Colorful Strips
Abstract
We study the following geometric hypergraph coloring problem: given a planar point set and an integer , we wish to color the points with colors so that any axisaligned strip containing sufficiently many points contains all colors. We show that if the strip contains at least points, such a coloring can always be found. In dimension , we show that the same holds provided the strip contains at least points. We also consider the dual problem of coloring a given set of axisaligned strips so that any sufficiently covered point in the plane is covered by colors. We show that in dimension the required coverage is at most . This complements recent impossibility results on decomposition of strip coverings with arbitrary orientations.
From the computational point of view, we show that deciding whether a threedimensional point set can be 2colored so that any strip containing at least three points contains both colors is NPcomplete. This shows a big contrast with the planar case, for which this decision problem is easy.
1 Introduction
There is a currently renewed interest in coloring problems on geometric hypergraphs, that is, set systems defined by geometric objects. This interest is motivated by applications to wireless and sensor networks [6]; conflictfree colorings [8], chromatic numbers [20], covering decompositions [18, 3], or polychromatic (colorful) colorings of geometric hypergraphs [4] have been extensively studied in this context.
In this paper, we are interested in coloring finite point sets in so that any region bounded by two parallel axisaligned hyperplanes, that contains at least some fixed number of points, also contains a point of each color.
An axisaligned strip^{1}^{1}1From here on, unless otherwise specified, a strip is always assumed to be axisaligned is the area enclosed between two parallel axisaligned hyperplanes. A coloring of a finite set assigns one of colors to each element in the set. Let be a colored set of points in . A strip is said to be polychromatic with respect to if it contains at least one element of each color class. We define the function as the minimum number for which there always exists a coloring of any point set in such that every strip containing at least points is polychromatic. This is a particular case of the general framework proposed by Aloupis, Cardinal, Collette, Langerman, and Smorodinsky in [4].
Note that the problem does not depend on whether the strips are open or closed, since the problem can be seen in a purely combinatorial fashion: an axisaligned strip isolates a subsequence of the points in sorted order with respect to one of the axes. Therefore, the only thing that matters is the order in which the points appear along each axis. We can thus rephrase our problem, considering dimensional points sets, as finding the minimum value such that the following holds: For permutations of a set of items , it is always possible to color the items with colors, so that in all permutations every sequence of at least contiguous items contains one item of each color.
We also study circular permutations, in which the first and the last elements are contiguous. We consider the problem of finding a minimum value such that, for any circular permutations of a set of items , it is possible to color the items so that in every permutation, every sequence of contiguous items contains all colors.
A restricted geometric version of this problem in consists of coloring a point set with respect to wedges. For our purposes, a wedge is any area delimited by two halflines with common endpoint at one of given apices. Each apex induces a circular ordering of the points in . This is illustrated in Figure 1. We aim at coloring so that any wedge containing at least points is polychromatic. In , the noncircular case corresponds to wedges with apices at infinity, hence the circular case can be seen as a generalization.
We then study a dual version of the problem, in which a set of axisaligned strips is to be colored so that sufficiently covered points are contained in strips from all color classes. For instance, in the planar case we study the following function . Let be a colored set of strips in . A point is said to be polychromatic with respect to if it is contained in strips of all color classes. The function is the minimum number for which there always exists a coloring of any set of strips in such that every point of contained in at least strips is polychromatic.
Note that the functions , and are monotone and nondecreasing (in particular, they all go to infinity when either or goes to infinity). Since we are interested in arbitrarily large pointsets, we always consider the set that we color to be “large enough” (that is, unbounded in terms of ).
Previous results.
A hypergraph is defined by a set (called the ground set) and a set of subsets of . The main problem studied here is the coloring of geometric hypergraphs where the ground set is a finite set of points, and the set of ranges consists of all subsets of that can be isolated by a single strip. In the dual case the ground set is a finite set of geometric shapes and the ranges are points contained in the common intersection of a subset of . In some places in the literature, finite geometric hypergraphs are also referred to as geometric range spaces.
Several similar problems have been studied in this context [14, 22, 4], where the range space is not defined by strips, but rather by halfplanes, triangles, disks, pseudodisks or translates of a centrally symmetric convex polygon. The problem was originally stated in terms of decomposition of covers (or fold coverings) in the plane: A cover of the plane by a convex body ensures that every point in the plane is covered by at least translated copies of . In 1980, Pach [14] asked if, given , there exists a function such that every cover of the plane can be decomposed into disjoint covers. A natural extension is to ask if given , there exists a function such that every cover of the plane can be decomposed into disjoint covers. This corresponds to a coloring of the cover, such that every point of the plane is polychromatic.
Partial answers to this problem are known: Pach [15] referenced an unpublished manuscript by Mani and Pach [12] showing that any 33cover of the plane by unit disks can be decomposed into two covers. This would imply that the function exists for unit disks, but could still be exponential in . Recently, Tardos and Tóth [22] proved that any 43cover by translated copies of a triangle can be decomposed into two covers. For the case of centrally symmetric convex polygons, Pach [16] proved that is at most exponential in . More than 20 years later, Pach and Tóth [18] improved this by showing that , and was afterwards Aloupis et al. [3] proved that . Recently, Gibson and Varadarajan [10] showed that the same property also holds for any arbitrary convex polygon .
On the other hand, for the range space induced by arbitrary disks, Mani and Pach [12] (see also [17]) proved that is unbounded: for any constant , there exists a set of points that cannot be colored so that all open disks containing at least points are polychromatic. Pach, Tardos and Tóth [17] obtained a similar result for the range spaces induced by the family of either nonaxisaligned strips, or axisaligned rectangles. Specifically, for any integer there exist fold coverings with nonaligned strips that cannot be decomposed into two coverings (i.e., cannot be colored). The previous impossibilities constitute our main motivation for introducing the problem of coloring axisaligned strips, and strips with a bounded number of orientations.
Paper Organization.
In Section 2 we give constructive upper bounds on the functions and for . In Section 3 we consider higherdimensional cases, as well as the computational complexity of finding a valid coloring. Section 4 concerns the dual problem of coloring strips with respect to points. Our lower and upper bounds are summarized in Table 1.
upper bound  

( for )  ( for )  
lower bound 
2 Axisaligned strips and circular permutations for
We first consider upper bounds for the functions and .
2.1 Axisaligned strips: Upper bound on
We refer to a strip containing at least points as an strip. Our goal is to show that for any integer there is a constant such that any finite planar point set can be colored so that all strips are polychromatic.
For , there is a reduction to the recently studied problem of coloring graphs so that monochromatic components are small. Haxell et al. [11] proved that the vertices of any graph with maximum degree can be colored so that every monochromatic connected component has size at most . For a given finite point set in the plane, let be the set of all pairs of points such that there is a strip containing only and . The graph has maximum degree , as it is the union of two paths. By the results of [11], can be colored so that every monochromatic connected component has size at most . In particular every path of size at least contains points from both color classes. To finish the reduction argument one may observe that every strip containing at least points corresponds to a path (of size at least ) in . We improve and generalize this first bound in the following.
Theorem 1.
For any finite planar set and any integer , can be colored so that any strip is polychromatic. That is,
Proof.
Let be the points of sorted by (increasing) coordinates and let be the sorting by coordinates. We first assume that divides , and later show how to remove the need for this assumption. Let be the set of disjoint contiguous tuples in . Namely, . Similarly, let be the tuples defined by .
We define a bipartite multigraph as follows: For every pair of tuples , , we include an edge if there exists a point in both and . Note that an edge has multiplicity and that the number of edges is . The multigraph is regular because every tuple contains exactly points and every point determines exactly one incident edge labeled . It is well known that the chromatic index of any bipartite regular multigraph is (and can be efficiently computed, see e.g., [1, 7]). Namely, the edges of such a multigraph can be partitioned into perfect matchings. Let be such a partition and be the set of labels of the edges of . The sets form a partition (i.e., a coloring) of . We assign color to the points of .
We claim that this coloring ensures that any strip is polychromatic. Let be a strip and assume without loss of generality that is parallel to the axis. Then contains at least one tuple . By the properties of the above coloring, the edges incident to in are colored with distinct colors. Thus, the points that correspond to the labels of these edges are colored with distinct colors, and is polychromatic.
To complete the proof, we must handle the case where does not divide . Let . Let be an additional set of points, all located to the right and above the points of . We repeat our previous construction on . Now, any strip which is, say, parallel to the axis will also contain a tuple disjoint from . Thus our arguments follow as before. ∎
The proof of Theorem 1 is constructive and leads directly to an time algorithm to color points in the plane so that every strip is polychromatic. The algorithm is simple: we sort , construct , and color the edges of with colors. The time analysis is as follows: sorting takes time. Constructing takes time. As has vertices and is regular, it has edges; so this step takes time. Finding the edgecoloring of takes time [1]. The total running time is therefore .
2.2 Circular permutations: Upper bound on
We now consider the value of . Given circular permutations of a set , we color so that every sufficiently long subsequence in any of the circular permutations is polychromatic. The previous proof for (Theorem 1) does not hold when we consider circular permutations. However, a slight modification provides the same upper bound, up to a constant term.
Theorem 2.
Proof.
If divides , we separate each circular permutation into sets of size . We define a multigraph, where the vertices represent the sets of items, and there is an edge between two vertices if two sets share the same item. Trivially, this graph is regular and bipartite, and can thus be edgecolored with colors. Each edge in this graph corresponds to one item in the permutation, thus each set of items contains points of all colors.
If does not divide , let , and . If divides , we separate each of the two circular permutations into sets, of alternating sizes and . Otherwise, the even sets will also alternate between size and , instead of . We extend both permutations by adding dummy items to each set of size less than , so that we finally have only sets of size . Dummy items appear in the same order in both permutations. We can now define the multigraph just as before.
If we remove the dummy nodes, we deduce a coloring for our original set. As each color appears in every set of size , the length of any subsequence between two items of the same color is at most . Therefore, .
Finally, if , then , and , we know that , and thus . ∎
3 Higher dimensional strips
In this section we study the same problem for strips in higher dimensions. We provide upper and lower bounds on . We then analyze the coloring problem from a computational viewpoint, and show that deciding whether a given instance can be 2colored such that every 3strip is polychromatic is NPcomplete.
3.1 Upper bound on strip size,
Theorem 3.
Any finite set of points can be colored so that every axisaligned strip containing points is polychromatic, that is,
Proof.
The proof uses the probabilistic method. Let denote the set of colors. We randomly color every point in independently so that a point gets color with probability for . For a strip , let be the “bad” event where is not polychromatic. It is easily seen that . Moreover, depends on at most other events. Indeed, depends only on strips that share points with . Assume without loss of generality that is orthogonal to the axis. Then has a nonempty intersection with at most other strips which are orthogonal to the axis. For each of the other axes, can intersect at most strips since every point in can belong to at most other strips.
By the Lovász Local Lemma, (see, e.g., [2]) we have that if satisfies
(where is the basis of the natural logarithm), then
In particular, this means that there exists a coloring for which every strip is polychromatic. It can be verified that satisfies the condition. ∎
The proof of Theorem 3 is nonconstructive. However, we can use known algorithmic versions of the Local Lemma (see for instance [13]) to obtain a constructive proof. Also note that Theorem 3 holds in the more general case where the strips are not necessarily axisaligned. In fact, one can have a total of arbitrary strip orientations in some fixed arbitrary dimension and the proof will hold verbatim. Finally, we note that the same proof also works for the case of circular permutations, yielding the same upper bound:
Theorem 4.
3.2 Lower bound on
We first introduce a wellknown result on the decomposition of complete graphs:
Lemma 1.
The edges of can be decomposed into pairwise edgedisjoint Hamiltonian paths.
This result follows from a special case of the Oberwolfach problem [5]. An explicit proof of this lemma can also be found in [21].
Note that if the vertices of are labeled , each path can be seen as a permutation of elements. Using Lemma 1 we obtain:
Theorem 5.
For any fixed dimension and number of colors , let . Then,
Proof.
The first inequality comes from the fact that any polychromatic coloring with respect to circular permutations is also polychromatic with respect to strips. We now focus on showing the second inequality: let be any decomposition of into paths: we construct the set , where . Note that the ordering of , when projected to the th axis, gives permutation . Since the elements decompose , in particular for any there exists a permutation in which and are adjacent.
We replace each point by a set of points arbitrarily close to . By construction, for any , there exists a strip containing exactly . Consider any possible coloring of the sets : since and we are using colors, there are at least colors not present in any set .
Since , we conclude that . That is, each set is missing strictly more than colors. By the pigeonhole principle, there exist and such that the set is missing a color (otherwise there would be more than colors). In particular, the strip that contains set is not polychromatic, thus the theorem is shown.
We gave a set of bounded size reaching the lower bound, but we can easily create larger sets reaching the same bound: we can add as many dummy points as needed at the end of every permutation, which does not decrease the value of . ∎
Note that, assymptotitically speaking, the lower bound does not depend on . However, by the negative result of [17], we know that when .
3.3 Computational complexity
In Section 2, we provided an algorithm that finds a coloring such that every planar strip is polychromatic. Thus for and , this yields a coloring such that every strip is polychromatic.
Note that in this case , but the minimum required size of a strip for a given instance can be either or . Testing if it is equal to 2 is easy: we can simply alternate the colors in the first permutation, and check if they also alternate in the other. Hence the problem of minimizing the size of the largest monochromatic strip on a given instance is polynomial for and . We now show that it becomes NPhard for and . The same problem for is left open.
Theorem 6.
The following problem is complete:
Input: permutations of an element set .
Question: Is there a 2coloring of , such that every 3
elements of that are consecutive according to one of the
permutations are not monochromatic?
Proof.
We show a reduction from NAE 3SAT (notallequal 3SAT) which is the
following complete problem [9]:
Input: A 3CNF Boolean formula .
Question: Is there a assignment to ? An
assignment is called if every clause has at least one
literal assigned True and at least one literal assigned False.
We first transform into another instance in which all variables are nonnegated i.e., we make the instance monotone; this part of the proof is folklore (see e.g., [19] for similar techniques). We then show how to realize using three permutations .
To transform into , for each variable , we first replace the th occurrence of in its positive form by a variable , and the th occurrence of in its negative form by . The index varies between 1 and the number of occurrences of each form (the maximum of the two). We also add the following consistencyclauses, for each variable and for all :
where and are new variables. This completes the construction of . Note that is monotone, as every negated variable has been replaced.
Moreover, has a assignment if and only if has a assignment. To see this, note that a assignment for can be translated to a assignment to as follows: for every variable of and every , set , , .
On the other hand, if has a assignment, then, by the consistency clauses, the variables in corresponding to any variable of are assigned a consistent value. Namely, for every we have and . This assignment naturally translates to a assignment for , by setting .
We next show how to realize by a set and three permutations . The elements of the set are the variables of , together with some additional elements that are described below. Permutation realizes the clauses of corresponding to the original clauses of , while and realize the consistency clauses of .
The additional elements in are clause elements (two elements and for every clause of ) and dummy elements (the dummy elements are not indexed for the ease of presentation, but they appear in the same order in all three permutations).
Permutation encodes the clauses of corresponding to original clauses of as follows (note that all these clauses involve different variables). For each such clause , permutation contains the following sequence:
At the end of , for every variable of we have the sequence:
Permutation contains, for every variable of , the sequences:
At the
end of we have the clauseelements and
remaining dummy elements:
Similarly, permutation contains, for every variable of , the sequences:
and at the end of
we have the clauseelements and
remaining dummy elements:
This completes the construction of and . Note that for every clause of (whether it is derived from or is a consistency clause), the elements corresponding to its three variables appear in sequence in one of the three permutations. Therefore, if there is a 2coloring of , such that every 3 elements of that are consecutive according to one of the permutations are not monochromatic, then there is a assignment to : each variable of is assigned True if its corresponding element is colored ‘1’, and False otherwise.
For the other direction, consider a assignment for . Observe that there is always a solution where and are assigned opposite values. Then assign color ‘1’ to elements corresponding to variables assigned with True, and assign color ‘0’ to elements corresponding to variables assigned with False. For the clause elements and appearing in the subsequence , assign to the color opposite to , and to the color opposite to . Finally, assign colors ‘0’ and ‘1’ to each pair of consecutive dummy elements, respectively. It can be verified that there is no monochromatic consecutive triple in any permutation. ∎
Approximation.
Note that the general minimization problem (find a coloring that minimizes the number of required points) can be approximated using the constructive version of the Lovász Local Lemma (as mentioned in Section 3.1). Since is a trivial lower bound for any problem instance, this guarantees an approximation factor of . In particular, there exists a constant factor approximation for the problem introduced in Theorem 6 (since it fixes and ).
4 Coloring strips
In this section we prove that any finite set of strips in can be colored so that every “deep” point is polychromatic. For a given set of strips (or intervals, if ), we say that a point is deep if it is contained in at least of the strips. We begin with the following easy lemma:
Lemma 2.
Let be a finite set of intervals. Then for every , can be colored so that every deep point is polychromatic, while any point covered by fewer than intervals will be covered by distinct colors.
Proof.
We use induction on . Let be the interval with the leftmost right endpoint. By induction, the intervals in can be colored with the desired property. Sort the intervals intersecting according to their left endpoints and let be the first intervals in this order. It is easily seen that coloring with a color distinct from the colors of those intervals produces a coloring with the desired property, and hence a valid coloring. ∎
Theorem 7.
For any and , one can color any set of axisaligned strips in so that every deep point is polychromatic. That is,
Proof.
We start by coloring the strips parallel to any given axis () separately using the coloring described in Lemma 2. We claim that this procedure produces a valid polychromatic coloring for all deep points. Indeed assume that a given point is deep and let be the set of strips covering . Since there are possible orientations for the strips in , by the pigeonhole principle at least of the strips in are parallel to the same axis. Then by property of the coloring of Lemma 2, is polychromatic. ∎
The above proof is constructive. By sorting the intervals that correspond to any of the given directions, one can easily find a coloring in time.
We now give a lower bound on . For that, we define strips as follows: strip is defined as (where is the coordinate of th dimension). Analogously, we define strip as . The main property of these strips is that we can always find a point covered by any subset of the strips:
Lemma 3.
For any , there exists a point such that , .
Proof.
Note that whether or not point is covered by strips or only depends in the th coordinate of . Thus, we define the th coordinate of point as follows:

if but

if both

if both
Since the choice is independent on each dimension, the construction of is valid and is only covered by strips in . ∎
We use these strips to find a lower bound on as follows:
Theorem 8.
For any fixed dimension and integer , it holds that
Proof.
Consider the strips defined above. We replace each strip with a cluster of overlapping strips , so that a point defined in Lemma 3 is in strip if and only if it is in . This can be obtained, say, by perturbing a boundary of the strip around (or around ).
Consider any coloring of the above problem instance with at most colors. As in the proof of Theorem 5, we can use the pigeonhole principle and the handshake lemma to conclude that there is at a color that is missing in at least clusters. Let be the set of at least indices whose clusters are missing the same color. By Lemma 3, point is covered only by the strips in clusters , for all . In particular, is at least deep and not colorful. ∎
acknowledgements
This research was initiated during the WAFOL’09 workshop at Université Libre de Bruxelles (U.L.B.), Brussels, Belgium. The authors want to thank all other participants, and in particular Erik D. Demaine, for helpful discussions. Also, the authors would like to thank the anonymous referees for pointing out useful references.
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