The Maximum Exposure Problem
TThe Maximum Exposure Problem
Neeraj Kumar ∗ , Stavros Sintos † , and Subhash Suri ‡ Department of Computer Science, University of California, Santa Barbara, USA Duke University, Durham, NC, USA
Abstract
Given a set of points P and axis-aligned rectangles R in the plane, a point p ∈ P is called exposed if it lies outside all rectangles in R . In the max-exposure problem , given an integerparameter k , we want to delete k rectangles from R so as to maximize the number of exposedpoints. We show that the problem is NP -hard and assuming plausible complexity conjecturesis also hard to approximate even when rectangles in R are translates of two fixed rectangles.However, if R only consists of translates of a single rectangle, we present a polynomial-timeapproximation scheme. For range space defined by general rectangles, we present a simple O ( k )bicriteria approximation algorithm; that is by deleting O ( k ) rectangles, we can expose at leastΩ(1 /k ) of the optimal number of points. Let S = ( P, R ) be a geometric set system, also called a range space , where P is a set of pointsand each R ∈ R is a subset of P , also called a range. We are primarily interested in range spacesdefined by a set of points in two dimensions and ranges defined by axis-aligned rectangles. We saythat a point p ∈ P is exposed if no range in R contains p . The max-exposure problem is definedas follows: given a range space ( P, R ) and an integer parameter k ≥
1, remove k ranges from R so that a maximum number of points are exposed. That is, we want to find a subfamily R ∗ ⊆ R with |R ∗ | = k , so that the number of exposed points in the (reduced) range space ( P, R \ R ∗ ) ismaximized.The max-exposure problem arises naturally in many geometric coverage settings. For instance,if points are the location of clients in the two-dimensional plane, and ranges correspond to coverageareas of facilities, then exposed points are those not covered by any facility. The max-exposureproblem in this case gives a worst-case bound on the number of clients that can be exposed if anadversary disables k facilities. Similarly, in distributed sensor networks, ranges correspond to sensingzones , points correspond to physical assets being monitored by the network, and the max-exposureproblem computes the number of assets exposed when k sensors are compromised.More broadly, the max-exposure problem is related to the densest k -subgraph problem in hypergraphs . In the densest k -subhypergraph problem, we are given a hypergraph H = ( X, E ), andwe want to find a set of k vertices with a maximum number of induced hyperedges. In generalhypergraphs, finding k -densest subgraphs is known to be (conditionally) hard to approximate within ∗ [email protected] † [email protected] ‡ [email protected] a r X i v : . [ c s . C G ] F e b factor of n − (cid:15) , where n is the number of vertices. The max-exposure problem is equivalent tothe densest k -subhypergraph problem on a dual hypergraph, whose vertices X corresponds to theranges R , and whose hyperedges correspond to the set of points P . Specifically, each point p ∈ P corresponds to a hyperedge adjacent to the set of ranges containing the point p . In the rest of thepaper, we will use n = |R| for the number of ranges in R and m = | P | for the number of points.We show that if the range space is defined by convex polygons , then the max-exposure problem isjust as hard as the densest k -subhypergraph problem. However, for ranges defined by axis-alignedrectangles , one can achieve better approximation. In particular, we obtain the following results. • We show that the max-exposure problem is NP -hard and assuming the dense vs random conjecture [1], it is also hard to approximate better than a factor of O ( n / ) even if the rangespace is defined by only two types of rectangles in the plane. For range space defined by convexpolygons, we show that max-exposure is equivalent to densest k -subhypergraph problem,which is hard to approximate within a factor of O ( n − (cid:15) ). • When ranges are defined by translates of a single rectangle, we give a polynomial-timeapproximation scheme (PTAS) for max-exposure. The PTAS stands in sharp contrast tothe inapproximability of ranges defined by two types of rectangles. Moreover, as an easyconsequence of this result, we obtain a constant approximation when the ratio of longestand smallest side of rectangles in R is bounded by a constant. However, we do not knowif max-exposure with translates of a single rectangle can be solved in polynomial time or isNP-hard. • For ranges defined by arbitrary rectangles, we present a simple greedy algorithm that achievesa bicriteria O ( k )-approximation. That is, if the optimal number of points exposed is m ∗ , thealgorithm picks a subset of k rectangles such that the number of points exposed is at least m ∗ /ck , for some constant c . No such approximation is possible for general hypergraphs. Ifrectangles in R have a bounded aspect ratio, the approximation improves to O ( √ k ). Forpseudodisks with bounded-ply (no point in the plane is contained in more than a constantnumber of ranges), this algorithm achieves a constant approximation.The PTAS is obtained by first optimally solving a restricted max-exposure instance where allpoints are contained in a unit square using dynamic programming in polynomial time. Next, wecarefully combine them to obtain an optimal solution in ( nm ) O ( h ) time for the case when inputpoints lie in a h × h square. Applying well known shifting techniques on this gives us the PTAS.Both bicriteria algorithms are obtained by carefully assigning the points to ranges and applyinggreedy strategies. Related Work
Coverage and exposure problems have been widely studied in geometry andgraphs. In the classical set cover problem, we want to select a subfamily of k sets that cover themaximum number of items (points) [2, 3]. For the set cover problem, the classical greedy algorithmachieves a factor log n approximation for the number of sets needed to cover all the items, orfactor (1 − /e ) approximation for the number of items covered by using exactly k sets. Similarly,in geometry, the art gallery problems explore coverage of polygons using a minimum number ofguards. Unlike coverage problems where greedy algorithms deliver reasonably good approximation,the exposure problems turn out to be much harder. Specifically, choosing k sets whose union isof minimum size is much harder to approximate with a conditional inapproximability of O ( n − (cid:15) )where n is the number of elements, or O ( m / − (cid:15) ) where m is the number of sets [1]. This so-called min-union problem is essentially the complement of the densest k -subhypergraph problem on2ypergraphs [4]. The densest k -subgraph problem for graphs has a long history [5–8]. The paper [4]also studies the special case of an interval hypergraph H = ( V, E ), whose vertices V is a finite subsetof N and for each edge e ∈ E there are values a e , b e ∈ N such that e = { i ∈ V : a e ≤ i ≤ b e } . That is,vertices are integer points and edges are intervals containing them. They show that this restrictedcase can be solved in polynomial time. The corresponding max exposure instance is when ranges R are intervals R i = ( a i , b i ) on the real line. As discussed later, this 1-D case can also be solvedin polynomial time. Moreover, we show that good approximations can also be obtained for somegeometric objects in two dimensions.The coverage problems have also been studied for geometric set systems where improvedapproximation bounds are possible using the V C dimension [9–11]. Multi-cover variants, where eachinput point must be covered by more than one set, are studied in [12, 13]. The geometric constraintremoval problem [14, 15], where given a set of ranges, the goal is to expose a path between two givenpoints by deleting at most k ranges (a path is exposed if it lies in the exterior of all ranges), is alsoclosely related to the max-exposure problem. Even for simple shapes such as unit disks (or unitsquares) [16, 17], no PTAS is known for this problem.The remainder of the paper is organized as follows. In Section 2, we discuss our hardness resultsfollowed by the bicriteria O ( k )-approximation in Section 3. In Section 4, we study the case when R consists of translates of a fixed rectangle and describe a PTAS for it. Finally, in Section 5, we usethese ideas to obtain a bicriteria O ( √ k )-approximation when the aspect ratio of rectangles in R isbounded by a constant. We show that the max-exposure problem for geometric ranges is both NP -hard, and inapproximable.We begin by reducing the densest k -subgraph on bipartite graphs ( bipartite-DkS ) to the max-exposure problem; the known NP -hardness of biparite-DkS then implies the hardness for max-exposure. Moreover, we show that bipartite-DkS is hard to approximate assuming the dense vsrandom conjecture, thereby establishing the inapproximability of max-exposure.In the bipartite-DkS problem, we are given a bipartite graph G = ( A, B, E ), an integer k , andwe want to compute a set of k vertices such that the induced subgraph on those k vertices hasthe maximum number of edges. Given an instance G = ( A, B, E ) of bipartite-DkS, we construct amax-exposure instance as follows.Let R = [0 , (cid:15) ] × [0 , n ] be a thin vertical rectangle and R = [0 , n ] × [0 , (cid:15) ] be a thin horizontalrectangle. For each vertex v i ∈ A , we create a copy R i of R , and place it such that its lower-leftcorner is at ( i, v j ∈ B , we create a copy R j of R , and place it suchthat its lower-left corner is at (0 , j ). These | A | + | B | rectangles create a checkerboard arrangement,with | A | × | B | cells of intersection. For each edge ( v i , v j ) ∈ E , we place a single point in the cellcorresponding to intersection of R i and R j . It is now easy to see that G has a k -subgraph with m ∗ edges if and only if we can expose m ∗ points in this instance by removing k rectangles: the removedrectangles are exactly the k vertices chosen in the graph, and each exposed point corresponds to theedge included in the output subgraph. (See also Figure 1.) We will later make use of this reduction,and therefore state it as the following lemma. Lemma 1.
The max-exposure problem is at least as hard as bipartite-DkS.
Since bipartite-DkS is known to be NP -hard [18], we have the following. Theorem 1.
Max-exposure problem with axis-aligned rectangles is NP -hard. igure 1: Reducing bipartite-DkS to max-exposure with axis-aligned rectangles. AB a b c d eabcde AB
Figure 2: Reducing densest k -subhypergraph problem to max-exposure. Hypergraph vertices A, B shown as convex ranges.
The construction in the preceding proof shows that max-exposure with rectangles is at least ashard as bipartite-DkS problem. Moreover, the geometric construction uses translates of only tworectangles R , R . In the following, we show that even with such a restricted range space, the problemis also hard to approximate. To that end we prove that bipartite-DkS cannot be approximatedbetter than a factor O ( n / ), where n is the number of vertices in this graph. More precisely, if thedensest subgraph over k vertices has m ∗ edges, it is hard to find a subgraph over k vertices thatcontains Ω( m ∗ /n − (cid:15) ) edges in polynomial time. This hardness of approximation is conditioned onthe so-called dense vs random conjecture [1] stated as follows.Given a graph G , constants 0 < α, β <
1, and a parameter k , we want to distinguish betweenthe following two cases.1. ( Random ) G = G ( n, p ) where p = n α − , that is, G has average degree approximately n α .2. ( Dense ) G is adversarially chosen so that the densest k -subgraph of G has average degree k β .The conjecture states that for all 0 < α <
1, sufficiently small (cid:15) >
0, and for all k ≤ √ n , one cannotdistinguish between the dense and random cases in polynomial time (w.h.p), when β ≤ α − (cid:15) .In order to obtain hardness guarantees using the above conjecture, one needs to find the‘distinguishing ratio’ r , that is the least multiplicative gap between the optimum solution for theproblem on the dense and random instances. If there exists an algorithm with an approximationfactor significantly smaller than r , then we would be able to use it to distinguish between the denseand random instances, thereby refuting the conjecture. We obtain the following result for densest k -subgraph problem on bipartite graphs. Lemma 2.
Assuming that dense vs random conjecture is true, the densest k -subgraph problem onbipartite graphs is hard to approximate better than a factor O ( n / ) of optimum.Proof. Let G (cid:48) = ( V (cid:48) , E (cid:48) ) be a graph sampled either from the dense or from the random instances.We construct a bipartite graph G = ( A, B, E ) as follows. For every vertex v ∈ V (cid:48) , add a vertex v a to A and v b to B . For every edge e = ( u, v ) ∈ E (cid:48) , we add the pair of edges e = ( u a , v b ) and e = ( v a , u b ) to E . That is, every edge e ∈ E (cid:48) is mapped to two copies e , e ∈ E and we define e to be their parent edge as par ( e ) = par ( e ) = e . Similarly, for a vertex u ∈ V (cid:48) and its two copies u a , u b ∈ V , we define par ( u a ) = par ( u b ) = u . We say that G is dense if the underlying graph G (cid:48) wassampled from the dense case, otherwise we say that G is random .Consider a set of k ∗ = 2 k vertices in G . If G came from the dense case, there must be a set of2 k vertices that have 2 k β +1 edges between them. So the number of edges in dense case m ∗ d ≥ k β +1 .Otherwise, we are in the random case. Consider the optimal set of k ∗ = 2 k vertices V ∗ thatmaximizes the set E ∗ of edges in the induced subgraph G [ V ∗ ]. Now consider the correspondingset of vertices V p = { par(v) | v ∈ V ∗ } of the original graph G (cid:48) and the set of edges E p in the4nduced subgraph G (cid:48) [ V p ]). We have that | V p | ≤ | V ∗ | = 2 k and | E p | ≥ | E ∗ | / e = ( u, v ) ∈ E ∗ , we will have the edge par ( e ) = ( par ( u ) , par ( v )) ∈ E p . Since | V p | ≤ k and weare in the random case, we can upperbound the number of edges in E p as the number of edges inthe densest subgraph of G ( n, n α − ) over 2 k vertices. This is ˜ O (max(2 k, k n α − )) w.h.p. where˜ O ignores logarithmic factors. Therefore the optimum number of edges in the random case is m ∗ r = | E ∗ | ≤ | E p | = ˜ O (max( k, k n α − )) w.h.p.Choosing k = n / , α = , β = − (cid:15) , gives us m ∗ r = ˜ O ( n / ) w.h.p. and m ∗ d = ˜Ω( n − (cid:15) ). Ifwe could approximate this problem within a factor O ( n / − (cid:15) ), then in the dense case, the numberof edges computed by this approximation algorithm is ˜Ω( n (cid:15) ) which is strictly more than themaximum possible edges in the random case. Therefore, we would be able to distinguish betweendense and random cases, and thereby refute the conjecture for these values of α, β and k .Using the same construction as in Lemma 1, we obtain the following. Corollary 2.
Assuming the dense vs random conjecture, max-exposure with axis-aligned rectanglesis hard to approximate better than factor O ( n / ) of optimum. We now show that the max-exposure problem is equivalent to the densest k -subhypergraph problemfor general hypergraphs when the range space ( P, R ) is defined by convex polygons. In one direction,the max-exposure instance ( P, R ) naturally corresponds to a hypergraph H = ( R , P ) whose verticesare the ranges and the edges correspond to points and are defined by the containment relationship.Clearly, the densest k -subhypergraph corresponds to the set of k ranges deleting which exposesmaximum number of points. For the other direction, we have the following lemma. (See alsoFigure 2.) Lemma 3.
Given a hypergraph H = ( X, E ) , one can construct a max-exposure instance with convexranges R and points P such that the densest k -subhypergraph of H corresponds to a solution ofmax-exposure.Proof. For each edge e ∈ E of the hypergraph, add a point p e ∈ P . We place all the points of P in convex position. Let v ∈ X be a vertex and E v be the set of hyperedges adjacent to v . Sincepoints in P are in convex position, any subset of P forms a convex polygon. Therefore, for every v ∈ V , we can draw a convex polygon R v ∈ R whose corners are the point set corresponding to thehyperedges E v . The polygons will likely overlap in the convex region but for every point p e ∈ P , thepolygons containing p e are precisely the ones that have p e as its corner. Therefore, p e is exposed ifand only if all vertices of the hyperedge e are selected. O ( k )-approximation Algorithm In this section, we present a simple approximation algorithm for the max-exposure problem thatachieves bicriteria O ( k )-approximation for range spaces defined by arbitrary axis-aligned rectangles.Specifically, if the optimal number of points exposed is m ∗ , the algorithm picks a subset of k rectangles such that the number of points exposed is at least m ∗ /ck , for some constant c . In fact,the results hold for any polygonal range with O (1) complexity.This bicriteria approximation should be contrasted with the fact that no such approximationis possible for the densest k -subhypergraph problem: that is, one cannot compute a set of O ( k b )vertices for any constant b such that the number of edges in the induced subhypergraph is at least5ptimal. Thus the geometric properties of the range space have a significant impact on the problemcomplexity. In particular, if R consists of rectangle ranges, we show that the following strategypicks a subset of αk ranges such that the number of points exposed is at least αm ∗ / ( ck ), for aparameter 1 ≤ α ≤ k and constant c that will be fixed later. Choosing α = k gives us the claimedbound.Our algorithm is essentially greedy. We divide the points into maximal equivalence classes,where each class is the maximal subset of points belonging to the same subset of ranges. We define R ( p ) as the set of ranges that contain a point p ∈ P , and remove all points that are contained inmore than k ranges, since they can be never exposed in the optimal solution. Therefore, withoutloss of generality, we can assume that |R ( p ) | ≤ k for all points p ∈ P . The rest of the algorithms isas follows. Algorithm 1
Greedy-Bicriteria1. Partition P into a set G of groups where each group G i ∈ G is an equivalence class of pointsthat are contained in the same set of ranges. That is, for any p ∈ G i , p (cid:48) ∈ G j , we have R ( p ) = R ( p (cid:48) ) if i = j and R ( p ) (cid:54) = R ( p (cid:48) ), otherwise.2. Sort the groups in G by decreasing order of their size | G i | and select the ranges in first α groups for deletion.3. Return m (cid:48) = (cid:80) ≤ i ≤ α | G i | as the number of points exposed.In Algorithm 1, observe that every point in the i th group G i is contained in the same set ofranges, which we denote by R ( G i ). Moreover, we have |R ( G i ) | ≤ k . Therefore, the total numberof ranges that we delete in Step 2 is at most αk . It remains to show that the number of pointsexposed m (cid:48) is at least αm ∗ /ck . Lemma 4.
Let m (cid:48) be the number of points exposed by the algorithm Greedy-Bicriteria , and let m ∗ be the optimal number of exposed points, Then, m (cid:48) ≥ αm ∗ /ck .Proof. Consider the optimal set R ∗ of k ranges that are deleted, and let P ∗ be the set of exposedpoints. We partition the set of points P ∗ into groups G ∗ as before, such that each group G ∗ i ∈ G ∗ is identified by the range set R ( G ∗ i ) = R ( p ), for any p ∈ G ∗ i . Since P ∗ ⊆ P , we must have that G ∗ ⊆ G . This holds because for every group G ∗ i ∈ G ∗ there must be a group G j ∈ G such that R ( G ∗ i ) = R ( G j ). Moreover since P ∗ is the maximum set of points that can be exposed, we musthave that G ∗ i = G j . Finally, we note that the number of groups |G ∗ | is bounded by the numberof cells in the arrangement of ranges in R ∗ which is at most ck for some fixed constant c , for all O (1)-complexity ranges. If the groups in G are arranged by decreasing order of their sizes, we havethat m ∗ = (cid:88) ≤ i ≤|G ∗ | | G ∗ i | ≤ (cid:88) ≤ i ≤|G ∗ | | G i | ≤ (cid:88) ≤ i ≤ ck | G i |≤ ck α (cid:88) ≤ i ≤ α | G i | = ck α · m (cid:48) α can be tuned to improve the approximation guarantee with respect to onecriterion (say the number of exposed points) at the cost of other. With α = k , the algorithm exposesat least Ω( m ∗ /k ) by removing k ranges. As for the running time, a simple implementation of thealgorithm can be made to run in O ( mn log m ) time: we can build the point-range containmentrelation in O ( mn ) time, partitioning the point set into groups takes an additional O ( mn log m )time. If the range space R consists of pseudodisks of bounded-ply (no point in the plane is containedin more than a constant number ρ pseudodisks), then the algorithm Greedy-Bicriteria achieves aconstant approximation. Due to the bounded-ply restriction, we have that the number of pseudodiskscontaining the points of group G i is |R ( G i ) | ≤ ρ , and therefore number of pseudodisks that areremoved in Step 2 of the algorithm is also at most αρ . Moreover, the number of cells in anarrangement of k pseudodisks with depth at most ρ is O ( ρk ) [19]. Therefore, we can bound thenumber of groups of the optimal solution |G ∗ | in the proof for Lemma 4 to be at most cρk . Thisgives us that the number of points exposed m (cid:48) ≥ αm ∗ /cρk , where m ∗ is the number of pointsexposed by the optimal solution. Lemma 5.
If the range space R consists of pseudodisks of bounded-ply ρ , then algorithm Greedy-Bicriteria exposes at least αm ∗ /cρk points by deleting at most αρ pseudodisks, where ≤ α ≤ k . Choosing α = k , the algorithm achieves a bicriteria O ( ρ )-approximation. With α = k/ρ , thealgorithm exposes at least 1 /cρ fraction of the optimal number of points by deleting k ranges. We have seen that max-exposure is hard to approximate even if the ranges are translates of twotypes of rectangles. We now describe an approximation scheme when the ranges are translates of a single rectangle. In this case, we can scale the axes so that the rectangle becomes a unit square without changing any point-rectangle containment. Therefore, we can assume that our ranges areall unit squares. The problem is non-trivial even for unit square ranges, and as a warmup we firstsolve the following special case: all the points lie inside a unit square.
We develop a dynamicprogramming algorithm to solve this case exactly, and then use it to design an approximation forthe general set of points.
We are given a max-exposure instance consisting of unit square ranges R and a set of points P ina unit square C . Without loss of generality, we can assume that the lower left corner of C lies atorigin (0 ,
0) and all ranges in R intersect C . We classify the ranges in R to be one of the two types:(See also Figure 3). Type-0:
Unit square ranges that intersect x = 0. Type-1:
Unit square ranges that intersect x = 1.(A unit square range coincident with both x = 0 and x = 1 is assumed to be Type-0). We drawtwo parallel horizontal lines (cid:96) : y = 0 and (cid:96) : y = 1 coincident with bottom and top horizontalsides of C respectively. We say that a range R ∈ R is anchored to a line (cid:96) if it intersects (cid:96) . Note7hat every R ∈ R is anchored to exactly one of (cid:96) or (cid:96) . (When R is coincident with both (cid:96) and (cid:96) ,we say that it is anchored to (cid:96) ).Moreover, for the rest of our discussion, let x = x i be a vertical line and define P i ⊆ P to be theset of points that have x -coordinate at least x i . In other words, P i is the set of active points at x = x i . Similarly, define R i ⊆ R to be the set of active ranges that have at least one corner to theright of x = x i . That is, R ∈ R i either intersects x = x i or lies completely to the right of it.In order to gain some intuition, we will first consider the following two natural dynamicprogramming formulations for the problem. DP-template-0
Suppose that the points in P are ordered by their increasing x -coordinatesand let x i be the x -coordinate of the i th point p i . We define a subproblem as S ( i, k (cid:48) , R d ) whichrepresents the maximum number of points in P i that can be exposed by removing k (cid:48) ranges fromthe range set R i \ R d . If we define x = 0, then S (0 , k, ∅ ) gives the optimal number of exposedpoints for our problem.Let k i = |R ( p i ) \ R d | be the number of new ranges in R i that contain p i . Then, we can canexpress the subproblems at i in terms of subproblems at i + 1 as follows. S ( i, k (cid:48) , R d ) = max ® S ( i + 1 , k (cid:48) − k i , R d ∪ R ( p i )) + 1 expose p i S ( i + 1 , k (cid:48) , R d ) p i not exposed (cid:96) (cid:96) C Figure 3: Max-exposure in a unit square C . Type 0ranges are drawn with solid lines, Type 1 ranges aredash-dotted. p (cid:96) (cid:96) p (cid:48) R R (cid:48) d ( R (cid:48) , (cid:96) ) Figure 4: An example of closer relationship.Point p is closer to (cid:96) than p (cid:48) . R is closer to (cid:96) than R (cid:48) . Roughly speaking, at x = x i which is the event corresponding to a point p i ∈ P , we have twochoices : expose p i or do not expose p i . If we expose p i , we pay for deleting the ranges in R i \ R d that contain p i and mark them as deleted by adding to the deleted range set R d . It is easy tosee that this correctly computes the optimal number of exposed points since we charge for everydeletion exactly once. However, there is one complication: a priori it is not clear how to bound thenumber of range subset R d used by this dynamic program. We later argue that the geometry ofrange space for Type-0 ranges allows us to use only a polynomial number of choices. DP-template-1
An alternative approach is to consider both point and begin-range events.That is, x = x i is either incident to a point p i ∈ P or to the left vertical side of a range R i ∈ R .Then, we can define a subproblem by the tuple S ( i, k (cid:48) , P f ) which represents the maximum numberof points in ( P i \ P f ) that can be exposed by removing k (cid:48) ranges in R i . If we define x = 0, then S (0 , k, ∅ ) gives the optimal number of exposed points. Let P ( R i ) ⊆ P be the set of points containedin the range R i , then we have the following recurrence.8 ( i, k (cid:48) , P f ) = max ® S ( i + 1 , k (cid:48) − , P f ) delete range R i S ( i + 1 , k (cid:48) , P f ∪ P ( R i )) R i not deleted (event x = x i was beginning of a range R i ∈ R i ) = max ® S ( i + 1 , k (cid:48) , P f ) if p i ∈ P f , cannot expose p i S ( i + 1 , k (cid:48) , P f ) + 1 otherwise, expose p i (otherwise, event x = x i was a point p i ∈ P i ) In the above formulation, at each begin-range event for some R i ∈ R i , we have two choices: delete R i or do not delete R i . If R i was deleted, we reduce the budget k (cid:48) by one. Otherwise, if R i wasnot deleted, we can never expose the points in P ( R i ), and therefore we add P ( R i ) to the forbiddenpoint set P f . The correctness of the dynamic program follows from the fact that for every point p i ,all the ranges containing it must begin before x = x i , and we expose p i only if those ranges were deleted . Again, it is not obvious how many different subsets P f are needed by the dynamic program.However, we will later show that by keeping track of polynomial number of sets P f , we can solvemax-exposure with Type-1 ranges.We note that the Type-0 and Type-1 ranges may superficially seem symmetric but once we fixthe order of computing subproblems, they become structurally different. Therefore, we would needslightly different techniques to handle each type. For the ease of exposition, we present dynamicprograms for Type-0 and Type-1 ranges separately and finally combine them. Also note that ifthe ranges in R are intervals on the real line (max exposure in 1D), then both DP-template-0 andDP-template-1 can be easily applied to obtain a polynomial time algorithm.We will now define the following ordering relations that will be useful later. Let (cid:96) be a horizontalline, and let d ( p, (cid:96) ) denote the orthogonal distance of p ∈ P from (cid:96) . If p, p (cid:48) ∈ P are two points, wesay that p is closer to (cid:96) than p (cid:48) if d ( p, (cid:96) ) < d ( p (cid:48) , (cid:96) ). Similarly, for a range R ∈ R that is anchored to (cid:96) , let d ( R, (cid:96) ) be the vertical distance inside the unit square C between (cid:96) and the side of R parallelto (cid:96) . If R, R (cid:48) ∈ R are two ranges, we say that R is closer (or equivalently R (cid:48) is farther ) from (cid:96) ifboth R, R (cid:48) are anchored to (cid:96) and d ( R, (cid:96) ) < d ( R (cid:48) , (cid:96) ). (See Figure 4.) Recall that Type-0 ranges intersect the vertical lines x = 0 and are anchored to either (cid:96) or (cid:96) . Wewill apply the formulation discussed in DP-template-0 . The key challenge here is to bound thenumber of possible deleted range sets R d . Towards that end, we make the following claim. Recallthat R i is the set of active ranges at x = x i . Lemma 6.
Let q , q be the two exposed points strictly to the left of x = x i that are closest to (cid:96) and (cid:96) respectively. Then our dynamic program only needs to consider the set of deleted ranges R d = R i ∩ ( R ( q ) ∪ R ( q )) at x = x i conditioned on q , q .Proof. Observe that since R consists of Type-0 ranges, every range in R i must intersect the verticalline x = x i . Suppose we partition R i into ranges R i that are anchored to (cid:96) and R i that areanchored to (cid:96) . Let P (cid:48) ⊆ P be the set of all exposed points strictly to the left of x = x i . Observethat for all p ∈ P (cid:48) , any range R ∈ R i that contains p must also contain q . Therefore, we must have R i ∩ R ( p ) ⊆ R i ∩ R ( q ), for all p ∈ P (cid:48) . Similarly, R i ∩ R ( p ) ⊆ R i ∩ R ( q ), for all p ∈ P (cid:48) . Thisgives us (cid:83) p ∈ P (cid:48) R i ∩ R ( p ) = R i ∩ ( R ( q ) ∪ R ( q )). Therefore, the set R d consists of all the activeranges that contain at least one exposed point and were therefore deleted to the left of x = x i .9herefore, if our dynamic program remembers the exposed points q , q , then we can compute thedeleted range set R d = R i ∩ ( R ( q ) ∪ R ( q )) at x = x i . There are O ( m ) choices for the pair q , q ,so the number of possible sets R d is also O ( m ). We can therefore identify our subproblems by thetuple S ( i, k (cid:48) , q , q ) which represents the maximum number of exposed points with x -coordinates x i or higher using k (cid:48) rectangles from the set R i \ R d . With k i = |R ( p i ) \ R d | , we obtain the followingrecurrence: S ( i, k (cid:48) , q , q ) =max ® S ( i + 1 , k (cid:48) − k i , closer ( q , p i ) , closer ( q , p i )) + 1 expose p i S ( i + 1 , k (cid:48) , q , q ) p i not exposedwhere the function closer ( q , p i ) returns whichever of q , p i is closer to (cid:96) , and closer ( q , p i ) returnswhichever of q , p i is closer to (cid:96) . The optimal solution is given by S (0 , k, q ∗ , q ∗ ), where q ∗ = (0 , q ∗ = (0 ,
0) are two artificial points with R ( q ∗ ) = R ( q ∗ ) = ∅ (not contained in any range). Thebase case is defined by the rightmost event at vertical line x = 1 and is initialized with zeroes for all q , q and k (cid:48) ≥
0. Any subproblem with k (cid:48) < −∞ . Next we consider the case when we only have Type-1 ranges in R . Unfortunately in this case, ourprevious dynamic program does not work and we need to remember a different set of parameters.More precisely, we will apply the formulation discussed in DP-template-1 , and bound the numberof possible forbidden point sets P f . Recall that P i is the set of active points at x = x i (with x -coordinate x i or higher). Lemma 7.
Let Q , Q be two ranges that begin to the left of x = x i and were not deleted . Moreover, Q is anchored to and is farthest from (cid:96) . Similarly Q is anchored to and is farthest from (cid:96) (Figure 5). Then the forbidden point set at x = x i is given by P f = P i ∩ ( P ( Q ) ∪ P ( Q )) , where P ( Q ) is the set of points contained in range Q .Proof. Recall that the set R i consists of ranges that have at least one corner to the right of thevertical line x = x i . Since we are dealing with Type-1 ranges, every range that begins to the left of x = x i lies in R i . Now let R (cid:48) ⊆ R i be the set of ranges that begin to the left of x = x i and were not deleted . Here P i is the set of points in P that have x -coordinate x i or higher. Now consider anyrange R ∈ R (cid:48) . Recall that R must be anchored to either (cid:96) or (cid:96) . If R was anchored to (cid:96) , thenevery point of P i that lies in R also lies in Q . Otherwise R was anchored to (cid:96) , so every point of P i that lies in R also lies in Q . Therefore, (cid:83) R ∈R (cid:48) ( P i ∩ P ( R )) = P i ∩ ( P ( Q ) ∪ P ( Q )), which isprecisely the forbidden point set P f . (cid:96) (cid:96) x i Q Q Figure 5: Undeleted ranges Q and Q far-thest from (cid:96) and (cid:96) respectively. R R p R R (cid:48) pp (cid:48) p p (a) (b) Figure 6: Remembering one of R , R in (a) or one of p , p in (b) is not sufficient. R ( p ) ranges containing point pP ( R ) points contained in range Rq , q exposed points closest to (cid:96) , (cid:96) Q , Q undeleted ranges farthest from (cid:96) , (cid:96) P i points with x -coordinate at least x i ( active points ) R i ranges with at least one corner to the right of x = x i ( active ranges ) R i subset of R i that are Type-0 ( active Type-0 ranges at x = x i ) P f forbidden point set given by P f = P i ∩ ( P ( Q ) ∪ P ( Q )) R d deleted range set given by R d = R i ∩ ( R ( q ) ∪ R ( q )) Table 1: A table of commonly used notations and their explanations.
Therefore, if our dynamic program remembers the ranges Q and Q , we can compute theforbidden point set P f = P i ∩ ( P ( Q ) ∪ P ( Q )) at x = x i . Since there are O ( n ) choices for the pair Q , Q , the number of possible sets P f is also O ( n ). We can now identify the subproblems by thetuple S ( i, k (cid:48) , Q , Q ) which represents the maximum number of points in P i \ P f that are exposedby deleting k (cid:48) ranges that begin on or after x = x i . This gives us the following recurrence. S ( i, k (cid:48) , Q , Q ) =max ® S ( i + 1 , k (cid:48) − , Q , Q ) delete range R i S ( i + 1 , k (cid:48) , farther ( Q , R i ) , farther ( Q , R i )) R i not deleted( event x = x i was beginning of a range R i ∈ R )max ® S ( i + 1 , k (cid:48) , Q , Q ) if p i ∈ P f , cannot expose p i S ( i + 1 , k (cid:48) , Q , Q ) + 1 otherwise, expose p i (otherwise, event x = x i was a point p i ∈ P ) Here, the function farther simply updates the ranges Q , Q with R i if needed. More precisely, if R i is anchored to (cid:96) and is farther from (cid:96) than Q , then farther ( Q , R i ) returns R i , otherwise itreturns Q . Similarly, if R i is anchored to (cid:96) , and is farther from (cid:96) than Q , then farther ( Q , R i )returns R i , otherwise it returns Q .The optimal solution is given by P (0 , k, Q ∗ , Q ∗ ), where Q ∗ , Q ∗ are two artificial ranges of zero-width : Q ∗ is anchored to (cid:96) and is defined by corners (0 ,
0) and (0 , Q ∗ is anchored to (cid:96) and is defined by corners (0 ,
1) and (1 , Remark 1.
We note that remembering a constant number of exposed points q , q (DP-template-0)or a constant number of undeleted ranges Q , Q (DP-template-1) by themselves cannot solve both Type-0 and Type-1 ranges. For instance, in Figure 6(a) with Type-0 ranges, if R , R were both notdeleted but we remembered one of them, then we will incorrectly expose one of p, p (cid:48) . Similarly inFigure 6(b) with Type-1 ranges, if p , p were both exposed but we only remembered one of them,we will pay for one of the ranges R, R (cid:48) again when we expose p . However, since both the dynamicprograms for Type-0 and Type-1 ranges express subproblems at event i in terms of subproblems atevent i + 1, we can easily combine them with minor adjustments.11 .1.3 Combining them together In the following, we will combine the dynamic programs for Type-0 and Type-1 ranges to obtain adynamic program for max-exposure in a unit square C . We will need a couple of changes. First,the events at x = x i are now defined by either a point p i ∈ P or beginning of a Type-1 range R i . Next, the deleted range set R d at x = x i will only consist of Type-0 ranges and is definedas R d = R i ∩ ( R ( q ) ∪ R ( q )) where R i ⊆ R i is the set of Type-0 ranges that intersect thevertical line x = x i . The forbidden point set P f = P i ∩ ( P ( Q ) ∪ P ( Q )) stays the same. Here q , q , Q , Q are same as defined before. (For the sake of convenience, Table 1 lists these notationswith explanation.)The subproblems represent the maximum number of points in P i \ P f that can be exposed bydeleting k (cid:48) ranges from R i \ R d . If k i = | ( R ( p i ) ∩ R i ) \ R d | , then we obtain the following combinedrecurrence. S ( i, k (cid:48) , q , q , Q , Q ) =max S ( i + 1 , k (cid:48) , q , q , Q , Q ) if p i ∈ P f , cannot expose p i S ( i + 1 , k (cid:48) , q , q , Q , Q ) choose to not expose p i S ( i + 1 , k (cid:48) − k i , closer ( q , p i ) , closer ( q , p i ) , Q , Q ) + 1 expose p i ( event x = x i was a point p i ∈ P i )max ® S ( i + 1 , k (cid:48) − , q , q , Q , Q ) delete Type-1 range R i S ( i + 1 , k (cid:48) , q , q , farther ( Q , R i ) , farther ( Q , R i )) R i not deleted( event x = x i was beginning of a Type-1 range R i ∈ R i )The optimal solution is given by S (0 , k, q ∗ , q ∗ , Q ∗ , Q ∗ ). The correctness of the aboveformulation follows from the fact that when we choose to expose p i , we are guaranteed that allType-1 ranges in R ( p i ) have already been deleted, and the expression k i only charges for Type-0ranges containing p i . As for the running time, for each event x = x i , we compute O ( kn m ) entriesand computing each entry takes constant time. Since there are O ( n + m ) events, we obtain thefollowing. Lemma 8.
Given a set P of m points in a unit square C and a set of n unit square ranges R , wecan compute their max-exposure in O ( k ( n + m ) n m ) time. We now use the preceding algorithm to solve the max-exposure problem for general set of pointsand unit square ranges within a factor 4 of optimum. In particular, we compute a set of 4 k rangesin R such that the number of points exposed in P by deleting them is at least the optimal numberof points. Suppose we embed the ranges R on a uniform unit-sized grid G , and define C as thecollection of all cells in G that contain at least one point of P . Then we can solve exactly for eachcell in C and combine them using dynamic programming as described in Algorithm 2 (DP-Approx).See also Figure 7. Lemma 9. If P ∗ ⊆ P is the optimal set of exposed points, then global (1 , k ) ≥ | P ∗ | , that is , thealgorithm DP-Approx achieves a -approximation and runs in O ( k ( n + m ) n m ) time.Proof. Consider the optimal set of ranges R ∗ ⊆ R . Observe that each range R ∈ R ∗ intersects atmost four grid cells. Let R i = R ∩ C i be the rectangular region defined by intersection of R and C i .12 igure 7: Embedding a max-exposure instance with unit square ranges on a unit-sized grid. Optimal solutionin each grid cell can be computed exactly using Lemma 8. Algorithm 2
DP-Approx1. Apply Lemma 8 to solve max-exposure locally in every cell C i ∈ C for all 0 ≤ k i ≤ k . Callthis a local solution denoted by local ( P ( C i ) , R ( C i ) , k i ), where P ( C i ) ⊆ P is the set of pointscontained in cell C i and R ( C i ) is the set of ranges intersecting C i .2. Process cells in C in any order C , C , . . . , C g , and define global ( i, k (cid:48) ) as the maximum numberof points exposed in the cells C i through C g using k (cid:48) ranges. Combine local solutions to obtain global ( i, k (cid:48) ) as follows. global ( i, k (cid:48) ) = max ≤ k i ≤ k (cid:48) global ( i + 1 , k (cid:48) − k i ) + local ( P ( C i ) , R ( C i ) , k i )3. Return global (1 , k ) as the number of exposed points.Clearly, there are at most four regions R i for each R ∈ R ∗ and therefore 4 k in total. At this point,the regions in cell C i are disjoint from regions in some other cell C j ∈ C . Therefore, optimal solutionexposes | P ∗ | points over a set of cells C ∗ such that the set R ∗ has at most 4 k disjoint componentsin the cells C ∗ . Since we can solve the problem exactly for each cell and can combine them usingthe above dynamic program, we have that global (1 , k ) ≥ | P ∗ | and we achieve a 4-approximation.For the running time, we observe that solving max-exposure locally in a cell C i takes O ( k ( n i + m i ) n i m i ) time, where n i is the number of ranges that intersect C i and m i is the number of pointsin P that lie in C i . Summed over all cells, we get the following bound. (cid:88) i k ( n i + m i ) n i m i ≤ k (cid:88) i ( n i + m i ) (cid:88) i n i (cid:88) i m i ≤ k ( n + m ) ( (cid:88) i n i ) ( (cid:88) i m i ) = O ( k ( n + m ) n m )Once the local solutions are computed, the dynamic program that merges them into a global solutionhas O ( k |C| ) subproblems and computing each subproblem takes O ( k ) time. Recall that every cell in C contains at least one point, so |C| ≤ n and the merge step takes an additional O ( k n ) time. In this section, we will show how to extend the exact algorithm for the restricted max-exposureinstance where all points lie inside a unit square (Lemma 8) to obtain an exact solution for themax-exposure instance where all points are contained in a h × h square C . Without loss of generality,we can assume that the lower left corner of C is at the origin (0 ,
0) and C is subdivided into h unit-sized grid cells. 13bserve that a major hurdle in generalizing the dynamic program from Section 4.1 for max-exposure in a unit square cell to the grid C is that a range R can be double counted in multiplecells. Specifically, range R may contain exposed points in at most four cells of C and can be countedin each one of them. (See also Figure 8.) Indeed a natural generalization of the earlier dynamicprogram to h anchor lines avoids double counting of ranges in the same column of C (verticalneighbors). However, some additional work is required to avoid double counting in adjacent columns(horizontal and diagonal neighbors). (c)(a) R R (b) R Figure 8: Examples where a deleted range R can potentially be counted in two cells that are: (a) verticalneighbors (b) horizontal neighbors (c) diagonal neighbors To handle this, we first apply the following transformation which we call flattening of the grid C . Flattening the grid C Intuitively, the flattening process transforms a h × h grid into a h × shifting the i -th column and aligning it on top of the ( i − column by column from left to right and bottom to top in each column. That is,cells of the column 1 are labeled as 1 , , . . . , h and the cells of column 2 are labeled h + 1 , . . . , h and so on. Then, flattening refers to simply stacking all the cells in their numbered order. Inother words, we shift the coordinates of all points and parts of ranges in column i of the grid C by( − ( i − , ( i − h ), for all 2 ≤ i ≤ h . (See also Figure 9.) a bcd ad bc Figure 9: Flattening a × grid containing one unit square range that is split into Type-0 and Type-1components. After this transformation, all x -coordinates are within the range [0 ,
1] and y -coordinates arewithin the range [0 , h ]. Moreover, every range R is split into two possibly disconnected half-rangeswhich preserve the following important property that follows readily from the fact that the rangesare unit squares. Lemma 10.
Let R be a range and C i , C i +1 be the two consecutive columns of the grid C intersectedby range R . Then, R is Type-1 with respect to cells in C i and Type-0 with respect to cells in C i +1 ,and after the flattening transformation, the x -coordinate at which R begins as a Type-1 range in C i is the same as the x -coordinate at which R finishes as Type-0 range in C i +1 .Proof. The range R intersects the vertical line x = i which is coincident with the right (resp. left)boundary of cells in C i (resp. C i +1 ). Therefore, R is Type-1 in cells of C i and Type-0 in cells of C i +1 .14et the x -coordinate of left boundary of R (that lies in i -th column) be ( i −
1) + δ . Therefore,the x -coordinate of right boundary of R would be ( i −
1) + δ + 1 = i + δ , and it will lie in ( i + 1)-thcolumn. After the transformation both these coordinate values would be δ .From the above lemma, it follows that every range R ∈ R has a Type-0 component and a Type-1component which may lie in non-consecutive cells. In the rest of the discussion, we will refer to thesecomponents by their type as prefix. For example, Type-0 range R refers to the Type-0 componentof R .Once we have flattened the grid C , our algorithm is an almost straightforward extension ofthe dynamic program from Section 4.1 to h + 1 anchor lines (cid:96) , (cid:96) , (cid:96) h . Same as before, weprocess the two types of events : x = x i is a point p i and x = x i is beginning of Type-1 range R i .However at every x = x i , we will now need to remember the set q = { q +0 , q − , . . . , q + h , q − h } of O ( h )points consisting of closest exposed points q + j , q − j respectively above and below every anchor line (cid:96) j .Similarly, we will need to remember the set Q = { Q +0 , Q − , . . . , Q + h , Q − h } of O ( h ) ranges consistingof farthest undeleted Type-1 ranges Q + j , Q − j respectively above and below every anchor line (cid:96) j .Then at x = x i , we extend the definitions from Table 1 to obtain the forbidden point set P f = P i ∩ (cid:83) Q ∈ Q P ( Q ) and the deleted range set R d = R i ∩ (cid:83) q ∈ q R ( q ). Recall that P i is theset of points with x -coordinate at least x i and R i is the set of Type-0 ranges that are active at x = x i . Also recall that P ( Q ) denotes the set of points contained in range Q and R ( q ) denotes theset of ranges containing point q . This gives us the following dynamic program which we will refer toas DP-Flattened . Same as before, we have k i = | ( R ( p i ) ∩ R i ) \ R d | . S ( i, k (cid:48) , q , Q ) =max S ( i + 1 , k (cid:48) , q , Q ) if p i ∈ P f , cannot expose p i S ( i + 1 , k (cid:48) , q , Q ) choose to not expose p i S ( i + 1 , k (cid:48) − k i , closer ( q , p i ) , Q ) + 1 expose p i ( event x = x i was a point p i ∈ P i )max S ( i + 1 , k (cid:48) , q , Q ) if R i ∈ R d , already deleted S ( i + 1 , k (cid:48) − , q , Q ) delete Type-1 range R i S ( i + 1 , k (cid:48) , q , farther ( Q , R i )) R i not deleted( event x = x i was beginning of a Type-1 range R i ∈ R i )Here, closer ( q , p i ) denotes the operation of updating the appropriate closest exposed pointin q with point p i . More precisely, let C j be the cell bounded by anchor lines (cid:96) j − and (cid:96) j thatcontains the exposed point p i . We update q such that q + j − = closer ( q + j − , p i ) and q − j = closer ( q − j , p i ).Similarly, let (cid:96) j be the anchor line intersecting R i , then farther ( Q , R i ) denotes the operation ofupdating Q with the farthest undeleted range on both sides of (cid:96) j as Q + j = farther ( Q + j , R i ) and Q − j = farther ( Q − j , R i ). The optimal solution is given by S (0 , k, q ∗ , Q ∗ ), where q ∗ , Q ∗ consist ofthe initial values for each anchor line.At any event x = x i , the above dynamic program accounts for the cost of deleting a range R inone of two ways: either as a Type-0 range included in the term k i or as Type-1 range by payingunit cost. In the next lemma, we show that every deleted range is counted exactly once and use itto establish the correctness. Lemma 11.
The dynamic program
DP-flattened computes an optimal solution for max-exposureinstance ( R , P, k ) in an h × h grid and runs in O ( k ( nm ) O ( h ) ) time. roof. The running time bound follows from the number of exposed points and undeleted rangeswe need to remember.To prove correctness, consider an optimal set of deleted ranges R ∗ and its exposed points P ∗ .Let the value of the solution returned by the dynamic program be the number of points it choses toexpose and the cost of the solution is the total cost of ranges it deletes. First, we claim that thereexists a sequences of choices at events x = x i where the dynamic program selects points and rangesconsistent with the optimal solution, that is, chooses to only expose points in P ∗ and to only deleteType-1 ranges in R ∗ . This is easy to verify because because P ∗ ∩ P f = ∅ , so the dynamic programcan choose to expose point p i ∈ P ∗ when x = x i is a point-event. Indeed the value of the solution is | P ∗ | . Next we will show that every range in R ∗ is counted exactly once, and therefore the cost ofthe solution is also k .We claim that at every point-event x = x i where we expose the point p i ∈ P ∗ , all ranges in R ( p i ) are deleted and counted exactly once. To see this, let R ∈ R ( p i ) be a range containing p i andlet x = x r be the x -coordinate at which R finishes as a Type-0 range and starts as a Type-1 range.We have two disjoint cases.1. p i is contained in Type-0 component of R . Let (cid:96) j be the line to which Type-0 range R isanchored. We have two subcases.(a) R does not contains any exposed point to the left of x = x i . In this case, we chargefor R and remember that R has already been counted using the closest exposed points q + j , q − j above and below (cid:96) j . Therefore, we will have R ∈ R d at least until x = x r , (whenit switches from being Type-0 to Type-1). Since R cannot be charged at x > x r , it ischarged exactly once in total.(b) R contains an exposed point to the left of x = x i . Then we will have R ∈ R d , and asdiscussed above R was already counted and would not be charged again.2. p i is contained in Type-1 component of R . Since p i is exposed, it is not contained in theforbidden point set P f . Therefore, R must be deleted when it began as a Type-1 event at x = x r or else we would have p i ∈ P f . As discussed above, if R was deleted as a Type-0 rangeto the left of x = x r , we must have R ∈ R d at x = x r , so it would not be charged again. If R was not deleted as a Type-0 range, then it would be charged at x = x r as a Type-1 range andis never charged again.Therefore, the solution returned by dynamic program S (0 , k, q ∗ , Q ∗ ) has value at least optimal. (1 + (cid:15) ) -Approximation Algorithm We will now apply grid shifting technique by Hochbaum and Maas [20] to obtain an (1 + (cid:15) )-approximation . In particular, if P ∗ is the optimal set of exposed points, then we show how tocompute a set of (1 + (cid:15) ) k ranges deleting which will expose at least | P ∗ | points. Using similar ideasbut with small adjustments, we also show how to expose at least (1 − (cid:15) ) | P ∗ | points by deletingexactly k ranges. Theorem 3.
There exists an algorithm for max-exposure with unit-square ranges running in k ( mn ) O (1 /(cid:15) ) time that exposes at least optimal number of points by deleting (1 + (cid:15) ) k ranges. The PTAS presented here simplifies and corrects an error in the PTAS that appeared in the conference version [21]of the paper. roof. For a given shift value a, b ∈ { , . . . , h − } , we compute the optimal solution inside every h × h cell C ij = [ a + ih, a + ( i + 1) h ] × [ b + jh, b + ( j + 1) h ] for all i, j ∈ Z . Using the exact solutionin each cell as local solution , we use the algorithm DP-Approx (from Section 4.2) to combine theminto a global solution for the entire grid given by S ab = global (1 , k (1 + (cid:15) )), with (cid:15) = (cid:100) /h (cid:101) . Werepeat this for every shift a, b , and return S ab that achieves the maximum value.To see why this exposes at least optimal number of points, consider an optimal set of deletedranges R ∗ and sets R ∗ a , R ∗ b ⊆ R ∗ intersected by boundary grid lines x = a + ih and y = b + jh respectively, for all i, j . These grid lines split the intersected ranges into at most Z ab = 2 |R ∗ a | + 2 |R ∗ b | disjoint components. (cid:88) ≤ a,b There exists an algorithm for max-exposure with unit-square ranges running in k ( mn ) O (1 /(cid:15) ) time that exposes at least (1 − (cid:15) ) fraction of optimal number of points by deleting k ranges.Proof. For a given shift value a, b ∈ { , . . . , h − } , we first preprocess the input by discarding pointsso that the set of ranges intersecting the boundary grid lines x = a + ih and y = b + jh do notcontain any point. Specifically, for every shift value a, b , discard the points that are within a unitdistance from grid boundary lines x = a + ih or y = b + jh for all i, j . On the modified input, werun the exact solution in each cell as local solution , and then use DP-Approx (from Section 4.2) tocombine them into a global solution for the entire grid given by S ab = global (1 , k ). We repeat thisfor every shift a, b , and return S ab that achieves the maximum value.Let P ∗ be the optimal set of exposed points. It remains to show that the above algorithmexposes at least (1 − (cid:15) ) | P ∗ | points. To see this, for the shift value a, b , consider the set of discardedpoints P ∗ a , P ∗ b ⊆ P ∗ that are within a unit distance from x = a + ih and y = b + jh respectively.These | P ∗ a | + | P ∗ b | will not be exposed by our algorithm (cid:88) ≤ a,b Given a set of points P , a set of rectangle ranges R such that the ratio of thelargest to the smallest side in R is bounded by a constant, then there exists a polynomial time O (1) -approximation algorithm for max-exposure. We now consider the case when rectangles in R have bounded aspect ratio. That is for all rectangles R ∈ R , the ratio of its two sides is bounded by a constant c . We transform the input ranges R to obtain a modified set of ranges R (cid:48) as follows. For each rectangle R ∈ R , let x be the lengthof the smaller side of R . Then we replace R by at most (cid:100) c (cid:101) squares each of sidelength x . If m ∗ is the optimal number of points exposed by deleting k ranges from R , then there exists a set of O ( k ) ranges in R (cid:48) deleting which will expose at least m ∗ points. Observe that the set R (cid:48) consists of square ranges, of possibly different sizes. Therefore, if we can obtain an f -approximation for squareranges, we can easily obtain O ( f )-approximation with fat rectangles. O ( √ k ) -approximation for Squares We will describe an approximation algorithm for the case when the set of ranges R consists ofaxis-aligned squares. We achieve an approximation algorithm in three steps. First, we partition thepoint set by assigning the points to one of the input squares. Next, we solve the problem exactly fora fixed input square. Finally, we combine these solutions to achieve a good approximation to theoptimal solution.We define A : P → R to be a function that assigns a point in P to exactly one range in R .If R ( p i ) is the set of squares that contain p i , then A ( p i ) is the smallest square in R ( p i ). Thisassignment scheme ensures the following property. Lemma 12. Let R ∈ R be a square and let P R = A − ( R ) be the set of points assigned to it.Moreover, let R (cid:48) ⊆ R be the set of squares that intersect R and contain at least one point in P R .Then, every square R (cid:48) ∈ R (cid:48) must have sidelength bigger than that of R , and therefore contains atleast one corner of R . Now suppose we fix a square R , and consider a restricted max-exposure instance with the set ofits assigned points P R . Since, ranges that contain a point in P R are all bigger then R , this case isessentially the same as points inside a unit square, and therefore Lemma 8 can be easily extendedto solve it exactly. This gives us the following algorithm. Here 1 ≤ α ≤ k is a parameter.18 lgorithm 3 Greedy-Squares1. For every square R ∈ R , apply Lemma 8 over the point set P R to expose the maximum set ofpoints P ( R, k ) ⊆ P R by deleting k ranges.2. Order squares in R by decreasing | P ( R, k ) | values, and pick the set S ⊆ R of first α squares.3. Return (cid:83) R ∈S P ( R, k ) as the set of exposed points. Lemma 13. Let m ∗ be the optimal number of points exposed using k squares, then algorithm Greedy-Squares computes a set of at most αk squares that expose at least αm ∗ /k points.Proof. It is easy to see that the number of squares is at most αk . To show the bound on number ofpoints exposed, consider the optimal set R ∗ of k ranges and let the optimal set of points exposedby R ∗ to be P ∗ . We will now use the same assignment procedure A ∗ : P ∗ → R ∗ to assign pointsin P ∗ to a square in R ∗ . That is, A ∗ ( p i ) is the smallest square in R ∗ that contains p i . We claimthat A ∗ ( p i ) = A ( p i ) for all p i ∈ P ∗ since every square that contains p i lies in R ∗ . Moreover, let P ∗ R denote the set of points of P ∗ assigned to R .Let m (cid:48) be the number of points exposed by the algorithm and assume that the squares in R areordered such that | P ( R i , k ) | ≥ | P ( R j , k ) | for all i < j . Then, we have the following. m ∗ = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:91) R ∈R ∗ P ∗ R (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:88) R ∈R ∗ |P ∗ R |≤ (cid:88) ≤ i ≤ k | P ( R i , k ) | ≤ kα (cid:88) ≤ i ≤ α | P ( R i , k ) | = kα m (cid:48) For α = √ k , the above algorithm achieves a bicriteria O ( √ k )-approximation. Since an f -approximation for square ranges gives an O ( f )-approximation for fat rectangles, we obtain thefollowing. Theorem 6. Given a set of points P and a set of ranges R consisting of rectangles of boundedaspect ratio, then one can obtain a bicriteria O ( √ k ) -approximation for max-exposure in polynomialtime. In this paper, we introduced the max-exposure problem, proved its hardness, and explored approxi-mation schemes for it. We showed that the problem is hard to approximate even when the rangespace R consists of two types of rectangles. When the ranges are defined by translates of a singlerectangle, we presented a polynomial-time approximation scheme (PTAS). Some natural questionsto explore in the future include better approximation algorithms, and simpler range spaces such asthose defined by axis-aligned squares. For instance, can one achieve a constant factor approximationfor axis-aligned squares? 19 eferences [1] E. Chlamt´aˇc, M. Dinitz, Y. Makarychev, Minimizing the union: Tight approximations for smallset bipartite vertex expansion, in: Proceedings of the 28th SODA, 2017, pp. 881–899. 2, 4[2] U. 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