1 Introduction
A range capturing hypergraph is a geometric hypergraph defined by a finite point set in the plane and a family of subsets of , called ranges. Possible ranges are for example the family of all axisaligned rectangles, all horizontal strips, or all translates of the first (northeast) quadrant. Given the points and ranges , the hypergraph has as its vertex set and a subset forms a hyperedge whenever there exists a range with . That is, we have points in the plane, and a subset of points forms a hyperedge whenever these vertices and no other vertices are captured by a range.
For a positive integer , we are then interested in the uniform subhypergraph given by all hyperedges of size exactly . In particular, we investigate polychromatic vertex colorings in colors of for different families of ranges , different values of . A vertex coloring is proper if every hyperedge contains at least two vertices of different colors. A vertex coloring is polychromatic if every hyperedge contains at least one vertex of each color. Note that a coloring is proper if and only if it is polychromatic. This case is sometimes called property B for the hypergraph in the literature. Further, a uniform hypergraph (hence, graph) admits a polychromatic 2coloring if and only if it is bipartite.
In this paper, we mostly focus on polychromatic colorings for range capturing hypergraphs with given range family . In particular, we investigate the following question.
Question 1.
Given and , what is the smallest such that for every finite point set the hypergraph admits a polychromatic coloring?
Of course, , while is also possible. Indeed, for all range families considered here, it holds that and we either show that for every or already . Note that in the latter case, there are range capturing hypergraphs that are not properly colorable, even for arbitrarily large uniformity .
The motivation for studying polychromatic colorings of such geometric hypergraphs comes from questions of coverdecomposability and conflictfree colorings. The interested reader is refered to the survey article [13] and the excellent website [1] which contains numerous references.
1.1 Related Work
There is a rich literature on range capturing hypergraphs, their polychromatic colorings, and answers to creftype 1. Let us list the positive results (meaning for all ) that are relevant here, whilst defining the respective families of ranges.
On the contrary, let us also list the negative results (meaning for some ) that are relevant here. In all cases, it is shown that already , meaning that there is a sequence of hypergraphs such that for each we have for some point set (in this case we say that can be realized with ) and admits no polychromatic coloring. One such sequence are the ary tree hypergraphs, defined on the vertices of a complete ary tree of depth , where for each nonleaf vertex, its children form a hyperedge, and for each leaf vertex, its ancestors form a hyperedge (introduced by Pach et al. [12]). A second such sequence is due to Pálvölgyi [14] (published in [11]), for which we do not repeat the formal definition here and simply refer to them as the 2size hypergraphs as their inductive construction involves hyperedges of two possibly different sizes.
Finally, for axisaligned rectangles it is also known that . See Theorem 2 below. However, the only known proof of Theorem 2 is a probabilistic argument and we do not have any explicit construction of a sequence of uniform hypergraphs defined by axisaligned rectangles that admit no polychromatic coloring.
Theorem 2 (Chen et al. [7]).
For the family of all axisaligned rectangles it holds that .
That is, for every there exists a finite point set such that for every coloring of some axisaligned rectangle contains points of , all of the same color.
1.2 Our Results
In this paper we consider range families that are the union of two range families , . The corresponding hypergraph is then the union of the hypergraphs and on the same vertex set . Clearly, if is polychromatic colorable, then so are and . But the converse is not necessarily true and this shall be the subject of our investigations.
In Section 2 we show that if and admit socalled hitting cliques, then we can conclude that for . This is for example the case for all horizontal (respectively vertical) strips, but already fails for all southwest quadrants. In Section 3 we then consider all possible families of unbounded axisaligned rectangles, such as axisaligned strips, all four types of quadrants, or bottomless rectangles. We determine exactly for which subset of those, when taking as their union, it holds that .
In particular, we show in Section 3.1 that for all when is the union of all bottomless rectangles and all horizontal strips. Our proof gives a new sequence of uniform hypergraphs that admit a geometric realization for simple ranges, but do not admit any polychromatic coloring. On the positive side, we show in Section 3.2 that (up to symmetry) all other subsets of unbounded axisaligned rectangles (excluding the above pair) admit polychromatic colorings for every . Here, our proof relies on socalled shallow hitting sets and in particular a variant in which a subset of hits every hyperedge defined by at least once and every hyperedge defined by at most a constant (usually or ) number of times.
Assumptions and Notation.
Before we start, let us briefly mention some convenient facts that are usually assumed, and which we also assume throughout our paper: Whenever a range family is given, we only consider points set that are in general position with respect to . For us, this means that the points in have pairwise different coordinates, pairwise different coordinates, and also pairwise different sums of  and coordinates. Secondly, all range families that we consider here are shrinkable, meaning that whenever a set of points is captured by a range in , then also some subset of points of is captured by a range in . This means that for every polychromatic coloring of , every range in capturing or more points of , contains at least one point of each color, thus giving as mentioned above. Finally, for every set captured by a range in , we implicitly associate to one arbitrary but fixed such range with . In particular, we shall sometimes consider the range for a given hyperedge of .
2 Polychromatic Colorings for Two Range Families
Let be two families of ranges, for each of which it is known that for any . We seek to investigate whether also for we have . First, we identify a simple sufficient condition.
For fixed , we say that we have hitting cliques if the following holds. For every there exist pairwise disjoint subsets of such that every hyperedge of fully contains at least one such subset. Clearly, if we have hitting cliques, then since we can simply use all colors on each such subset (and color any remaining vertex arbitrarily).
Theorem 3.
For fixed , suppose that we have hitting cliques with respect to and hitting cliques with respect to . Then for it holds that .
Proof.
Consider , a set of hitting cliques of and a set of hitting cliques of . Let be the hypergraph with one hyperedge for each vertex such that for , we have if and only if . Removing the empty hyperedges from , we have that is a bipartite multigraph together with some additional loops. As every vertex of has degree exactly (loops contribute only once to a vertex degree), it follows that admits a proper edgecoloring [5], which interpreted as a coloring of gives the result since every clique gets all colors. ∎
As a corollary, we easily reprove the upper bound on from [3] when consists of all axisparallel strips (or more generally, all cones with apex at one of two fixed points).
Corollary 4.
For the family of all axisaligned strips and any we have .
Proof.
For , the family of all vertical strips has hitting cliques by grouping always points with consecutive coordinates, leaving out the last up to points. Symmetrically, we have hitting cliques for the family of all horizontal strips. As , we have by Theorem 3. ∎
Somewhat unfortunately, hitting cliques appear to be very rare. Already for the range family of all southwest quadrants, for which one can easily show that , we do not even have hitting cliques for any . This will follow from the following result, which will also be useful later.
Lemma 5.
Let be a rooted tree, and be the hypergraph on where for each leaf vertex its ancestors (including itself) form a hyperedge. Then can be realized with the family of all southwest quadrants.
Moreover, the root is the bottommost and leftmost point and the children of each nonleaf vertex lie on a diagonal line of slope .
Proof.
We do induction on the height of , with height being a trivial case of a single vertex. For height at least , remove the root from to obtain new trees , each of smaller height and rooted at a child of . By induction, there are point sets in the plane for each , , with each respective root being bottommost and leftmost. We scale each of these points sets uniformly until the bounding box of each of them has width as well as height less than . For every , we put the point set for into the plane so that the root of has the coordinate . Finally, we place in the origin. This gives the desired representation. ∎
Corollary 6.
For , the family of all southwest quadrants, and any , we do not have hitting cliques.
Proof.
Take the rooted complete ary tree of depth , for which is realizable with southwest quadrants by Lemma 5. By induction on , we show that does not have hitting cliques. This is trivial for . Otherwise, any collection of disjoint subsets either avoids the root , or pairs with a vertex in one of the subtrees of below . In any case, there is a subtree below , none of whose vertices is paired with and hence, there exist hitting 2cliques of . Note that is a complete ary tree of depth so it contains as a subtree. But then admits hitting cliques too. A contradiction to induction hypothesis.
∎
Corollary 7.
For , the family of all halfplanes, and any , we do not have hitting cliques.
Proof.
By a result of Middendorf and Pfeiffer [10], every range capturing hypergraph for southwest quadrants can also be realized by halfplanes and the result follows from Corollary 6. ∎
To summarize, parallel strips have hitting cliques, but quadrants do not. Hence, we can not apply Theorem 3 to conclude that maybe we have when we consider to be the union of all quadrants of one direction and all parallel strips of one direction. In Section 3 we shall prove that indeed for the union of all quadrants and strips, however only provided that the strips are axisaligned. In fact, if they are not, this is not necessarily true.
Corollary 8.
Let be the family of all southwest quadrants and be the family of all diagonal strips of slope .
Then for we have .
Proof.
Given , consider the rooted complete ary tree of depth . By Lemma 5, can be placed in the plane such that for each leaf vertex, its ancestors (including itself) are captured by a southwest quadrant, and for each nonleaf vertex, its children are captured by a diagonal strip. Using a slight pertubation of the points, we can ensure that the lines of slope guaranteed by Lemma 5 are pairwise distinct. Hence, every ary tree hypergraph can be represented with . By [12] the hypergraph does not admit a polychomatic coloring for every , which gives the result. ∎
3 Families of Unbounded Rectangles
In this section we consider the following range families of unbounded rectangles:

all (axisaligned) southwest quadrants ,

similarly all southeast , northeast , northwest quadrants,

all horizontal strips , vertical strips , diagonal strips of slope ,

all bottomless rectangles , and finally all topless rectangles .
Observe that if a point set is captured by a southeast quadrant , then it is also captured by a bottomless rectangle having the same top and left sides as and whose right side lies to the right of every point in the vertex set. Analogous statements hold for other quadrants and vertical strips. Further, note that each of the above range families, except the diagonal strips , is a special case of the family of all axisaligned rectangles. Recall that for the family of all axisaligned rectangles, it is known that [7]. Here we are interested in the maximal subsets of so that for the union of all these ranges it still holds that for all . In fact, we shall show that for we have , strengthening the result for axisaligned rectangles [7]. On the other hand, for , i.e., the union of all quadrants and axisaligned strips, we have for all , strengthening the results for southwest quadrants [8] and axisaligned strips [3]. Secondly, for , i.e., the union of bottomless and topless rectangles (which also contains all quadrants and all vertical strips), we again have for all , thus strengthening the result for bottomless rectangles [4]. Using symmetries, this covers all cases of the considered unbounded axisaligned rectangles. We complement our results by also considering the diagonal strips and recall that we already know by Corollary 8 that for we have .
3.1 The Case with no Polychromatic Coloring: Bottomless Rectangles and Horizontal Strips
Definition 9.
For every , the uniform hypergraph is defined as follows. First we define a rooted forest consisting of trees whose vertices are partitioned into a set of the socalled stages. The vertices of a stage will be totally ordered and we denote this ordering by . All vertices of a stage will have the same distance to the root of the corresponding tree, we refer to this distance as the level of . Every stage on level will consist of vertices.
To define the rooted forest and the stages, we start with roots, one for each tree in . They build the unique stage on level and they are ordered in an arbitrary but fixed way. After that, for every already defined stage on level and every subset , we add a new stage on level consisting of new vertices so that every vertex in gets exactly one child from and the vertices of are ordered by as their parents by . Informally speaking, every vertex in gets a child for every subset of in which it occurs.
Now we can define the hypergraph . The vertex set is exactly . There are two types of hyperedges. First, stagehyperedges : for every stage , each consecutive vertices in constitute a stagehyperedge. Second, the pathhyperedges : every roottoleaf path in forms a pathhyperedge. Then, the set of hyperedges is defined as . Note that is indeed uniform. For a vertex , let denote the root of the tree in containing , and denote the vertices on the path from to in .
Theorem 10.
For every the uniform hypergraph admits no polychromatic coloring with 2 colors.
Proof.
We show that every coloring of that makes all stagehyperedges polychromatic necessarily produces a monochromatic pathhyperedge. Let be such a coloring. The key observation is that a stage on level (i.e., one that contains vertices) can be partitioned into disjoint stagehyperedges and hence, it contains at least red vertices.
We prove for that there is a stage on level and a subset such that and for every vertex , the vertices in are all red. For , the stage consisting of roots contains at least red roots and these vertices have the desired property. Suppose the statement holds for some . Consider the stage and a set of red points in it. By definition, each of these points has its parent in and hence, all vertices in are red. So the statement holds for , too. By induction, it holds for and hence, there is a vertex on level (i.e., a leaf) whose roottoleaf path is completely red. So admits no polychromatic 2coloring. ∎
Theorem 11.
For every the uniform hypergraph admits a realization with bottomless rectangles and horizontal strips.
Proof.
For a point , let and denote its  respectively coordinate. We call a sequence of points ascending (respectively descending) if and (respectively and ). Writing about the vertices of a stage , we always refer to their ordering in . We shall embed each stage of into a closed horizontal strip, denoted , in such a way that whenever . Note that this way, the embedded stages are vertically ordered with some available space between any two consecutive ones.
First, we embed the roots of , i.e., the unique stage on level , as an ascending sequence in a horizontal strip for this stage. After that, we choose some stage that has already been embedded but the stages containing its children not yet. In one step we embed as follows. We pick a thin horizontal strip between and the strip above (if it exists) and within identify disjoint horizontal strips . Then, every is embedded inside so that every vertex gets initially the same coordinate as its parent and the vertices of build an ascending sequence in . After that, for every we slightly shift all children of to the right so that they build a descending sequence but the ordering of coordinates relative to all other points remains unchanged.
The arising embedding ensures the following two properties. First, every stagehyperedge is captured by a horizontal strip. Second, for every vertex , the bottomless rectangle with topright corner and on the left side captures exactly . The first property holds since every stage is embedded as an ascending sequence (along ) in a thin horizontal strip and these horizontal strips are pairwise disjoint. We prove the second property by induction on the level of the unique stage containing . If is the stage on level , then is a root and clearly contains only . Otherwise, let be the parent of and be the stage containing . Then arises from by extending its topright corner from to , giving their difference an Lshape as illustrated in Figure 1. We claim that contains only the point from , which by induction then gives , as desired.
To see that , first consider the step in which was embedded. Since lies slighty to the right of and the vertices of form an increasing sequence, no further vertex of belongs to . All vertices lying vertically between and also lie between and so in this step all such vertices have their parent in (or they belong to but we have already excluded this case). Each of these vertices lies slightly to the right of its parent. Since , the only vertices that could lie in above are the children of . But they form a decreasing sequence so is the only such vertex. Since lies slightly to the right of , there is also no further point in below in this step. Finally, each further point (i.e., embedded after this step) is embedded above and ever so slightly to the right of its parent. As every vertex in has all its children already embedded, no further points are embedded into . So the hypergraph can be realized by bottomless rectangles and horizontal strips. ∎
3.2 The Cases with Polychromatic Colorings
First, recall the result of Ackerman et al. [2] that for the range family of all axisaligned squares, we have . This already seals the deal for bottomless and topless rectangles.
Theorem 12.
For the family of all bottomless and topless rectangles, we have for all .
Proof.
Let be a finite point set and let be arbitrary. For every bottomless (respectively topless) rectangle capturing a hyperedge of , we introduce a bottom (respectively top) side below the bottommost (respectively above the topmost) point in so that these rectangles are bounded now. After that we stretch the plane horizontally until the width of every aforementioned rectangle becomes larger than its height and obtain the point set . This stretching preserves the ordering of  and coordinates of the points so that the set of hyperedges captured by remains the same. Finally, we pick every (now bounded) bottomless (respectively topless) rectangle capturing a hyperedge of and shift its bottom (respectively top) side down (respectively up) until it becomes a square. Now for every hyperedge in , there is an axisaligned square capturing it and hence, a hyperedge in . Thus, each polychromatic coloring of yields a polychromatic coloring of and this concludes the proof. ∎
For the remaining cases, we utilize socalled shallow hitting sets. For a positive integer , a subset of vertices of a hypergraph is a shallow hitting set if every hyperedge of contains at least one and at most points from . It is known for example that for being the family of all halfplanes, every range capturing hypergraph admits a shallow hitting set [15], which implies that in this case. In general, we have the following.
Lemma 13 (Keszegh and Pálvölgyi [9]).
Suppose that for a shrinkable range family , every hypergraph admits a shallow hitting set. Then .
Proof.
Doing induction on , we first observe that the claim holds for , as in this case . For let and consider a shallow hitting set of . Then every hyperedge contains at least points of . Since is shrinkable, for every hyperedge of , there is a hyperedge of with . So if is polychromatic in some coloring, then is as well. By induction, admits a polychromatic coloring in colors . Taking as the th color, gives a polychromatic coloring of . ∎
Remark 14.
Lemma 13 states that if shallow hitting sets exist (for a global constant ), then . However, it is not clear whether the converse is also true, for example when is the family of all bottomless rectangles. Keszegh and Pálvölgyi [9] construct for this family range capturing hypergraphs without shallow hitting sets, but their constructed hypergraphs are not uniform. In fact, it follows from the proof of Corollary 4 that axisaligned strips do in fact admit shallow hitting sets, since every hyperedge of of size or is hit by at most three of the hitting cliques, and thus contains color of the resulting coloring at least once and at most three times. To the best of our knowledge, it is open whether all admit shallow hitting sets for the bottomless rectangles .
Recall that for the family of all northwest quadrants we have , and observe that in such a polychromatic coloring, every color class is a shallow hitting set. Besides northwest quadrants, we want to consider other range families. Thus, we are interested in shallow hitting sets for that additionally do not hit other ranges, such as axisparallel strips or other quadrants, too often.
Lemma 15.
For the family of all northwest quadrants, every hypergraph admits a shallow hitting set such that the points in have decreasing coordinates and decreasing coordinates, and

[(i)]

is the leftmost point of the topmost points in ,

the hyperedge consisting of the topmost vertices of is hit exactly once,

for any two consecutive points in , the bottomless rectangle with topright corner and on the left side satisfies , and

for any three consecutive points in , the axisaligned rectangle with topright corner and bottomleft corner satisfies .
Proof.
For each hyperedge of consider a fixed northwest quadrant capturing these points of . These quadrants can be indexed along their apices with decreasing coordinates (and hence also coordinates). I.e., contains the topmost points of , while contains the leftmost points of . See Figure 2 for an illustrative example.
Starting with , we go through the northwest quadrants from to , and whenever does not contain any point of , we add the leftmost point of to . Label the points in by in the order in which they were added to . Along this order, the points have decreasing coordinates and decreasing coordinates. Clearly, is a hitting set of and satisfies Item 1.
Since is the leftmost point of , does not belong to for every . Since contains exactly the topmost vertices, the corresponding hyperedge is hit exactly once. This proves Item 2.
For any two consecutive points in , consider the bottomless rectangle with topright corner and on the left side. Then as contains and all points of the northwest quadrant for which we added to . This proves Item 3.
Moreover, every point in lies above (if it exists), as is leftmost in . Thus, the axisaligned rectangle with topright corner and bottomleft corner contains , all the points in , and . This proves Item 4 and also implies that is shallow. ∎
Let (respectively ) denote the set of topmost (respectively bottommost) points in . For symmetry reasons, statements analogous to the above lemma hold for other types of quadrants so we obtain the following corollary:
Corollary 16.
For a point set and , a hypergraph
admits a hitting set
such that:

It satisfies the properties of (the symmetrical version of) Lemma 15.

Every hyperedge of is hit by / / / at most twice / not once / once / once.

Every hyperedge of is hit by / / / at most not once / twice / once / once.

Every hyperedge of is hit by / / / at most once / once / twice / not once.

Every hyperedge of is hit by / / / at most once / once / not once / twice.

The set is hit by / / / at most once / once / once / once.

The set is hit by / / / at most once / once / once / once.
Intuitively speaking, Lemma 15 allows us to color some points in , in such a way that every northwest quadrant already contains all colors, while other ranges, such as bottomless rectangles or diagonal strips, have most of their points still uncolored.
Now we provide a framework which can then be applied to color various range families.
Lemma 17.
Let be shrinkable range families, be a function, and be such that:

For every , it holds that .

For every point set and every , the hypergraph admits a shallow hitting set such that every hyperedge of it hit at most times by .
Then for every , we have
Proof.
Let be a point set and let . Let . We construct a polychromatic coloring of with colors in two steps. First, similarly to Lemma 13 we iteratively identify shallow hitting sets of to obtain a coloring of a subset of that is already polychromatic for . Second, we show that every hyperedge captured by contains at least (uncolored) points in and hence, there is a coloring of that is polychromatic for too. Now we formalize this procedure.
We start with , and for repeat the following. Let be a shallow hitting set of as given by Item 2. We set and . Note that for all and in particular .
We claim that for every , every hyperedge of , and every hyperedge of the following holds:

,

,

.
We prove the statement by induction on . For we have , . Since satisfies Item 2, is hit by and both and are hit at most times by , so the claim holds. Now suppose it holds for some . Then we know that
Since and are shrinkable range families, there exist hyperedges in and in . For the shallow hitting set of we have
First, this implies . Second, we have:
Similarly, we have:
Therefore, the claimed properties hold for too and by induction they hold for every . First of all, this implies that each of is a hitting set of , where by construction, are pairwise disjoint. Now we color all points in with color for every . This is a coloring of that is polychromatic for . For the remaining points we take a polychromatic coloring of , which exists as . Then every hyperedge of is polychromatic since and thus, as is shrinkable, there exists a hyperedge in . Altogether, the arising coloring is a polychromatic coloring of and this concludes the proof. ∎
With Lemma 17 in place, we can prove the upper bounds for several range families:
Theorem 18.

For the range family , we have for all .

For the range family , we have for all .

For the range family , we have for all .
Proof.
In all cases, the proof combines Corollary 16 and Lemma 17. To obtain the desired bounds, we only need to choose suitable parameters.

By Corollary 16, for every point set and every , there exist subsets such that is a 2shallow hitting set of and every hyperedge of
Comments
There are no comments yet.