The Parameterized Hardness of the k-Center Problem in Transportation Networks

# The Parameterized Hardness of the k-Center Problem in Transportation Networks

Andreas Emil Feldmann111Supported by project CE-ITI (GAČR no. P202/12/G061) of the Czech Science Foundation. Department of Applied Mathematics, Charles University, Prague, Czechia
feldmann.a.e@gmail.com
Dániel Marx222Supported by ERC Consolidator Grant SYSTEMATICGRAPH (No. 725978) Institute for Computer Science and Control, Hungarian Academy of Sciences
(MTA SZTAKI) dmarx@cs.bme.hu
###### Abstract

In this paper we study the hardness of the -Center problem on inputs that model transportation networks. For the problem, an edge-weighted graph and an integer are given and a center set needs to be chosen such that . The aim is to minimize the maximum distance of any vertex in the graph to the closest center. This problem arises in many applications of logistics, and thus it is natural to consider inputs that model transportation networks. Such inputs are often assumed to be planar graphs, low doubling metrics, or bounded highway dimension graphs. For each of these models, parameterized approximation algorithms have been shown to exist. We complement these results by proving that the -Center problem is W-hard on planar graphs of constant doubling dimension, where the parameter is the combination of the number of centers , the highway dimension , and even the treewidth . Moreover, under the Exponential Time Hypothesis there is no time algorithm for any computable function . Thus it is unlikely that the optimum solution to -Center can be found efficiently, even when assuming that the input graph abides to all of the above models for transportation networks at once!

Additionally we give a simple parameterized -approximation algorithm for inputs of doubling dimension with runtime . This generalizes a previous result, which considered inputs in -dimensional metrics.

## 1 Introduction

Given a graph with positive edge lengths , the -Center problem asks to find center vertices such that every vertex of the graph is as close as possible to one of centers. More precisely, if denotes the length of the shortest path between and according to the edge lengths , let be the ball of radius  around . A solution to -Center is a set of centers such that , and the objective is to minimize the cost of the solution, which is the smallest value for which . This problem has numerous applications in logistics where easily accessible locations need to be chosen on a map under a budget constraint. For instance, a budget may be available to build hospitals, shopping malls, or warehouses. These should be placed so that the distance from each point on the map to the closest facility is minimized.

The -Center problem is NP-hard , and so approximation algorithms [27, 28] as well as parameterized algorithms [8, 11] have been developed for this problem. The former are algorithms that use polynomial time to compute an -approximation, i.e., a solution that is at most times worse than the optimum. For the latter, a parameter is given as part of the input, and an optimum solution is computed in time for some computable function independent of the input size . The rationale behind such an algorithm is that it solves the problem efficiently in applications where the parameter is small. If such an algorithm exists for a problem it is called fixed-parameter tractable (FPT) for . Another option is to consider parameterized approximation algorithms [20, 21], which compute an -approximation in time for some parameter .

By a result of Hochbaum and Shmoys , on general input graphs, a polynomial time -approximation algorithm exists, and this approximation factor is also best possible, unless P=NP. A natural parameter for -Center is the number of centers , for which however the problem is W-hard . In fact it is even W-hard  to compute a -approximation for any , and thus parametrizing by does not help to overcome the polynomial-time inapproximability. For structural parameters such as the vertex-cover number or the feedback-vertex-set number the problem remains W-hard , even when combining with the parameter . For each of the more general structural parameters treewidth and cliquewidth, an efficient parameterized approximation scheme (EPAS) was shown to exist , i.e., a -approximation can be computed in time for any , if is either the treewidth or the cliquewidth, and is the number of vertices.

Arguably however, graphs with low treewidth or cliquewidth do not model transportation networks well, since grid-like structures with large treewidth and cliquewidth can occur in road maps of big cities. As we focus on applications for -Center in logistics, here we consider more natural models for transportation networks. These include planar graphs, low doubling metrics such as the Euclidean or Manhattan plane, or the more recently studied low highway dimension graphs. Our main result is that -Center is W-hard on all of these graph classes combined, even if adding and the treewidth as parameters. Before introducing these graph classes, let us formally state our theorem.

###### Theorem 1.

Even on weighted planar graphs of doubling dimension , the -Center problem is W-hard for the combined parameter , where is the treewidth and the highway dimension of the input graph. Moreover, under ETH there is no time algorithm333Here means any function such that . for the same restriction on the input graphs, for any computable function .

A planar graph can be drawn in the plane without crossing edges. Such graphs constitute a realistic model for road networks, since overpasses and tunnels are relatively rare. It is known  that also for planar graphs no -approximation can be computed in polynomial time, unless P=NP. On the positive side, -Center is FPT  on unweighted planar graphs for the combined parameter and . However, typically if is small then  is large and vice versa, and thus the applications for this combined parameter are rather limited. If the parameter is only , then an XP-algorithm exists for planar graphs . By a very recent result  the -Center problem on weighted planar graphs admits an efficient polynomial-time bicriteria approximation scheme, which for any in time computes a solution that uses at most centers and approximates the optimum with at most centers within a factor of . This algorithm implies an EPAS for parameter  on weighted planar graphs, since setting forces the algorithm to compute a -approximation in time using at most centers, i.e., at most centers as is an integer. This observation is complemented by our hardness result by showing that it is necessary to approximate the solution when parametrizing by in weighted planar graphs.

###### Definition 2.

The doubling dimension of a metric is the smallest such that for any , every ball of radius is contained in the union of at most  balls of radius . The doubling dimension of a graph is the doubling dimension of its shortest-path metric.

Since a transportation network is embedded on a large sphere (namely the earth), a reasonable model is to assume that the shortest-path metric abides to the Euclidean -norm. In cities, where blocks of buildings form a grid of streets, it is reasonable to assume that the distances are given by the Manhattan -norm. Every metric for which the distance function is given by the -norm in -dimensional space has doubling dimension . Thus a road map, which is embedded into can reasonably be assumed to have constant doubling dimension. It is known  that -Center is W-hard for parameter in two-dimensional Manhattan metrics. Also, no polynomial time -approximation algorithm exists for -Center in two-dimensional Manhattan metrics , and no -approximation for two-dimensional Euclidean metrics . On the positive side, Agarwal and Procopiuc  showed that for any metric in dimensions, the -Center problem can be solved optimally in time , and an EPAS exists for the combined parameter . We generalize the latter to any metric of doubling dimension , as formalized by the following theorem.

###### Theorem 3.

Given a metric of doubling dimension , a -approximation for -Center can be computed in time.

Theorem 1 complements this result by showing that it is necessary to approximate the cost of the solution if parametrizing by and .

###### Definition 4.

The highway dimension of a graph is the smallest such that, for some universal constant , for every and every ball of radius , there is a set of hubs such that and every shortest path of length more than  lying in contains a hub of .

The highway dimension was introduced by Abraham et al.  as a formalization of the empirical observation by Bast et al. [6, 5] that in a road network, starting from any point and travelling to a sufficiently far point along the quickest route, one is bound to pass through some member of a sparse set of “access points”, i.e., the hubs. In contrast to planar and low doubling graphs, the highway dimension has the potential to model not only road networks, but more general transportation networks such as those given by air-traffic or public transportation. This is because in such networks longer connections tend to be serviced through larger and sparser stations, which act as hubs. Abraham et al.  were able to prove that certain shortest-path heuristics are provably faster in low highway dimension graphs than in general graphs. They specifically chose the constant in their original definition, but later work by Feldmann et al.  showed that when choosing any constant in the definition, the structure of the resulting graphs can be exploited to obtain quasi-polynomial time approximation schemes for problems such as Travelling Salesman or Facility Location. Note that increasing the constant in creftype 4 restricts the class of graphs further. Other definitions of the highway dimension exist as well [3, 2, 1, 13] (see [14, Section 9] for a detailed discussion).

Later, Becker et al.  used the framework introduced by Feldmann et al.  to show that whenever there is an EPAS for -Center parameterized by , , and . Note that the highway dimension is always upper bounded by the vertex-cover number, as every edge of any non-trivial path is incident to a vertex cover. Hence the aforementioned W-hardness result by Katsikarelis et al.  for the combined parameter and the vertex-cover number proves that it is necessary to approximate the optimum when using and as the combined parameter. When parametrizing only by the highway dimension but not , it is not even known if a parameterized approximation scheme (PAS) exists, i.e., an time -approximation algorithm for some functions . However, under the Exponential Time Hypothesis (ETH) , by  there is no algorithm with doubly exponential runtime to compute a -approximation for any . The same paper  also presents a -approximation for -Center with runtime for a more general definition of the highway dimension than the one given in creftype 4. In contrast to the result of Becker et al. , it is not known whether a PAS exists when combining this more general definition of with as a parameter. Theorem 1 complements these results by showing that even of planar graphs of constant doubling dimension, for the combined parameter  no FPT algorithm exists, unless FPT=W. Therefore approximating the optimum is necessary, regardless of whether  is according to creftype 4 or the more general one from , and regardless of how restrictive creftype 4 is made by increasing the constant .

###### Definition 5.

A tree decomposition of a graph is a tree each of whose nodes  is labelled by a bag of vertices of , and has the following properties: (a) , (b) for every edge there is a node such that contains both and , (c) for every the set induces a connected subtree of . The width of the tree decomposition is . The treewidth of a graph  is the minimum width among all tree decompositions for .

As mentioned above, arguably, bounded treewidth graphs are not a good model for transportation networks. Also it is already known that -Center is W-hard for this parameter, even when combining it with  . We include this well-studied parameter here nonetheless, since the reduction of our hardness result in Theorem 1 implies that -Center is W-hard even for weighted planar graphs when combining any of the parameters , , , and . As noted in , these parameters are not bounded in terms of each other, i.e., they are incomparable. Furthermore, the doubling dimension is in fact bounded by a constant in Theorem 1. Hence, even if one were to combine all the models presented above and assume that a transportation network is planar, embeddable into some constant dimensional metric, has bounded highway dimension, and even has bounded treewidth, it is unlikely that the -Center problem can be solved efficiently. Thus it seems unavoidable to approximate the problem in transportation networks, when developing fast algorithms.

### 1.1 Related work

The -Center problem is a fairly general clustering problem and therefore finds further applications in, for instance, image processing and data-compression. The above mentioned very recent efficient bicriteria approximation scheme  improves on a previous (non-efficient) bicriteria approximation scheme [eisenstat2014planar], which for any and weighted planar input graph computes a -approximation with at most centers in time for some function (note that in contrast to above, such an algorithm does not imply a PAS for parameter ). The paper by Demaine et al.  on the -Center problem in unweighted planar graphs also considers the so-called class of map graphs, which is a superclass of planar graphs that is not minor-closed. They show that the problem is FPT on unweighted map graphs for the combined parameter . Also for the tree-depth, -Center is FPT . A closely related problem to -Center is the -Dominating Set problem, in which is given and the number of centers covering a given graph with balls of radius needs to be minimized. As this generalizes the Dominating Set problem, no -approximation is possible in polynomial time , unless P=NP, and computing an -approximation is W-hard  when parametrizing by , for any computable function .

## 2 The reduction

In this section we give a reduction from the Grid Tiling with Inequality (GT) problem, which is defined as follows. Given non-empty sets of pairs of integers, where , the task is to select one pair for each set such that

• if and then , whenever , and

• if and then , whenever .

The GT problem is W-hard  for parameter , and moreover, under ETH has no time algorithm for any computable function .

#### Construction

Given an instance of GT with sets, we construct the following graph . First, for each set , where , we fix an arbitrary order on its elements, so that , where . We then construct a gadget for , which contains a cycle  of length for which each edge has length (see Figure 1(a)). Additionally we introduce five vertices , , , , and . If then we connect these five vertices to the cycle as follows. The vertex is adjacent to the four vertices , , , and , with edges of length each. For every and , if we add the four edges

• of length ,

• of length ,

• of length , and

• of length .

We say that the element corresponds to the four vertices , , , and . Note that (which always exists) corresponds to the four vertices adjacent to . Note also that , since .

The gadgets are now connected to each other in a grid-like fashion (see Figure 1(b)). That is, for we introduce a path between and that has edges, each of length . Analogously, for we add a path between and with edges of length each. Note that these paths all have length .

The resulting graph forms an instance of -Center with . We claim that the instance of GT has a solution if and only if the optimum solution to -Center on has cost at most . We note at this point that the reduction would still work when removing the vertices and decreasing to . However their existence will greatly simplify analysing the doubling dimension of in Section 3.

#### A solution to a Gt≤ instance implies cost at most 2n2 for the k-Center instance

Recall that we fixed an order of each set , so that each element corresponds to four equidistant vertices on cycle with distance between consecutive such vertices on the cycle. If is in the solution to the GT instance , let contain the vertices of corresponding to in addition to . The solution to the -Center instance is given by the union , which are exactly centers in total.

Let us denote the set containing the four vertices of by , and note that each of these four vertices covers vertices of with balls of radius , as each edge of has length . Since the distance between any pair of centers in is at least , these four sets of covered vertices are pairwise disjoint. Thus the total number of vertices covered by on is , i.e. all vertices of the cycle are covered. Recall that for the lengths of the edges between the vertices , , , and and the cycle we have . Hence the centers in also cover , , , and by balls of radius .

Now consider a path connecting two neighbouring gadgets, e.g., let be the path connecting and . The center sets and contain vertices corresponding to the respective elements and of the solution to the GT instance. This means that if and then . Thus the closest centers of and are at distance from each other, as has length . From we get \@checkenddoitall

 dccclxxxviiℓb+1+ℓ′b′=2n2+bn+1−1+1+2n2−b′n+1≤4n2.

Therefore all vertices of are covered by the balls of radius around the two closest centers of and . Analogously, we can also conclude that any path connecting some vertices and are covered, using the fact that if and are in the solution to the GT instance then .

Finally, the remaining center vertices in cover the additional vertex  in each gadget .

#### A k-Center instance with cost at most 2n2 implies a solution to the Gt≤ instance

We first prove that in any solution to the -Center instance of cost at most , each cycle must contain exactly four centers. Recall that , that is incident to four edges of length each, and that each edge of has length . Now consider the vertices , , , and , each of which is not connected by an edge to any vertex , where , nor to . Thus each of these four vertices must be covered by centers on the cycle if the radius of each ball is at most . Furthermore, the distance between each pair of these four vertices is at least , which means that any solution of cost at most needs at least four centers on to cover these four vertices.

Each vertex must be contained in any solution of cost at most , since the distance from to any other vertex is more than . This already uses of the available centers. Since there are cycles and only remaining available centers, we proved that each cycle  contains exactly four centers, and no other centers exist in the graph . Let be the set of four centers contained in . As each center of covers at most vertices of by balls of radius at most , to cover all vertices of these four centers must be equidistant with distance exactly between consecutive centers on . Furthermore, since and each edge of has length , to cover  for any some center of must lie on a vertex of adjacent to . This means that the four centers of are exactly those vertices corresponding to element of .

It remains to show that the elements corresponding to the centers in form a solution to the GT instance . For this, consider two neighbouring gadgets and , and let and be the respective elements corresponding to the center sets and . Note that for any we have and . Since every edge of the cycles and has length , this means that the distance from the closest centers and to and , respectively, is determined by the edges of length and incident to and , respectively. In particular, the distance between and is , as the path connecting and has length . Assume now that , which means that since and are integer. Hence this distance is \@checkenddoitall

 dccclxxxviiℓb+1+ℓ′b′=2n2+bn+1−1+1+2n2−b′n+1≥4n2+1n+1.

As the centers and only cover vertices at distance at most each, while the edges of the path have length , there must be some vertex of that is not covered by the center set. However this contradicts the fact that the centers form a feasible solution with cost at most , and so .

An analogous argument can be made for neighbouring gadgets and , so that for the elements and corresponding to the centers in and , respectively. Thus a solution to of cost at most implies a solution to .

## 3 Properties of the constructed graph

The reduction of Section 2 proves that the -Center problem is W-hard for parameter , since the reduction can be done in polynomial time and is function of . Since this function is quadratic, we can also conclude that, under ETH, there is no algorithm for -Center. We will now show that the reduction has various additional properties from which we will be able to conclude Theorem 1. First of all we prove that any constructed graph for an instance of GT is planar and has bounded doubling dimension.

###### Lemma 6.

The graph is planar and has doubling dimension at most for .

###### Proof.

It is obvious from the construction of the graph that it is planar. To bound its doubling dimension, consider the shortest-path metric on the vertex set given by the distances between these vertices in . As these vertices are arranged in a grid-like fashion, this shortest-path metric on approximates the -metric. We consider a set of index pairs, for which the corresponding vertices in roughly resemble a ball in the shortest-path metric on . That is, consider the set of index pairs , and let contain all vertices of gadgets such that in addition to the vertices of paths of length connecting these gadgets. We would like to bound the diameter of the graph induced by in , for which we need the following claim, which we will reuse later.

###### Claim 7.

For any gadget and with , the distance between and lies between and .

###### Proof.

The distance between and is less than , via the path passing through and the two vertices of adjacent to , , and . Note that the shortest path between and inside the gadget does not necessarily pass through , but may pass along the cycle instead. This is because the set of the GT instance may contain up to elements, which would imply a direct edge from to on . Thus we can give a lower bound of for the distance between and inside of . This is also the shortest path between these vertices in , since any other path needs to pass through at least three gadgets.

By creftype 7, the diameter of is at most as one needs to traverse at most gadgets with and the paths of length connecting them, in order to reach any vertex of from any other. Assuming , i.e., the set contains the maximum number of index pairs, the diameter of is at least , since any path between two points of at maximum distance must pass through at least gadgets with and the paths of length connecting them.

Consider a ball around a vertex of radius in , and let be the closest vertex of to . The distance between and is at most , whether lies on or on one of the paths of length connecting with an adjacent gadget. Hence the ball is contained in a ball of radius around . The latter ball is in turn contained in the set centered at if its diameter is at most the diameter of , i.e., . This for instance is true if , , and . From the upper bound on the diameter of a set , we also know that is contained in a ball of radius around if , which is true if . However we also want to be non-empty, i.e., , which by the latter equality means that . We may cover all vertices of with sets , since in these sets correspond to “squares rotated by 45 degrees of diameter and , respectively”. Thus we can cover with sets , i.e., if and we can cover a ball of radius in with balls of radius .

If , a ball has radius less than . Consider the case when lies in a gadget of . The distance from to any cycle for which is at least , as : to reach a path from first needs to traverse a neighbouring gadget of , which we know has diameter at least by creftype 7, and the vertex has distance more than from . Thus contains at most the four neighbouring gadgets of and the paths of length connected to these. On each of the five cycles that intersect , at most balls of radius  are needed to cover all vertices in the intersection of and : as the edges of all have length we may choose vertices equidistantly at every -th vertex on the part of in . As long as , any path of length that intersects can be covered by one ball of radius . Any vertex that lies in can also be covered by one ball of radius . As intersects at most cycles , as well as at most vertices , and at most paths of length , these amount to at most balls of radius .

If does not lie in any gadget , then it lies on some path of length connecting two gadgets. Given that , in this case the ball intersects at most cycles and vertices , and at most paths of length , since by creftype 7 the diameter of a gadget is at least if . Thus in this case at most balls of radius suffice.

Finally, if , then a ball contains only a subpath of some cycle , a subpath of a path of length connecting two gadgets, or a single vertex , since any edge connecting a cycle to or some has length more than if . In this case at most  balls of radius suffice to cover all vertices of . ∎

We next show that we can bound the parameters and , i.e. the treewidth and highway dimension of , as a function of the parameter . Note that the following lemma bounds the highway dimension in terms of , no matter how restrictive we make creftype 4 by increasing the constant .

###### Lemma 8.

For any constant of creftype 4, the graph has highway dimension at most .

###### Proof.

For any scale and universal constant we will define a hub set hitting all shortest paths of length more than in , such that for any ball of radius in . This bounds the highway dimension to according to creftype 4.

Let , i.e. it contains all vertices connecting gadgets to each other in addition to the vertices . If then . Any shortest path containing only vertices of a cycle has length at most , since the cycle has length . Any (shortest) path that is a subpath of a path connecting two gadgets has length at most . Hence any shortest path of length more than must contain some vertex of . The total size of is , and so any ball, no matter its radius, also contains at most this many hubs of .

If then any path of length more than but not containing any vertex of  must lie on some cycle . We define the set , i.e. it contains every -th vertex on the cycle after rounding down. This means that every path on of length more than contains a vertex of . Thus for these values of we set . Any ball of radius  contains hubs of any . By creftype 7, the distance between any pair of the four vertices , where , that connect a gadget to other gadgets, is more than . This means that can only intersect gadgets, since and the gadgets are connected in a grid-like fashion. Hence the ball only contains hubs for each of the sets for which intersect the respective gadget . At the same time each gadget contains only  vertices of . Thus if is a constant, then the number of hubs of in is constant.

If , a path of length more than may be a subpath of a path connecting two gadgets. Let be the path connecting and for , and let be the path connecting and for . Recall that each of these paths consists of edges of length  each. If , we define the set , and if , we define the set , i.e. these sets contain vertices of consecutive distance on the respective paths, after rounding down. Now let , so that every path of length more than  contains a hub of . Any ball of radius intersects only gadgets , as observed above. As the edges of a cycle have length , the ball contains only vertices of . Thus contains hubs of for each of the gadgets it intersects. For constant , this proves the claim. ∎

To bound the treewidth of we use so-called cops and robber games. Given a graph and , a state of the cops and robber game on is a pair where with , and . The set encodes the positions of the cops, while is the position of the robber. The game proceeds in rounds, where each round is associated with a state . Initially, in round the cops first choose positions and then the robber chooses a position . In each round , first the cops choose a new position , after which the robber can choose a position , such that and lie in the same connected component of . The cops win the game if after a finite number of rounds, the robber has no position to choose, i.e., the robber is caught. By , a graph has treewidth  if and only if cops can win in .

###### Lemma 9.

The graph has treewidth at most .

###### Proof.

We prove that cops can win the cops and robber game on . For each we define the sets and