6 research outputs found
On Hamiltonian alternating cycles and paths
We undertake a study on computing Hamiltonian alternating cycles and paths on bicolored point sets. This has been an intensively studied problem, not always with a solution, when the paths and cycles are also required to be plane. In this paper, we relax the constraint on the cycles and paths from being plane to being 1-plane, and deal with the same type of questions as those for the plane case, obtaining a remarkable variety of results. For point sets in general position, our main result is that it is always possible to obtain a 1-plane Hamiltonian alternating cycle. When the point set is in convex position, we prove that every Hamiltonian alternating cycle with minimum number of crossings is 1-plane, and provide O(n) and O(n2) time algorithms for computing, respectively, Hamiltonian alternating cycles and paths with minimum number of crossings.Peer ReviewedPostprint (author's final draft
-covering red and blue points in the plane
We say that a finite set of red and blue points in the plane in general
position can be -covered if the set can be partitioned into subsets of
size , with points of one color and 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
of red points and a set of blue points in the plane in general
position, how many points of can be -covered? and we prove
the following results:
(1) If and , for some non-negative integers and ,
then there are point sets , like -equitable sets (i.e.,
or ) and linearly separable sets, that can be -covered.
(2) If , and the points in are in convex position,
then at least points can be -covered, and this bound is tight.
(3) There are arbitrarily large point sets in general position,
with , such that at most points can be -covered.
(4) If , then at least points of
can be -covered. For , there are too many red points and at
least of them will remain uncovered in any -covering.
Furthermore, in all the cases we provide efficient algorithms to compute the
corresponding coverings.Comment: 29 pages, 10 figures, 1 tabl
Polynomially solvable cases of the bipartite traveling salesman problem
Given two sets, R and B, consisting of n cities each, in the bipartite traveling salesman problem one looks for the shortest way of visiting alternately the cities of R and B, returning to the city of origin. This problem is known to be NP-hard for arbitrary sets R and B. In this paper we provide an O(n6) algorithm to solve the bipartite traveling salesman problem if the quadrangle property holds. In particular, this algorithm can be applied to solve in O(n6) time the bipartite traveling salesman problem in the following cases: S=R¿B is a convex point set in the plane, S=R¿B is the set of vertices of a simple polygon and V=R¿B is the set of vertices of a circular graph. For this last case, we also describe another algorithm which runs in O(n2) time
K1,3-covering red and blue points in the plane
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
On Hamiltonian alternating cycles and paths
We undertake a study on computing Hamiltonian alternating cycles and paths on bicolored point sets. This has been an intensively studied problem, not always with a solution, when the paths and cycles are also required to be plane. In this paper, we relax the constraint on the cycles and paths from being plane to being 1-plane, and deal with the same type of questions as those for the plane case, obtaining a remarkable variety of results. For point sets in general position, our main result is that it is always possible to obtain a 1-plane Hamiltonian alternating cycle. When the point set is in convex position, we prove that every Hamiltonian alternating cycle with minimum number of crossings is 1-plane, and provide O(n) and O(n2) time algorithms for computing, respectively, Hamiltonian alternating cycles and paths with minimum number of crossings.Peer Reviewe