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Abstract. In the TSP with neighborhoods problem we are given a set of n regions (neighborhoods) in the plane, and seek to find a minimum length TSP tour that ...
On approximating the TSP with intersecting neighborhoods Khaled Elbassioni1 , Aleksei V. Fishkin2 , and Ren´e Sitters1 1

Max-Planck-Institut f¨ ur Informatik, Saarbr¨ ucken, Germany; {elbassio,sitters}@mpi-sb.mpg.de 2 University of Liverpool, UK; [email protected]

Abstract. In the TSP with neighborhoods problem we are given a set of n regions (neighborhoods) in the plane, and seek to find a minimum length TSP tour that goes through all the regions. We give two approximation algorithms for the case when the regions are allowed to intersect: We give the first O(1)-factor approximation algorithm for intersecting convex fat objects of comparable diameters where we are allowed to hit each object only at a finite set of specified points. The proof follows from two packing lemmas that are of independent interest. For the problem in its most general form (but without the specified points restriction) we give a simple O(log n)-approximation algorithm.

1

Introduction

In the TSP with neighborhoods problem we are given a set of n subsets of the Euclidean plane and we have to find a tour of minimum length that visits at least one point from each subset. This generalizes both the classical Euclidean TSP and the group Steiner tree problems [6] with applications in VLSI-design, and other routing-related applications (see e.g. [10, 13]). Although the problem has been extensively studied in the last decade after Arkin and Hassin [1] introduced it in 1994, still large discrepancies remain between known inapproximability and approximation ratios for various cases. Safra and Schwartz [14] showed that the problem in the general case is NP-hard to approximate within any constant factor, and is APX-hard if each set forms a connected region in the plane. For connected polygonal regions, Mata and Mitchell [9] gave an O(log n)approximation in O(N 5 )-time based on ”guillotine rectangular subdivisions”, where N is the total number of vertices of the polygons. Gudmundsson and Levcpoulos [7] reduced the running time to O(N 2 log N ). In another interesting discrete variant of the problem, sometimes called the group-TSP problem, we are given n connected subsets of the plane, often refred to as regions or objects, and one set of points P . The TSP-tour must hit each region in one or more of the points of P . Typically, this restriction makes the problem harder. For the most general version in which the subsets are unrestricted, and the metric is not necessarily Euclidean, the gap is almost closed: Garg et al. [6] gave a randomized O(log N log log N log k log n)-approximation algorithm for the

group Steiner tree problem, the variant in which we are given a graph with N vertices and a set of n groups with at most k vertices per group, and seek a minimum cost Steiner tree connecting to at least one point from each group. Slav´ık [15] showed that the problem can be approximated within O(k). On the negative side, Halperin and Krauthgamer [8] gave an inapproximability threshold of Ω(log2−ǫ n) for any fixed ǫ > 0. With this being the situation for the general case, recent research has considered the cases where the given subsets are connected regions in the plane. We speak of the continues case if we can hit any point of the connected region, and we speak of the discrete case if we are only allowed to hit the region in one of the specified points. Previous results also distinguish between fat regions, such as disks, and non-fat regions, such as line-segments, and between instances with disjoint regions and intersecting regions. Non-fatness and intersections seem to make the problem much harder and, in fact, no constant factor approximation algorithm is known for the general case of intersecting non-fat regions. For the continuous case when the regions are translates of disjoint convex polygons, and for disjoint unit disks, Arkin and Hassin [1] presented constantfactor approximations. Dumitrescu and Mitchell [4] gave an O(1)-approximation algorithm for intersecting unit disks. For disjoint varying-sized convex fat regions, de Berg et al. [3] presented an O(α3 )-approximation algorithm, where α is a measure of fatness of the regions. A much simpler algorithm was given in [5] with an improved approximation factor of O(α), where also an O(1)approximation algorithm was given for the discrete case with intersecting unit disks. Very recently, Mitchell [12] gave a PTAS for the continuous case with disjoint fat neighborhoods, even under a weaker notion of fatness than the one used in this paper. Perhaps the two most natural extensions for which no constant factor algorithm is known, are that of non-fat disjoint objects, and that of fat intersecting objects. In Section 2 we consider the discrete version of the latter problem and give an O(α3 )-approximation algorithm for intersecting convex α-fat objects of comparable size. The proof follows from two lemmas given in Section 2.2. Lemma 1) is interesting on its own and gives a relation between the length of the optimal TSP tour inside a square and the distribution of the points. Additionally, we give in Section 3 a simple alternative O(log n)-algorithm for the general problem of connected regions, which does not require the regions to be simple polygons. There are several definitions of fatness in the literature and the following is commonly used for the problem we consider [3, 5, 16]. Definition 1. An object O ⊆ R2 is said to be α-fat if for any disk D which does not fully contain O and whose center lies in O, the area of the intersection of O and D is at least 1/α times the area of D. Notice for example that the plane R2 has fatness 1, a halfspace has fatness 2 and a disk has fatness 4. 2

2

Intersecting convex fat objects - the discrete case

We denote the given set of objects by O = {O1 , . . . , On }. In this section, we assume that each object can be hit only at specified points, i.e., we are given a set of points P and the required TSP tour must hit Oi at some point in Si ≡ P ∩ Oi . We consider the case when O1 , . . . , On are intersecting convex αfat objects of the same (or comparable) diameter δ. We assume P = S1 ∪. . .∪Sn . We first present the algorithm. In subsection 2.2, we derive a packing lemma that will be used in analyzing the performance of the algorithm. We briefly comment on the analysis of the approximation ratio in subsection 2.3. 2.1

The algorithm

A subset P ′ ⊆ P is called a hitting pointset for O if P ′ ∩ Si 6= ∅ for i = 1, . . . , n and a minimal hitting pointset if for every x ∈ P ′ there exists an i ∈ [n] such that (P ′ \ {x}) ∩ Si = ∅. A minimal hitting set can be found by the natural greedy algorithm: Set P ′ = P , and keep deleting points from P ′ as long as it is still a hitting set. An axis-aligned square B is called a covering box for the set of objects O if B contains a hitting pointset for O, and a minimum covering box if it is smallest size amongst all such covering boxes. Since a minimum covering box is determined by at most three points of P on its boundary, there are only O(|P |3 ) such candidates. By enumerating over all such boxes, and verifying if they contain a hitting set, one can compute a minimum covering box. Consider the following algorithm for the Group TSP problem on sets S1 , . . . , Sn (which is essentially the same as the one used in [5] for unit disks): Algorithm B: (1) Compute a minimum covering box B of O. (2) Find a minimal hitting pointset P ′ ⊆ P for O inside B. (3) Compute a (1 + ǫ)-approximate TSP tour on P ′ . The last step can be done efficiently for any ǫ > 0 using techniques from [2] and [11]. Theorem 1. Algorithm B is an O(α3 )-approximation algorithm for the Group TSP problem, with convex and α-fat neighborhoods of the same diameter. To analyze the performance of the algorithm, we need to show that, even though a collection of convex fat objects, with exactly one point in each, might be intersecting, they still exhibit a packing property that admits a ”short tour” visiting all the points. This is the content of the packing lemmas, stated and proved in the next section. 2.2

Two packing lemmas

We give two lemmas which are of independent interest. The first relates the length of a TSP tour through a set of points in the plane to the distribution of the points. Call a circular sector with head angle θ ≤ π, and radius γ a (γ, θ)-sector. 3

ξ4

∆′(p)

ξ3

ξ2

θ

ξ5

ξ2

ξ1

θ

p ∆′′ (p)

ξ6

ξ7

ξ8 jth slap

Fig. 1. (a) Partitioning the covering box. (b) A (βL, θ)-sector S(p) ∈ Si,j .

Lemma 1. Let P = {p1 , . . . , pn } ⊆ R2 be a a set of points with covering box of size L, and β > 0 and 0 < θ < π/2 be two constants. Then there exists an absolute constant c = c(β, θ) such that the following holds: If for every point p ∈ P there is a (βL, θ)-sector centered at p which contains no other point from P , then the optimum TSP tour on P has length |Opt| ≤ cL. Proof. Let P be a point set satisfying the conditions of the lemma, i.e. for every point p ∈ P there is a (βL, θ)-sector S(p), centered at p, which contains no other point from P . We begin by partitioning the set of sectors S = {S(p) : p ∈ P } into k = hg groups, √ depending on their orientations and locations, where h = ⌈2π/θ⌉ and g = ⌈ 2/(β cos θ)⌉. The precise partitioning is done as follows. Fix h directions, ξ1 , . . . , ξh , where ξi , for i ∈ [h], makes an angle of (i − 1)θ with the horizontal direction. For each direction ξi , we partition the covering box, into at most g parallel slabs ρi,j (j = 1 . . . g), along the direction ξi⊥ orthogonal to ξi . See Figure 1-(a) for an example. For i ∈ [h] and j ∈ [g], let Si,j ⊆ S be the set of sectors S(p) with the following two properties (see Figure 1-(b)): (i) the line through p with direction ξi intersects the circular arc part of S(p), and (ii) p lies in the jth slab with respect to the direction of ξi⊥ . Since h = ⌈2π/θ⌉ we can find for each p ∈ P a direction ξi such that (i) is satisfied. S Clearly (ii) is satisfied for some value j given the direction ξi for p. Hence, i∈[h], j∈[g] Si,j = S. Claim 1 Fix i ∈ [h] and j ∈ [g]. Then there √exists a path T√ on the set of points {p ∈ P : S(p) ∈ Si,j } of length at most (4 2/ sin(θ/2) + 2)L.

Proof. By performing the appropriate rotation, we may assume without loss of generality that ξi is the vertical direction, and thus the slab ρi,j determined by 4

the pair of the covering box is at √ √ (i, j) is horizontal. Note that, since the diameter most 2L, the width of such a slab is at most 2L/g ≤ Lβ cos θ. In particular, if we consider any point p ∈ P such that S(p) ∈ Si,j , then the circular arc of S(p) lies completely outside ρi,j (see Figure 1-(b)), and thus the boundary of the intersection of S(p) and ρi,j is a triangle ∆(p), with head angle θ. A line passing through p parallel to the direction ξi divides this triangle into two, one on the left ∆′ (p) and one on the right ∆′′ (p) of the line (see Figure 1-(b)). Clearly, the angle with head p in one of these triangles is at least θ/2. Now we partition ′ Si,j further into two groups of sectors: Si,j is the set of sectors S(p) whose left ′ triangle ∆ (p) makes an angle of at least θ/2 with the vertical direction, and ′′ ′ Si,j = S \ Si,j .

We claim that there is a path λ connecting all the points in P ′ = {p ∈ ′ P : S(p) ∈ Si,j }, with total length   √ √ 1 + cos(θ/2) (1) |λ| ≤ 2L ≤ 2 2L/ sin(θ/2). sin(θ/2)

To see this, we may assume without loss of generality that each triangle ∆′ (p), for p ∈ P ′ makes an angle of exactly θ/2 with the vertical direction. The path λ is obtained by traversing the boundary of these triangles from left to right as shown in Figure 2-(a). By projecting the sides of each such triangle on the big dotted triangle ∆0 containing all of them (see Figure 2-(a)), we observe that the sum of all these lengths is at most the sum of the two non-horizontal sides of ∆0 , ′′ which in turn implies 1. Applying the same for Si,j and connecting both paths √ ⊓ ⊔ by a segment of length at most 2L implies Claim 1. Now construct an Eulerian graph by taking two copies of the minimum covering box, together with the hg paths of the claim above, but extended to start and end at the covering box, which adds at most L for each slab. The total length is at most  √  √ 8L + 4 2/ sin(θ/2) + 2 + 1 Lhg, √ where h = ⌈2π/θ⌉ and g = ⌈ 2/(β cos θ)⌉.

⊓ ⊔

Notice that the upper bound in the previous lemma does not depend on the number of points. An infinite set of points could still satisfy the condition of the lemma. The next lemma is the analogue for the TSP with intersecting neighborhoods. Lemma 2. Let B be a box of size L containing a set of points P = {p1 , . . . , pn } ⊆ R2 . Assume that there is a collection of n convex α-fat objects O = {O1 , . . . , On }, each of diameter δ, such that (i) each point p ∈ P is contained in exactly one object O(p) ∈ O (ii) each object O contains exactly one point p(O) ∈ P . Then there exists a tour T on P with length O(L2 α2 /δ). Proof. Consider an object O with its unique point p = p(O) ∈ P . We will prove that there is (βL, θ)-sector with center p that lies completely inside O, with θ = 2π/(3α) and β = δ/(4L). 5

√ x≤L 2

θ 2

θ 2

O(p)

p

Tour λ u

z≤

√ L 2 sin(θ/2)

y≤

s

v

p′

R 2

R

√ L 2 tan(θ/2)

∆0 θ 2

D

Fig. 2. (a) Bounding the tour length in the proof of Lemma 1. (b) Illustration for the proof of Lemma 2.

Let p′ be a point in O at maximum distance, say R, from p (See Figure 2(b)). Obviously, R ≥ δ/2. Let s be the point in the middle of line segment pp′ and consider a disk D with center p′ and radius R/2. Let u and v be points in O on the circumference of D such that the angle hupvi is maximum. Finally, denote by S the (R, φ)-sector passing through u and v, with center p, radius R, and head angle upv, where φ = hupvi. Denote by A(U ) the area of a given region U of the plane. By definition of fatness we have A(O ∩ D) ≥ π(R/2)2 /α. Further, A(S ∩ D) ≤ 3/4 · A(S) = 3/4 · R2φ/2. It follows from the definition of u, v, and p′ , and the convexity of O that O ∩ D ⊆ S ∩ D. Therefore, π(R/2)2 /α ≤ A(O ∩ D) ≤ A(S ∩ D) ≤ 3R2 φ/8. Hence, π/α ≤ 3φ/2, implying φ ≥ 2π/(3α). The sector with center p, radius R/2 and head angle upv is contained in O. Now we apply Lemma 1 with θ = 2π/(3α) and β = δ/(4L). Note that if α → ∞, then cos(θ/2) → 1 and sin(θ/2) = O(1/α). We conclude that the length of the optimum tour is O(L2 α2 /δ). ⊓ ⊔ 2.3

Analysis of the approximation ratio

The analysis is similar to the one given in [5] modulo Lemma 2 stated above. We only give a sketch here and leave the details for the full version. Take a maximal independent set of objects O′ ⊆ O, i.e., a maximal collection of k = |O′ | pairwise disjoint objects. If we pick an arbitrary point in each of these objects and consider an optimal TSP tour on this set, then its length is at most 6

OPT+O(k), where δ is hidden in the constant. Since the independent set is maximal we can partition O in k clusters and by Lemma 2 we can connect all objects in one cluster by a tour of length O(α2 ). The lemma below is taken from [5] and will also be used in the proof of Theorem 3, part (ii). It sates that OPT= Ω(|O′ |/α). Hence, Theorem 1 follows. Lemma 3 ([5]). The length of the shortest path connecting k disjoint α-fat objects in R2 is at least (k/α − 1)πδ/4, where δ is the diameter of the smallest object.

3

General objects - the continuous case

De Berg et al. [3] prove that the TSP with connected neighborhoods problem is APX-hard. The constant was raised to 2 − ε by Safra and Schwartz [14]. Both their reduction use curved objects of varying sizes. We can show that the problem is even APX-hard for the very restricted case where all objects are line-segments of nearly the same length. The proof is omitted in this abstract. Theorem 2. The TSP with neighborhood problem is APX-hard, even if all objects are line segments of approximately the same length, i.e, the lengths differ by an arbitrarily small constant factor. If we do not restrict the shape of the objects then no better approximation algorithm than O(log n) is known. Gudmundsson and Levcpoulos [7] used a guillotine subdivision of the plane to obtain an algorithm which runs in time N 2 log N , where N is the total number of vertices of the polygonal objects. Here, we give a very simple approximation algorithm, with the same approximation guarantee, which does not require the objects to be simple polygons. As before, we denote the objects by O1 , . . . , On and their respective diameters by δ1 , . . . , δn . Lemma 4. Let O be a set of connected objects with diameter at least δ and a minimum covering box of size at most L. Then  √there exists a TSP tour T connecting all the objects in O of length |T | ≤ 4 Lδ 2 + 1 L. √ Proof. Consider covering box B of size at most L. A grid G of granularity δ/ 2 in B covers √ all the objects in O, and the total length of all the lines in G is at most 2(L 2/δ + 1)L. By doubling each line of G we can build a TSP tour with length as stated in the lemma. ⊓ ⊔

For two objects O, O′ ∈ O, define their distance d(O, O′ ) = min{kx − yk : x ∈ O, y ∈ O′ }, and for an object O ∈ O and r ∈ R+ , define the r-neighborhood of O to be N (O, r) = {O′ ∈ O : d(O, O′ ) ≤ r}. Finally, we fix a constant c and for i = 1, . . . , n, define the neighborhood of Oi as N (Oi ) = N (Oi , cδi ).

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Algorithm A: (1) For i = 1, 2, . . ., let Qi be the smallest object in O not belonging to N (Q1 ) ∪ N (Q2 ) ∪ . . . ∪ N (Qi−1 ). Let k be the largest value of i for which such Qi exists. (2) Let B be a minimum covering box for O, and Bi be a minimum covering box for {O ∩ B : O ∈ N (Qi )}, for i = 1, . . . , k. (3) Pick arbitrary points pi ∈ Qi ∩ Bi , for i = 1, . . . , k. Construct a (1 + ǫ)approximate TSP tour T0 on {p1 , . . . pk }. (4) For i = 1, . . . , k, let Ti be the TSP tour guaranteed by Lemma 4 on the set of objects N (Qi ) with the covering box Bi . (5) Combine the tours T0 , T1 , . . . , Tk into a single TSP tour T . (6) Output the minimum of T and the tour implied by Lemma 4 on the set O. Step (5) above can be done by adding two copies of a line segment connecting pi to the closest point of Ti , for i = 1, . . . , k. This yields an Eulerian tour that can be shortcut to a TSP tour T . Theorem 3. (i) Algorithm A gives an O(log n)-approximate solution for the TSP with connected neighborhoods. (ii) If all the neighborhoods have the same (or comparable) diameters, then A is an O(1)-approximation algorithm. Proof. Let Opt and Opt′ be respectively optimal TSP tours on O and O′ = {Q1 , . . . , Qk }. Note that δ1 ≤ δ2 ≤ . . . ≤ δk . Let L, L1 , . . . , Lk be respectively the sizes of the minimum covering box B, B1 , . . . , Bk . (i) We first establish the following claim. Pk−1 Claim 2 If k ≥ 1 then |Opt′ | ≥ c i=1 δi / log k.

Proof. Fix an orientation of Opt′ and define Wi as the arc of this directed tour (that connects exactly 2 objects in O′ ) starting from Qi . For any i = 1, . . . , k, let Qh(i) be an object with smallest diameter among the two objects on the arc Wi . Then for i = 1, . . . , k, |Wi | ≥ cδh(i) . (This follows from the fact that if Wi connects Qh(i) and Qj then Qj 6∈ N (Qh(i) ).) Consequently, |Opt′ | = Pk Pk ′′ ′ i=1 δh(i) . Let O = O \ {Qh(i) : i = 1, . . . , k}. Note that i=1 |Wi | ≥ c ′′ ′′ ′ |O | ≤ |O |/2. Moreover, if Opt is an optimum TSP tour connecting the objects in O′′ then |Opt′′ |P≤ |Opt′ |. Thus applying the above argument to O′′ , we obtain |Opt′′ | ≥ c i:Qi ∈O′′ δh′ (i) , where h′ (i) is the index of the object with smallest diameter among the two consecutive objects on the ith part of optimal path Opt′′ . Next we let O′′′ = O′′ \ {Qh′ (i) : i s.t. Qi ∈ O′′ } and note again that |O′′′ | ≤ |O′′ |/2. This process can continue for at most log k steps leading to at most log k lower bounds on |Opt|. Adding together these inequalities and noting that |Opt′ | ≥ |Opt′′ | ≥ |Opt′′′ | ≥. . . , we arrive at the bound stated in the claim. ⊓ ⊔ h Pk−1 i Claim 3 |T0 | ≤ (1 + ǫ) 4|Opt| + 2 i=1 δi . 8

Proof. Let Opt′′ be an optimum tour on {Q1 , . . . , Qk−1 } and qi be a point in Qi ∩ Opt′′ , for i = 1, . . . , k − 1. Then the union of Opt′′ with two copies of each of the segments pi qi , for i = 1, . . . , k − 1, and p1 pk , forms a connected Eulerian graph that can be shortcut to a TSP tour T ′ on {p1 , . . . , pk } of length |T ′ | ≤ |Opt′′ | + 2

k−1 X i=1

|pi qi | + 2|p1 pk | ≤ |Opt′′ | + 2

k−1 X i=1

δi + 2|p1 pk |

k−1 X √ δi , ≤ (1 + 2 2)|Opt| + 2 i=1

′′

since |Opt √ √ | ≤ |Opt|, pi , qi ∈ Qi , p1 , pk ∈ B, and thus |pi qi | ≤ δi and |p1 pk | ≤ 2L ≤ 2|Opt|. ⊓ ⊔ From the definition of Q1 , . . . , Qk , it follows that the minimum diameter of objects in the set N (Qi ) is δi . It also follows from the definition of N (Qi ) that Li ≤ (2c + 1)δi . Thus Lemma 4 gives, for i = 1, . . . , k,   √ Li √ 2 + 1 Li ≤ 4[(2c + 1) 2 + 1]Li = O(Li ). (2) |Ti | ≤ 4 δi The length of the tour T returned by the algorithm can be bounded as follows: |T | ≤ |T0 | +

k X i=1

|Ti | + 2

k X i=1

k−1 X

Li ≤ O(1)|Opt| + O(

δi ) = O(log k)|Opt|,

i=1

using Claims 2 and 3, Li ≤ (2c + 1)δi , and Lk ≤ L ≤ |Opt|. (ii) Let δ be the diameter of the smallest object, and assume that the diameters of all other objects are bounded by ρδ, for some constant ρ ≥ 1. We require that the constant c used in the definition of the neighborhood is at least √ 2(1 + ρ). In case the objects have comparable diameters, we can strengthen Claim 2 as follows.   √ Claim 4 |Opt′ | ≥ k4 − 1 πδ 2. 4 √ Proof. Let Di , for i = 1, . . . , k, be a disk of diameter δi 2, enclosing object Qi . We observe that, since c ≥ 1+ρ, all the disks are disjoint. (Otherwise, √ there exist 2(δi +δj ) ≤ two indices i < j, such that D ∩D = 6 ∅, implying that d(Q , Q ) ≤ i j i j √ 2(1 + ρ)δi ≤ cδi , and contradicting the fact that Qj 6∈ N (Qi ).) Thus we can apply Lemma 3 to the disks D1 , . . . , Dk , (using α = 4 for disks) to conclude that any TSP tour connecting these disks, and hence connecting the objects inside them, must have length bounded as stated in the claim. ⊓ ⊔ To bound the tour T returned by the algorithm, we observe that |T0 | ≤ P (1 + ǫ)(|Opt| + 2 ki=1 δi ), and combine this with 2 and Claim 4 to get |T | ≤ |T0 | +

k X i=1

|Ti | + 2 9

k X i=1

Li = O(ρ)|Opt|,

√ assuming that L ≥ δ (otherwise, a tour of length at most 4( 2 + 1)|Opt| is guaranteed by Lemma 4 and Step (6) of the algorithm). ⊓ ⊔

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