Edge Deletion to Restrict the Size of an Epidemic
aa r X i v : . [ c s . D S ] F e b Edge Deletion to Restrict the Size of anEpidemic
Ajinkya Gaikwad and Soumen Maity
Indian Institute of Science Education and Research, Pune, India [email protected]; [email protected]
Abstract.
Given a graph G = ( V, E ), a set F of forbidden subgraphs,we study F -Free Edge Deletion , where the goal is to remove a min-imum number of edges such that the resulting graph does not containany F ∈ F as a subgraph. For the parameter treewidth, the question ofwhether the problem is FPT has remained open. Here we give a negativeanswer by showing that the problem is W[1]-hard when parameterized bythe treewidth, which rules out FPT algorithms under common assump-tion. Thus we give a solution to a conjecture posted by Jessica Enrightand Kitty Meeks in [Algorithmica 80 (2018) 1857-1889]. We also provethat the F -Free Edge Deletion problem is W[2]-hard when parame-terized by the solution size k , feedback vertex set number or pathwidthof the input graph. A special case of particular interest is the situationin which F is the set T h +1 of all trees on h + 1 vertices, so that we deleteedges in order to obtain a graph in which every component contains atmost h vertices. This is desirable from the point of view of restricting thespread of disease in transmission network. We prove that the T h +1 -FreeEdge Deletion problem is fixed-parameter tractable (FPT) when pa-rameterized by the vertex cover number. We also prove that it admitsa kernel with O ( hk ) vertices and O ( h k ) edges, when parameterized bycombined parameters h and the solution size k . Keywords:
Parameterized Complexity · FPT · W[1]-hard · treewidth · feedback vertex set number Animal diseases pose a risk to public health and cause damage to businesses andthe economy at large. Among different reasons for livestock disease, livestockmovements constitute major routes for the spread of infectious livestock disease[8]. For example, the long-range movement of sheep in combination with localtransmission resulted in the FMD epidemic in the UK in 2001 [8,14]. Livestockmovements could, therefore, provide insight into the structure of the underlyingtransmission network and thus allow early detection and more effective man-agement of infectious disease [11]. To do this, mathematical modelling has beenemployed widely to describe contact patterns of livestock movements and analysetheir potential use for designing disease control strategies [11]. For the purpose of
A. Gaikwad and S. Maity modelling disease spread among farm animals, it is common to consider a trans-mission network with farms as nodes and livestock movement between farms asedges.In order to control or limit the spread of disease on this sort of transmis-sion network, we focus our attention on edge deletion, which might correspondto forbidden trade partners or more reasonably, extra vaccinations or diseasesurveillance along certain trade routes. Introducing extra control of this kindis costly, so it is important to ensure that this is done as efficiently as possible.Many properties that might be desirable from the point of view of restricting thespread of disease can be expressed in terms of forbidden subgraphs: delete edgesso that each connected component in the resulting graph has at most h vertices,is equivalent to edge-deletion to a graph avoiding all trees on h + 1 vertices. Weare therefore interested in solving the following general problem: F -Free Edge Deletion Input:
A graph G = ( V, E ), a set F of forbidden subgraphs and a positiveinteger k . Question:
Does there exist E ′ ⊆ E ( G ) with | E ′ | = k such that G \ E ′ doesnot contain any F ∈ F as a subgraph?A special case of particular interest is the situation in which F is the set T h +1 of all trees on h + 1 vertices, so that we delete edges in order to obtain a graphin which every component contains at most h vertices, so this special case is theproblem T h +1 -Free Edge Deletion . T h +1 -Free Edge Deletion Input:
A graph G = ( V, E ), and two positive integers k and h . Question:
Does there exist E ′ ⊆ E ( G ) with | E ′ | = k such that each connectedcomponent in G \ E ′ has at most h vertices, that is, the graph G \ E ′ does notcontain any tree on h + 1 vertices as a subgraph?A problem with input size n and parameter k is said to be ‘fixed-parametertractable (FPT)’ if it has an algorithm that runs in time O ( f ( k ) n c ), where f issome (usually computable) function, and c is a constant that does not depend on k or n . What makes the theory more interesting is a hierarchy of intractable pa-rameterized problem classes above FPT which helps in distinguishing those prob-lems that are not fixed parameter tractable. Closely related to fixed-parametertractability is the notion of preprocessing. A reduction to a problem kernel, orequivalently, problem kernelization means to apply a data reduction process inpolynomial time to an instance ( x, k ) such that for the reduced instance ( x ′ , k ′ )it holds that ( x ′ , k ′ ) is equivalent to ( x, k ), | x ′ | ≤ g ( k ) and k ′ ≤ g ( k ) for somefunction g only depending on k . Such a reduced instance is called a problemkernel. We refer to [2,3] for further details on parameterized complexity. Our results:
Our main results are the following: – The F -Free Edge Deletion problem is W[1]-hard when parameterizedby treewidth. dge Deletion to Restrict the Size of an Epidemic 3 – The F -Free Edge Deletion problem is W[2]-hard when parameterizedby the solution size k , the feedback vertex set number or pathwidth of theinput graph. . – The T h +1 -Free Edge Deletion problem is fixed-parameter tractable (FPT)when parameterized by the vertex cover number of the input graph. – The T h +1 -Free Edge Deletion problem admits a kernel with O ( hk ) ver-tices and O ( h k ) edges, when parameterized by combined parameters h andthe solution size k . Previous Work: If π is a graph property, the general edge-deletion problem can bestated as follows: Find the minimum number of edges, whose deletion results ina subgraph satisfying property π . Yannakakis [19] showed that the edge-deletionproblem is NP-complete for several common properties, for example, planar,outer-planar, line-graph, and transitive digraph. Watanabe, Ae, and Nakamra[18] showed that the edge-deletion problem is NP-complete if π is finitely charac-terizable by 3-connected graphs. Natanzon, Shamir and Sharan [15] proved theNP-hardness of edge-deletion problems with respect to some well-studied classesof graphs. These include perfect, chordal, chain, comparability, split and aster-oidal triple free graphs. This problem has also been studied in generality underparadigms like approximation [6,13] and parameterized complexity [1,9]. FPTalgorithms have been obtained for the problem of determining whether there are k edges whose deletion results in a split graph [7] and to chain, split, threshold,and co-trivially perfect graphs [9]. Enright and Meeks [4] gave an algorithm forthe F -Free Edge Deletion problem with running time 2 O ( |F| w r ) n where w is the treewidth of the input graph and r is the maximum number of verticesin any element of F . This is a significant improvement on Cai’s algorithm butdoes not lead to a practical algorithm for addressing real world problems. Thespecial case of this problem in which F is the set of all trees on at most h + 1vertices is of particular interest from the point of view of the control of diseasein livestock, and they have derived an improved algorithm for this special case,running in time O (( wh ) w n ). F -Free Edge Deletion parameterized bytreewidth In this section we show that F -Free Edge Deletion is W[1]-hard parameter-ized by treewidth, via a reduction from Minimum Maximum Outdegree . Thuswe give a solution to a conjecture posted by Jessica Enright and Kitty Meeks [4].Let G = ( V, E ) be an undirected and edge weighted graph, where V , E ,and w denote the set of nodes, the set of edges and a positive integral weightfunction w : E → Z + , respectively. An orientation Λ of G is an assignment of adirection to each edge { u, v } ∈ E ( G ), that is, either ( u, v ) or ( v, u ) is containedin Λ . The weighted outdegree of u on Λ is w u out = P ( u,v ) ∈ Λ w ( { u, v } ). We define Minimum Maximum Outdegree problem as follows:
A. Gaikwad and S. Maity
Minimum Maximum Outdegree
Input:
A graph G , an edge weighting w of G given in unary, and a positiveinteger r . Question:
Is there an orientation Λ of G such that w u out ≤ r for each u ∈ V ( G )?It is known that Minimum Maximum Outdegree is W[1]-hard when param-eterized by the treewidth of the input graph [17]. In this section, we prove thefollowing theorem:
Theorem 1.
The F -Free Edge Deletion problem is W[1]-hard when pa-rameterized by the treewidth of the graph. Proof.
Let G = ( V, E, w ) and a positive integer r ≥ I of Minimum Maximum Outdegree . We construct an instance I ′ = ( G ′ , k, F ) of F -Free Edge Deletion the following way. See Figure 1 for an illustration. Foreach edge ( u, v ) ∈ E ( G ), we introduce the following sets of new vertices V uv = { u v , . . . , u vw ( u,v ) } , V ′ uv = { u ′ v , . . . , u ′ vw ( u,v ) } V vu = { v u , . . . , v uw ( u,v ) } and V ′ vu = { v ′ u , . . . , v ′ uw ( u,v ) } . We make u (resp. v ) adjacent to all the vertices in V uv ∪ V ′ uv (resp. V vu ∪ V ′ vu ). Let E u,u v = n ( u, x ) | x ∈ V uv o , E ′ u,u ′ v = n ( u, x ) | x ∈ V ′ uv o , E v,v u = n ( v, x ) | x ∈ V vu o and E ′ v,v ′ u = n ( v, x ) | x ∈ V ′ vu o . Let ω = P e ∈ E w ( e )and N = n + 3 ω + 1. Let δ ( v ; E ) denote the set of edges in E incident to v ∈ V .The weighted degree d w ( v ; G ) of a vertex v ∈ V is defined as P e ∈ δ ( v,E ) w ( e ). Theweighted maximum degree ∆ w ( G ) of G is defined as max v ∈ V d w ( v ; G ). For everyvertex u ∈ V ( G ), we also add a set V (cid:3) u of ∆ w ( G ) − d w ( v ; G ) many one degreevertices. We define two sets of pair of vertices: C = n { ( u ′ vi , v ′ ui ) } | ( u, v ) ∈ E ( G ) , ≤ i ≤ w ( u, v ) o[ n ( u ′ vi , v ′ ui +1 ) , ( u ′ vw ( u,v ) , v ′ u ) | ( u, v ) ∈ E ( G ) , ≤ i ≤ w ( u, v ) − o ,C = n { ( u vi , v ui ) } | ( u, v ) ∈ E ( G ) , ≤ i ≤ w ( u, v ) o[ n ( u vi , v ui +1 ) , ( u vw ( u,v ) , v u ) | ( u, v ) ∈ E ( G ) , ≤ i ≤ w ( u, v ) − o . For every pair of vertices ( u ′ v , v ′ u ) ∈ C , we add a 4 N − P ′ ( u ′ v ,v ′ u ) joining u ′ v and v ′ u , whose internal vertices were not originally part of G . Similarly, for every pair of vertices ( u v , v u ) ∈ C , we add an N length redpath P ( u v ,v u ) joining u v and v u , whose internal vertices were not originally part dge Deletion to Restrict the Size of an Epidemic 5 of G . Now, we define the unweighted graph G ′ as follows: V ( G ′ ) = V ( G ) [ u ∈ V ( G ) V (cid:3) u [ ( u,v ) ∈ E ( G ) ( V uv ∪ V ′ uv ∪ V vu ∪ V ′ vu ) [ ( u ′ v ,v ′ u ) ∈ C V ( P ′ ( u ′ v ,v ′ u ) ) [ ( u v ,v u ) ∈ C V ( P ( u v ,v u ) )and E ( G ′ ) = [ u ∈ V ( G ) { ( u, α ) | α ∈ V (cid:3) u } [ ( u,v ) ∈ E ( G ) E u,u v ∪ E v,v u ∪ E ′ u,u ′ v ∪ E ′ v,v ′ u [ ( u ′ v ,v ′ u ) ∈ C E ( P ′ ( u ′ v ,v ′ u ) ) [ ( u v ,v u ) ∈ C E ( P ( u v ,v u ) )where V ( P ) and E ( P ) denote the set of vertices and edges of P respectively. We a ad ad a ′ d a ′ d da da d ′ a d ′ a b bc bc bc b ′ c b ′ c b ′ c ba b ′ a ccb cb cb c ′ b c ′ b c ′ b cd cd c ′ d c ′ d ddc dc d ′ c d ′ c ab a ′ b a db c
231 2
Fig. 1.
Result of our reduction on a
Minimum Maximum Outdegree instance G with r = 2. The graph G long with its orientation is shown at the left; and G ′ is shown atthe right. Complementary vertex pairs are shown using dashed lines. The vertices ofthe set V △ are filled with red color whereas the vertices of the set V (cid:3) are filled with bluecolor. The vertices in the first part of satisfactory partition ( V , V ) of G ′ are shown inred label and vertices of V are shown in blue label for the given orientation of G . Here ω = 6 and V contains 64 isolated vertices. set k = ω and F = { S ∆ w ( G )+ r +1 , C N +2 } where S ∆ w ( G )+ r +1 is the star graph A. Gaikwad and S. Maity or the complete bipartite graph K ,∆ w ( G )+ r +1 and C N +2 is the cycle of length5 N + 2. We observe that the gadget replacing every edge ( u, v ) ∈ E ( G ) hastreewidth at most eleven because deleting the set n u, u v , u vw ( u,v ) , u ′ v , u ′ vw ( u,v ) , v, v u , v uw ( u,v ) , v ′ u , v ′ uw ( u,v ) o of vertices makes it a forest. This implies that the treewidth of G ′ is a at mosttreewidth of G plus eleven. Now we show that our reduction is correct. That is,we prove that ( G, w, r ) is a yes instance of
Minimum Maximum Outdegree ifand only if I ′ is a yes instance of F -Free Edge Deletion .Let D be the directed graph obtained by an orientation of the edges of G such that for each vertex the sum of the weights of outgoing edges is at most r .We claim that the set of edges E ′ = [ ( u,v ) ∈ E ( D ) (cid:8) ( x, v ) | x ∈ V vu (cid:9) is a solution of I ′ . Clearly, we have | E ′ | = k . We need to show that, f G ′ = G ′ \ E ′ does not contain any forbidden graph as a subgraph. First we show that ev-ery vertex has degree at most ∆ w ( G ) + r in f G ′ . It is clear from constructionthat if x ∈ V ( G ′ ) \ V ( G ) then d f G ′ ( x ) ≤
3. Let w x out and w x in denote the sum ofthe weights of outgoing and incoming edges of vertex x , respectively. Note that d w ( x ; G ) = w x out + w x in and x is adjacent to ∆ w ( G ) − d w ( x ; G ) + w x in + 2 w x outmany vertices in f G ′ . This implies that d f G ′ ( x ) ≤ ∆ w ( G ) + r as w x out ≤ r . There-fore, f G ′ does not contain S ∆ w ( G )+ r +1 as a subgraph. Next, we prove that f G ′ does not contain C N +2 as a subgraph. Suppose, for the sake of contradiction, f G ′ contains C N +2 as a subgraph. We make two cases based on whether thecycle contains some original vertex u from V ( G ) or not. Case 1:
Let us assume that the cycle includes at least one original vertex u ∈ V ( G ). Further, we make two subcases based on whether the cycle con-tains a blue edge or not. Subcase 1.1:
Let us assume that the cycle includes at least one blue edge from ablue path P ′ ( u ′ v ,v ′ u ) . Then the cycle includes all the blue edges of P ′ ( u ′ v ,v ′ u ) andreaches the vertex v ′ u . Without loss of generality, we assume that the directionof edge { u, v } is from u to v in D . Then the edges in E v,v u = n ( v, x ) | x ∈ V vu o are not present in f G ′ . Therefore, the only way to return from v ′ u to u is to takeanother blue path, which makes the length of the cycle at least 8 N − > N + 2.This implies that a cycle of length 5 N + 2 does not exist in this case. Subcase 1.2:
Let us assume that the cycle does not contain any blue edge. Let usassume that the cycle starts at u ∈ V ( G ). In this case, the cycle starts with anedge e ∈ E u,u v for some v ∈ N G ( u ). Next, it must continue with a red edge. Wealso observe that if a cycle includes a red edge from P u v ,v u then it must includeall the red edges of the path and reaches v u . Again, it must take a N length rededge path as edges in E v,v u are not present in f G ′ . In this way, we observe that dge Deletion to Restrict the Size of an Epidemic 7 a path of length 5 N + 1 will end up at a vertex in the set V vu . Since there is nopath of length 1 from a vertex in V vu to u , we show that such a cycle does notexist. Case 2:
If the cycle does not include any original vertex u ∈ V ( G ) then it alsodoes not include any edge from S ( u,v ) ∈ E ( G ) E u,u v ∪ E v,v u ∪ E ′ u,u ′ v ∪ E ′ v,v ′ u . Further,we make two subcases based on whether the cycle contains a blue edge or not. Subcase 2.1:
Let us assume that the cycle contains a blue edge. In this case, weobserve that the length of the cycle is at least 8 N − > N + 2. Subcase 2.2:
We observe that since the blue edges and the original vertices in V ( G ) are not allowed, the cycle must contain only red edges. In this case we canget cycles of even length 2 w ( u, v ) N = 5 N + 2 only.Conversely, suppose E ′ is a solution of the instance I ′ . First, we show thatthe set E ′ must contain exactly one of the following four sets E u,u v , E v,v u , E ′ u,u ′ v or E ′ v,v ′ u for every ( u, v ) ∈ E ( G ). For each edge ( u, v ) ∈ E ( G ), there are 2 w ( u, v )distinct ( u, v ) paths of length 4 N through the blue edges. We call such pathsthe paths of type A. Similarly, for each edge ( u, v ) ∈ E ( G ), there are 2 w ( u, v )distinct ( u, v ) paths of length N through the red edges. We call such paths thepaths of type B. We observe that a combination of type A and type B pathsform a cycle of length 5 N + 2. Therefore, to avoid such a cycle, the solution mustdestroy all the paths of type A or all the paths of type B. Since the maximumnumber of edge-disjoint ( u, v ) path of type A (resp. B) is w ( u, v ), the minimumnumber of edges whose deletion destroys all ( u, v ) paths of type A (resp. B)is w ( u, v ). We must add at least w ( u, v ) many edges to the solution for eachedge ( u, v ) and since k = ω , it implies that the solution E ′ must include exactly w ( u, v ) many edges corresponding to each edge ( u, v ) ∈ E ( G ). Case 1:
Let us assume that the solution is targeting to destroy all the type Bpaths. In that case, we observe that the solution cannot involve any red edgesbecause if we delete a red edge there are still at least w ( u, v ) many edge dis-joint paths of type B left. It implies that solution must contain edges from E u,u v ∪ E v,v u . We first observe that we cannot add edges ( u, u vi ) and ( v, v ui ) forany 1 ≤ i ≤ w ( u, v ) in the solution. As otherwise, we will still have w ( u, v ) − u, v ) paths of type B. However, now we are only allowed todelete w ( u, v ) − u, v ). This is a contradic-tion as we cannot get rid of all the 5 N + 2 length cycles corresponding to theedge ( u, v ). Without loss of generality, we assume that ( u, u v ) is not part of thesolution, that is, we are not deleting ( u, u v ) from the graph G ′ . It forces ( v, v u )and ( v, v u ) to be inside the solution. As ( v, v u ) is part of the solution impliesthat ( u, u v ) is not part of the solution. Again, it will force ( v, v u ) to be part ofthe solution. Applying this argument repetitively, we see that E ′ contains E v,v u and since | E v,v u | = w ( u, v ) implies that no edge from set E u,u v can be part ofthe solution. This shows that E ′ contains either E u,u v or E v,v u . Case 2:
Let us assume that the solution is targeting to destroy all the type Apaths. Using the same arguments, we can prove that E ′ contains either E u,u ′ v A. Gaikwad and S. Maity or E v,v ′ u .We define a directed graph D by V ( D ) = V ( G ) and E ( D ) = n ( u, v ) | E v,v u or E ′ v,v ′ u ⊆ E ′ o [ n ( v, u ) | E u,u v or E ′ u,u ′ v ⊆ E ′ o . Suppose there is a vertex x in D for which w x out > r . In this case, we observethat x is adjacent to more than ∆ w ( G ) + r vertices in graph f G ′ . This is acontradiction as vertex x and its neighbours form the star graph S ∆ w ( G )+ r +1 ,which is a forbidden graph in I ′ . F -Free Edge Deletion parameterized bysolution size In this section we show that F -Free Edge Deletion is W[2]-hard parameter-ized by the solution size k , via a reduction from Hitting Set . In the
HittingSet problem, we are given a universe U = { , , . . . , n } , a family A of sets over U , and a positive integer k . The objective is to decide whether there is a sub-set H ⊆ U of size at most k such that H contains at least one element fromeach set in A . It is known that the Hitting Set problem is W[2]-hard whenparameterized by solution size [2]. We prove the following theorem:
Theorem 2.
The F -Free Edge Deletion problem is W[2]-hard when pa-rameterized by the solution size k , the feedback vertex set number or pathwidthof the input graph. Proof.
Let ( U, A , k ) be an instance I of the Hitting Set problem and let U = { , , . . . , n } . We construct an instance I ′ = ( G, F , k ′ ) of the F -Free EdgeDeletion problem as follows. We first introduce a central vertex v . For every i ∈ U , we attach to this vertex a cycle C i of length 2 i +2. Note that C , C , . . . , C n have only one vertex v in common. We define G as follows V ( G ) = [ i ∈ U V ( C i ) and E ( G ) = [ i ∈ U E ( C i ) . We observe that the graph G contains a unique cycle C i of length 2 i + 2 for each i ∈ U . It is clear that { v } is a feedback vertex set of G . Now, we define a family F of forbidden subgraphs. For every set A ∈ A , we add a graph F A in F , where F A is defined as follows: V ( F A ) = [ i ∈ A V ( C i ) and E ( F A ) = [ i ∈ A E ( C i ) . We take k ′ = k . Next, we show that I is a yes instance if and only if I ′ is a yesinstance. Let H be a solution for the instance I . We see that by deleting onearbitrary edge from every cycle C i , i ∈ H , we can avoid all the forbidden graphs dge Deletion to Restrict the Size of an Epidemic 9 v C C C Fig. 2.
The graph G of the F -Free Edge Deletion problem instance constructedin the reduction of Theorem 2 for n = 3. in F . Therefore, we have a solution E ′ ⊆ E ( G ) for the instance I ′ such that | E ′ | ≤ k ′ .Conversely, suppose E ′ ⊆ E ( G ) with | E ′ | ≤ k is a solution for the instance I ′ . We see that H = { i | E ( C i ) ∩ E ′ = ∅} is a hitting set for the instance I . Wealso observe that | H | ≤ k as | H | ≤ | E ′ | . Corollary 1.
The F -Free Edge Deletion problem is W[2]-hard when pa-rameterized by the feedback vertex set number, pathwidth of the input graph andsolution size even when restricted to planar, outerplanar, bipartite and planarbipartite graphs. In this section, we present an FPT algorithm for the T h +1 -Free Edge Dele-tion problem parameterized by the vertex cover number. A set C ⊆ V ( G ) is avertex cover of G = ( V, E ) if each edge in E has at least one endpoint in C . Inother words, C is a vertex cover of G if and only if I = V \ C is an independentset of G . The size of a smallest vertex cover of G is the vertex cover number of G . Theorem 3.
The T h +1 -Free Edge Deletion problem is FPT when param-eterized by the vertex cover number of the input graph. Proof.
Without loss of generality we assume that the graph has no isolated ver-tices. Let S be a vertex cover of G = ( V, E ) of size k . We denote by I theindependent set V \ S . We partition the independent set I into at most 2 k twinclasses I , I , . . . , I k , where some of them can also be empty. Two vertices u and v are in the same twin class if N ( u ) = N ( v ). Our goal is to minimize the size of E ′ ⊆ E ( G ) such that after deleting E ′ from G , each connected component of theresulting graph has at most h vertices. First, we guess the intersection of S withthe connected components in e G = G \ E ′ . It is clear that the number of guessesis equal to the number of different partitions of the k -element set S , which is equal to the Bell number B k . For every guess, we will reduce our problem to aninteger linear programming (ILP) where the number of variables is a function ofthe vertex cover number k . Since integer linear programming is fixed-parametertractable when parameterized by the number of variables, we will conclude thatour problem is fixed-parameter tractable when parameterized by the vertex covernumber. Let us consider a particular partition P = { S , S , . . . , S ℓ } , ℓ ≤ k , of S .For a given partition P of S , we call an edge a cross edge if both endpoints ofthat edge are in S but one endpoint is in S i and other is in S j such that i = j .We denote the number of cross edges of partition P by cr ( P ). ILP Formulation:
Given a partition P = { S , S , . . . , S ℓ } of S , let C i be thecomponent of e G such that S ∩ C i = S i for 1 ≤ i ≤ ℓ . Let C ℓ +1 be the collectionof size one components in e G such that S ∩ C ℓ +1 = ∅ . For each I i and C j , weassociate a variable x ij that indicates | I i ∩ C j | = x ij , that is, x ij denotes thenumber of vertices in twin class I i that goes to C j . Because the vertices in I i havethe same neighbourhood, the variables x ij determine the components uniquelyand hence determine the required set of edges E ′ ⊆ E ( G ). We add the followingconstraints to ILP. The vertices of each twin class I i is distributed among thecomponents C , C , . . . , C ℓ and C ℓ +1 . Thus we have the following constraints: ℓ +1 X j =1 x ij = | I i | for all 1 ≤ i ≤ k (1)We want each connected component C j in the resulting graph e G has at most h vertices. Thus we have the following constraint: k X i =1 x ij + | S i | ≤ h for all 1 ≤ j ≤ ℓ (2)Note that every vertex in I i has the same set of neighbours in S . Thus if a vertex v ∈ I i goes to C j then we have to remove all edges between v and S \ S j , so that C , C , . . . , C ℓ and C ℓ +1 remains distinct components. Therefore, if x ij verticesof I i go to C j , then we need to remove total | N S \ S j ( v ) | × x ij edges, where v isa vertex in I i . Hence we want to minimize the following objective function: cr ( P ) + k X i =1 ℓ +1 X j =1 | N S \ S j ( v i ) | × x ij (3)where S ℓ +1 = ∅ , cr ( P ) is the number of cross edges of partition P and v i is avertex in the twin class I i . Solving the ILP:
Lenstra [12] showed that the feasibility version of p -ILP isFPT with running time doubly exponential in p , where p is the number of vari-ables. Later, Kannan [10] proved an algorithm for p -ILP running in time p O ( p ) .In our algorithm, we need the optimization version of p -ILP rather than thefeasibility version. We state the minimization version of p -ILP as presented by dge Deletion to Restrict the Size of an Epidemic 11 Fellows et. al. [5]. p -Variable Integer Linear Programming Optimization ( p -Opt-ILP) :Let matrices A ∈ Z m × p , b ∈ Z p × and c ∈ Z × p be given. We want to find avector x ∈ Z p × that minimizes the objective function c · x and satisfies the m inequalities, that is, A · x ≥ b . The number of variables p is the parameter. Thenthey showed the following: Lemma 1. [5] p -Opt-ILP can be solved using O ( p . p + o ( p ) · L · log ( M N )) arith-metic operations and space polynomial in L . Here L is the number of bits in theinput, N is the maximum absolute value any variable can take, and M is anupper bound on the absolute value of the minimum taken by the objective func-tion.In the formulation for T h +1 -Free Edge Deletion problem, we have atmost 2 k ( k + 1) variables. The value of objective function is bounded by n andthe value of any variable in the integer linear programming is bounded by n . Theconstraints can be represented using at most O (2 k log n ) bits. Lemma 1 impliesthat we can solve the problem with the guess P in FPT time. There are at most B k choices for P , and the ILP formula for a guess can be solved in FPT time.Thus Theorem 3 holds. k and h In this section we give a kernelization algorithm for the T h +1 -Free Edge Dele-tion problem based on a reduction rule. For a given instance ( G, k, h ) of the T h +1 -Free Edge Deletion problem if G has a component of size at most h ,then its removal does not change the solution. This shows that the following ruleis safe. Reduction 1: If G contains a component C of size at most h , then delete C from G , the new instance is ( G − C, k, h ).This leads to the following lemma.
Lemma 2. If ( G, k, h ) is a yes-instance and Reduction rule 1 is not applicableto G , then | V ( G ) | ≤ kh and | E ( G ) | ≤ kh + k .Proof. Because we cannot apply Reduction rule 1, G has no components of sizeat most h . Since ( G, k, h ) is a yes-instance, there is a subset E ′ ⊆ E ( G ) suchthat | E ′ | = k and every component of G \ E ′ has at most h vertices. If we putback the k edges, as one edge can join two components, k edges of E ′ can join atmost 2 k components. This implies that the number of connected components of G is bounded by 2 k . As there are at most 2 k components, we get | V ( G ) | ≤ kh .Since each component can have at most h edges and there are 2 k components,we get | E ( G ) | ≤ kh + k . Finally, we remark that the Reduction rule 1 is applicable in O ( V + E ) time.Thus we obtain the following theorem Theorem 4.
The T h +1 -Free Edge Deletion problem admits a kernel with O ( hk ) vertices and O ( h k ) edges. The main contributions in this paper are that the F -Free Edge Deletion problem is W[1]-hard when parameterized by treewidth; it is W[2]-hard whenparameterized by the solution size, pathwidth or feedback vertex set number;the T h +1 -Free Edge Deletion problem is FPT when parameterized by ver-tex cover number; and it is FPT when parameterized by combined parameters k and h . We list some nice problems emerge from the results here: does T h +1 -Free Edge Deletion admit a polynomial kernel in vertex cover? Also, notingthat the problem is FPT in vertex cover, it would be interesting to considerthe parameterized complexity with respect to twin cover. The modular widthparameter also appears to be a natural parameter to consider here. The parame-terized complexity of the problem remains unsettle when parameterized by otherimportant structural graph parameters like clique-width. As mentioned in [4],one problem of practical relevance to epidemiology would be the complexity ofthe problems on planar graph; this would be relevant for considering the spreadof a disease based on the geographic location of animal holdings. References
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A Preliminaries
Unless otherwise stated all graphs are simple, undirected, and loopless. For graph G = ( V, E ), V = V ( G ) is the vertex set of G , and E = E ( G ) the edge set of G .We now recall some graph parameters used in this paper. The graph parameterswe explicitly use in this paper are feedback vertex set and treewidth. Definition 1.
A feedback vertex set in an undirected graph G is a subset of ver-tices whose removal results in an acyclic graph. The minimum size of a feedbackvertex set in G is the feedback vertex set number of G , denoted by fvc ( G ).We now review the concept of a tree decomposition, introduced by Robertsonand Seymour in [16]. Definition 2. A tree decomposition of a graph G is a pair ( T, { X t } t ∈ V ( T ) ),where T is a tree and each node t of the tree T is assigned a vertex subset X t ⊆ V ( G ), called a bag, such that the following conditions are satisfied:1. Every vertex of G is in at least one bag.2. For every edge uv ∈ E ( G ), there exists a node t ∈ T such that bag X t contains both u and v .3. For every u ∈ V ( G ), the set { t ∈ V ( T ) | u ∈ X t } induces a connected subtreeof T . Definition 3.