Minimum Manhattan network problem in normed planes with polygonal balls: a factor 2.5 approximation algorithm
aa r X i v : . [ c s . C G ] J un Minimum Manhattan network problem in normed planeswith polygonal balls: a factor 2.5 approximation algorithm
N. Catusse, V. Chepoi, K. Nouioua, and Y. Vax`es
Laboratoire d’Informatique Fondamentale de Marseille,Facult´e des Sciences de Luminy, Aix-Marseille Universit´e,F-13288 Marseille Cedex 9, France { catusse,chepoi,nouioua,vaxes } @lif.univ-mrs.fr Abstract
Let B be a centrally symmetric convex polygon of R and || p − q || be the distancebetween two points p , q ∈ R in the normed plane whose unit ball is B . For a set T of n points (terminals) in R , a B - Manhattan network on T is a network N ( T ) = ( V, E ) withthe property that its edges are parallel to the directions of B and for every pair of terminals t i and t j , the network N ( T ) contains a shortest B -path between them, i.e., a path oflength || t i − t j || . A minimum B -Manhattan network on T is a B -Manhattan network ofminimum possible length. The problem of finding minimum B -Manhattan networks hasbeen introduced by Gudmundsson, Levcopoulos, and Narasimhan (APPROX’99) in thecase when the unit ball B is a square (and hence the distance || p − q || is the l or the l ∞ -distance between p and q ) and it has been shown recently by Chin, Guo, and Sun[6] to be strongly NP-complete. Several approximation algorithms (with factors 8,4,3,and 2) for the minimum Manhattan problem are known. In this paper, we proposea factor 2.5 approximation algorithm for the minimum B -Manhattan network problem.The algorithm employs a simplified version of the strip-staircase decomposition proposedin our paper [5] and subsequently used in other factor 2 approximation algorithms for theminimum Manhattan problem. Keywords.
Normed plane, distance, geometric network design, Manhattan network, approxi-mation algorithms.
Given a compact, centrally symmetric, convex set B in the plane R , one can define a norm ||·|| := ||·|| B : R → R + by setting || v || = λ, where v = λ u and u is a unit vector belonging tothe boundary of B . We can then define a metric d := d B on R by setting d ( p , q ) = || p − q || . The resulting metric space ( R , d B ) is called a normed (or Minkowski ) plane with unit disk(gauge) B [2, 24]. In this paper, we consider normed planes in which the unit ball B is acentrally symmetric convex polygon (i.e., a zonotope ) of R . We denote by b , . . . b m − the2 m vertices of B (in counterclockwise order around the circle) as well as the 2 m unit vectors1hat define these vertices. By central symmetry of B , b k = − b k + m for k = 0 , . . . , m − . A legal k -segment of ( R , d B ) is a segment pq lying on a line parallel to the line passing via b k and b k + m . A legal path π ( p , q ) between two points p , q of R is any path connecting p and q in which all edges are legal segments. The length of π ( p , q ) is the sum of lengths of its edges.A shortest B -path between p and q is a legal ( p , q )-path of minimum length. The best knownexample of normed planes with polygonal unit balls is the l -plane (also called the rectilinearplane) with norm || v || = | x ( v ) | + | y ( v ) | . The unit ball of the l -plane is a square whose twodiagonals lie on the x -axis and y -axis, respectively. The l -distance between two points p and q is d ( p , q ) := || p − q || = | x ( p ) − x ( q ) | + | y ( p ) − y ( q ) | . The legal paths of the rectilinear planeare the paths consisting of horizontal and vertical segments, i.e., rectilinear paths. Anotherimportant particular case of polygonal norms is that of λ -norms (alias uniform norms) [4, 3]for which the unit ball B is a regular polygon. B -Manhattan network problems A rectilinear network N = ( V, E ) in R consists of a finite set V of points and horizontal andvertical segments connecting pairs of points of V. The length of N is the sum of lengths of itsedges. Given a finite set T of points in the plane, a Manhattan network [15] on T is a rectilinearnetwork N ( T ) = ( V, E ) such that T ⊆ V and for every pair of points in T, the network N ( T )contains a shortest rectilinear path between them. A minimum Manhattan network on T isa Manhattan network of minimum possible length and the Minimum Manhattan Networkproblem ( MMN problem ) is to find such a network.More generally, given a zonotope
B ⊂ R , a B -network N = ( V, E ) consists of a finite set V of points and legal segments connecting pairs of points of V (the edges of N ). The length l ( N ) of N is the sum of lengths of its edges. Given a set T = { t , . . . , t n } of n points (called terminals ), a B -Manhattan network on T is a B -network N ( T ) = ( V, E ) such that T ⊆ V andfor every pair of terminals in T, the network N ( T ) contains a shortest B -path between them(see Fig. 1). A minimum B -Manhattan network on T is a B -Manhattan network of minimumpossible length and the Minimum B -Manhattan Network problem ( B -MMN problem ) is to findsuch a network. Fig. 3 illustrates the evolution of a minimum B -Manhattan network definedon the same set of terminals when the number of directions in the unit ball B is increasing(the directions of the unit ball are indicated at the upper left corner of each figure). The minimum Manhattan network problem has been introduced by Gudmundsson, Levcopou-los, and Narasimhan [15]. Gudmundsson et al. [15] proposed an O ( n )-time 4-approximationalgorithm, and an O ( n log n )-time 8-approximation algorithm. They also conjectured thatthere exists a 2-approximation algorithm for this problem and asked if this problem is NP-complete. Quite recently, Chin, Guo, and Sun [6] solved this last open question from [15] andestablished that indeed the minimum Manhattan network problem is strongly NP-complete.Kato, Imai, and Asano [17] presented a 2-approximation algorithm, however, their correct-ness proof is incomplete (see [1]). Following [17], Benkert, Wolff, Shirabe, and Widmann2igure 1: A B -Manhattan network in the normedplane whose unit ball is depicted in Fig. 4 Figure 2: The unique optimal solu-tion for this instance does not belongto the grid Γ (the unit ball B is ahexagon)[1] described an O ( n log n )-time 3-approximation algorithm and presented a mixed-integerprogramming formulation of the MMN problem. Nouioua [20] and later Fuchs and Schulze[12] presented two simple O ( n log n )-time 3-approximation algorithms. The first correct 2-approximation algorithm (thus solving the first open question from [15]) was presented byChepoi, Nouioua, and Vax`es [5]. The algorithm is based on a strip-staircase decompositionof the problem and uses a rounding method applied to the optimal solution of the flow basedlinear program described in [1]. In his PhD thesis, Nouioua [20] described a O ( n log n )-time2-approximation algorithm based on the primal-dual method from linear programming andthe strip-staircase decomposition. In 2008, Guo, Sun, and Zhu [13, 14] presented two com-binatorial factor 2 approximation algorithms, one with complexity O ( n ) and another withcomplexity O ( n log n ) (see also the PhD thesis [22] of Schulze for yet another O ( n log n )-time2-approximation algorithm). Finally, Seibert and Unger [21] announced a 1.5-approximationalgorithm, however the conference format of their paper does not permit to understand thedescription of the algorithm and to check its claimed performance guarantee (a counterexam-ple that an important intermediate step of their algorithm is incorrect was given in [12, 22]).Quite surprisingly, despite a considerable number of prior work on minimum Manhattan net-work problem, no previous paper, to our knowledge, consider its generalization to normedplanes.Gudmundsson et al. [15] introduced the minimum Manhattan networks in connectionwith the construction of sparse geometric spanners. Given a set T of n points in a normedplane and a real number t ≥
1, a geometric network N is a t -spanner for T if for each pair ofpoints p , q ∈ T, there exists a ( p , q )-path in N of length at most t times the distance k p − q k between p and q . In the Euclidian plane and more generally, for normed planes with roundballs, the line segment is the unique shortest path between two endpoints, and therefore theunique 1-spanner of T is the complete graph on T. On the other hand, if the unit ball ofthe norm is a polygon, the points are connected by several shortest B -paths, therefore theproblem of finding the sparsest 1-spanner becomes non trivial. In this connection, minimum B -Manhattan networks are precisely the optimal 1-spanners. Sparse geometric spanners have3igure 3: “Morphing” a minimum B -Manhattan networkapplications in VLSI circuit design, network design, distributed algorithms and other areas,see for example the survey of [10] and the book [18]. Lam, Alexandersson, and Pachter[19] suggested to use minimum Manhattan networks to design efficient search spaces for pairhidden Markov model (PHMM) alignment algorithms.Algorithms for solving different distance problems in normed spaces with polygonal andpolyhedral balls were proposed by Widmayer, Wu, and Wang [25] (for more references anda systematic study of such problems, see the book by Fink and Wood [11]). There is also anextensive bibliography on facility location problems in normed spaces with polyhedral balls,see for example [9, 23]. Finally, the minimum Steiner tree problem in the normed planes wasa subject of intensive investigations, both from structural and algorithmic points of view;[4, 3, 8] is just a short sample of papers on the subject. We continue by setting some basic definitions, notations, and known results. Let B be azonotope of R with 2 m vertices b , . . . b m − having its center of symmetry at the originof coordinates (see Fig. 4 for an example). The segment s k := b k b k +1(mod 2 m ) is a side of B . We will call the line ℓ i passing via the points b k and b k + m an extremal line of B .4 b s s b b b b b s s s b s s s s ′ k − C ( q ) C ( p ) s ′′ j pq I ( p,q ) Figure 4: A unit ball B and an interval I ( p , q )Two consecutive extremal lines ℓ k and ℓ k +1 defines two opposite elementary k -cones C k and C k + m = − C k containing the sides s k and s k + m , respectively. We extend this terminology,and call elementary k -cones with apex v the cones C k ( v ) = C k + v and − C k ( v ) = C k + m + v obtained by translating the cones C k and C k + m by the vector v . We will call a pair ofconsecutive lines D k = { ℓ k , ℓ k +1 } a direction of the normed plane. Denote by B ( v , r ) = r ·B + v the ball of radius r centered at the point v . Let I ( p , q ) = { z ∈ R : d ( p , q ) = d ( p , z ) + d ( z , q ) } be the interval between p and q . Theinclusion pq ⊆ I ( p , q ) holds for all normed spaces. If B is round, then pq = I ( p , q ) , i.e.,the shortest path between p and q is unique. Otherwise, I ( p , q ) may host a continuous setof shortest paths. The intervals I ( p , q ) in a normed plane (and, more generally, in a normedspace) can be constructed in the following pretty way, described, for example, in the book[2]. If pq is a legal segment, then pq is the unique shortest path between p and q , whence I ( p , q ) = pq . Otherwise, set r = d ( p , q ) . Let s ′ k be the side of the ball B ( p , r ) containingthe point q and let s ′′ j be the side of the ball B ( q , r ) containing the point p . Notice thatthese sides are well-defined, otherwise q is a vertex of B ( p , r ) and pq is a legal segment. Thesegments s ′ k and s ′′ j are parallel, thus | k − j | = m, say k ≤ m and j = k + m. Then I ( p , q )is the intersection of the elementary cones C k ( p ) and C k + m ( q ) = − C k ( q ) (see Fig. 4 for anillustration): Lemma 2.1 [2] I ( p , q ) = C k ( p ) ∩ ( − C k ( q )) . An immediate consequence of this result is the following characterization of shortest B -paths between two points p and q . Lemma 2.2 If pq is a legal segment, then pq is the unique shortest B -path. Otherwise,if I ( p , q ) = C k ( p ) ∩ ( − C k ( q )) , then any shortest B -path π ( p , q ) between p and q has only k -segments and ( k + 1) -segments as edges. Moreover, π ( p , q ) is a shortest B -path if and only f it is monotone with respect to ℓ k and ℓ k +1 , i.e., the intersection of π ( p , q ) with any line ℓ parallel to the lines ℓ k , ℓ k +1 is empty, a point, or a (legal) segment. Proof.
The first statement immediately follows from Lemma 2.1. Suppose that pq is not alegal segment and I ( p , q ) = C k ( p ) ∩ C k + m ( q ) . Let uv be the first edge on a shortest path π ( p , q ) from p to q which is neither a k -segment nor a ( k + 1)-segment. Since u ∈ I ( p , q ) = C k ( p ) ∩ C k + m ( q ) , the point q belongs to the cone C k ( u ) and the point u belongs to thecone C k + m ( q ) , whence I ( u , q ) = C k ( u ) ∩ C k + m ( q ) . Obviously, the point v belongs to I ( u , q ) . However, by the choice of the segment uv and the fact that ℓ k and ℓ k +1 are consecutivelines that forms a direction, the point v cannot belong C k ( u ) , a contradiction. This showsthat any shortest legal path π ( p , q ) between p and q has only k - and ( k + 1)-segments asedges. Additionally, the intersection of π ( p , q ) with any line ℓ parallel to ℓ k or ℓ k +1 is empty,a point, or a (legal) segment. Indeed, pick any two points in this intersection. Since thelegal segment defined by these points is the unique shortest path between them, it must alsobelong to the intersection of π ( p , q ) with ℓ . Conversely, consider a monotone path π ( p , q )between p and q , namely suppose that the intersection of π ( p , q ) with any line ℓ parallel tothe lines ℓ k or ℓ k +1 is empty, a point, or a (legal) segment). We proceed by induction onthe number of edges of π ( p , q ) . The monotonicity of π ( p , q ) implies that π ( p , q ) lies entirelyin the interval I ( p , q ) . In particular, the neighbor u of p in π ( p , q ) belongs to I ( p , q ) . Thesubpath π ( u , q ) of π ( p , q ) between u and q is monotone, therefore by induction assumption, π ( u , q ) is a shortest path between u and q . Since pu is a legal segment and u ∈ I ( p , q ) , weimmediately conclude that π ( p , q ) is also a shortest path between p and q . (cid:3) We continue with some notions and notations about the B -MMN problem. Denote byOPT( T ) the length of a minimum B -Manhattan network for a set of terminals T . For adirection D k = { ℓ k , ℓ k +1 } , denote by F k the set of all pairs { i, j } (or pairs of terminals { t i , t j } ) such that any shortest B -path between t i and t j uses only k -segments and ( k + 1)-segments. Equivalently, by Lemma 2.2, F k consists of all pairs of terminals which belong totwo opposite elementary cones C k ( v ) and − C k ( v ) with common apex. For each direction D k and the set of pairs F k , we formulate an auxiliary problem which we call Minimum 1-Directional Manhattan Network problem (or ( F k ) problem ): find a network N optk ( T )of minimum possible length such that every edge of N optk ( T ) is an k -segment or an ( k + 1)-segment and any pair { t i , t j } of F k is connected in N optk ( T ) by a shortest B -path. We denoteits length by OPT k ( T ) . We continue by adapting to 1-DMMN the notion of a generatingset introduced in [17] for MMN problem: a generating set for F k is a subset F of F k withthe property that a B -Manhattan network containing shortest B -paths for all pairs in F is a1-Directional Manhattan network for F k . Let N ∗ ( T ) be a minimum B -Manhattan network, i.e., a B -Manhattan network of total length l ( N ∗ ( T )) = OPT( T ) . For each direction D k , let N ∗ k ( T ) be the set of k -segments and ( k + 1)-segments of N ∗ ( T ) . The network N ∗ k ( T ) is an admissible solution for 1-DMMN( F k ) , thus6he length l ( N ∗ k ( T )) of N ∗ k ( T ) is at least OPT k ( T ) . Any k -segment of N ∗ ( T ) belongs totwo one-directional networks N ∗ k ( T ) and N ∗ k − ( T ) . Vice-versa, if N k ( T ) , k = 0 , . . . , m − , are admissible solutions for the 1-DMMN( F k ) problems, since S m − k =0 F k = T × T, thenetwork N ( T ) = S m − k =0 N k ( T ) is a B -Manhattan network. Moreover, if each N k ( T ) isan α -approximation for respective 1-DMMN problem, then the network N ( T ) is a 2 α -approximation for the minimum B -Manhattan network problem. Therefore, to obtain afactor 2.5-approximation for B -MMN, we need to provide a 1.25-approximation for the 1-DMMN problem. The remaining part of our paper describe such a combinatorial algorithm.The 1-DMMN problem is easier and less restricted than the B -MMN problem because wehave to connect with shortest paths only the pairs of terminals of the set F k correspondingto one direction D k , while in case of the MMN problem the set T × T of all pairs is parti-tioned into two sets corresponding to the two directions of the l -plane. For our purposes,we will adapt the strip-staircase decomposition of [5], by considering only the strips and thestaircases which “are oriented in direction D k ”. In the next two sections, we assume that D k = { l k , l k +1 } is a fixed but arbitrary direction ofthe normed plane. We recall the definitions of vertical and horizontal strips and staircasesintroduced in [5]. Then we consider only those of them which correspond to pairs of terminalsfrom the set F k , which we call one-directional strips and staircases. We formulate severalproperties of one-directional strips and staircases and we prove those of them which do nothold for usual strips and staircases.Denote by L k and L k +1 the set of all lines passing via the terminals of T and parallel tothe extremal lines ℓ k and ℓ k +1 , respectively. Let Γ k be the grid defined by the lines of L k and L k +1 . The following lemma can be proved in the same way as for rectilinear Steiner trees orManhattan networks (quite surprisingly, this is not longer true for the B -MMN problem: Fig.2 presents an instance of B -MMN for which the unique optimal solution does not belong tothe grid Γ := S m − k =0 Γ k ): Lemma 3.1
There exists a minimum 1-Directional Manhattan Network for F k contained inthe grid Γ k . For two terminals t i , t i ′ , set R i,i ′ := I ( t i , t i ′ ) . A pair t i , t i ′ defines a k -strip R i,i ′ if either(i) (degenerated strip) t i and t i ′ are consecutive terminals belonging to the same line of L k or (ii) t i and t i ′ belong to two consecutive lines of L k and the intersection of R i,i ′ with anydegenerated k -strip is either empty or one of the terminals t i or t i ′ ; see Fig. 6 of [5]The two k -segments of R i,i ′ are called the sides of R i,i ′ . The ( k + 1) -strips and their sides are definedanalogously (with respect to L k +1 ). With some abuse of language, we will call the k -strips horizontal and the ( k + 1)-strips vertical . If a pair { t i , t i ′ } defining a horizontal or a verticalstrip R i,i ′ belongs to the set F k , then we say that R i,i ′ is a one-directional strip or a ,for short. Denote by F ′ k the set of all pairs of F k defining one-directional strips.7 j ′ T i l j T i j t i t j t ′ i t i l t ′ i l t l t i t j π S ij | i ′ j ′ S i ′ j ′ | ij Q ′ Q ′′ t i ′ s ′ j,j ′ t j ′ o ′ o Figure 5: Strips, staircases, and completion
Lemma 3.2 If R i,i ′ and R j,j ′ are two horizontal 1-strips or two vertical 1-strips, then R i,i ′ ∩ R j,j ′ = ∅ if { i, i ′ } ∩ { j, j ′ } = ∅ and R i,i ′ ∩ R j,j ′ = { t i } if { i, i ′ } ∩ { j, j ′ } = { i } . Proof.
From the definition follows that if R i,i ′ and R j,j ′ are both degenerated or one isdegenerated and another one not, then they are either disjoint or intersect in a single terminal.If R i,i ′ and R j,j ′ are both non-degenerated and intersect, then from the definition immediatelyfollows that the intersection is one point or a segment belonging to their sides. However, if R i,i ′ and R j,j ′ intersects in a segment, then one can easily see that at least one of R i,i ′ and R j,j ′ cannot be a 1-strip. (cid:3) We say that a vertical 1-strip R i,i ′ and a horizontal 1-strip R j,j ′ (degenerated or not) forma crossing configuration if they intersect (and therefore cross each other). Lemma 3.3 If R i,i ′ and R j,j ′ form a crossing configuration, then from the shortest B -pathsbetween t i and t i ′ and between t j and t j ′ one can derive shortest B -paths connecting t i , t j ′ and t i ′ , t j , respectively. For a crossing configuration defined by the 1-strips R i,i ′ , R j,j ′ , denote by o and o ′ the twoopposite corners of the parallelogram R i,i ′ ∩ R j,j ′ , such that the cones C k ( o ) and − C k ( o ′ ) donot intersect the interiors of R i,i ′ and R j,j ′ . Additionally, suppose without loss of generality,8hat t i and t j belong to the cone C k ( o ) , while t i ′ and t j ′ belong to the cone − C k ( o ′ ) . Denoteby T i,j the set of all terminals t l ∈ ( T \{ t i , t j } ) ∩ C k ( o ) such that ( − C k ( t l )) \ ( − C k ( o )) does notcontain any terminal except t l . Denote by S i,j | i ′ ,j ′ the region of C k ( o ) which is the union of theintervals I ( t l , o ) , t l ∈ T i,j , and call this polygon an one-directional staircase or a ,for short; see Fig. 5 and Figures 7,8 of [5] for an illustration. Note that S i,j | i ′ ,j ′ is boundedby the 1-strips R i,i ′ and R j,j ′ and a legal path between t i and t j passing via all terminalsof T i,j and consisting of k -segments and ( k + 1)-segments. The point o is called the origin and R i,i ′ and R j,j ′ are called the basis of this staircase. Since I ( t l , o ) ⊂ ( − C k ( t l )) \ ( − C k ( o ))for all t l ∈ T i,j , I ( t l , o ) ∩ T = { t l } and therefore S i,j | i ′ ,j ′ ∩ T = T i,j . For the same reason,there are no terminals of T located in the regions Q ′ and Q ′′ depicted in Fig. 5 ( Q ′ is theregion comprised between the leftmost side of R i,i ′ , the highest side of R j,j ′ , and the line of L k passing via the highest terminal of T i,j , while Q ′′ is the region comprised between therightmost side of R i,i ′ , the lowest side of R j,j ′ , and the line of L k +1 passing via the rightmostterminal of T i,j ). Analogously one can define the set T i ′ ,j ′ and the staircase S i ′ ,j ′ | i,j withorigin o ′ and basis R i,i ′ and R j,j ′ . Lemma 3.4
If a 1-strip R l,l ′ intersects a 1-staircase S i ′ ,j ′ | i,j and R l,l ′ is different from the1-strips R i,i ′ and R j,j ′ , then R l,l ′ ∩ S i ′ ,j ′ | i,j is a single terminal. Proof.
If a 1-strip R l,l ′ traverses a staircase S i ′ ,j ′ | i,j , then one of the terminals t l , t l ′ mustbe located in one of the regions Q ′ and Q ′′ , which is impossible because ( Q ′ ∪ Q ′′ ) ∩ T = ∅ . Thus, if R l,l ′ and S i ′ ,j ′ | i,j intersect more than in one point, then they intersect in a segment s which belongs to one side of R l,l ′ and to the boundary of S i ′ ,j ′ | i,j . If say the 1-strip R l,l ′ ishorizontal, then necessarily s is a part of the lowest side of R l,l ′ and of the highest horizontalside of S i ′ ,j ′ | i,j . Let t be the highest terminal of T i,j . Then either t belongs to R l,l ′ and isdifferent from t l , t l ′ , contrary to the assumption that R l,l ′ is a strip, or t together with thelowest terminal t l ′ of R l,l ′ define a degenerated strip with t l ′ belonging to Q ′ , contrary to theassumption that Q ′ ∩ T = ∅ . (cid:3) Lemma 3.5
Two 1-staircases either are disjoint or intersect only in common terminals.
Proof.
From the definition of a staircase follows that the interiors of two staircases aredisjoint (for a short formal proof of this see [5]). Therefore two staircases may intersect onlyon the boundary. In this case, the intersection is either a subset of terminals of both staircasesor a single edge. In the second case, one of the two staircases necessarily is not a 1-staircasewith respect to the chosen direction. (cid:3)
Let F ′′ k be the set of all pairs { t j ′ , t l } such that there exists a 1-staircase S i,j | i ′ ,j ′ with t l belonging to the set T i,j . The proof of the following essential result is identical to the proofof Lemma 3.2 of [5] and therefore is omitted.
Lemma 3.6 F := F ′ k ∪ F ′′ k is a generating set for F k . The algorithm
We continue with the description of our factor 1.25 approximation algorithm for 1-DMMNproblem. Let F hk and F vk denote the pairs of F ′ k defining horizontal and vertical 1-strips,respectively. Let S h and S h be the networks consisting of lower sides and respectively uppersides of the horizontal 1-strips of F hk . Analogously, let S v and S v be the networks consistingof rightmost sides and respectively leftmost sides of the vertical 1-strips of F vk . The algorithmcompletes optimally each of the networks S h , S h , S v , and S v , and from the set of four com-pletions N h , N h , N v , N v , the algorithm returns the shortest one, which we will denote by N k ( T ) (in this respect, our algorithm has some similarity with the approach of Benkert et al.[1]). We will describe now the optimal completion N h for the network S h , the three othernetworks are completed in the same way (up to symmetry).An optimal completion of S h is a subnetwork N h of Γ k extending S h ( S h ⊆ N h ) of smallesttotal length such that any pair of terminals of F can be connected in N h by a shortest path.By Lemma 3.6, to solve the completion problem for S h , it suffices to (i) select a shortest path π ( t i , t i ′ ) of Γ k between each pair t i , t i ′ defining a vertical 1-strip R i,i ′ , (ii) for each horizontal1-strip R j,j ′ find a shortest path π ( t j , t j ′ ) between t j and t j ′ subject to the condition thatthe lowest side s ′ j,j ′ of R j,j ′ is already available, (iii) for each staircase S i,j | i ′ ,j ′ whose sidesare R i,i ′ and R j,j ′ select shortest paths from the terminals of T i,j to the terminal t j ′ subjectto the condition that the lowest side s ′ j,j ′ of R j,j ′ is already available. We need to minimizethe total length of the resulting network N h over all vertical 1-strips and all 1-staircases. Tosolve the issue (ii) for a horizontal 1-strip R j,j ′ , we consider the rightmost 1-staircase S i,j | i ′ ,j ′ having R j,j ′ as a basis, set T i,j := T i,j ∪ { t j } , and solve for this staircase the issue (iii) forthe extended set of terminals. For all other 1-staircases S i,j | i ′ ,j ′ and S i ′ ,j ′ | i,j having R j,j ′ as abasis, we will solve only the issue (iii) for T i,j and T i ′ ,j ′ , respectively.To deal with (iii), for each vertical 1-strip R i,i ′ , we pick each shortest path π of Γ k between t i and t i ′ , include it in the current completion, and solve (iii) for all 1-staircases having R i,i ′ as a vertical base and taking into account that π is already present. We have to connect theterminals of T i,j by shortest paths of Γ k of least total length to the terminal t j ′ subject tothe condition that the union π ∪ s ′ j,j ′ is already available; see Fig. 5. For a fixed path π, this task can be done by dynamic programming in O ( | T i,j | ) time. For this, notice that inan optimal solution (a) either the highest terminal of T i,j is connected by a vertical segmentto s ′ j,j ′ , or (b) the lowest terminal of T i,j is connected by a horizontal segment to π , or (c) T i,j contains two consecutive (in the staircase) terminals t l , t l +1 , such that t l is connectedto π by a horizontal segment and t l +1 is connected to s ′ j,j ′ by a vertical segment. In eachof the three cases and subsequent recursive calls, we are lead to solve subproblems of thefollowing type: given a set T ′ of consecutive terminals of T i,j , the path π and a horizontalsegment s ′ , connect to t j ′ the terminals of T ′ by shortest paths of least total length if theunion π ∪ s ′ is available. We define by C πi,i ′ the optimal completion obtained by solving bydynamic programming those problems for all staircases having R i,i ′ as a vertical basis (notethat π ⊆ C πi,i ′ however S h ∩ C πi,i ′ = ∅ ). For each vertical 1-strip R i,i ′ , the completion algorithmreturns the partial completion C opti,i ′ of least total length, i.e, C opti,i ′ is the smallest completion of10he form C πi,i ′ taken over all O ( n ) shortest paths π running between t i and t i ′ in Γ k . Finally,let N h be the union of all C opti,i ′ over all vertical 1-strips R i,i ′ and S h . The pseudocode of thecompletion algorithm is presented below (the total complexity of this algorithm is O ( n )). Algorithm 1
Optimal completion ( S h ) N h ← S h for each vertical 1-strip R i,i ′ do for each shortest path π of Γ k connecting the terminals t i and t i ′ do compute the partial completion C πi,i ′ in the following way: C πi,i ′ ← π \ S h for each 1-staircase S i,j | i ′ ,j ′ and each 1-staircase S i ′ ,j ′ | i,j do if S i,j | i ′ ,j ′ is the rightmost staircase having R j,j ′ as a base, then set T i,j ← T i,j ∪{ t j } compute by dynamic programming the subset C of edges of Γ k of least total lengthsuch that C ∪ ( π ∪ s ′ j,j ′ ) contains a shortest path of Γ k from each terminal of T i,j to t j ′ or from each terminal of T i ′ ,j ′ to t j C πi,i ′ ← C πi,i ′ ∪ C end for end for let C opti,i ′ be the partial completion of least total length, i.e, C opti,i ′ is the smallest comple-tion C πi,i ′ over all shortest paths π between t i and t i ′ N h ← N h ∪ C opti,i ′ end for return N h Lemma 4.1
The network N h returned by the algorithm Optimal completion is an optimalcompletion for S h . Proof.
We described above how to compute for each 1-staircase S i,j | i ′ ,j ′ a subset C of edgesof Γ k of minimum total length such that C ∪ ( π ∪ s ′ j,j ′ ) contains a shortest path of Γ k from eachterminal of T i,j to t j ′ . This standard dynamical programming approach explores all possiblesolutions and therefore achieves optimality for this problem. Next, we assert that, for eachvertical 1-strip R i,i ′ , the subset of edges C opti,i ′ computed by our algorithm, is an optimalcompletion of S h for the strip R i,i ′ and the staircases having R i,i ′ as vertical bases. Indeed,our algorithm considers every possible shortest path π of Γ k between t i and t i ′ . Once thepath π is fixed, the subproblems related to distinct staircases become independent and canbe solved optimally by dynamic programming. The problems arising from distinct vertical1-strips are also disjoint and independent (according to Lemmas 3.4 and 3.5). Therefore thesolution N h obtained by combining the optimal solutions C opti,i ′ of every vertical 1-strip R i,i ′ is an optimal completion of S h . It remains to show that to obtain a completion satisfying the conditions (i),(ii), and (iii),it suffices for each horizontal 1-strip R j,j ′ to add t j to the set T i,j of terminals of the rightmost11taircase S i,j | i ′ ,j ′ having R j,j ′ as a basis and to solve (iii) for this extended set of terminals.Indeed, in any completion any shortest path between t j and t j ′ necessarily makes a verticalswitch either before arriving at the origin o of S i,j | i ′ ,j ′ or this path traverses the vertical basisof this staircase. Since the completion contains a shortest path connecting the terminals ofthe vertical basis of S i,j | i ′ ,j ′ , combining these two paths, we can derive a shortest path between t j and t ′ j which turns in R i,i ′ ∩ R j,j ′ . As a result, we conclude that at least one shortest pathbetween t j and t j ′ passes via o ′ . This shows that indeed it suffices to take into account thecondition (ii) only for each rightmost staircase. (cid:3)
Lemma 4.2
The network N k ( T ) is an admissible solution for the problem 1-DMMN ( F k ) . Proof.
By Lemma 4.1, N h is a completion of S h and thus contains a shortest path betweenevery pairs of vertices from F. By symmetry, we get the same result for N h , N v and N v . Since N k ( T ) is one of these networks, by Lemma 3.6, it is admissible solution for the problem1-DMMN( F k ) . (cid:3) In this section, we will prove the following main result:
Theorem 5.1
The network N k ( T ) is a factor 1.25 approximation for 1-DMMN ( F k ) problemfor k = 0 , . . . , m − . The network N ( T ) := S m − k =0 N k ( T ) is a factor 2.5 approximation forthe B -MMN problem and can be constructed in O ( mn ) time.Proof of Theorem 5.1. First we prove the first assertion of the theorem. Let Λ h = l ( S h ) = l ( S h ) and Λ v = l ( S v ) = l ( S v ) . Further, we suppose that Λ h ≤ Λ v . Assume N optk be anoptimal 1-restricted Manhattan network for F k . Let M be a subnetwork of N optk ∩ ( S h ∪ S h )of minimum total length which completed with some vertical edges of N optk contains a shortestpath between each pair of terminals defining a horizontal 1-strip of F k . Such M exists becausethe network N optk ∩ ( S h ∪ S h ) already satisfies this requirement. Further, we assume that l ( M ∩ S h ) ≥ l ( M ∩ S h ) . Lemma 5.2 l ( M ) = Λ h . Proof.
By Lemma 3.2, two horizontal 1-strips either are disjoint or intersect only in commonterminals, thus any horizontal 1-strip R i,i ′ contributes to M separately from other horizontal1-strips. Since the terminals t i and t ′ i defining R i,i ′ are connected in N optk by a shortest pathconsisting of two horizontal segments of total length equal to the length of a side of R i,i ′ anda vertical switch between these segments, from the optimality choice of M we conclude thatthe contribution of R i,i ′ to M is precisely the length of one of its sides. (cid:3) Lemma 5.3 l ( M ∩ S h ) ≥ l ( M ) . roof. The proof follows from the assumption l ( M ∩ S h ) ≥ l ( M ∩ S h ) and the fact that M ∩ S h and M ∩ S h form a partition of M. (cid:3) Lemma 5.4 l ( S h \ M ) ≤ l ( N optk ) . Proof.
Since l ( S h \ M ) = l ( S h ) − l ( M ∩ S h ) , by Lemma 5.2 and 5.3 we get l ( S h \ M ) ≤ l ( M ) = Λ h ≤ l ( N optk ) . The last inequality follows from l ( N optk ) ≥ Λ h + Λ v (aconsequence of Lemma 3.2) and the assumption Λ h ≤ Λ v . (cid:3) Now, we complete the proof of Theorem 5.1. Note that l ( N k ( T )) ≤ l ( N h ) = l ( S h ∪ N h ) ≤ l ( S h ∪ N optk ) = l ( S h \ N optk ) + l ( N optk ) ≤ l ( N optk ) . The first inequality follows from the choice of N k ( T ) as the shortest network among the fourcompletions N h , N h , N v , N v . The second inequality follows from Lemma 4.1 and the factthat N optk (and therefore S h ∪ N optk ) is an admissible solution for the completion problem for S h . Finally, the last inequality follows from Lemma 5.4 by noticing that M ⊆ N optk and thus l ( S h \ N optk ) ≤ l ( S h \ M ) ≤ l ( N optk ) . This concludes the proof of the first assertion ofTheorem 5.1.Let N ∗ ( T ) be a minimum B -Manhattan network. For each direction D k , let N ∗ k ( T ) bethe set of k -segments and ( k + 1)-segments of N ∗ ( T ) . By Lemma 2.2 the network N ∗ k ( T ) is anadmissible solution for 1-DMMN( F k ) problem, thus l ( N ∗ k ( T )) ≥ OPT k ( T ) . Any k -segmentof N ∗ ( T ) belongs to exactly two one-directional networks N ∗ k ( T ) and N ∗ k − ( T ) , we concludethat P m − k =0 OPT k ( T ) ≤ P m − k =0 l ( N ∗ k ( T )) ≤ l ( N ∗ ( T )) = OPT( T ) . The first assertion ofTheorem 5.1 implies that l ( N k ( T )) ≤ l ( N optk ) = OPT k ( T ) for all k = 0 , . . . , m − , hence l ( N ( T )) ≤ m − X k =0 l ( N k ( T )) ≤ m − X k =0 OPT k ( T ) ≤ OPT( T ) . This concludes the proof that the approximation factor of N ( T ) := S m − k =0 N k ( T ) is 2.5.To finish the proof of Theorem 5.1, it remains to analyze the complexity of the algorithm.First, we use a straightforward analysis to establish a O ( mn ) bound on its running time.Then we show that this bound can be reduced to O ( mn ) by using a more advanced imple-mentation. The time complexity of Optimal completion ( S h ) is dominated by the execution ofthe dynamic programming algorithm that computes an optimal completion for each staircase S i,j | i ′ ,j ′ . The staircase S i,j | i ′ ,j ′ is processed O ( | T i,j | ) times (once for each shortest ( t i , t i ′ )-pathin Γ k ) using a O ( | T i,j | )-time dynamic programming algorithm (each of the O ( | T i,j | ) entriesof the dynamic programming table is computed in time O ( | T i,j | )). Therefore, each stair-case S i,j | i ′ ,j ′ contributes O ( | T i,j | ) to the execution of the algorithm Optimal completion ( S h ).Since each terminal belongs to at most two staircases, the overall complexity of Optimalcompletion ( S h ) is O ( n ) . This algorithm is processed to compute four optimal completionsfor each of the m directions. Therefore the total complexity of our 2.5-approximation algo-rithm for the B -MMN problem is O ( mn ) . S i,j | i ′ ,j ′ to O ( | T i,j | ) instead of O ( | T i,j | ) , leading to a total complexity of O ( mn ) . First, note thatamong all O ( | T i,j | ) subproblems, whose optima are stored in the dynamic programmingtable, only O ( | T i,j | ) are affected by the choice of the π (those are the subproblems containingthe highest and the rightmost terminal of T i,j ). Therefore, instead of solving each of O ( | T i,j | )subproblems O ( | T i,j | ) times, we solve the subproblems not affected by the choice of π onlyonce. Now, consider the number of subproblems obtained by taking into account the choice of π, then it is easy to verify that the total number of subproblems encountered is not O ( | T i,j | )but only O ( | T i,j | ) . Since each entry of the dynamic programming table is computed in time O ( | T i,j | ) , we obtain a contribution of O ( | T i,j | ) for each staircase S i,j | i ′ ,j ′ , and thus a totalcomplexity of O ( mn ) . (cid:3) In this paper, we presented a combinatorial factor 2.5 approximation algorithm for NP-hardminimum Manhattan network problem in normed planes with polygonal unit balls (the B -MMN problem). Its complexity is O ( mn ) , where n is the number of terminals and 2 m isthe number of extremal points of the unit ball B . Any B -Manhattan network N ( T ) can bedecomposed into m subnetworks, one for each direction of the normed plane. Each suchsubnetwork N k ( T ) ensures the existence of shortest paths between the pairs of terminals forwhich all legal paths use only k - and ( k + 1)-segments. We presented a factor 1.25 O ( n )algorithm for computing one-directional Manhattan networks, which lead to a factor 2.5algorithm for minimum B -Manhattan network problem. One of the open questions is whether the one-directional Manhattan network problem is NP-complete? Another open question is designing a factor 2 approximation algorithm for B -MMN, thus meeting the current bestapproximation factor for the classical MMN problem. Notice that polynomial time algorithmfor 1-DMMN problem will directly lead to a factor 2 approximation for B -MMN.Notice some similarity between the 1-DMMN problem and the oriented minASS probleminvestigated in relationship with the minimum stabbing box problem [16], alias the minimumarborally satisfied superset problem (minASS) [7]. 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