8 research outputs found

    Covering many points with a small-area box

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    Let PP be a set of nn points in the plane. We show how to find, for a given integer k>0k>0, the smallest-area axis-parallel rectangle that covers kk points of PP in O(nk2log⁥n+nlog⁥2n)O(nk^2 \log n+ n\log^2 n) time. We also consider the problem of, given a value α>0\alpha>0, covering as many points of PP as possible with an axis-parallel rectangle of area at most α\alpha. For this problem we give a probabilistic (1−Δ)(1-\varepsilon)-approximation that works in near-linear time: In O((n/Δ4)log⁥3nlog⁥(1/Δ))O((n/\varepsilon^4)\log^3 n \log (1/\varepsilon)) time we find an axis-parallel rectangle of area at most α\alpha that, with high probability, covers at least (1−Δ)Îș∗(1-\varepsilon)\mathrm{\kappa^*} points, where Îș∗\mathrm{\kappa^*} is the maximum possible number of points that could be covered

    Covering Points by Disjoint Boxes with Outliers

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    For a set of n points in the plane, we consider the axis--aligned (p,k)-Box Covering problem: Find p axis-aligned, pairwise-disjoint boxes that together contain n-k points. In this paper, we consider the boxes to be either squares or rectangles, and we want to minimize the area of the largest box. For general p we show that the problem is NP-hard for both squares and rectangles. For a small, fixed number p, we give algorithms that find the solution in the following running times: For squares we have O(n+k log k) time for p=1, and O(n log n+k^p log^p k time for p = 2,3. For rectangles we get O(n + k^3) for p = 1 and O(n log n+k^{2+p} log^{p-1} k) time for p = 2,3. In all cases, our algorithms use O(n) space.Comment: updated version: - changed problem from 'cover exactly n-k points' to 'cover at least n-k points' to avoid having non-feasible solutions. Results are unchanged. - added Proof to Lemma 11, clarified some sections - corrected typos and small errors - updated affiliations of two author

    Diseño de un algoritmo de minería metaheurístico para separar puntos de dos colores en un entorno bidimensional

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    The separation of color points is one of the important issues in computational geometry, which is used in various parts of science; it can be used in facility locating, image processing and clustering. Among these, one of the most widely used computational geometry in the real-world is the problem of covering and separating points with rectangles. In this paper, we intend to consider the problemof separating the two-color points sets, using three rectangles. In fact, our goal is to separate desired blue points from undesired red points by three rectangles, in such a way that these three rectangles contain the most desire points. For this purpose, we provide a metaheuristic algorithm based on the simulated annealing method that could separates blue points from input points, , in time order O (n) with the help of three rectangles. The algorithm is executed with C# and also it has been compared and evaluated with the optimum algorithm results. The results show that our recommended algorithm responses is so close to optimal responses, and also in some cases we obtains the exact optimal response

    Finding Axis-Parallel Rectangles of Fixed Perimeter or Area Containing the Largest Number of Points

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    Let P be a set of n points in the plane in general position, and consider the problem of finding an axis-parallel rectangle with a given perimeter, or area, or diagonal, that encloses the maximum number of points of P. We present an exact algorithm that finds such a rectangle in O(n^{5/2} log n) time, and, for the case of a fixed perimeter or diagonal, we also obtain (i) an improved exact algorithm that runs in O(nk^{3/2} log k) time, and (ii) an approximation algorithm that finds, in O(n+(n/(k epsilon^5))*log^{5/2}(n/k)log((1/epsilon) log(n/k))) time, a rectangle of the given perimeter or diagonal that contains at least (1-epsilon)k points of P, where k is the optimum value. We then show how to turn this algorithm into one that finds, for a given k, an axis-parallel rectangle of smallest perimeter (or area, or diagonal) that contains k points of P. We obtain the first subcubic algorithms for these problems, significantly improving the current state of the art

    On k-enclosing objects in a coloured point set

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    We introduce the exact coloured k -enclosing object problem: given a set P of n points in R 2 , each of which has an associated colour in f 1 ;:::;t g , and a vec- tor c = ( c 1 ;:::;c t ), where c i 2 Z + for each 1 i t , nd a region that contains exactly c i points of P of colour i for each i . We examine the problems of nd- ing exact coloured k -enclosing axis-aligned rectangles, squares, discs, and two-sided dominating regions in a t -coloured point setPostprint (published version

    Enclosing K Points in the Smallest Axis Parallel Rectangle

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    We consider the following clustering problem. Given a set S of n points in the plane, and given an integer k, n 2 ! k n, we want to find the smallest axis parallel rectangle (smallest perimeter or area) that encloses exactly k points of S. We present an algorithm which runs in time O(n + k(n \Gamma k) 2 ) improving previous algorithms which run in time O(k 2 n) and do not perform well for larger k values. We present an algorithm to enclose k of n given points in an axis parallel box in d-dimensional space which runs in time O(dn+dk(n \Gamma k) 2(d\Gamma1) ) and occupies O(dn) space. We slightly improve algorithms for other problems whose runtimes depend on k

    Large bichromatic point sets admit empty monochromatic 4-gons

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    We consider a variation of a problem stated by Erd˝os and Szekeres in 1935 about the existence of a number fES(k) such that any set S of at least fES(k) points in general position in the plane has a subset of k points that are the vertices of a convex k-gon. In our setting the points of S are colored, and we say that a (not necessarily convex) spanned polygon is monochromatic if all its vertices have the same color. Moreover, a polygon is called empty if it does not contain any points of S in its interior. We show that any bichromatic set of n ≄ 5044 points in R2 in general position determines at least one empty, monochromatic quadrilateral (and thus linearly many).Postprint (published version
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