Approximating Geometric Coverage Problems

Transcription

1 Approximating Geometric Coverage Problems Thomas Erlebach Erik Jan van Leeuwen Abstract We present the first study on the approximability of geometric versions of the unique coverage problem and the minimum membership set cover problem. In the former problem, one is given a family of sets of elements from some universe and aims to select sets that maximize the number of elements contained in precisely one selected set. Unique Coverage has important applications in wireless networks and it is thus natural to consider it in a geometric setting. We use the well-known (unit) disk model for wireless networks and show that Unique Coverage remains NP-hard in this case and present a polynomial-time 1/18-approximation algorithm. This algorithm is extended to the budgeted lowcoverage problem, where covering an element multiple times yields less profit and we have a fixed budget to buy sets. We give an asymptotic FPTAS in case the disks have arbitrary size, but bounded ply. For the case that the geometric objects are arbitrary fat objects, we show that these problems are as hard to approximate as in the general case. In the minimum membership set cover problem, the goal is to cover all elements while minimizing the maximum number of sets in which any element is contained. For unit squares and unit disks, we show that the problem remains NP-hard and does not admit a polynomial-time approximation algorithm with ratio smaller than 2 unless P=NP. For unit squares, we give a 5-approximation algorithm for instances where the optimum objective value is bounded by a constant. 1 Introduction Let U be a universe, P some (finite) subset of U, and F a family of subsets of U. Given F F, we say that an element u P is uniquely covered by F if there is precisely one F F containing u. The (maximum) unique coverage problem is to find an F F maximizing the number of uniquely covered elements of P. One can generalize this problem to Budgeted (Maximum) Unique Coverage, where each element u Phas a profit p(u), each F Facostc(F ), there is a budget B, and one has to find an F F of Department of Computer Science, University of Leicester, University Road, Leicester LE1 7RH, United Kingdom. T.Erlebach@mcs.le.ac.uk CWI, Kruislaan 413, 1098 SJ Amsterdam, the Netherlands. E.J.van.Leeuwen@cwi.nl. Partially supported by the Bsik project BRICKS. total cost at most B that maximizes the total profit of the uniquely covered elements of P. The (budgeted) unique coverage problem was proposed by Demaine et al. [7] and is mainly motivated by its applicability in wireless networks. Providers of wireless communication networks should provide service for their customers. This can be achieved by placing a number of base stations that cover customer locations. However, if too many base stations cover a certain customer location, the resulting interference might cause this customer to receive no service at all. Ideally, each customer is serviced by exactly one base station and service is provided to as many customers as possible. Of course, it might be more profitable to provide service to certain customers. Furthermore, placing a base station is costly and providers generally have a limited budget. This gives rise to the budgeted unique coverage problem. In practice, mobile devices can distinguish between signals from different base stations. However, this capability is limited and decreases with the number of base stations in range. Demaine et al. model this by satisfactions s 0 = 0, s 1 s 2 0, where a customer u yields satisfaction modulated profit s i p(u) if it receives service from exactly i base stations. The aim of Budgeted Low-Coverage is to maximize this satisfaction modulated profit over all customers, given the cost function and a limited budget. The applicability of these problems to wireless networks research is the main reason to study them in a geometric setting. The universe is the plane R 2, the set P of points corresponds to customer locations, and each F F is a geometric object modeling the broadcasting range of the corresponding base station. If all base stations are equivalent and we ignore obstacles to the signal, these geometric objects are unit disks. We prove Unique Coverage (and thus Budgeted Low- Coverage as well) remains NP-hard for unit disks and give polynomial time, constant-factor approximation algorithms for both problems on unit disks. One can make the problem more realistic by assuming the base stations may have different broadcasting ranges and they are hindered by obstacles, but the overlap of the broadcasting ranges of the potential base station locations is bounded. The latter assumption is rea- 1267

2 sonable, as in practice there are usually very few spots where a base station can or may be placed. We model this situation by a set of fat objects where any point in the plane is overlapped by a bounded number of objects, i.e. a set of fat objects of bounded ply. We show Budgeted Low-Coverage has an asymptotic FPTAS on fat objects of bounded ply, but prove that without the bounded ply assumption, the problem is as hard to approximate as in its general setting. Related to Budgeted Low-Coverage is Minimum Membership Set Cover. One has to cover every point of P minimizing the maximum overlap on any point. It has an O(ln n)-approximation algorithm and has no (1 ɛ)lnn-approximation for any ɛ>0, unless NP DTIME(n O(log log n) ) [16]. Again, as the problem is motivated by wireless network design, it is natural to consider the geometric case. We prove approximating the problem with ratio less than 2 is NP-hard for unit disks and unit squares, and give a 5-approximation algorithm for unit squares provided that the optimal objective value is bounded by a constant. 1.1 Related Work. Demaine et al. [7] formulated the unique coverage problem and studied it in its general setting. They present a polynomial time Ω(1/ log ρ) = Ω(1/ log n) approximation algorithm using a greedy method, where n is the number of elements and ρ is one plus the ratio of the maximum number and the minimum number of sets in which an element is contained. They give hardness results to show this algorithm is near-optimal. For any ɛ>0, it is hard to approximate Unique Coverage with ratio Ω(1/ log σ(ɛ) n), assuming NP BPTIME(2 nɛ ), where σ(ɛ) is some constant dependent on ɛ. Demaine et al. give a gap preserving reduction from a variant of Balanced Binary Independent Set, so any (in)approximability result for this problem holds for Unique Coverage as well. Moser et al. [18] study the parameterized complexity of Unique Coverage. The unique hitting set problem, where one tries to select elements to uniquely hit as many sets as possible, is equivalent to Unique Coverage. If sets have size at most k, Guruswami and Trevisan [13] give an Ω(1/ log k)-approximation algorithm. They also consider the more general problem 1-in-k SAT and give a 1 e -approximation algorithm for satisfiable instances. To our knowledge, the unique coverage problem and its extensions have not been studied in a geometric setting, although several related problems have. If one tries to maximize the number of points that are uniquely covered subject to the constraint that the selected objects are disjoint, this is a maximum weight independent set problem, which has a PTAS for fat objects of arbitrary ply [4, 9, 19]. For a grid-based version of Unique Coverage with unit disks, where the disks are restricted to lie at grid points, a PTAS has been announced by Lev-Tov and Peleg [17]. Our asymptotic FPTAS for fat objects of bounded ply is significantly more general than this result. The algorithm by Brönnimann and Goodrich for geometric Set Cover yields an O(1)- approximation algorithm for unit disks [2] (see [3] for recent improvements). Clarkson and Varadarajan [6] present an O(1)-approximation algorithm for pseudodisks. A 72-approximation algorithm for the weighted version of this problem for unit disks was presented by Ambühl et al. [1]. There are also several papers that consider base station placement problems. Glaßer, Reith, and Vollmer [12] give approximation algorithms and PTASs for several base station placement problems, focusing mainly on various interference models. Lev-Tov and Peleg [17] give a PTAS for the variation where one has to minimize the total sum of the radii of the selected base stations. 1.2 Our Results. We begin in 2 by showing Unique Coverage on unit disks is NP-hard. We then give a polynomial time 1/18-approximation algorithm. This algorithm is extended to a (1 ɛ)/18-approximation for Budgeted Low-Coverage on unit disks for any ɛ>0, where the running time is polynomial in 1/ɛ. The ratio improves to (1 ɛ)/4 for unit squares. In 3 we consider Budgeted Low-Coverage on fat objects of bounded ply. We start by giving an asymptotic FPTAS for Budgeted Unique Coverage on disks of bounded ply, relying on an advanced variation of the shifting technique. This approximation scheme can then easily be extended to fat objects of bounded ply. In 4, we prove that for fat objects of arbitrary ply, Unique Coverage is as hard to approximate as in its general version. Finally, in 5 we deal with geometric Minimum Membership Set Cover. We show NP-hardness for approximating the problem with ratio less than 2 on unit disks and unit squares, and we give a 5-approximation algorithm for unit squares provided that the optimal objective value is bounded by a constant. 1.3 Definitions. For a maximization (minimization) problem, an algorithm is an α-approximation algorithm if the ratio of the value of the solution output by the algorithm and the value of the optimum solution is at least α (at most α). If the approximation factor α is some positive constant, we call it a constant-factor approximation algorithm. For a maximization problem, an algorithm is a polynomial-time approximation scheme or PTAS if it gives a (1 ɛ) approximation in time polynomial in the size of the input, for any (fixed) ɛ>0. If the running time is polynomial in 1/ɛ as well, we speak of a 1268

3 u v By scaling, w.l.o.g. all disks in D have radius 1/2. We aim to find D Dmaximizing the number of uniquely covered points of P. We apply the well-known shifting technique [14] in a novel way. Figure 1: The edge gadget for edge (u, v), shown together with the vertex disks for u and v. fully polynomial-time approximation scheme or FPTAS. If the algorithm runs in time polynomial in the size of the input and 1/ɛ, but guarantees a (1 ɛ) approximation only if the input size is at least c ɛ, where c ɛ is a constant only dependent on ɛ, we call it an asymptotic FPTAS or FPTAS ω. 2 Unique Coverage on Unit Disks This section studies the complexity and approximability of Unique Coverage on unit disks. Theorem 2.1. Unique Coverage on unit disks is NPhard. Proof. (Sketch) We reduce from the NP-hard Independent Set on planar graphs of maximum degree 4 [11]. The starting point of our construction is similar to one used to prove NP-hardness of Independent Set on unit disks graphs [5]. For a planar graph G of degree 4, create a rectilinear embedding of G. Replace each vertex by a disk with a point in its center and each edge by a gadget, shown in Fig. 1. The even-sized set of (solid) disks connecting u and v for each edge (u, v) E(G) is denoted D (u,v). A single (dashed) disk contains the middle black blob. The small grey points correspond to a single point. Each disk of the construction, except the middle disk of each edge gadget (drawn dashed in Fig. 1), contains a unique grey point. Each big black blob corresponds to a collection of t = V (G) + (u,v) E(G) D(u,v) points. We claim in the constructed instance of Unique Coverage, a set of unit disks can uniquely cover at least k := k+ 1 ( 2 (u,v) E(G) D (u,v) +2t( D (u,v) +1) ) points if and only if G has an independent set of size at least k. For the only if part, note that any set of disks uniquely covering at least k points must uniquely cover all black blobs. This can only be done using exactly 1 2 D(u,v) disks of each edge gadget (which uniquely cover 1 2 (u,v) E(G) D(u,v) grey points in total) and at least k vertex disks. By the construction of the edge gadgets, these vertex disks form an independent set, which indeed has size at least k. We can now easily verify the if part of the claim. 2.1 Approximation Algorithm. Let D be a set of equally sized disks and P a set of points, both in R 2. Lemma 2.1. Suppose for points in a square of size δ δ, 0 <δ 1, the unique coverage problem on unit disks has a polynomial time 1/c-approximation algorithm. Then there is a polynomial time 1/c δ 2 /(1+δ) 2 -approximation algorithm for the general unique coverage problem on unit disks. Proof. Let OPT be an optimal solution to the unique coverage problem on unit disks for some set of disks D and set of points P. Pick two numbers a, b uniformly at random from [0, 1 + δ). Consider the set of squares S = {[a + h + hδ, a + h +(h +1)δ) [b + v + vδ, b + v+ (v +1)δ) h, v Z}. As these squares have pairwise distance at least one, no unit disk can cover a point in more than one square of this set. Hence we may consider these squares to be independent. The probability that a point of P is in a square of S is δ 2 /(1 + δ) 2. Hence E [ S S uc S(OPT ) ] = uc(opt ) δ 2 /(1 + δ) 2, where uc(opt ) is the set of points uniquely covered by OPT and uc S (OPT )is the set of points in S uniquely covered by OPT. By assumption, we can find a 1/c-approximation of the unique coverage problem on unit disks for each of the squares in S in polynomial time. Let size(s) denote the number of points uniquely covered by the solution of the algorithm for a particular square S S. Then size(s) 1/c uc S (OPT ). We can assume the solutions produced by the 1/c-approximation algorithm contain only disks intersecting the square it was invoked on, and we have E [ S S size(s)] 1/c E [ S S uc S(OPT ) ] = 1/c uc(opt ) δ 2 /(1 + δ) 2. This approach can be derandomized. Choices of a, b for which the same set of points is in the squares of S give an approximation of the same quality. Hence it suffices to look at the O( 2 ) values of a, b for which a square boundary hits a point of P and thus we can consider all values in polynomial time. The solution with the highest S S size(s) isa1/c δ2 /(1 + δ) 2 - approximation of the optimum. The proof of the next theorem uses some ideas from Ambühl et al. [1]. Theorem 2.2. Unique Coverage on unit disks has a O( 3 D 8 ) time 1/18-approximation algorithm. Proof. We prove there is a polynomial time 1/2- approximation algorithm for Unique Coverage on unit disks for size 1/2 1/2 squares. Together with Lemma 2.1, this proves the theorem. 1269

4 Consider a size 1/2 1/2 squares containing a set of points P S and intersected by a set of disks D S of radius 1/2. We may assume no disk covers all points of P S, for such a disk would trivially constitute an optimal solution. Construct a mapping of disks in D S to one of the four boundaries of S. The mapping assigns a disk to the boundary of S it overlaps most, breaking ties arbitrarily. If we can solve the unique coverage problem on unit disks optimally for both pairs of opposing boundaries of S, the best solution gives a 1/2-approximation of the optimum for Unique Coverage on unit disks for S. Let OPT S be an optimal solution to the unique coverage problem on unit disks for S. Consider two opposing boundaries b, b of S (say b is the top boundary, b the bottom boundary) and the sets of disks D b, D b assigned to them. Observe that the projection of d S onto b is equal to d b for any disk d D b,ord would overlap another boundary of S more and the mapping would assign it to that boundary. Hence any disk in D b OPT S meets the lower envelope of D b OPT S (for a set of disks, we say that a disk d from the set meets or is on the lower envelope of the set if part of that envelope is formed by the boundary of d), or any point of P S it covers would already be covered by a disk on the lower envelope. Furthermore, for any disk d on the lower envelope of D b OPT S, all points of P S uniquely covered by d are in d d d, where d (d ) is the disk lying directly to the left (right) of d on the lower envelope of D b OPT S. The same properties hold for disks in D b and the upper envelope of D b OPT S. Consider a triple (d, d,d ) such that d D b {s, b}, d D b {b}, d D b {t, b}, d, d,d appear in this order on the lower envelope of d d d,andd d and d d. Here s (t) is a dummy disk that always appears before (after) the first (last) disk on the lower envelope. Such triples are called proper. Ifd t for some proper triple (d, d,d ), then a proper triple (d,d,d )isa successor of (d, d,d ). We also define (d, d,d )tobe successor of itself. We define similar notions for triples (e, e,e ) w.r.t. b. Order the points in P S from left to right w.r.t. non-decreasing x-coordinate, with u 1 the leftmost and u m the rightmost point. Define for any point u i and any pair of proper triples ((d, d,d ), (e, e,e )) with the intersection point of d (e ) and the sweep-line through u i on the lower envelope of d d d (respectively on the upper envelope of e e e ), a function h i via h i [(d, d,d ), (e, e,e )] ι i [(d, d,d ), (e, e,e )] if i =1; = ι i [(d, d,d ), (e, e,e )]+ max h i 1 [( d, d, d ), (ē, ē, ē )] otherwise, where ι i [(d, d,d ), (e, e,e )] is 1 if u i is in d d d or in e e e but not in both, and 0 otherwise. The maximum is over all proper triples ( d, d, d )and(ē, ē, ē ) that are valid parameters of h i 1 and are succeeded by (d, d,d ) and (e, e,e ) respectively. Finally, the maximum value of h m over all (d, d,d ), (e, e,e )isthe optimal solution for this pair of opposing boundaries of S. Ash can be computed in O( P S D S 8 ) time, the theorem follows. 2.2 Budgets and Satisfactions. The above algorithm extends easily to the case where points have associated profits and we aim to maximize the total profit of uniquely covered points. However, for Budgeted Low- Coverage on unit disks, we need to change our approach. In the following, let p max and c max denote the maximum profit and the maximum cost respectively. We may assume the budget is at least the maximum cost and the optimum has profit at least s 1 p max. Theorem 2.3. If s z > 0 and s z+1 =0for some fixed z, then for any ɛ>0, there is a O( 9 D 4z+4 /ɛ 4 ) time (1 ɛ)/18-approximation algorithm for Budgeted Low-Coverage on unit disks. Proof. First assume profits and satisfactions are integers. Recall the proof of Lemma 2.1. Let S be the set of squares under consideration. We use similar ideas as in Theorem 2.2 to compute for both pairs of opposing boundaries of each square S Sand for each r =0,...,p max s 1 a set of disks (if it exists) such that the total satisfaction modulated profit of covered points is at least r and the total cost is minimized. For fixed S, r, the minimum cost over both pairs of opposing boundaries is denoted cost(s, r) and can be computed for all S, r in O( 2 p max s 1 D 4z+4 ) time. Now max{ S S r S S S cost(s, r S) B} gives a 1/2-approximation of S S sp S(OPT ), where sp S (OPT ) is the satisfaction modulated profit accrued by OPT in square S. This maximum is just an instance of Multiple-Choice Knapsack, which we have to solve in a way that avoids having B in the running time. We first compute for each r =0,...,p max s 1 a function g(r) = min{ S S cost(s, r S) S S r S = r} and then choose the largest r such that g(r) B. Using dynamic programming, this can be done in O((p max s 1 ) 2 3 ) time, as the number of non-empty squares of S is at most. Applying the shifting technique as in Lemma 2.1 yields a 1/18-approximation in O((p max s 1 ) 2 5 D 4z+4 ) time. To extend the algorithm to deal with non-integer profits and satisfactions and to achieve polynomial 1270

5 running-time, we apply scaling. Scale the profits by ɛ p max, i.e. define p (u) = p(u) ɛ p max, and scale the satisfactions by ɛ s1, i.e. define s i = si ɛ s 1. We first give an auxiliary inequality. For any u P and any i, ( ɛ s1 s i p(u) ɛ p ) max s i p (u) s i p(u) 2 ɛ s 1 p max. ( s i ɛ s 1 ) ( p(u) ɛ p max For a set of disks D,letsp(D ) denote the satisfaction modulated profit achieved by D under the original profits and satisfaction and let s p (D ) be the satisfaction modulated profit achieved by D under the scaled profits and satisfaction. Apply the above algorithm to obtain a set of disks D that is a 1/18-approximation of the optimum under the scaled profits and satisfactions. Let OPT denote a set of disks giving an optimum satisfaction modulated profit under the budget and the original profits and satisfactions. Then sp(d ) ɛ s 1 ɛ s ɛ p max s p (D ) ɛ p max 1 18 s p (OPT ) ( sp(opt ) 2 ɛ s 1 p max (sp(opt ) 2 ɛ sp(opt )). Observe that s 1,p max /ɛ. The theorem follows. If the satisfactions are different for each client, but all still non-increasing, a similar algorithm may be used. It remains an interesting open problem if Budgeted Low- Coverage on unit disks is approximable for arbitrary z and/or satisfactions that are not non-increasing. When looking at unit squares instead of unit disks, we can find an optimal solution for 1 1 squares. Theorem 2.4. If s z > 0 and s z+1 =0for some fixed z, then for any ɛ>0, there is a polynomial-time (1 ɛ)/4- approximation algorithm for Budgeted Low-Coverage on unit squares. 3 Improved Algorithm for Unique Coverage on Disks of Bounded Ply If the disks have arbitrary size, but have bounded ply, we can improve on the algorithm of the previous section and give an FPTAS ω for Unique Coverage on disks. The ply of a set of disks is the maximum over all points in the plane of the number of disks containing this point. ) ) To obtain the FPTAS ω, we apply the shifting technique [14]. This technique has been used before in the case of bounded disk ply [19]. However, the complexity of the budgeted unique coverage problem on disks forces major changes in this approach. In particular, it is nontrivial to enforce the global budget constraint; we handle this by creating dynamic programming tables that are additionally indexed by profit values and contain the cheapest cost for achieving a certain profit in a given square. The best choice of profit values for disjoint squares can then be addressed as a multiple-choice knapsack problem. Furthermore, relating the profit of the algorithm s solution to that of a modified optimal solution (Lemma 3.1) is significantly more difficult than for the problems studied in [19]. In the proofs below we mainly focus on those aspects different from the standard approach found in [19]. The setup of the algorithm is as follows. By scaling, we may assume all disks have radius at least 1/2. Partition the disks into levels, where a disk with radius r has level j Z 0 if 2 j 1 r<2 j. Then we can define D =j as the set of disks having level exactly j, D j as the set of disks having level at least j, etc.weusel to denote the level of the largest disk. For each level j, partition the plane using a grid induced by horizontal lines y = hk2 j and vertical lines x = vk2 j (h, v Z), where k 5 is an odd integer determined later. The squares of this partition for level j are called j-squares. Anyj-square is contained in precisely one j + 1-square, while each j + 1-square contains exactly four j-squares, denoted S 1,...,S 4 and called siblings. Foraj-square S, letd S denote the set of disks intersecting S and let D b(s) denote the set of disks intersecting the boundary of S. As a shorthand, let D i(s) = D S D b(s) be the set of disks fully contained inside S and D +(S) = 4 i=1 Db(Si) D b(s) be the set of disks intersecting the boundary of a child S i of S, but not the boundary of S itself. Combinations such as D b(s) >j should be self-explanatory. Let P S denote the set of points contained in a j-square S. Finally, j(s) denotes the level of a square S. The following theorem shows that we can efficiently compute a set of disks of cost at most B such that the profit of the uniquely covered points is at least p(uc( i(s) S OPT =j(s) )), the profit obtained by the disks in the optimal solution that are fully contained in a square on their own level. Theorem 3.1. Let D be a set of disks of ply γ, k 5 an odd integer, and OPT a subset of D such that p(uc(opt )) is maximum under c(opt ) B. Then in time O(k 3 D ( D + )2 32kγ/π (p max ) 2 ),wecan findasetd Dsuch that c(d ) B and p(uc(d )) p(uc( i(s) S OPT =j(s) )). 1271

6 To prove this theorem, we apply dynamic programming on the squares. Define for each j-square S, eachset W D b(s) >j, and each r {0,..., p max } the following function cost(s, W, r) min { c(t ) p(uc S (T W )) r; T D i(s) } j if j =0; { = min c(u)+ 4 i=1 cost( ) S i, (U W ) b(si),r i } 4 i=1 r i = r; U D +(S) >j 1 if j>0. The minimum over an empty set is defined to be. Let sol(s, W, r) denote the subset of D attaining cost(s, W, r) ifcost(s, W, r), or otherwise. Note that actually we would only need to define cost(s, W, r) and sol(s, W, r) for subsets W of D b(s) >j ) S S Di(S =j(s ). This will turn up in the analysis of the approximation factor, but it is not very important in the analysis of the (worst-case) running time. We start by proving the function cost is indeed close to the optimum. Lemma 3.1. It holds that max { r S cost(s,,r S ) B } S; j(s)=l p ( ( uc S S; j(s)=l OPT i(s) =j(s) )). Proof. (Sketch) Define up(s) = S S OPT i(s ) =j(s ) and down(s) = S S OPT i(s ) =j(s ). To prove the lemma, we claim that for any j-square S ( ( cost S, (up(s)) b(s),p (uc S (up(s)) b(s) ))) OPT i(s) >j down(s) ( ) c OPT i(s) >j + ( ) S S c OPT i(s ) =j(s ). The intuition behind this formula is that if we consider the set W the optimum uses and the profit r attained by the optimum for S, the cost attained by cost(s, W, r) is at most the cost needed by the optimum. It can be proved by induction on j. As l is the largest level, up(s) = and OPT i(s) >l = for any l-square S. Hence cost (S,,p(uc S (down(s)))) S; j(s)=l and thus S; j(s)=l S S ( ) c OPT i(s ) =j(s ) c(opt ) B max{ S; j(s)=l r S S; j(s)=l cost(s,,r S) B } S; j(s)=l p (uc S (down(s))) ( ( )) i(s) = p uc SOPT =j(s). This proves the lemma. We say a j-square S is non-empty if it contains a level j disk and empty otherwise. A j-square S is said to be relevant if one of its three siblings is non-empty or there is a non-empty S S of level at most j + log k. This implies that any non-empty square is relevant, and there are at most O(k 2 D ) relevant squares so far. In addition, 0-squares containing at least one point in P are relevant. A relevant square S is a relevant child of another relevant square S if S S and there is no relevant square S with S S S. We show that it is sufficient to consider only relevant squares. Lemma 3.2. For each relevant level 0 square S, allcostand sol-values can be computed in time O(k D S 2 p max 2 32kγ/π ). Proof. Since 16kγ/π [19], enumerating all D b(s) >0 W D b(s) >0 takes O(2 16kγ/π ) time. We show how to compute min{c(t ) p(ucs (T W )) r; T D i(s) 0 } for a particular set W D b(s) >0 and each r =0,..., p max.(ifd i(s) 0 =, this computation is trivial and takes only O( p max ) time.) Partition S into k vertical slabs of width exactly one and assign a point of P to a slab if the point is contained in the slab (w.l.o.g. no point lies on a slab boundary). Observe that any disk of D i(s) 0 covering a point in a certain slab must intersect the left or the right boundary of this slab. Now order the slabs from left to right. Define for each slab i = 1,...,k, each subset X of disks intersecting the right boundary of slab i, and each r =0,..., p max a function a(i, X, r) = min { c(x Y )+a(i 1,Y,r p(uc i (X Y W ))) Y disks intersecting left boundary of i, r p(uc i (X Y W )) }. Here uc i (X Y W ) is the set of points in slab i uniquely covered by X Y W. We set a(0,, 0) = 0 and a(0,,r)=+ for each r 0 (as we only consider disks in D i(s), no disk intersects the left boundary of the first slab). The minima we are looking for are in a(k,,r). We can compute them in O(k D S 2 p max 2 16kγ/π ) time, as computing p(uc i (X Y W )) takes O(( X + Y + W ) ) time and a vertical line in S intersects at most 8kγ/π disks of D i(s) [19]. We require the following auxiliary proposition. It is an instance of Multiple-Choice Knapsack, solved similarly as in the proof of Theorem

7 Proposition 3.1. For any j-square S, given W D b(s) >j and U D +(S) >j 1, we can compute for all r = 0,..., p max, min{ 4 i=1 cost(s i, (U W ) b(si),r i ) 4 i=1 r i = r} in O(( p max ) 2 ) time. Proof. We can compute for all r = 0,.., P p max the values g(1,r )=cost(s 1, (U W ) b(s1),r ) and, for i =2, 3, 4, the values g(i, r ) = min{g(i 1,r r )+ cost(s i, (U W ) b(si),r ) r =0,...,r }. This takes time O(( P p max ) 2 ). The values we need are the values g(4,r)forr =0,..., p max. Lemma 3.3. For each relevant j-square S (j > 0), all cost- and sol-values can be computed in total time O(( D + )2 32kγ/π (p max ) 2 ). Proof. (Sketch) If S has relevant level j 1 children, this follows immediately from D b(s) >j, D +(S) >j 1 16kγ/π [19] and Proposition 3.1. If S has no relevant level j 1 children, we can show that cost(s, W, r) = min{ relevant child S of S cost(s,w b(s ),r S ) relevant child S of S r S = r} for any W D b(s) >j and any r = 0,..., p max. Furthermore, D b(s) >j = D b(s) D j+ log k and thus b(s) >j = D b(s) j+ log k 32γ/π. For a fixed W, we follow the approach of Proposition 3.1 and extend it to deal with all relevant children of S. The increase in time complexity is linear in the number of relevant children of S. As a relevant square has O( D + ) relevant children, the time bound follows. Proof. (of Theorem 3.1) Using Lemmas 3.2 and 3.3, we can compute the cost- and sol-values for all relevant squares in O(k 3 D ( D + )2 32kγ/π (p max ) 2 ) time. By using an extension to Proposition 3.1, similarly as in Lemma 3.3, the value of the maximum max { S; j(s)=l r S S; j(s)=l cost(s,,r S) B } is computable in O(( p max ) 2 D ) time. The theorem follows from Lemma 3.1. Given this algorithm, the shifting technique is applied as follows. Given an integer a (0 a k 1), a line of level j is active if it is of the form y =(hk + a2 l j )2 j or x =(vk + a2 l j )2 j for h, k Z. Active lines partition the plane into j-squares as before, shifted with respect to a. Hence we can still apply Theorem 3.1. Let D a denote the set of disks output by the algorithm for the j-squares induced by a and let D max be the set among such sets with the maximum profit of uniquely covered points of P. Lemma 3.4. p(uc(d max )) (1 4/k) p(uc(opt )). Proof. (Sketch) Let Da b = S Db(S) =j(s) be the set of disks of level j intersecting the boundary of a j-square S induced by a. By Lemma 3.1, p(uc(d a )) p(uc(opt Da)). b So for a fixed value of a, we might lose the profit of points uniquely covered by OPT by disks in OPT Da. b AdiskisinDa b for at most 4 values of a [19], so we lose the profit of a certain point for at most 4 values of a. Hence there is a value of a for which we lose at most a 4/k-fraction of the optimal profit. Therefore p(uc(d max )) (1 4/k) p(uc(opt )). In the following, we denote by x the given instance of Budgeted Unique Coverage and by x the length of some natural encoding of x. We can clearly assume that x D +, where D and P denote the given sets of disks and points, respectively. Theorem 3.2. There is an FPTAS ω for Budgeted Unique Coverage on disks of bounded ply, i.e. of ply γ = γ( x ) =o(log x ). Proof. Consider any ɛ > 0. Choose k as the largest odd integer such that 32 π kγ δ log x for some constant δ>0. If k<5, output (or any other arbitrary solution of cost at most B). Otherwise, scale the profits by ɛ p max, similar to Theorem 2.3. Following the analysis of this theorem and applying Theorem 3.1 and Lemma 3.4, we obtain a (1 4/k) (1 ɛ)-approximation of the optimum in O(k 4 D ( D +)2 32kγ/π ( 2 /ɛ) 2 ) time. By the choice of k, this is bounded by O( x 1+δ D (log 4 x ) ( 2 /ɛ) 2 ). Hence in time polynomial in the size of the input and 1 ɛ, a feasible solution is computed. Furthermore, there is a t ɛ such that k 4 ɛ and k 5 for all x >t ɛ. Therefore if x >t ɛ, we obtain a (1 ɛ) 2 - approximation. This theorem can be easily extended to Budgeted Low- Coverage on fat objects of bounded ply. Theorem 3.3. There is an FPTAS ω for Budgeted Low-Coverage on fat objects of bounded ply, i.e. of ply γ = γ( x ) =o(log x ). If the objects are arbitrary unit disks, but the density of the set of points (i.e. the maximum number of points of P in any 1 1 box) is bounded, we can reduce the set of unit disks to a set of unit disks of bounded ply. We can then improve on the results of 2. Theorem 3.4. There is an FPTAS ω for Budgeted Low-Coverage on unit disks and point sets of bounded density, i.e. of density d = d( x ) =o( log x ). 1273

8 4 Hardness of Unique Coverage For arbitrary ply, we show that the (budgeted) unique coverage problem is as hard to solve for fat objects as it is for general set systems. A convex subset S of R 2 is α-fat [8] for some α 1 if the ratio between the radii of the smallest disk circumscribing S and the largest disk inscribed in S is at most α. Theorem 4.1. For any α>1, there is a gap preserving reduction from general Unique Coverage to Unique Coverage on α-fat objects. Proof. (Sketch) The construction is similar to the one used in [10] for proving approximation hardness of the dominating set problem for fat objects. Each element u corresponds to a distinct point p u on the boundary of a fixed disk D, and for each set S j we add a fat object d j obtained from D by cutting off (using intersection with half-planes) the parts that contain points representing elements that are not in S j. This implies Demaine et al. s [7] hardness results for general Unique Coverage carry over to Unique Coverage on α-fat objects. The algorithms of 3 and the approximation hardness results above extend to a related problem, Budgeted Maximum Coverage, a budgeted version of Maximum Set Cover. In general, Budgeted Maximum Coverage has a (1 1/e)-approximation algorithm and this is best possible unless NP DTIME(n O(log log n) ) [15]. In the geometric version, the approximation bound can be improved for disks and other fat objects of bounded ply by changing Lemma 3.2 to include the profit of all covered points, not only of uniquely covered points. Budgeted Maximum Coverage on fat objects of arbitrary ply is as hard as it is in general. Theorem 4.2. Budgeted Maximum Coverage has an FPTAS ω on fat objects of bounded ply, i.e. of ply γ = γ( x ) = o(log x ), where x is the size of the input. For any α > 1 and c (1 1/e, 1], there is no polynomial-time c-approximation algorithm for Budgeted Maximum Coverage on α-fat objects, unless NP DTIME(n O(log log n) ). 5 Minimum Membership Set Cover In this section we deal with a geometric version of Minimum Membership Set Cover (MMSC). Formally, the input of the problem consists of a finite set P of points and a set S of geometric objects, whose union contains all points in P. For a subset S S, the membership mem S (p) of a point p is defined as the number of objects in S that contain p. Themaximum membership is mem S (P) = max p P mem S (p). The goal is to compute a subset S Sthat covers all points in P and minimizes mem S (P). We consider MMSC on unit squares (squares with side length 1) and unit disks (disks with radius 1 2 ). Theorem 5.1. For Geometric MMSC on unit squares or unit disks, it is NP-hard to decide if a solution with maximum membership 1 exists or not. The proof of this theorem is omitted due to space limitations. The theorem implies that there is no polynomialtime approximation algorithm with ratio smaller than 2 for MMSC on unit squares or unit disks, unless P=NP. Now we present an approximation algorithm for MMSC on unit squares, achieving constant approximation ratio. Its running-time is polynomial only if the optimum objective value is bounded by a constant. We partition the plane into horizontal slabs of height 1 and compute a separate solution for each such slab. The solution for a slab must cover all points in the slab while ensuring that the maximum membership of points inside and outside the slab is bounded. In the end, the union of the solutions for the different slabs is output. To solve the problem for a slab M, we observe that the unit squares of any minimal solution consist of squares intersecting the top side of M, all of which are on the lower envelope of their union, and of squares intersecting the bottom side of M, all of which are on the upper envelope of their union. This enables a sweep-line approach in which we maintain 2l + 1 squares around the current position on each of the two envelopes, where l is a given bound on the maximum membership. The details are presented in the following lemma. Lemma 5.1. Let P be a set of points and S be a set of unit squares. Let M denote the slab contained between the lines y =0(inclusive) and y =1(exclusive). Let P M = P M and P M = P P M. For any constant l, there is a polynomial time algorithm that either asserts that there is no S S that covers P with mem S (P) l, or computes a subset S that covers all points in P M and satisfies mem S (P M ) l (points inside M are covered at most l times) and mem S (P M ) 2l (points outside M are covered at most 2l times). Proof. Let S t (S b ) be the set of squares in S that intersect the top (bottom) side of M. Squares that intersect both sides are assigned to S t. All squares that overlap M are in either S t or S b. We use a vertical sweep-line that moves from left to right and stops at all x-positions of points in P M = P M. Let P M = {p 1,...,p k }, with points indexed in order of non-decreasing x-positions. For a given position of 1274

9 the sweep-line, a (2l + 1)-tuple s t = (s 1,...,s 2l+1 ) of squares from S t is a proper tuple if all squares are on the lower envelope of their union (in the order s 1,s 2,...,s 2l+1 ) and the intersection of the sweepline with this envelope is with s l+1. (We allow any prefix and/or suffix of a proper tuple to consist of dummy objects that do not contain any points in order to represent the case that fewer than l squares appear before or after the current position on the lower envelope. We ignore this technicality below.) Proper tuples s b from S b are defined analogously. For the sweep-line at p i, a pair ( s t, s b )ofproper tuples s t and s b is called admissible if the following two conditions hold: (1) The point p i is covered by the union of the squares in s t and s b. (If there are several points in P M with the same x-coordinate as p i, this must hold for all of them.) (2) The maximum membership of the union of the squares in s t and s b (taking into account all points from P, also the ones outside M) is at most l. For each position of the sweep-line, we compute the set A i of all admissible pairs ( s t, s b ), simply by enumerating all subsets of cardinality at most 2l + 1 from S t and S b. Then we build a directed graph G whose vertex set is the disjoint union of the sets A i.we add a directed edge from ( s t, s b ) A i to ( u t, u b ) A i+1 if u t = s t or u t is obtained from s t =(s 1,...,s 2l+1 )by discarding s 1 and appending a new square s 2l+2 at the end, i.e., u t =(s 2,...,s 2l+1,s 2l+2 ), and the analogous condition holds for s b, u b. Finally, we check whether there is a directed path π from some element ( s t, s b ) A 1 to some element ( u t, u b ) A k. If there is such a path π, the union of the squares from all the tuples on the path is output as the solution S. Otherwise, the algorithm outputs that there is no solution with maximum membership at most l. We see that the algorithm is correct as follows. If the algorithm outputs a solution S, it is clear that S covers P M. To bound the maximum membership, note that the solution consists of a set U t of squares from S t that all meet the lower envelope of their union, and a set U b of squares from S b that all meet the upper envelope of their union, and S = U t U b. We imagine the squares in U t to be ordered from left to right as they appear on their lower envelope, and similarly for U b. Terms such as before, after, and between refer to this order. We now argue separately about the maximum membership for points in P M and in P M. For a point p i in P M, note that the squares in U t that contain p i are consecutive on the lower envelope of U t ; if there were two squares containing p i and another square in between that does not contain p i, then the latter square would not be on the lower envelope. Furthermore, if U t contains any square covering p i, then also the square that is on the envelope at the x-position of p i must contain p i. Hence, if g squares from U t contain p i, then the proper tuple s t of the admissible pair ( s t, s b ) that was chosen for the sweep-line at p i must contain at least min{g, l+1} squares containing p i.the analogous statement holds for U b. Therefore, if more than l squares from U t U b were to contain p i, then at least l + 1 of the squares in the pair ( s t, s b ) would contain p i, and thus the pair would not be admissible. Hence, we have that points in P M are covered at most l times, or mem S (P M ) l. Now consider a point p in P M that lies below M. (The reasoning for points above M is analogous.) The point p cannot be contained in any square in U t. Consider the set U p of squares in U b that contain p. We claim that U p consists of at most two subsets of contiguous (i.e., contiguous on the upper envelope of U b ) squares in U b. This can be seen as follows. Consider the first (i.e., leftmost) square x in U b that contains p. Let y be the first square after x in U b that does not contain p. If y is entirely to the right of p, then no further square in U b can contain p, and thus U p is one contiguous subset of U b (starting with x and ending with the square just before y). Assume now that there is a square after y in U b that contains p. Then y must be entirely above p. Let z be the first square after y that contains p. Clearly, z must be to the lower right of y (i.e., z can be obtained from y by shifting y right and down). All further squares (after z) whose x-range contains the x-coordinate of p must also be to the lower right of z and hence contain p; if one of them, say u, was to the upper right of z, then z could not be on the upper envelope of U b,asz would be below the upper envelope of {y, u}. Thus, U p consists of two contiguous subsets of U b. By construction, the number of consecutive squares from U b containing p is bounded by l; otherwise, the path π would pass through a pair of tuples containing l+1 consecutive squares containing p, but such a tuple would not be admissible. As there are only two contiguous subsets of U b containing p, wehave that p is contained in at most 2l squares from U b. This shows mem S (P M ) 2l. We have shown that if the algorithm outputs a solution S, then S covers all points in P M and satisfies mem S (P M ) l and mem S (P M ) 2l. On the other hand, if there is a solution that covers P M and has maximum membership at most l, then the (2l + 1)- tuples of consecutive squares on the two envelopes allow us to construct a valid candidate path for our algorithm, and thus the algorithm will indeed output a solution. This implies that if the algorithm does not output a solution, then there is no solution that covers P M and has membership at most l (and thus also no solution 1275

10 that covers P and has membership at most l). For the running time, note that each A i contains O( S 4l+2 ) pairs of tuples, and there are O( S 4l+4 ) directed edges between A i and A i+1. So the running time is O( S 4l+4 ). Theorem 5.2. There is a polynomial-time 5-approximation algorithm for instances of Geometric MMSC on unit squares if the optimal objective value is bounded by an arbitrary constant L. Proof. For a given constant l, the following procedure either computes a solution with maximum membership at most 5l or asserts that no solution with maximum membership at most l exists. Partition the plane into horizontal slabs of unit height. For each slab M that contains at least one point from P, run the algorithm of Lemma 5.1 to compute a cover S M S for the points inside M with maximum membership at most l for points in M and at most 2l for points outside M. If for one of the slabs the algorithm of Lemma 5.1 outputs that there is no cover with maximum membership at most l, return that the whole instance has no solution with objective value at most l. Otherwise, return the union of the solutions S M computed for all slabs M. Note that the squares in the solution computed for a slab M can cover points only in M and in the slabs directly above and below M. A point in slab M is covered at most l times by squares in the solution computed for M, at most 2l times by squares in the solution computed for the slab directly above M, and at most 2l times by squares in the solution computed for the slab directly below M. This shows that every point in P is covered at most 5l times. Now run the above procedure for l =1, 2,...,L. The first time the procedure returns a cover, we output that cover and terminate. If the procedure does not return a cover for any of the calls, we output that the instance does not have a solution with maximum membership at most L. The approximation algorithm does not seem to extend to unit disks directly. References [1] Ambühl, C., Erlebach, T., Mihalák, M., Nunkesser, M., Constant-Factor Approximation for Minimum- Weight (Connected) Dominating Sets in Unit Disk Graphs in Proc. APPROX-RANDOM 2006, LNCS 4110, Springer-Verlag, Berlin, 2006, pp [2] Brönnimann, H., Goodrich, M.T., Almost Optimal Set Covers in Finite VC-Dimension, Discrete Comput. Geom (1995), pp [3] Carmi, P., Katz, M.J., Lev-Tov, N., Covering Points by Unit Disks of Fixed Location in Proc. ISAAC 2007, LNCS, Springer, To appear. [4] Chan, T.M., Polynomial-time approximation schemes for packing and piercing fat objects, J. Algorithms, 46 (2003), pp [5] Clark, B.N., Colbourn, C.J., Johnson, D.S., Unit Disk Graphs, Discrete Math (1990), pp [6] Clarkson, K.L., Varadarajan, K.R., Improved Approximation Algorithms for Geometric Set Cover, Discrete Comput. Geom (2007), pp [7] Demaine, E.D., Hajiaghaji, M.T., Feige, U., Salavatipour, M.R., Combination Can Be Hard: Approximability of the Unique Coverage Problem in Proc. SODA 06, ACM Press, New York, pp [8] Efrat, A., Sharir, M., The Complexity of the Union of Fat Objects in the Plane, Discrete Comput. Geom (Feb. 2000), pp [9] Erlebach, T., Jansen, K., Seidel, E., Polynomial-time Approximation Schemes for Geometric Intersection Graphs, SIAM J. Comput (2005), pp [10] Erlebach, T., van Leeuwen, E.J., Domination in Geometric Intersection Graphs, Manuscript, 2007 [11] Garey, M.R., Johnson, D.S., Computers and Intractability A Guide to the Theory of NP- Completeness, W.H. Freeman and Company, San Francisco, [12] Glaßer, C., Reith, S., Vollmer, H., The Complexity of Base Station Positioning in Cellular Networks, Discrete Appl. Math (2005), pp [13] Guruswami, V., Trevisan, L., The Complexity of Making Unique Choices: Approximating 1-in-k SAT in Proc. APPROX-RANDOM 2005, LNCS 3624, Springer-Verlag, Berlin, 2005, pp [14] Hochbaum, D.S., Maass, W., Approximation Schemes for Covering and Packing Problems in Image Processing and VLSI, J. ACM 32 1 (1985), pp [15] Khuller, S., Moss, A., Naor, J., The Budgeted Maximum Coverage Problem, Inform. Process. Lett (April 1999), pp [16] Kuhn, F., von Rickenbach, P., Wattenhofer, R., Welzl, E., Zollinger, A., Interference in Cellular Networks: The Minimum Membership Set Cover Problem in Proc. COCOON 2005, LNCS3595, Springer-Verlag, Berlin, 2005, pp [17] Lev-Tov, N., Peleg, D., Exact Algorithms and Approximation Schemes for Base Station Placement Problems in Proc. SWAT 2002, LNCS 2368, Springer- Verlag, Berlin, 2002, pp [18] Moser, H., Raman, V., Sikdar, S., The Parameterized Complexity of the Unique Coverage Problem in Proc. ISAAC 2007, LNCS, Springer, To appear. [19] van Leeuwen, E.J., Better Approximation Schemes for Disk Graphs in Proc. SWAT 2006, LNCS 4059, Springer-Verlag, Berlin, 2006, pp