ON THE DISCRETE UNIT DISK COVER PROBLEM

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1 ON THE DISCRETE UNIT DISK COVER PROBLEM GAUTAM K. DAS,1, ROBERT FRASER,2, ALEJANDRO LÓPEZ-ORTIZ,2, and BRADFORD G. NICKERSON,3 1 Department of Mathematics, Indian Institute of Technology Guwahati, Guwahati , India 2 David R. Cheriton School of Computer Science, University of Waterloo, 200 University Ave. West, Waterloo, Ontario, N2L3G1, Canada 3 Faculty of Computer Science, University of New Brunswick, P.O. Box 4400, 540 Windsor Street, Fredericton, New Brunswick, E3B 5A3, Canada Abstract Given a set P of n points and a set D of m unit disks on a 2-dimensional plane, the discrete unit disk cover (DUDC) problem is (i) to check whether each point in P is covered by at least one disk in D or not and (ii) if so, then find a minimum cardinality subset D D such that the unit disks in D cover all the points in P. The discrete unit disk cover problem is a geometric version of the general set cover problem which is NP-hard. The general set cover problem is not approximable within c log P, for some constant c, but the DUDC problem was shown to admit a constant factor approximation. In this paper, we provide an algorithm with constant approximation factor 18. The running time of the proposed algorithm is O(n log n + m log m + mn). The previous best known tractable solution for the same problem was a 22-factor approximation algorithm with running time O(m 2 n 4 ). Keywords: Discrete Unit Disk Cover; Geometric Set Cover; Approximation Algorithms. 1 Introduction Research on geometric set cover problems is often motivated by applications in wireless networking or facility location problems. Our interest in the problem arose from data management problems upon terrains, in bathymetric applications in particular. Suppose that we have a survey of a terrain with data represented as points in the xy-plane, and the height data is stored as an attribute of each point. Call this point set X. Given a new survey of the same area, we A preliminary version of this paper appeared in the 2011 Workshop on Algorithms and Computation (WAL- COM 2011) This work has been carried out when this author was a postdoctoral fellow at Faculty of Computer Science, University of New Brunswick, Fredericton, New Brunswick, E3B 5A3, Canada gkd@iitg.ernet.in r3fraser@cs.uwaterloo.ca alopez-o@cs.uwaterloo.ca bgn@unb.ca 1

2 Table 1: Summary of discrete unit disk cover (DUDC) algorithms Approximation factor Reference O(1) Brönnimann and Goodrich, 1995 [4] 108 Calinescu et al., 2004 [6] 72 Narayanappa and Vojtěchovský, 2006 [17] 38 Carmi et al., 2007 [7] 22 Claude et al., 2010 [5] 1+ε (PTAS) Mustafa and Ray, 2010 [16] 18 this paper obtain a new point set Y. We wish to update our data set by treating the new data set Y as the standard, but we wish to maintain some of the old data for completeness. Our restrictions are that each new point in Y must have at least one old point from X within unit distance, and that a minimum number of points from X are maintained. This problem is an instance of the discrete unit disk cover (DUDC) problem. Other applications of DUDC that are commonly used include selecting locations for wireless servers from a set of candidate locations to cover a set of wireless clients, or positioning municipal service centers (e.g. fire stations, mail collection hubs, garbage collection hubs, bus stations) from a set of candidate sites so that all points of interest (houses, etc.) are within a predefined maximum distance of the service centers. 1.1 Our Results Given a set P of n points and a set D of m unit disks in a 2-dimensional plane, we first check the feasibility of the DUDC problem i.e., check whether each point in P is covered by at least one disk in D or not. If the answer to the first step is yes, then we propose an 18-factor approximation algorithm for the DUDC problem. The running time of the proposed algorithm is O(n log n + m log m + mn). 1.2 Related Work The DUDC problem is a geometric version of the general set cover problem which is NP-hard [14]. The general set cover problem is not approximable within c log P, for some constant c, assuming P NP [18]. Obviously the general approximation algorithms for set cover apply to DUDC to get an easy O(log n) approximation (i.e. [8]), but a series of constant factor approximation algorithms and a PTAS have been presented for DUDC, mostly published within the past few years (see Table 1 for a summary 1 ). 1 In the table, we list the approximation factor for [6] as 108, although the original publication presented the factor as 102. Narayanappa and Vojtěchovský [17] noticed a small bug in the analysis, and corrected the factor to that listed here. 2

3 Using local search, Mustafa and Ray [16] developed a PTAS for the DUDC problem. Their algorithm runs in O(m 2(c/ε)2 +1 n) time, where c 4γ [16], and γ can be bounded from above by 2 2 [10, 15]. The depth of the local search is bounded by (c/ε) 2 4γ 2, so ε 2. Thus, the fastest operation of this algorithm is obtained by setting ε = 2 for a 3-factor approximation result, and this will run in O(m 2 (8 2/2) 2 +1 n) = O(m 65 n) time in the worst case. Clearly, this algorithm is not practical for m 2. Our present work is directed towards finding approximation algorithms for DUDC that are tractable. Minimum Geometric Disk Cover. In the minimum geometric disk cover problem, the input consists of a set of points in the plane, and the problem is to find a set of unit disks of minimum cardinality whose union covers the points. Unlike DUDC, disk centers are not constrained to be selected from a given discrete set, but rather may be centered at arbitrary points in the plane. Again, this problem is NP-hard [9] and has a PTAS solution [11, 12]. Discrete k-center. Given two sets of points in the plane P and Q and an integer k, find a set of k disks centered on points in P whose union covers Q such that the radius of the largest disk is minimized. Observe that set Q has a discrete unit disk cover consisting of k disks centered on points in P if and only if Q has a discrete k-center centered on points in P with radius at most one. This problem is NP-hard if k is an input variable [2], but when k is fixed, a solution can be found in m O( k) time [13], or m O(k1 1/d) time for points in R d [1]. 2 Preliminaries Here, we describe two restricted discrete unit disk cover problems. The first problem is called the Restricted Line Separable Discrete Unit Disk Cover (RLSDUDC) problem and the second one is called the Line Separable Discrete Unit Disk Cover (LSDUDC) problem. Both problems are defined on the 2-dimensional plane, and depend on the plane being divided into two half-planes l + and l defined by a line l. The definition of RLSDUDC and LSDUDC are given below: RLSDUDC: Given a set of n points P and a set D of m unit disks such that the points in P and the center of disks in D are separated by the line l, and each point in P is covered by at least one disk in D, find a minimum size set D D such that each point in P is covered by at least one disk in D. LSDUDC: Given a set of n points P in l + and a set D of m unit disks with centers in l + l such that each point in P is covered by at least one disk centered in l, find a minimum size set D D such that each point in P is covered by at least one disk in D. Theorem 1. [5] The RLSDUDC problem can be solved optimally whereas we solve the LSDUDC problem with a 2-factor approximation. The running time for both solutions is O(n log n + mn) where m = D and n = P. 3 Testing Feasibility Historically, discussions of set cover problems typically assume that a feasible solution exists, both to simplify the presentation and because the test is simple. We include a brief discussion 3

4 here for completeness, but furthermore to demonstrate that the incorporation of a feasibility test does not affect the final running time of our algorithm. For the feasibility check, given a point set P and a unit disk set D, the test that all points in P are covered by at least one disk in D can be done in O(m log m + n log m) time as follows: Let D center be the set of centers of the unit disks in D. Draw the Voronoi diagram V D(D center ) of the points in D center [3]. For each point p P, find its nearest point q p D center using the point location algorithm in the planar subdivision V D(D center ). If δ(p, q p ) 1 for all p P we can say that all the points in P are covered by at least one unit disk in D, where δ(a, b) = the Euclidean distance between points a and b. The time complexity follows from the following facts: (i) constructing the Voronoi diagram takes O(m log m) time, and (ii) for each point p P, the planar point location algorithm takes O(log m) time. Thus, we have the following lemma: Lemma 1. The feasibility of the DUDC problem can be determined in O(m log m + n log m) time. 4 Discrete Unit Disk Cover Problem Given the feasibility test, for the remainder of the discussion we will assume that all points in P are covered by at least one disk in D. Let R be an axis aligned rectangle such that all points in P and all centers of the disks in D are inside R. In the DUDC algorithm, we first divide R by horizontal line segments l 1, l 2,..., l t 1 from top to bottom such that δ(l i, l i+1 ) = 1 2 for each i {0, 1,..., t 1} where l 0 and l t are the top and bottom boundaries of the rectangle R and δ(a, b) is the Euclidean distance between two horizontal line segments a and b. We denote the horizontal strip bounded by the lines l i and l i+1 as [l i, l i+1 ]. The proposed algorithm for the discrete unit disk cover problem has two fundamental components. In the first step, we describe an algorithm for covering all points in P which are line-separable from a disk in D by any line in the set {l 1,..., l t 1 }. By carefully partitioning the set P, we obtain a 12-approximation algorithm for this setting, which we call the outside strip discrete unit disk cover problem. In the second step, we present an algorithm for covering points in P [l i, l i+1 ], which are covered only by the disks centered inside the strip [l i, l i+1 ], for each i {0, 1,..., t 1} (this is the case for all points remaining after the first step). We present a 6-approximation algorithm for this problem, which we denote the within strip discrete unit disk cover problem. 4.1 Outside Strip Discrete Unit Disk Cover This subsection presents a 12-approximation algorithm for finding a cover for all points P o P, where for each p P o, if p is contained in the strip H then p may be covered by a disk centered outside of H 2. 2 Other points not having this property (i.e. those points only covered by a disk centered in H) may be covered when we use the outside strip discrete unit disk cover algorithm, but this has no effect on the running time or 4

5 Let Di U and Di L be the set of disks centered in the horizontal strips [l i 1, l i ] and [l i, l i+1 ] (defined at the beginning of Section 4) respectively. Let Pi L ( P) be the set of points below the line l i and covered by the disks centered in the horizontal strip [l i 1, l i ] but not by any disk centered above the line l i 1 (such points belong to Pi 1 L ). Similarly, let PU i ( P) be the set of points above the line l i and covered by the disks centered in the horizontal strip [l i, l i+1 ] but not by any disk centered above the line l i 1 (these will be in either Pi 2 L or PL i 1 ) or below the line l i+1 (if not covered from above, such points belong to Pi+1 U ). Compute the sets P1 L, PL 2,..., PL t, Pt U, Pt 1 U,..., PU 1 P in order as follows: 1. Set P = P. 2. For each i {1, 2,..., t} compute P L i = P D U i and set P = P \ P L i. 3. For each i {t, t 1,..., 1} compute P U i = P D L i and set P = P \ P U i. Note that all sets Pi L are covered by the disks centered inside the horizontal strip [l i 1, l i ]. Similarly, all sets Pi U are covered by the disks in D whose centers are inside the horizontal strip [l i, l i+1 ] but no points in Pi U are covered by a disk centered above the line l i 1. Without loss of generality assume that t = 4s. Now, consider six sets of sets related to P i for i {1, 2,... 6} defined below: S 1 = {Q 1j (= P1+4j L ) j = 0, 1,..., s 1} S 2 = {Q 2j (= P2+4j L ) j = 0, 1,..., s 1} S 3 = {Q 3j (= P3+4j L ) j = 0, 1,..., s 1} S 4 = {Q 4j (= P4+4j L ) j = 0, 1,..., s 1} S 5 = {Q 5j (= P1+2j U ) j = 0, 1,..., 2s 2} S 6 = {Q 6j (= P2+2j U ) j = 0, 1,..., 2s 2} We characterize the elements in each of the sets S i for i {1, 2,..., 6} using the following lemma: Lemma 2. If p Q uv and p Q uw are two points such that for u {1, 2, 3, 4}, v w and v, w {1, 2,... s 1}, and for u {5, 6}, v w and v, w {1, 2,..., 2s 2}, then a single disk in D cannot cover both the points p and p. Proof. We prove here the lemma for u = 1 and u = 5. A proof for the other cases arises in a similar fashion. Case u = 1: From the definition of Q 1v and Q 1w ; Q 1v = P1+4v L and Q 1w = P1+4w L. Now, from the definition of P1+4v L and PL 1+4w each element of PL 1+4v and PL 1+4w is covered by the disks centered inside the horizontal strip [l 4v, l 1+4v ] and [l 4w, l 1+4w ], respectively. Since v w and 1 the height of each horizontal strip is 2, so the distance between a point inside the strip [l 4v, approximation factor. 5

6 l 1+4v ] and a point inside the strip [l 4w, l 1+4w ] is greater than 2. Thus, in this case the lemma follows from the fact that the disks are unit radius disks. Case u = 5: From the definition of Q 5v and Q 5w ; Q 5v = P1+2v U and Q 5w = P1+2w U. Now, from the definition of P1+2v U each element of PU 1+2v is covered by the disks centered inside the horizontal strip [l 1+2v, l 2+2v ], but no points in P1+2v U are covered by a disk centered above the line l 2v. Similarly, from the definition of P1+2w U, each element of PU 1+2w is covered by the disks centered inside the horizontal strip [l 1+2w, l 2+2w ], but no points in P1+2w U are covered by a disk centered above the line l 2w. Since v w, there does not exist a disk which covers one point in P1+2v U and a point in PU 1+2w. The OSDUDC algorithm for covering the points in P1 L, PL 2,..., PL t, Pt U, Pt 1 U,..., PU 1 described in Algorithm 1. is Algorithm 1 OSDUDC(P, D) 1: Input: A point set P and a unit disk set D. 2: Output: A set D1 D. 3: D1 4: Compute the sets P1 L, PL 2,... PL t, Pt U, Pt 1 U,... PU 1 as defined at the beginning of section : for (i = 1, 2,..., t) do 6: Run the LSDUDC algorithm on the point set Pi L based on the line l i. Let D D be the output of the algorithm. Set D1 D 1 D. 7: end for 8: for (i = t, t 1,..., 1) do 9: Run the LSDUDC algorithm on the point set Pi U based on the line l i. Let D D be the output of the algorithm. Set D1 D 1 D. 10: end for 11: return D 1 Lemma 3. Algorithm 1 computes a 12-factor approximation in O(n log n + mn) time. Proof. The sets of points P1 L, PL 2,..., PL t, Pt U, Pt 1 U,..., PU 1 can be partitioned into 6 sets of sets such that a disk cannot cover a point from each of Q ij S i and Q ij S i, j j, by Lemma 2. Thus, the approximation factor of the lemma follows from the fact that the LSDUDC algorithm produces a 2-factor approximation result (see Theorem 1) and can participate in at most 6 sets. Since we can divide the plane into strips such that each strip contains at least one point in P, t = O(n). The time complexity of the lemma follows from the fact that (i) each disk in D can participate 6 times in the LSDUDC algorithm and (ii) the running time of the LSDUDC algorithm is O(n log n + mn) (see Theorem 1). 4.2 Within Strip Discrete Unit Disk Cover In this subsection, we propose a 6-factor approximation algorithm for the discrete unit disk cover problem with a restricted configuration. The points in P are inside a horizontal strip H with 6

7 1 height 2 and the centers of the unit disks in D are also inside the same strip 3. The objective is to cover all the points in P using the minimum number of disks from D. Without loss of generality, assume that the length of H is k 2 for some positive integer k. Before describing the covering algorithm we partition the rectangle H into k squares of equal size by introducing vertical line segments L 0, L 1,..., L k. Assume L 0 and L k are the left and right vertical boundaries of the rectangle. We denote the squares as σ 1, σ 2,..., σ k in left to right order. Let Di σ( D) be the set of disks centered in the square σ i, and Pi σ ( P) be the set of points in the square σ i. A block is the union of a maximal run of consecutive squares σ i,..., σ j, such that no disk is centered inside any square of the block, i.e. i {i,..., j}, Di σ =. Let B 1, B 2,..., B l be the blocks from left to right order in the horizontal strip H, where B i is bounded by the vertical line segments L b and L e (0 b < e k). The within strip discrete unit disk cover (WSDUDC) algorithm for P works as follows: In the first step, we cover all points within the blocks B 1, B 2,..., B l. To cover points in the block B i, we first apply the LSDUDC algorithm based on the line L b to cover the points in B i that are covered by disks centered left of L b (although the resulting cover may consist of disks centered on either side of L b, since we use LSDUDC). Next, the RLSDUDC algorithm is used on the line L e to cover any remaining points in B i. Note that since the latter points are only covered by disks centered right of L e, the RLSDUDC algorithm computes a globally optimal cover of these points. All points in B i will be covered by the above two algorithms, if the problem instance has a feasible solution. In the second step of the algorithm, we scan all σ i s from left to right order. For any square σ i, if there are points in Pi σ not covered by any disk from the solution set found in the first step, then any disk from Di σ covers all points in Pi σ. One such disk is chosen arbitrarily and added to the solution set. The pseudocode for the within strip discrete unit disk cover (WSDUDC) algorithm for P is described in Algorithm 2. Lemma 4. Algorithm 2 never calls LSDUDC with respect to two consecutive lines L i 1 and L i for i {1, 2,... k}. Furthermore, the algorithm never calls RLSDUDC with respect to two consecutive lines L i 1 and L i for i {1, 2,... k}. Proof. The LSDUDC algorithm is only ever applied with respect to a line which is the left boundary of a block (line 23 of Algorithm 2), say w.l.o.g. one such line is L i 1. By definition, this means that Di 1 σ and Dσ i =, which together imply that neither L i 2 nor L i can form the left boundary of a block. Therefore, LSDUDC is never applied to two consecutive lines. Analogously, the RLSDUDC is only ever applied to a line which forms the right boundary of a block (line 17 of Algorithm 2), again say w.l.o.g. one such line is L i 1. It follows that in this case Di 1 σ = and Dσ i, which together imply that neither L i 2 nor L i can form the right boundary of a block. Therefore, Algorithm 2 never applies RLSDUDC to two consecutive lines. Lemma 5. Algorithm 2 produces a 6-factor approximation result in O(n log n + mn) time. 3 Recall that when used as a subroutine of the DUDC algorithm, the set of points P consists only of those points not covered by any disk centered outside of the strip. 7

8 Algorithm 2 WSDUDC(P, D) 1: Input: A set P of points in a horizontal strip H of dimensions 1 2 k 2, and a set D of unit disks centered in H. 2: Output: A set D2 D of disks covering all the points in P. 3: D 1 σ, Dσ 0, Pσ 1, Pσ 0, Pσ k+1, Pσ k+2 and D 2. 4: for (i = 1, 2,..., k) do 5: Compute the sets Di σ and Pi σ 6: end for 7: Set i 1 8: while (i k) do 9: while ((i k) and (Di σ = )) do 10: i i + 1 /* block searching */ 11: end while 12: if Di 2 σ = then 13: P left ( Pi 2 σ ( i 1) Pσ D σ i Di+1) σ (see Figure 1(a)); 14: else 15: P left Pi 1 σ ( Di σ i+1) Dσ (see Figure 1(b)). 16: end if 17: Apply the RLSDUDC algorithm based on the line L i 1 to cover the points in P left. Let D be the set of disks in this solution. 18: Set D2 D 2 D and i i : while ((i k) and (Di σ )) do 20: i i + 1 /* skip σ i s to find the next block */ 21: end while 22: P right ( Pi σ Pi+1) σ ( D σ i 2 Di 1) σ (see Figure 1(c)). 23: Apply the LSDUDC algorithm based on the line L i 1 to cover the points in P right. Let D be the set of disks in this solution. 24: Set D2 D 2 D and i i : end while 26: Set j 1 27: while (j k) do 28: If Dj σ, Pσ j but all points of Pσ j are not covered by D 2 (i.e., Dσ j D 2 = ) then add an arbitrary disk from Dj σ to D 2. 29: Set j j : end while 31: return D 2 8

9 σ i-2 σ i-1 σ i σ i L i-3 L i-2 L i-1 L i L i+1 (a) σ i-2 σ i-1 σ i σ i L i-3 L i-2 L i-1 L i L i+1 (b) σ i-1 σ i L i-2 L i-1 L i (c) Figure 1: Demonstration of Algorithm 2. Shaded circles indicate the points in P and empty circles indicate disk centers. Proof. To prove that Algorithm 2 gives a 6-factor approximation, we consider the following four cases, which describe the various ways that disks from Di σ may be used: Case 1: Used by LSDUDC on line L j, where j < i. Since the distance between the lines L i 1 and L i 3 is less than 2, a disk d σ i may cover points that are covered by disks with centers left of the lines L i 1, L i 2 and L i 3. Therefore, the disk d may appear in a solution to the LSDUDC algorithm with respect to the lines L i 1, L i 2 and L i 3. Since Di σ (as d σ i ), Algorithm 2 never calls LSDUDC with respect to the line L i 1. From Lemma 4, Algorithm 2 never calls the LSDUDC algorithm with respect to two consecutive lines. Therefore, d may appear in the solution of LSDUDC with respect to either L i 2 or L i 3. Case 2: Used by LSDUDC on line L j, where j i. Again, since the distance between the lines L i and L i+1 is less than 1, a disk d σ i may cover points in σ i+1 and σ i+2. Therefore, the disk d may appear in a solution to the LSDUDC algorithm with respect to the lines L i and L i+1. From Lemma 4, Algorithm 2 never calls the LSDUDC algorithm with respect to two consecutive lines. Therefore, d may appear in the solution of LSDUDC with respect to either L i or L i+1. Case 3: Used by RLSDUDC. Since the distance between the lines L i 1 and L i 2 is less than 1, a disk d σ i may cover points in σ i 1 and σ i 2. Therefore, the disk d may appear in a solution to the RLSDUDC algorithm with respect to the lines L i 1 and L i 2. However, by 9

10 Lemma 4, Algorithm 2 never calls the RLSDUDC algorithm with respect to two consecutive lines. Therefore, d may appear in the solution of RLSDUDC with respect to either L i 1 or L i 2. Case 4: Used to cover Pi σ. A disk d σ i may be selected in the solution if some points in Pi σ are not covered by any disk in the solutions of the LSDUDC and RLSDUDC algorithms (line 28 of Algorithm 2). Since the size of all squares σ i (1 i k) is , one disk centered anywhere in σ i is sufficient to cover all the points in Pi σ. There are two basic properties that are fundamental to our analysis. First, a disk can cover points in at most four squares. Second, Case 4 is only ever used for a particular square if no disk was used for Cases 1-3 (since any disk centered in σ i covers all points in Pi σ). Suppose that there are four squares σ i 2,..., σ i+1, all of whose points may be covered by a single disk from Di σ. We may choose poorly so that Case 4 is applied to each square, and so a disk is chosen from each of Di 2 σ,dσ i 1,Dσ i, and Dσ i+1. In this example, we have used four disks to cover the points rather than one, so the approximation factor of Algorithm 2 is at least 4. By substituting some of these squares with blocks, we can force disks from Di σ to fall into Cases 1-3 to increase the approximation factor. Specifically, suppose there exists a block B j which consists of a single square lying to the left of σ i, and a block B j+1 which begins to the right of σ i so that the left boundary lines of B j and B j+1 may be intersected by disks from Di σ. In this setting, disks from Di σ could be used in Cases 1 and 3 to cover the points of B j, and also by Case 2 to cover the points of B j+1. These applications lead to at least a 5-approximation, since Cases 1 and 2 lead to 2-approximations and Case 3 is an optimal algorithm. Disks from Di σ may also cover points in two other squares outside of these blocks, say σ i and σ i, i < i < i. If these squares are adjacent to σ i, then if applicable, their disks are used in the same computations of LSDUDC and RLSDUDC as the disks of Di σ, so this would have no effect on the approximation factor. However, it is possible that σ i and σ i fall into Case 4, and that a set of disks of equal cardinality for the solutions to Cases 1-3 would also cover all the points in σ i and σ i. Therefore, the overall approximation factor is five times the size of optimal set of disks, plus two. Note that since disks from Di σ are covering points from five squares in this setting (specifically σ i 2,..., σ i+2 ), the optimal solution requires at least two disks to cover the same set of points. Let Opt be the size of the optimal solution, then note that 5 Opt Opt when Opt 2, so we have our bound. If all disks from Di σ cover points from at most four squares, then the approximation factor of 6 is immediate. Thus, the approximation factor of the lemma follows. The time complexity of the lemma follows from the fact that each disk in D can participate a constant number of times in the LSDUDC and RLSDUDC algorithms and the running time of both the LSDUDC and RLSDUDC algorithms is O(n log n + mn) (see Theorem 1). 4.3 The Discrete Unit Disk Cover Algorithm We describe the algorithm for the DUDC problem in Algorithm 3, using Algorithms 1 and 2 as subroutines. Theorem 2. Algorithm 3 produces an 18-approximation result in O(m log m + n log n + mn) time for the discrete unit disk cover problem. 10

11 Algorithm 3 DUDC(P, D) 1: Input: A set of points P and a set of unit disks D. 2: Output: A set D D. 3: D and D1. 4: if (Some point in P is not covered by any disk from D) then 5: return D (This indicates that no feasible solution exists) 6: end if 1 7: Let H i [l i, l i+1 ] for i = 0, 1,..., t 1 be the horizontal strips of height 2 defined in Section 4. 8: D1 OSDUDC(P, D) (Algorithm 1) 9: P P \ (D1 P) 10: for (i=0,1,..., t-1) do 11: Let D i D be the set of disks having centers in H i. 12: Let P i P H i be the set of points in H i. 13: D2 i WSDUDC(P i, D i ) (Algorithm 2) 14: end for 15: D 2 D0 2 D Dt : D D 1 D 2 17: return D Proof. The approximation result follows from Lemma 3 and Lemma 5, which established a 12- approximation and a 6-approximation for OSDUDC and WSDUDC, respectively. The time complexity of Algorithm 3 follows from the following: (i) In line number 4, the running time of the feasibility test for a DUDC instance is O(m log m+ n log m) (Lemma 1). (ii) In line number 8, the running time of OSDUDC(P, D) is O(n log n + mn) (Lemma 3). (iii) In line number 13, the running time of each call to WSDUDC(P i, D i ) is O( P i log P i + D i P i ) (Lemma 5). Therefore, the running time of the for loop (lines 10-14) is O(n log n+ mn). 5 Conclusion We have considered the discrete unit disk cover (DUDC) problem, where a set P of n points and a set D of m unit disks in the 2-dimensional plane are given, the objective is (i) to check whether each point in P is covered by at least one disk in D or not and (ii) if so, then find a minimum cardinality subset D D such that unit disks in D cover all the points in P. We provide an 18-factor approximation algorithm for the DUDC problem. The running time of the proposed algorithm is O(n log n + m log m + mn). The previous best known solution for the same problem 11

12 was a 22-factor approximation algorithm with running time O(m 2 n 4 ). Therefore, our solution is a significant improvement over the best known solution in terms of approximation factor as well as running time of the algorithm. Acknowledgements Funding for this project was provided by the Natural Sciences and Engineering Research Council of Canada (NSERC), partially under the NSERC Strategic Grant on Optimal Data Structures for Organization and Retrieval of Spatial Data. The authors would like to thank the anonymous reviewers for their valuable comments which helped to improve the quality of this paper. References [1] P. K. Agarwal and C. M. Procopiuc, Exact and approximation algorithms for clustering, Algorithmica, 33, (2002) [2] P. K. Agarwal and M. Sharir, Efficient algorithms for geometric optimization, ACM Computing Surveys, 30, (1998) [3] M. de Berg, M. V. Kreveld, M. Overmars and O. Schwarzkopf, Computational Geometry Algorithms and Applications, 3rd edition, Springer-Verlag, (2008). [4] H. Brönnimann and M. Goodrich, Almost optimal set covers in finite VC-dimension, Discrete & Computational Geometry, 14(1), (1995) [5] F. Claude, G. K. Das, R. Dorrigiv, S. Durocher, R. Fraser, A. López-Ortiz, B. G. Nickerson and A. Salinger, An improved line-separable algorithm for discrete unit disk cover, Discrete Mathematics, Algorithms and Applications, 2, (2010) [6] G. Călinescu, I. Măndoiu, P.J. Wan and A. Zelikovsky, Selecting forwarding neighbours in wireless ad hoc networks, Mobile Networks and Applications, 9, (2004) [7] P. Carmi, M. J. Katz and N. Lev-Tov, Covering points by unit disks on fixed location, Proc. of the 18th International Symposium on Algorithms and Computation, LNCS , (2007) [8] V. Chvátal, A greedy heuristic for the set-covering problem, Mathematics of Operations Research, 4(3), (1979) [9] R. Fowler, M. Paterson and S. Tanimoto, Optimal packing and covering in the plane are NP-complete, Information Processing Letters, 12, (1981) [10] G. Frederickson, Fast algorithms for shortest paths in planar graphs, with applications, SIAM Journal on Computing, 16, (1987)

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