On balanced 4holes in bichromatic point sets
^{1}
Abstract
Let be a point set in the plane in general position such that each of its elements is colored either red or blue, where and denote the points colored red and the points colored blue, respectively. A quadrilateral with vertices in is called a hole if its interior is empty of elements of . We say that a hole of is balanced if it has red and blue points of as vertices. In this paper, we prove that if and contain points each then has at least balanced holes, and this bound is tight up to a constant factor. Since there are twocolored point sets with no balanced convex holes, we further provide a characterization of the twocolored point sets having this type of holes.
1 Introduction
Let be a set of points in the plane in general position. A hole of is a simple polygon with vertices in and with no element of in its interior. If has vertices, it is called a hole of . Note that we allow for a hole to be nonconvex. We will refer to a hole that is not necessarily convex as general hole, and to a hole that is convex as convex hole. The study of convex holes in point sets has been an active area of research since Erdős and Szekeres [6, 7] asked about the existence of points in convex position in planar point sets. It is known that any point set with at least ten points contains convex holes [10]. Horton [11] proved that for there are point sets containing no convex holes. The question of the existence of convex holes remained open for many years, but recently Nicolás [15] proved that any point set with sufficiently many points contains a convex hole. A second proof of this result was subsequently given by Gerken [9].
Recently, the study of general holes of colored point sets has been started [1, 2]. Let be a finite set of points in general position in the plane. The elements of and will be called, respectively, the red and blue elements of , and will be called a bicolored point set. A hole of is balanced if it has two blue and two red vertices.
In this paper, we address the following question: Is it true that any bicolored point set with at least two red and two blue points always has a balanced hole? We answer this question in the positive by showing that any bicolored point set with always has a quadratic number of balanced holes. We further characterize bicolored point sets that have balanced convex holes.
The study of convex holes in colored point sets was introduced by Devillers et al. [5]. They obtained a bichromatic point set with points that contains no convex monochromatic hole. Huemer and Seara [12] obtained a bichromatic point set with points containing no monochromatic holes. Later, Koshelev [13] obtained another such point set with elements. Devillers et al. [5] also proved that every colored Horton set with at least elements contains an empty monochromatic convex hole. In the same paper the following conjecture is posed: Every sufficiently large bichromatic point set contains a monochromatic convex hole. This conjecture remains open, and on the other hand Aichholzer et al [2] have proved that any bicolored point set always has a monochromatic general hole. Recently, a result well related with balanced holes was proved by Aichholzer et al [3]: Every twocolored linearlyseparable point set with contains at least balanced general holes. In a forthcoming paper, the same authors proved the lower bound on such holes in the case where and are not necessarily linearlyseparable. One can note that a balanced hole with vertices (even if and are linearly separable) does not always imply a balanced hole with vertices (see, e.g., Figure 1).
Our results: For balanced general holes, that is, balanced holes not necessarily convex, we first show that every bicolored point set with has at least one balanced hole. We then prove that if then has at least balanced holes (Theorem 1 of Section 2), and show that this bound is tight up to a constant factor. This lower bound is improved to in the case where and are linearly separable (Theorem 5 of Section 2.1). On the other hand, for balanced convex holes, we provide a characterization of the bicolored point sets having at least one such hole (Theorem 10 of Section 3.1, and Theorem 13 of Section 3.2). Finally, in Section 4, we discuss extensions of our results such as generalizing the above lower bounds for point sets in which , proving the existence of convex holes either balanced or monochromatic, deciding the existence of balanced convex holes, and others.
General definitions: Given any two points of the plane, we denote by the straight segment connecting and , by the line passing through and , and by the ray that emanates from and contains . For every three points of the plane, we denote by the open triangle with vertex set . Given , let denote the convex hull of .
Given three noncollinear points , , and , we denote by the open convex region bounded by the rays and . Given a set , let denote a point minimizing the area of over all points of .
2 Lower bounds for general balanced holes
It is not hard to see that if , then contains a balanced hole. To prove this, observe that for every set of four points there always exists a simple polygon whose vertices are the elements of . Let be a subset of containing exactly two red points and two blue points, such that the area of the convex hull of is minimum. Clearly, any simple polygon whose vertex set is contains no element of in its interior, and thus it is a balanced hole of .
On the other hand, if has exactly two points of one color and many points of the other color, then might contain only a constant number of balanced holes. For example, the reader may verify that the point set of Figure 2 contains exactly five balanced holes.
In the case where , has (at least) a linear number of balanced holes. Indeed, by applying the hamsandwich theorem recursively, we can partition into a linear number of constant size disjoint subsets whose convex hulls are pairwise disjoint, and each of them contains at least two red points and two blue points, and has thus a hole.
In this section we prove the following stronger result:
Theorem 1.
Let be a set of points in general position in the plane such that . Then has at least balanced holes.
We consider some definitions and preliminary results to prove Theorem 1. In the rest of this section we will assume that .
Given two points with different colors, let be the set of the at most four points obtained by taking the first point found in each of the next four rotations: the rotation of around clockwise; the rotation of around counterclockwise; the rotation of around clockwise; and the rotation of around counterclockwise.
We classify (or color) the edge with one of the following four colors: green, black, red, and blue. We color green if it is an edge, or a diagonal, of some balanced hole. If is an edge of the convex hull of and is not green, then is colored black. If is neither green nor black, then all the points in must have the same color and there are elements of to each side of . We then color with the color of the points in .
Lemma 2.
The number of red edges and the number of blue edges are each at most .
Proof.
Let be any red point. Sort the elements radially around in counterclockwise order, and label them in this order. Subindices are taken modulo .
Suppose that the edge is red, , and the angle needed to rotate the ray counterclockwise around in order to reach is less than . If does not contain elements of , then there must exist a red point in . Then, the quadrilateral with vertex set is a balanced hole, where , which contradicts that is red (see Figure a). Hence, must contain red points. In fact, contains at least three red points in order to avoid that , , and , joint with some red point in , form a balanced hole with edge (see Figure b and Figure c). By symmetry, for every red edge , if is reached by rotating clockwise by an angle less than , then the triangle contains at least three red points. These observations imply that the number of red edges among (i.e. the number of red edges incident to ) is at most . Summing over all the red points, the total number of red edges is at most .
Analogously, the total number of blue edges is also at most . ∎
Lemma 3.
The number of green edges is at least .
Proof.
There are bichromatic edges in total. By Lemma 2, at most of them are red and at most are blue. Further observe that at most edges are black. Then the number of green edges is at least:
∎
Observe now that any balanced hole defines at most four green edges as polygonal edges or diagonals. Thus, by Lemma 3, the number of balanced general holes is at least , and Theorem 1 thus follows.
2.1 The separable case
We now improve our bounds of the previous section for the case where and are linearly separable. Suppose without loss of generality that there is a horizontal line such that the elements in are above , and those in are below . Further assume that no two elements in have the same coordinate.
Lemma 4.
If and are linearly separable then both the number of red edges and the number of blue edges are each at most .
Proof.
Label the red points in the ascending order of the coordinates. Let be any red point, . Sort the blue points radially around in counterclockwise order and label them in this order. Similarly as in the proof of Lemma 2, if is red, , then among the triangle contains at least three elements if , and the triangle contains at least three elements if . Then the number of red edges incident to is at most , and over all the red points, the number of red edges is at most
If , for some integer , then:
If , then:
Finally, if , then:
Therefore, we have that the number of red edges is at most . Analogously, there are at most blue edges in total. ∎
Theorem 5.
If and are linearly separable then the number of balanced holes is at least .
Proof.
Since and are linear separable, the number of black edges is at most . Using Lemma 4, we can ensure that the number of green edges is at least
Then the number of balanced holes is at least . ∎
We observe that our lower bounds are asymptotically tight for point sets with . For example, if and are far enough from each other (i.e. any line passing through two points of does not intersect , and vice versa), is a concave chain, and a convex chain (see Figure 5), then the number of balanced holes is precisely ; each of them convex and formed by two consecutive red points and two consecutive blue points. This point set (without the colors) was called the double chain [8].
3 Balanced convex holes
In this section we characterize bicolored point sets that contain balanced convex holes. To start with, we point out that in general does not necessarily have balanced convex holes. The point sets shown in Figure 6 do not have balanced convex holes. Observe that the number of blue points in the interior of the convex hull of the blue points in Figure a and Figure b can be arbitrarily large. A more general example with eight points, 4 red and 4 blue linearly separable, is shown in Figure d, which can be generalized to point sets with points, , red and blue.
Let be two points of the same color. If and are red, will be called a redred edge. Otherwise, if and are blue, we call it a blueblue edge.
3.1 and are not linearly separable
We proceed now to characterize bicolored point sets , not linearly separable, which contain balanced convex holes. We assume .
Lemma 6.
If contains a redred edge and a blueblue edge that intersect each other, then contains a balanced convex 4hole.
Proof.
Choose a redred edge and a blueblue edge such that and the convex quadrilateral with vertex set is of minimum area among all possible convex quadrilaterals having a redred diagonal and a blueblue diagonal. Observe that is balanced and assume that is not a hole. Then contains a point of in its interior. Suppose w.l.o.g. that there is a red point in the interior of . Then we have that intersects , or intersects . Suppose w.l.o.g. the former case. Hence, is the vertex set of a balanced convex quadrilateral with a redred diagonal and a blueblue diagonal with area smaller than that of , a contradiction. ∎
Lemma 7.
If the boundaries of and intersect each other, then contains a balanced convex 4hole.
Proof.
Observe that there exist a redred edge and a blueblue edge that intersect each other. Therefore, the result follows from Lemma 6. ∎
Lemma 8.
Let be a bichromatic point set such that and are not linearly separable, , , and . Then contains a balanced convex hole if and only if there is a blueblue edge of such that one of the open halfplanes bounded by contains exactly 2 red points and no blue point.
Proof.
Let denote the three elements of . Suppose that there exists an edge of such that and belong to one of the two open halfplanes bounded by and that the elements of belong to the other open halfplane (see Figure 7). Then the quadrilateral with vertex set is a balanced convex hole.
Suppose now that has a balanced convex hole. Assume w.l.o.g. that this hole has vertex set , where intersects at the point , and intersects at the point (see Figure 7). We select the points and so that the sum of the distance of from the segment and the distance of from the segment is minimized. Observe from the choice of and that does not contain blue points. Hence, the result follows. ∎
Lemma 9.
Let be a bicolored point set such that and are not linearly separable, , , and . Then has a balanced convex hole.
Proof.
Let be a triangulation of . If there are two blue points that belong to different triangles of , then there exist a redred edge and a blueblue edge intersecting each other, and the result thus follows from Lemma 6. Suppose then that is completely contained in a single triangle of , with vertices in counterclockwise order.
If , there exists an edge of which is not intersected by the line through the two blue points. Then the two red points of that edge, joint with the two blue points, form a balanced convex hole (Lemma 8).
Suppose then that , thus has at least three vertices. Since there exists a triangle of sharing an edge with . Assume w.l.o.g. that such an edge is , and denote by the other vertex of . Further assume w.l.o.g. that is horizontal, and is below .
Let . Observe that is a vertex of . Let denote the vertex succeeding in in the counterclockwise order, and denote the vertex succeeding in in the clockwise order. Both and are not below the horizontal line through by the definition of . If either or does not intersect , then there is a balanced convex hole by Lemma 8. Suppose then that both and intersect . Refer to Figure 8.
We consider the following four cases according to the possible locations of point , by assuming w.l.o.g. that point is to the left of . The other symmetric cases arise when is to the right of .
Case 1: (see Figure a). The quadrilateral with vertex set is a balanced convex hole.
Case 2: (see Figure b). The quadrilateral with vertex set is a balanced convex hole, where .
Case 3: and (see Figure c). The quadrilateral with vertex set is a balanced convex hole, where .
Case 4: and (see Figure d). The quadrilateral with vertex set is a balanced convex hole, where and .
Since any location of is covered by one of the above cases (or by one of their symmetric ones), there exists a balanced convex hole. The result follows. ∎
By combining Lemma 7, Lemma 8, and Lemma 9, we obtain the following result that completely characterizes the nonlinearly separable bichromatic point sets that have a balanced convex hole.
Theorem 10.
Let be a bichromatic point set such that and are not linearly separable. Then has a balanced convex hole if and only if one of the following conditions holds:

, , , and there is a blueblue edge of such that one of the open halfplanes bounded by contains exactly 2 red points and no blue point.

, , , and there is a redred edge of such that one of the open halfplanes bounded by contains exactly 2 blue points and no red point.

, , ,

, , ,

The boundaries of and intersect each other.
3.2 and are linearly separable
In the rest of this section, we will assume that and are linearly separable. At first glance, one might be tempted to think that if the cardinalities of and are large enough, then always contains balanced convex holes. This certainly happens in the point set of Figure 5, in which and are far enough from each other. There are, however, examples of linearly separable bicolored point sets with an arbitrarily large number of points that do not contain any balanced convex hole. For instance, the point set shown in Figure d has no balanced convex hole. Observe in this example that if we choose a redred edge and a blueblue edge, the convex hull of their vertices is either a triangle or a convex quadrilateral that contains at least one other point in its interior.
Given an edge of and an edge of , we say that and see each other if the union of the sets of their vertices defines a balanced convex hole whose interior intersects with neither nor . We assume that there exists a nonhorizontal line such that the elements of are located to the left of and the elements of are located to the right.
Definition 11.
Let be a bicolored point set such that and are linearly separable. Conditions C1 and C2 are defined as follows:

There exist an edge of and an edge of such that and see each other.

There exists an edge of and points such that , , and ; or this statement holds if we swap and .
Lemma 12.
Let be a bicolored point set such that and are linearly separable. If there exist a point , a point , an edge of , and an edge of , such that the interiors of and intersect with the interior of , then C1 or C2 holds.
Proof.
Let and be the endpoints of and and the endpoints of . Assume w.l.o.g. that is horizontal, and are above , and then and are below . If and see each other (see Figure a), then C1 holds. Otherwise, assume w.l.o.g. that is contained in (see Figure b). We have because lies between the intersections of with and , which both are in the closure of . This implies that and . Then C2 is satisfied.
∎
Theorem 13.
A bichromatic point set , such that and are linearly separable, has a balanced convex hole if and only if C1 or C2 holds.
Proof.
If condition C1 holds then has trivially a balanced convex hole. Then suppose that condition C2 holds. Let and observe that since . Let be any red point in (see Figure a). Observe that we have either or . Assume w.l.o.g. the former case. Then the quadrilateral with vertex set is a balanced convex hole, where and .
Suppose now that has a balanced convex hole with vertices in counterclockwise order, where and . Let and be the edges of and , respectively, that intersect with both and (note that and might share vertices with and , respectively). If we have that and then and see each other, and thus C1 holds. Otherwise, if and then the interior of some edge among , , , and intersects the interiors of both and . Then, by Lemma 12, we have that C1 or C2 holds. Otherwise, there are two cases to consider: (1) and ; and (2) and . Consider case (1), case (2) is analogous. Let . If and see each other, then C1 holds. Otherwise (up to symmetry), belongs to (see Figure b). Since and , we have that C2 is satisfied. ∎
4 Discussion
A better counting of black edges: In the proof of our lower bounds, we considered the edges colored black, as those being edges of the convex hull of () that connect a red point with a blue point and are neither an edge nor a diagonal of any balanced hole. Specifically, in the proof of Lemma 3, we gave the simple upper bound for the number of black edges, but one can note that this bound can be improved. Nevertheless, any upper bound must be at least since the following bicolored point set has precisely black edges.
Let and consider a regular gon . Put a colored point at each vertex of such that the colors of its vertices alternate along its boundary. Orient the edges of counterclockwise. Then for each edge of put in the interior of three points of the color of the origin vertex of such that they are close enough to and ensure that there is no balanced hole with as edge. In total we have points, consisting of red points (i.e. red points in vertices of and red points in the interior of ) and blue points. See for example Figure 11, in which . Then, all the edges of are black.
Generalization of the lower bound for nonbalanced point sets: Let be a redblue colored point set such that . Let and . Using arguments similar to the ones used in Section 2, it can be proved that has at least
balanced holes. Observe that this bound is positive if and only if is at least roughly and is at least . Therefore, we leave as an open problem to obtain a lower bound for the cases in which the number of points of one color is below times the number of points of the other color.
Existence of convex holes, either balanced or monochromatic: Combining the characterization given by Theorem 10 and by Theorem 13, we obtain the following result:
Proposition 14.
Let a bicolored point set in the plane. If then always has a convex hole either balanced or monochromatic.
Proof.
If and are not linearly separable, then has a balanced convex hole by Theorem 10. Otherwise, consider that and are linearly separable. If the convex hull of contains a red point and the convex hull of contains a blue point in their interiors, then has a balanced convex hole by Lemma 12. Otherwise, at least one between and is in convex position and then has a monochromatic convex hole. ∎
Deciding the existence of balanced convex holes: Using the characterization Theorems 10 and 13, arguments similar to those given in Sections 3.1 and 3.2, and wellknown algorithmic results of computational geometry, we can decide in time if a given bicolored point set () of total points has a balanced convex hole.
We first compute the convex hulls and of and , respectively. After that, we decide if and are linearly separable. If they are not, we can decide in time whether one of the conditions (15) of Theorem 10 holds. Otherwise, if and are linearly separable, we proceed with the following steps, each of them in time. If the decision performed in any of these steps has a positive answer, then a balanced convex hole exists:

Decide whether the next two conditions hold: (1) contains red points in the interior or has at least three red vertices; and (2) contains blue points in the interior or has at least three blue vertices. If the answer is positive then the conditions of Lemma 12 are met and there thus exists a balanced convex hole in . Otherwise, if the answer is negative, assume w.l.o.g. that is in convex position.

Decide whether the conditions of Lemma 12 hold for at least one red point . Fixing a red point , those conditions can be verified in time as follows: Let be all the blue points labelled clockwise along the boundary of (subindices are taken modulo ). Let and be the two blue points such that and are tangent to , and let be the points between and which are on the side of the line opposite to the side containing . Note that the boundary of intersects the interior of for every . If is a vertex of , then it suffices to verify the existence of a blue point in , where and are the vertices preceding and succeeding , respectively, in the boundary of . Otherwise, if belongs to the interior of , then it suffices to verify the existence of a blue point in , that is, is not empty. Both and can be found in time, as well the existence of such a point can be decided in time by applying binary search over the points .

Decide whether Condition C1 holds. This can be done in time by simultaneously traversing the boundaries of and .

Decide whether Condition C2 holds. Using the fact that neither condition C1 nor the conditions of Lemma 12 hold, we claim that condition C2 can be decided by assuming that segment is an edge of and that point is the only blue point in the triangle (the condition C2 with and swapped is similar to decide). Namely, let be an edge of and be points such that , , and . Let , and observe that at least one neighbor of in the boundary of , say , satisfies and is the only one blue point in . The fact implies that we can verify condition C2 with being and being , where is an edge of (see Figure a). The claim thus follows. Therefore, there is a linearsize set of wedges of the form to consider, and we need to check if there is an incidence between any red point and an element of . Note that the elements of can be divided into two groups, such that in each group the intersections of the wedges with the interior of are pairwise disjoint (see Figure b). The wedge goes to the first group when is the clockwise neighbor of in the boundary of , and to the other group otherwise. Then, for each red point , one can decide in time such an incidence.
Counting balanced holes: Adapting the algorithm of Mitchell et al. [14] for counting convex polygons in planar point sets, we can count the balanced holes of a bicolored point set of points in time, where is the number of empty triangles of .
Existence of balanced holes in balanced point sets: The arguments used to prove the existence of at least one balanced hole in any point set with (at the beginning of Section 2) do not directly apply to prove the existence of balanced holes in point sets with . However, we can prove the following:
Proposition 15.
For all and , every point set with contains a balanced hole.
Proof.
If is in convex position then the result follows. Then, suppose that is not in convex position. For every point let , and for every let . W.l.o.g. let be a point in the interior of , and denote the elements of sorted radially in clockwise order around . For , let , where subindices are taken modulo . Notice that all ’s are odd, and implies that the points form a balanced hole.
We have that , which implies (given that ) that not all ’s can be greater than and that not all ’s can be smaller than . Suppose for the sake of contradiction that none of the ’s is equal to . Then, there exist an and an . Since we further have that for all , there must exist an element among which is equal to , and the result thus follows. ∎
Open problems: As mentioned above, we leave as open the problem of obtaining a lower bound for the number of balanced holes in point sets in which either or . Another open problem is to study lower bounds on the number of balanced holes, for even .
Acknowledgements
The authors would like to thank the two anonymous referees for their useful comments and suggestions.
Footnotes
 thanks: This is an arxiv version of [4].
 footnotemark:
References
 O. Aichholzer, R. FabilaMonroy, H. GonzálezAguilar, T. Hackl, M. A. Heredia, C. Huemer, J. Urrutia, and B. Vogtenhuber. 4holes in point sets. In 27th European Workshop on Computational Geometry EuroCG’11, pages 115–118, Morschach, Switzerland, 2011.
 O. Aichholzer, T. Hackl, C. Huemer, F. Hurtado, and B. Vogtenhuber. Large bichromatic point sets admit empty monochromatic 4gons. SIAM J. Discret. Math., 23(4):2147–2155, January 2010.
 O. Aichholzer, J. Urrutia, and B. Vogtenhuber. Balanced 6holes in linearly separable bichromatic point sets. Electronic Notes in Discrete Mathematics, 44(0):181–186, 2013.
 S. Bereg, J. M. DíazBáñez, R. FabilaMonroy, P. PerezLantero, A. RamirezVigueras, T. Sakai, J. Urrutia, and I. Ventura. On balanced 4holes in bichromatic point sets. Comput. Geom. Theory Appl., 48(3):169–179, 2015.
 O. Devillers, F. Hurtado, G. Károlyi, and C. Seara. Chromatic variants of the ErdősSzekeres theorem on points in convex position. Comput. Geom. Theory Appl., 26(3):193–208, November 2003.
 P. Erdős. Some more problems on elementary geometry. Austral. Math. Soc. Gaz., 5:52–54, 1978.
 P. Erdős and G. Szekeres. A combinatorial problem in geometry. Compositio Math., 2:463–470, 1935.
 A. García, M. Noy, and J. Tejel. Lower bounds on the number of crossingfree subgraphs of k. Comput. Geom. Theory Appl., 16(4):211–221, 2000.
 T. Gerken. Empty convex hexagons in planar point sets. Discrete Comput. Geom., 39(1):239–272, March 2008.
 H. Harborth. Konvexe fünfecke in ebenen punktmengen. Elem. Math., 33:116–118, 1978.
 J. D. Horton. Sets with no empty convex 7gons. Canad. Math. Bull., 26:482–484, 1983.
 C. Huemer and C. Seara. 36 twocolored points with no empty monochromatic convex fourgons. Geombinatorics, XIX(1):5–6, 2009.
 V. Koshelev. On ErdősSzekeres problem and related problems. ArXiv eprints, 2009.
 J. S. B. Mitchell, G. Rote, G. Sundaram, and G. J. Woeginger. Counting convex polygons in planar point sets. Inf. Process. Lett., 56(1):45–49, 1995.
 C. M. Nicolas. The empty hexagon theorem. Discrete Comput. Geom., 38(2):389–397, September 2007.