109 research outputs found

    Few Cuts Meet Many Point Sets

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    We study the problem of how to breakup many point sets in Rd\mathbb{R}^d into smaller parts using a few splitting (shared) hyperplanes. This problem is related to the classical Ham-Sandwich Theorem. We provide a logarithmic approximation to the optimal solution using the greedy algorithm for submodular optimization

    K1,3K_{1,3}-covering red and blue points in the plane

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    We say that a finite set of red and blue points in the plane in general position can be K1,3K_{1,3}-covered if the set can be partitioned into subsets of size 44, with 33 points of one color and 11 point of the other color, in such a way that, if at each subset the fourth point is connected by straight-line segments to the same-colored points, then the resulting set of all segments has no crossings. We consider the following problem: Given a set RR of rr red points and a set BB of bb blue points in the plane in general position, how many points of RBR\cup B can be K1,3K_{1,3}-covered? and we prove the following results: (1) If r=3g+hr=3g+h and b=3h+gb=3h+g, for some non-negative integers gg and hh, then there are point sets RBR\cup B, like {1,3}\{1,3\}-equitable sets (i.e., r=3br=3b or b=3rb=3r) and linearly separable sets, that can be K1,3K_{1,3}-covered. (2) If r=3g+hr=3g+h, b=3h+gb=3h+g and the points in RBR\cup B are in convex position, then at least r+b4r+b-4 points can be K1,3K_{1,3}-covered, and this bound is tight. (3) There are arbitrarily large point sets RBR\cup B in general position, with r=b+1r=b+1, such that at most r+b5r+b-5 points can be K1,3K_{1,3}-covered. (4) If br3bb\le r\le 3b, then at least 89(r+b8)\frac{8}{9}(r+b-8) points of RBR\cup B can be K1,3K_{1,3}-covered. For r>3br>3b, there are too many red points and at least r3br-3b of them will remain uncovered in any K1,3K_{1,3}-covering. Furthermore, in all the cases we provide efficient algorithms to compute the corresponding coverings.Comment: 29 pages, 10 figures, 1 tabl

    Projective center point and Tverberg theorems

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    We present projective versions of the center point theorem and Tverberg's theorem, interpolating between the original and the so-called "dual" center point and Tverberg theorems. Furthermore we give a common generalization of these and many other known (transversal, constraint, dual, and colorful) Tverberg type results in a single theorem, as well as some essentially new results about partitioning measures in projective space.Comment: 10 page

    Projective center point and Tverberg theorems

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    K1,3-covering red and blue points in the plane

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    We say that a finite set of red and blue points in the plane in general position can be K1, 3-covered if the set can be partitioned into subsets of size 4, with 3 points of one color and 1 point of the other color, in such a way that, if at each subset the fourth point is connected by straight-line segments to the same-colored points, then the resulting set of all segments has no crossings. We consider the following problem: Given a set R of r red points and a set B of b blue points in the plane in general position, how many points of R ¿ B can be K1, 3-covered? and we prove the following results: (1) If r = 3g + h and b = 3h + g, for some non-negative integers g and h, then there are point sets R ¿ B, like {1, 3}-equitable sets (i.e., r = 3b or b = 3r) and linearly separable sets, that can be K1, 3-covered. (2) If r = 3g + h, b = 3h + g and the points in R ¿ B are in convex position, then at least r + b - 4 points can be K1, 3-covered, and this bound is tight. (3) There are arbitrarily large point sets R ¿ B in general position, with r = b + 1, such that at most r + b - 5 points can be K1, 3-covered. (4) If b = r = 3b, then at least 9 8 (r + b- 8) points of R ¿ B can be K1, 3-covered. For r > 3b, there are too many red points and at least r - 3b of them will remain uncovered in any K1, 3-covering. Furthermore, in all the cases we provide efficient algorithms to compute the corresponding coverings
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