Drawing the Horton Set in an Integer Grid of Minimum Size

# Drawing the Horton Set in an Integer Grid of Minimum Size

## Abstract

In 1978 Erdős asked if every sufficiently large set of points in general position in the plane contains the vertices of a convex -gon, with the additional property that no other point of the set lies in its interior. Shortly after, Horton provided a construction—which is now called the Horton set—with no such -gon. In this paper we show that the Horton set of points can be realized with integer coordinates of absolute value at most . We also show that any set of points with integer coordinates combinatorially equivalent (with the same order type) to the Horton set, contains a point with a coordinate of absolute value at least , where is a positive constant.

## 1 Introduction

Although the number of distinct sets of points in the plane is infinite, for most problems in Combinatorial Geometry only a finite number of them can be considered as essentially distinct. Various equivalence relations on point sets have been proposed by Goodman and Pollack [12, 14, 15, 13]. One of these is the order type; it is defined on a set of points, , as follows. To every triple of points of , assign: a if lies to the left of the oriented line from to ; a if lies to the right of this line; and a if , , and are collinear. This assignment is also called the orientation of the triple, which may be negative, positive or zero, respectively. The set of all triples together with their orientation is the order type of ; two sets of points have the same order type if there is a bijection between them that preserves orientations.

Since their inception, order types were defined with computational applications in mind (see [13] for example). The orientation of a triple is determined by the sign of a determinant; many algorithms use precisely this determinant as their geometric primitive. Given that the determinant of an integer valued matrix is an integer, for numerical computations it is best if a point set has integer coordinates. Two main reasons are that integer arithmetic is much faster than floating point arithmetic, and that floating point arithmetic is prone to rounding errors. The latter is easily taken care of with an integer representation that can handle arbitrarily large numbers.

If a set of points has already integer coordinates, it is best if these coordinates have as small absolute value as possible—again, for computational reasons. Even though rounding errors can be avoided using arbitrarily large integers, the cost of computation increases as the numbers get larger. Also, if we wish to store the point set, the number of bits needed depend on the size of the coordinates.

Let be a set of points in general position in the plane. A drawing of is a set of points with integer coordinates and with the same order type as . The size of a drawing is the maximum of the absolute values of its coordinates. For the reasons mentioned above, it is of interest to find the drawing of of minimum size. In [16] Goodman, Pollack and Sturmfels presented sets of points in general position whose smallest drawings have size , and proved that every point set has a drawing of size at most (where and are positive constants).

Aichholzer, Aurenhammer and Krasser [1] have assembled a database of drawings. For , the database contains a drawing of every possible set of points in general position in the plane. The main advantage of having these drawings is that one can use them to compute certain combinatorial parameters of all point sets up to eleven points. The order type data base stops at eleven because the size of the database grows prohibitively fast. Thus, we cannot hope to store drawings for all point sets beyond small values of ; it is convenient however, to have programs that generate small drawings of infinite families of point sets which are of known interest in Combinatorial Geometry.

In this direction, Bereg et al. [6] provided a linear time algorithm to generate a drawing of a set of points called the Double Circle3. Their drawing has size ; they also proved a lower bound of on the size of every drawing of the Double Circle. In this paper we do likewise for a point set called the Horton Set [18]. In section 2 we provide a drawing of size of the Horton set of points; our drawing can be easily constructed in linear time. We also show in Section 3, a lower bound of (for some ) on the minimum size of any drawing of the Horton set. As a corollary, bits are necessary and sufficient to store a drawing of the Horton set.

We are mainly interested in having an algorithm that generates small drawings of the Horton set. However, the problem of finding small drawings also raises interesting theoretical questions. For example, after learning of our lower bound, Alfredo Hubard posed the following problem.

###### Problem 1.

Does every sufficiently large set of points, for which there exist a drawing of polynomial size, contains an empty -hole?

A -hole of a point set , is a subset of points of that form a convex polygon, with no other point of in its interior. Horton sets were constructed as an example of arbitrarily large point sets without -holes. In particular our lower bound implies that any set of points that has a drawing of polynomial size, cannot have large copies of the Horton set. We also believe that the machinery developed to prove Theorem 3.6 will be useful for analyzing Horton sets in other settings.

A preliminary version of this paper appeared in CCCG’14 [5]. In this paper all point sets are in general position and all logarithms are base 2.

### 1.1 The Horton Set(s)

In 1978 Erdős [9] asked if for every , any sufficiently large set of points in the plane contains a -hole. Shortly after, Harborth [17] showed that every set of points contains a -hole. The case of empty triangles (-holes) is trivial; the case of -holes was settled in the affirmative in another context by Esther Klein long before Erdős posed his question (see [10]). Horton [18] constructed arbitrarily large point sets without -holes, and thus without -holes for larger values of . His construction is now known as the Horton set. The case of -holes remained open for almost 30 years, until Nicolás [21], and independently Gerken [11], proved that every sufficiently large set of points contains a -hole.

Since its introduction, the Horton set has been used as an extremal example in various combinatorial problems on point sets. For example, a natural question is to ask: What is the minimum number of -holes in every set of points in the plane? The case of empty triangles was first considered by Katchalski and Meir [19]—they constructed a set of points with empty triangles and showed that every set of points contains of them. This bound was later improved by Bárány and Füredi [3], who showed that the Horton set has empty triangles. The Horton set was then used in a series of papers as a building block to construct sets with fewer and fewer -holes. The first of these constructions was given by Valtr [24]; it was later improved by Dumitrescu [8] and the final improvement was given by Bárány and Valtr [4].

Devillers et al. [7] considered chromatic variants of these problems. In particular, they described a three-coloring of the points of the Horton set with no empty monochromatic triangles. Since every set of points contains a -hole, every two-colored set of at least points contains an empty monochromatic triangle. The first non trivial lower bound of , on the number of empty monochromatic triangles in every two-colored set of points, was given by Aichholzer et al [2]. This was later improved by Pach and Tóth [22] to . The known set with the least number of empty monochromatic triangles is given in [2]; it is based on the known set with the fewest number of empty triangles, which in turn is based on the Horton set.

We now define 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, and be the subset of the odd-indexed points. That is, and . Let and be two 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.

The Horton set is a set of points, with no two points having the same -coordinate, that satisfies the following properties.

1. is a Horton set;

2. both and are Horton sets ();

3. is high above ().

This definition is very similar to the one given in Matoušek’s book [20] (page 36). The only difference is that in that definition either is high above or is high above ; i.e. this relationship is allowed to change at each step of the recursion. As a result, for a fixed value of , one gets a family of “Horton sets” (with different order types), rather than a single Horton set. Normally, this does not affect the properties that make Horton sets notable. For example, none of the them have empty heptagons. In some circumstances it does; as is the case of the constructions with few -holes [24, 8, 4]. We fixed one of these two options in order to make the proof of our lower bound more readable, but our results should hold for the general setting. Note that this choice fixes the order type of the Horton set. However, an arbitrary drawing of the Horton set need not satisfy Definition 1.

Horton described his set in a concrete manner with specific integer coordinates. Another description given in [3], is the following.

• .

• .

• .

This drawing and the original due to Horton have exponential size; we have not seen in the literature a drawing of subexponential size. Then again, to the best of our knowledge nobody has tried to find small drawings of the Horton set.

## 2 Upper bound

In this section we construct a small drawing of the Horton set of points. First, we define the following two functions.

 f(i)= {0ifi=1.2i(i−1)2−1ifi≥2. g(i)= {0ifi=1.f(i)−f(i−1)ifi≥2.

Afterwards, we use and to construct our drawing recursively as follows.

• ;

• ;

• ;

• .

###### Theorem 2.1.

There exist a drawing of the Horton set of points of size for .

###### Proof.

We prove by induction on that is the desired drawing. It can be verified by hand that has size equal to ; assume that . By induction and are Horton sets; it only remains to show that is high above . We only prove that every point of is above every line through two points of ; the proof that every point of is below every line through two points of is analogous.

Let be the points of sorted by their -coordinate. Let be two even integers, and let be the directed line from to . By definition is above the vertical line passing through ; in particular is above the line segment joining and . Since the smallest -coordinate of is equal to , and are above the line segment joining the points and . Therefore, it suffices to show that , and are above .

We define a line with the property that if and are above , then , and are above . Afterwards we show that indeed and are above .

If the slope of is non-positive, define to be the line passing through the points and ; if the slope of is positive, define to be the line passing through the points and . Note that the largest -coordinate of is equal to . Therefore the slope of is at least and at most ; in particular the absolute value of the slope of is larger or equal to the absolute value of the slope of . The farthest point of to the right that can contain while having non-positive slope is (which has -coordinate equal to ); the farthest point of to the left that can contain while having positive slope is . Therefore in both cases if and are above , then they are also above ; see Figure 1.

If has non-positive slope and is above , then is also above since has larger -coordinate. If has positive slope and is above , then must also be above . Otherwise intersects the line segment joining and ; this line segment has slope equal to , since the -coordinate of is equal to . This in turn would imply that has slope larger than —a contradiction.

Suppose has non-positive slope. Then it suffices to show that is above . This is the case since:

 ∣∣ ∣ ∣∣n−6f(k−1)1n−4011g(k)1∣∣ ∣ ∣∣ = 2g(k)−(n−5)f(k−1) = 2f(k)−(n−3)f(k−1) = 2f(k)−2kf(k−1)+3f(k−1) = 3f(k−1) > 0.

Suppose has positive slope. Then it suffices to show that is above . This is the case since:

 ∣∣ ∣ ∣∣0012f(k−1)1n−3g(k)1∣∣ ∣ ∣∣ = 2g(k)−(n−3)f(k−1) = 2f(k)−(n−1)f(k−1) = 2f(k)−2kf(k−1)+f(k−1) = f(k−1) > 0.

Finally the largest -coordinate of is equal to , and the largest -coordinate of is equal to

 k∑i=1g(i)=f(k)=2k(k−1)2−1=12n12log(n/2),

since . Therefore, is a drawing of the Horton set of points of size . ∎

## 3 Lower bound

In this section we prove a lower bound on the size of any drawing of the Horton set. As mentioned before, a drawing of the Horton set might not satisfy Definition 1; we call a drawing that does, an isothetic drawing of the Horton set. We first show a lower bound on the size of isothetic drawings of the Horton set (Theorem 3.5); afterwards, we consider the general case (Theorem 3.6). Throughout this section is an isothetic drawing of the Horton set of points, and are the points of sorted by their -coordinate.

As an auxiliary structure, we recursively define a complete rooted binary tree , as follows. is the root of ; and if is a vertex of , of at least two points, then and are its left and right children, respectively. Furthermore, for each vertex in , label the edge incident to its left child with a “0” and the edge incident with its right child with a“1”; the labels encountered in a path from a leaf to the root are precisely the bits in the binary expansion of ; see Figure 2.

By construction, the vertices of are sets of points of (for some ). Let be the set of vertices of that consist of exactly points of : we call it the -th4 level of . The first level, , are the vertices of that consist of a pair of points of . For each such pair, we consider the line through them as defined by them.

Let be the closed vertical slab bounded by the vertical lines through and . Let be a vertex at the first level of and let and be its leftmost and rightmost points respectively. Suppose that is a left child. Then the two most significant bits in the binary expansion of are “00”, and the two most significant bits in the binary expansion of are “10”. This implies that and ; in particular is contained in , while is to the left of . In this case, we say that is left-to-right crossing. By similar arguments if is a right child, then is contained in , while is to the right of . In this case we say that is right-to-left crossing. Note that the vertices in the first level of , in their left to right order in , are alternatively left-to-right and right-to-left crossing (see Figure 2). The following lemma relates the left to right order of these vertices in , to the bottom-up order of their corresponding pairs of points of .

###### Lemma 3.1.

The lines defined by the vertices of the first level of do not intersect inside . In particular, the bottom-up order of convex hull of these vertices corresponds to their left to right order in .

###### Proof.

Let and be two vertices in the first level of such that is a left child and is a right child. Without loss of generality assume that in the left to right order in , is before . Then, is left-to-right crossing and is right-to-left crossing. Let and be the lines defined by and , respectively. If and intersect inside , then the leftmost point of is above or the rightmost point of is below —a contradiction to property 3 of Definition 1. Since between every two left children there is a right child, and between every two right children there is a left child, the result follows. ∎

By construction every vertex of is an isothetic drawing of the Horton set. The main idea behind the proof of the lower bound on the size of isothetic drawings of the Horton set is to lower bound the size of these drawings in terms of the size of their children. We define some parameters on the vertices of , that make this idea more precise.

Let be an integer. Let and be four vertical lines sorted from left to right, such that:

• All of them are contained in the interior of .

• There are exactly points of between both pairs () and ).

Let be a vertex of with more than two points. For each of the , we define two parameters of . Let be the line defined by the leftmost descendant of in . Let be the line defined by the rightmost descendant of in . Note that is bounded from below by and from above by (Lemma 3.1). Let and be the left and right children of , respectively. Define as the distance between the points and , and as the distance between the points and ; see Figure 3.

We lower bound the girth of a vertex of in terms of the girth of one of its children. This bound is expressed in Lemma 3.2. Before proceeding we need one more definition. Let be a vertex of with more than two points and let be its parent. If is the left child of , let be the right child of ; otherwise let be the left child of .

###### Lemma 3.2.

Let be a vertex at the -th level of , for some . If the distance between and is , and the distance between and is , then:

• and,

• .

###### Proof.

We will prove inequality ; the proof of is analogous. Assume that is the left child of and let be the right child of ; the case when is the right child of can be proven with similar arguments. Note that is is the right child, , of .

Let and be two consecutive points in lying between and at a horizontal distance of at most from each other; such a pair exists as there are points of between and . Let be the point in that lies between and (in the -coordinate order). Let be the line through and . Let ; note that the slope of is at most . Recall that by Lemma 3.1, and do not intersect between and ; this implies that , in particular . Therefore, the slope of is at most

 −Δy/Δx=−(d1d1+d2girth4(Q))/Δx=d1(d1+d2)d22l−t−1girth4(Q).

Define the following points , and (see Figure 4). Note that the leftmost point of is to the left of ; since this point is above , cannot be above . Therefore, the distance from to is at most the distance from to ; the distance from to is precisely . We now show that the distance from to is at least —this completes the proof of .

Let be the line parallel to and passing through the intersection point of and . Note that is below . Therefore, the distance from to is at least the distance of to the intersection point of : this is at least . ∎

Two obstacles prevent us from directly applying Lemma 3.2. One is that the difference between and may be too big and in consequence or too small. This situation can be fixed with following Lemma.

###### Lemma 3.3.

For and , has size at least or can be chosen so that the ratio between and is at least and at most .

###### Proof.

Let be consecutive vertical lines such that:

• all of them lie in the interior of and,

• between every pair of two consecutive lines there are exactly points of .

For , let be the distance between and ; let be these distances sorted by size. We look for a pair , such that one is at most two times the other. Suppose there is no such pair; then . Since between the two lines defining there are exactly points of , and no three of them have the same integer -coordinate, . Therefore,

 Δ′2t−1−1≥2k−t−1⋅22t−1−2≥212k2+k−t−3≥212k2+k−2log(k)−4≥n12logn.

The latter part of the inequality follows from our assumption that . Therefore, if there is no such pair, has size at least .

The second obstacle is that the second term in the right hand sides of inequalities (1) and (2) of Lemma 3.2 may be too large. In this case, we prune to get rid of vertices of large width; this is done by choosing an integer and then removing from all the points are contained in either: all the vertices of that are a left child to their parent, or all the the vertices of that are a right child to their parent (see Figure 5). We call this operation pruning the -th level of . The resulting set is a drawing of the Horton set, as shown by the following lemma.

###### Lemma 3.4.

Let be the subset of that results from pruning the -th level of . Then:

• is an isothetic drawing of the Horton set of points.

• Suppose that . Let be the tree associated to , and be any vertex at the -th level of (for some ). Then there exist a vertex at the -level of that contains . Moreover, .

###### Proof.

Assume without loss of generality that the left children are removed when pruning . If , (2) holds trivially, and (1) holds because in that case . Assume that , and let ; we proceed by induction on .

Note that and are each an isothetic drawing of the Horton set of points. Moreover, their corresponding trees, and , are the subtrees of rooted at and , respectively. Therefore, when we prune the -th level of , we also prune the -th level of and . By induction and (1), this produces two isothetic drawings of the Horton set of points; let and be these drawings, respectively.

We first prove that

can be constructed from by, starting at , alternatively removing and keeping intervals of consecutive points of .

()

For , this is trivial since and . Thus, by induction, and are constructed from and by, starting at their leftmost point, alternatively removing and keeping intervals of consecutive points of and , respectively. Let and be these intervals (in order). Finally, follows by letting .

We now prove 1 and 2.

• Note that implies that and . Thus and ; in particular is high above . Therefore, is an isothetic drawing of the Horton set of points.

• Consider the following algorithm. Remove from the subtrees rooted at the vertices in the -th level of that are a left child to their parent; afterwards, remove from each vertex of the points in . After this last step, each vertex at the -th level of that was not removed is equal to its parent—producing a loop; remove these loops. We claim that this algorithm produces . For , this follows from . Let and be the left and right subtrees of the root of , respectively. By induction and can constructed from and with the above algorithm, respectively. Since the root of is precisely the root of minus the points in , the algorithm produces .

Now, suppose that and let be a vertex at the -th level of , for some . By the algorithm, there is a vertex such that ; this vertex is in the -level of . Finally, also by the algorithm we have that .

We are now ready to prove our lower bound on the size of isothetic drawings of the Horton set.

###### Theorem 3.5.

For a sufficiently large value of , every isothetic drawing of the Horton set of points has size at least .

###### Proof.

Set and assume that . By Lemma 3.3 and can be chosen so that, the ratio of the distance between and is at least and at most . Without loss of generality assume that . Let be the distance between and . We may assume that

 D

as otherwise we are done.

Let be a vertex in the -th level of . Note that between two consecutive points in every vertex at the -th level of there are exactly points of . This trivially holds for ; it holds for smaller values of , by induction on . In particular, there are points of between the leftmost and rightmost point of .

This implies that there are exactly two points of between and , and exactly two points of between and .

Suppose that there were less than two points of between and , then the number of points of would be at least the sum of the following.

• The number of points of between and ; recall that this is equal to .

• The number of points of between the leftmost and rightmost point of that are not between and ; since there are exactly points of between two consecutive points of , and at most one point of between and , this is at least .

• Two, for the leftmost and rightmost point of .

In total this is at least —a contradiction; similar arguments hold for and .

Suppose that there are more than two points of between and , then the number of points of between and is at least ; this is a contradiction to the assumption that there are exactly points of between and . The same argument holds for and .

The two points of between and , and the two points of between and , have integer coordinates. Therefore, by Pick’s theorem [23] the area of their convex hull is at least one. Since these points are contained in trapezoid bounded by , , and , the area of this trapezoid is also at least one. But this area is at most . Therefore

 max{width1(Q),width4(Q)}≥1/D. (1)

This bound also holds for every vertex at a level higher than (since all of these vertices contain vertices at the -level as subsets).

Let be the largest positive integer such that there exists a vertex in the -th level of that satisfies:

 max{width1(S(R)),width4(S(R))}≥2(l−t−6)(l−t−7)/2D. (2)

Such an and exist since (2) holds for every vertex at the -th level of . Indeed if is a vertex at the -th level of , then is in the level of and by (1):

 max{width1(S(Q)),width4(S(Q))}≥1/D=2((t+6)−t−6)((t+6)−t−7)/2D.

Without loss of generality assume that and that is a left child. We may assume that , otherwise (2) implies that has size at least (for a sufficiently large value of ).

We now apply Lemma 3.4 to prune of all the vertices of large width (that, is that satisfy (2)). Prune the -th level of by removing all the vertices that are a left child to their parent. Let be the resulting point set and its corresponding tree. No vertex of in a level higher than satisfies (2); otherwise, by part of Lemma 3.4 there would be a vertex at level of higher than that satisfies (2).

Let be the path from to the root of . We prove inductively for , that:

 girth1(Q′m) ≥2(m−t−6)(m−t−7)/2Difm≡lmod2, (3) girth4(Q′m) ≥2(m−t−6)(m−t−7)/2Difm≢lmod2 (4)

(3) holds for since . Assume that and that both (3) and (4) hold for smaller values of . Suppose that has the same parity as . Then by inequality (1) of Lemma 3.2 and inequalities (2) and (3):

 girth1(Q′m) ≥((d1)2(d1+d2)d2)2m−t−2girth4(Q′m−1)−width1(S(Q′m−1)) ≥2m−t−5girth4(Q′m−1)−2(m−t−7)(m−t−8)/2D ≥2m−t−52(m−t−7)(m−t−8)/2D−2(m−t−7)(m−t−8)/2D ≥2m−t−62(m−t−7)(m−t−8)/2D =2(m−t−6)(m−t−7)/2D

Therefore has size at least . This at least , for a sufficiently large value of . Since , the result follows. The proof when has different parity as is similar, but uses inequality (2) of Lemma 3.2 instead. ∎

To prove the general lower bound we do the following. Take a drawing of the Horton set; find a subset of half of its points, for which we know that there exists a linear transformation that maps it into an isothetic drawing; afterwards, apply Lemma 3.5 to the image and use the obtained lower bound to lower bound the size of original drawing.

###### Theorem 3.6.

Every drawing of the Horton set of points has size at least , for a sufficiently large value of and some positive constant .

###### Proof.

Let be a (not necessarily isothetic) drawing of the Horton set of points. As and have the same order type we can label with the same labels as , such that corresponding triples of points in and have the same orientation. Let be with these labels.

Note that the clockwise order by angle of around is , and that lies in an unbounded cell of the line arrangement of the lines defined by every pair of points of ; thus, point can be moved towards infinity without changing this radial order around . Therefore, there is a direction in which if is projected orthogonally the order of the projection is precisely . We may rotate as long as it does not coincide with a direction defined by a pair of points of and the order of in this projection does not change. Let and be the first vectors, defined by pairs of points of , encountered when rotating to the left and to the right, respectively; let .

We may assume that ; otherwise one of and has length at least , and therefore a coordinate of value at least . Let . Consider a change of basis from the standard basis to . Note that under this transformation is mapped to . We multiply the image of under this mapping by , to obtain an isothetic drawing of the Horton set on points. By Theorem 3.5, this drawing has size at least . Therefore, has size at least . ∎

We point out that the constants in the exponent of the lower bounds of Theorems 3.5 and 3.6 can be improved. We simplified the exposition at the expense of these worse bounds.

Acknowledgments. We thank Dolores Lara, Gustavo Sandoval and Andrés Tellez for various helpful discussions.

### Footnotes

1. footnotemark:
2. footnotemark:
3. The Double Circle of points is constructed as follows. Start with a convex -gon; arbitrarily close to the midpoint of each edge, place a point in the interior of this polygon; finally place a point at each vertex of the polygon.
4. In the literature the -th level of a binary tree are those vertices at distance from the root; we have precisely the opposite order.

### References

1. O. Aichholzer, F. Aurenhammer, and H. Krasser.