aa r X i v : . [ c s . C G ] J u l Dispersion in disks ∗ Adrian Dumitrescu † Minghui Jiang ‡ November 1, 2018
Abstract
We present three new approximation algorithms with improved constant ratios for selecting n points in n disks such that the minimum pairwise distance among the points is maximized.(1) A very simple O ( n log n )-time algorithm with ratio 0 .
511 for disjoint unit disks.(2) An LP-based algorithm with ratio 0 .
707 for disjoint disks of arbitrary radii that uses alinear number of variables and constraints, and runs in polynomial time.(3) A hybrid algorithm with ratio either 0 . . O ( n log n )-time algorithm or the LP-based algorithm.The LP algorithm can be extended for disjoint balls of arbitrary radii in R d , for any (fixed)dimension d , while preserving the features of the planar algorithm. The algorithm introducesa novel technique which combines linear programming and projections for approximating Eu-clidean distances. The previous best approximation ratio for dispersion in disjoint disks, evenwhen all disks have the same radius, was 1 /
2. Our results give a partial answer to an openquestion raised by Cabello, who asked whether the ratio 1 / Keywords : Dispersion problem, linear programming, approximation algorithm.
Let R be a family of n subsets of a metric space. The problem of dispersion in R is that ofselecting n points, one in each subset, such that the minimum inter-point distance is maximized.This dispersion problem was introduced by Fiala et al. [8] as “systems of distant representatives”,generalizing the classic problem “systems of distinct representatives”. An especially interestingversion of the dispersion problem, which has natural applications to wireless networking and maplabeling, is in a geometric setting where R is a set of disks in the plane.Let D be a set of n disks in the plane. Dispersion in disks is the problem of selecting n points, one from each disk in D , such that the minimum pairwise distance of the selected points ismaximized. Dispersion in disks is a hard problem. Fiala et al. [8] showed that dispersion in unitdisks is already NP-hard. It is not difficult to modify their construction, which gives a reduction ∗ A preliminary version of this paper [5] appeared in the Proceedings of the 27th International Symposium onTheoretical Aspects of Computer Science, Nancy, France, March 2010. † Department of Computer Science, University of Wisconsin–Milwaukee, WI 53201-0784, USA. Email: [email protected] . Supported in part by NSF CAREER grant CCF-0444188. Part of the research by this au-thor was done at Ecole Polytechnique F´ed´erale de Lausanne. ‡ Department of Computer Science, Utah State University, Logan, UT 84322-4205, USA. Email: [email protected] . Supported in part by NSF grant DBI-0743670.
Placement with ratio 3 / Centers with ratio 1 / Placement and
Centers and taking the better solution yields an approximation algorithm with ratio about0 . / . p = ( x p , y p ) and q = ( x q , y q ) in the plane,let | pq | denote the Euclidean distance between them: | pq | = p ( x p − x q ) + ( y p − y q ) . A unit disk is a disk of radius one. The distance between two disks is the distance between their centers; e.g.,the distance between two tangent disks of radii r and r is r + r . Throughout this paper, disjoint means interior-disjoint .We now review the previous best -approximation for dispersion in disjoint disks, which isachieved by a naive algorithm Centers that simply selects the centers of the given disks as thepoints [3]. Let D = { Ω , . . . , Ω n } be a set of n disjoint disks of arbitrary radii in the plane. Foreach i , let r i be the radius of Ω i . For i = j , let δ ij be the distance between Ω i and Ω j . Let δ bethe minimum pairwise distance of the disks in D , i.e., δ = min i = j δ ij . Let OPT denote an optimalsolution and CEN denote the solution returned by Centers . We clearly have CEN = δ . Since thedisks are disjoint, r i + r j ≤ δ ij , i = j. It follows that OPT ≤ min i = j ( δ ij + r i + r j ) ≤ i = j δ ij = 2 δ. (1)Consequently, the algorithm Centers achieves an approximation ratio ofCENOPT ≥ δ δ = 12for disjoint disks of arbitrary radii. Cabello asked whether this trivial -approximation can beimproved, even when all disks are not only disjoint but also have the same radius [3, p. 72].We start with a very simple and efficient algorithm that achieves a ratio better than 1 / Theorem 1.
There is an O ( n log n ) -time approximation algorithm with ratio . for dispersionin n disjoint unit disks. Using linear programming, we then obtain the following substantially better approximation fordispersion in disjoint disks of arbitrary radii.
Theorem 2.
There is an LP-based approximation algorithm, with O ( n ) variables and constraints,and running in polynomial time, that achieves approximation ratio . , for dispersion in n disjointdisks of arbitrary radii. Moreover, the algorithm can be extended for disjoint balls of arbitrary radiiin R d , for any (fixed) dimension d , while preserving the same features. We next improve the 0 . Placement of Cabello in combination with eitherthe simple O ( n log n )-time algorithm in Theorem 1 or the LP-based algorithm in Theorem 2.2 heorem 3. In combination with an algorithm of Cabello, the simple O ( n log n ) -time algorithmin Theorem 1 yields an O ( n ) -time algorithm with ratio . , and the LP-based algorithm inTheorem 2 yields a polynomial-time algorithm with ratio . , for dispersion in n (not necessarilydisjoint) unit disks. It is likely that our method for proving Theorem 2, which uses projections for approximat-ing distances, and linear programming for optimization, is also applicable to other optimizationproblems involving distances.
Related work.
The dispersion problem in disks we study here is related to a few other problemsin computational geometry. We mention several results that are more closely related to ours:1. For labeling n points with n disjoint congruent disks, each point on the boundary of a dis-tinct disk, such that radius of the disks is maximized, Jiang et al. [10] presented a . ε -approximation algorithm, and proved that the problem is NP-hard to approximate with ratiomore than . .2. For packing of n axis-parallel congruent squares (congruent disks in the L ∞ metric) in thesame rectilinear polygon such that the side length of the squares is maximized, Baur andFekete [1] presented a -approximation algorithm, and proved that the problem is NP-hardto approximate with ratio more than . A -approximation algorithm for a related problemof packing n unit disks in a rectangle without overlapping an existent set of m unit disks inthe same rectangle, has been obtained by Benkert et al. [2].3. Given n points in the plane, Demaine et al. [4] considered the problem of moving them to an independent set in the unit disk graph metric: that is, each point has to move to a positionsuch that all pairwise distances are at least 1, and such that the maximum distance a pointmoved is minimized. They presented an approximation algorithm, which achieves a goodratio if the points are initially “far from” an independent set. However the approximationratio becomes unbounded for instances that are “very close to” an independent set. Observethat in this problem, the optimum may be arbitrarily small, i.e., arbitrarily close to 0.4. Dumitrescu and Jiang [6] obtained hardness results and approximation algorithms for tworelated geometric problems involving movement. The first is a constrained variant of the k -center problem, arising from a geometric client-server problem. The second is the problemof moving points towards an independent set, discussed previously. In this section we present a very simple approximation algorithm A1 with ratio 0 .
511 for disjointunit disks, and thereby prove Theorem 1.The idea of the algorithm is as follows. Recall that δ is the minimum pairwise distance amongthe unit disks. Let σ = σ ( δ ) be a positive parameter to be specified; in particular, at the thresholddistance δ = 2 for disjoint unit disks, we have σ (2) = 2 . . . . , which is only slightly larger than δ . Consider the distance graph of the unit disks for the parameter σ , which has a vertex for eachdisk, and an edge between two vertices if and only if the corresponding disks have distance at most σ . If there is a vertex of degree at least two in the distance graph, that is, if there is a disk close totwo other disks, then a packing argument shows that the minimum pairwise distance of any threepoints in the three disks must be small. Thus simply placing the points at the disk centers already3chieves a good approximation ratio. Otherwise, every vertex in the distance graph has degree atmost one, and the edges form a matching. In this case, the disks that are close to each other aregrouped into pairs. The distance between the two points in each pair can be slightly increased bymoving them away from the disk centers, at the cost of possibly decreasing the distances betweenpoints in different pairs.Let D be a set of n (not necessarily disjoint) unit disks in the plane. The algorithm A1 consistsof three steps:1. Compute the minimum pairwise distance δ of the disks in D , and for each disk, find the twodisks closest to it.2. If the distance from some disk to its second closest disk is at most σ = σ ( δ ), return the n disk centers as the set of points. Otherwise, proceed to the next step.3. Place a point at the center of each disk. Then, for each disk, if the distance from the diskto its closest disk is at most σ , move the point away from the closest disk for a distance of( σ − δ ) /
4, so that the two points in each close pair of disks are moved in opposite directions;we will show that δ < σ < δ + 4, thus the distance ( σ − δ ) / Algorithm analysis.
The bottleneck for the running time of the algorithm A1 is simply thecomputation of the two closest disks from each disk in step 1, which takes O ( n log n ) time [7,p. 306]. The other two steps of the algorithm can clearly be done in O ( n ) time. For the proof ofthe approximation ratio, define the following function f ( s ) for s ≥ f ( s ) = q (1 + s ) + 1 / p s ) − / . (2)The function f ( · ) is increasing and f (0) = √
3. The justification for step 2 of the algorithm A1 isthe following packing lemma. Here the disk with center O is close to two other disks with centers P and Q , respectively; see Figure 1.PSfrag replacements A AB B CC OOP PQ Q ss tt t (a) (b)Figure 1: (a) A linkage of the five segments
AP, BQ, CO, OP, OQ for three points
A, B, C in three unitdisks with centers
P, Q, O , respectively. (b) The extreme configuration:
A, P, O are collinear,
B, Q, O arecollinear, | AP | = | BQ | = | CO | = 1, | OP | = | OQ | = s , | AC | = | BC | = | AB | = t . emma 1. Let
A, B, C be three points in three unit disks with centers
P, Q, O , respectively. Let s = max {| OP | , | OQ |} and t = min {| AC | , | BC | , | AB |} . Then t ≤ f ( s ) .Proof. Refer to Figure 1(a) for a linkage of the five segments
AP, BQ, CO, OP, OQ under thelength constraints max {| OP | , | OQ |} = s and max {| AP | , | BQ | , | CO |} ≤
1. Then, by a continuousmovement argument (with the point O fixed), it follows that the minimum pairwise distance t ofthe three points A, B, C is maximized by the extreme configuration in Figure 1(b).Now consider the extreme configuration in Figure 1(b). Let θ = ∠ AOB . In △ AOB , we have t = 2(1 + s ) sin θ = ⇒ sin θ = t s ) . In △ AOC , we have t = 1 + (1 + s ) − s ) cos( π − θ ) = ⇒ cos θ = t − (1 + s ) − s ) . From sin θ + cos θ = 1, it follows that t + (cid:0) t − (1 + s ) − (cid:1) = 4(1 + s ) . Substitute x = (1 + s ) and y = t , and we have y + ( y − x − = 4 x, which solves to y + y + x + 1 − xy − y + 2 x = 4 xy − (2 x + 1) y + x − x + 1 = 0 y = x + 1 / ± p x − / y are valid, but since we are deriving an upper bound for t , we take the largersolution y = x + 1 / p x − / . Thus, in this extreme case, we have t = (1 + s ) + 1 / p s ) − / f ( s ) . It follows that, in general, t ≤ f ( s ).Consider the following equation in σ : δf ( σ ) = σ + δ δ + 2) . (3)The next lemma confirms that σ exists and lies in the desired range: Lemma 2.
There is a unique solution σ to (3) . Moreover, δ < σ < δ + 4 . roof. We first show that, for any s ≥ s + 1 < f ( s ) < s + 2: f ( s ) = q (1 + s ) + 1 / p s ) − / > p (1 + s ) = s + 1 ,f ( s ) = q (1 + s ) + 1 / p s ) − / < q (1 + s ) + 1 + p s ) = s + 2 . Since the function f ( s ) is increasing in s for s ≥
0, the left-hand side of (3) is decreasing in σ .On the other hand, the right-hand side of (3) is increasing in σ . If σ ≤ δ , then we would have theinequality δf ( σ ) ≥ δf ( δ ) > δδ + 2 = δ + δ δ + 2) ≥ σ + δ δ + 2) . Similarly, if σ ≥ δ + 4, then we would have the inequality δf ( σ ) ≤ δf ( δ + 4) < δ ( δ + 4) + 1 < δ + 4) + δ δ + 2) ≤ σ + δ δ + 2) . Therefore, there is a unique solution σ to (3), and δ < σ < δ + 4.We now analyze the approximation ratio of the algorithm A1 . Let ALG be the minimumpairwise distance of the points returned by the algorithm. Let OPT be the minimum pairwisedistance of the optimal set of points. Let c = c ( δ ) = δf ( σ ) = σ + δ δ + 2) . (4)We next prove that the approximation ratio of the algorithm A1 is at least c , namely that ALG ≥ c · OPT, by considering two cases: • If the algorithm returns the n disk centers as the set of points in step 2, then there is a disksuch that the distances from the disk to its two closest disks are at most σ . By Lemma 1, wehave OPT ≤ f ( σ ). Since ALG = δ , it follows thatALGOPT ≥ δf ( σ ) . (5) • If the algorithm proceeds to step 3, then the distance from each disk to its second closest diskis more than σ . If two disks have distance at most σ , then they must be the closest disks ofeach other, and the movements of points in step 3 ensure that their two points have distanceat least δ + 2( σ − δ ) / σ + δ ) /
2. On the other hand, if two disks have distance more than σ ,then after the movements their two points have distance at least σ − σ − δ ) / σ + δ ) / ≥ ( σ + δ ) /
2. Since OPT ≤ δ + 2, it follows thatALGOPT ≥ σ + δ δ + 2) . (6)By (4), (5), and (6), the algorithm A1 achieves an approximation ratio of c ( δ ) for δ ≥
0. It canbe verified that c ( δ ) is an increasing function of δ for δ ≥
0. Thus, for dispersion in disjoint unitdisks, the approximation ratio is c ( δ ) ≥ c (2) = 0 . . . . , for δ ≥ . This completes the proof of Theorem 1. 6
An LP-based approximation algorithm for disjoint disks
Warm-up: one dimension.
As a warm-up exercise, we first study the dispersion problem on theline and on a closed curve. In these two settings the problem can be solved exactly in polynomialtime.
Proposition 1.
There exists a polynomial-time exact algorithm based on linear-programming for
Dispersion in disjoint intervals on the line and on a closed curve.Proof.
For the line, the input consists on n interior-disjoint intervals [ a , b ] , . . . , [ a n , b n ], where(i) a i ≤ b i , for i = 1 , . . . , n , and (ii) b i ≤ a i +1 , for i = 1 , . . . , n −
1. Computing an optimal solution x i ∈ [ a i , b i ], i = 1 , . . . , n , amounts to solving the following linear program with the n variables x i and 3 n − z (LP1)subject to (cid:26) a i ≤ x i ≤ b i ≤ i ≤ nx i +1 − x i ≥ z, ≤ i ≤ n − L , the input consists on n interior-disjoint intervals [ a , b ] , . . . ,[ a n , b n ], where (i) a i ≤ b i , for i = 1 , . . . , n , (ii) b i ≤ a i +1 , for i = 1 , . . . , n −
1, and (iii) 0 ≤ a , b n ≤ L . Computing an optimal solution x i ∈ [ a i , b i ], i = 1 , . . . , n , amounts to solving the followinglinear program with the n variables x i and 3 n constraints:maximize z (LP2)subject to a i ≤ x i ≤ b i ≤ i ≤ nx i +1 − x i ≥ z, ≤ i ≤ n − x + L − x n ≥ z Two dimensions.
Next we present and analyze the approximation algorithm A2 and therebyprove Theorem 2. We first introduce some definitions and notations. Let Ω , . . . , Ω n be n pairwisedisjoint disks of radii r , . . . , r n , and centers o , . . . , o n . Let 0 ≤ λ < λ ′ < λ = 1 / λ ′ = 3 / i = 1 , . . . , n , let ω i and ω ′ i be two disks of radii λ · r i and λ ′ · r i , respectively, that are concentricwith Ω i . Note that ω i ⊂ ω ′ i ⊂ Ω i . Let α ij ∈ [ − π/ , π/
2) be the direction (or angle) of the linedetermined by o i and o j . For α ∈ [ − π/ , π/ ℓ α be any line of direction α . For two vectors u = ( u , u ), and v = ( v , v ), their dot product is h u, v i = u v + u v . The scalar projection of v onto u is the length of the orthogonal projection of the vector v onto u , with a minus sign if thedirection is opposite. It is given by the formulaproj u v = h u, v i| u | . (7)For two points, p and q , let proj α ( p, q ) denote the length of the projection of the segment pq ontoa line ℓ α of direction α , i.e., onto the vector (cos α, sin α ). For simplicity, we write proj ij ( q i , q j )instead of proj α ij ( q i , q j ).The idea of the algorithm is as follows: Suppose we restrict the feasible region of each point p i from the given disk Ω i to the smaller concentric disk ω i of radius λ · r i . We then show the7xistence of a good approximation for the dispersion problem constrained to the smaller size disks.First, observe that the centers of the original disks Ω i are still in the feasible regions for each ofthe n points. So the -approximation that we could easily achieve earlier, is still attainable. Forinstance, setting λ = 0 yields the algorithm Centers discussed earlier. Second, observe that if λ is sufficiently small, then the distance between two points (in two smaller disks) can be wellapproximated by the projection of the segment connecting the two points onto the line connectingthe centers of the two disks. Enclose each smaller disk ω i in a suitable convex polygon Q i , where ω i ⊂ Q i ⊂ ω ′ i ⊂ Ω i . The length of each such projection can be expressed as a linear combinationof the coordinates of the two points, and we can use linear programming to maximize the smallestprojection length of an inter-point distance. All the constraints in the dispersion problem will beexpressed as linear inequalities, at the cost of finding only an approximate solution. We now presentthe technical details.We start with a technical lemma that will be used in establishing the approximation ratio ofAlgorithm A2 . Lemma 3.
Let λ = 1 / , a ∈ [0 , , and α ∈ [0 , π ) . Then aλ cos α √ a + 2 a cos α ≥ √ . (8) Proof.
Since the numerator of the left hand side in (8) is non-negative, (8) is equivalent to1 + a λ cos α + 2 aλ cos α ≥ λ (1 + a + 2 a cos α ) . (9)After reducing the term 2 aλ cos α , (9) is equivalent to1 + a λ cos α ≥ λ + λa . (10)This follows easily from the following chain of inequalities λ (1 + a ) = 1 + a ≤ ≤ a λ cos α, (11)as required.A key fact relating projections to distances is: Lemma 4.
Let λ = 1 / . Consider two disjoint disks Ω i and Ω j at distance δ ij = | o i o j | . Let p i ∈ Ω i and p j ∈ Ω j be two points. Let q i ∈ ω i be the point on o i p i at distance λ | o i p i | from o i . Similarlydefine q j ∈ ω j as the point on o j p j at distance λ | o j p j | from o j . Then proj ij ( q i , q j ) | p i p j | ≥ √ . (12) Proof.
We can assume w.l.o.g. that o i = (0 ,
0) and o j = (1 , δ ij = 1. To represent points,we use complex numbers in the proof. Let Re( z ) denote the real part of a complex number z ∈ C .The point p i is represented by z i , where z i ∈ C , with | z i | ≤ r i ; hence q i is represented by λz i .Since the disks Ω i and Ω j are disjoint, we have r i + r j ≤
1. The point p j is represented by 1 + z j ,where z j ∈ C , with | z j | ≤ r j ; hence q j is represented by 1 + λz j . Write z = z j − z i , and note that | z | ≤ | z i | + | z j | ≤ r i + r j ≤
1. Let z = a (cos α + i sin α ) be the complex number representationof z , where 0 ≤ a ≤
1, and α ∈ [0 , π ). The segments q i q j and p i p j are then represented by1 + λ ( z j − z i ) = 1 + λz and 1 + z , respectively. 8ote that proj ij ( q i , q j ) = Re(1 + λz ) = 1 + aλ cos α, and | p i p j | = | z | = | a (cos α + i sin α ) | = p a + 2 a cos α . Hence by Lemma 3 we have proj ij ( q i , q j ) | p i p j | = 1 + aλ cos α √ a + 2 a cos α ≥ √ , as required.We now outline our LP-based algorithm A2 for disjoint disks. Conveniently select disjointconvex polygons Q i , i = 1 , . . . , n , such that ω i ⊂ Q i ⊂ ω ′ i ⊂ Ω i , for each i = 1 , . . . , n . For instancelet Q i be an axis-aligned square of side length r i concentric with ω i . Since Ω i are pairwise disjoint, Q i are also pairwise disjoint. Moreover, since Q i and Q j are separated by the perpendicular bisectorof o i o j , for any q i ∈ Q i and q j ∈ Q j , the dot product h q i q j , o i o j i we use in formulating the LP isnon-negative. We are lead to the following linear program, LP3, with the constraints expressedsymbolically at this point. A set { q , . . . , q n } of n points is sought, where q i = ( x i , y i ) ∈ Q i ,for i = 1 , . . . , n . LP3 maximizes the minimum pairwise projection on the line connecting thecorresponding centers of the disks; that is, for each pair ( i, j ), the length of the projection of thesegment connecting the two points q i and q j , on the line connecting the corresponding disk centers o i and o j . maximize z (LP3)subject to (cid:26) q i ∈ Q i , ≤ i ≤ n proj ij ( q i , q j ) ≥ z, ≤ i < j ≤ n Writing the linear constraints.
Each symbolic constraint q i ∈ Q i is implemented as fourlinear inequalities, one for each side of Q i . Implement each symbolic constraint proj ij ( q i , q j ) ≥ z asfollows: Let o i = ( ξ i , ν i ) be coordinates of o i , for i = 1 , . . . , n (part of the input). Consider a pair( i, j ), where i < j . Recall that α ij ∈ [ − π/ , π/
2) is the angle of the line determined by o i and o j .Assuming that ξ ≤ ξ ≤ . . . ≤ ξ n , we havecos α ij = ξ j − ξ i | o i o j | , sin α ij = ν j − ν i | o i o j | . (13)Let a ij = (cos α ij , sin α ij ), so that | a ij | = 1. Let s ij = ( x j − x i , y j − y i ). According to (7),proj ij ( q i , q j ) = h a ij · s ij i| a ij | = h a ij · s ij i = ( x j − x i ) cos α ij + ( y j − y i ) sin α ij . As noted earlier the above expression for the projection is always non-negative. Consequently,for each pair ( i, j ), where i < j , generate the constraint:( x j − x i ) cos α ij + ( y j − y i ) sin α ij ≥ z, where cos α ij and sin α ij are as in (13). 9 olving the linear program.Lemma 5. For any given ε > , a (1 − ε ) -approximation of the solution of LP3 can be obtained inpolynomial time.Proof.
Recall that the linear program LP3 finds a point set { q i = ( x i , y i ) , i = 1 , . . . , n } , q i ∈ Q i ⊂ Ω i , for which the minimum projection is maximized. The constraints of the linear program LP3involve irrational numbers, and hence it cannot be claimed that the original LP is solvable inpolynomial time. However, it is enough to solve the LP up to some precision. For this, it is enoughto approximate the numbers involved in the constraints up to some precision, which is polynomialin the error of the output [12]. Consequently, we can encode each coefficient into a rational numberwith (1 /ε ) O (1) bits. Then, for a constant ε , each coefficient has a constant number of bits, and theLP algorithm runs in polynomial time; e.g., O ( n ) or O ( n . ) using interior point methods. Establishing the approximation ratio.Lemma 6.
For any given ε > , the approximation algorithm A2 can achieve a ratio at least − ε √ for pairwise disjoint disks.Proof. Let OPT = ζ denote an optimal solution, given by n points p , . . . , p n , where p i ∈ Ω i , suchthat | p i p j | ≥ ζ , for all i = j , and | p i p j | = ζ for at least one pair ( i, j ). Let q i ∈ ω i be the point on o i p i at distance λ | o i p i | from o i . By Lemma 3 and 4, for all i = j ,proj ij ( q i , q j ) | p i p j | = 1 + aλ cos α √ a + 2 a cos α ≥ √ . Since | p i p j | ≥ ζ , for all i = j , we immediately conclude thatproj ij ( q i , q j ) ≥ ζ √ , (14)for all i = j .Since ω i ⊂ Q i , by Lemma 5, the point set found by the linear program satisfies q i ∈ Q i ⊂ Ω i ,such that proj ij ( q i , q j ) ≥ − ε √ · ζ. (15)Since obviously, | q i q j | ≥ proj ij ( q i , q j ), we have found points q i ∈ Q i ⊂ Ω i , such that | q i q j | ≥ − ε √ · ζ = 1 − ε √ · OPT . Hence the approximation ratio is at least − ε √ , as claimed. For instance, when setting ε = 10 − , weget a 0 .
707 approximation.
Reducing the number of constraints to O ( n ) . Recall that OPT ≤ δ , as shown in (1).Observe that the LP solution, z ∗ , is bounded from above as (recall that λ ′ = 3 / z ∗ ≤ δ + 3( r i + r j )4 ≤ δ , i, j ) are a closest pair of disks. It follows that there is no need to write any constraints forpairs of disks at distance larger than 7 δ . Indeed, if now ( i, j ) is such a pair, the distance betweentwo points, one in Q i and one in Q j , is at least δ ij − r i + r j )4 ≥ δ ij − δ ij δ ij > δ > z ∗ . An easy packing argument shows that the number of pairs of disks at distance at most 7 δ is only O ( n ): this is the same as the number of pairs of points at distance at most 7 times the minimumpairwise distance among n points. Extension to any (fixed) dimension d . We briefly sketch the differences. The balls ω i and ω ′ i are two smaller balls of radii λ · r i and λ ′ · r i concentric with Ω i , where λ = 1 / λ ′ = 3 / Q i is any suitable convex polytope in R d such that ω i ⊂ Q i ⊂ ω ′ i ⊂ Ω i . The proof of the followinglemma we need, closely mimics the proof of Lemma 4. Lemma 7.
Let λ = 1 / . Consider two disjoint balls Ω i and Ω j at distance δ ij = | o i o j | . Let p i ∈ Ω i and p j ∈ Ω j be two points. Let q i ∈ ω i be the point on o i p i at distance λ | o i p i | from o i . Similarlydefine q j ∈ ω j as the point on o j p j at distance λ | o j p j | from o j . Then proj ij ( q i , q j ) | p i p j | ≥ √ . (16) Proof.
We can assume w.l.o.g. that o i = (0 , . . . ,
0) and o j = (1 , , . . . , e = (1 , , . . . , | e | = 1. The point p i is represented by v i , where | v i | ≤ r i ; hence q i is represented by λv i . Since the two balls Ω i and Ω j are disjoint, wehave r i + r j ≤
1. The point p j is represented by e + v j , where | v j | ≤ r j ; hence q j is representedby e + λv j . Write v = v j − v i , and note that | v | ≤ | v i | + | v j | ≤ r i + r j ≤
1. Let a = | v | ; clearly0 ≤ a ≤
1. Let α ∈ [0 , π ) be the angle made by v with e . We have | p i p j | = | e + v j − v i | = | e + v | . | q i q j | = | e + λv j − λv i | = | e + λv | . By (7), we have proj ij ( q i , q j ) = proj e ( e + λv ) = h e , e + λv i| e | = h e , e + λv i = h e , e i + h e , λv i = 1 + λ h e , v i = 1 + aλ cos α, that is, the same expression as in the planar case. Further, by the cosine formula we have | p i p j | = | e + v | = p | e | + | v | + 2 | e || v | cos α = p a + 2 a cos α, matching again the expression from the planar case.We conclude in the same way by using Lemma 3:proj ij ( q i , q j ) | p i p j | = 1 + aλ cos α √ a + 2 a cos α ≥ √ , as required. 11he symbolic constraints are implemented analogous to those for the planar case. There existsa convex polytope Q ⊂ R d and a function f ( d ) such that ω ⊂ Q ⊂ ω ′ ⊂ Ω, where Q has f ( d )facets, and ω , ω ′ and Ω are concentric balls of radii 1 /
2, 3 / d = 5, a concentric unit hyper-cube is not contained in the unit ball!). The polytope Q i isa translate of r i Q placed at o i , so that ω i ⊂ Q i ⊂ ω ′ i ⊂ Ω i . Each symbolic constraint q i ∈ Q i is implemented as f ( d ) linear inequalities, one for each facet of Q i . Each symbolic constraintproj ij ( q i , q j ) ≥ z implements the dot products from the proof of Lemma 7. Since (again) thereis no need to write any constraints for pairs of balls at distance larger than 7 δ , and the numberof such pairs is linear in n for fixed d [11, p. 211], the total number of constraints is O ( n ). Theapproximation ratio remains the same as for the planar case, namely − ε √ , for any given ε >
0, e.g.,0 .
707 for ε = 10 − . This completes the proof of Theorem 2. In this section we prove Theorem 3. For dispersion in (not necessarily disjoint) unit disks, Cabello [3]presented a hybrid algorithm that applies two different algorithms
Placement and
Centers andthen returns the better solution. We first briefly review Cabello’s analysis for his hybrid algorithm,then present an improved hybrid algorithm that uses the algorithm
Placement in combinationwith either the simple O ( n log n )-time algorithm in Theorem 1 or the LP-based algorithm in Theo-rem 2. In the following, let OPT = 2 x and δ = 2 µ . We can assume w.l.o.g. that δ ≤
2, as otherwisethe unit disks are disjoint. We also record the obvious inequalities: δ ≤ OPT ≤ δ + 2 ≤ ⇔ µ ≤ x ≤ µ ≤ . (17)The algorithm Placement , which runs in O ( n ) time, achieves a ratio of c ( x ) = −√ √ x + √ x − x x , for 1 ≤ x ≤ , (18)and a ratio of at least for 0 ≤ x ≤ Centers achieves a ratio of c ( x ) = x − x , for x ≥ , (19)which is at least for x ≥ c ( x ) is decreasing in x and c ( x ) is increasing in x , the minimum approximation ratio ofthe hybrid algorithm occurs at the intersection of the two curves c ( x ) and c ( x ) for 1 ≤ x ≤
2: moreprecisely, c ( x ) = c ( x ) = √ − √ = 0 . . . . (1 / . . . . ) for x = 1 + √ − √ = 1 . . . . .Recall the approximation ratio c ( δ ) of our algorithm A1 as a function of the minimum pairwisedistance δ of the disks; see Equation (4). Now define a ( x ) = c (2 x − , for x ≥ . (20)Then we have the following lemma: Lemma 8.
For any x ≥ , the algorithm A1 achieves an approximation ratio at least a ( x ) fordispersion in unit disks. roof. Recall that the function c ( δ ) is increasing in δ . From (17) we deduce that δ ≥ OPT − x − A1 achieves an approximation ratio of at least c ( δ ) ≥ c (2 x −
2) = a ( x )for x ≥ x ≥ a ( x ) is an increasing functions of x . If we replace thealgorithm Centers by our algorithm A1 in the hybrid algorithm, then the two curves c ( x ) and a ( x ) intersect at x = 1 . . . . and, correspondingly, the approximation ratio of the new hybridalgorithm is at least 0 . . . . (1 / . . . . ).We next discuss the hybrid algorithm that runs Placement and A2 . Obviously the n disksof radius µ ≤ n input unit disks are pairwise-disjoint. The hybrid algorithmruns Placement on the given unit disks and A2 on the disks of radius µ and then returns thebetter solution. Clearly the solution is valid, and it remains to analyze the approximation ratio. Wewill show that for any given ε >
0, it can achieve a ratio at least (1 − ε ) √ √ − √ = (1 − ε ) · . . . . .By taking then ε = 10 − , we get a 0 . µ for the smaller disjoint disks. Lemma 9.
For a problem instance with µ ∈ [0 , , we have OPT µ ≥ OPT − − µ ) .Proof. Consider an optimal solution given by n points p , . . . , p n , where p i ∈ Ω i , such that | p i p j | ≥ OPT, for all i = j , and | p i p j | = OPT for at least one pair ( i, j ). Let q i ∈ Ω i be the point on o i p i atdistance µ | o i p i | from o i . Obviously, the set { q i : i = 1 , . . . , n } is a valid solution for dispersion inthe disks of radius µ concentric with the unit disks Ω , . . . , Ω n . Moreover, since | p i q i | ≤ − µ , for i = 1 , . . . , n , by the triangle inequality we have | q i q j | ≥ OPT − − µ ), for all i = j . Consequently,OPT µ ≥ OPT − − µ ), as claimed.Now define a ( x, µ ) = x − µx √ , for x ∈ [1 , , and µ ∈ [0 , . (21)Observe that a ( x, µ ) is an increasing function in both arguments. Then we have the followinganalogous lemma for our algorithm A2 . Lemma 10.
Consider a problem instance with µ ∈ [0 , and x ∈ [1 , µ ] as in (17) . For anygiven ε > , the algorithm A2 can achieve an approximation ratio at least (1 − ε ) · a ( x, µ ) fordispersion in unit disks.Proof. By Lemma 9, we have OPT µ ≥ OPT − − µ ). By Lemma 6, for any given ε >
0, thealgorithm A2 can achieve an approximation ratio at least − ε √ , and therefore find a point set in thedisks of radius µ with minimum inter-point distance at least1 − ε √ · OPT µ ≥ − ε √ · (OPT − − µ )) = 1 − ε √ · (2 x − − µ )) . Equivalently, the approximation ratio obtained is at least1 − ε √ · x − − µ )2 x = (1 − ε ) · x − µx √ − ε ) · a ( x, µ ) . µ ∈ [0 , x ( µ ) be the solution in [1 , ∞ ) of the equation c ( x ) = a ( x, µ ), or −√ √ x + √ x − x x = x − µx √ . (22)For solving (22), make the substitution y = x −
1, and get √ y + p − y = 2 √ y + µ )(3 − √ y + (4 − √ µy + (2 µ −
1) = 0 y , = − (4 − √ µ ± p − √ − µ − √ . Since the solution with a minus sign is negative, hence infeasible, we get y ( µ ) = − (4 − √ µ + p − √ − µ − √ , x ( µ ) = 1 + y ( µ ) . (23)Observe that y ( µ ) and x ( µ ) are both decreasing functions of µ for 0 ≤ µ ≤
1, with y (0) = √ −√ = 1 . . . . , and y (1 / √
2) = 0.Let µ be the solution in [0 ,
1] of the equation y ( µ ) = µ , or − (4 − √ µ + p − √ − µ − √
6) = µ. (24)After rearranging terms and squaring, we get µ = s √
657 = 1 p − √ . . . . . (25) Lemma 11. c (1 + µ ) = a (1 + µ , µ ) = √
21 + p − √ . . . . . (26) Proof.
By definitions we have c (1 + µ ) = c (1 + y ( µ )) = a (1 + y ( µ ) , µ ) = a (1 + µ , µ )= (1 + µ ) − µ √ µ + 1) = √ µ µ + 1 = √
21 + µ = √
21 + p − √ . . . . , as required.It suffices to prove the following lemma and then apply Lemma 10 to complete the proof of thetheorem. Lemma 12.
For every µ ∈ [0 , we have min x ∈ [1 , µ ] max { c ( x ) , a ( x, µ ) } ≥ √
21 + p − √ . . . . . (27)14 roof. Let h ( x, µ ) = max { c ( x ) , a ( x, µ ) } . We distinguish two cases. Case 1 : µ ≤ µ . By the definition of µ we have µ = y ( µ ) ≤ y ( µ ). Observe that 1 ≤ x ≤ µ ≤ µ . By (26), and the monotonicity of c ( · ), we have h ( x, µ ) ≥ c ( x ) ≥ c (1 + µ ) ≥ c (1 + µ ) = √
21 + p − √ . . . . . Case 2 : µ ≥ µ . If x ≤ µ then 1 ≤ x ≤ µ ≤ µ . By (26) and the monotonicity of c ( · ), h ( x, µ ) ≥ c ( x ) ≥ c (1 + µ ) = √
21 + p − √ . . . . . If x ≥ µ then 1 ≤ µ ≤ x ≤ µ . By (26) and the monotonicity of a ( · , µ ) and a ( x, · ), h ( x, µ ) ≥ a ( x, µ ) ≥ a (1 + µ , µ ) ≥ a (1 + µ , µ ) = √
21 + p − √ . . . . . Hence the inequality (27) is satisfied in both cases.This completes the proof of Theorem 3.
Recall our two algorithms A1 and A2 and the two algorithms Placement and
Centers byCabello [3]. In conclusion, we summarize the current best approximation ratios for the threevariants of dispersion in disks: • Arbitrary (not necessarily unit or disjoint): 3 / .
375 by
Placement . • Unit (not necessarily disjoint): 0 . A2 plus Placement , which improves 0 . Centers plus
Placement . • Disjoint (not necessarily unit): 0 .
707 by A2 , which improves 0 . Centers . Acknowledgment.
The authors would like to thank Andreas Bj¨orklund for pertinent commentson an earlier version of the manuscript.
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