W[1]-Hardness of the k-Center Problem Parameterized by the Skeleton Dimension
WW[1]-Hardness of the k -Center ProblemParameterized by the Skeleton Dimension Johannes BlumUniversity of Konstanz, Germany [email protected]
Abstract
In the k -Center problem, we are given a graph G = ( V, E ) with positive edge weightsand an integer k and the goal is to select k center vertices C ⊆ V such that the maximumdistance from any vertex to the closest center vertex is minimized. On general graphs, theproblem is NP-hard and cannot be approximated within a factor less than 2.Typical applications of the k -Center problem can be found in logistics or urban planningand hence, it is natural to study the problem on transportation networks. Such networksare often characterized as graphs that are (almost) planar or have low doubling dimension,highway dimension or skeleton dimension. It was shown by Feldmann and Marx that k -Center is W[1]-hard on planar graphs of constant doubling dimension when parameterizedby the number of centers k , the highway dimension hd and the pathwidth pw [10]. Weextend their result and show that even if we additionally parameterize by the skeletondimension κ , the k -Center problem remains W[1]-hard. Moreover, we prove that underthe Exponential Time Hypothesis there is no exact algorithm for k -Center that has runtime f ( k, hd, pw, κ ) · | V | o ( pw + κ + √ k + hd ) for any computable function f . The k -Center problem consists of the following task: Given a graph G = ( V, E ) with positiveedge weights (cid:96) : E → Q + and some k ∈ N , choose k center vertices C ⊆ V that minimize themaximum distance from any vertex of the graph to the closest center. Formally, if the shortestpath distances in G are given by dist : V → Q + and B r ( v ) = { w ∈ V | dist( v, w ) ≤ r } denotesthe ball of radius r around v , we aim for a solution C ⊆ V of size | C | ≤ k that has minimumcost, which is the smallest radius r ≥ V = (cid:83) v ∈ C B r ( v ).On general graphs, the k -Center problem is NP-complete [14], as well as on planargraphs [13] and geometric graphs using L -, L - or L ∞ -distances [8]. On the positive sidethere is a general 2-approximation algorithm by Hochbaum and Shmoys [11], i.e. an efficientalgorithm that computes a solution which deviates from the optimum at most by a factor of 2.This factor is tight, as for any (cid:15) >
0, it is NP-hard to compute a (2 − (cid:15) )-approximation, evenwhen considering planar graphs [13] or graphs with L - or L ∞ -distances [8].However, common applications of the k -Center problem arise in domains like logisticsor urban planning. For instance, one might want to place a limited number of warehouses,hospitals or police stations on a map such that the distance from any point to the closest facilityis minimized. Hence, it is natural to study the problem on transportation networks. Commoncharacterizations of such networks are graphs that are planar or have low doubling dimension,highway dimension, or skeleton dimension. For formal definitions of these parameters, seeSection 2. Usually, it is assumed that in transportation networks the mentioned parameters arebounded by O (polylog | V | ) or O ( (cid:112) | V | ). It was shown that on graphs of maximum degree ∆and highway dimension hd , the skeleton dimension is at most (∆ + 1) · hd [12]. The relationshipbetween highway dimension hd and skeleton dimension κ was also evaluated experimentally1 a r X i v : . [ c s . CC ] A ug n several real-world road networks and it turned out that κ (cid:28) hd [5]. Moreover, it wasconjectured that on road networks the skeleton dimension is a constant whereas the highwaydimension grows faster than O (polylog | V | ).Still, a low highway dimension or skeleton dimension does not suffice to overcome the generalinapproximability bound of k -Center . In particular, it was shown that for any (cid:15) >
0, thereis no (2 − (cid:15) )-approximation algorithm for graphs of highway dimension hd ∈ O (log | V | ) [9] orskeleton dimension κ ∈ O (log | V | ) [4], unless P=NP.Apart from approximation, a common way of dealing with NP-hard problems is the useof fixed-parameter algorithms. Such an algorithm computes an exact solution in time f ( p ) · n O (1) , where f is a computable function and p a parameter of the problem instance whichis independent of the problem size n . In other words, if a problem admits a fixed-parameteralgorithm, the complexity of the problem can be captured through some parameter p . If thisis the case, we call the problem fixed-parameter tractable (FPT). A natural parameter for k -Center is the number of center vertices k . However, it was shown that in general, k -Center isW[2]-hard for parameter k , and hence it is not fixed-parameter tractable unless W[2] = FPT [7].Feldmann and Marx studied the fixed-parameter tractability of k -Center on transportationnetworks [10]. They showed that k -Center is W[1]-hard even if the input is restricted to planargraphs of constant doubling dimension and the parameter is a combination of k , the highwaydimension hd and the pathwidth pw . Moreover, they proved that under the Exponential TimeHypothesis (ETH) there is no exact algorithm with runtime f ( k, pw, hd ) · | V | o ( pw + √ k + hd ) . Inthe present paper we extend their result and show that one can additionally parameterize bythe skeleton dimension κ without affecting W[1]-hardness. Formally, we show the followingtheorem. Theorem 1.
On planar graphs of constant doubling dimension, the k -Center problem is W [1] -hard for the combined parameter ( k, pw, hd, κ ) where pw is the pathwidth, hd the high-way dimension and κ the skeleton dimension of the input graph. Assuming ETH there is no f ( k, pw, hd, κ ) · | V | o ( pw + κ + √ k + hd ) time algorithm for any computable function f . The reduction of Feldmann and Marx produces a graph where the maximum degree ∆ canbe quadratic in the input size. As we have ∆ ≤ κ , it does not imply any hardness for theskeleton dimension. Our new construction yields a graph of constant maximum degree, whichenables us to bound the skeleton dimension as well as the highway dimension and the pathwidth.The results reported by Blum and Storandt [5] indicate that in real-world road networks, theskeleton dimension κ is significantly smaller than the highway dimension, which motivates theuse of κ as a parameter. Note that in general, the parameters pw , hd and κ are incomparable [4].Still, our main result shows that combining all these parameters and the number of centers k doesnot allow a fixed-parameter algorithm unless FPT=W[1]. However, for the combined parameters( k, hd ) [9] and ( k, κ ) [10], the existence of a fixed-parameter approximation algorithm was shown,i.e. an approximation algorithm with runtime f ( p ) · n O (1) for parameter p . Theorem 1 indicatesthat apart from approximation there is not much hope for efficient algorithms. For n ∈ N , let [ n ] = { , . . . , n } . Addition modulo 4 is denoted by (cid:1) . For ( a, b ) , ( a (cid:48) , b (cid:48) ) ∈ N let( a, b ) ≤ ( a (cid:48) , b (cid:48) ) iff a < a (cid:48) or a = a (cid:48) and b ≤ b (cid:48) .In a graph G = ( V, E ) we denote the shortest s - t path by π ( s, t ) and the length of a path P by | P | . The concatenation of two paths P and P (cid:48) is denoted by P ◦ P (cid:48) .A graph G is planar if it can be embedded into the plane without crossing edges, and d -doubling if for any r >
0, any ball B r ( v ) of radius 2 r in G is contained in the union of d balls Here o ( pw + κ + √ k + hd ) stands for g ( pw + κ + √ k + hd ) where g is a function with g ( x ) ∈ o ( x )
2f radius r . If d is the smallest integer such that G is d -doubling, the graph G has doublingdimension log d .For the highway dimension several slightly different definitions can be found in the litera-ture [3, 2, 1]. Here we use the one given in [3]. Definition 2.
The highway dimension of a graph G is the smallest integer hd such that for anyradius r and any vertex v there is a hitting set S ⊆ B r ( v ) of size hd for the set of all shortestpaths π satisfying | π | > r and π ⊆ B r ( v ) . To define the skeleton dimension, which was introduced in [12], we need to consider thegeometric realization ˜ G of a graph G . Intuitively, ˜ G is a continuous version of G where everyedge is subdivided into infinitely many infinitely short edges. For a vertex s ∈ V , let T s bethe shortest path tree of s . We assume that in G every shortest path is unique, which can beachieved, e.g., by slightly perturbing the edge weights, and it follows that T s is also unique.The skeleton T ∗ s is defined as the subtree of ˜ T s induced by all v ∈ ˜ V , for which there is somevertex w such that v is contained in π ( s, w ) and moreover, we have dist( s, v ) ≤ · dist( v, w ). Definition 3.
For a skeleton T ∗ s = ( V ∗ , E ∗ ) and a radius r > , let Cut rs = { v ∈ V ∗ | dist( s, v ) = r } . The skeleton dimension of a graph G is κ = max s,r | Cut rs | . We conclude this section with a definition of the pathwidth.
Definition 4.
A path decomposition of a graph G = ( V, E ) is a sequence ( X , . . . , X r ) whereevery X i (also called bag ) is a subset of V and the following properties are satisfied:(i) (cid:83) ri =1 X i = V ,(ii) for every edge { u, v } ∈ E there is a bag X i containing both u and v , and(iii) for every three indices i ≤ j ≤ we have X i ∩ X k ⊆ X j .The width of a path decomposition is the size of the largest bag minus one, i.e. max ri =1 ( | X i | − .The pathwidth pw of a graph G = ( V, E ) is defined as the minimum width of all path decompo-sitions of G . Following the idea of Feldmann and Marx [10], who showed that on planar graphs of constantdoubling dimension, k -Center is W [1]-hard for parameter ( k, pw, hd ), we present a reductionfrom the Grid Tiling with Inequality ( GT ≤ ) problem. This problem asks the followingquestion: Given χ sets S i,j ⊆ [ n ] of pairs of integers, where ( i, j ) ∈ [ χ ] , is it possible to chooseone pair s i,j ∈ S i,j from every set, such that • if s i,j = ( a, b ) and s i +1 ,j = ( a (cid:48) , b (cid:48) ) we have a ≤ a (cid:48) , and • if s i,j = ( a, b ) and s i,j +1 = ( a (cid:48) , b (cid:48) ) we have b ≤ b (cid:48) .It is known that the GT ≤ problem is W [1]-hard for parameter χ and, unless the ExponentialTime Hypothesis (ETH) fails, it has no f ( χ ) · n o ( χ ) time algorithm for any computable f [6]. In [10] the following graph H I is constructed from an instance I of GT ≤ . For any of the χ sets S i,j , the graph H I contains a gadget H i,j that consists of a cycle O i,j = v v . . . v n +4 v andfive additional vertices x i,j , x i,j , x i,j , x i,j and y i,j . Every edge contained in some cycle O i,j hasunit length and every vertex y i,j is connected to O i,j via edges to v , v n +2 , v n +3 and v n +4 ,which all have length 2 n + 1. Moreover, for every pair ( a, b ) ∈ S i,j and τ = ( a − · n + b , thegadget H i,j contains the four edges 3 { x i,j , v τ } of length 2 n − an +1 , • { x i,j , v τ +4 n +1 } of length 2 n + bn +1 − • { x i,j , v τ +8 n +2 } of length 2 n + an +1 −
1, and • { x i,j , v τ +12 n +3 } of length 2 n − bn +1 .Finally, the individual gadgets are connected in a grid-like fashion, which means that thereis a path from x i,j to x i,j +1 and from x i,j to x i +1 ,j . Each of these paths has length 1 and consistsof n + 2 edges of length n +2 .Feldmann and Marx showed that the given GT ≤ instance I has a solution if and only ifthe k -Center problem in the graph H I has a solution of cost 2 n using k = 5 χ centers.Moreover, the graph H I is planar and has doubling dimension O (1), highway dimension O ( χ )and pathwidth O ( χ ). Observe that the degree of any vertex x hi,j is | S i,j | . This means that theskeleton dimension of H I might be as large as Ω( n ), as the maximum degree of H I is a lowerbound on its skeleton dimension. We show now how to construct a graph G I that resembles H I , but has skeleton dimension O ( χ ) and fulfills the other mentioned properties. We assume that in the given GT ≤ -instance, for all ( i, j ) ∈ [ χ ] and every b ∈ [ n ], there issome a ∈ [ n ] such that ( a, b ) ∈ S i,j . This is a valid assumption, as from an instance I ofordinary GT ≤ , we can construct the following instance I (cid:48) . For i ∈ [ χ −
1] and j ∈ [ χ ] we addthe pairs { ( n + χ − i, b ) | b ∈ [ n ] } to S i,j . Moreover, we add the pairs { (0 , b ) | b ∈ [ n ] } to every S χ,j . Clearly, every solution for I is also a solution for I (cid:48) . Consider now a solution for I (cid:48) .For ( i, j ) ∈ [ χ − × [ χ ] we cannot choose a dummy pair s i,j = ( n + χ − i, b ), as there is no( a (cid:48) , b (cid:48) ) ∈ S i +1 ,j such that a (cid:48) ≥ n + χ − i . Moreover, it is not possible to choose s χ,j = (0 , b (cid:48) ) as S χ − ,j contains no pair ( a, b ) satisfying a ≤
0. Hence, I has a solution if and only if I (cid:48) has asolution.Given a GT ≤ -instance I we construct the following graph G I (cf. Figure 1). Like in [10], wecreate a gadget G i,j for every set S i,j . Any G i,j contains a cycle O i,j , which initially consists offour edges that have length 2 n +2 + / n . Denote the four vertices of the cycle O i,j by z i,j , . . . , z i,j and for h ∈ [4] let O hi,j = π (cid:16) z hi,j , z h (cid:1) i,j (cid:17) . Now, for any pair ( a, b ) ∈ S i,j and any h ∈ [4] we inserta vertex v h ( a,b ) into the path O hi,j and place it such that its distance to z hi,j is d ( a,b ) = 2 b − an . It follows that the distance between v h ( a,b ) and v h (cid:1) a,b ) is 2 n +2 + / n . Moreover, for ( a (cid:48) , b (cid:48) ) ≤ ( a, b ),the distance from v h ( a (cid:48) ,b (cid:48) ) to v h ( a,b ) is 2 b − b (cid:48) + ( a − a (cid:48) ) / n .Additionally, for any pair ( a, b ) ∈ S i,j and any h ∈ [4], we insert two vertices ψ h ( a,b ) and ψ (cid:48) h ( a,b ) into the path O hi,j such that their distance from v h ( a,b ) is 2 n +1 and 2 n +1 + / n , respectively. Thisimplies that dist (cid:16) ψ (cid:48) h ( a,b ) , v h (cid:1) a,b ) (cid:17) = 2 n +1 and dist (cid:16) ψ h ( a,b ) , v h (cid:1) a,b ) (cid:17) = 2 n +1 + / n .Any gadget G i,j also contains a central vertex y i,j that is connected to each z hi,j through anedge of length 2 n +1 + 1. Finally, we add four vertices x i,j , . . . , x i,j to every gadget G i,j , throughwhich we will connect the individual gadgets. For ( a, b ) ∈ S i,j and h ∈ [4] denote the distancebetween v h ( a,b ) and x hi,j by d h ( a,b ) . The idea of our reduction is that we attach every x hi,j to thecycle O i,j such that for every pair ( a, b ) ∈ S i,j and h ∈ { , } , the distance d h ( a,b ) reflects thevalue of b , whereas for h ∈ { , } , the distance d h ( a,b ) reflects the value of a .4 i,j z i,j z i,j z i,j z i,j x i,j x i,j x i,j x i,j ρ b u b ψ a,b ) ψ (cid:48) a,b ) v a,b ) (a) A gadget G i,j . G , G χ,χ (b) The graph G I . Figure 1: A single gadget G i,j and the whole graph G I .For the latter, we simply add an edge between x i,j and the vertex v a ∗ ,b ∗ ) where ( a ∗ , b ∗ ) =min S i,j . The length of this edge is chosen as d a ∗ ,b ∗ ) = 2 n + 1 + d ( a ∗ ,b ∗ ) = 2 n + 2 b ∗ + a ∗ n . Similarly we add the edge (cid:110) x i,j , v a ∗ ,b ∗ ) (cid:111) and set its length to d a ∗ ,b ∗ ) = 2 n +1 − d ( a ∗ ,b ∗ ) = 2 n +1 + 1 − b ∗ − a ∗ n . It follows that for all ( a, b ) ∈ S i,j we have d a,b ) = d a ∗ ,b ∗ ) + d ( a,b ) − d ( a ∗ ,b ∗ ) = 2 n + 2 b + an and d a,b ) = d a ∗ ,b ∗ ) + d ( a ∗ ,b ∗ ) − d ( a,b ) = 2 n +1 + 1 − b − an . Attaching x i,j and x i,j to G i,j is slightly more elaborate. We want to ensure that for anytwo pairs ( a, b ) , ( a, b (cid:48) ) ∈ S i,j that agree on the first component, we have d a,b ) = d a,b (cid:48) ) . Forthat purpose, we add a path U i,j = u . . . u n and set the length of every edge { u λ , u λ +1 } to2 λ . Moreover, we add the edge { u n , x i,j } of length 2 n . For every b ∈ [ n ], consider the vertex v a ∗ ,b ) that is furthest from z i,j . We call it also the b -portal ρ b . We attach it to u b throughan edge of length 2 b − a ∗ / n , the so called b -portal edge . It follows that for ( a, b ) ∈ S i,j we havedist( v a,b ) , u b ) = 2 b − a ∗ / n + d ( a ∗ ,b ) − d ( a,b ) = 2 b − a / n and dist( u b , x i,j ) = (cid:80) nλ = b λ = 2 n +1 − b , and hence we have d a,b ) = 2 n +1 − an . Similarly we proceed with the vertices contained in O i,j . We add a path U i,j = u . . . u n , setthe length of every edge { u λ , u λ +1 } to 2 λ and add the edge { u n , x i,j } of length 2 n . For b ∈ [ n ]we use the vertex v a ∗ ,b ) that is closest to z i,j as the b -portal ρ b and attach it to u b trough aportal edge of length 2 b − a ∗ / n . It follows that d a,b ) = 2 n +1 − an . here the minimum is taken w.r.t the lexical order as defined in the preliminaries b u b u u n v a,b ) x i,j (a) The shortest path tree of a vertex v a,b ) . ρ b u b u u n x i,j (b) The shortest path tree of a vertex u b . Figure 2: Illustration of the shortest path structure as shown in Lemma 5.To complete the construction, we connect the individual gadgets in a grid-like fashion. For i ∈ [ n −
1] we connect x i,j and x i +1 ,j through a path P i,j of length 1 that consists of ( n + 1)edges of length / ( n +1) each. Moreover, for j ∈ [ n −
1] we connect x i,j and x i,j +1 through a path P (cid:48) i,j = w . . . w n where w = x i,j +1 and w n = x i,j . We set the length of every edge { w λ +1 , w λ } to2 λ which implies that | P (cid:48) i,j | = 2 n −
2. The resulting graph G I can be constructed in polynomialtime from the given GT ≤ -instance I . We now show some basic properties of G I that will be useful to prove the correctness of ourreduction and to obtain bounds on several graph parameters. We first observe that all shortestpaths between the cycle O i,j and a path U hi,j have a certain structure (cf. Figure 2). Lemma 5.
Let a, b, b (cid:48) ∈ [ n ] and h ∈ { , } . For β ∈ [ n ] denote the path π (cid:16) v h ( a,b ) , ρ hβ (cid:17) ◦ (cid:110) ρ hβ , u hβ (cid:111) ◦ π (cid:16) u hβ , u hb (cid:48) (cid:17) by P β .(a) If b (cid:48) ≥ b , the shortest path from v h ( a,b ) to u hb (cid:48) is P b .(b) If b (cid:48) < b , the shortest path from v h ( a,b ) to u hb (cid:48) is P b (cid:48) .Proof. Any shortest path from v h ( a,b ) to u hb (cid:48) needs to contain some portal edge (cid:110) ρ hβ , u hβ (cid:111) . Weonly prove case (a) for h = 1, the remaining cases can be shown similarly.Let β ∈ [ n ] and let ρ β = v α,β ) be the β -portal. The path P β has length dist( v a,b ) , ρ β ) +dist( ρ β , u β ) + dist( u β , u b (cid:48) ) = (cid:12)(cid:12) β − b + ( α − a ) / n (cid:12)(cid:12) + 2 β − α / n + (cid:12)(cid:12)(cid:12) b (cid:48) − β (cid:12)(cid:12)(cid:12) .This means that | P b | = 2 b (cid:48) − a / n , and for β < b we have | P β | = 2 b (cid:48) +2 b − β + ( a − α ) / n > b (cid:48) − a / n .Let now β > b . If β ≤ b (cid:48) , we obtain that | P β | = 2 b (cid:48) + 2 β − b − α / n while for β > b (cid:48) , we have | P β | = 3 · β − b − b (cid:48) − a / n , which is both greater than 2 b (cid:48) − a / n . Hence, P b is the shortest pathfrom v h ( a,b ) to u hb (cid:48) .Moreover, it holds that for any vertex v of the graph G I , there is some central vertex y i,j not too far away. Lemma 6.
For every vertex v ∈ V , we have min ( i,j ) dist( v, y i,j ) ≤ n +2 + 2 n +1 .Proof. Assume first that v is contained in some gadget G i,j . If v is contained in the cycle O i,j , the distance to the closest vertex z hi,j is at most 2 n +1 as every edge length is a multipleof / n and the subpath O hi,j between z hi,j and z h (cid:1) i,j has length 2 n +2 + / n . Moreover, we havedist( z hi,j , y i,j ) = 2 n +1 + 1, and hence, the distance between any v ∈ O i,j and y i,j is bounded by2 n +2 + 1.Consider now some vertex x hi,j . The distance from x hi,j to any vertex v h ( a,b ) is d h ( a,b ) and thelength of the path from d h ( a,b ) to y i,j via z hi,j is d ( a,b ) + 2 n +1 + 1. It follows, thatdist( x hi,j , y i,j ) ≤ d h ( a,b ) + d ( a,b ) + 2 n +1 + 1 ≤ n +2 + 2 n + 2 , (1)6here the last inequality follows from the fact that d h ( a,b ) + d ( a,b ) ≤ n +1 + 2 n + 1, which is easyto verify. Assume now that v ∈ u hb for some h ∈ { , } and b ∈ [ n ]. The shortest path π ( u hb , y i,j )passes through the portal edge { u hb , ρ hb } of length at most 2 b and the vertex z hi,j . The distancefrom ρ hb to z hi,j is at most 2 b and it follows thatdist( u hb , y i,j ) ≤ b + 2 b + 2 n +1 + 1 ≤ n +2 + 1 . It remains to consider the case where v is not contained in any gadget. If this holds, v iscontained in some path P i,j or P (cid:48) i,j between two gadgets. The lengths of these paths is boundedby 2 n − x hi,j such that dist( v, x hi,j ) ≤ n −
2. It follows fromEquation (1), that dist( v, y i,j ) ≤ n +2 + 2 n +1 . We show now that the GT ≤ -instance I has a solution if and only if the k -Center instance G I has a solution of cost at most 2 n +1 for k = 5 χ centers. Lemma 7.
A solution for the GT ≤ -instance I implies a solution for the k -Center instance G I of cost at most n +1 .Proof. For ( i, j ) ∈ [ n ] let s i,j be the pair from S i,j that is chosen in a solution of I . For the k -Center instance G I , we choose a center set C of size 5 χ by selecting from every gadget G i,j the central vertex y i,j and the four vertices v s i,j , . . . , v s i,j . We show that C has cost at most2 n +1 .Consider a gadget G i,j and the four chosen centers v a,b ) , . . . , v a,b ) . It holds that the distancebetween any two neighboring centers v h ( a,b ) and v h (cid:1) a,b ) is 2 n +2 + / n and moreover, the lengthof every edge of the cycle O i,j is a multiple of / n . Hence, it follows that for every vertex v ∈ O i,j there is some center vertex v h ( a,b ) at distance at most 2 n +1 . Consider some vertex u hb (cid:48) for h ∈ { , } . It follows from Lemma 5, that dist( v h ( a,b ) , u hb (cid:48) ) ≤ n +1 . Finally, the vertex y i,j ischosen as a center. This means that the complete gadget G i,j is contained in the five balls ofradius 2 n +1 around y i,j and v a,b ) , . . . , v a,b ) .It remains to show that the chosen centers cover all paths P i,j and P (cid:48) i,j that connect theindividual gadgets. Consider two neighboring gadgets G i,j and G i +1 ,j and let s i,j = ( a, b ) and s i +1 ,j = ( a (cid:48) , b (cid:48) ) be the corresponding pairs from the solution of I . We have a ≤ a (cid:48) . From G i,j we have chosen a center v a,b ) that has distance d a,b ) to x i,j . Similarly, we have chosen some v a (cid:48) ,b (cid:48) ) from G i +1 ,j whose distance to x i +1 ,j is d a (cid:48) ,b (cid:48) ) . The path P i,j between x i,j and x i +1 ,j haslength 1, and hence the distance between the two considered centers is d a,b ) + | P i,j | + d a (cid:48) ,b (cid:48) ) = 2 n +1 − an + 1 + 2 n +1 − a (cid:48) n = 2 n +2 + a − a (cid:48) n ≤ n +2 . This means that P i,j can be covered with balls of radius 2 n +1 around v a (cid:48) ,b (cid:48) ) and v a,b ) . Similarly, b ≤ b (cid:48) yields d a,b ) + | P (cid:48) i,j | + d a (cid:48) ,b (cid:48) ) = 2 n + 2 b + an + 2 n − n +1 + 1 − b (cid:48) − a (cid:48) n = 2 n +2 − b − b (cid:48) + a − a (cid:48) n < n +2 Hence, any vertex contained in a path P (cid:48) i,j has distance at most 2 n +1 from a chosen center.Moreover, every solution for G I of cost at most 2 n +1 contains four equidistant vertices v a,b ) , . . . , v a,b ) from every G i,j , which yield a solution for I . The following lemma completesour correctness proof. 7 emma 8. A solution for the k -Center instance G I of cost at most n +1 implies a solutionfor the GT ≤ -instance I .Proof. Let C be a solution for G I of cost at most 2 n +1 . Consider a gadget G i,j . The centralvertex y i,j has distance at least 2 n +1 + 1 to any other vertex. Hence we have y i,j ∈ C . Let C i,j be the remaining centers from C that have distance at most 2 n +1 from any vertex of G i,j . As k = 5 χ , there are at most 4 χ such centers in total. We first show that every C i,j consistsof exactly 4 vertices contained in the cycle O i,j , such that any two consecutive vertices havedistance 2 n +2 + / n . Claim 9.
For ( i, j ) ∈ [ n ] we have C i,j ⊆ O i,j and | C i,j | = 4 .Proof. Let (
A, B ) = max S i,j and let h ∈ [4]. We show that ψ h ( A,B ) can only be covered throughvertices from G i,j \ { x i,j , . . . , x i,j } .Consider some vertex x h (cid:48) i,j . The shortest path from x h (cid:48) i,j to ψ h ( A,B ) has to pass trough either v h ( A,B ) or z hi,j . The distance from ψ h ( A,B ) to v h ( A,B ) is 2 n +1 whereas the distance from ψ h ( A,B ) to z hi,j is 2 n +1 + / n − B + 1 − A / n > n . Moreover, the distance from x h (cid:48) i,j to any vertex in the cycle O i,j is at least 2 n . It follows that dist( x h (cid:48) i,j , ψ h ( A,B ) ) > n +1 and hence, ψ h ( A,B ) cannot be coveredthrough x h (cid:48) i,j or any vertex not contained in the gadget G i,j .Moreover, any two of the vertices ψ A,B ) , . . . , ψ A,B ) have distance at least 2 n +2 + / n andhence we need at least 4 centers to cover them with balls of radius 2 n +1 . This implies that C i,j ⊆ G i,j \ { x i,j , . . . , x i,j } and | C i,j | = 4.Assume now that C i,j (cid:54)⊆ O i,j , which means that some vertex u hb ∈ C i,j was chosen as acenter. Let v h ( a,b ) be the corresponding b -portal. Lemma 5 implies that the distance from u hb to any of the vertices ψ a,b ) , . . . , ψ a,b ) is more than 2 n +1 . Moreover, the pairwise distance of ψ a,b ) , . . . , ψ a,b ) is at least 2 n +2 + / n . This means that apart from u hb , the set C i,j needs tocontain 4 more centers, which contradicts | C i,j | = 4. Hence we obtain C i,j ⊆ O i,j . (cid:67) We now show, that every C i,j contains four equidistant centers v h ( a,b ) . Claim 10.
For ( i, j ) ∈ [ n ] we have C i,j = (cid:110) v a,b ) , . . . , v a,b ) (cid:111) for ( a, b ) ∈ S i,j .Proof. Let ( α, β ) be the minimum of S i,j . Consider the vertex x i,j . Its distance to z i,j , ψ α,β ) and any vertex of O i − ,j is more than 2 n +1 . Hence, it must be covered through some vertex v a,b ) where ( a, b ) ∈ S i,j . Consider the vertices ψ (cid:48) a,b ) , ψ a,b ) , ψ (cid:48) a,b ) , ψ a,b ) , ψ (cid:48) a,b ) , ψ a,b ) . None ofthem is contained in the ball of radius 2 n +1 around v a,b ) . Moreover, for h ∈ { , , } , thedistance between ψ (cid:48) h ( a,b ) and ψ h (cid:1) a,b ) is 2 n +2 , whereas the distance between ψ (cid:48) a,b ) and ψ (cid:48) a,b ) andthe distance between ψ a,b ) and ψ a,b ) are both 2 n +2 + / n . This means that complete cycle O i,j can only be covered with 4 balls of radius 2 n +1 if we have (cid:110) v a,b ) , v a,b ) , v a,b ) (cid:111) ⊆ C i,j . (cid:67) Finally we show that the sets C i,j yield a solution for the GT ≤ -instance I . Claim 11.
For ( i, j ) ∈ [ n ] choosing s i,j = ( a, b ) where v a,b ) ∈ C i,j yields a solution for the GT ≤ -instance I .Proof. Let s i,j = ( a, b ) and s i +1 ,j = ( a (cid:48) , b (cid:48) ) and assume that a > a (cid:48) . Consider the path P i,j connecting the vertices x i,j and x i +1 ,j . As the path P i,j consists of n + 1 edges of length / ( n +1) ,8t contains a vertex w that has distance 1 − a / ( n +1) from x i,j and distance a / ( n +1) from x i +1 ,j .It follows that the distances from w to the closest centers in G i,j and G i +1 ,j are d a,b ) + 1 − an + 1 = 2 n +1 − an + 1 − an + 1 > n +1 and d a (cid:48) ,b (cid:48) ) + an + 1 = 2 n +1 − a (cid:48) n + an + 1 > n +1 , respectively. This contradicts the fact that C is a solution for the k -Center instance, andhence a ≤ a (cid:48) . Similarly, let s i,j = ( a, b ) and s i,j +1 = ( a (cid:48) , b (cid:48) ) and assume that b > b (cid:48) . Considerthe path P (cid:48) i,j = w . . . w n connecting x i,j and x i,j +1 . Recall that every edge { w λ +1 , w λ } haslength 2 λ and hence we have dist( x i,j , w b ) = 2 n − b and dist( w b , x i,j +1 ) = 2 b −
2. It followsthat the distances from w b to the closest centers in G i,j and G i,j +1 are d a,b ) + 2 n − b = 2 n + 2 b + an + 2 n − b = 2 n +1 + an > n +1 and d a (cid:48) ,b (cid:48) ) + 2 b − n +1 + 1 − b (cid:48) + a (cid:48) n + 2 b − ≥ n +1 + a (cid:48) n > n +1 , respectively, which gives a contradiction. It follows that b ≤ b (cid:48) and hence, choosing s i,j = ( a, b )for v a,b ) ∈ C i,j yields a solution for I . (cid:67) This completes the proof as any solution C of cost at most 2 n +1 for the k -Center instance G I implies a solution for the GT ≤ -instance I . In this section we show bounds on the doubling dimension, the highway dimension, the skeletondimension and the pathwidth of the graph G I , which imply Theorem 1. To bound the doublingdimension, we exploit the fact that the individual gadgets G i,j are connected in a grid-likefashion. This means that we can bound the diameter of balls within this grid. For that purpose,let A i,j ( d ) = { ( i (cid:48) , j (cid:48) ) ∈ [ χ ] | | i (cid:48) − i | + | j (cid:48) − j | ≤ d } . Moreover, let V i,j ( d ) be the vertices of allgadgets G i (cid:48) ,j (cid:48) satisfying ( i (cid:48) , j (cid:48) ) ∈ A i,j ( d ) and the vertices on the paths P i (cid:48) ,j (cid:48) and P (cid:48) i (cid:48) ,j (cid:48) betweenthese gadgets. We now bound the diameter of the graph induced by V i,j ( d ). Lemma 12.
Consider the graph induced by V i,j ( d ) . Its diameter is at most (2 n +3 + 2 n +1 + 2 n +2) · (2 d + 1) . Moreover, if | A i,j ( d ) | = (2 d + 1) , i.e. A i,j ( d ) contains all possible index pairs, thediameter is at least (2 n +2 + 2 n ) · (2 d + 1) .Proof. Let ( i, j ) ∈ [ χ ]. We first bound the distance between any x hi,j and x h (cid:48) i,j . Claim 13.
For h, h (cid:48) ∈ [4] where h (cid:54) = h (cid:48) we have n +2 + 2 n < dist( x hi,j , x h (cid:48) i,j ) ≤ n +3 + 2 n +1 + 4 .Proof. The upper bound follows directly from Equation (1) in the proof of Lemma 6. For thelower bound, observe that the shortest path between x hi,j and x h (cid:48) i,j needs to pass through twovertices v h ( a,b ) and v h (cid:48) ( a (cid:48) ,b (cid:48) ) of the cycle O i,j . It holds that the distance from x hi,j to v h ( a,b ) andfrom x h (cid:48) i,j to v h (cid:48) ( a (cid:48) ,b (cid:48) ) are d h ( a,b ) and d h (cid:48) ( a (cid:48) ,b (cid:48) ) , which are both at least 2 n . Moreover, the distancefrom v h ( a,b ) to v h (cid:48) ( a (cid:48) ,b (cid:48) ) is minimized, if v h ( a,b ) = v h ( n,n ) and v h (cid:48) ( a (cid:48) ,b (cid:48) ) = v h (cid:1) , . As dist( v h ( n,n ) , v h (cid:1) , ) =2 n +2 + / n + d (1 , − d ( n,n ) = 2 n +1 +2 n +1+ / n , a lower bound of 2 · n +2 n +1 +2 n +1+ / n > n +2 +2 n on dist( x hi,j , x h (cid:48) i,j ) follows. (cid:67) Consider the graph induced by V i,j ( d ). Any shortest path in this graph traverses at most 2 d + 1gadgets and contains at most 2 d paths between two gadgets. These paths have length at most2 n −
2. Moreover, it follows from the proof of Lemma 6 that the diameter of a single gadget is9t most 2 n +3 + 2 n +1 + 4. This means that the distance of any shortest path is upper boundedby (2 d + 1) · (2 n +3 + 2 n +1 + 2 n + 2).If | A i,j ( d ) | = (2 d + 1) , the shortest path from x i − d,j to x i + d,j has to traverse 2 d + 1 gadgetshence a lower bound of (2 d + 1) · (2 n +2 + 2 n ) on the diameter of the graph induced by V i,j ( d )follows.This allows us to show that the doubling dimension of G I is constant. Lemma 14.
The graph G I is planar and has constant doubling dimension.Proof. It can be seen easily that G I is planar. Recall that a graph has doubling dimension atmost d if any ball of radius 2 r can be covered with 2 d balls of radius r .To bound the doubling dimension of G I , consider a ball B r ( v ) of radius 2 r around somevertex v ∈ V . Lemma 6 implies that there is a vertex y i,j satisfying dist( v, y i,j ) ≤ n +2 +2 n +1 . Itfollows that the ball B r ( v ) is contained in the ball around y i,j that has radius 2 n +2 + 2 n +1 + 2 r .Moreover, Lemma 12 implies that the latter ball in turn is contained in V i,j ( d ) if 2 · (2 n +2 +2 n +1 + 2 r ) ≤ (2 n +2 + 2 n ) · (2 d + 1). This is true for 2 d + 1 = 6 r/ n +2 and r ≥ n +2 + 2 n +1 .We now show that we can cover the vertices V i,j ( d ) through a constant number of balls thathave radius r and are centered at vertices y i (cid:48) ,j (cid:48) . Lemma 12 implies that for every ( i (cid:48) , j (cid:48) ) ∈ [ χ ] ,the ball B r ( y i (cid:48) ,j (cid:48) ) contains the set V i (cid:48) ,j (cid:48) ( d (cid:48) ) if 2 r ≥ (2 n +3 + 2 n +1 + 2 n + 2) · (2 d (cid:48) + 1). This isthe case for 2 d (cid:48) + 1 = 2 r/ n +4 . As we want V i (cid:48) ,j (cid:48) ( d (cid:48) ) to be nonempty, we require d (cid:48) ≥
0, whichholds for r ≥ n +3 . Hence it suffices to show that there is a constant number of sets V i (cid:48) ,j (cid:48) ( d (cid:48) )whose union contains V i,j ( d ).As it was observed in [10], the index set A i,j ( d ) is contained in the union of (cid:100) d +12 d (cid:48) +1 (cid:101) indexsets A i (cid:48) ,j (cid:48) ( d (cid:48) ). It follows that we can cover the vertices V i,j ( d ) through (cid:100) d +12 d (cid:48) +1 (cid:101) vertex sets V i (cid:48) ,j (cid:48) ( d (cid:48) ). Hence, for r ≥ n +3 we can cover B r ( v ) with (cid:100) d +12 d (cid:48) +1 (cid:101) = (cid:100) r/ n +2 r/ n +4 (cid:101) = 144 balls ofradius r .Assume now that r < n +3 . We already showed that B r ( v ) is contained in V i,j ( d ) if2 d + 1 = 6 r/ n +2 <
12, which implies d <
6. Hence, the ball B r ( v ) intersects at most | A i,j (5) | ≤ (2 · = 121 gadgets G i (cid:48) ,j (cid:48) . We show that we can cover any of these gadgets G i (cid:48) ,j (cid:48) and the paths to its neighboring gadgets through a constant number of ball B r ( w ).If r ≥ n +1 , we can choose the 9 balls centered at y i (cid:48) ,j (cid:48) , z hi (cid:48) ,j (cid:48) and x hi (cid:48) ,j (cid:48) where h ∈ [4], as forevery w ∈ O i (cid:48) ,j (cid:48) there is some z hi (cid:48) ,j (cid:48) satisfying dist( z hi (cid:48) ,j (cid:48) , w ) ≤ n +1 , for h ∈ { , } and b ∈ [ n ]it holds that dist( x hi (cid:48) ,j (cid:48) , u hb ) ≤ n +1 and the length the paths to the neighboring gadgets havelength at most 2 n − r < n +1 . If v = y i (cid:48) ,j , i.e. the ball B r ( v ) is centered at y i (cid:48) ,j , we can choose X i (cid:48) ,j (cid:48) = { y i (cid:48) ,j (cid:48) , z hi (cid:48) ,j (cid:48) | h ∈ [4] } . Otherwise, B r ( v ) ∩ O i (cid:48) ,j (cid:48) is a subpath of O i (cid:48) ,j (cid:48) that has length atmost 4 r , which can be covered by 4 balls of radius r . Similarly, we can also cover B r ( v ) ∩ U hi (cid:48) ,j (cid:48) , B r ( v ) ∩ P i (cid:48) ,j (cid:48) and B r ( v ) ∩ P (cid:48) i (cid:48) ,j (cid:48) with 4 balls of radius r each. This means that we can cover B r ∩ G i (cid:48) ,j (cid:48) through a constant number of balls of radius r .It follows that we can cover any ball B r ( v ) for any v ∈ V and any r > r , which completes the proof.We next bound the highway dimension of G I . Lemma 15.
The graph G I has highway dimension hd ∈ O ( χ ) .Proof. For any radius r > H r such that every shortest path π satisfying | π | > r intersects H r and moreover, for every vertex v ∈ V we have | H r ∩ B r ( v ) | ∈ O ( χ ).Let X = { y i,j , x hi,j , z hi,j | ( i, j ) ∈ [ χ ] , h ∈ [4] } . For r ≥ n +2 we choose H r = X . We have | H r | = 9 χ and hence for every vertex v ∈ V we have | H r ∩ B r ( v ) | ∈ O ( χ ). We show now thatany shortest path of length more than r intersects H r . Clearly, all shortest paths that are notcompletely contained within one single gadget are hit by H r as all x hi,j are contained in H r and10he paths P i,j and P (cid:48) i,j between the individual gadgets have length at most 2 n −
2. Considersome gadget G i,j . All edges of the cycle O i,j have length at least / n and for any h ∈ [4] we havedist( z hi,j , z h (cid:1) i,j ) = 2 n +2 + / n . Hence, any subpath of O i,j that has length at least 2 n +2 intersects H r . Moreover, for h ∈ { , } , the path U hi,j has length 2 n − π ( s, t ) where s ∈ O i,j and t ∈ U hi,j . Let t = u hb .According to Lemma 5, the shortest path π ( s, t ) traverses exactly one portal edge { ρ hβ , u hβ } where β ∈ [ b ]. This means that dist( s, t ) = dist( s, ρ hβ ) + dist( ρ hβ , u hb ) ≤ dist( s, ρ hβ ) + 2 b . Thevertex s is contained in the shortest path π ( z hi,j , ρ hβ ) or in π ( ρ hβ , z h (cid:1) i,j ). In the first case we havedist( s, ρ hβ ) < dist( z hi,j , ρ hβ ) ≤ β . This implies that dist( s, t ) < β + 2 b ≤ n +1 . In the secondcase we have dist( s, ρ hβ ) ≤ n +2 − β and moreover Lemma 5 implies that β = b . Hence weobtain dist( s, t ) ≤ n +2 − β + 2 β = 2 n +2 . This means that every shortest path of length morethan r ≥ n +2 is hit by H r .Let now r < n +2 . For a shortest path p = v . . . v ν and q > p (cid:104) q (cid:105) be a q -cover of p ,i.e. we have p (cid:104) q (cid:105) ⊆ { v , . . . , v ν } such that any subpath of p that has length at least q containssome node from p (cid:104) q (cid:105) . We consider q -covers p (cid:104) q (cid:105) that are constructed greedily, i.e. we start with p (cid:104) q (cid:105) = { v } and iteratively add the closest vertex that has distance at least q . For ( i, j ) ∈ [ χ ] let X i,j = (cid:91) h ∈ [4] O hi,j (cid:104) r/ (cid:105) ∪ (cid:91) h ∈{ , } U hi,j (cid:104) r/ (cid:105) ∪ (cid:8) u n , u n (cid:9) ∪ P (cid:104) r/ (cid:105) i,j ∪ P (cid:48) i,j (cid:104) r/ (cid:105) and choose H r = X ∪ (cid:83) ( i,j ) ∈ [ χ ] X i,j . Consider some shortest path π ( s, t ) that has length morethan r . Clearly, π ( s, t ) is hit by H r if it contains some node from X or it is a subpath of somecycle O i,j , some path U hi,j or some path P i,j or P (cid:48) i,j . It remains to be shown that π ( s, t ) is alsohit by H r if s ∈ O i,j and t ∈ U hi,j . Let t = u hb . Lemma 5 implies that π ( s, t ) consists of a subpath p of O i,j , a portal edge { ρ hβ , u hβ } and a subpath p (cid:48) of U hi,j . Assume that π ( s, t ) is not hit by H r . By the choice of X i,j we have | p | < r / and | p (cid:48) | < r / . This means that dist( ρ hβ , u hβ ) > r / .By construction of the graph G I we have dist( ρ hβ , u hβ ) ≤ β and hence 2 β > r / . As we have u hβ (cid:54)∈ X i,j , it holds that β (cid:54)∈ { , n } and moreover it follows from the choice of U hi,j (cid:104) r/ (cid:105) , thatdist( u hβ − , u hβ ) ≤ r / . However, by construction of G I we have dist( u hβ − , u hβ ) = 2 β − , whichimplies 2 β ≤ r / , a contradiction to 2 β > r / . This means that every shortest path of lengthmore than r is hit by H r .Finally we have to show that for every vertex v ∈ V we have | H r ∩ B r ( v ) | ∈ O ( χ ). As forthe r / -cover of some shortest path p we have | B r ( v ) ∩ p (cid:104) r/ (cid:105) | ∈ O (1), it follows that for every( i, j ) ∈ [ χ ] we have | B r ( v ) ∩ X i,j | ∈ O (1). Moreover there are χ different sets X i,j and wehave | X | = 9 χ , which implies | H r ∩ B r ( v ) | ∈ O ( χ ).Observe, that for any graph G of highway dimension hd and maximum degree ∆, an upperbound of (∆ + 1) hd on the skeleton dimension of G follows [12]. As the graph G I has maximumdegree ∆ = 4, it follows that the skeleton dimension of G I is bounded by O ( χ ).However, with some more effort, we can show a stronger bound of O ( χ ). We will use thefollowing lemma, which was shown in [5]. Lemma 16.
Consider vertices u, v, w ∈ V such that v is contained in π ( u, w ) . If w is containedin the skeleton of u , it is also contained in the skeleton of v . We now bound the size of a skeleton within a single gadget. For simplicity, in the followingwe confuse a graph G and its geometric realization ˜ G . Lemma 17.
For any ( i, j ) ∈ [ χ ] and any vertex s contained in G i,j , the subtree of the skeleton T ∗ s induced by the vertices of G i,j is the union of a constant number of paths. roof. We first show that every skeleton contains only a limited number of portal edges. Recallthat the skeleton of a shortest path tree is defined on the geometric realization, where everyedge is subdivided into infinitely many infinitely short edges. We refer to vertices that wereintroduced during this subdivision as interior vertices . Claim 18.
Consider a vertex s = v h ( a,b ) for ( a, b ) ∈ [ n ] and h ∈ { , } . If the skeleton T ∗ s of s contains an interior vertex of a portal edge { ρ hβ , u hβ } , we have β ∈ { b, b − } .Proof. Assume h = 1. For β > b , it follows from Lemma 5 that { ρ hβ , u hβ } is not contained in theshortest path tree of s and hence, no interior vertex of { ρ hβ , u hβ } can be contained in T ∗ s . Let β < b − ρ hβ = v h ( α,β ) . Lemma 5 implies that u hβ is the furthest descendant of ρ hβ in theshortest path tree T s , and we have dist( ρ hβ , u hb ) < β . Moreover, the distance from v h ( a,b ) to ρ hβ is d ( a,b ) − d ( α,β ) = 2 b − β + ( a − α ) / n > β +1 > / · dist( ρ hβ , u hb ). This means that no interior vertexof { ρ hβ , u hβ } can be contained in T ∗ s . The case h = 3 can be shown similarly. (cid:67) Claim 19.
Consider a vertex s = u hb for b ∈ [ n ] and h ∈ { , } . If the skeleton T ∗ s of s containsan interior vertex of a portal edge { ρ hβ , u hβ } , we have β ∈ { b, b − , } .Proof. Assume h = 1. For β > b , it follows from Lemma 5 that { ρ hβ , u hβ } is not contained inthe shortest path tree of s and hence, no interior vertex of { ρ hβ , u hβ } can be contained in T ∗ s .Let 1 < β < b − ρ hβ = v h ( α,β ) . It follows from Lemma 5 that the furthest possibledescendant of u hβ within the shortest path tree of s is v h (1 ,β ) , which has distance 2 β − / n from u hβ . The distance from u hb to u hβ is dist( u hb , u hβ ) = 2 b − β > β +1 > dist( u hβ , v h (1 ,β ) ) and hence, nointerior vertex of { ρ hβ , u hβ } can be contained in T ∗ s . The case h = 3 can be shown similarly. (cid:67) Now, let T s be the shortest path tree of s and T s [ G i,j ] be the subtree of T s induced by thevertices of G i,j . If we disregard all portal edges { ρ hb , u hb } , it follows from Lemma 5 that T s [ G i,j ]consists of a constant number of subpaths of the cycle O i,j , of the two paths U i,j and U i,j , the 4edges incident to x i,j , . . . , x i,j and some of the edges incident to y i,j . Hence, if we do not countthe portal edges, the subtree T ∗ s [ G i,j ] of the skeleton T ∗ s induced by G i,j consists of a constantnumber of paths. It remains to be shown that T ∗ s [ G i,j ] intersects only a constant number ofportal edges.Assume that s = ρ hb for h ∈ { , } and b ∈ [ n ]. It follows from Claim 18 that only { ρ hb − , u hb − } and { ρ hb , u hb } can intersect T ∗ s [ G i,j ]. Consider now a portal edge { ρ h (cid:1) β , u h (cid:1) β } on the oppositeside of the cycle. It follows from Lemma 16 that it can only intersect T ∗ s [ G i,j ], if it also intersectsthe skeleton of ρ h (cid:1) or ρ h (cid:1) n , which according to Claim 19 holds only for β ∈ { , n − , n } . Thismeans that T ∗ s [ G i,j ] consists of a constant number of paths. If s ∈ U h for h ∈ { , } , we canshow the same using Claim 19 and Lemma 16.Assume now that s is contained in the cycle O i,j , but not a portal. Let u and v be the twoclosest portals such that s is contained in π ( u, v ). It follows from Lemma 16, that T ∗ s [ G i,j ] isa subgraph of T ∗ u [ G i,j ] ∪ T ∗ v [ G i,j ] ∪ π ( u, v ) and hence, it is the union of a constant number ofpaths. For similar reasons, the same holds if s = y i,j or s = x hi,j .Moreover we can show that every cut in any skeleton of G I intersects at most O ( χ ) differentgadgets and connecting paths between two gadgets. Lemma 20.
For every vertex s ∈ V and every radius r > , Cut rs intersects O ( χ ) gadgets G i,j and O ( χ ) paths P i,j and P (cid:48) i,j .Proof. It can be shown that for any ( i, j ) ∈ [ χ ] , we have dist( y i,j , y i +1 ,j ) = 2 n +3 + 4 + / n anddist( y i,j , y i,j +1 ) = 2 n +3 + 3 + / n . This means that for any ( i, j ) , ( i (cid:48) , j (cid:48) ) ∈ [ χ ] we havedist( y i,j , y i (cid:48) ,j (cid:48) ) = | i − i (cid:48) | · (2 n +3 + 4 + / n ) + | j − j (cid:48) | · (2 n +3 + 3 + / n ) . (2)12et r > , s ∈ V and consider a vertex v ∈ Cut rs . It holds that dist( s, v ) = r . Accordingto Lemma 6 there are two central vertices y i,j and y i (cid:48) ,j (cid:48) satisfying dist( s, y i,j ) ≤ n +2 + 2 n +1 and dist( v, y i (cid:48) ,j (cid:48) ) ≤ n +2 + 2 n +1 . Using the triangle inequality we obtain that dist( y i,j , y i (cid:48) ,j (cid:48) ) ∈ [ r − , r + ] where r − = r − (2 n +3 + 2 n +2 ) and r + = r + 2 n +3 + 2 n +2 . Moreover, the ball around y i (cid:48) ,j (cid:48) of radius 2 n +2 + 2 n +1 intersects O (1) gadgets G i (cid:48)(cid:48) ,j (cid:48)(cid:48) and O (1) paths P i (cid:48)(cid:48) ,j (cid:48)(cid:48) and P (cid:48) i (cid:48)(cid:48) ,j (cid:48)(cid:48) . Thismeans that any bound on the size of the set Y = { y i (cid:48) ,j (cid:48) | dist( y i,j , y i (cid:48) ,j (cid:48) ) ∈ [ r − , r + ] } yields abound on the number of gadgets and paths intersecting Cut rs .Consider now a vertex y i (cid:48) ,j (cid:48) ∈ Y . Assume that i (cid:48) ≥ i and consider some i ∗ ≥ i (cid:48) + 4. Itfollows from Equation (2) and dist( y i,j , y i (cid:48) ,j (cid:48) ) ≥ r − thatdist( y i,j , y i ∗ ,j (cid:48) ) ≥ dist( y i,j , y i (cid:48) ,j (cid:48) ) + 4 · (2 n +3 + 4 + / n ) > r + . This means that y i ∗ ,j (cid:48) (cid:54)∈ Y and it follows that for any j (cid:48) ∈ [ χ ] we have |{ i ∗ ≥ i | y i ∗ ,j (cid:48) ∈ Y }| ≤ |{ i ∗ ≤ i | y i ∗ ,j (cid:48) ∈ Y }| ≤ j (cid:48) ∈ [ χ ] . This implies | Y | ∈ O ( χ ),which completes the proof.Combining Lemmas 16, 17 and 20, we obtain that the skeleton dimension of G I is boundedby O ( χ ). Lemma 21.
The graph G I has skeleton dimension κ ∈ O ( χ ) .Proof. Let s ∈ V, r > rs . Any vertex v ∈ Cut rs is either contained in somegadget G i,j or some connecting path P i,j or P (cid:48) i,j .We start with bounding the number of vertices that are contained in Cut rs and some P i,j or P (cid:48) i,j . For any ( i, j ) ∈ [ χ ] we have | Cut rs ∩ P i,j | ≤ P i,j contains at most two distinct verticesthat have the same distance from s . For the same reason we have | Cut rs ∩ P (cid:48) i,j | ≤
2. HenceLemma 20 implies that the size of Cut rs ∩ { P i,j , P (cid:48) i,j | ( i, j ) ∈ [ χ ] } is bounded by O ( χ ).Consider now some gadget G i,j . We show that | Cut rs ∩ G i,j | ∈ O (1). If s is contained in G i,j this follows immediately from Lemma 17, as Cut rs intersects any path in T ∗ s at most twice. If s is not contained in G i,j , Lemma 16 implies that Cut rs ∩ G i,j is a subset of (cid:26) Cut r ( h ) x hi,j | h ∈ [4] and r ( h ) = r − dist( s, x hi,j ) (cid:27) ∩ G i,j . Observe that every x hi,j is contained in G i,j , which means that | Cut r ( h ) x hi,j ∩ G i,j | ∈ O (1). This meansthat the size of Cut rs ∩ { G i,j | ( i, j ) ∈ [ χ ] } is bounded by O ( χ ). Hence we have | Cut rs | ∈ O ( χ )and it follows that G I has skeleton dimension κ ∈ O ( χ ).Finally we bound the pathwidth of the graph G I . Lemma 22.
The graph G I has pathwidth pw ∈ O ( χ ) .Proof. Consider the graph ˆ G I that arises when we contract all vertices of degree 2 except thevertices x hi,j . It suffices to show that ˆ G I has pathwidth at most O ( χ ). For ( i, j ) ∈ [ χ ] denotethe gadget G i,j and the cycle O i,j after the contraction by ˆ G i,j and ˆ O i,j , respectively. We firstconstruct a path decomposition of constant width for every ˆ G i,j . To this end, consider the cycleˆ O i,j , which (as every cycle) has a path decomposition where every bag has size at most 3. For h ∈ { , } and b ∈ [ n ], add u hb to every bag containing the portal ρ hb . Finally, add y i,j and x i,j , . . . , x i,j to every bag. This yields a path decomposition of ˆ G i,j which has constant width.We now combine the path decompositions of the gadgets ˆ G i,j to a path decomposition of ˆ G I .For ( i, j ) ∈ [ χ ] , consider the path decomposition of ˆ G i,j and add the vertices { x i (cid:48) ,j (cid:48) , . . . x i (cid:48) ,j (cid:48) | ≤ ( i (cid:48) − i ) · χ + ( j (cid:48) − j ) ≤ χ } to every bag. According to Figure 1, these are the vertices x hi (cid:48) ,j (cid:48) of the χ gadgets after ˆ G i,j when considering the gadgets row-wise from left to right. Denotethe resulting path decomposition by P ( i − · χ + j . We can observe, that its width is bounded O (1) + 4 χ . Concatenating all these path decompositions as P P . . . P χ then yields a pathdecomposition of ˆ G I of width O (1) + 4 χ , which concludes the proof.13 Conclusion
The properties shown in the previous section now imply Theorem 1. As the GT ≤ problemis W [1]-hard for parameter χ and we have k = 5 χ , hd ∈ O ( χ ) , κ ∈ O ( χ ) and pw ∈ O ( χ ),it follows that on planar graphs of constant doubling dimension, k -Center is W [1]-hard forparameter ( k, pw, hd, κ ). Assuming ETH there is no f ( χ ) · n o ( χ ) time algorithm for GT ≤ andhence, k -Center has no f ( k, hd, pw, κ ) · | V | o ( pw + κ + √ k + h ) time algorithm unless ETH fails.It follows that on planar graphs of constant doubling dimension, k -Center has no fixed-parameter algorithm for parameter ( k, pw, hd, κ ) unless FPT=W[1]. Moreover, it was shownthat k -Center has no efficient (2 − (cid:15) )-approximation algorithm for graphs of highway dimension hd ∈ O (log | V | ) [9] or skeleton dimension κ ∈ O (log | V | ) [4].Still, combining the paradigms of approximation and fixed-parameter algorithms allows oneto compute a (2 − (cid:15) ) approximation for k -Center on transportation networks. For instance,there is a 3 / O ( k · hd log hd ) · n O (1) for highway dimen-sion hd [9] and a (1 + (cid:15) )-approximation algorithm with runtime ( k k /(cid:15) O ( k · d ) ) · n O (1) for doublingdimension d [10]. As the doubling dimension is bounded by O ( κ ), the latter result implies a(1 + (cid:15) )-approximation algorithm that has runtime ( k k /(cid:15) O ( k · κ ) ) · n O (1) .On the negative side, there is no (2 − (cid:15) )-approximation algorithm with runtime f ( k ) · n O (1) for any (cid:15) > f unless W[2]=FPT [10]. It remains open, to what extent thepreviously mentioned algorithms can be improved. References [1] Ittai Abraham, Daniel Delling, Amos Fiat, Andrew V. Goldberg, and Renato F. Werneck.Highway dimension and provably efficient shortest path algorithms.
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