54 research outputs found

    Linear transformation distance for bichromatic matchings

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    Let P=B∪RP=B\cup R be a set of 2n2n points in general position, where BB is a set of nn blue points and RR a set of nn red points. A \emph{BRBR-matching} is a plane geometric perfect matching on PP such that each edge has one red endpoint and one blue endpoint. Two BRBR-matchings are compatible if their union is also plane. The \emph{transformation graph of BRBR-matchings} contains one node for each BRBR-matching and an edge joining two such nodes if and only if the corresponding two BRBR-matchings are compatible. In SoCG 2013 it has been shown by Aloupis, Barba, Langerman, and Souvaine that this transformation graph is always connected, but its diameter remained an open question. In this paper we provide an alternative proof for the connectivity of the transformation graph and prove an upper bound of 2n2n for its diameter, which is asymptotically tight

    Quasi-Parallel Segments and Characterization of Unique Bichromatic Matchings

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    Given n red and n blue points in general position in the plane, it is well-known that there is a perfect matching formed by non-crossing line segments. We characterize the bichromatic point sets which admit exactly one non-crossing matching. We give several geometric descriptions of such sets, and find an O(nlogn) algorithm that checks whether a given bichromatic set has this property.Comment: 31 pages, 24 figure

    Bichromatic compatible matchings

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    Abstract For a set R of n red points and a set B of n blue points, a BR-matching is a non-crossing geometric perfect matching where each segment has one endpoint in B and one in R. Two BRmatchings are compatible if their union is also non-crossing. We prove that, for any two distinct BRmatchings M and M , there exists a sequence of BR-matchings M = M 1 , . . . , M k = M such that M i−1 is compatible with M i . This implies the connectivity of the compatible bichromatic matching graph containing one node for each BR-matching and an edge joining each pair of compatible BR-matchings, thereby answering the open problem posed by Aichholzer et al. in [6]

    Bichromatic compatible matchings

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    ABSTRACT For a set R of n red points and a set B of n blue points, a BR-matching is a non-crossing geometric perfect matching where each segment has one endpoint in B and one in R. Two BR-matchings are compatible if their union is also noncrossing. We prove that, for any two distinct BR-matchings M and M , there exists a sequence of BR-matchings M = M1, . . . , M k = M such that Mi−1 is compatible with Mi. This implies the connectivity of the compatible bichromatic matching graph containing one node for each BR-matching and an edge joining each pair of compatible BR-matchings, thereby answering the open problem posed by Aichholzer et al. in [5]

    Disjoint compatibility graph of non-crossing matchings of points in convex position

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    Let X2kX_{2k} be a set of 2k2k labeled points in convex position in the plane. We consider geometric non-intersecting straight-line perfect matchings of X2kX_{2k}. Two such matchings, MM and M′M', are disjoint compatible if they do not have common edges, and no edge of MM crosses an edge of M′M'. Denote by DCMk\mathrm{DCM}_k the graph whose vertices correspond to such matchings, and two vertices are adjacent if and only if the corresponding matchings are disjoint compatible. We show that for each k≥9k \geq 9, the connected components of DCMk\mathrm{DCM}_k form exactly three isomorphism classes -- namely, there is a certain number of isomorphic small components, a certain number of isomorphic medium components, and one big component. The number and the structure of small and medium components is determined precisely.Comment: 46 pages, 30 figure

    Flip Distance to some Plane Configurations

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    We study an old geometric optimization problem in the plane. Given a perfect matching M on a set of n points in the plane, we can transform it to a non-crossing perfect matching by a finite sequence of flip operations. The flip operation removes two crossing edges from M and adds two non-crossing edges. Let f(M) and F(M) denote the minimum and maximum lengths of a flip sequence on M, respectively. It has been proved by Bonnet and Miltzow (2016) that f(M)=O(n^2) and by van Leeuwen and Schoone (1980) that F(M)=O(n^3). We prove that f(M)=O(n Delta) where Delta is the spread of the point set, which is defined as the ratio between the longest and the shortest pairwise distances. This improves the previous bound for point sets with sublinear spread. For a matching M on n points in convex position we prove that f(M)=n/2-1 and F(M)={{n/2} choose 2}; these bounds are tight. Any bound on F(*) carries over to the bichromatic setting, while this is not necessarily true for f(*). Let M\u27 be a bichromatic matching. The best known upper bound for f(M\u27) is the same as for F(M\u27), which is essentially O(n^3). We prove that f(M\u27)<=slant n-2 for points in convex position, and f(M\u27)= O(n^2) for semi-collinear points. The flip operation can also be defined on spanning trees. For a spanning tree T on a convex point set we show that f(T)=O(n log n)

    Flip Distance to a Non-crossing Perfect Matching

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    A perfect straight-line matching MM on a finite set PP of points in the plane is a set of segments such that each point in PP is an endpoint of exactly one segment. MM is non-crossing if no two segments in MM cross each other. Given a perfect straight-line matching MM with at least one crossing, we can remove this crossing by a flip operation. The flip operation removes two crossing segments on a point set QQ and adds two non-crossing segments to attain a new perfect matching M′M'. It is well known that after a finite number of flips, a non-crossing matching is attained and no further flip is possible. However, prior to this work, no non-trivial upper bound on the number of flips was known. If g(n)g(n) (resp.~k(n)k(n)) is the maximum length of the longest (resp.~shortest) sequence of flips starting from any matching of size nn, we show that g(n)=O(n3)g(n) = O(n^3) and g(n)=Ω(n2)g(n) = \Omega(n^2) (resp.~k(n)=O(n2)k(n) = O(n^2) and k(n)=Ω(n)k(n) = \Omega (n))
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