M. H. Overmars and C.-K. Yap. New upper bounds in Klee's measure problem. SIAM J. Comput., 20:1034--1045, 1991.  Home/Search   Document Not in Database   Summary   Related Articles   Check

....(and at times, creative) ways. Most of our results are therefore theoretical. It is assumed that the input is in general position, without loss of generality, by known perturbation schemes. We mention some applications in Section 8, including improved time bounds for Klee s measure problem [26] in the case of 4 dimensional unit hypercubes, and for the minimum diameter spanning tree problem, which are of independent interest. 2 The strategy for semi online dynamization The most common dynamization strategy [5, 24, 25] is based on decomposing a set of objects into subsets, solving the ....

.... the union of n unit axis parallel cubes in three dimensions, a structure also known to have linear complexity [6] O( n log n) dynamic algorithms for measuring the union of squares (in fact, arbitrary axis parallel rectangles) were previously known by simple variants of the k d tree (see [26]) Theorem 6.1 We can maintain the volume of the union of a collection C of n unit axis parallel cubes in IR in e O( n) time per semi online update. Proof: We show how to store C so that the volume of (C [C ) can be computed quickly given an additional set C of b unit cubes. ....

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M. Overmars and C.-K. Yap, New upper bounds in Klee's measure problem, SIAM J. Comput., 20:1034-1045, 1991.

....the plane. Our goal is to maintain the approximation error of Rs with respect to R under insertions into and deletions from R and Rs. Ve use n to denote the total number of rectangles in the current sets. Our structure uses a partitioning of the plane similar to the one used by Overnmrs and Yap [13]. More precisely, we partition the plane into vertical slabs by drawing O(v ) vertical lines such that in between any two consecutive lines there are at most vertices of rectangles (in the current set) A rectangle is said to belong to a slab if any vertex of the rectangle is contained within the ....

M.H. Overmars and C.K. Yap. New upper bounds in Klee's Measure Problem. SIAM d. Cornput. 20:1034 1045 (1991).

....stores the union in an implicit way; it uses O(n log n) storage, the amortized insertion time is O(vlog n) and the query time is O( l Jr 1) log n) where 1 is the complexity of the union inside the query rectangle. Our structure is closely related to a structure described by Overmars and Yap [17]. They have shown how to maintain the measure of a set of rectangles, using O(vlog n) time per update. We want to use the structure for answering union range queries, where the number of queries can be much larger than the number of updates. We therefore do not want to spend, say, v log n) time ....

....can be inserted into the structure in O(vflog n) amortized time, assuming we know all rectangles to be inserted in advance. 3.2.1 The structure Let us first describe the structure for a static set R. The basic ingredient that we use is a parti tioning of the plane described by Overmars and Yap [17]. This partitioning is as follows. Sort the vertical 2 boundary edges of the rectangles in R by x coordinate, and draw a vertical line through every x d th vertical boundary edge. This partitions the plane into O( vertical slabs. Each slab is divided further into O(x d ) rectangular cells by ....

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M.H. Overmars and C.K. Yap, New Upper Bounds in Klee's Measure Problem, SlAM J. Cornput. 20 (1991), pp. 1034-1045.

....among the possible L1 diameters. Lemma 5.7. We can find the minimum L1 diameter k point subset of a set of n ) Proof: Finding the optimal placementofahypercube is equivalent to finding the deepest point in an arrangementofhypercubes. We can easily adapt an algorithm of Overmars and Yap [32], originally applied to Klee s measure problem, to find the deepest point in an arrangement of axis aligned boxes in time O(n log n) and ) To find the optimal hypercube size, we search along each coordinate axis as follows. We sort the points by the appropriate coordinate, and define a ....

M. H. Overmars and C.-K. Yap. New upper bounds in Klee's measure problem. SIAM J. Comput., 20:1034--1045, 1991.

....(and at times, creative) ways. Most of our results are therefore theoretical. It is assumed that the input is in general position, without loss of generality, by known perturbation schemes. We mention some applications in Section 8, including an improved time bound for Klee s measure problem [25] in the case of 4 dimensional unit hypercubes, of independent interest. 2 The strategy for semi online dynamization The most common dynamization strategy [5, 23, 24] is based on decomposing a set of objects into subsets, solving the problem on each subset, and combining the answers. An update ....

.... the union of n unit axis parallel cubes in three dimensions, a structure also known to have linear complexity [6] O( p n log n) dynamic algorithms 10 for measuring the union of squares (in fact, arbitrary axis parallel rectangles) were previously known by simple variants of the k d tree (see [25]) Theorem 6.1 We can maintain the volume of the union of a collection C of n unit axis parallel cubes in IR 3 in e O( p n) time per semi online update. Proof: We show how to store C so that the volume of S (C [C 0 ) can be computed quickly given an additional set C 0 of b unit cubes. ....

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M. Overmars and C.-K. Yap. New upper bounds in Klee's measure problem. SIAM J. Comput., 20:1034{ 1045, 1991. 16

....many enter and or leave bundle B j between t l and t r . of the slice S i on the two planes. As previously, we store a set of critical levels j of A(S x ) within the projected window S x i in a persistent B tree T . However, inspired by previous approaches for two dimensional range searching [19, 23], instead of storing every B i th level, we now only store every ( p NB i )th level of A(S x ) Refer to Figure 6 (i) We have p N=B i critical levels, and hence we define bundle B j to be the p NB i levels between critical level j Gamma1 and j . Using Lemma 3.4, we can prove ....

M. H. Overmars and C.-K. Yap. New upper bounds in Klee's measure problem. SIAM J. Comput., 20:1034--1045, 1991.

....turn equivalent to the element uniqueness problem which has a lower bound of Omega Gamma n log n) 10] 5 Extensions to Higher Dimensions In this section, we discuss the extension of our algorithm to higher dimensions. It is assumed that the reader is familiar with the work of Overmars and Yap [9], since we will use their data structure extensively in this section. 22 We first discuss the one separability problem in 3 dimensions and we will indicate briefly how this problem can be solved in higher dimensions with some modifications of the three dimensional solution. Our algorithm is ....

....discussed in the next subsection. After the sweep plane has crossed the minimum z coordinate facet of the collection of polyhedra, our data structure is ready. 23 5.1. 1 The data structure Our primary data structure is the so called orthogonal partition tree first introduced by Overmars and Yap [9]. This data structure is used to maintain a two dimensional cross section during a space sweep of a collection of three dimensional orthogonal objects. Note that, there may be O(n 2 ) rectangular regions in such a two dimensional cross section. This is due to the presence of trellises. In a ....

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M. H. Overmars and C. K. Yap. "New upper bounds in Klee's measure problem", SIAM J. Computing, Vol. 20, No. 6, (1991), pp. 1034-1045.

....of the data structure after the plane sweep has occurred. We solve this persistent one dimensional problem using a data structure combining ideas from B trees and segment trees. Our approach to the filter conflict detection problem uses a technique related to an algorithm by Overmars and Yap [11] for Klee s measure problem (determining the volume of a union of rectangular blocks) we use a kD tree [5] to divide the plane into rectangular cells, not containing any rectangle vertex, so that the rectangles intersecting any cell form stripes (i.e. rectangles that are unbounded in one ....

M. H. Overmars and C. K. Yap. New upper bounds in Klee's measure problem. SIAM J. Comput., 20:1034--1045, 1991.

....call. But, quite obviously, if one can do those incremental operations in O(f(n) steps in d dimensions, one can derive an algorithm in d 1 dimensions that computes the maximum intersection in O(nf(n) time. Since the best algorithm known in d 1 dimensions takes O(n (d 1) 2 log(n) time [9], we cannot expect to derive from that a better than O( p n log(n) increment complexity in 2D, unless the original static algorithm can be improved. 3.2 Taking the seam into account The seam appears on constellations as a line where links are interrupted. Due to earth self rotation, its ....

M. H. Overmars and C.-K. Yap. New upper bounds in Klee's measure problem. SIAM J. Comput., 20:10341045, 1991.

....and then using it as an oracle in the Frederickson and Johnson [10] optimization scheme. For the oracle in 3 space we prove that a set of n unit cubes can be decomposed into O(n) disjoint axis parallel boxes. We then apply the orthogonal partition trees (OPTs) described by Overmars and Yap [16] to find the maximal depth of disjoint axis parallel boxes. We show that this suffices to answer the Hausdorff decision problem in 3 space. For d 3 there is a super linear lower bound on the number of boxes obtained by disjoint decomposition of a union of boxes (see [4] thus we cannot use a ....

....In Section 2 we define the minimum Hausdorff distance problem, and describe the Hausdorff distance decision problem. In Section 3 we show that the union of n axis parallel unit cubes in 3 space can be decomposed into O(n) disjoint axis parallel boxes, and use the orthogonal partition trees of [16] to solve the Hausdorff distance decision problem in 3 space. For d 3, our algorithm is more involved and hence its description is separated into two sections; Section 4 contains a relaxed version of our data structures and an oracle which runs in time O(n 3d=2 Gamma1 log n) In Section 5 we ....

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M. Overmars and C.K. Yap, New Upper Bounds in Klee's Measure Problem, SIAM Journal of Computing, 20:6 (1991), 1034--1045.

....section) and then using it as an oracle in the Frederickson and Johnson [8] optimization scheme. For the oracle in 3 space we prove that a set of n unit cubes can be decomposed into O(n) disjoint axis parallel boxes. We then apply the orthogonal partition trees (OPTs) described by Overmars and Yap [13] to find the maximal depth of disjoint axis parallel boxes. We show that this suffices to answer the Hausdorff distance decision problem in 3 space. For d 3 there is a super linear lower bound on the number of boxes obtained by disjoint decomposition of a union of boxes (see [3] thus we cannot ....

....In Section 2 we define the minimum Hausdorff distance problem, and describe the Hausdorff distance decision problem. In Section 3 we show that the union of n axis parallel unit cubes in 3 space can be decomposed into O(n) disjoint axis parallel boxes, and use the orthogonal partition trees of [13] to solve the Hausdorff distance decision problem in 3 space. For d 3, our algorithm is more involved and hence its description is separated into two sections: Section 4 contains a relaxed version of our data structures and an oracle which runs in time O(n 3d=2 Gamma1 log n) in Section 5 we ....

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M. Overmars and C.K. Yap, New Upper Bounds in Klee's Measure Problem, SIAM Journal of Computing, 20 (1991), 1034--1045.

....the problem of placing a fixed radius circle to contain the largest subset of a given set S. Their algorithm requires O(n 2 ) time. Eppstein and Erickson [5] as a substep of their algorithm to find the minimum L1 diameter k subset of a given set S, note that an algorithm of Overmars and Yap [12] can be modified to find the maximum depth of an arrangement of axis aligned rectangles. This approach solves in O(n log n) time the problem of finding an optimal translation of a rectangle to cover the maximum sized subset of S. That is, it solves Problem 1 in O(n log n) time in the special ....

M.H. Overmars and C.K. Yap "New upper bounds in Klee's measure problem," SIAM J. Computing 20 (1991) 1034--1045.

....rings into 4n rectangles and applying the segment tree [3] to compute the depth of the set of rectangles in O(n log n) time. Our method, that is easily extendable to higher dimensions, also uses the Klee measure (the depth and the union of a set of rectangles are examples for the Klee measure, see [3, 4]) First we use the algorithm from Section 2 to find whether the set R r of the external rectangles defining the given rings is 1 pierceable. If R r is not 1 pierceable then neither is the set R. Otherwise, we find the region Q (also a rectangle) where all the rectangles from R r intersect. In our ....

....boundaries of the rectangles in R, covers Q. If it does not cover Q, then R is 1 pierceable; otherwise it is not 1 pierceable. In higher dimensional space Q is easily found as above in time O(dn logn) In order to find the union of the internal rectangles we use the algorithm of Overmars and Yap [4] who solve the Klee measure problem in higher dimensions in time O(n b d 2 c log n) Thus, this is the runtime of our 1 piercing algorithm for rings for d 2. 2 and 3 piercing For two and three pierceability problems a non trivial but quadratic algorithm is as follows. We first check ....

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M. Overmars, C. Yap, "New upper bounds in Klee's measure problem", In Proc. 29 Annual IEEE Symp. on the Found. of Comput. Sci., 1988.

....n) time per site. In total, this comes down to O(n 2 log n) time, whereas plane sweep only takes O(n log n) time. Some variants of the standard sweep are versions where a line rotates about a point [9, 15, 35] and the three dimensional version where a horizontal plane translates through space [39, 41]. There are also sweep algorithms where a topological line is used instead of a straight line [12] Implementation of plane sweep algorithms is not difficult, certainly not if existing code for balanced search trees, sorting, and geometric primitives is used. A major advantage of sweeping over ....

M. H. Overmars and C.-K. Yap. New upper bounds in Klee's measure problem. SIAM J. Comput., 20:1034--1045, 1991.

....and O(n 1 ffl ) space, 0 ffl 1, when k ffl O(n) 4.3 Minimizing L1 diameter We can solve the k point clustering problem minimizing L1 diameter in the same way as we have done for the k point clustering problem minimizing euclidean diameter. However, we can do much better. Overmars and Yap [10] presented an O(k d=2 log k) time and O(k d=2 ) space algorithm (based on space sweep) for finding an optimal placement of an L1 box of size c (u) in S i . This problem is equivalent to the problem of finding the deepest point in an arrangement of hypercubes where each point x of S i is ....

....(based on space sweep) for finding an optimal placement of an L1 box of size c (u) in S i . This problem is equivalent to the problem of finding the deepest point in an arrangement of hypercubes where each point x of S i is replaced by a square box of size c (u) with x as its center. However, [10] cannot be used for the planar point set. In 2 dimensional space the algorithm of Lee [7] also based on plane sweep) can be used to solve the problem in O(k log k) time and O(k) space. The probe sequence can be generated in O(n log n) time (Lemma 9) Therefore from Lemma 3 it follows that ....

M. H. Overmars and C.-K. Yap. "New upper bounds in Klee's measure problem", SIAM J. Comput., Vol. 20, 1991, 1034-1045.

....L1 diameter k point subset of a set of n points in IR d , in time O(n d=2 log 2 n) and space O(n d=2 ) Proof: Finding the optimal placement of a hypercube is equivalent to finding the deepest point in an arrangement of hypercubes. We can easily adapt an algorithm of Overmars and Yap [32], originally applied to Klee s measure problem, to find the deepest point in an arrangement of axis aligned boxes in time O(n d=2 log n) and space O(n d=2 ) To find the optimal hypercube size, we search along each coordinate axis as follows. We sort the points by the appropriate coordinate, ....

M. H. Overmars and C.-K. Yap. New upper bounds in Klee's measure problem. SIAM J. Comput., 20:1034--1045, 1991.

....y Department of Computer Science, Cornell University, Ithaca, NY 14853; schar cs.cornell.edu. Supported by NSF grant IRI9057928. Eppstein and Erickson [6] as a substep of their algorithm to find the minimum L1 diameter k subset of a given set S, note that an algorithm of Overmars and Yap [9] can be modified to find the maximum depth of an arrangement of axis aligned rectangles. This approach solves in O(n log n) time the problem of finding an optimal translation of a rectangle to cover the maximum sized subset of S. That is, it solves Problem 1 in O(n log n) time in the special case ....

M. Overmars and C. Yap. New upper bounds in Klee's measure problem. SIAM J. Computing, 20:1034--1045, 1991.

....is investigated by Chew et al. CDEK95] For the decision problem in case of the L1 metric, the space of feasible translations is an intersection of unions of unit boxes. This space is maintained using a modification of the data structure of orthogonal partition trees by Overmars and Yap [OY91] This gives algorithms for the decision problem which are used to solve the optimization problem by parametric search [Meg83] Col84] In particular, for the L1 metric an algorithm of running time O(n (4d Gamma2) 3 log 2 n) is obtained where n is the number of points in both patterns. ....

M. H. Overmars and C.-K. Yap. New upper bounds in Klee's measure problem. SIAM J. Comput., 20:1034--1045, 1991.

....naive O(n k ) bound, and provides the best algorithm we know of for dimensions 2 and 3. In higher dimensions, we can further improve this bound using an alternative approach which we now outline. Our algorithm for this case is based on an technique of Overmars and Yap for Klee s measure problem [23], Instead of directly searching for the maximum discrepancy orthant, we find for each value of i from 1 to n the orthants with minimum and maximum area that contain exactly i of the n points. Once these are found, the discrepancy can then be computed in linear time. We dualize the problem by ....

....relation between points and orthants is preserved by this dualization. The problem thus becomes one of finding, in this dual arrangement of orthants, the point contained in exactly i orthants and minimizing or maximizing the product of its coordinates. We now apply a technique of Overmars and Yap [23], to subdivide space into O(n k=2 ) boxes, the positions of which depend on the input orthant arrangement. We say that an orthant of our dual problem crosses a box of the partition if some portion of the orthant boundary is interior to the box. We say that an orthant of our dual problem covers a ....

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M. Overmars and C.K. Yap. New upper bounds in Klee's measure problem. SIAM J. Comput. 20 (1991) 1034--1045.

....plane, the strategy is to track depth information as a hyperplane is swept along x d . A higher dimensional data structure to play the role of the segment tree is needed for this task. The authors use a modified version of the Orthogonal Partition Tree (OPT) first introduced by Overmars and Yap [14]. They prove the following result: Theorem 6 Given point sets A and B in d space and using the L1 metric, the minimum Hausdorff distance under translation between A and B can be computed in time O(n 3 log 2 n) if d = 3 and O(n (4d Gamma2) 3 log 2 n) if d 3, where n = max(jAj; jBj) 6 A ....

Mark H. Overmars and Chee-Keng Yap. New upper bounds in klee's measure problem. SIAM Journal of Computing, 20(6):1034--1045, December 1991.

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M. H. Overmars and C.-K. Yap. New upper bounds in Klee's measure problem. SIAM J. Comput., 20:1034--1045, 1991.

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M.H. Overmars and C.K. Yap. New upper bounds in Klee's Measure Problem. SIAM J. Comput. 20:1034--1045 (1991). 25

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M. Overmars and C.K. Yap, New Upper Bounds in Klee's Measure Problem, SIAM Journal of Computing, 20 (1991), 1034--1045.

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M. H. Overmars and C.-K. Yap. New upper bounds in Klee's measure problem. SIAM J. Comput., 20:1034-1045, 1991.

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