The alternating path problem revisited

It is well known that, given n red points and n blue points on a circle, it is not always possible to find a plane geometric Hamiltonian alternating path. In this work we prove that if we relax the constraint on the path from being plane to being 1-plane, then the problem always has a solution, and even a Hamiltonian alternating cycle can be obtained on all instances. We also extend this kind of result to other configurations and provide remarks on similar problems.Ministerio de Economía y CompetitividadGeneralitat de CatalunyaEuropean Science FoundationMinisterio de Ciencia e InnovaciónJunta de Andalucía (Consejería de Innovación, Ciencia y Empresa

K1,3K_{1,3}-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 $K_{1,3}$-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 $R\cup B$ can be $K_{1,3}$-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 $R\cup B$, like $\{1,3\}$-equitable sets (i.e., $r=3b$ or $b=3r$) and linearly separable sets, that can be $K_{1,3}$-covered. (2) If $r=3g+h$, $b=3h+g$ and the points in $R\cup B$ are in convex position, then at least $r+b-4$ points can be $K_{1,3}$-covered, and this bound is tight. (3) There are arbitrarily large point sets $R\cup B$ in general position, with $r=b+1$, such that at most $r+b-5$ points can be $K_{1,3}$-covered. (4) If $b\le r\le 3b$, then at least $\frac{8}{9}(r+b-8)$ points of $R\cup B$ can be $K_{1,3}$-covered. For $r>3b$, there are too many red points and at least $r-3b$ of them will remain uncovered in any $K_{1,3}$-covering. Furthermore, in all the cases we provide efficient algorithms to compute the corresponding coverings.Comment: 29 pages, 10 figures, 1 tabl

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

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

Hamiltonian orthogeodesic alternating paths

AbstractLet R be a set of red points and let B be a set of blue points. The point set P=R∪B is called equitable if ||B|−|R||⩽1 and it is called general if no two points are vertically or horizontally aligned. An orthogeodesic alternating path on P is a path such that each edge is an orthogeodesic chain connecting points of different color and such that no two edges cross. We consider the problem of deciding whether a set of red and blue points admits a Hamiltonian orthogeodesic alternating path, that is, an orthogeodesic alternating path visiting all points. We prove that every general equitable point set admits a Hamiltonian orthogeodesic alternating path and we present an O(nlog2n)-time algorithm for finding such a path, where n is the number of points. On the other hand, we show that the problem is NP-complete if the path must be on the grid (i.e., vertices and bends have integer coordinates). Further, we show that we can approximate the maximum length of an orthogeodesic alternating path on the grid by a factor of 3, whereas we present a family of point sets with n points that do not have a Hamiltonian orthogeodesic alternating path with more than n/2+2 points. Additionally, we show that it is NP-complete to decide whether a given set of red and blue points on the grid admits an orthogeodesic perfect matching if horizontally aligned points are allowed. This contrasts a recent result by Kano (2009) [9] who showed that this is possible on every general point set

The Chromatic Number of the Disjointness Graph of the Double Chain

Let $P$ be a set of $n\geq 4$ points in general position in the plane. Consider all the closed straight line segments with both endpoints in $P$. Suppose that these segments are colored with the rule that disjoint segments receive different colors. In this paper we show that if $P$ is the point configuration known as the double chain, with $k$ points in the upper convex chain and $l \ge k$ points in the lower convex chain, then $k+l- \left\lfloor \sqrt{2l+\frac{1}{4}} - \frac{1}{2}\right\rfloor$ colors are needed and that this number is sufficient

Cell-paths in mono- and bichromatic line arrangements in the plane

We show that in every arrangement of n red and blue lines | in general position and not all of the same color | there is a path through a linear number of cells where red and blue lines are crossed alternatingly (and no cell is revisited). When all lines have the same color, and hence the preceding alternating constraint is dropped, we prove that the dual graph of the arrangement always contains a path of length (n2).Peer ReviewedPostprint (author’s final draft