will spread to every vertex that is not fully behind the k vertices of thefirebreak. This is the essence of the Firebreak decision problem, which is the focus ofthis paper. We establish that the problem is intractable on the class of split graphs aswell as on the class of bipartite graphs, but can be solved in linear time when restrictedto graphs having constant-bounded treewidth, or in polynomial time when restrictedto intersection graphs. We also consider some closely related problems.
Key words: firebreak, separation, connectivity, computational complexityAMS subject classifications: 05C85, 05C40, 68Q25 ∗ Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John’s, NL,Canada. † Department of Mathematics and Statistics, University of New Brunswick, Saint John, NB, Canada. [email protected] ‡ School of Computing Science, University of Glasgow, Glasgow, Scotland.
[email protected] § School of Science and the Environment, Grenfell Campus, Memorial University of Newfoundland, CornerBrook, NL, Canada. [email protected] ¶ Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John’s, NL,Canada. [email protected] k Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John’s, NL,Canada. Introduction
In this paper we consider the
Firebreak decision problem, which asks whether it is possibleto establish a firebreak of a given size in a network represented by a graph and thereby protecta desired number of other vertices from being reached by a fire that breaks out at a specifiedvertex of the graph. The problem is formally stated as follows:
Firebreak
Instance: A graph G , an integer k , an integer t , and a vertex v f ∈ V ( G ).Question: Does V ( G ) contain a k -subset S such that v f / ∈ S and the numberof vertices of G − S that are separated from v f is at least t ?There are similarities between this problem and the well-known Firefighting problem,which itself takes the form of a game with two players (fire and firefighters). The gamebegins with fire starting at a vertex. Thereafter, in each round of the game each firefighteris able to designate one unburnt vertex as a permanent firebreak, and then the fire spreadsfrom each of its vertices to all of their unprotected neighbours. The game concludes whenthe fire can spread no more (which, in the case of infinite graphs, may result in a game thatnever terminates). Depending on context, the goal of the firefighters may be to minimise thenumber of vertices that are scorched, or to minimise the time in which the fire is contained.The
Firefighting problem was introduced in 1995 by Hartnell [22] and has since attractedconsiderable attention. For a survey of results and open questions, see [18].The
Firebreak problem could be viewed as variant of the
Firefighting problem inwhich the firefighters are active for only the initial round of the game, after which thefire spreads without further intervention. The particular nature and formulation of the
Firebreak problem lends itself to several applications of practical interest. Although wemodel the problem in terms of fire, it readily applies to the spread of any contagion froma point of infection in a network and where a one-time response is able to be deployed inimmediate reaction to the outbreak.In the course of our investigation into the
Firebreak problem, we noted that it is alsoclosely related to what we will call the
Key Player decision problem that pertains to thenumber of connected components that can be created by the removal of a set of vertices. Bydefining c ( G ) to be the number of connected components of a graph G and G − S to be thesubgraph of G that is induced by the vertices of V ( G ) \ S , the Key Player problem canbe formally described as follows:
Key Player
Instance: A graph G , an integer k and an integer t .Question: Does V ( G ) contain a k -subset S such that c ( G − S ) > t ?The Key Player problem also models various real-world scenarios and applications ofpractical interest in networks. If we have the means to inoculate k nodes which then becomeimpenetrable to the contagion, we can ask which k nodes to inoculate in order to create thegreatest number of segregated quarantine zones. As another scenario, G might represent acommunications network that we wish to disrupt by selectively disabling k of its nodes, with2he goal being to maximise the number of subnetworks that become unable to communicatewith other subnetworks.A related problem is one of dissemination rather than separation, whereby instead ofselecting a k -set S so that c ( G − S ) is maximised, we wish to selected a k -set S that canmost efficiently reach the other vertices of G . One such scenario could be the spread ofnews: we wish to directly inform k individuals, who then propagate the news to their neigh-bours, so that the news spreads to everybody as quickly as possible. Instead of forwardinginformation, we might care about influence and opinion (such as might be the case in a mar-keting campaign, whereby k individuals are selected to receive a new commercial productin the hope that they will exert influence among their friends to acquire the product). Thisdissemination problem and its related separation problem appear to have been first jointlydescribed by Borgatti in 2002 [12]. In [13] he refers to the dissemination scenario as the“Key Player Problem / Positive” and uses the phrase “Key Player Problem / Negative” forthe problem involving the deletion of k nodes.This “negative” variant corresponds to our interest. Some of the early papers aboutthis problem in the literature attest to its applicability to network tolerance and robust-ness [5, 30, 34]). More recent results have considered it from a computational complexityperspective, establishing that it is NP -hard for various classes of graphs yet being solvablein polynomial time for graphs having bounded treewidth [1, 7, 28, 35]. A recent survey byLalou et al. on the topic of detecting critical nodes in networks also touches on this prob-lem [26]. Incidentally, our literature search revealed that there is also an edge-based versionof the problem; although the edge version is beyond the scope of the present paper, wenevertheless provide a few references for the interested reader (see [6, 23, 31]).In this paper we concentrate on the Firebreak decision problem, while also presentingsome new results about the
Key Player problem. In Section 2 we establish that bothproblems are NP -complete when restricted to the class of split graphs. We also find thatthe Firebreak problem is NP -complete when restricted to the class of bipartite graphs.Although the Key Player problem is NP -complete (and it remains so for planar cubicgraphs), in the situation where the number of vertices to be removed from the graph coincideswith the graph’s connectivity the problem is found to be solvable in polynomial time. InSection 3 we consider graphs having treewidth that is bounded by a constant and for suchgraphs we show that the Firebreak problem can be solved in linear time, and in Section 4we show that it can be solved in polynomial time for some classes of intersection graphs.Before continuing, we establish some basic notation and terminology. For a graph G =( V ( G ) , E ( G )), we let deg G ( u ) (or just deg( u ) if the graph G is unambiguously implicit)denote the degree of vertex u , and we use the notation u ∼ v to indicate that vertex u isadjacent to vertex v . The maximum degree of G is ∆( G ) = max { deg( u ) : u ∈ V ( G ) } . By N G ( v ) (or just N ( v ) if there is no ambiguity) we denote the open neighbourhood of a vertex v , so that N ( v ) = { u ∈ V ( G ) : u ∼ v } ; the closed neighbourhood N ( v ) ∪ { v } will be denotedby N [ v ]. The order of a graph G is the cardinality | V ( G ) | of its vertex set, and its size is thecardinality | E ( G ) | of its edge set. If A is a subset of V ( G ) then the subgraph of G inducedby A , denoted G [ A ], is the graph with vertex set A and edge set E ( G ) ∩ (cid:8) { u, v } : u, v ∈ A (cid:9) .Throughout this paper we limit ourselves to finite undirected graphs without loops andwithout parallel edges, and so it follows that for the size of a graph G on n vertices we have | E ( G ) | (cid:0) n (cid:1) ∈ O ( n ). 3or an instance ( G, k, t, v f ) of the Firebreak problem we define F ( G, k, v f ) as themaximum number of vertices of G − S that are not in the same connected component as v f ,where this maximum is taken over all choices for k -subsets S ⊆ V ( G ) \ { v f } . Any k -subset S ⊆ V ( G ) \ { v f } that separates F ( G, k, v f ) vertices from v f will be called an optimal set. Clearly the
Firebreak and
Key Player decision problems are in NP since any given k -set S can be easily validated to determine whether S separates at least t vertices from v f (forthe Firebreak problem) or c ( G − S ) > t (for the Key Player problem). In this sectionwe show that these two decision problems are both NP -complete.A split graph is any graph G that admits a vertex partition V ( G ) = A ∪ B such that A ∩ B = ∅ , G [ A ] is a maximum clique, and B is an independent set ( i.e. , B is a set ofpairwise non-adjacent vertices). To prove that the Firebreak and
Key Player decisionproblems are NP -complete, we will show that they are NP -complete even when the inputgraph G is restricted to the class of split graphs. We first note that under certain conditionsthe problems can be easily solved. Lemma 1.
The
Firebreak problem can be solved in linear time when | N ( v f ) | k . Proof.
Let (
G, k, t, v f ) be an instance of the Firebreak problem where G is a graph and | N ( v f ) | k . We now choose S to consist of every neighbour of v f plus any choice of k −| N ( v f ) | additional vertices selected from V ( G ) \ N [ v f ]. Clearly no vertex in V ( G ) \ ( S ∪ { v f } )will be in the same connected component of G − S as v f . Since there are | V ( G ) | − k − G − ( S ∪ { v f } ), the problem is equivalent to asking if | V ( G ) | − k − > t . Henceone simply needs to count the vertices in G , which can be done in linear time. ✷ Lemma 2.
The
Key Player problem on a split graph G can be solved in linear time when k is at least the size ω ( G ) of a maximum clique in G . Proof.
Let G be a split graph with V ( G ) = A ∪ B , where A induces a maximum cliqueand B is an independent set, and let ( G, k, t ) constitute an instance of the
Key Player problem. Observe that the maximum number of connected components will be produced bydeleting A and k − | A | of the | V ( G ) | − | A | vertices of B . Thus the given instance of the KeyPlayer problem has an affirmative answer if and only if | V ( G ) | − k > t . To solve this, onesimply needs to count the vertices of G . This can clearly be done in linear time. ✷ With the next result we show that when restricted to split graphs, the
Key Player problem is no more difficult than the
Firebreak problem.
Lemma 3.
The
Key Player decision problem on split graphs can be solved in polynomialtime with an oracle for the
Firebreak decision problem on split graphs.
Proof.
Let ( G , k , t ) constitute an instance of the Key Player problem, where G =( A ∪ B, E ) is a split graph with a maximum clique on A and independent set B . Without4oss of generality we may assume that k < | A | for otherwise the problem is easily solved byLemma 2.We proceed to formulate an instance ( G , k , t , v f ) of the Firebreak problem as follows.Construct G from G by adding a new vertex named v f and adding an edge { u, v f } for each u ∈ A . Let k = k and t = t − k -subset S of V ( G ) such that c ( G − S ) > t . Since G is a split graph, this means G − S must have at least t = t − A . Thus we let S = S . Then the t isolated vertices of G − S are alsoisolated vertices of G − S . As none of these vertices are members of the clique induced by A , none of them are in the same connected component as v f . Thus S is a k -subset of V ( G )such that v f / ∈ S and there are at least t vertices in G − S not in the same component as v f . Hence an affirmative answer to the Key Player problem implies an affirmative answerto the associated
Firebreak problem ( G , k , t , v f ).Conversely, suppose that the Firebreak problem ( G , k , t , v f ) has an affirmative an-swer, namely a k -subset S of V ( G ) \ { v f } such that there are at least t vertices of G − S that are not in the same connected component as v f . Since v f is part of the clique of G ,these t vertices must be isolated vertices in G − S . Let S = S . Then by the construc-tion of G , these same t vertices are also isolated in G − S and so they form t distinctcomponents in G − S . Since by assumption k < | N ( v f ) | , then there must be at least oneother vertex remaining in the clique with v f . This vertex also remains in the clique of G , so c ( G − S ) = t + 1 = t and hence the Key Player problem has an affirmative solution.Since the instance ( G , k , t ) of the Key Player problem has an affirmative answerif and only if the associated instance ( G , k , t , v f ) of the Firebreak problem has anaffirmative answer, and this associated instance can clearly be constructed in polynomialtime, then the
Key Player problem can be solved in polynomial time with the availabilityof an oracle for the
Firebreak problem. ✷ We now proceed to show that the
Firebreak and
Key Player decision problems areboth NP -complete, even when restricted to the class of split graphs. To do so we will referto the t - Way Vertex Cut problem studied by Berger et al. [7], expressed as a decisionproblem as follows: t - Way Vertex Cut
Instance: A graph G , an integer k and an integer t .Question: Does V ( G ) contain a subset S such that | S | k and c ( G − S ) > t ? Theorem 1.
When restricted to split graphs, the
Firebreak and
Key Player decisionproblems are both NP -complete. Proof.
Relying on a construction of Marx [28] that is restricted to split graphs, Berger et al. show that the t - Way Vertex Cut problem is NP -complete when restricted to splitgraphs [7]. By using an oracle for the Key Player problem, it is straightforward to answerany given instance (
G, k, t ) of the t - Way Vertex Cut problem. In particular, for each k ′ ∈ { , , . . . , k } present the Key Player oracle with (
G, k ′ , t ). The t - Way Vertex Cut problem has an affirmative answer if and only if one or more of the k answers provided by5he Key Player oracle is affirmative. Hence the
Key Player problem is NP -completewhen restricted to split graphs. It then follows from Lemma 3 that the Firebreak problemis also NP -complete when restricted to split graphs. ✷ Corollary 1.
The
Firebreak and
Key Player decision problems are W [ ] -hard on splitgraphs with parameters k and t . Proof.
For the
Key Player problem, the result has already been proved by Theorem 16of [28]. The reduction in Lemma 3 is clearly polynomial in k and t and is thus parameterizedin k and t . Thus the Firebreak problem also is W [ ]-hard on split graphs with parameters k and t . ✷ We can also use split graphs to show that the
Firebreak problem is NP -complete forbipartite graphs. Theorem 2.
When restricted to bipartite graphs, the
Firebreak decision problem is NP -complete. Proof.
Suppose (
G, k, t, v f ) is an arbitrary instance of the Firebreak problem on a splitgraph G , where V ( G ) = A ∪ B , A induces a maximum clique in G , and B is an independentset in G . Without loss of generality we may assume that k < | N ( v f ) | , as otherwise theproblem is easily solved by Lemma 1. We may further assume that k < | A ∩ N ( v f ) | forotherwise an optimal set S can be obtained by selecting each vertex of A ∩ N ( v f ) along with k − | A ∩ N ( v f ) | vertices of V ( G ) \ (( A ∩ N ( v f )) ∪ { v f } ) such that vertices of B ∩ N ( v f ) arepreferentially selected prior to selecting any other vertices of V ( G ) \ (( A ∩ N ( v f )) ∪ { v f } ).Let G ′ be the graph obtained from G by subdividing every edge of G [ A ], so that | V ( G ′ ) | = | V ( G ) | + (cid:0) | A | (cid:1) and | E ( G ′ ) | = | E ( G ) | + (cid:0) | A | (cid:1) . For convenience we let C denote the set V ( G ′ ) \ V ( G ) of (cid:0) | A | (cid:1) vertices that are newly created in this process, and for any two distinctvertices a , a ∈ A let c ( a , a ) be the vertex of C that is a common neighbour of a and a .Observe that G ′ is a bipartite graph with vertex bipartition ( A, B ∪ C ). We will show thatthe instance ( G, k, t, v f ) of the Firebreak problem has an affirmative answer if and only ifthe instance ( G ′ , k, t + (cid:0) k (cid:1) , v f ) also has an affirmative answer.First suppose that S ⊆ ( V ( G ) \ { v f } ) is an optimal set for the instance ( G, k, t, v f ). Since k < | A ∩ N G ( v f ) | , in G − S none of the vertices of ( A \ ( S ∪ { v f } )) are separated from v f and so if there should exist some vertex u ∈ S ∩ B then for each v ∈ A \ ( S ∪ { v f } ) the k -set ( S \ { u } ) ∪ { v } is also optimal. From an iterated application of this observation itfollows that there must exist an optimal set S such that S ∩ B = ∅ . In G ′ , the numberof vertices that are separated from v f by S is F ( G, k, v f ) + (cid:0) | S ∩ A | (cid:1) = F ( G, k, v f ) + (cid:0) k (cid:1) andhence F ( G ′ , k, v f ) > F ( G, k, v f ) + (cid:0) k (cid:1) .Now let S ′ ⊆ ( V ( G ′ ) \ { v f } ) be an optimal set for the instance ( G ′ , k, t + (cid:0) k (cid:1) , v f ) suchthat among all optimal sets, S ′ has the least intersection with C . Since k < | A ∩ N G ( v f ) | ,it follows that the set A ′ = ( A \ ( S ′ ∪ { v f } )) is not empty. Recall that v f ∈ V ( G ), so either v f ∈ A or v f ∈ B .Consider the situation in which v f ∈ B . If there should exist a vertex y ∈ A ′ that isseparated from v f in G ′ − S ′ , then it is necessary that c ( y, z ) ∈ S ′ for each z ∈ ( A ∩ N G ( v f )) \ S ′ k = | S ′ | > | A ∩ N G ( v f ) ∩ S ′ | + | ( A ∩ N G ( v f )) \ S ′ | . However, k < | A ∩ N G ( v f ) | and so none of the vertices of A ′ are separated from v f in G ′ − S ′ .If v f ∈ A then the set A ′ equals ( A ∩ N G ( v f )) \ S ′ , which we partition into subsets Y and Z such that Y consists of all vertices of A ′ that are separated from v f in G ′ − S ′ , and Z consists of all vertices of A ′ that are not separated from v f in G ′ − S ′ . Since k < | A ∩ N G ( v f ) | then for some a ∈ A ′ neither a nor c ( a, v f ) is in S ′ , and hence Z = ∅ . For each vertex y ∈ Y it is necessary that S z ∈ Z ∪{ v f } { c ( y, z ) } ⊆ S ′ . Consequently k = | S ′ | > | A ∩ N G ( v f ) ∩ S ′ | + | Y | (1 + | Z | ), which is at least | A ∩ N G ( v f ) | when Y = ∅ and thus it must be that Y = ∅ .Therefore, regardless of whether v f is in A or B , no vertex of A ′ is separated from v f in G ′ − S ′ .Let w ∈ A ′ . If there should exist some vertex x ∈ S ′ ∩ C then the k -set ( S ′ \ { x } ) ∪ { w } would contradict the selection of the set S ′ as an optimal set having minimum intersectionwith C . Hence S ′ ∩ C = ∅ . If there should exist some vertex u ∈ S ′ ∩ B then the k -set( S ′ \ { u } ) ∪ { w } is also optimal. Iterated application of this observation ensures that thereis an optimal set S ′′ such that S ′′ ∩ B = S ′′ ∩ C = ∅ . In the graph G , the number of verticesthat are separated from v f by S ′′ is F ( G ′ , k, v f ) − (cid:0) | S ′′ ∩ A | (cid:1) = F ( G ′ , k, v f ) − (cid:0) k (cid:1) and hence F ( G, k, v f ) > F ( G ′ , k, v f ) − (cid:0) k (cid:1) .So when k < | A ∩ N G ( v f ) | we conclude that F ( G ′ , k, v f ) = F ( G, k, v f ) + (cid:0) k (cid:1) . Thus the Firebreak instance (
G, k, t, v f ) for the split graph G has an affirmative answer if and onlyif the instance ( G ′ , k, t + (cid:0) k (cid:1) , v f ) also has an affirmative answer. ✷ Key Player problem
While the
Firebreak problem is the main focus of this paper, we do have some additionalresults pertaining to the
Key Player problem. In Theorem 1 it was established that the
Key Player problem is not only NP -complete, but also that it remains so when restrictedto the class of split graphs. It happens that the problem is also NP -complete when restrictedto cubic planar graphs, in contrast to the Firebreak problem which we now show can besolved in polynomial time when restricted to graphs of constant-bounded degree (includingcubic graphs).
Lemma 4.
The
Firebreak problem can be solved in polynomial time on graphs of constant-bounded degree.
Proof.
Let m be a fixed integer and let ( G, k, t, v f ) be an instance of the Firebreak problem where G is a graph on n vertices such that ∆( G ) m . If k > deg( v f ) then theanswer is affirmative if and only if t > n − − k because separating the maximum numberof vertices from v f is accomplished by deleting N ( v f ) plus k − | N ( v f ) | other vertices. If k < deg( v f ) then the answer can be computed by exhaustively considering all (cid:0) nk (cid:1) k -subsetsof V ( G ) \ { v f } and determining whether any of them separate at least t vertices from v f .Since k is bounded by the constant m , this computation can be done in polynomial time. ✷ Showing that the
Key Player problem is intractable on cubic planar graphs involvesconsideration of the well-known
Independent Set problem.7 ndependent Set
Instance: A graph G and an integer m .Question: Does G contain an independent set of at least m vertices? Theorem 3.
The
Key Player decision problem on cubic planar graphs is NP -complete. Proof.
We employ a reduction from the
Independent Set problem, which Garey andJohnson established in 1977 to be NP -complete on 3-regular planar graphs [19]. Given aninstance ( G, m ) of the
Independent Set problem, construct an instance ( G, | V ( G ) | − m, m ) of the Key Player problem. It is easy to see that
Independent Set has anaffirmative answer if and only if this instance of the
Key Player problem has an affirmativeanswer. Thus an oracle for the
Key Player problem can be used to efficiently solve the
Independent Set problem. ✷ Having established that the
Key Player problem is intractable for a variety of classes ofgraphs, we now proceed to consider the special case in which the parameter k is restricted tobeing the connectivity of the graph in question. For any nontrivial graph G , its connectivity κ ( G ) is the size of a smallest set S of vertices such that G − S is not a connected graph;such a set S is called a cut (or k - cut when we wish to explicitly mention the size of the cut).As we shall see, when k = κ ( G ) the Key Player problem can be solved in polynomialtime. An initial thought for how to potentially prove that this is so is to enumerate all ofthe κ ( G )-cuts of G and if they are polynomial in number then simply calculate c ( G − S ) foreach κ ( G )-cut S . However, unlike edge cuts of size κ ′ ( G ) (of which there are at most (cid:0) n (cid:1) ;see Section 4.3 of [29] for a proof), there can be exponentially many κ ( G )-cuts in a graph G , as is the case with the graph illustrated in Figure 2.1. Hence the na¨ıve idea of examiningeach κ ( G )-cut individually will not serve as a valid approach. K t K t K t K t Figure 1: A graph G on n = 4 t vertices. G consists of four copies of K t that arejoined by t disjoint paths of length 3. The graph has connectivity κ ( G ) = t andmore than 2 t cut sets of size κ ( G ).Instead, we consider the notion of a k - shredder in a k -connected graph, which is defined tobe a set of k vertices whose removal results in at least three components being disconnectedfrom one another; note that it is necessary here that k > κ ( G ) since G must be k -connectedand each k -shredder is also a k -cut. An algorithm that is capable of finding all of the k -shredders of a graph G on n vertices in polynomial time is presented in [15]. The actual8umber of k -shredders is determined in [17] to be at most n when k >
4. We can thus solvethe
Key Player problem in polynomial time when k = κ ( G ) by following the steps of thealgorithm presented below. Algorithm 1.
1. Let k = κ ( G ).2. If k > k -shredders of G .As [17] asserts that there are at most n of them, we then exhaustively checkto see which k -shredder S maximises c ( G − S ). If there should happen tobe no k -shredders, then it must be that every k -cut produces exactly twocomponents.3. If k k -subset S of V ( G ) to see which k -setsare cut sets, and then determine which k -cut maximises c ( G − S ). The numberof k -subsets that must be checked is (cid:0) nk (cid:1) , which is polynomial in n since k iseither 1, 2 or 3.4. Having determined the maximum number of components that can result fromthe deletion of any k -cut, compare this quantity with t to answer the giveninstance of the Key Player problem.
Before we review the technical details of treewidth, we first observe that the
Firebreak problem can be easily solved in the case where the graph in question is a tree.
Theorem 4.
The
Firebreak problem can be solved in polynomial time on trees.
Proof.
Suppose (
T, k, t, v f ) is an instance of the Firebreak problem where T is a tree on n vertices. If | N ( v f ) | k then a polynomial time solution follows from Lemma 1, so wehenceforth assume that | N ( v f ) | > k .Root the tree T at v f . For each vertex v of T define T ( v ) to be the subtree of T rooted at v and let T ( v ) denote the number of vertices in T ( v ). If v is a leaf in the treethen clearly T ( v ) = 1. For any other vertex v with children x , . . . , x m , we have that T ( v ) = T ( x ) + · · · + T ( x m ) + 1. The computation of each T ( v ) can be clearly done in O ( n )time.To find an optimal k -set S (namely one that separates the most vertices from v f ) simplyselect k vertices in N ( v f ) having the k greatest subtree sizes. If v , . . . , v k are these k vertices,then there are T ( v ) + · · · + T ( v k ) − k vertices not in the same connected component of v f .To answer the given instance of the Firebreak problem, it now suffices to ask if T ( v ) + · · · + T ( v k ) − k > t . Since both the computation of each T ( v ) and the identification of the k largest values of T ( v ) can be done in polynomial time, the problem can therefore be answeredin polynomial time. ✷ Firebreak problem can besolved in linear O ( n ) time for a tree on n vertices, except that the task of selecting the k neighbours of v f with the greatest subtree sizes may require a nonlinear sort to be per-formed. However, it will be shown later in this section that a linear time solution doesnevertheless exist. Our approach will be to consider the effect that bounded treewidth hason the complexity of the problem. The treewidth parameter, defined below, was introducedby Robertson and Seymour [32]. Definition 1. A tree decomposition of a graph G is a pair ( X, T = (
I, F )) where X = { X i : i ∈ I } is a family of subsets of V ( G ), and T is a tree whose vertices are the subsets X i suchthat:1. S i ∈ I X i = V ( G ).2. For every edge uv ∈ E ( G ), both u and v are in some X i , i ∈ I .3. If i, j, k are vertices of T , and k lies on the (unique) path from i to j , then X i ∩ X j ⊆ X k .The width of a tree decomposition is max {| X i | − i ∈ I } . The treewidth tw( G ) of a graph G is the minimum width of all tree decompositions of G .It is easy to see that trees have treewidth at most 1. Other graphs with small treewidthare, in a sense, tree-like. For instance, if the treewidth of a graph G is bounded by aconstant ( i.e. , tw( G ) w ), then it follows from Lemma 3.2 of [33] that | E ( G ) | = O ( n )where n = | V ( G ) | . Graphs that are in some way similar to trees often lend themselves totractable solutions for problems that are intractable for graphs in general (see [2, 4, 8] fordetails of several examples).Moreover, Bodlaender has presented an algorithm that finds a tree decomposition of agraph G in time that is linear in the number of vertices and exponential in the cube of thetreewidth [9]. For graphs having constant-bounded treewidth, it is therefore possible to findtree decompositions in linear time.Theorem 4 demonstrated that the Firebreak problem is easily solved for trees. Toshow that it is also tractable for graphs for which the treewidth is bounded by a constant,we will rely on a powerful result that is based on work of Courcelle, independently proved byBorie, Parker and Tovey, and further extended by Arnborg, Lagergren and Seese [3, 14, 16].A survey by Langer et al. [27] presents it in a slightly more general form than we require.For our purposes, the following will suffice:
Theorem 5 (see Theorem 30 of [27]) . Let G be a graph on n vertices, let w be a constant,and let P be a graph theoretic decision problem that can be expressed in the form of extendedmonadic second-order logic. If tw( G ) w then determining whether G has property P canbe accomplished in time O ( f P ( w ) · n ) where f P is a function that depends on the property P . Although the function f P may not be polynomial, if w is a constant then so too is f P ( w ).Hence decision problems that have extended monadic second-order (EMSO) formulationsare fixed-parameter tractable. Monadic second-order (MSO) logic expressions for graphs arebased on 10 variables for vertices, edges, sets of vertices and sets of edges, • universal and existential quantifiers, • logical connectives of conjunction, disjunction and negation, • and binary relations to assess set membership, adjacency of vertices, incidence of edgesand vertices, and equality for vertices, edges and sets.We will only need to consider vertices and sets thereof, which will be respectively denotedby lower case and upper case variable names. Predicates can be constructed from the basicones and incorporated into expressions (in this manner a predicate for implication can bebuilt). To provide an illustrative example, the expression ∃ X ∃ Y (cid:0) ∀ u (( u ∈ X ) ∧ ( u Y )) ∨ (( u X ) ∧ ( u ∈ Y )) (cid:1) ∧ (cid:0) ∀ u ∀ v ((adj( u, v )) ⇒ (( u ∈ X ) ∧ ( v ∈ Y )) ∨ (( u ∈ Y ) ∧ ( v ∈ X ))) (cid:1) encodes whether a given graph is bipartite, where adj( u, v ) represents a Boolean predicatethat evaluates whether vertices u and v are adjacent.Extended MSO logic has additional features that enable set cardinalities to be considered.The survey by Langer et al. [27] provides an excellent overview, to which we direct readersfor more details.Since the factor f P ( w ) in the conclusion of Theorem 5 effectively becomes a hiddenconstant, it follows that deciding whether a graph G has the property P can be done inlinear time when the hypothesis of the theorem is satisfied. With this in mind, we now showthat the Firebreak problem is tractable when restricted to graphs having treewidth atmost a constant w . Theorem 6.
The
Firebreak problem can be solved in linear time for graphs with constant-bounded treewidth.Proof.
Let ϕ represent the following logic expression with two set variables ( S and X ). ϕ = ( v f S ) ∧ ( v f X ) ∧ (cid:0) ∀ y ( y ∈ S ) ⇒ ( y X ) (cid:1) ∧ (cid:0) ∀ x ∀ y (cid:0) ( x ∈ X ) ∧ (adj( x, y )) ∧ ( y S ) (cid:1) ⇒ ( y ∈ X ) (cid:1) Observe that ϕ encodes whether the set S separates the set X from a designated vertex v f . To take into consideration the cardinalities of the sets S and X , we now describe howto construct an evaluation relation ψ as indicated by Definition 18 of [27]. Following thenotation of [27], given that we have two sets ( S and X ) and two integer input values ( k and t ), choosing m = 1 will result in ψ having four variables: y = X u ∈ S w ( u ) y = X u ∈ X w ( u ) y = k y = t Define the weight function w : V ( G ) → R such that w ( v ) = 1 for each v ∈ V ( G ). Now, let ψ be the evaluation relation ( y = y ) ∧ ( y > y )We have adhered to Definition 18 of [27]. Hence we have created an EMSO expression thatencodes the Firebreak decision problem. The result now follows from Theorem 5.11
Intersection Graphs
Given a family of sets S = { S , S , . . . , S k } , the intersection graph of S is a graph G = ( V, E )for which there exists a bijection f between V and S such that u is adjacent to v in G if andonly if f ( u ) ∩ f ( v ) = ∅ , that is, f ( u ) intersects f ( v ). We say that the bijective assignment andthe family of sets are a representation of G . When we restrict the nature of the representingsets, we can restrict the class of representable graphs, and structured representations haveprovided a wide variety of tractability results (many examples are listed in [20]).Here, we give polynomial-time algorithms for two classes of intersection graphs: subtreeintersection graphs of limited leafage and permutation graphs. In both cases, we use anapproach that sweeps the representation for separators, allowing us to exhaustively checkthese separators for firebreak feasibility. In this section we focus on the intersection graphs of subtrees in a tree of constant boundedleafage, for which we show that the
Firebreak problem can be solved in polynomial time.For our purposes, two subtrees are considered to intersect if they share at least one vertex.The intersection graphs of a tree are the chordal graphs, and the intersection graphs of treeswith a constant bounded number of leaves (the leafage ) can be recognised and a represen-tation constructed in time O ( n ) [21] (where n is the number of vertices in the graph to berepresented). Because these are a subfamily of chordal graphs, for which tw( G ) = ω ( G ) − Theorem 7.
The
Firebreak problem on a graph G = ( V, E ) that is the intersection graphof subtrees of a tree of leafage ℓ can be solved in time O ( | V | ℓ +1 ) .Proof. Let (
G, k, t, v f ) be an instance of the Firebreak problem where G is the intersectiongraph of subtrees T of a tree T = ( V T , E T ) with constant bounded number of leaves ℓ , anddenote by T ( v ) the subtree of T that represents vertex v ∈ V ( G ). For convenience, let n = | V | . Recall that we denote by S the set of k vertices that we remove from the graph G in order to form a firebreak.We make a simplifying assumption that we should not place in S a neighbour of v f thatis adjacent only to other neighbours of v f , as such a vertex being included in S can only everprotect from burning that single vertex and no others; there is always a non-worse choice ofvertex for inclusion in S . Note also that we assume that v f has more than k neighbours, forif not then we apply Lemma 1 to resolve the question in O ( n ) time.We argue that we can find a polynomially-bounded number of useful minimal separators,that any solution to the Firebreak problem will place in S at most a constant boundednumber of them, and that we can check each candidate set of separators efficiently.Given the representation T , which by [21] we can construct in time O ( n ), we know from[25] that there are at most O ( n ) minimal vertex separators in our graph G , and that theycorrespond to the vertices of T : specifically, there is one for each vertex u of T , and it iscomposed of the vertices of G that are represented by subtrees that contain u . Note that(also due to [25]) there are at most O ( n ) vertices in T . Any candidate separating set S that12ill serve as a certificate to an affirmative answer to our problem instance ( G, k, t, v f ) mustbe composed of the union of a set of these minimal separators.There is a unique path from each leaf to the closest vertex of T ( v f ). Let v i , v j be verticeson that path in the tree, and let S i , S j be their corresponding minimal vertex separators in G . Without loss of generality, let v i be closer to T ( v f ) than v j is. Then the set of verticesseparated from v f in G − S i is at least as large as the set of vertices separated from v f in G − ( S i ∪ S j ). Thus in a candidate separating set S we need include only at most oneminimal separator corresponding to a vertex on the unique path from each leaf to the subtree T ( v f ). There are at most ℓ such paths, so there are (cid:0) n ℓ (cid:1) = O ( n ℓ ) possible combinationsof minimal separators to consider when constructing candidate solutions to our probleminstance ( G, k, t, v f ).Given a particular candidate separator S , we can check if it provides a certificate for anaffirmative answer to ( G, k, t, v f ) by checking to see if | S | k , and if the number of verticesseparated from v f in G − S is at least t in O ( n ) time.Thus we can generate all candidate solutions in time O ( n ℓ ), and check the feasibility ofeach in O ( n ), giving an overall running time of O ( n ℓ +1 ).As a special case for leafage ℓ = 2 this argument gives us an algorithm to solve the Firebreak problem in interval graphs in time O ( n ). We can do somewhat better inthis case using a representation-construction algorithm due to Booth and Lueker [11], whogive an O ( | V | + | E | ) algorithm, which, using the reasoning above, we can use to give an O (( | V | + | E | ) ) algorithm. There are a large variety of types of intersection graphs (in fact, every graph is an intersectiongraph of some set of objects). While Theorem 7 applies to intersection graphs of paths ina tree (which we note includes interval graphs), permutation graphs are not a subclass ofthis class of intersection graphs. By using a sweeping-for-separators approach we are ableto show that the
Firebreak problem is tractable on permutation graphs as well. We makeuse of one of the many characterisations of a permutation graph, as in [36]: a permutationgraph is the intersection graph of straight line segments between two parallel lines.We give an example of a permutation graph and a corresponding representation in Fig-ure 2. We will make use of several results on these representations to address the
Firebreak problem on permutation graphs.
Theorem 8.
An instance ( G, k, t, v f ) of the Firebreak problem where G is a permutationgraph on n vertices can be solved in time O ( n k ) .Proof. As noted by Lemma 1, the case in which | N ( v f ) | k can be solved in O ( n ) time, sowe will now consider the case where | N ( v f ) | > k . A permutation graph has a representation in the form of line segments between two parallel horizontal lines, which itself can be createdin O ( n ) time, and in which we can assume without loss of generality that there are novertical line segments [36].This representation gives a partition of vertices: those corresponding to line segments tothe left of the line segment representing v f , those corresponding to line segments to the right13 v v v v v v v v v v v v v v Figure 2: An example of a permutation graph and an associated representation.of the line segment representing v f , and those adjacent to v f ( i.e. , the lines that intersectthe line representing v f ). In this representation there are n line-segment endpoints on thetop horizontal line, and n on the bottom (there is one of each for each vertex). If we segmentthe horizontal lines into portions between line segment endpoints, there are therefore n − cut-line is a line between the two horizontal walls with one endpoint on the top horizontalline and one on the bottom, and we consider one cut-line for each pair of top segment andbottom segment. There are thus O ( n ) such cut-lines, and again due to [10] we know thatevery minimal separator in the permutation graph consists of the vertices corresponding tothe line segments crossed by one of these cut-lines in the representation.Of these O ( n ) minimal separators, there are O ( nk ) that are of interest to us, namelythose that contain at most ( k −
1) vertices; for convenience we call minimal vertex separatorsthat meet this size constraint ( k − -small . Given any ( k − S defined by the cut-line s , let S left be the subset of vertices to the left of s and let S right bethe subset of vertices to the right of s .Consider the following algorithm that searches for firebreaks ( i.e. , k -subsets of V ( G ) thatseparate v f from some other vertices) of the form S ∪ T . Algorithm 2.
1. Find all minimal separators S with v f ∈ S left (resp. S right ) which have size atmost k , and denote this set as L (resp. R ).2. Exhaustively search for all pairs ( S, T ) such that S ∈ L , T ∈ R , | S | + | T | k and | S right | + | T left | > t .If such a pair ( S, T ) is disjoint and | S | + | T | = k , then S ∪ T is a firebreak.Otherwise, consider the component C in G − ( S ∪ T ) containing v f . If | V ( C ) | − | S right | + | T left | − t + | S ∪ T | > k then there is a firebreak as | V ( C ) | − | S right | + | T left | − t verticescan be removed from other components to produce a firebreak of size k byadding them to S ∪ T .3. If no firebreak was found during the exhaustive search, then no firebreak exists.14uppose some firebreak exists but this algorithm found none. There exist t vertices thatcan be separated from v f , possibly some to the left, say L , and some to the right, say R , of v f . Note that at most one of L and R can be empty. Since L and R are separated from v f ,there must be cut-lines between them and v f that define minimal separators and hence thealgorithm must have found a firebreak.Since there are O ( nk ) minimal separators that are ( k − O ( n k ) time. For each pair ( S, T ) we may have to find G − ( S ∪ T ) and count | V ( C ) | , | S right | , and | T left | . This adds a factor of n and thus the problem can be solved in O ( n k ) time. Authors Burgess and Pike acknowledge research grant support from NSERC. Ryan acknowl-edges support from an NSERC Undergraduate Student Research Award.
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