Recognition of Largest Empty Orthoconvex Polygon in a Point Set

CCCG 2008, Montréal, Québec, August 13–15, 2008Recognition of Largest Empty Orthoconvex Polygon in a Point SetSubhas C. Nandy ∗ Krishnendu Mukhopadhyaya ∗ Bhargab B. Bhattacharya ∗AbstractAn algorithm for computing the maximum area emptyisothetic orthoconvex polygon among a set of n pointsin a rectangular region, is presented. The worst casetime and space complexities of the proposed algorithmare O(n 3 ) and O(n 2 ) respectively.1 IntroductionThe problem of finding an empty convex k-gon of maximumarea or perimeter amidst a point set [4] has severalapplications. A survey paper [2] has been publishedvery recently, which elaborates several optimization issuesrelated to this problem. In VLSI layout design,document image processing and shape description, isotheticpolygons play a major role. A polygon is said tobe isothetic if its sides are parallel to coordinate axes.The problem of identifying the largest empty isotheticrectangle among a set of points has been studied extensively.The best known algorithm for this problem runsin O(n log 2 n) time [1]. The same time complexity holdsif the obstacles are arbitrary polygons [6, 9]. Recentlyit is shown that the largest isothetic rectangle inside asimple polygon can be obtained in O(n log n) time [5].In isothetic domain, the generalization of this problemis recognizing the largest empty orthoconvex polygon.An isothetic polygon is said to be orthoconvex if theintersection of the polygon with a horizontal or a verticalline is a single line segment. Orthoconvexity hasimportance in robotic visibility, and also its discretevariant appears in other areas like digital geometry [3]and discrete tomography [8]. Datta and Ramkumar[7] proposed algorithms for recognizing largest emptyorthoconvex polygon of some specified shapes amidsta 2D point set. These include (i) L-shape, (ii) crossshape, (iii) point visible, and (iv) edge visible polygons.The time complexity of these algorithms are allO(n 2 ). Another variation in this class of problems isrecognizing the largest empty staircase polygon amongpoint and isothetic polygonal obstacles, which can alsobe solved in O(n 2 ) time and space [10]. But, to thebest of our knowledge, the problem of recognizing anempty orthoconvex polygon of arbitrary shape maximizingarea/perimeter is not studied yet. In this paper,we propose an algorithm of recognizing an empty orthoconvexpolygon of maximum area, that runs in O(n 3 )time using O(n 2 ) space.∗ Indian Statistical Institute, Kolkata - 700 108, India2 PriliminariesLet R be a rectangular region containing a set of npoints P = {p 1 , p 2 , . . . , p n }. We will assume a coordinatesystem with bottom and left boundaries of R asthe x- and y-axes respectively. The coordinates of apoint α are denoted as (x(α), y(α)). We assume thatthe points in P are in general positions, i.e., for everytwo points p i and p j , x(p i ) ≠ x(p j ) and y(p i ) ≠ y(p j ).Henceforth, we shall use H i and V i to denote a horizontaland a vertical line passing through the point p i .Definition 1 An isothetic curve is a rectilinear pathconsisting of alternately horizontal and vertical line segments.An isothetic curve is a monotonically risingstaircase (R-stair) if for every pair of points α and βon the curve, x(α) ≤ x(β) implies y(α) ≤ y(β). Similarly,for every pair of points α and β on a monotonicallyfalling staircase (F-stair), we have x(α) ≤ x(β)implies y(α) ≥ y(β). An isothetic polygon is a regionbounded by a closed isothetic curve.Definition 2 An isothetic polygon Π is said to be orthoconvexif for any horizontal or vertical line l, the intersectionof Π with l is a line segment of length greaterthan or equal to 0. In other words, l either intersectsno edge or exactly two edges of Π.An orthoconvex polygon is empty if it does not containany member of P in its interior. Our objective is toidentify the largest empty orthoconvex polygon in R.Definition 3 An empty orthoconvex polygon Π is saidto be maximal empty orthoconvex polygon (MEOP) ifthere exists no other empty orthoconvex polygon Π ′ thatproperly encloses Π.It is easy to observe that an MEOP is bounded bytwo R-stairs R tl and R br , and two F-stairs F tr andF bl , where R tl spans from the left boundary to the topboundary, R br spans from the bottom boundary to theright boundary, F tr spans from the top boundary to theright boundary and F bl spans from the left boundaryto the bottom boundary of R, and each concave vertexof these stairs must coincide with a member in P (seeFigure 1). It may be observed that an R-stair (or anF-stair) may degenerate to a corner point of R.The number of maximal empty staircase polygonsamidst a point set of size n may be exponential in n;however, the maximum-area empty staircase polygoncan be computed in O(n 2 ) time [10]. Since a maximalempty staircase polygon is an MEOP as well, the same

20th Canadian Conference on Computational Geometry, 2008RtlFtrRtlp jFtrFblp jp i Rbr Fblp iRbr(a)Figure 1: Orthoconvex polygons(b)combinatorial explosion holds for MEOP also. We nowpresent a polynomial time algorithm for computing themaximum-area MEOP.3 AlgorithmWe shall consider all possible pairs of points p i , p j ∈ P,and identify the maximum area MEOP with p i ∈ F blas the closest point of the bottom boundary of R, andp j ∈ F tr as the closest point of the top boundary of R.The points p i and p j are said to be the bottom-pivot andtop-pivot respectively, and the corresponding MEOP isdenoted by MEOP(p i , p j ). We will use S to denote thevertical slab bounded by V i and V j . The projections ofa point p k ∈ S on V i , V j , H i and H j are denoted by q k ,qk ′ , r k and rk ′ respectively. We now separately considertwo cases: (i) x(p i ) < x(p j ), and (ii) x(p i ) > x(p j ).In Case (i), the lines V i and V j split the point set Pinto three parts, P 1 , P 2 and P 3 , where P 1 and P 3 arethe set of points to the left of V i and to the right of V jrespectively, and the points in P 2 lie inside the verticalslab S. If V i hits the top and bottom boundaries ofR at t 1 and b 1 respectively, and V j hits the top andbottom boundaries of R at t 2 and b 2 respectively. Now,the portion of the MEOP inside the vertical slab S,denoted by M 2 (b 1 , t 2 ), is an empty staircase polygonwith diagonally opposite corners b 1 and t 2 among thepoints in P 2 . The two stairs of M 2 (b 1 , t 2 ) are parts ofthe rising stairs R br and R tl respectively. If R tl hitsV i at q α , then the portion of the MEOP to the left ofV i , denoted by M 1 (q α ), is an empty edge-visible polygonwith base [p i , q α ] among the points in P 1 such that everypoint inside the polygon is visible from its base [p i , q α ].Similarly, if R br hits V j at q β ′ then M 3(q β ′ ) is an emptyedge-visible polygon with base [p j , q β ′ ] among the pointsin P 3 .In Case (ii), V j is to the left of V i . Here M 2 is an emptystaircase polygon from p i to p j , and these are the partsof F bl and F tr of the MEOP respectively. If F bl (resp.F tr ) hits V j (resp. V i ) at q ′ α (resp. q β), then M 1 (q ′ α )(portion to the left of V j ) is an edge-visible polygon withbase [q ′ α , p j], and M 3 (q β ) (portion to the right of V i ) isan edge-visible polygon with base [q β , p i ]. After fixing p iand p j as the bottom-pivot and top-pivot respectively,we need to choose M 1 , M 2 and M 3 such that the sumof areas of these three polygons is maximum among allsuch polygons. We shall describe our algorithm for Case(i) only. Case (ii) can easily be handled using a similarmethod. For Case (i), we explain the method of computingthe desired M 1 and M 2 . The computation of M 3is the same as that of M 1 .3.1 Computation of M 1Let us consider a point p i ∈ P. Let P 1 = {p k |x(p k ) x(p i ) and y(p k ) > y(p i )}.Q includes the top-right corner of R, and |Q| = m +1. Let q 0 , q 1 , q 2 , . . . , q m denote the projections of thepoints in Q on the vertical line V i in decreasing orderof their y-coordinates. We create an array EV L(p i )whose elements are the maximum area empty edgevisiblepolygon M 1 (q k ) with [p i , q k ] as the base for allk = 0, 1, 2, . . ., m.We use vertical line sweep among the points in P 1 startingfrom the position of V i to create a height-balancedbinary tree T . Its each node v is represented as a 5-tuple(I, x val, y val, ∆, δ). I is the base of the edge-visiblepolygons attached to node v. (x val, y val) is the pointwhere the node v is generated, and ∆ contains the areaof the largest edge visible polygon rooted at that node.The ∆ parameters are computed in two passes. In theforward pass during the sweep, the ∆ parameter of anode contains the area of the edge visible polygon thatis computed so far at that node. At the end of thesweep, a backward pass is executed from the leaf levelof T up to its root, and ∆ value of each node is properlyset. The δ value of all nodes are 0 at the time of creationof T ; it will be set and used during the computation ofM 1 (q k ) for different q k .Creation of TThe root r of T corresponds to the entire intervalI = [p i , q 0 ]; its x val and ∆ parameters are set tox(p i ) and 0 respectively. A vertical line sweep is performedfrom x = x(p i ) towards left. When a pointp = (x(p), y(p)) ∈ P 1 is faced by the sweep line, theleaf nodes in T are searched. If y(p) lies in the inter-

CCCG 2008, Montréal, Québec, August 13–15, 2008val [α, β] of a node v = ([α, β], µ, ν, ∆, δ), we compute∆ ∗ = ∆+(µ−x(p))×(β −α). Next, we create two childrenof v, namely v a = ([α, y(p)], x(p), y(p), ∆ ∗ , 0) andv b = ([y(p), β], x(p), y(p), ∆ ∗ , 0). Finally, the backwardpass is executed from leaves towards the root in postordermanner to set the ∆ values as described above.Computation of M 1 (q k )M 1 (q 0 ) = ∆ attached to the root node r. While processingq k , we assume that q k−1 is already processed. Westart scaning from the root of T . At a particular nodev = ([α, β], µ, ν, ∆(v), δ(v)) on the search path, one ofthe following two situations may happen: (i) ν ≥ y(q k )and (ii) ν < y(q k ).In Case (i), we compute A = (x(p i ) − µ) × (y(q k ) −y(q k−1 )). The area A is to be subtracted from all theedge-visible polygons stored in the the right-child v b ofthe node v. We subtract A from ∆(v b ). Without entirelytraversing the subtree rooted at v b , we add A inδ(v b ). The motivation is that, while processing someother q l , if the subtree rooted at v b is traversed, δ(v b )will be subtracted from the ∆ value of those nodes. Thesearch proceeds towards the left child of the node v. Ateach move from a node v to its children v ′ , δ(v) is subtractedfrom ∆(v ′ ), and added to δ(v ′ ), and then δ(v)is set to 0.In Case (ii), the edge visible polygon in the left child v aof the node v does not exist; so we delete the subtreerooted at v a and the node v also; the search proceedstowards the right child of v. The propagation of δ isto be performed at each step, but the computation ofexcess area A is to be performed when Case (i) arises.Next, a backward pass is needed to set the ∆ field of allthe nodes in the updated T . Finally, M 1 (q k ) is set withthe ∆ value of the root of the updated T .Lemma 1 The computation of M 1 (q k ) for all k =0, 1, . . .,m needs O(n 2 ) time.Proof. The creation of T needs O(n log n) time. Thelemma follows from the fact that the combinatorial complexityof M 1 (q k ) is O(n) for all k = 0, 1, . . .,m. ✷Similarly, with each point p i ∈ P, an array EV R(p i ) isattached. If the projections of the points {p k |x(p k )

20th Canadian Conference on Computational Geometry, 2008In our problem, computing the maximum area emptystaircase polygon among the points in P 2 will not suffice.Suppose MEOP(p i , p j ) consists of a staircase polygonMESP(p i , p j ), that has edges (p i , q α ) along V i , (p j , q α ′)along V j , (p i , r β ) along H i and (p j , r β ′) along H j , thenit includes (i) an edge-visible polygon with base (p i , q)to the left of V i , (ii) an edge visible polygon with base(p j , q ′ ) to the right of V j , (ii) an L-polygon with base(p i , r) and (iv) an L-polygon with base (p j , r ′ ). Let q, q ′ ,r and r ′ correspond to p α , p α ′, p β , p β ′ ∈ P 2 respectively.Thus, in order to compute the MEOP of maximumarea, we need to modify the weight of some edges of thegraph SG as follows, and then compute the maximumweighted path in the graph SG.For each α such that p α ∈ P 2 , change the weight of itseach outgoing edge e to w(e) + area(M 1 (p i , q α )).For each edge e ′ = (p i , β k ′), change the weight of e ′ tow(e ′ ) + area(LB(p i , r β )).For each edge α ′ such that p α ′ ∈ P 2 , if there existsan edge e ∗∗ from a footprint of α ′ to a footprintof p j , then change its weight to w(e ∗∗ ) +area(M 3 (p j , q ′ α ′)).For each incoming edge e ′ on j β ′), change the weightof e ′ to w(e ′ ) + area(LA(p j , r β ′)).Theorem 2 The largest MEOP among a set of npoints can be computed in O(n 5 ) time and O(n 2 ) space.In [10], it is also shown that the geometric propertiesof the problem can be exploited to design an algorithmfor computing the empty staircase polygon of maximumarea in O(|P 2 | 2 ) time and space. Here the results of procesinga point p j for computing the MESP(o, p j ) areused to compute MESP(o, p k ), where x(p k ) > x(p j )and y(p k ) > y(p j ); The point o is the bottom left cornerof the rectangular floor. We will use the same principleto reduce the time complexity of the problem ofcomputing the maximum area MEOP to O(n 3 ).4 Further improvementWe will fix a point p i ∈ P, and consider all p j ∈ Pwith x(p j ) > x(p i ) and y(p j ) > y(p i ). We also use thethe notion of complete processing of a point p j [10]. Apoint p j is said to be completely processed if all the edgesincident to p j in the graph G are processed. When apoint p j is completely processed, the weights of differentpaths in the staircase graph (SG) with bottom-pivotand p i and p j respectively, are available at the footprintsof p j . Each of these polygons has included thecorresponding M 1 (p i , q α ) and LB(p i , r β ) for some appropriatep α , p β ∈ S. In order to get the maximum areaMEOP with a MESP(p i , p j ), we need to add the areaof the appropriate M 3 (p j , q α ′) and LA(p j , r β ′), wherep α ′, p β ′ ∈ S. We can compute the array L containingLA(p j , r k ) for all the points p k ∈ S above H j in O(n)time. The values of area(M 3 (p j , q k )) are all availablein the array EV R(p j ). Now we can use EV R(p j ), L,and the area attached to the different footprints of p jto compute the largest MEOP(p i , p j ) in O(n) time.We process the points above H i in S by sweeping ahorizontal line upwards. After complete processing ofp j , we compute MEOP(p i , p j ) as described above, andthen process the outgoing edges of p j in G. Thus, wehave the following theorem:Theorem 3 The largest MEOP among a set of npoints can be computed in O(n 3 ) time and O(n 2 ) space.Proof. For a fixed p i , the generation of footprints forall p j ∈ P with x(p j ) > x(p i ) and y(p j ) > y(p i ) needsO(n 2 ) time [10]. 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