An abstract approach to polychromatic coloring: shallow hitting sets in ABA-free hypergraphs and pseudohalfplanes

The goal of this paper is to give a new, abstract approach to cover-decomposition and polychromatic colorings using hypergraphs on ordered vertex sets. We introduce an abstract version of a framework by Smorodinsky and Yuditsky, used for polychromatic coloring halfplanes, and apply it to so-called ABA-free hypergraphs, which are a generalization of interval graphs. Using our methods, we prove that (2k-1)-uniform ABA-free hypergraphs have a polychromatic k-coloring, a problem posed by the second author. We also prove the same for hypergraphs defined on a point set by pseudohalfplanes. These results are best possible. We could only prove slightly weaker results for dual hypergraphs defined by pseudohalfplanes, and for hypergraphs defined by pseudohemispheres. We also introduce another new notion that seems to be important for investigating polychromatic colorings and epsilon-nets, shallow hitting sets. We show that all the above hypergraphs have shallow hitting sets, if their hyperedges are containment-free.


Introduction
The study of proper and, more generally, polychromatic colorings of geometric hypergraphs has attracted much attention, not only because this is a very basic and natural theoretical problem but also because such problems often have important applications.One such application area is resource allocation, e.g., battery consumption in sensor networks.Moreover, the coloring of geometric shapes in the plane is related to the problems of cover-decomposability, conflict-free colorings and ǫ-nets; these problems have applications in sensor networks and frequency assignment as well as other areas.For surveys on these and related problems see [17,24].
A very general definition of polychromatic coloring of geometric hypergraphs is that given a set of points and a collection of regions, k-color the points such that every region that contains at least m(k) points contains points of all k colors.In the dual version, we color the regions such that every point which is contained by at least m(k) regions is contained by a region of every color.This dual version is equivalent to decomposing an m(k)-fold covering into k coverings.For more about this notion, see e.g., the survey [17].
Geometric hypergraphs defined by translates of a given shape are usually investigated in the context of cover-decomposability (into multiple coverings).In this case the primal and dual versions are equivalent.One of the most important results in coverdecomposability is equivalent to the statement that any point set can be polychromatically k-colored such that any translate of a fixed convex polygon that contains at least m(k) = c • k points (where c depends only on the polygon) contains points with all colors [8].Polygons for which m(2) (and also m(k)) is finite have been classified [20,23].
As it was shown recently [22], there is no finite m(2) for most convex sets, e.g., for the translates of a disc.However, it was also shown in the same paper that for the translate of any unbounded convex set such a 2-coloring exists for m(2) = 3-fold coverings.In this paper we extend this result to every k, showing that m(k) = 2k − 1 is an optimal function for unbounded convex sets.
For homothets of a given shape the primal and dual cases are not equivalent.For homothets of a triangle (a case closely related to the case of translates of octants [12,13,3]), there are several results, the current best are m(k) = O(k 4.53 ) in the primal point-coloring case [14] and m(k) = O(k 5.53 ) in the dual region-coloring case [3].For the homothets of other convex polygons, in the dual case there is no finite m(2) [16], and in the primal case only conditional results are known [14] (namely, the existence of a finite m(2) implies the existence of an m(k) that grows at most polynomially in k).
For other shapes, cover-decomposability has been less investigated, in these cases the investigation of polychromatic-colorings is motivated rather by conflict-free colorings or ǫ-nets.Most closely to our paper, coloring halfplanes for small values were investigated in [10], [11] and [7], and polychromatic k-colorings in [25].We generalize all the (primal and dual) results of the latter paper to pseudohalfplanes.Note that translates of an unbounded convex set form a set of pseudohalfplanes, thus the above mentioned result about unbounded convex sets is a special case of this generalization to pseudohalfplanes.
Axis-parallel rectangles are usually investigated from the ǫ-net point of view (e.g., [4,18]), for which the coloring function f is not independent of the number of points/regions.Motivated by them, bottomless rectangles are regarded for small values in [10], [11] and polychromatic k-colorings in [1].In this paper we place bottomless rectangles in our abstract context and pose some further problems about them.
Besides improving earlier results, our contribution is a more abstract approach to the above problems.Namely, we introduce the notion of collections of sets which do not contain an independent alternating sequence of length t, shallow hitting sets and balanced polychromatic colorings, and discuss their relevance.

Definitions and results
We start with a definition, resembling the notion of a Davenport-Schinzel sequence [5].
Definition 1.1.Given two sets of real numbers, A and B, we say that they contain an independent alternating sequence of length s, if there is a sequence x 1 < . . .< x s such that either of the below conditions holds.
A hypergraph whose vertices are real numbers is t-ias-free if it does not contain an independent alternating sequence of length t.For a hypergraph with an unordered vertex set we say that it can be made t-ias-free if its vertices have an ordering without an independent alternating sequence of length t.
Remark 1.2.ABA-free hypergraphs were first defined in [22] under the name special shift-chains, as they are a special case of shift-chains introduced in [21].
Example 1.3 ([22]).Let S be a set of points with different x-coordinates and let C be a convex set that contains a vertical halfline.Define a hypergraph H whose vertex set is the x-coordinates of the points of S. A set of numbers H is a hyperedge of H if there is a translate of C such that the x-coordinates of the points of S contained in the translate is exactly H.The hypergraph H defined this way is ABA-free.
The above example shows how to reduce geometric problems to abstract problems about t-ias-free hypergraphs.We are especially interested in polychromatic coloring problems1 where our goal is to color a set of vertices/points with k colors such that every color appears in every hyperedge/geometric range of some given collection.
To study these, we also introduce the following definition, which is implicitly used in [25], but deserves to be defined explicitly as it seems to be important in the study of polychromatic colorings.
We denote the smallest (resp.largest) element of an ordered set H (if exists) by min(H) (resp.max(H)).
Observation 1.5.An induced subhypergraph of a t-ias-free hypergraph is also t-ias-free.
Our main results and the organization of the rest of this paper is as follows.In Section 2 we prove (following closely the ideas of Smorodinsky and Yuditsky [25]) that every (2k − 1)-uniform ABA-free hypergraph has a polychromatic coloring with k colors.
In Section 3 we prove that given a set of points S and a pseudohalfplane arrangement H, we can color S with k colors such that any pseudohalfplane in H that contains at least 2k − 1 points of S contains all k colors.These two theorems are best possible.We also prove that given a pseudohalfplane arrangement H, we can color H with k colors such that if a point p belongs to a subset H p of at least 3k − 2 of the pseudohalfplanes of H then H p contains a pseudohalfplane from each of the k color classes.We also discuss consequences about ǫ-nets on pseudohalfplanes.
Finally, in Section 4 we discuss ABAB-free hypergraphs and related problems.

ABA-free hypergraphs
Suppose we are given an ABA-free hypergraph H on n vertices.As the hypergraph is ABA-free, for any pair of sets A, B ∈ H either there is an occurrence of AB (i.e., an a < b such that a ∈ A \ B and b ∈ B \ A) or there is an occurrence of BA (i.e., a b < a such that a ∈ A \ B and b ∈ B \ A), or none of them, but not both (as that would yield an occurrence of the forbidden pattern ABA or BAB).
Define A < B if and only if there is an occurrence of AB.By the above observation this is well-defined and it is also easy to check that it gives a partial ordering of the sets.
If H is finite ABA-free, then every A ∈ H contains an unskippable vertex.
Remark 2.3.Note that finiteness is needed, as the hypergraph whose vertex set is Z and edge set is {Z \ {n} : n ∈ Z} is ABA-free without unskippable vertices.
Proof of Lemma 2.2.Take an arbitrary set A ∈ H, suppose that it does not contain an unskippable vertex, we will reach a contradiction.Call a ∈ A rightskippable if there is a B ∈ H rightskipping a, that is for which a ∈ A \ B and there are b If A contains no unskippable vertex, max(A) must be rightskippable (the set skipping max(A) must also rightskip max(A)).Also, min(A) cannot be rightskippable, as otherwise A and the set B rightskipping min(a) would violate ABA-freeness (we would get b Therefore we can take the largest a ∈ A that is not rightskippable.By the assumption, it is skipped by a set, call it Recall that a hypergraph is called sperner if no two of its sets (i.e., hyperedges) contain each other.
Lemma 2.4.If H is finite, ABA-free and sperner, then any minimal hitting set of H that contains only unskippable vertices is 2-shallow.
Proof.Let R be a minimal hitting set of unskippable vertices.Assume to the contrary that there exists a set A such that |A ∩ R| ≥ 3. Let l = min(A ∩ R) and r = max(A ∩ R).There exists a third vertex l < a < r in A ∩ R. We claim that R ′ = R \ {a} hits all sets of H, contradicting its minimality.Assume on the contrary that R ′ is disjoint from some B ∈ H.As R must hit B, we have R ∩ B = {a}.If there is a b ∈ B \ A such that l < b < r, that would contradict the ABA-free property.If there is a b ∈ B such that b < l < a or a < r < b, that would contradict that l and r are unskippable.Thus B ⊂ A, contradicting that H is sperner.
Our strategy to give a polychromatic coloring using k colors of the vertices of an ABA-free hypergraph with edges of size at least 2k − 1 is as follows.If any edge contains another, then we delete the bigger edge, thus making our hypergraph sperner.Next, using Lemma 2.4 we select a 2-shallow hitting set R and color its vertices with the first color.We delete these vertices.Using Observation 1.5 we still have an ABA-free hypergraph with edges of size at least 2k − 3, thus using induction on k we have proved Theorem 2.5.Given a finite ABA-free H we can color its vertices with k colors such that every A ∈ H whose size is at least 2k − 1 contains all k colors.Notice that the above theorem is sharp, as taking H to be all subsets of size 2k − 2 from 2k − 1 vertices, in any coloring of the vertices, one color must occur at most once and is thus missed by some edge.
We prove another interesting property of ABA-free hypergraphs before which we need a definition.
Definition 2.7.The dual of a hypergraph H, denoted by H * , is such that its vertices are the hyperedges of H and its hyperedges are the vertices of H with the same incidences as in H.
Claim 2.8.If H is ABA-free, then its dual, H * , can be made ABA-free.
Proof.First define the partial order < on the hyperedges of H as A < B if there are vertices i < j such that i ∈ A \ B and j ∈ B \ A. ABA-freeness guarantees that this is indeed an order.Then we extend this arbitrarily into a total order, this will show that H * can be made ABA-free.

Pseudohalfplanes
Here we extend a result of Smorodinsky and Yuditsky [25].A pseudoline arrangement is a finite collection of simple curves in the plane such that any two intersect at most once and there they cross.A graphic pseudoline arrangement is such that every curve is the graph of a function, i.e., an x-monotone curve without ends, and any two curves intersect exactly once.A pseudohalfplane arrangement is a graphic pseudoline arrangement, with a side of each pseudoline selected.We say that two pseudoline arrangements are equivalent if there is a bijection between their pseudolines such that the order in which a pseudoline intersects the other pseudolines remains the same.
Notice that an ABA-free hypergraph naturally corresponds to an x-monotone pseudoline arrangement and a set of points S such that each hyperedge corresponds to the points of S above a pseudoline.(Similarly, an (t + 2)-ias-free hypergraph would correspond to an arrangement of x-monotone curves that intersect at most s times.)We now show that given a finite set of points, S, and a pseudohalfplane arrangement, we can always color the points of S with k colors such that any pseudohalfplane containing at least 2k − 1 points contains all k colors, extending Theorem 2.5.The proof will be a combination of some simple geometric observations and the proof from Section 2.
We start with some well-known results about pseudoline arrangements, which can be found in [2].We also recommend [6] where generalizations of classical theorems are proved for topological affine planes.
Facts about pseudoline arrangements.
I. Given a pseudoline arrangement, any two points of the plane can be connected by a new pseudoline.
II.Given a pseudoline arrangement, we can find an equivalent graphic pseudoline arrangement.
III.Given a pseudohalfplane arrangement and a set of points S all contained in the pseudohalfplane H, we can add a new pseudohalfplane H ′ contained completely in H that contains all but one of the points of S.
Now we show how to extend Theorem 2.5 to hyperhalfplane arrangements, i.e., to the case when the points of S below a line also define a hyperedge.The proof uses the same ideas.
Theorem 3.1.Given a set of points S and a pseudohalfplane arrangement H, we can color S with k colors such that any pseudohalfplane in H that contains at least 2k − 1 points of S contains all k colors.Proof.By repeatedly using III, we add new pseudohalfplanes until every pseudohalfplane contains a pseudohalfplane that contains only one point.If a point is the only one contained in some pseudohalfplane, then we call it unskippable (these are the vertices of the convex hull if the pseudohalfplanes are halfplanes).We also call the set of points of S contained in an upward pseudohalfplane the upset and the set of points of S contained in a downward pseudohalfplanes the downset and the respective unskippable points top and bottom.Note that the upsets and the downsets both form an ABA-free hypergraph.There is a relation among these ABA-free hypergraphs (in fact the union of two arbitrary ABA-free hypergraphs might not be 2-colorable, see [22] for such a construction).For example the following observation holds, as we can use that pseudolines intersect at most once.
Observation 3.2.If x is top and x ∈ X is a downset, then X contains all points that are bigger or all points that are smaller than x.The same holds if x is bottom and X is an upset.Lemma 3.3.If H is a sperner pseudohalfplane family and R is a minimal hitting set of unskippable points, then R is a 2-shallow hitting set.
Proof.Suppose for a contradiction that {a < b < c} ⊂ R ∩ X for some X ∈ H. Without loss of generality, suppose that b is top.As R is minimal, let B be a set for which B ∩ R = {b}.From Observation 3.2 it follows that B is an upset.First suppose that X is an upset.As B ⊂ X, take a b 2 ∈ B \ X.As B and X are both upsets and thus have the ABA-free property, we have b 2 < a or c < b 2 .Without loss of generality, we can suppose c < b 2 .If c is top, {c} and B violate ABA-freeness.See Figure 1(a).If c is bottom, then using Observation 3.2, X contains all the points that are smaller than c.Take a set A ⊂ X for which A ∩ R = {a}.This set must contain an a 2 ∈ A \ X and so we must have c < a 2 .If A is an upset, as it does not contain b (recall a < b < a 2 ), A and {b} violate ABA-freeness.See Figure 1(b).If A is a downset, as it does not contain c (recall a < c < a 2 ), A and {c} violate ABA-freeness, both cases lead to a contradiction.
The case when X is a downset is similar.Using Observation 3.2 for X and {b} we can suppose without loss of generality that X contains all points that are smaller than b.Take a set A ⊂ X for which A ∩ R = {a} and an a 2 ∈ A \ X.A cannot be an upset, as then it would contain b, so it is a downset.As A and X are ABA-free, we must have c < a 2 .This means c cannot be bottom, so it is top.Using Observation 3.2, X contains all the points that are smaller than c.But then B \ X must have an element that is bigger than c, contradicting the ABA-freeness of B and {c}.
From here the rest of the proof is the same.To give a k-coloring of the points of a pseudohalfplane arrangement such that every pseudohalfplane of size at least 2k − 1 contains all k colors is as follows.Using III, it is enough to consider pseudohalfplanes of size exactly 2k − 1, thus we can use Lemma 3.3 to select a 2-shallow hitting set R and color its points with the first color.We delete these points and we are done by induction on k.
We can also generalize the dual result about coloring halfplanes of [25] from halfplanes to pseudohalfplanes: Theorem 3.4.Given a pseudohalfplane arrangement H, we can color H with k colors such that if a point p belongs to a subset H p of at least 3k − 2 of the pseudohalfplanes of H then H p contains a pseudohalfplane from each of the k color classes.
Proof.If every point is covered by a pseudohalfplane, then there is a set H k of at most 3 pseudohalfplanes covering the whole plane, as we can use Helly's theorem (which also holds for pseudoconvex sets, see [6]) for the complements of the pseudohalfplanes.Color these to color k and remove them.Apply induction for k − 1 on H ′ = H \ H k .Thus, if a point p belongs to at least 3k − 2 pseudohalfplanes, then it is covered by one of the pseudohalfplanes with color k and as in H ′ it belongs to at least 3(k − 1) − 2 pseudohalfplanes, by induction it is covered by a pseudohalfplane with color i for each color i ≤ k − 1.
Otherwise, if there is a point not covered by any pseudohalfplane, we can dualize.(For details about dualization, see e.g., [25].)In the dual setting, by Theorem 3.1 we can color the points (corresponding to the pseudohalfplanes in the original setting) such that any pseudohalfplane (corresponding to the points in the original setting) containing at least 2k − 1 points contains points of each color.This coloring clearly gives us a coloring of the original pseudohalfplanes such that if a point p belongs to a subset H p of at least 2k − 1 ≤ 3k − 2 of the pseudohalfplanes then H p contains a pseudohalfplane from each of the k color classes.
Note that the 3k − 2 in the above theorem is not tight, from below we know only that by the dual of the primal construction it cannot be changed to anything smaller than 2k − 1.In case of halfplanes, it was shown by Fulek [7] that for k = 2 it is sufficient to consider a 3-fold covering.We are not aware of any nice characterization for the dual of (pseudo)halfplane arrangements, like we had for ABA-free hypergraph in Claim 2.8.

Small epsilon-nets for pseudohalfplanes
We conclude this section by briefly mentioning the consequences of our results to ǫnets of hypergraphs defined by pseudohalfplanes.We omit proofs as they are not hard and can be obtained exactly as the corresponding results in [25].
Let H = (V, E) be a hypergraph where V is a finite set.Let ǫ ∈ (0, 1] be a real number.A subset N ⊆ V is called an ǫ-net if for every hyperedge S ∈ E such that |S| ≥ ǫ|V |, we also have S ∩ N = ∅, i.e., N is a hitting set for all "large" hyperedges.It is known that hypergraphs with VC-dimension d have small ǫ-nets (of size O(d/ǫ log(1/ǫ)) [9] and r Figure 2: H 3 in general this is best possible [15].However, for geometric hypergraphs this is usually not optimal, in particular for halfplanes the following is true.Consider a hypergraph H = (P, E) where P is a finite set of points in the plane and E = {P ∩ h: h is a halfplane}.For this hypergraph there is an ǫ-net of size 2/ǫ−1 for every ǫ [26,25].Theorem 3.1 implies that the same bound holds if the hypergraph is defined by pseudohalfplanes instead of halfplanes.Also, for the dual hypergraph H = (P, E) where P is a finite set of pseudohalfplanes and E = {{P : v ∈ P }: v is a point in the plane}, Theorem 3.4 implies that for ǫ ≤ 2/3 there exists an ǫ-net of size 2/ǫ.Note that our results are in fact stronger as in the appropriate polychromatic coloring each color class intersects the large enough hyperedges, thus we get a partition of the vertices into ǫ-nets (and at least one of them is a small ǫ-net by the pigeonhole principle).

ABAB-free hypergraphs that are not two-colorable
We show that the vertices of a non-2-colorable hypergraph often used in counterexamples (e.g., [19]) can be ordered in a tricky way to give an ABAB-free hypergraph.First we define this hypergraph H k .Definition 4.1.Let G k be the complete k-ary tree of depth k, i.e., the rooted tree such that its root r has k children, each vertex of G k in distance at most k − 2 from r has exactly k children and the vertices in distance k from r are the leafs (have no children).
H k is the k-uniform hypergraph which has two types of hyperedges.First, for every nonleaf vertex the set of its children form an edge.Second, the vertices of every descending path starting in r and ending in a leaf form an edge.It is easy to see that H k is not two-colorable.Now we show how to realize H k such that its vertices correspond to points in the plane and its hyperedges correspond to the points above pseudoparabolas (simple curves such that any two intersect at most twice).We fix k and define H ′ l (resp.G ′ l ) to be the hypergraph (resp.graph) induced by H k (resp.G k ) and the subset of the vertices that are in distance at most l − 1 from the root r in G k (H ′ l is a simple hypergraph, i.e., if multiple edges induce the same edge, we take it only once).Thus in particular G ′ 1 has one vertex and no edges while H ′ 1 has one vertex and one edge containing it, while Note that in G ′ l every non-leaf vertex has k children, and H ′ l has hyperedges of size l corresponding to descending paths (which we will usually denote by H i for some i) and hyperedges of size k corresponding to the set of children of some vertex (which we will usually denote by J i for some i).See Figure 2.
In our realization, to simplify the presentation, points corresponding to vertices will be denoted with the same label, and similarly hyperedges and the corresponding pseudoparabolas will have the same label.
We will recursively realize H ′ l , for an illustration see Figure 4. We will additionally maintain that each hyperedge (pseudoparabola) H i corresponding to a descending path has a vertical strip S i associated to it, such that inside S i there are no points and H i has the lowest boundary (thus no other hyperedge intersects H i inside S i ).For l = 1, this is trivial as H ′ 1 has no edges, we just put a point on the plane.For l = 2, Figure 3 shows a way to achieve this (for k = 3).Now suppose that for some l we have H ′ l and we want to construct H ′ l+1 .Take the construction of H ′ l , and for each hyperedge H i corresponding to a descending path P i with endvertex p i , do the following.First make k vertically translated copies of H i very close to each other.Denote these by K 1 , K 2 , . . .K k .Next, using these k copies of H i , realize H ′ 2 (except the root r) in an appropriately small area inside S i , by adding k more points k 1 , k 2 , . . .k k such that for every i, k i is above Finally, define the pseudoparabola J i , which corresponds to the hyperedge containing all the k i 's but no other vertex, as a parabola very close to the vertical strip containing the k i 's.For each i, the vertical strip that belongs to K i in the inner copy of H ′ 2 is the strip corresponding to the descending hyperedge that ends at k i .Therefore all properties are maintained, and by repeating the above procedure for each of the leafs p i of H ′ l we get a realization of H ′ l+1 .We are not aware of any nice characterization for the dual of an ABAB-free hypergraph, like we had for ABA-free hypergraph in Claim 2.8.

Bottomless rectangles and balanced colorings
Every hypergraph given by a set of points and a collection of bottomless rectangles is ABAB-free, but not necessarily ABA-free.In fact, it is not hard to see that such hypergraphs would correspond exactly to "aBAb"-free hypergraphs, which can be defined similarly to Definition 1.1 as follows.
Definition 4.2.A hypergraph whose vertices are real numbers is aBAb-free if for any two of its hyperedges, A and B, and vertices x 1 < x 2 < x 3 < x 4 it does not hold that x 1 ∈ A, x 2 ∈ B \ A, x 3 ∈ A \ B, x 4 ∈ B.
It was shown in [1] that any finite set of points can be colored with k colors such that any bottomless rectangle that contains at least 3k − 2 points contains all k colors.Unfortunately, we were not able to reprove this using our methods.We also do not know the answer to the following problem.Problem 4.3.Is it possible to find a c-shallow hitting set for every sperner bottomless rectangle family for some suitable constant c?
We also want to point out a nice property of the colorings that we get by repeatedly finding c-shallow hitting sets and making them the next color class.In the proofs in earlier sections, after we have made k such steps, we did not care about the remaining points, we could color them arbitrarily.Modifying this, we can start over with the first color after k steps, then the second color, etc.In the i-th step the c-shallow hitting set we get we color with color i (mod k), where color 0 and color k denote the same color.This way we achieve a coloring that is not just polychromatic, but also balanced in the sense that if we denote the size of any two color class in any given set of our family by n 1 and n 2 , then we have n 1 ≤ c(n 2 + 1).Problem 4.4.Is there a balanced coloring for any family of bottomless rectangles?
We also do not know whether a c-shallow or a balanced coloring exists for other families such as homothets of a triangle or translates of an octant in R 3 .
then C would rightskip a, a contradiction, thus we must have a ≤ c 1 .By our choice of b 2 , we must have c 1 / ∈ B and as b 1 < a ≤ c 1 = min(C), we also have b 1 / ∈ C. Putting all together, we get that b 1 < c 1 < b 2 , where b 1 , b 2 ∈ B \ C and c 1 ∈ C \ B, so B and C contradict ABA-freeness.
A cannot be an upset

Figure 4 :
Figure 4: Recursive realization of H ′ l : adding k children to a leaf