1 Introduction
We consider geometric hypergraphs that we call Delaunayhypergraphs and are defined as follows: given a family of regions and a set of points , the vertex set of the hypergraph is and a subset of is a hyperedge of if there exists a region such that . A special case that has attracted attention in the literature is the case in which we only consider hyperedges of size two, called Delaunayedges. In this case is the socalled Delaunaygraph of with respect to . This notion generalizes the wellknown concept of Delaunay triangulation of planar point sets.
The Delaunay triangulation is of course planar. It is also known that the Delaunaygraph is planar when it is defined with respect to a planar set of points and a family of pseudodisks [buzaglo]. (A family of compact regions in the plane, each of which is bounded by a Jordan curve, is called a family of pseudodisks if every two boundary curves are disjoint or properly cross twice.) Moreover, the second and third authors have recently proved that there is a plane embedding of this graph such that every edge lies within all the pseudodisks that contain its endpoints [PDplane]. Note that halfplanes, disks, or more generally, homothets^{1}^{1}1The homothetic copy of a set is its scaled and translated copy (rotations are not allowed). of a certain convex shape with a smooth^{2}^{2}2Any Delaunayhypergraph that can be defined using the homothets of a convex set can also be defined as the Delaunayhypergraph of a convex set with a smooth boundary. boundary, are all families of pseudodisks. However, not every Delaunaygraph with respect to a planar point set is planar. For example, it is easy to see that if is the family of all axisparallel rectangles, then the Delaunaygraph may contain as a subgraph for every integer .
The problem of coloring the vertices of Delaunayhypergraphs has attracted a lot of attention. Specifically, for a Delaunayhypergraph and an integer , one wishes to determine the smallest integer such that the vertices of can be colored with colors such that every Delaunayhyperedge of size at least contains vertices of different (or all ) colors. The former is known as proper, the latter as polychromatic coloring.
In this paper we introduce a generalization of the problem of coloring the vertices of a (geometric) hypergraph. In particular, we wish to color (unordered) tuples of vertices such that every Delaunayhyperedge with enough vertices contains tuples of different colors (or all colors). Note that when , we get the usual (proper or polychromatic) coloring of vertices. For we identify two interesting variants: In the first one, we color all the possible tuples of vertices, whereas in the second variant we only color tuples that are themselves Delaunayhyperedges. For example, for we have a coloring of the Delaunayedges.
1.1 Our contribution
In Section 2 we consider proper coloring of tuples that are themselves Delaunayhyperedges and prove the following results.
Theorem 1.1
Let be a finite set of points in the plane, let be a family of pseudodisks that is shrinkable with respect to (see below). Then it is possible to color the Delaunayedges of with respect to with four colors, such that every pseudodisk that contains at least three points from contains Delaunayedges of two different colors.
Theorem 1.1 is proved in Section 2.1. Roughly speaking, a family is shrinkable with respect to if for every and a point it holds that contains a hyperedge that still contains and one less point from and . Note that we do not require the defining this new hyperedge to be a subset of . For example, disks are shrinkable with respect to any finite point set, but unit disks are not. Theorem 1.1 also holds whenever is a family of homothets of a compact convex set and is in general position with respect to . However, we can get better results in terms of the number of colors for two specific families.
Theorem 1.2
Let be a finite set of points in the plane and let be the family of halfplanes. Then the Delaunayedges of with respect to can be colored such that every halfplane that contains at least three Delaunayedges contains two differently colored edges. This statement is false if we replace three with two.
We remark that a halfplane may contain an arbitrary number of points from while containing only one Delaunayedge; this is why we consider halfplanes that contain enough Delaunayedges instead of halfplanes that contain enough points.
Theorem 1.3
Let be a finite set of points in the plane and let be the family of bottomless rectangles. Then the Delaunayedges of with respect to can be colored such that every bottomless rectangle that contains at least four points from contains two differently colored edges. This statement is false if we replace four with three.
We also consider coloring Delaunayedges with respect to axisparallel rectangles. However, in this case we only managed to show that by using colors one can guarantee that every rectangle that contains at least three points from contains two differently colored Delaunayedges (proved in Section 2.4). We leave it as an open problem whether constantly many colors also suffice.
Considering coloring of all tuples of vertices of a hypergraph, we first study in Section 3 the case of abstract
hypergraphs (that is, which are not necessarily defined by geometric means). In particular, we prove some results about how polychromatic colorability of
tuples implies polychromatic colorability of tuples, for . Note that it is an open problem whether colorability of vertices implies polychromatic colorability of vertices, for , see [PPT14].Finally, in Section 4 we consider colorings of all tuples of vertices of a hypergraph defined by a point set and a family of geometric regions. Given a finite family of halfspaces we call a region an region if it is the intersection of finitely many halfspaces, such that each of them is a translate of one of the halfspaces in . Our main result about coloring all tuples is the following.
Theorem 1.4
For every dimension and integers , there exists an integer with the following property: given a finite family of halfspaces in , the tuples of every finite set of points in can be colored such that every region that contains at least points from contains a tuple of points of each of the colors.
As a corollary, similar results hold for dimensional axisparallel boxes and homothets of a dimensional convex polytope.
1.2 Related work
We are not aware of any work that considers coloring of Delaunayhyperedges or tuples of vertices as defined in this paper. There is however vast literature on coloring points with respect to Delaunayhyperedges. Here we briefly describe the results that are most relevant to our work.
As we mention above, it is known that we can properly color the points with respect to pseudodisks since their Delaunaygraph is planar. On the other hand, there is no constant for which we can properly color the points with respect to (unit/pseudo)disks containing at least points [PP13, PTT05]. It is an open problem whether a proper coloring is always possible for a large enough for (unit/pseudo)disks [K11, KP16].
Considering the family of axisparallel rectangles, recall that the Delaunaygraph in this case is not necessarily planar. Estimating the chromatic number of the Delaunaygraph in this case is a famous open problem with a large gap between the best known lower bound of
[chenpach] and the best known upper bound of [chan12] (where is the size of ). There is however a proper coloring with respect to rectangles that contain at least points from [eyalrom].There is a proper coloring with respect to bottomless rectangles containing points [K11], and more generally, for bottomless rectangles containing points there is a polychromatic coloring [A13]. For halfplanes containing points there is a coloring [K11], and more generally, for halfplanes containing points there is a polychromatic coloring [SY10]. We remark that it is an open problem whether colorability for some implies polychromatic colorability for some ; see [PPT14]. For more results, see the survey [PPT14] or the webpage [cogezoo].
The geometric hypergraphs that we consider are usually called primal hypergraphs. We briefly mention that one can define dual hypergraphs where the vertices correspond to regions from and the hyperedges correspond to points of the plane. One can even generalize these notions further, in particular one can define intersection hypergraphs where both the vertices and hyperedges correspond to regions from (or from two different families). All of these variants have a vast literature with connections to the notions of coverdecomposability and conflictfree coloring, that are not relevant to this work and hence are not mentioned.
2 Coloring Delaunayedges
Let be a finite set of points in the plane and let be a family of regions. Recall that is the Delaunayhypergraph whose vertex set is and whose hyperedge set consists of all subsets for which there exist a region such that .
Definition 1
The Delaunayedges of with respect to are all the hyperedges of size two in . The Delaunaygraph is the graph whose vertex set is and whose edge set consists of the Delaunayedges of with respect to .
Our goal is to find a coloring such that every region in that contains points from contains two differently colored Delaunayedges. The families that we consider are psuedodisks, axisparallel rectangles, bottomless rectangles and halfplanes.
2.1 Coloring Delaunayedges with respect to pseudodisks
We say that a family of pseudodisks is saturated with respect to a set of points , if there is no region such that is a family of pseudodisks and . Note that every family of pseudodisks can be extended greedily to a saturated family with respect to .
2.1.1 The Delaunaygraph with respect to pseudodisks
Let be a set of points in the plane and let be a family of regions.
It is wellknown [Kvoronoi] that when is the set of homothets of a convex shape (or any other convex pseudodisk family),
then embedding the Delaunaygraph such that every two adjacent points in are connected by a straightline segment results in a plane graph.
For Delaunaygraphs with respect to (nonconvex) pseudodisks this might not hold, but using the HananiTutte theorem it is possible to show that the Delaunaygraph is always planar [buzaglo].
We need the following stronger result, which is proved in a companion paper [PDplane].
Theorem 2.1 ([PDplane])
Given a finite pseudodisk family and a point set , there is a drawing of the Delaunaygraph such that every edge lies inside every pseudodisk that contains both of its endpoints and no two edges cross.
A useful ingredient of the proof of Theorem 2.1 is the following lemma of Buzaglo et al. [buzaglo].
Lemma 1 ([buzaglo])
Let and be two pseudodisks from a pseudodisk family and let and be two curves such that contains and does not contain the endpoints of , for . Then and cross each other an even number of times.
We will use this lemma as well as the next one due to Pinchasi [Pin14], whose proof applies a result of Snoeyink and Hershberger [SH89].
Lemma 2 ([Pin14])
Let be a family of pairwise disjoint sets in the plane and let be a family of pseudodisks. Let be a member of and suppose that intersects exactly members of , one of which is . Then for every integer there exists such that intersects and exactly other sets from , and is a family of pseudodisks.
Definition 2
We say that a family of regions is shrinkable with respect to if for every region with and every there exists another region such that and .
By setting to be , Lemma 2 implies that a family of pseudodisks which is saturated with respect to is also shrinkable with respect to .
2.1.2 Coloring Delaunayedges with respect to pseudodisks
Let be a finite set of points in the plane, let be a family of pseudodisks that is shrinkable with respect to . Theorem 1.1 states that it is possible to color the edges of the Delaunaygraph with four colors, such that every pseudodisk that contains at least three points from contains Delaunayedges of two different colors.
Proof (Proof of Theorem 1.1)
Let be a plane drawing of such that every Delaunayedge lies inside every pseudodisk that contains its endpoints. By Theorem 2.1 such a drawing exists.
Let be a subgraph of the line graph of that is defined as follows. The vertex set of is the set of Delaunayedges, and two Delaunayedges and are adjacent in if they share an endpoint and there is a pseudodisk such that is exactly the three endpoints of and .
Using we draw as follows (we do not distinguish between points and edges and their embeddings). For every Delaunayedge we choose arbitrarily a point in the interior of this edge. Suppose that and are two Delaunayedges that are adjacent in , and let be the common endpoint of and . Then the drawing of the edge of consists of the segment of between and and the segment of between and .
Note that the drawing of does not contain crossing edges, since is a plane graph. However, contains several overlaps between the drawn edges.
Proposition 1
The graph can be redrawn as a plane graph.
Proof
In order to remove overlaps between the edges we redraw the edges with the following procedure (see Figure 1 for an illustration).
First, we slightly inflate every edge and every point in , such that every edge has a very small width and every point is replaced by a very small disk . Denote the neighbors of some point by in clockwise order around . We choose points, called ports, on the intersection of the inflated edge and for each (so in total ports). We denote the ports by , where the ports () are ordered in counterclockwise order for every . Every edge of is drawn as follows: is connected to with a curve inside the inflated , then is connected to inside with a straightline segment, and finally is connected to with a curve inside the inflated . Due to the ordering of the ports, it is easy to achieve that for every the curves do not cross inside the inflated . It is also easy to achieve that these curves do not cross the curves that go towards from (which correspond to the edges of the line graph between the edges incident to ). Also, due to the ordering of the ports, drawings of and (for any ) do not intersect inside .
We also need to show that no segments cross inside . Suppose that this is not the case and denote the endpoints of the respective edges (after a suitable reindexing) by . Let and be the pseudodisks in whose intersections with is and , respectively. Since the Delaunayedges are drawn such that they are contained in every pseudodisk that contains their endpoints, it follows that our drawn curves lie inside and , respectively. Moreover, by Lemma 1 they must cross an even number of times. Since the parts outside are disjoint, and inside the two segments can cross at most once, we get a contradiction. Therefore, we have proved that with this drawing is a plane graph.
Since is a planar graph, it is colorable. We use a coloring of to color the Delaunayedges of . Now suppose that is a pseudodisk that contains at least three points. Since is shrinkable, it follows that there is a pseudodisk that contains exactly three points from . Lemma 2 also implies that each of the three points from inside is connected by a Delaunayedge to one of the other two points. Therefore, contains two Delaunayedges that share a common endpoint and thus are colored with different colors.
Note that Theorem 1.1 implies the same for saturated pseudodisk families as a saturated family is always shrinkable. The following result concerning homothets of a convex compact set is also implied by Theorem 1.1.
Let be a convex compact set and let denote its boundary. We say that a set of points is in general position with respect to , if there is no homothet of that contains more than three points of on its boundary and no two points in define a line that is parallel to a linesegment which is contained in .
Corollary 1
Let be a convex compact set and let be a finite planar set of points that is in general position with respect to . Then it is possible to color the Delaunayedges of with respect to the family of homothets of such that every homothet of that contains at least three points from contains Delaunayedges of different colors.
Proof
Let be a finite subfamily of the set of homothets of such that if is a homothet of , then there is such that . One can show along the lines of the proof of Lemma 2.3 in [AKV15] that this subfamily is shrinkable with respect to .
We may assume that for every the boundaries and intersect a finite number of times. Indeed, if there are overlaps between boundaries, then we can slightly inflate one of the homothets, say , without changing since is closed.
It follows from the next folklore lemma (see, e.g., [Ma2000, Corollary 2.1.2.2]), that is a family of pseudodisks.
Lemma 3
Let be a convex and compact set and let and be homothets of . Then if and intersect finitely many times, then they intersect in at most two points.
Since is also shrinkable, the corollary now follows from Theorem 1.1.
2.2 Coloring Delaunayedges with respect to halfplanes
Let be a finite set of points in the plane and let be the family of halfplanes. Note that a halfplane may contain an arbitrary number of points from while containing only one Delaunayedge of , see Figure 1(a) for an example.
Therefore a claim along the lines of, e.g., Theorem 1.3 is impossible. Thus, Theorem 1.2 states that the edges of the Delaunaygraph can be colored such that every halfplane that contains at least three Delaunayedges contains two differently colored edges..
Proof (Proof of Theorem 1.2)
For simplicity we assume that the points are in general position, as it is easy to modify the proof so that it works for points that are not in general position.
Let be a finite set of points in the plane and let be the family of halfplanes. The convex hull of , , is a convex polygon whose vertices are points from . Note that for each of the Delaunayedges in the Delaunaygraph at least one of the two endpoints is a vertex of . Thus, consists of the sides of and stars centered at each of the vertices of .
Starting from an arbitrary vertex of we go over all the Delaunayedges adjacent to this vertex in a counterclockwise order and color them alternatively with red and blue. Then we move to the next vertex of in a clockwise order. Repeat the same process (the first edge adjacent to it is already colored), and so on until all the edges are colored (see Figure 1(b) for an example).
Let be the Delaunayedges listed in the order they were visited by the coloring algorithm. Note that every pair of consecutive edges are of different colors, except perhaps and which are of the same color if
is odd.
Proposition 2
If a halfplane contains two Delaunayedges and , , then contains the edges or the edges .
Proof
Suppose first that and share a vertex of as an endpoint and denote it by . Assume without loss of generality that all the edges are incident to and let denote the other endpoint of , . Assume for contradiction that does not contain for some . It follows that are in convex position such that this is also their clockwise or counterclockwise order around their convex hull. However, then there cannot be a halfplane that separates from contradicting the fact that is a Delaunayedge.
Suppose now that and are disjoint and let and be vertices of such that is an endpoint of and is an endpoint of . Let be the vertices of in the clockwise order from to , and assume without loss of generality that contains all of them (this can be assumed since if does not contain all of them, then it must contain all the vertices of from to ). We may also assume that are incident to the vertices in .
We claim that must contain each of these edges. Indeed, suppose that does not contain , for some . If none of and is an endpoint of , then let be one of its endpoints and let be its other endpoint. As in the previous case, it follows that are in convex position such that this is also their clockwise order around their convex hull. However, then there cannot be a halfplane that separates from . Finally, suppose that one endpoint of is, without loss of generality, and let be its other endpoint. Denote also by the other endpoint of . Then, as before, it follows that are in convex position in this cyclic order. However, there should be a halfplane that separates from the other two vertices which is impossible.
Suppose that is a halfplane that contains at least three Delaunayedges. Then it follows from Proposition 2 that contains at least three consecutive Delaunayedges and therefore it contains two edges of different colors.
To see that the constant three in the theorem is tight, note that when is a set of an odd number of points in convex position, then it is impossible to color the Delaunayedges of such that every halfplane that contains two Delaunayedges contains edges of both colors.
2.3 Coloring Delaunayedges with respect to bottomless rectangles
A bottomless rectangle is the set of points for some numbers such that . Let be the set of bottomless rectangles and let be a finite set of points in the plane. As in the previous section, we may assume that no two points in share the same  or coordinates. It is known [K11] that it is possible to color the points in with two colors such that every bottomless rectangle that contains at least four points from contains two points of different colors. This easily implies that the Delaunayedges of can be colored with two colors such that every bottomless rectangle that contains at least five points from contains two Delaunayedges of different colors. Indeed, color first the points of such that every bottomless rectangle that contains four points from is nonmonochromatic. Then color every Delaunayedge with the color of its leftmost endpoint. Suppose that contains at least five points from and let be the th point in from left to right. There is a bottomless rectangle such that , therefore there are two differently colored points among these points. Since is a Delaunayedge for every , it follows that contains two Delaunayedges of different colors.
Next we prove Theorem 1.3. Namely, that containing four points from suffices for a bottomless rectangle to contain Delaunayedges of different colors and this is tight. The proof is similar to proof of the aforementioned statement in [K11].
Proof (Proof of Theorem 1.3)
We add the points in onebyone from bottom to top. Note that any Delaunayedge that is introduced when a point is added, remains a Delaunayedge when adding the remaining points.
At any moment call an edge
neighborly if it connects points that are next to each other in the horizontal order of points. Denote the points added so far by their order from left to right, , and let be the last point to be added. Thus, the neighborly edges are , for . We maintain the invariant that there are no three consecutive neighborly edges of the same color. This is trivial to obtain for , so assume that . When is added, it introduces either one (if or ) or two new neighborly edges (and causes an existing edge to cease being a neighborly edge). In the first case we color the new edge with a color that is different from the color of the single neighborly edge that shares an endpoint with the new edge. In the second case, if (resp., ), then we color the new edges and (resp., and ) with a color that is different from the color of (resp., ). When there are two cases to consider: If and are of the same color, then we color the new edges and with the opposite color. Otherwise, if and are of different colors, then is colored with the color of and is colored with the color of . It is easy to verify that the invariant that there are no three consecutive neighborly edges of the same color is maintained in all cases.Suppose now that is a bottomless rectangle that contains at least four points. When the topmost point from was added, then all the points in formed consecutive neighborly edges. Therefore, not all of these edges are of the same color. Since these edges remain Delaunayedges (perhaps nonneighborly) when the remaining points are added, contains Delaunayedges of both colors.
To see that it is not enough to consider bottomless rectangles that contain three points from , consider the point set in Figure 3 and assume it is colored as required.
Assume without loss of generality that is red. Then and must be both blue. This implies that must be red, and so must be blue. Therefore, all the Delaunayedges in the bottomless rectangle that contains , and are blue, a contradiction.
2.4 Coloring Delaunayedges with respect to axisparallel rectangles
For a partially ordered set we say that is an immediate predecessor of if and there is no such that (where means that and ). The directed Hasse diagram of is the directed graph is an immediate predecessor of . The following is easy to prove.
Lemma 4
Let be a partially ordered set such that is finite. Then it is possible to color the edges of the directed Hasse diagram of with colors such that there is no monochromatic path of length two.
Proof
Denote . It is enough to prove the lemma for , . We prove by induction on . For the lemma holds. Let be a poset where and . Let be a linear extension of and let be the elements of ordered according to . Set and . By the induction hypothesis the edge set of each of the directed Hasse diagrams of and can be colored with (the same set of) colors such that there is no monochromatic path of length two.
By using the same coloring for the edges of the directed Hasse diagram of and introducing a new color for the edges we obtain the required coloring.
Remark.
The number of colors in Lemma 4 cannot be an absolute constant. In fact, for any integer there is no integer such that it is possible to color the edges of the directed Hasse diagram of any finite poset with colors such that there is no monochromatic path of length . To see this, observe that if such an integer would exist, then it would be possible to properly color the vertices of the Hasse diagram of any poset with colors which is known to be false [chenpach, KN91]. Indeed, it is easy to verify that by assigning each vertex the color where is the length of the longest monochromatic path ending at (in the directed Hasse diagram of ) and is the color of that path one obtains a proper coloring with colors.
Corollary 2
Let be a finite set of points in the plane and let be the family of axisparallel rectangles. Then it is possible to color the edges of the (nonplanar) Delaunaygraph with colors such that every axisparallel rectangle that contains at least three points from contains Delaunayedges of different colors.
Proof
We may assume without loss of generality that no two points in share the same  or coordinates.^{3}^{3}3Note that we cannot just take an arbitrary perturbation, as we did for points. For example, if we have 3 collinear points inside a rectangle (thus two Delaunayedges) we might introduce a new Delaunayedge with an arbitrary perturbation. Indeed, this can be obtained by a mapping for a very small (that depends on ) without changing the hyperedges of .
Orient every edge of from left to right. Let where if and . Let where if and . Note that the edge set of the oriented Delaunaygraph is the union of the (disjoint) edge sets of the directed Hasse diagrams of and . Therefore, it follows from Lemma 4 that we can color these edges with colors such that there is no monochromatic path of length two.
Let be an axisparallel rectangle that contains at least three points from and let be the leftmost points in , ordered from left to right. Then and are edges in the oriented Delaunaygraph . If they belong to different Hasse diagrams, then they are of different colors. Otherwise, if they belong to the same Hasse diagram, then they are of different colors since they form a path of length two.
3 Coloring all tuples of abstract hypergraphs
Henceforth we consider colorings of all the (unordered) tuples of points. In this section we show some general results that do not rely on geometry assumptions. First, intuitively, as the number of all tuples increases considerably by increasing , the coloring problem should become easier or at least not become (significantly) harder. The next claim confirms this intuition in the following sense: proper colorability of tuples implies proper colorability of tuples for (while the corresponding might increase).
Proposition 3
Let be a hypergraph and suppose that for some positive integers , and the tuples of the vertices of can be colored such that every hyperedge that contains at least vertices contains two differently colored tuples. Then for every integer there exists an integer such that the tuples of the vertices of can be colored such that every hyperedge that contains at least vertices contains two differently colored tuples.
Proof (Proof of Proposition 3)
We construct a coloring of tuples from the coloring of tuples. The color of a tuple is red if every tuple of the vertices of is colored with the same color by , and blue otherwise. By Ramsey’s Theorem (for hypergraphs) we can choose such that in any coloring of all tuples of a set of size there is a monochromatic tuple, i.e., whose every subset of size got the same color. This way in every set of size at least (and so also every hyperedge of size at least ) contains a red tuple. Also, in the coloring every hyperedge in of size was not monochromatic, thus such a hyperedge also contains a blue tuple.
One can of course can get better bounds on if we construct a coloring more carefully instead of using Ramsey’s theorem. The following is such a claim for about polychromatic colorability. We denote by the set and by the element subsets of a set .
Proposition 4
Let be a hypergraph and let and be integers such that the vertices of can be colored such that every hyperedge of that contains at least vertices contains a vertex of each of the colors. Then for every integer there are integers and such that there exists a coloring of the tuples of the vertices of such that every hyperedge of that contains at least vertices contains a tuple of each of the colors. In particular, the above holds for and (note that and if then ).
Proof
Let be a coloring of the vertices of as above, and set . We will construct a coloring of the tuples of vertices of such that is a mapping to the union of the element subsets of and the element subsets of , for . Thus, there are indeed colors. The color of each tuple will be determined by the multiset of the colors of its vertices, , as follows:

If no color appears twice in (only possible if ), then is the element subset of defined by those colors, i.e., .

If exactly one color (say, ) appears at least twice in and each of the other colors in appears once, then will be an element subset of , determined as follows. Define a bijection between and such that for and for . Then .

Otherwise, define arbitrarily.
Now we need to show that all colors appear in a hyperedge of size . Note that among the vertices of we have all colors from since . Let be a set of vertices such that . If , then the tuples of vertices from are colored by all colors in .
Now let be an element subset of , for some . By the pigeonhole principle there is a subset of vertices that are colored by the same color . Suppose that is a tuple that consists of vertices from and every vertex such that . Since is a bijection there are such vertices and it follows that .
Surprisingly, such simple arguments fail for . Below we give an example for , and .
Suppose we are given a coloring of every pair of the vertex set of a hypergraph such that every hyperedge that contains at least vertices contains a pair of each of the colors, and we want to construct a coloring of the triples of vertices with the same property (possibly for a larger ). We show that this is not possible if the color of each triple depends only on the colors of the three pairs in it:
Proposition 5
There is no mapping from the element multisets of to such that the following holds for some . For every hypergraph if is a coloring of the pairs of vertices of such that every hyperedge that contains at least vertices contains pairs of all three colors, then and defines a coloring of the triples of vertices of such that every hyperedge that contains at least vertices contains triples of all three colors (where for every triple of vertices ).
Proof
Suppose for contradiction that and as above exist. Note that there are ten element multisets of , that is, there are ten possible colorings of the three pairs of a triple by colors: , , , , , , , , and .
Let be a hypergraph such that and let and be two sets of pairs of vertices, such that no vertex belongs to two pairs in . Let consists of the hyperedge (all the vertices) and all the subsets of of size that contain a pair from and a pair from .
Suppose that is a coloring of the pairs in such that the color of every pair in is , for , and the color of every other pair is . Then clearly every hyperedge of size at least contains a pair of each of the colors. Note that the colors of the three pairs defined by a triple of vertices is either , or . Consider the coloring of the triples as defined by and . Since the hyperedge must contain a triple of every color, it follows that maps each of the above multisets of colors to a different color.
Suppose now that is a hypergraph such that and let and be two disjoint subsets of vertices. Let consists of the hyperedge (all the vertices) and all the hyperedges and , for some . Let be the pairs of vertices of the form or (for ), Let be the pairs of vertices of the form , and let be the (remaining) pairs of vertices of the form (for ).
Suppose that is a coloring of the pairs in such that the color of every pair in is , for . Then clearly every hyperedge of size at least contains a pair of each of the colors. Note that the colors of the three pairs defined by a triple of vertices is either , or . Consider the coloring of the triples as defined by and . Since the hyperedge must contain a triple of every color, it follows that maps each of the above multisets of colors to a different color. Therefore, is mapped to a color different from the colors of and . Similarly, if we color the pairs in with , the pairs in with and the pairs in with , then we conclude that maps to a color different from the colors of and . And if we color the pairs in with , the pairs in with and the pairs in with , then we conclude that maps to a color different from the colors of and . Thus, according to the color of is different from the colors of , and . However, we observed before that maps these three multisets of colors to different colors, a contradiction.
It is possible to change the constructions in the proof of Proposition 5 such that every hyperedge that contains at least vertices contains many pairs of each of the colors, so even such an assumption is not strong enough to guarantee a coloring for triples. One might try to get a coloring of tuples from stronger conditions on the coloring of the tuples, however, we do not further study this question.
4 Coloring all tuples of geometric hypergraphs
In this section we consider mainly three (closely related) types of families of geometric regions, regions, homothets of a convex polytope and axisparallel boxes.
Given and , we want to color all tuples of such that no member of that has many points of is monochromatic, i.e., it contains two tuples that are colored with different colors.
Definition 3
Given a finite family of halfspaces we call a region an region if it is the intersection of finitely many halfspaces, such that each of them is a translate of one of the halfspaces in .
Note that every such region can be expressed as an intersection of at most halfplanes (at most one translate of each ).
Throughout this section we assume, without loss of generality, that any given point set is in general position in the sense that there are no two points in that are contained in for some , where
denotes the hyperplane that bounds the halfspace
. Indeed, otherwise after a small perturbation of the points of for every subset that is the intersection of and some region, the perturbed subset is still the intersection of (the perturbed) and some (possibly different) region.As mentioned in the Introduction, the requirement that in Theorem 1.4 is essential. Indeed, Chen et al. [chenpach] proved that there is no absolute constant with the property that every set of points in the plane can be colored such that every axisparallel rectangle that contains at least points from contains points of both colors. Note that axisparallel rectangles can be defined as the regions of four halfplanes. Several other constructions about polychromatic colorings/coverdecomposition can also be realized as the regions of four halfplanes; see [PTT05, P10]. If there are only three halfspaces in that are in general position, then after an affine transformation the problem becomes equivalent to coloring points with respect to octants (nongeneral position and different dimension can only make the problem easier). For this case, it was shown in [KP11] that exists for every (which implies the same for and although these can be shown directly as well).
Note that a homothet of a convex polytope is a region when is the family of all the supporting halfspaces of . Thus Theorem 1.4 implies:
Theorem 4.1
For every dimension , integers , and a convex polytope in there exists an integer with the following property: the tuples of every finite point set in can be colored such that every homothet of that contains at least points from contains a tuple of points of each of the colors.
As mentioned above, it is impossible to color points (that is, the case ) with respect to regions in the plane such that every region that contains a constant number of points (for some absolute constant) contains a point of each color. However, for homothets of a convex polygon it might be possible. The following is an equivalent restatement (by the selfcoverability of convex polygons [KP13]) of a conjecture that first appeared in [AKV15].
Conjecture 1
For every convex polygon and a positive integer there exists an integer with the following property: the points of every finite point set in the plane can be colored such that every homothet of that contains at least points from contains a point of each of the colors.
We note that Theorem 4.1 for would be implied by Conjecture 1 using Theorem 4.2 (see below), however, Conjecture 1 was verified only for triangles [K13] and parallelograms [AKV15]. Recently, a relaxation of this conjecture was proved in [KP16], namely, that it is possible to color the points such that each homothet with enough points contains points of at least two colors. The conjecture fails in for every polytope [P10].
We have seen that for general hypergraphs we could not show that polychromatic colorability of tuples implies the same for tuples, . On the other hand we give a simple argument proving this for regions. This argument can also be adopted to work for most geometric families.
Theorem 4.2
Given a family of halfspaces in and a positive integer , suppose that for some positive integer there exists an integer with the following property: the tuples of every finite point set in the plane can be colored such that every region that contains at least points from contains a tuple of each of the colors. Then the same property holds when exchanging with an integer and with .
Proof
Let be a halfspace in and let be a translate of that contains all the points in . We fix an order of the points in in decreasing order of their distance from (recall that the points are in general position and therefore no two points can be at the same distance). Consider a coloring of the tuples having the property assumed by the theorem and let be a tuple of points, where the points are listed according to the determined order (that is, is the farthest from and so on). Set the color of to be the color of the tuple .
Let and and suppose that is an region that contains at least points from , denote them by according to their order. Since is in general position, it is possible to continuously translate such that it is still a translate of , and obtain a translate of such that . Note that is an region that contains at least points and therefore for each of the colors it contains a tuple of that color. Let be such a tuple. Then is a tuple of the same color that is contained in . Therefore, contains a tuple of each of the colors.
Remark.
Note that in the above proof we have only used that the points of can be ordered as such that if for some , then also for some . This implies that similar arguments work not only for regions but for other families as well, especially if we do not insist on .
In order to prove Theorem 1.4, we reduce the problem of coloring with respect to regions to coloring with respect to axisparallel boxes (which are possibly higher dimensional). Note that axisparallel boxes are also regions.
For a point we denote by the th coordinate of . An axisparallel dimensional (closed) halfspace is a halfspace of the form or for some and a number . A dimensional box is the intersection of dimensional axisparallel halfspaces. Theorem 1.4 follows from the next result.
Theorem 4.3
For every integer and positive integers there exists an integer
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