On the Parameterized Complexity of the Expected Coverage Problem

The Maximum Covering Location Problem (MCLP) is a well-studied problem in the field of operations research. Given a network with positive or negative demands on the nodes, a positive integer k, the MCLP seeks to find k potential facility centers in the network such that the neighborhood coverage is maximized. We study the variant of MCLP where edges of the network are subject to random failures due to some disruptive events. One of the popular models capturing the unreliable nature of the facility location is the linear reliability ordering (LRO) model. In this model, with every edge e of the network, we associate its survival probability 0 ≤ pe ≤ 1, or equivalently, its failure probability 1 − pe. The failure correlation in LRO is the following: If an edge e fails then every edge e′\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$e^{\prime }$\end{document} with pe′≤pe\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$p_{e^{\prime }} \leq p_{e}$\end{document} surely fails. The task is to identify the positions of k facilities that maximize the expected coverage. We refer to this problem as Expected Coverage problem. We study the Expected Coverage problem from the parameterized complexity perspective and obtain the following results. 1. For the parameter pathwidth, we show that the Expected Coverage problem is W[1]-hard. We find this result a bit surprising, because the variant of the problem with non-negative demands is fixed-parameter tractable (FPT) parameterized by the treewidth of the input graph. 2. We complement the lower bound by the proof that Expected Coverage is FPT being parameterized by the treewidth and the maximum vertex degree. We give an algorithm that solves the problem in time 2O(twlogΔ)nO(1)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$ 2^{{\mathcal {O}}({\textbf {tw}} \log {\varDelta })} n^{{\mathcal {O}}(1)}$\end{document}, where tw is the treewidth, Δ is the maximum vertex degree, and n the number of vertices of the input graph. In particular, since Δ ≤ n, it means the problem is solvable in time nO(tw)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$ n^{{\mathcal {O}}({\textbf {tw}})} $\end{document}, that is, is in XP parameterized by treewidth.


Introduction
The MAXIMUM COVERING LOCATION PROBLEM (MCLP) is a well-studied problem in the field of operations research [8]. Given a network with demands on the nodes, a positive integer budget k, the MCLP asks to find k potential facility centers in the network such that the neighborhood coverage is maximized. We are interested in investigating the unreliable nature of the MCLP. Unreliability is introduced by associating survival probabilities on the edges of the input network. The notion of unreliability is used in disaster management, surviving network design and influence maximization. Assume that the network is subjected to a disaster event. During the course of disaster, some link may become non-functional. This yields a structural change in the underlying graph of the network. The resulting graph is an edgeinduced subgraph of the original graph. In certain cases, the resulting graph can have multiple connected components. The real challenge is to place a limited number of potential facility centers a priori such that the expected coverage after an event of disaster is maximized. See [9, 13-15, 19, 32] for further references on unreliable MCLP.
In this paper, we study the following model of the MCLP with edge failures. Let G = (V , E, w) be a vertex weighted underlying graph of the MCLP. On each edge e ∈ E, let p e > 0 be the survival probability associated with e such that the edge e can survive in the network with probability p e . Under the assumption that edges fail independently, the input graph can be rendered into one of 2 m edge subgraphs called realization, where m is number of edges in the graph. Each realization will have a non-zero probability of occurrence. Since the number of realizations is exponential and many of them occur with close to zero probability, Hassin et al. [26,27] formulated a dependency model for edge failure in unreliable facility networks called linear reliability ordering (LRO). In LRO model, for each pair of edges e = e ∈ E, p(e) = p(e ), and for any pair of edges e i and e j with p e i > p e j , the Pr[e j fails | e i fails ] = 1. More precisely, if an edge e fails then every edge e with p e < p e surely fails. The LRO model is defined on graphs with distinct edge probabilities. It is clear that, in this model, we have exactly m + 1 edge subgraphs. We consider the LRO model with a relaxation that the edges can have the same probability. If the probabilities of two edges are the same, then either both or neither will survive. In this case, the number of subgraph realization will be at most m + 1.
While in most articles dealing with maximum coverage problems the weights are assumed to be positive, there are situations when the weights can be negative. Such mixed-weight coverage problems are useful for modeling situations when some of the demand nodes are obnoxious and their inclusion in the coverage area may be detrimental [4,5]. Nodes with a negative demand are nodes we do not wish to cover. If a node has negative demand, then we wish to cover as little as possible. For example, opening a new facility (grocery store) close to many positive weighted modes (customers) seems as an excellent opportunity but the proximity of a big supermarket (a neighbor with negative weight) could decrease the expected profit.
Problem Statement Let G = (V , E) be a vertex-weighted undirected graph with a weight function w : V → R and a probability function p : E → Q [0, 1] , and k be a positive integer. Assume that the edges are ordered using p in descending order. That is, p 1 > p 2 > · · · > p m . In the LRO model, let G 0 G 1 · · · G m be the linear ordering of the realizations of G, that occur with probability P (G i ) for 0 ≤ i ≤ m. The value of P (G i ) can be written as follows: The EXPECTED COVERAGE problem asks to find a k-sized vertex set S such that the expected coverage by S on the distribution {G i | 0 ≤ i ≤ m} is maximized. We use the expected coverage function C defined by Narayanaswamy et al. [33]. Given a pair of sets S, T ⊆ V , the expected coverage of T by S is Further, if S or T is a singleton set, we just write the element of the set instead of the set notation. An instance of the optimization version of the EXPECTED COVERAGE problem is denoted by the tuple (G, w, p, k). The decision version of the problem is defined as follows.
An instance of the decision version of the EXPECTED COVERAGE problem is denoted by the tuple (G, w, p, k, t).

Related Works
The facility location problems can take many forms, depending on the objective function. In the most facility location problems, the objective function focuses on comforting the clients. For example, in the k-center problem, the goal is minimizing the maximum distance of each client from its nearest facility center [7]. The facility location problem has received a good deal of attention in the parameterized perspective [1,6,20,21].
The MCLP with edge failure is studied with various constraints. Eiselt et al. [19] considered the problem with a single edge failure. In this case, exactly one edge would have failed after a disaster and the objective is to place k facility centers such that the expected weight of non-covered vertices is minimized. If the number of facility centers is k = 1, and the facility center can cover all the vertices in the connected component, then the problem is studied as MOST RELIABLE SOURCE (MRS) problem. In this problem, the edges fail independently. The MRS problem has received a good deal of attention in literature [9,13,15,32]. Hassin et al. [26] studied the problem with edge failure follows LRO failure model. The problem is referred as MAX-EXP-COVER-R problem. An additional input radius of coverage R is also given such that any facility center can cover a vertex at distance at most R. The MAX-EXP-COVER-R problem is shown to be NP-hard even when R = 1. When R = ∞ (it is sufficient to say R > n), the problem is polynomial time solvable [26].
In the BUDGETED DOMINATING SET problem, we are given a graph G and a positive integer k, and asked to find a set of at most k vertices S maximizing the value w (N[S]) in G. Set theoretic version of the BDS problem is studied as budgeted maximum coverage in [28,29]. The EXPECTED COVERAGE problem can be viewed as a generalization of the BDS problem. When we have probability 1 on all the edges, then both these problems are the same. The BDS problem generalizes PARTIAL DOMINATING SET (PDS) problem, where one seeks a set of size at most k vertices dominating at least t vertices [31]. Of course, all these problems also generalize the fundamental DOMINATING SET problem, where the task is to find a set of at most k vertices dominating all remaining vertices of the graph.
The DOMINATING SET problem parameterized by k (solution size) on general graphs is W[2]-hard [16]. However, on planar graphs it is FPT [25]. Moreover, on planar, and more generally on H -minor-free graphs it is solvable in sub-exponential time [3,11]. It also admits a linear kernel on planar graphs, H -minor-free graphs and graphs of bounded expansion [2,18,23,24,34]. Sub-exponential parameterized algorithm for the PDS problem on planar graphs, and more generally, apex-minorfree graphs, was given in [22].
On graphs of bounded treewidth, the classical dynamic programming, see e.g. [10], shows that the DOMINATING SET problem is FPT parameterized by the treewidth of an input graph. The FPT algorithm for the DOMINATING SET problem can be adapted to solve the BDS problem in FPT time. Further, when we have mixed vertex weights on the BDS problem, the above modified algorithm will work. Narayanaswamy et al. [33] gave an FPT algorithm parameterized by treewidth of the input graph to solve the EXPECTED COVERAGE problem with non-negative weights. 1

Our Results
Since the EXPECTED COVERAGE problem (with mixed-weights) generalizes both the BDS problem and the EXPECTED COVERAGE problem with non-negative weights, it is also natural to ask what algorithmic results for these problems can be extended to the EXPECTED COVERAGE problem. We obtain the following results.
1. For the parameter pathwidth, we show that the EXPECTED COVERAGE problem is W[1]-hard. Moreover, the problem remains W[1]-hard for any combination of parameters pathwidth pw, solution size k and value of coverage t. This is interesting because as it was shown by Narayanaswamy et al. [33], the variant of the problem with only non-negative weight is FPT parameterized by the treewidth. Thus the results for non-negative weights cannot be (unless FPT=W [1]) extended to the mixed-weight model.

2.
We complement the lower bound by the proof that EXPECTED COVERAGE is FPT being parameterized by the treewidth and the maximum vertex degree. We give an algorithm that solves the problem in time 2 O(tw log Δ) n O (1) , where tw is the treewidth, Δ is the maximum vertex degree, and n the number of vertices of the input graph. In particular, since Δ ≤ n, it means the problem is solvable in time n O(tw) , that is, is in XP parameterized by treewidth.

Preliminaries
We recall in this section some notations and definitions used throughout this article. . Other than this, we follow standard graph theoretic notations based on Diestel [12].
A tree decomposition of an undirected graph G = (V , E) is a pair (T, X) where T is a tree whose vertices are called nodes and X = {X i ⊆ V | i ∈ V (T)} such that 1. for each vertex u ∈ V , there is a node i ∈ V (T) such that u ∈ X i , 2. for each edge uv ∈ E, there is a node i ∈ V (T) such that u, v ∈ X i , and 3. for each vertex v ∈ V , the set {i ∈ V (T) | v ∈ X i } forms a subtree of T .
The width of a tree decomposition (T, X) equals max i∈V (T) |X i | − 1. The treewidth of a graph G is the minimum width over all tree decompositions of G. For a node i ∈ V (T), let T i be the subtree rooted at i and X i + = ∪ j∈V (T i ) {X j }. The graph induced by the vertices X i + is G[X i + ] and it is denoted by G i . A tree decomposition (T, X) is said to be a path decomposition if T is a path. The pathwidth of a graph G is minimum width over all possible path decompositions of G. Let pw(G) and tw(G) denote the pathwidth and treewidth of the graph G, respectively.
We give a dynamic programming algorithm working on a so-called nice tree decomposition of the input graph G. A tree decomposition (T, X) is a nice tree decomposition if T is rooted by a node r with X r = ∅ and every node in T is either an insert node, forget node, join node or leaf node. Thereby, a node i ∈ V (T) is an insert node if i has exactly one child j such that X i = X j ∪ {v} for some v / ∈ X j ; it is a forget node if i has exactly one child j such that X i = X j \ {v} for some v ∈ X j ; it is a join node if i has exactly two children j and h such that X i = X j = X h ; and it is a leaf node if X i = ∅. Given a tree decomposition of width tw, a nice tree decomposition of width tw can be obtained in linear time [30]. For a node i ∈ V (T), let T i be a subtree rooted at i and X i + = ∪ j∈V (T i ) {X j }. We will also use the parameters vertex cover, feedback vertex set and distance to star forest of a graph G. For a graph G, by vc(G) we mean the size of minimum vertex set whose deletion leaves the graph edgeless, and by fvs(G) we mean the size of the minimum vertex set whose deletion leaves the graph acyclic. In this article, we consider that the trivial graph structure is a forest that consists of only star trees. Then, for a graph G, by dsf(G) we mean the size of the minimum vertex set whose deletion leaves the graph disjoint union of star trees (a forest of stars). For all these structural properties, we will omit G if it is clear from the context.
We refer to the recent books of Cygan et al. [10] and Downey and Fellows [17] for detailed introductions to parameterized complexity.

Parameterized Intractability: The EXPECTED COVERAGE problem is W[1]-hard for the Parameter Pathwidth
In this section, we show that the EXPECTED COVERAGE problem is W[1]-hard parameterized by the pathwidth. We reduce from the MULTI-COLORED CLIQUE problem which is defined as follows. Given a k- , and a positive integer k, the MULTI-COLORED CLIQUE problem seeks to decide whether there exists a k-clique with exactly one vertex from each part.

Construction
Given an instance (G, k) of the MULTI-COLORED CLIQUE problem, we construct an . Now we describe the construction of the graph H , the function w : V (H ) → Q and the probability function p : For each i ∈ [k], we construct a vertex-partition gadget H i corresponding to the vertex partition V i as follows. For each vertex u ∈ V i , add a vertex a u with w(a u ) = 0 in the gadget H i . Add two more vertices t i with w(t i ) = k 2 , and q i with w(q i ) = k 2 to the gadget H i . For each vertex u ∈ V i , the vertex a u is made adjacent to the vertices t i and q i . For each edge e ∈ E(H i ), we define the survival probability p(e) = 1. Thus, the gadget H i has |V i | + 2 vertices and 2|V i | edges.
For each 1 ≤ i < j ≤ k, we construct an edge-partition gadget H i,j corresponding to the edge partition E i,j as follows. For each edge e ∈ E i,j , add a vertex a e with w(a e ) = 0 in the gadget H i,j . Add two more vertices t i,j and q i,j with w(t i,j ) = k 2 = w(q i,j ) = k 2 to the gadget H i,j . For each edge e ∈ E i,j , the vertex a e is made adjacent to the vertices t i,j and q i,j . For each edge e ∈ E(H i,j ), we define the survival probability p(e) = 1. Thus, the gadget H i has |E i,j | + 2 vertices and 2|E i,j | edges.
Next, we introduce connector vertices to connect the edge-partition gadgets and vertex-partition gadgets.
To establish the edges between the gadgets and the connector vertices, we define a probability function z : V → Q [0,1] such that for any two vertices x, y ∈ V with x = y, z(x) = z(y). For each i ∈ [k], the gadget H i is connected to the set R as follows. For each vertex u ∈ V i and for each j ∈ [k] with j = i, -if i < j then the vertex a u ∈ H i is made adjacent to the vertices s i i,j and r i i,j with survival probabilities z(u) and 1 − z(u), respectively, and -if j < i then the vertex a u ∈ H i is made adjacent to the vertices s i j,i and r i j,i with survival probabilities z(u) and 1 − z(u), respectively.
respectively. An illustration of a vertex-partition gadget and an edgepartition gadget connected to the connector vertices is given in Fig. 1. For clarity, we denote the vertices a u and a e in V (H ) for each u ∈ V and e ∈ E, as selection vertices. Thus, the graph H is constructed with N = n + m + 3k 2 − k vertices and M = 2kn + 6m edges.

Lemma 1 For each i ∈ k, the pathwidth of the gadget H i is two.
Proof We observe that the removal of the vertex t i from H i results a star tree. It is known that the pathwidth of a star tree is one. Let (T, X ) be a path decomposition of Similarly, for each 1 ≤ i < j ≤ k, pw(H i,j ) = 2. We bound some structural properties of the graph H in the following lemma.
Lemma 2 Some structural properties of the graph H are as follows: We prove each of the structural parameters mentioned above as follows.
(a) If we remove R from the graph H , then the resulting graph is a collection of disjoint vertex-partition gadgets and edge-partition gadgets. From Lemma 1, the pathwidth of a gadget is two. Let (T, X ) be a path decomposition of the graph H − R with pathwidth two. Thus, adding the vertex set R to all bags of (T, X ) gives a path decomposition for the graph H with pathwidth at most |R| + 2 = 4 k 2 + 2. (b) If we remove the set T ∪ Q ∪ R from the graph H , then the resulting graph is edgeless. Thus, the set T ∪ Q ∪ R is a vertex cover for H . Therefore, vc(H ) ≤ |T | + |Q| + |R| = 3k 2 − k. (c) If we remove the set T ∪ R from the graph H , then the resulting graph is a forest. Thus, the set T ∪ R is a feedback vertex set for H . Therefore, fvs(H ) ≤ |T | + |R| = 5 k 2 + k. (d) If we remove the set T ∪R from the graph H , then the resulting graph is a forest that consists of only star trees. Therefore, dsf(H ) ≤ |T | + |R| = 5 k 2 + k.

Properties of a feasible solution for the instance (H , w, p, k , t ) of the EXPECTED COVERAGE problem
Observe that any vertex in H can achieve an expected coverage of value at most 2k 2 . In particular, for each u ∈ V (or e ∈ E), the selection vertex a u (or a e ) can achieve an expected coverage of value at most 2k 2 . If a vertex u ∈ V (H ) is not a selection vertex, then u can achieve an expected coverage of value at most k 2 . In the following lemma, we show that the set S consists of only selection vertices.

Lemma 3 Every vertex in the set S is a selection vertex.
Proof We prove this by contradiction. Assume that there exists a vertex u ∈ S where u is not a selection vertex. We know that C(V (H ), u) ≤ k 2 . Then, the expected coverage by the set S is given as follows: This contradicts the feasibility of the set S. Therefore, every vertex in the set S is a selection vertex.
Then, we show that the set S has a non-empty intersection with each gadget in the graph H .

Lemma 4 For each
Proof By construction of the graph H , the vertex-partition gadgets and edgepartition gadgets are disjoint and connected through connector vertices. By contradiction, assume that there exists a gadget with no vertex from the gadget is in S. Since there are k 2 + k gadgets, at least one gadget should have two vertices from the set S. For any gadget, the expected coverage contribution by the vertices in the gadget is at most 2k 2 even if the gadget has more than one vertex from S. Then we have the following: This contradicts the feasibility of the set S. Therefore, the set S has one selection vertex in each gadget.
The Lemmas 3 and 4 together state that for each i ∈ [k], there exists a vertex v ∈ V i such that a v ∈ S, and for each 1 ≤ i < j ≤ k, there exists an edge e ∈ E i,j such that a e ∈ S. Lemma 5 For each 1 ≤ i < j ≤ k, let a u and a xy be the selection vertices in the set S for some u, x ∈ V i and y ∈ V j . Then, the expected coverage of the vertices {s i i,j , r i i,j } by {a u , a xy } is given as follows.

C(r
To maximize the expected coverage, we need the coverage of r i i,j and s i i,j by the pair of vertices a u and a xy is as maximum as possible. This implies the following corollary from Lemma 5.

Equivalence
Now we show the equivalence of both the problems. More precisely, the graph G has a k-clique if and only if H has a k sized vertex set that achieves the expected coverage of value at least t .

Lemma 6 If (G, k) is a YES-instance of the
We apply the Corollary 1 in the second step to replace the exact value of expected coverage. Thus, we showed that C(V (H ), S) = k 4 + k 3 − k 2 + k = t . Therefore, the set S is a feasible solution to the instance (H, w, p, k , t ) of the EXPECTED COVERAGE problem.
Now we prove the other direction of equivalence.

Lemma 7 If (H, w, p, k , t ) is a YES-instance of the EXPECTED COVERAGE problem then (G, k) is a YES-instance of the MULTI-COLORED CLIQUE problem.
Proof Let S be a feasible solution to the instance (H, w, p, k , t ) of the EXPECTED COVERAGE problem. The feasibility of S ensures that every gadget has a selection vertex from the set S. More specifically, each gadget contributes an expected coverage of 2k 2 . Then C(V (H ) \ R, S) = k (2k 2 ) = k 4 + k 3 since S is a feasible solution.
There are 2 k 2 pairs of s i i,j and r i i,j connector vertices in H . By Lemma 5, each pair can contribute at most −1. Then, the value k − k 2 can be achieved only when each pair is contributing exactly −1. From Corollary 1, for each 1 ≤ i < j ≤ k, the pair r i i,j and s i i,j together can contribute exactly −1 when a u ∈ S and a uy ∈ S for some u ∈ V i and y ∈ V j . By construction of H , there is an edge between the vertices u and y in G. Let K = {u ∈ V i | a u ∈ S} be a k sized vertex set from V . For every pair of distinct vertices in K will have en edge between them in G and thus form a k-clique in G.
Thus, we state the following theorem using the Lemmas 2, 6 and 7.

Theorem 2 The EXPECTED COVERAGE problem is W[1]-hard for the parameter pathwidth.
Proof Given an instance (G, k) of the MULTI-COLORED CLIQUE problem, the instance (H, w, p, k , t ) is constructed in polynomial time where k = k + k 2 and t = k 4 + k 3 − k 2 + k. From Lemma 2, we know that pw(H ) is quadratic function of k. Finally, Lemmas 6 and 7 it follows that the instance (H, w, p, k , t ) of the EXPECTED COVERAGE problem output by the reduction is equivalent to the instance (G, k) of the MULTI-COLORED CLIQUE problem that was input to the reduction. Since the MULTI-COLORED CLIQUE problem is W[1]-hard for the parameter k, it follows that the EXPECTED COVERAGE problem is W[1]-hard for the parameter pathwidth of the input graph.
Moreover, the parameterized reduction preserves the parameters k , t and pathwidth of the constructed graph as a functions of k. That is, k = k + k 2 , t = k 4 + k 3 − k 2 + k and pathwidth of the graph H is O(k 2 ). Further, observe that the number of negative demand vertices in the reduced graph is 4 k 2 = O(k 2 ). Thus, we conclude the section with following corollary.

FPT Algorithm for the EXPECTED COVERAGE problem Parameterized by Treewidth on Bounded Degree Graphs
While, as we have seen, the EXPECTED COVERAGE problem is W[1]-hard for the parameter pathwidth of the input graph, we complement the lower bound with an FPT algorithm for the combined treewidth and maximum vertex degree parameters.
Let (G, w, p, k) be an input to the optimization version of the EXPECTED COV-ERAGE problem. Let (T, X) be a nice tree decomposition of G with treewidth tw. Narayanaswamy et al. [33] introduced the notion of best neighbor to solve the EXPECTED COVERAGE problem with non-negative weights on bounded treewidth graphs. Consider a set S ⊆ V of size k. In the LRO model, for each vertex u (with u ∈ N[S]), there exists a unique vertex in S called best neighbor of u in S, denoted by bn(u, S) such that C(u, S) = C(u, bn(u, S)). Definition 1 (Narayanaswamy et al. [33]) Given a vertex u and a set S with u ∈ N[S], by bn(u, S) we denote the best neighbor of u in S defined as follows: If u / ∈ N[S] then bn(u, S) is undefined. We use the fact that the graph G has bounded maximum degree. We define a structural ordering called neighborhood indexing on the neighborhood of each vertex. This LRO specific intuitions "best neighbor" and "neighborhood indexing", help us to solve the problem efficiently in tree decomposition.

Solution structure
For each node i in T, we compute two tables Sol i and Val i . The rows of both tables are indexed by 3-tuple which we refer to as states. Let S i denote the set of all states associated with node i. For a state s ∈ S i , the DP formulation gives a recursive definition of the values Sol i [s] and Val i [s]. Sol i [s] is a set S ⊆ X i + , an optimal solution for the instance specified by the state s.
where α is an element of the state s defined below. A state s ∈ S i is a tuple ( , α, β), where -0 ≤ ≤ k is an integer and specify the size of Sol i [s], -α : X i → {0, 1} is an indicator function for the vertices in X i . This specifies the constraint that whether a vertex in X i is considered for the coverage. -β : X i → {−1, 0, 1, . . . , Δ} is a function. This specifies the constraint that ).
An instance of the EXPECTED COVERAGE problem at the state s is (G i , w i , p i , ) where the functions w i and p i are obtained from w and p for the domain X i + and E(G i ), respectively. Additionally, the solution should satisfy the constraints specified by the state s. A state s is said to be invalid if there is no feasible solution that satisfies the constraints specified by s. We define one more notion of validity of the states called "locally valid". A state s is said to be locally valid if the following properties are satisfied.
If a state s is not locally valid, then s is an invalid state.
State induced at a node in T For any set S ⊆ X i + of size at most k and a function α :  Proof The correctness follows from the fact that the graph G i is a null graph. Thus, for a null graph a valid state s, empty set with coverage value zero is the only optimal solution.
Introduce Node Let i be an introduce node with child j such that We define the state s j , and define Sol i [s] in terms of Sol j [s j ]. To define the state s j , we consider two cases depend on the value of β(v).
Case β(v) = 0 In this case, the desired solution for the state s must not contain the vertex v. We define the functions α j : X j → {0, 1} and β j → {−1, 0, 1, . . . , Δ} from α and β by excluding the vertex v in the domains of both functions, respectively. We define s j = ( , α j , β j ). If the state s j at the node j is invalid, then the state s at the node i is invalid. Therefore, we consider that the sate s j is valid. Then, the solution for the state s is defined as follows: and Case β(v) = 0 In this case, the desired solution for the state s must contain the vertex v. We have α(v) = 1 since s is locally valid. The key idea is to figure out the vertices those will have v as its best neighbor in the desired solution. Then, find a suitable optimal state at node j and complete the bottom-up computation. Next we find set of all vertices that will have v as its best neighbor in the desired solution. Let We enumerate all possible subsets of D v to find such a set. For each D ⊆ D v , we define the following: For a set D ⊆ D v and a function f ∈ F D , we define the following: ).
An optimal set D ⊂ D v and f ∈ F D can be computed as follows: If no such valid state found in j, then we mark s in i is also invalid.
Then, the solution for the state s is defined as follows: and The time to compute the state s j is depending on the size of F. Therefore, s j can be computed in O * (Δ tw ) time.

Lemma 9
Let i be an introduce node in T, and let s and s j be as defined above. If S is an optimal solution for the state s, then there exists a stateŝ at the node j such that the set S \ {v} is an optimal solution for the stateŝ, and Val j [s j ] = Val j [ŝ].
Proof Since S is an optimal solution to the state s, the pair (S, α) induces the state s at the node i. We consider two cases based on whether v ∈ S or not. First consider the case v ∈ S. In this case, observe that β(v) = 0. Consider the set D = {u ∈ X i | bn(u, S) = v} and the function α D j as defined above for the case β(v) = 0.
Then, the expected coverage C i (X i + \ α −1 (0), S) can be written as follows: Note that the term C(D, v) + w(v) is independent of the set S. Thus, the set S \ {v} is an optimal solution for the stateŝ at the node j.
Next we consider the case v / ∈ S. In this case, observe that β(v) = 0 and S \ {v} = S. Consider the α j as defined above for the case β(v) = 0. Letŝ = ( , α j ,β j ) ∈ S j be the state induced by the set S at the node j. Observe that the functions β j andβ j are same. Thus,ŝ and s j are same. Then, the expected coverage C(X i + \ α −1 (0), S) can be written as follows: Thus, the set S is an optimal optimal solution for the stateŝ at the node j.
Forget Node Let i be a forget node with child j such that X i = X j \ {v} for some v ∈ X j . Since i is a forget node, N(v) ⊆ X i + . We define the state s j = ( , α j , β j ) and define Sol i [s] in terms of Sol j [s j ]. The state s does not impose any constraint on v since v / ∈ X i . So, we try all possible values of α and β for v, and find an optimal one. We define α j : X j → {0, 1} such that for each u ∈ X j , Therefore, we consider all possible values of β j (v) to define the state s j . For each For each z ∈ {0, 1, . . . , deg(v), we define s z j = ( , α j , β z j ). If for each z ∈ {0, 1, . . . , deg(v)}, the state s z j at the node j is invalid then the state s at the node i is invalid. Therefore, we consider that there exists a z ∈ {0, 1, . . . , deg(v)} such that the state s z j at the node j is valid. Let Define s j = s z j = ( , α j , β z j ). Then, the solution for the state s is defined as follows: and The state s j can be computed in O(Δ) time.

Lemma 10
Let i be a forget node in T, and let s and s j be as defined above. If S is an optimal solution for the state s, then there exists a stateŝ at the node j such that the set S is an optimal solution for the stateŝ, and Val j [s j ] = Val j [ŝ].
Proof Since S is an optimal solution for the state s, the pair (S, α) induces the state s at the node i. Consider the function α j as defined above. Letŝ = ( , α j ,β j ) be the state induced by the pair (S, α j ) at the node j. Let z =β j (v). Note that the functions β z j andβ j are same, and thus the statesŝ and s z j are same. From (6) and the optimality of the set S for the state s, Val j [ŝ] = Val j [s j ]. Then, the expected coverage C(X i + \ α −1 (0), S) can be written as follows: Thus, the set S is an optimal solution for the stateŝ at the node j.
Join Node Let i be a join node with children j and h such that X i = X j = X h . We define two states s j = ( j , α j , β j ) and s h = ( h , α h , β h ) at nodes j and h, respectively, and define Sol i [s] in terms of Sol j [s j ] and Sol h [s h ]. Among vertices in the solution to be computed, |β −1 (0)| many vertices are taken from X i and −|β −1 (0)| many vertices will be chosen from X i + \ X i . Since X i + \ X i is a disjoint union of the sets X + j \ X j and X + h \ X h , we consider a parameter z to partition the value − |β −1 (0)|. For each 0 ≤ z ≤ − |β −1 (0)|, let z j = |β −1 (0)| + z and z h = b − z: we consider the states at nodes j and h with budget z j and z h , respectively. Let Note that the set D is partitioned into D j and D h . We define α j : X j → {0, 1} such that for each u ∈ X j , Similarly, we define α h : X h → {0, 1} such that for each u ∈ X h , α h (u) = 0 ifu ∈ D j and α(u) = 1, α(u) otherwise.
For each 0 ≤ z ≤ − |β −1 (0)|, a ∈ A and b ∈ B, let s z,a j = ( j z , α j , β j a ) and Define s j = s z ,a Then, the solution for the state s is defined as follows: and The subtracted term in the (11) is the over-counting term of the combined solution that is obtained from the states s j and s h . The time to compute the states s j and s h is depending on the sizes of the functions A and B. Therefore, the states s j and s h can be computed in |A| · |B| = O(Δ tw ) time.

Lemma 11
Let i be a join node in T, and let s, s j and s h be as defined above. Let S be an optimal solution for the state s. Let S j = S ∩ X + j and S h = S ∩ X + h . Then, there exists two statesŝ ands at nodes j and h, respectively, such that the sets S j and S h are optimal solutions for the statesŝ ands, respectively. Further, Proof Since S is an optimal solution for the state s, the pair (S, α) induces the state s at the node i. Consider the set D = D j ∪ D h , and the functions α j and α h as defined above. Letŝ = (ˆ j ,α j ,β j ) be the state at the node j induced by the pair (S j , α j ). Lets = (˜ h ,α h ,β h ) be the state at the node h induced by the pair (S h , α h ). Let z =ˆ j − |β −1 (0)|. Define a : D h → {−1, 0, 1, . . . , Δ} such that for each u ∈ D h , a(u) =β j (u). Then, define b : D j → {−1, 0, 1, . . . , Δ} such that for each u ∈ D j , a(u) =β h (u). Note that the functionsβ j andβ h are same as β a j and β b h , respectively. Thus, the statesŝ ands are same as s z,a j and s z,b h , respectively. From (9) (0)) can be written as follows: If either S j or S h is not optimal to the stateŝ ors, then it contradicts the optimality of the solution S for the state s. Thus, the sets S j and S h are optimal solutions for the statesŝ ands, respectively.

Bottom-up evaluation: Correctness of the DP formulation
Correctness Invariant For a node i and a valid state s = ( , α, β) at i, the recursive definition in Section 4.2 ensures that

Lemma 12
For each i ∈ V (T), and each valid state s ∈ S i , the correctness invariant is maintained for Sol i [s].
Proof The proof is by induction on the height of a node in T. The height of a node i in a rooted tree T is the distance to the farthest leaf in the subtree rooted at i. The base case is when i is a leaf node in T and height is zero, and the proof follows from Lemma 8. Let us assume that the claim is true for all nodes in T of height at most h − 1 ≥ 0. We now prove that if the claim is true for all nodes of height at most h − 1, then it is true for a node of height h. Let i be a node of height h ≥ 1. Since i is not a leaf node, its children are of height at most h − 1. Therefore, by the induction hypothesis, the correctness invariant is maintained at all children of i. We now prove that the correctness invariant is maintained at the node i. Let S be an optimal solution for the state s. Then, we show that Val i [s] = C(X i \ α −1 (0), S). If i is an introduce node then from Lemma 9, we have that the set S \ {v} is an optimal solution for the stateŝ ∈ S j . Since j is at height at most h − 1 and by induction hypothesis, the correctness invariant is maintained at state s of the node i. If i is a forget node then from Lemma 10, we have that the set S is an optimal solution for the stateŝ ∈ S j . Since j is at height at most h−1 and by induction hypothesis, the correctness invariant is maintained at state s of the node i. If i is a join node then from Lemma 11, we have that the sets S j and S h (as defined in Lemma 11) are optimal solutions for the stateŝ s ∈ S j ands ∈ S h , respectively. Since j and h are at height at most h − 1 and by induction hypothesis, the correctness invariant is maintained at state s of the node i. This completes the proof.
Thus we conclude the section with following theorem.

Theorem 3
The EXPECTED COVERAGE problem can be solved optimally in time 2 O(tw log Δ) n O (1) .
Proof An optimal solution can be obtained the state s = (k, ∅ → {0, 1}, ∅ → {−1, 0, 1, . . . , Δ}) at the node r. That is, the set Sol r [s] is an optimal solution to the input instance of the EXPECTED COVERAGE problem. The correctness of the tables computation is proved in Lemma 12. Note that every node has a table of size (k + 1)(2Δ + 4) tw and each entry can be updated in time O(Δ tw ). A nice tree decomposition with width tw and O(n · tw) nodes can be computed in polynomial time [30] on inputting a tree decomposition with width tw. Therefore, given a graph G and a tree decomposition of G with width tw, the EXPECTED COVERAGE problem can be solved in time 2 O(tw log Δ) n O(1) .

Conclusion
In this article, we considered the EXPECTED COVERAGE problem. We focus on structural parameterization, due to the EXPECTED COVERAGE problem is W[2]-hard for the solution size k. In particular, we show the parameterized complexity of the EXPECTED COVERAGE problem with respect to the well-known graph parameters treewidth tw, pathwidth pw, vertex cover vc, feedback vertex set fvs, and distance to star forest dsf. Further, we observed from our reduction that the EXPECTED COV-ERAGE problem with respect to number of negative demand vertices as parameter is W[1]-hard. Finally, for the combined parameters treewidth and maximum degree (tw + ), we give an FPT algorithm to the EXPECTED COVERAGE problem. Remaining open questions on the structurally parameterized complexity of the EXPECTED COVERAGE problem concern the tight fine-grained bounds. On the other hand, FPT approximation scheme for any of the above structural properties would be a good way to complement the parameterized hardness result of the EXPECTED COVERAGE problem.
Funding Open access funding provided by University of Bergen (incl Haukeland University Hospital).
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