Point Sets with Small Integer Coordinates and no Large Convex Polygons

# Point Sets with Small Integer Coordinates and with Small Convex Polygons

## Abstract

In 1935, Erdős and Szekeres proved that every set of points in general position in the plane contains the vertices of a convex polygon of vertices. In 1961, they constructed, for every positive integer , a set of points in general position in the plane, such that every convex polygon with vertices in this set has at most vertices. In this paper we show how to realize their construction in an integer grid of size .

## 1 Introduction

A set of points in the plane is in general position if no three of its points are collinear. Let be a set of points in general position in the plane. A convex -gon of is a convex polygon of vertices whose vertices are points of . Erdős and Szekeres [7] proved that every set of points in general position in the plane contains a convex -gon. Using Stirling’s approximation this bound can be rephrased as follows.

###### Theorem 1

(Erdős and Szekeres 1935)

Every set of points in general position in the plane has a convex -gon of at least vertices.

More than quarter of a century afterwards, Erdős and Szekeres [8] provided a construction of a set of points in general position such that every convex -gon of this set has logarithmic size. Specifically, they showed the following.

###### Theorem 2

(Erdős and Szekeres 1961)

For every integer , there exists a set of points in general position in the plane such that every -gon of this set has at most vertices.

Some inaccuracies in the construction of Erdős and Szekeres were corrected by Kalbfleisch and Stanton in [12]. The Erdős-Szekeres construction as described in [12] uses integer-valued coordinates. The size of these coordinates grows quickly with respect to . This has led some researchers to conjecture that the Erdős-Szekeres construction cannot be carried out with small integer coordinates.

As an example here are some excerpts from the book “Research Problems in Discrete Geometry” [4] by Brass, Moser and Pach regarding the Erdős-Szekeres construction.

“The complexity of this construction is reflected by the fact that none of the numerous papers on the Erdős-Szekeres convex polygon problem includes a picture of the -point set without a convex hexagon.”

“Kalbfleisch and Stanton [12] gave explicit coordinates for the points in the Erdős-Szekeres construction. However, even in the case of the coordinates are so large that they cannot be used for a reasonable illustration.”

“The exponential blowup of the coordinates in the above lower bound constructions may be necessary. It is possible that all extremal configurations belong to the class of order types that have no small realizations.”

Also, in the survey [16] on the Erdős-Szekeres problem by Morris and Soltan we find the following.

“The size of the coordinates of the points in the configurations given by Kalbfleisch and Stanton [12] that meet the conjectured upper bound on grows very quickly. A step toward showing that this is unavoidable was taken by Alon et al. [1].”

In this paper we prove that the Erdős-Szekeres construction can be realized in a rather small integer grid of size. Our main result is the following.

###### Theorem 15

The Erdős-Szekeres construction of points can be realized in an integer grid of size .

This solves an open problem of [4], which we discuss, together with other problems, in Section 4.

To finish this section we mention the rich history behind the improvements on the upper bound on the Erdős-Szekeres theorem. Let be the smallest integer such that every set of points in general position in the plane contains a convex -gon.

The upper bound given by Erdős and Szekeres is of

 n(k)≤(2k−4k−2)+1.

It took 63 years for an improvement to be found, but in the course of one year many of them followed: Chung and Graham [5] proved that ; Kleitman and Pachter [13] proved that Tóth and Valtr [20] improved this bound roughly by a factor of , they showed that Eight years later in 2006, Tóth and Valtr [21] further improved this bound by .

Very recently there has been a new set of improvements. Vlachos [24] proved that

 n(k)≤(2k−5k−2)−(2k−8k−3)+(2k−10k−7)+2.

This implies that

 limsupk→∞n(k)(2k−5k−2)≤2932.

Afterwards, Norin and Yuditsky [18] improved this to

 limsupk→∞n(k)(2k−5k−2)≤78.

Mojarrad and Vlachos [15] made a furthter improvement and showed that . Finally, in 2016, Suk [19] made a huge improvement; he proved that

 n(k)≤2k+o(k).

The best lower bound is the one given by the Erdős-Szekeres construction. It implies that

 n(k)≥2k−2+1.

This is conjectured to be value of .

This paper is organized as follows. In Section 2 we describe in detail the Erdős-Szekeres construction and show how to realize it in a small integer grid. Based on the description of Section 2, we implemented an algorithm to compute the Erdős-Szekeres construction. In Section 3 we discuss optimizations made on our implementation to further reduce the size of the integer coordinates; we also provide figures of the Erdős-Szekeres construction for and . In Section 4 we finalize the paper by proposing Erdős-Szekeres type problems for point sets with integer coordinates.

## 2 The Erdős-Szekeres Construction

The Erdős-Szekeres construction is made from smaller point sets which we now describe.

### 2.1 Cups and Caps

A -cup of is a convex -gon of bounded from above by a single edge; similarly a -cap of is a convex -gon of bounded from below by a single edge; see Figure 1. 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 .

The building blocks of the Erdős-Szekeres construction are point sets ; which are constructed recursively as follows.

• if or ;

• ;

where: {IEEEeqnarray}rCl L_k,l & :=& S_k-1,l;
R_k,l &:= &{(x+δ_k,l,y+δ_k,l’):(x,y) ∈S_k,l-1}; is chosen large enough so that is to the right of ;

and is chosen large enough with respect to so that is high above .

It can be shown by induction on that has points, and that does not contain a -cup nor a -cap.

In Matoušek’s book [14] it is left as an exercise to show that can be realized in an integer grid of polynomial size. According to Matoušek this was noted by Valtr. We follow a different approach; we prove that a superset of can be realized with small positive integer coordinates; even though for some values of and this implies a realization of with integer coordinates of exponential size.

### 2.2 A Superset of Sk,l

Let be an integer. In this section we construct a point set in general position in the plane with integer coordinates such that has points, and for , contains as a subset.4

We define recursively as follows.

• ;

• ;

where: {IEEEeqnarray}rCl L_r & :=& P_r-1
R_r &:= &{(x+δ_r,y+δ_r’):(x,y) ∈L_r}
δ_r & := & 3 ⋅4^r-1;
δ_r’ & := & (3r+1) ⋅4^r-1.

Let be the value of the largest -coordinate of ; note that for

 Xr=Xr−1+δr.

Since , by induction we have that

 Xr=4r−1. (1)

Let be the value of the largest -coordinate of ; note that for

 Yr=Yr−1+δ′r.

Since , by induction we have that

 Yr=r⋅4r. (2)

Since

 δr>Xr−1,

every point of is to the left of every point of .

For , let be the rightmost point of and let be the leftmost point of ; let be the straight line passing through and ; see Figure 2. By construction of , the point is the point of of largest -coordinate, and the point is the point of of smallest -coordinate; therefore, the slope of is given by {IEEEeqnarray}rCl m_r & = & Yr-2Yr-1Xr-2Xr-1
& = & r4r-2(r-1)4r-14r-1-2(4r-1-1)
& = & 2r ⋅4r-1+2⋅4r-12⋅4r-1+1
& = & r+1-r+12⋅4r-1+1 .

The are increasing, since

 mr−mr−1=1+r2⋅4r−2+1−r+12⋅4r−1+1>0;

the last inequality follows from the fact that for .

Let and be the translations of and in , respectively. Let be the line defined by the rightmost point of and the leftmost point of . Thus, is the translation of in . We now prove some properties of .

###### Lemma 3

Among the lines passing through two points of , is the line with the largest slope.

###### Proof.

We proceed by induction on . For and , the lemma holds trivially. So assume that and that the lemma holds for smaller values of . Let be a line passing through two points of .

Suppose that passes through two points of or through two points of . By induction the slope of is at most . Since , the slope of is larger than the slope of .

Suppose that passes through a point of and a point of . Consider the polygonal chain . Since is to the left of and is to the left of , the slope of is at most the maximum of the slopes of the edges of . By induction each of these edges has slope at most . Therefore, the slope of is at most the slope of . ∎

###### Lemma 4

The rightmost point of is above and the leftmost point of is below .

###### Proof.

The result holds trivially for and ; assume that .

First we prove that the rightmost point of is above . Note that . Let be the point in with -coordinate equal to ; note that since contains the point , the -coordinate of is equal to . Therefore, it is sufficient to show that:

 Yr>Yr−2+mr−1(Xr−Xr−2).

Equivalently that

 Yr−Yr−2Xr−Xr−2>mr−1.

This follows from {IEEEeqnarray}rCl Yr-Yr-2Xr-Xr-2 & =& r4r-(r-2)4r-24r-1-(4r-2-1)
& = & 15r⋅4r-2+2⋅4r-215⋅4r-2
& = & r+215, and that by (2.2)

 mr−1=r−r2⋅4r−2+1.

Now we prove that the leftmost point of is below . Note that is the leftmost point of . Let be the point in with -coordinate equal to ; note that since contains the point , the -coordinate of is equal to . Therefore, it is sufficient to show that:

 Yr−2+δ′r−mr−1(Xr−2+δr)>0.

Equivalently that

 Yr−2+δ′rXr−2+δr>mr−1.

This follows from {IEEEeqnarray}rCl Yr-2r’Xr-2r & =& (r-2)4r-2+(3r+1)4r-14r-2-1+3 ⋅4r-1
& = & 13r⋅4r-2+2⋅4r-213⋅4r-2-1
& = & r+2 ⋅4r-2+r13 ⋅4r-2-1, and that by (2.2)

 mr−1=r−r2⋅4r−2+1.

###### Lemma 5

The following properties hold.

• is above ;

• is below ;

• no point of is below ;

• no point of is below ;

• no point of is above ; and

• no point of is above .

###### Proof.

For the lemma can be verified directly or holds trivially; assume that .

• By Lemma 4, the rightmost point of is above . If a point of is below , then the line defined by and the rightmost point of has slope larger than — a contradiction to Lemma 3 and the fact that is a translation of .

• By Lemma 4, the leftmost point of is below . If a point of is above , then the line defined by and the leftmost point of has slope larger than the slope of — a contradiction to Lemma 3 and the fact that is parallel to .

• If a point of is below , then the line defined by and the rightmost point of has slope larger than — a contradiction to Lemma 3.

• Follows from (c) and the fact that is a translation of .

• If a point of is above , then the line defined by and the leftmost point of has slope larger than — a contradiction to Lemma 3.

• Follows from (e) and the fact that is a translation of .

is high above .

###### Proof.

For the lemma holds trivially or can be verified directly; assume that and that lemma holds for smaller values of . We proceed by induction on .

We first prove that is above every line defined by two points of . By (a) of Lemma 5, is above . By Lemma 3, the slope of is at most the slope of . Thus we may assume that does not contain the rightmost point of nor the leftmost point of . Suppose that passes through a point of ; then, by (e) of Lemma 5, this point is below . Since the slope of is at most the slope of , is above in this case. Suppose that passes through two points of . Then, by induction the leftmost point of is above . Since the slope of is at most the slope of , is above in this case.

We now prove that is below every line defined by two points of . By (b) of Lemma 5, is below . Since is a translation of , by Lemma 3, we have that the slope of is at most the slope of . Thus we may assume that does not contain the rightmost point of nor the leftmost point of . Suppose that passes through a point of ; then, by (d) of Lemma 5, this point is above . Since the slope of is at most the slope of , is below in this case. Suppose that passes through two points of . Since is translation of , by induction we have that the rightmost point of is below . Since the slope of is at most the slope of , is above in this case. The result follows. ∎

###### Proposition 7

can be realized with non-negative integer coordinates of size at most .

###### Proof.

This follows from and . ∎

###### Proposition 8

Let be positive integers. If then is a subset of .

###### Proof.

The result holds for or , since in these cases . Therefore, the result holds for . Assume that , and that the result holds for smaller values of . By induction and are subsets of . The result follows from Lemma 6 and by setting and . ∎

### 2.3 The Erdős-Szekeres Construction with Small Integer Coordinates

In this section we use the set of points described in Section 2.2 to realize, with small integer coordinates, the construction given by Erdős and Szekeres in [8]. The Erdős-Szekeres construction is made from a small number of translations of (for the pairs , where is fixed). We first describe these translations.

Let be an integer and let . For every integer we define the vector

 vi:=(3(t−i),−3i).

Using these vectors, we define a set of points recursively as follows.

• ;

• for .

For , let be the unit square whose lower left corner is equal to .

###### Lemma 9

The union, , of the squares lies in a integer grid.

###### Proof.

Note that the largest absolute value of the -coordinates of the ’s is equal to

 t−3∑i=0(3t−i)<3t2;

and the largest absolute value of the -coordinates of the ’s is equal to

 t−3∑i=03i<3t2.

Therefore, lies in a integer grid. ∎

Let be the square scaled by factor of , that is

 Di:={((t+1)4t+1x,(t+1)4t+1y):(x,y)∈Ci}.
###### Lemma 10

Let be three integers; let and be three points in and , respectively. Then is a right turn.

###### Proof.

For , let be the set of vectors of the form where is a point of and is a point of . Note that the endpoints of these vectors lie in a square centered at ; let be the smallest cone with apex at the origin and that contains . By the previous observation the only intersect at the origin; see Figure 3.

Let be the slope of a line passing through a point of and a point of and let be the slope of a line passing through a point of and a point of . The vector defining lies in and the vector defining lies in . This implies that . Let be three integers. Let be the slope of a line passing through a point of and a point of and let be the slope of a line passing through a point of and a point of . Thus, we have that

 mi,j>mj,k. (3)

Note that (3) also holds for the lines defined by pairs points in the ’s. Therefore, is a right turn. ∎

Let be the lower left corner of , and let be the translation of by . That is

 S′t−i,i+2:={p+qi:p∈St−i,i+2}.

The Erdős-Szekeres construction is given by

 St=t−2⋃i=0S′t−i,i+2.

Note that

 |St|=t−2∑i=0|St−i,i+2|=t−2∑j=0(t−2j)=2t−2=n.
###### Proposition 11

lies in an integer grid of size in an integer grid of size .

###### Proof.

Recall that is a scaling of by a factor of . Therefore, by Lemma 9, lies in an integer grid of size

 3t2(t+1)4t+1=192n2log2(4n)2log2(8n)=O(n2log2(n)3).

###### Proposition 12

is in general position.

###### Proof.

Let be three points of . If these three points are contained in a same , then they are not collinear since is in general position. If the three of them are in different , then by Lemma 10 they are not collinear. If two of them lie on a same and one of them in some , then they are not collinear since the slope of a line joining a point in and a point in is negative, while the slope of a line defined by two points in is greater or equal to zero. Therefore, does not contain three collinear points. ∎

###### Theorem 13

Every convex -gon of has at most vertices.

###### Proof.

Let be a convex -gon of . Let and be the upper and lower polygonal chains of , respectively. Let be the index such that the leftmost point of (and ) is in , and let be the index such that the rightmost point of (and ) is in .

Note that for all , the slopes of an edge joining a point of with a point of are negative; since the slope of an edge defined by a pair of points in is greater or equal to zero, neither nor contain two consecutive vertices in for . By Lemma 10 such a cannot contain a vertex of . Therefore, contains at most vertices not in .

The vertices of contained in must form a cap and thus consists of at most vertices. Similarly, the vertices of contained in must form a cup and therefore consists of at most vertices. Therefore, has at most vertices; the result follows. ∎

To summarize, we have the following.

###### Theorem 14

The Erdős-Szekeres construction of points can be realized in an integer grid of size .

## 3 Implementation

A direct implementation of Section 2.3 gives way to an efficient algorithm to compute the Erdős-Szekeres construction. By Proposition 14, the size of the grid needed by this algorithm is asymptotically small; however, there are large constants hidden in such an implementation. In this section we mention some optimizations we have done to further reduce the size of the integer grid needed for the Erdős-Szekeres construction.

• Decrease the horizontal distance between left and right parts of .

In Section 2.2, to construct , we gave explicit values to and . This allowed us to show that is to the left of and that is high above . However, it is enough to show that is to the left of and that the rightmost point of is above . That is that

 Yr>Yr−Yr−2Xr−Xr−2(Xr−Xr−2)+Yr−2. (4)

Let be a constant and set . It can be shown that if we replace inequality (4) by an equality and solve for , then is of order . Therefore, if we set , then both and are of order . In the actual implementation we choose so that

 Xr=⌈(2+√3)r⌉.

Then we choose so that

 Yr=⌈Yr−Yr−2Xr−Xr−2(Xr−Xr−2)+Yr−2+1⌉.

The addition of the ceiling functions has prevented us from proving that is of order . If this is the case then can be realized in an integer grid of size . In Section 2.2, we opted to avoid using ceiling functions at the expense of being able to show a slightly worse upper bound.

Inspired by this, we do likewise when constructing . First we construct and . Let be the horizontal length of . We choose

 δk,l:=⌈(1+√3)(Xk−1,l+Xk,l−12)⌉.

For any two positive integers and , let be the straight line passing through the rightmost point of and the leftmost point of the copy of in . (This definition is similar to the definition of for .) We choose so that; the rightmost point in the translation of is above ; and the leftmost point of is below the corresponding translation of in .

• Separate the left and right parts of by one in the last step of the recursion.

The reason for choosing a relatively large horizontal separation between the left and right parts of is so that the slope of does not increase too quickly. We do not need to do this in the last step of the construction. At each step in the construction of for , and , we separate the corresponding left and right parts as before. In the last step, when constructing , we separate from the copy of by one.

• Decrease the size of the squares (rectangles) .

In Section 2.3, for we defined a square , inside which we placed a copy of . was chosen large enough so that fits inside for . Since we only need to fit we replace by a rectangle of length and height . The definitions of the ’s and ’s are changed accordingly.

In Figure 4 we show in integer grid; in Figure 5 we show in a integer grid; and in Figure 6 we show in a integer grid. For a comparison, we note that Kalbfleisch and Stanton [12] realize in a integer grid.

Our implementations are freely available as part of “Python’s Discrete and Combinatorial Geometry Library” at www.pydcg.org. They are included in the points module. In that module the following constructions are also included: Convex Position and the Double Circle as described in [3]; the Horton Set as described in [2] and the Squared Horton set (this set was first defined by Valtr in [22]).

## 4 Erdős-Szekeres Type Problems in Restricted Planar Point Sets

In this section we propose some open Erdős-Szekeres type problems on point sets in integer grids. Let be the maximum distance between a pair of points of , and let be the minimum distance between a pair of points of . Alon, Katchalski and Pulleyblank [1], showed that if for some constant , satisfies

 diam(S)mindist(S)≤αn12,

then contains a convex -gon of vertices; in  [22], Valtr improved this bound to . He also showed that if for some constant , satisfies

 diam(S)mindist(S)≤α√n,

then contains a convex -gon of vertices. That is, metric restrictions on may force large convex polygons.

This prompted the following two problems in [4].

###### Problem 1

Does there exist, for every , a suitable constant with the following property: any set of of points in general position in the plane with contains a convex -gon?

###### Problem 2

Does there exist, for every , a suitable constant with the following property: any set of points in the general position in the plane with positive integer coordinates that do not exceed contains a convex -gon?

Valtr in his PhD thesis [23] showed that the answer for Problem 1 is “yes” for . He also noted, in passing, in page 55 of his thesis the following.

“If then it is possible to construct an -dense set of size which contains no more than vertices of a convex polygon. Such a set can be obtained by an affine transformation from the construction of Erdős and Szekeres [8]…”

is said to be -dense if . So this observation solves Problem 1 for as well. Problem 2 appeared first in Valtr’s thesis (Problem 10), where it is attributed to Welzl.

In this paper we have shown that the answer to Problem 2 is “No” for all . We conjecture that the answer to Problem 2 is “No” for all . However, we conjecture that there exists a such that the Erdős-Szekeres construction cannot be realized in an integer grid. We propose the following alternative to Problem 2.

###### Problem 3

Does there exist, for every and every , a suitable constant and a set of points in general position in the plane with the following property? has positive integer coordinates not exceeding , and does not contain a convex -gon.

### 4.1 Empty k-gons

A convex -gon of is empty if it does not contain a point of in its interior. In 1978, Erdős [6] asked whether an analogue of the Erdős-Szekeres theorem holds for empty convex -gons. That is, if for every , every sufficiently large point set in general position in the plane contains an empty -gon.

Every set of at least three points contains an empty triangle; Esther Klein [7] proved that every set of five points contains an empty convex -gon; Harborth [10] proved that every set of 10 points contains an empty convex -gon; and Horton [11] constructed arbitrarily large point sets without empty convex -gons. (His construction is now known as the Horton Set.) The question for convex -gons remained open for more than a quarter of a century until Nicolás [17] and independently Gerken [9] showed that every sufficiently large set of points contains a convex empty -gon.

Alon et al. posed a problem in [1], similar to Problem 1, but for empty convex -gons. They asked whether if for some constant every sufficiently large -dense set of points contains an empty convex -gon. Valtr in [22] showed that there exist arbitrarily large -dense point sets not containing an empty convex -gon. His construction is based on the Horton set.

The same question can be asked for point sets in an integer grid.

###### Problem 4

Does the following hold for every constant ? Every sufficiently large set of points in the general position in the plane with positive integer coordinates that do not exceed contains an empty convex -gon.

As far as we know all constructions without an empty convex -gons are based on the Horton set. This is particularly relevant for Problem 4 for the following reason. In [2], Barba, Duque, Fabila-Monroy and Hidalgo-Toscano proved that the Horton set cannot be realized in an integer grid of polynomial size.

### Acknowledgments

We thank Luis Felipe Barba for providing us with a copy of [12].

### Footnotes

1. footnotemark:
2. footnotemark:
3. footnotemark:
4. More accurately, contains a subset with the same order type as .

### References

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