# Empty Rainbow Triangles in k-colored Point Sets

Let S be a set of n points in general position in the plane. Suppose that each point of S has been assigned one of k ≥ 3 possible colors and that there is the same number, m, of points of each color class. A polygon with vertices on S is empty if it does not contain points of S in its interior; and it is rainbow if all its vertices have different colors. Let f(k,m) be the minimum number of empty rainbow triangles determined by S. In this paper we give tight asymptotic bounds for this function. Furthermore, we show that S may not determine an empty rainbow quadrilateral for some arbitrarily large values of k and m.

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## 1 Introduction

A set of points in the plane is in general position if no three of its points are collinear. In this paper all sets of points are in general position. The well known Erdős-Szekeres theorem [12] states that for every positive integer there exists a positive integer such that every set of (or more points) in the plane contains the vertices of a convex polygon of vertices.

Let be a set of points in the plane. A polygon with vertices on is said to be empty if it does not contain a point of in its interior. An -hole of is an empty convex polygon of sides with vertices on . In 1978, Erdős [13] asked if for every , every sufficiently large set of points in the plane contains an -hole. Klein [12] had already noted that every set of points contains a -hole. Harboth [17] showed that every set of points contains a -hole. Horton [18] constructed arbitrarily large sets of points without -holes. The case for -holes remained open until Nicolás [22] and Gerken [16], independently showed that every sufficiently large point set contains a -hole.

Once the existence of -holes for some given in every sufficiently large point set is established, it is natural to ask what is the minimum number of -holes in every set of points in the plane. Katchalski and Meir [20] first considered this question for triangles. They showed that every set of points determines empty triangles and provided an example of a point set determining empty triangles. The lower and upper bounds on this number have been improved throughout the years [5, 11, 24, 15, 9, 2]. The problem of determining the minimum number of -holes in every set of points in the plane has also been considered in these papers.

Colored variants of these problems where first studied by Devillers, Hurtado, Károlyi and Seara [10]. A point set is -colored if every one of its points is assigned one of available colors. We say that an -hole on is monochromatic if all its vertices are of the same color, and that it is rainbow111In [10] rainbow -holes are called heterochromatic. We prefer to use “rainbow”, because this term is used, with this meaning, in the more general setting of anti-Ramsey problems. if all its vertices are of different colors. Many chromatic variants on problems regarding -holes in colored points sets have been studied since; see  [6, 23, 1, 3, 7, 21, 4, 19, 8, 14, 25]. In particular, Aichholzer, Fabila-Monroy, Flores-Peñaloza, Hackl, Huemer, and Urrutia showed that every -colored set of points in the plane determines empty monochromatic triangles [1]. This was later improved to by Pach and Tóth [23]. The current best upper bound on this number is and this is conjectured to be the right asymptotic value.

In this paper we consider the problem of counting the number of empty rainbow triangles in -colored point sets in which there are the same number, , of points of each color class. Let be the minimum number of empty rainbow triangles in such a point set. We give the following tight asymptotic bound for .

###### Theorem 1.1.
 f(k,m)={Θ(k2m) if m

Note that in contrast to the number of empty monochromatic triangles, the number of empty rainbow triangles does not necessarily grow with the number of points.

## 2 Lower Bound

###### Proof of the lower bound in Theorem 1.1.

Let be a -colored set of points with points of each color class. Without loss of generality assume that no two points of have the same -coordinate. Assume that the set of colors is . For each , let be the leftmost point of color . Without loss of generality assume that when sorted by -coordinate these points are .

Let and let . We show that there are at least empty rainbow triangles having a point of color as its rightmost point. Let be the first points of color when sorted by -coordinate. For each do the following. Sort the points of to the left of counterclockwise by angle around . Note that any two consecutive points in this order define an empty triangle with as its rightmost point. Since the points are to the left of , there are at least of these empty triangles such that the first point is of a color distinct from , and the next point is of a color distinct from . Furthermore, for at least of these triangles the next point is not of color ; thus, they are rainbow. We have at least

 ri−1∑j=1i−j−1=(ri−1)(2i−ri−2)2 (1)

empty rainbow triangles with a point of color as its rightmost point. If then the right hand side of (1) is equal to

 i2−3i+22

Thus, if then determines at least

 k∑i=3i2−3i+22=16k3−12k2+13k=Ω(k3)

empty rainbow triangles; and if then determines at least

 k∑i=3(ri−1)(2i−ri−2)2 =m∑i=3i2−3i+22+k∑i=m+1(m−1)(2i−m−2)2 =12k2m−12km2+16m3−12k2+12k−16m =Ω(k2m)+Ω(km2+m3) =Ω(k2m)

empty rainbow triangles ∎

## 3 Upper Bound

In this section we construct a -colored point set which provides our upper bounds for .

### 3.1 The Empty Triangles of the Horton Set

As a building block for our construction we use Horton sets [18]; in this section we characterize the empty triangles of the Horton set. Let be a set of points in the plane with no two points having the same -coordinate; sort its points by their -coordinate so that . Let be the subset of the even-indexed points of , and

be the subset of the odd-indexed points of

. That is, and . Let and be two finite sets of points in the plane. We say that is high above if: every line determined by two points in is above every point in ; and every line determined by two points in is below every point in .

###### Definition 1.

is a Horton set if

1. ; or

2. ; and are Horton sets; and is high above .

Assume that is a Horton set. We say that an edge is a visible edge of if one of the following two conditions are met.

• Both and are even and for every even , the point is below the line passing through . In this case we say that is visible from above.

• Both and are odd and for every odd , the point is above the line passing through . In this case we say that is visible from below.

###### Lemma 3.1.

The number of visible edges of is less than .

###### Proof.

Let be a binary string starting with a and followed by a trail of ’s of length at most . Every consecutive pair of points of defines a visible edge from below of . Moreover, all visible edges from below of are of this form, for some . Note that . A similar analysis holds for the edges visible from above of , using the binary strings starting with a and followed by a trail of ’s of length at most . The number of visible edges of is at most

 2⌈log2(n)⌉∑i=1n2i<2n.

The visible edges of allows to characterize its empty triangles recursively as follows.

###### Lemma 3.2.

Let and be the vertices of a triangle of such that either

• is an edge visible from below and ; or

• is an edge visible from above and .

Then is empty. Moreover, every empty triangle of with at least one vertex in each of and is of one these forms.

###### Proof.

If is such a triangle then its emptiness follows from the definition of the Horton set. Suppose now that is an empty triangle of with and , or and . Then, for to be empty, must be an edge visible from above (resp. below). ∎

We can now get a good upper bound on the number of empty triangles of .

###### Corollary 3.3.

The number of empty triangles of is at most .

###### Proof.

Let be the number of empty triangles in a Horton set of points. Then is equal to the number of empty triangles with at least one vertex in each of and , plus the number of empty triangles with all their vertices in or all their vertices in . By the definition of Horton sets and Lemma 3.2 we have that

 T(n)

### 3.2 Blockers

Our strategy is to start with a Horton set of points and replace each point of with a cluster of points. All of the points of are of the same color and are at a distance of at most some from . We choose to be arbitrarily small. Let be the resulting set. Note that every rainbow triangle of must have all its vertices in different clusters. Moreover, since each is arbitrarily close to we have the following. If is an empty triangle of with vertices in different clusters , and then , and are the vertices of an empty triangle in . In principle, this gives empty rainbow triangles in per empty triangle of . However, we can place the points within each cluster in such a way so that only very few of these triangles are actually empty.

Let , and . In what follows we iteratively define real numbers

 ε=ε1>ε2>⋯>εr+1>0;

in the process we also place a subset of points of at some of these distances; we refer to the points in as blockers. For suppose that has been defined and possibly some points of have been placed. Consider every pair of points distinct from . Let be at distance or more from and such that is in the interior of every triangle with vertices and , where and are at a distance of at most of and , respectively. Let be the triangle with vertices and , where is any point at a distance of at most from . We define small enough so that every such is in the interior of every such . We say that blocks the triangle with vertices and .

We construct iteratively as follows. We say that a blocker point at distance from is at layer . Let be the binary strings such that:

• ;

• for every , or ; and

• .

By and we have that for every . Note that .

Sort the points of counterclockwise by angle around . For every and as long as we have placed at most blocker points, we place two blocker points at a distance from from as follows.

• Suppose that . Place one blocker point just after the leftmost point of in order by angle around ; place another blocker point just before the rightmost point of in order by angle around , as depicted in Figure 1(a).

• Suppose that . Place one blocker point just after the leftmost point of in order by angle around ; place another blocker point just before the rightmost point of in order by angle around .

Let the set of these blocker points. If then we proceed to place the remaining points of . If then place the remaining points of in any way at a distance of at most of ; in this case we have that . Suppose that . For every we place additional blocker points as follows.

• If then place a blocker point, at a distance of from , between any two consecutive points of in order by angle around ; see Figure 1(b).

• If then place a blocker point, at a distance of from , between any two consecutive points of in order by angle around .

Let the set of these blocker points. No more blocker points are added and . If then place the remaining points of in any way at a distance of at most from .

We are now ready to prove our upper bounds on .

###### Proof of the upper bound in Theorem 1.1.

We count the number of empty rainbow triangles determined by as constructed above. To every empty triangle of with vertices , and , we assign the empty triangle of with vertices , and . Let be the binary string such that the vertices of are contained in but not in and . We say that is in layer . Without loss of generality suppose that and are both contained in or are both contained in .

If contains a blocker point of each of and then these blocker points are at layer . In this case there at most possible choices for each of and . Otherwise, and there are at most possible choices for each of and . If then contains a point from ; and there at most possible choices for . Otherwise, and there at most possible choices for . Summarizing, is assigned to at most the following number of empty rainbow triangles of :

 m3 if m≤2|s|+1;\par4(|s|+1)2m if 2|s|+1

By Lemma 3.2, is a visible edge of . Since , by Lemma 3.1 there are at most empty triangles in . Thus, for every there at most empty triangles in at layer . Let . Therefore, the number of empty rainbow triangles determined by is at most

 ⌊m′/2⌋∑t=14t2m′(8k22t−1)+⌈log2(k)⌉∑t=⌊m/2⌋+1m3(8k22t−1), (2)

where the second term is set to if .

If then (2) is at most

 ⌊m/2⌋∑t=14t2m(8k22t−1)+⌈log2(k)⌉∑t=⌊m/2⌋+14t2m(8k22t−1), =⌈log2(k)⌉∑t=14t2m(8k22t−1) =32k2m⌈log2(k)⌉∑t=1(t22t−1) ≤384k2m =O(k2m).

If then (2) is at most

 ⌈log2(k)⌉∑t=14t2(k+2⌈log2(k)⌉)(8k22t−1) =32k2(k+2⌈log2(k)⌉)⌈log2(k)⌉∑t=1(t22t−1) ≤384k2(k+2⌈log2(k)⌉) =O(k3).

Therefore,

 f(k,m)={O(k2m) if m≤k,\parO(k3) if m>k

## 4 Empty Rainbow Quadrilaterals

A natural generalization is to consider empty rainbow polygons; we construct a -colored point set with the same number of points in each color class and that does not determine an empty rainbow quadrilateral. First, we observe the following.

###### Lemma 4.1.

The point set depicted in Figure 2 does not determine an empty rainbow quadrilateral.

###### Proof.

Let be a rainbow quadrilateral of the point set depicted in Figure 2. Note that must have and as vertices. Thus, at least two of , and are sides of . Assume without loss of generality that and are sides of . If the fourth vertex of is not one of the red points near then these points are inside ; and is not empty. If the fourth vertex of is one of the red points near then by construction the other red point near is inside ; and again is not empty. ∎

We use Lemma 4.1 to construct our point set. First we take a regular -gon, , with vertices ; we replace every point with a cluster of points of color . Let be a copy of with vertices , which is rotated by . So form a regular -gon. Let be sufficiently small. For every , we place the points of at a distance of at most from . For () we place at least points of color arbitrarily close to the line segment and so that the following holds. Let and be any two consecutive points of distinct from . In the triangle with vertices and there are at least two points on of color . Furthermore, these points are at a distance of at least to the lines and . Note that . Let be the set of points of color in this construction. The construction for is depicted in Figure 3.

###### Theorem 4.2.

The set is a -colored point set, with the same number of points of each color class, that does not determine an empty rainbow quadrilateral.

###### Proof.

Let be a rainbow triangle with vertices , and . We show that has a structure like the point set depicted in Figure 2. Consider the points with color near the line segment and that are between the line segments and . By construction, there are at least two points of color near between and . Since these points are at a distance of at most to either of these lines, they are also between and . By the same argument there are points of color on and inside . Therefore, by Lemma 4.1, the point set does not determine an empty rainbow quadrilateral. ∎

The construction described above determines many monochromatic quadrilaterals. This leads us to the following question.

###### Problem 1.

Does every sufficiently large -colored () point set with the same number of points in each color class determines an empty rainbow quadrilateral or an empty monochromatic quadrilateral?

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