On grids in point-line arrangements in the plane

TL;DR

This paper extends the Szemerédi-Trotter theorem by showing that arrangements avoiding a specific grid structure have fewer incidences, and constructs examples with many incidences that avoid such grids.

Contribution

It introduces a Turán-type bound for point-line arrangements avoiding a natural grid, and provides explicit constructions with high incidences without such grids.

Findings

Arrangements without a natural t×t grid have O(n^{4/3 - ε}) incidences.

Constructed arrangements without a 2×2 grid have at least Ω(n^{1+1/14}) incidences.

The results generalize incidence bounds by excluding specific grid configurations.

Abstract

The famous Szemer\'{e}di-Trotter theorem states that any arrangement of $n$ points and $n$ lines in the plane determines $O(n^{4/3})$ incidences, and this bound is tight. In this paper, we prove the following Tur\'an-type result for point-line incidence. Let $\mathcal{L}_1$ and $\mathcal{L}_2$ be two sets of $t$ lines in the plane and let $P=\{\ell_1 \cap \ell_2 : \ell_1 \in \mathcal{L}_1, \ell_2 \in \mathcal{L}_2\}$ be the set of intersection points between $\mathcal{L}_1$ and $\mathcal{L}_2$. We say that $(P, \mathcal{L}_1 \cup \mathcal{L}_2)$ forms a \emph{natural $t\times t$ grid} if $|P| =t^2$, and $conv(P)$ does not contain the intersection point of some two lines in $\mathcal{L}_i,$ for $i = 1,2.$ For fixed $t > 1$, we show that any arrangement of $n$ points and $n$ lines in the plane that does not contain a natural $t\times t$ grid determines $O(n^{\frac{4}{3}- \varepsilon})$ incidences, where $\varepsilon = \varepsilon(t)$. We also provide a construction of $n$ points and $n$ lines in the plane that does not contain a natural $2 \times 2$ grid and determines at least $\Omega({n^{1+\frac{1}{14}}})$ incidences.

Figures

Equations

$$|I(P,\mathcal{L})|\le O(m^{2/3}{n}^{2/3}+m+n).$$

$$c_{1}m\sqrt{r}=\sqrt{10}{c}_1{m}^{\frac{2t-2}{3t-2}}{n}^{\frac{2t-1}{3t-2}}<(c/2)m^{\frac{2t-2}{3t-2}}{n}^{\frac{2t-1}{3t-2}}.$$

$$\frac{c{m}^{\frac{2t-2}{3t-2}}{n}^{\frac{2t-1}{3t-2}}}{2r}=\frac{c{m}^{\frac{4t-2}{3t-2}}}{20n^{\frac{2t-1}{3t-2}}}.$$

$$E_{j}=\left\{\begin{array}[]{ll}\{(\ell_{1},\ell_{2}),(\ell_{3},\ell_{4}),\ldots,(\ell_{s-1},\ell_{s})\}&\text{if }s\text{ is even}.\\ \\ \{(\ell_{1},\ell_{2}),(\ell_{3},\ell_{4}),\ldots,(\ell_{s-2},\ell_{s-1})\}&\text{if }s\text{ is odd}.\end{array}\right.$$

$$|E(H)|\ge \frac{c{m}^{\frac{4t-2}{3t-2}}}{60n^{\frac{2t-1}{3t-2}}}.$$

$$|V(H)|=|{\mathcal{L}}_i|\le \frac{m^{\frac{2t}{3t-2}}}{5n^{\frac{t}{3t-2}}},$$

$$|E(H)|\ge \frac{c}{60\cdot 25}(V(H){)}^{2-\frac{1}{t}}.$$

$$u_1{x}_1+u_2{x}_2+u_3{x}_3+u_4{x}_4=0.$$

$$\{1,2,3,4\}=T_1\cup \cdots \cup T_d,$$

$$\sum _{i\in T_{\nu }}{u}_i=0,\nu =1,\dots ,d.$$

$$u_1{x}_1+u_2{x}_2+u_3{x}_3+u_4{x}_4=0,$$

$$a_m\le 2^8{k}^4{m}^3,$$

$$\displaystyle a_{m}\neq-\Big{(}\sum_{i\in S}u_{i}\Big{)}^{-1}\sum_{1\leq i\leq 4,i\notin S}u_{i}x_{i},$$

$$(2k+1{)}^4\sum _{j=1}^3(\genfrac{}{}{0ex}{}{4}{j})(m-1{)}^{4-j}=(2k+1{)}^4(m^4-(m-1{)}^4-1)<2^8{k}^4{m}^3.$$

$$|A|\ge c{n}^{1/14},$$

$$|P|$$

$$|\mathcal{L}|$$

$$|I(P,\mathcal{L})|$$

$${\ell }_1\cap {\ell }_2=p_1=(x_1,y_1),$$

$${\ell }_2\cap {\ell }_3=p_2=(x_2,y_2),$$

$${\ell }_3\cap {\ell }_4=p_3=(x_3,y_3),$$

$${\ell }_4\cap {\ell }_1=p_4=(x_4,y_4).$$

$$p_1$$

$$p_2$$

$$p_3$$

$$p_4$$

$$(m_1-m_2)x_1+(m_2-m_3)x_2+(m_3-m_4)x_3+(m_4-m_1)x_4=0.$$

$$u_1{x}_1+u_1{x}_2+u_3{x}_3+u_4{x}_4=0,$$

Full text

1 Introduction

Given a finite set $P$ of points in the plane and a finite set $\mathcal{L}$ of lines in the plane, let $I(P,\mathcal{L})=\{(p,\ell)\in P\times\mathcal{L}:p\in\ell\}$ be the set of incidences between $P$ and $\mathcal{L}$. The incidence graph of $(P,\mathcal{L})$ is the bipartite graph $G=(P\cup\mathcal{L},I)$, with vertex parts $P$ and $\mathcal{L},$ and $E(G)=I(P,\mathcal{L})$. If $|P|=m$ and $|\mathcal{L}|=n$, then the celebrated theorem of Szemerédi and Trotter [16] states that

[TABLE]

Moreover, this bound is tight which can be seen by taking the $\sqrt{m}\times\sqrt{m}$ integer lattice and bundles of parallel ”rich” lines (see [13]). It is widely believed that the extremal configurations maximizing the number of incidences between $m$ points and $n$ lines in the plane exhibit some kind of lattice structure. The main goal of this paper is to show that such extremal configurations must contain large natural grids.

Let $P$ and $P_{0}$ (respectively, $\mathcal{L}$ and $\mathcal{L}_{0}$) be two sets of points (respectively, lines) in the plane. We say that the pairs $(P,\mathcal{L})$ and $(P_{0},\mathcal{L}_{0})$ are isomorphic if their incidence graphs are isomorphic. Solymosi made the following conjecture (see page $291$ in [2]).

Conjecture 1.1**.**

For any set of points $P_{0}$ and for any set of lines $\mathcal{L}_{0}$ in the plane, the maximum number of incidences between $n$ points and $n$ lines in the plane containing no subconfiguration isomorphic to $(P_{0},\mathcal{L}_{0})$ is $o(n^{\frac{4}{3}}).$

In [15], Solymosi proved this conjecture in the special case that $P_{0}$ is a fixed set of points in the plane, no three of which are on a line, and $\mathcal{L}_{0}$ consists of all of their connecting lines. However, it is not known if such configurations satisfy the following stronger conjecture.

Conjecture 1.2**.**

For any set of points $P_{0}$ and for any set of lines $\mathcal{L}_{0}$ in the plane, there is a constant $\varepsilon=\varepsilon(P_{0},\mathcal{L}_{0})$, such that the maximum number of incidences between $n$ points and $n$ lines in the plane containing no subconfiguration isomorphic to $(P_{0},\mathcal{L}_{0})$ is $O(n^{4/3-\varepsilon}).$

Our first theorem is the following.

Theorem 1.3**.**

For fixed $t>1$, let $\mathcal{L}_{a}$ and $\mathcal{L}_{b}$ be two sets of $t$ lines in the plane, and let $P_{0}=\{\ell_{a}\cap\ell_{b}:\ell_{a}\in\mathcal{L}_{a},\ell_{b}\in\mathcal{L}_{b}\}$ such that $|P_{0}|=t^{2}$. Then there is a constant $c=c(t)$ such that any arrangement of $m$ points and $n$ lines in the plane that does not contain a subconfiguration isomorphic to $(P_{0},\mathcal{L}_{a}\cup\mathcal{L}_{b})$ determines at most $c(m^{\frac{2t-2}{3t-2}}n^{\frac{2t-1}{3t-2}}+m^{1+\frac{1}{6t-3}}+n)$ incidences.

See the Figure 1. As an immediate corollary, we prove Conjecture 1.2 in the following special case.

Corollary 1.4**.**

For fixed $t>1$, let $\mathcal{L}_{a}$ and $\mathcal{L}_{b}$ be two sets of $t$ lines in the plane, and let $P_{0}=\{\ell_{a}\cap\ell_{b}:\ell_{a}\in\mathcal{L}_{a},\ell_{b}\in\mathcal{L}_{b}\}.$ If $|P_{0}|=t^{2}$, then any arrangement of $n$ points and $n$ lines in the plane that does not contain a subconfiguration isomorphic to $(P_{0},\mathcal{L}_{a}\cup\mathcal{L}_{b})$ determines at most $O(n^{\frac{4}{3}-\frac{1}{9t-6}})$ incidences.

In the other direction, we prove the following.

Theorem 1.5**.**

Let $\mathcal{L}_{a}$ and $\mathcal{L}_{b}$ be two sets of $2$ lines in the plane, and let $P_{0}=\{\ell_{a}\cap\ell_{b}:\ell_{1}\in\mathcal{L}_{a},\ell_{b}\in\mathcal{L}_{b}\}$ such that $|P_{0}|=4.$ For $n>1,$ there exists an arrangement of $n$ points and $n$ lines in the plane that does not contain a subconfiguration isomorphic to $(P_{0},\mathcal{L}_{a}\cup\mathcal{L}_{b}),$ and determines at least $\Omega({n^{1+\frac{1}{14}}})$ incidences.

Given two sets $\mathcal{L}_{a}$ and $\mathcal{L}_{b}$ of $t$ lines in the plane, and the point set $P_{0}=\{\ell_{a}\cap\ell_{b}:\ell_{a}\in\mathcal{L}_{a},\ell_{b}\in\mathcal{L}_{b}\},$ we say that $(P_{0},\mathcal{L}_{a}\cup\mathcal{L}_{b})$ forms a natural $t\times t$ grid if $|P_{0}|=t^{2},$ and the convex hull of $P_{0},$ $conv(P_{0}),$ does not contain the intersection point of any two lines in $\mathcal{L}_{a}$ and does not contain the intersection point of any two lines in $\mathcal{L}_{b}.$ See Figure 2.

Theorem 1.6**.**

For fixed $t>1,$ there is a constant $\varepsilon=\varepsilon(t),$ such that any arrangement of $n$ points and $n$ lines in the plane that does not contain a natural $t\times t$ grid determines at most $O(n^{\frac{4}{3}-\varepsilon})$ incidences.

Let us remark that $\varepsilon=\Omega(1/t^{2})$ in Theorem 1.6, and can be easily generalized to the off-balanced setting of $m$ points and $n$ lines.

We systemically omit floor and ceiling signs whenever they are not crucial for the sake of clarity of our presentation. All logarithms are assumed to be base $2.$ For $N>0,$ we let $[N]=\{1,\ldots,N\}.$

2 Proof of Theorem 1.3

In this section we will prove Theorem 1.3. We first list several results that we will use. The first lemma is a classic result in graph theory.

Lemma 2.1** (Kövari-Sós-Turán [10]).**

Let $G=(V,E)$ be a graph that does not contain a complete bipartite graph $K_{r,s}$ $(1\leq r\leq s)$ as a subgraph. Then $|E|\leq c_{s}|V|^{2-\frac{1}{r}},$ where $c_{s}>0$ is constant which only depends on $s.$

The next lemma we will use is a partitioning tool in discrete geometry known as simplicial partitions. We will use the dual version which requires the following definition. Let $\mathcal{L}$ be a set of lines in the plane. We say that a point $p$ crosses $\mathcal{L}$ if it is incident to at least one member of $\mathcal{L},$ but not incident to all members in $\mathcal{L}.$

Lemma 2.2** (Matousek [12]).**

Let $\mathcal{L}$ be a set of $n$ lines in the plane and let $r$ be a parameter such that $1<r<n.$ Then there is a partition on $\mathcal{L}=\mathcal{L}_{1}\cup\cdots\cup\mathcal{L}_{r}$ into $r$ parts, where $\frac{n}{2r}\leq|\mathcal{L}_{i}|\leq\frac{2n}{r},$ such that any point $p\in\mathbb{R}^{2}$ crosses at most $O(\sqrt{r})$ parts $\mathcal{L}_{i}.$

Proof of Theorem 1.3..

Set $t\geq 2.$ Let $P$ be a set of $m$ points in the plane and let $\mathcal{L}$ be a set of $n$ lines in the plane such that $(P,\mathcal{L})$ does not contain a subconfiguration isomorphic to $(P_{0},\mathcal{L}_{a}\cup\mathcal{L}_{b}).$

If $n\geq m^{2}/100,$ then (1.1) implies that $|I(P,\mathcal{L})|=O(n)$ and we are done. Likewise, if $n\leq m^{\frac{t}{2t-1}},$ then (1.1) implies that $|I(P,\mathcal{L})|=O(m^{1+\frac{1}{6t-3}})$ and we are done. Therefore, let us assume $m^{\frac{t}{2t-1}}<n<m^{2}/100.$ In what follows, we will show that $|I(P,\mathcal{L})|=O(m^{\frac{2t-2}{3t-2}}n^{\frac{2t-1}{3t-2}}).$ For sake of contradiction, suppose that $I(P,\mathcal{L})\geq cm^{\frac{2t-2}{3t-2}}n^{\frac{2t-1}{3t-2}},$ where $c$ is a large constant depending on $t$ that will be determined later.

Set $r=\lceil 10{n^{\frac{4t-2}{3t-2}}}/{m^{\frac{2t}{3t-2}}}\rceil.$ Let us remark that $1<r<n/10$ since we are assuming $m^{\frac{t}{2t-1}}<n<m^{2}/100.$ We apply Lemma 2.2 with parameter $r$ to $\mathcal{L},$ and obtain the partition $\mathcal{L}=\mathcal{L}_{1}\cup\cdots\cup\mathcal{L}_{r}$ with the properties described above. Note that $|\mathcal{L}_{i}|>1.$ Let $G$ be the incidence graph of $(P,\mathcal{L}).$ For $p\in P,$ consider the set of lines in $\mathcal{L}_{i}.$ If $p$ is incident to exactly one line in $\mathcal{L}_{i},$ then delete the corresponding edge in the incidence graph $G.$ After performing this operation between each point $p\in P$ and each part $\mathcal{L}_{i},$ by Lemma 2.2, we have deleted at most $c_{1}m\sqrt{r}$ edges in $G,$ where $c_{1}$ is an absolute constant. By setting $c$ sufficiently large, we have

[TABLE]

Therefore, there are at least ${(c/2)m^{\frac{2t-2}{3t-2}}}{n^{\frac{2t-1}{3t-2}}}$ edges remaining in $G.$ By the pigeonhole principle, there is a part $\mathcal{L}_{i}$ such that the number of edges between $P$ and $\mathcal{L}_{i}$ in $G$ is at least

[TABLE]

Hence, every point $p\in P$ has either [math] or at least $2$ neighbors in $\mathcal{L}_{i}$ in $G.$ We claim that $(P,\mathcal{L}_{i})$ contains a subconfiguration isomorphic to $(P_{0},\mathcal{L}_{a}\cup\mathcal{L}_{b}).$ To see this, let us construct a graph $H=(\mathcal{L}_{i},E)$ as follows. Set $V(H)=\mathcal{L}_{i}.$ Let $Q=\{q_{1},\ldots,q_{w}\}\subset P$ be the set of points in $P$ that have at least two neighbors in $\mathcal{L}_{i}$ in the graph $G.$ For $q_{j}\in Q,$ consider the set of lines $\{\ell_{1},\ldots,\ell_{s}\}$ from $\mathcal{L}_{i}$ incident to $q_{j},$ such that $\{\ell_{1},\ldots,\ell_{s}\}$ appears in clockwise order. Then we define $E_{j}\subset\binom{\mathcal{L}_{i}}{2}$ to be a matching on $\{\ell_{1},\ldots,\ell_{s}\},$ where

[TABLE]

Set $E(H)=E_{1}\cup E_{2}\cup\cdots\cup E_{w}.$ Note that $E_{j}$ and $E_{k}$ are disjoint, since no two points are contained in two lines. Since $|E_{j}|\geq 1,$ we have

[TABLE]

Since

[TABLE]

this implies

[TABLE]

By setting $c=c(t)$ to be sufficiently large, Lemma 2.1 implies that $H$ contains a copy of $K_{t,t}.$ Let $\mathcal{L}^{\prime}_{1},\mathcal{L}^{\prime}_{2}\subset\mathcal{L}_{i}$ correspond to the vertices of this $K_{t,t}$ in $H,$ and let $P^{\prime}=\{\ell_{1}\cap\ell_{2}\in P:\ell_{1}\in\mathcal{L}^{\prime}_{1},\ell_{2}\in\mathcal{L}^{\prime}_{2}\}.$ We claim that $(P^{\prime},\mathcal{L}^{\prime}_{1}\cup\mathcal{L}^{\prime}_{2})$ is isomorphic to $(P_{0},\mathcal{L}_{a}\cup\mathcal{L}_{b}).$ It suffices to show that $|P^{\prime}|=t^{2}.$ For the sake of contradiction, suppose $p\in\ell_{1}\cap\ell_{2}\cap\ell_{3},$ where $\ell_{1},\ell_{2}\in\mathcal{L}^{\prime}_{1}$ and $\ell_{3}\in\mathcal{L}^{\prime}_{2}.$ This would imply $(\ell_{1},\ell_{3}),(\ell_{2},\ell_{3})\in E_{j}$ for some $j$ which contradicts the fact that $E_{j}\subset\binom{\mathcal{L}_{i}}{2}$ is a matching. Same argument follows if $\ell_{1}\in\mathcal{L}^{\prime}_{1}$ and $\ell_{2},\ell_{3}\in\mathcal{L}^{\prime}_{2}.$ This completes the proof of Theorem 1.3. ∎

3 Natural Grids

Given a set of $n$ points $P$ and a set of $n$ lines $\mathcal{L}$ in the plane, if $|I(P,\mathcal{L})|\geq cn^{\frac{4}{3}-\frac{1}{9k-6}}$, where $c$ is a sufficiently large constant depending on $k,$ then Corollary 1.4 implies that there are two sets of $k$ lines such that each pair of them from different sets intersects at a unique point in $P.$ Therefore, Theorem 1.6 follows by combining Theorem 1.3 with the following lemma.

Lemma 3.1**.**

There is a natural number $c$ such that the following holds. Let $\mathcal{B}$ be a set of $ct^{2}$ blue lines in the plane, and let $\mathcal{R}$ be a set of $ct^{2}$ red lines in the plane such that for $P=\{\ell_{1}\cap\ell_{2}:\ell_{1}\in\mathcal{B},\ell_{2}\in\mathcal{R}\}$ we have $|P|=c^{2}t^{4}.$ Then $(P,\mathcal{B}\cup\mathcal{R})$ contains a natural $t\times t$ grid.

To prove Lemma 3.1, we will need the following lemma which is an immediate consequence of Dilworth’s Theorem.

Lemma 3.2**.**

For $n>0,$ let $\mathcal{L}$ be a set of $n^{2}$ lines in the plane, such that no two members intersect the same point on the $y$-axis. Then there is a subset $\mathcal{L}^{\prime}\subset\mathcal{L}$ of size $n$ such that the intersection point of any two members in $\mathcal{L}^{\prime}$ lies to the left of the $y$-axis, or the intersection point of any two members in $\mathcal{L}^{\prime}$ lies to the right of the $y$-axis.

Proof.

Let us order the elements in $\mathcal{L}=\{\ell_{1},\ldots,\ell_{n^{2}}\}$ from bottom to top according to their $y$-intercept. By Dilworth’s Theorem [5], $\mathcal{L}$ contains a subsequence of $n$ lines whose slopes are either increasing or decreasing. In the first case, all intersection points are to the left of the $y$-axis, and in the latter case, all intersection points are to the right of the $y$-axis. ∎

Proof of Lemma 3.1..

Let $(P,\mathcal{B}\cup\mathcal{R})$ be as described above, and let $\ell_{y}$ be the $y$-axis. Without loss of generality, we can assume that all lines in $\mathcal{B}\cup\mathcal{R}$ are not vertical, and the intersection point of any two lines in $\mathcal{B}\cup\mathcal{R}$ lies to the right of $\ell_{y}.$ Moreover, we can assume that no two lines intersect at the same point on $\ell_{y}.$

We start by finding a point $y_{1}\in\ell_{y}$ such that at least $|\mathcal{B}|/2$ blue lines in $\mathcal{B}$ intersect $\ell_{y}$ on one side of the point $y_{1}$ (along $\ell_{y}$) and at least $|\mathcal{R}|/2$ red lines in $\mathcal{R}$ intersect $\ell_{y}$ on the other side. This can be done by sweeping the point $y_{1}$ along $\ell_{y}$ from bottom to top until $ct^{2}/2$ lines of the first color, say red, intersect $\ell_{y}$ below $y_{1}.$ We then have at least $ct^{2}/2$ blue lines intersecting $\ell_{y}$ above $y_{1}.$ Discard all red lines in $\mathcal{R}$ that intersect $\ell_{y}$ above $y_{1},$ and discard all blue lines in $\mathcal{B}$ that intersect $\ell_{y}$ below $y_{1}.$ Hence, $|\mathcal{B}|\geq ct^{2}/2.$

Set $s=\lfloor ct^{2}/4\rfloor.$ For the remaining lines in $\mathcal{B},$ let $\mathcal{B}=\{b_{1},\ldots,b_{2s}\},$ where the elements of $\mathcal{B}$ are ordered in the order they cross $\ell_{y},$ from bottom to top. We partition $\mathcal{B}=\mathcal{B}_{1}\cup\mathcal{B}_{2}$ into two parts, where $\mathcal{B}_{1}=\{b_{1},\ldots,b_{s}\}$ and $\mathcal{B}_{2}=\{b_{s+1},\ldots,b_{2s}\}.$ By applying an affine transformation, we can assume all lines in $\mathcal{R}$ have positive slope and all lines in $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ have negative slope. See Figure 3.

Let us define a $3$-partite $3$-uniform hypergraph $H=(\mathcal{R}\cup\mathcal{B}_{1}\cup\mathcal{B}_{2},E),$ whose vertex parts are $\mathcal{R},\mathcal{B}_{1},\mathcal{B}_{2},$ and $(r,b_{i},b_{j})\in\mathcal{R}\times\mathcal{B}_{1}\times\mathcal{B}_{2}$ is an edge in $H$ if and only if the intersection point $p=b_{i}\cap b_{j}$ lies above the line $r.$ Note, if $b_{i}$ and $b_{j}$ are parallel, then $(r,b_{i},b_{j})\notin E.$ Then a result of Fox et al. on semi-algebraic hypergraphs implies the following (see also [3] and [9]).

Lemma 3.3** (Fox et al. [8], Theorem 8.1).**

There exists a positive constant $\alpha$ such that the following holds. In the hypergraph above, there are subsets $\mathcal{R}^{\prime}\subseteq\mathcal{R},\mathcal{B}^{\prime}_{1}\subseteq\mathcal{B}_{1},\mathcal{B}^{\prime}_{2}\subseteq\mathcal{B}_{2},$ where $|\mathcal{R}^{\prime}|\geq\alpha|\mathcal{R}|,|\mathcal{B}^{\prime}_{1}|\geq\alpha|\mathcal{B}_{1}|,|\mathcal{B}^{\prime}_{2}|\geq\alpha|\mathcal{B}_{2}|,$ such that either $\mathcal{R}^{\prime}\times\mathcal{B}^{\prime}_{1}\times\mathcal{B}^{\prime}_{2}\subseteq E,$ or $(\mathcal{R}^{\prime}\times\mathcal{B}^{\prime}_{1}\times\mathcal{B}^{\prime}_{2})\cap E=\emptyset.$

We apply Lemma 3.3 to $H$ and obtain subsets $\mathcal{R}^{\prime},\mathcal{B}^{\prime}_{1},\mathcal{B}^{\prime}_{2}$ with the properties described above. Without loss of generality, we can assume that $\mathcal{R}^{\prime}\times\mathcal{B}^{\prime}_{1}\times\mathcal{B}^{\prime}_{2}\subset E,$ since a symmetric argument would follow otherwise. Let $\ell_{1}$ be a line in the plane such that the following holds.

The slope of $\ell_{1}$ is negative. 2. 2.

All intersection points between $\mathcal{R^{\prime}}$ and $\mathcal{B^{\prime}}_{1}$ lie above $\ell_{1}.$ 3. 3.

All intersection points between $\mathcal{R^{\prime}}$ and $\mathcal{B^{\prime}}_{2}$ lie below $\ell_{1}.$

See Figure 4.

Observation 3.4**.**

Line $\ell_{1}$ defined above exists.

Proof.

Let $U$ be the upper envelope of the arrangement $\bigcup_{\ell\in\mathcal{R}^{\prime}}\ell,$ that is, $U$ is the closure of all points that lie on exactly one line of $\mathcal{R}^{\prime}$ and strictly above exactly the $|\mathcal{R}^{\prime}|-1$ lines in $\mathcal{R}^{\prime}.$

Let $P_{1}$ be the set of intersection points between the lines in $\mathcal{B}^{\prime}_{1}$ with $U.$ Likewise, we define $P_{2}$ to be the set of intersection points between the lines in $\mathcal{B}^{\prime}_{2}$ with $U.$ Since $U$ is $x$-monotone and convex the set $P_{2}$ lies to the left of the set $P_{1}.$ Then the line $\ell_{1}$ that intersects $U$ between $P_{1}$ and $P_{2}$ and intersects $\ell_{y}$ between $\mathcal{B}^{\prime}_{1}$ and $\mathcal{B}^{\prime}_{2}$ satisfies the conditions above.∎

Now we apply Lemma 3.2 to $\mathcal{R}^{\prime}$ with respect to the line $\ell_{1},$ to obtain $\sqrt{\alpha c/2}\cdot t$ members in $\mathcal{R}^{\prime}$ such that every pair of them intersects on one side of $\ell_{1}.$ Discard all other members in $\mathcal{R}^{\prime}.$ Without loss of generality, we can assume that all intersection points between any two members in $\mathcal{R}^{\prime}$ lie below $\ell_{1},$ since a symmetric argument would follow otherwise. We now discard the set $\mathcal{B}^{\prime}_{2}.$

Notice that the order in which the lines in $\mathcal{R}^{\prime}$ cross $b\in\mathcal{B}^{\prime}_{1}$ will be the same for any line $b\in\mathcal{B}^{\prime}_{1}.$ Therefore, we order the elements in $\mathcal{R}^{\prime}=\{r_{1},\ldots,r_{m}\}$ with respect to this ordering, from left to right, where $m=\lceil\sqrt{\alpha c/2}\cdot t\rceil.$ We define $\ell_{2}$ to be the line obtained by slightly perturbing the line $r_{\lfloor m/2\rfloor}$ such that:

The slope of $\ell_{2}$ is positive. 2. 2.

All intersection points between $\mathcal{B}^{\prime}_{1}$ and $\{r_{1},\ldots,r_{\lfloor m/2\rfloor}\}$ lie above $\ell_{2}.$ 3. 3.

All intersection points between $\mathcal{B}^{\prime}_{1}$ and $\{r_{\lfloor m/2\rfloor+1},\ldots,r_{m}\}$ lie below $\ell_{2}.$

See the Figure 5.

Finally, we apply Lemma 3.2 to $\mathcal{B}^{\prime}_{1}$ with respect to the line $\ell_{2},$ to obtain at least $\sqrt{\alpha c}\cdot t/2$ members in $\mathcal{B}^{\prime}_{1}$ with the property that any two of them intersect on one side of $\ell_{2}.$ Without loss of generality, we can assume that any two such lines intersect below $\ell_{2}$ since a symmetric argument would follow. Set $\mathcal{B}^{\ast}\subset\mathcal{B}^{\prime}_{1}$ to be these set of lines. Then $\mathcal{B}^{\ast}\cup\{r_{1},\ldots,r_{\lfloor m/2\rfloor}\}$ and their intersection points form a natural grid. By setting $c=c(t)$ to be sufficiently large, we obtain a natural $t\times t$ grid. ∎

4 Lower Bound Construction

In this section, we will prove Theorem 1.5. First, let us recall the definitions of Sidon and $k$-fold Sidon sets.

Let $A$ be a finite set of positive integers. Then $A$ is a Sidon set if the sum of all pairs are distinct, that is, the equation $x+y=u+v$ has no solutions with $x,y,u,v\in A$, except for trivial solutions given by $u=x,y=v$ and $x=v,y=u.$ We define $s(N)$ to be the size of the largest Sidon set $A\subset\{1,\ldots,N\}.$ Erdős and Turán proved the following.

Lemma 4.1** (See [7] and [14]).**

For $N>1,$ we have $s(N)=\Theta(\sqrt{N}).$

Let us now consider a more general equation. Let $u_{1},\ldots,u_{4}$ be integers such that $u_{1}+u_{2}+u_{3}+u_{4}=0,$ and consider the equation

[TABLE]

We are interested in solutions to (4.1) with $x_{1},x_{2},x_{3},x_{4}\in\mathbb{Z}.$ Suppose $(x_{1},x_{2},x_{3},x_{4})=(a_{1},a_{2},a_{3},a_{4})$ is an integer solution to (4.1). Let $d\leq 4$ be the number of distinct integers in the set $\{a_{1},a_{2},a_{3},a_{4}\}.$ Then we have a partition on the indices

[TABLE]

where $i$ and $j$ lie in the same part $T_{\nu}$ if and only if $x_{i}=x_{j}.$ We call $(a_{1},a_{2},a_{3},a_{4})$ a trivial solution to (4.1) if

[TABLE]

Otherwise, we will call $(a_{1},a_{2},a_{3},a_{4})$ a nontrivial solution to (4.1).

In [11], Lazebnik and Verstraëte introduced $k$-fold Sidon sets which are defined as follows. Let $k$ be a positive integer. A set $A\subset\mathbb{N}$ is a $k$-fold Sidon set if each equation of the form

[TABLE]

where $|u_{i}|\leq k$ and $u_{1}+\cdots+u_{4}=0,$ has no nontrivial solutions with $x_{1},x_{2},x_{3},x_{4}\in A.$ Let $r(k,N)$ be the size of the largest $k$-fold Sidon set $A\subset\{1,\ldots,N\}.$

Lemma 4.2**.**

There is an infinite sequence $1=a_{1}<a_{2}<\cdots$ of integers such that

[TABLE]

and the system of equations (4.2) has no nontrivial solutions in the set $A=\{a_{1},a_{2},\ldots\}.$ In particular, for integers $N>k^{4}\geq 1,$ we have $r(k,N)\geq ck^{-{4}/{3}}{N^{{1}/{3}}},$ where $c$ is a positive constant.

The proof of Lemma 4.2 is a slight modification of the proof of Theorem $2.1$ in [14]. For the sake of completeness, we include the proof here.

Proof.

We put $a_{1}=1$ and define $a_{m}$ recursively. Given $a_{1},\ldots,a_{m-1},$ let $a_{m}$ be the smallest positive integer satisfying

[TABLE]

for every choice $u_{i}$ such that $|u_{i}|\leq k,$ for every set $S\subset\{1,\ldots,4\}$ of subscripts such that \Big{(}\sum_{i\in S}u_{i}\Big{)}\neq 0, and for every choice of $x_{i}\in\{a_{1},\ldots,a_{m-1}\}$, where $i\notin S.$ For a fixed $S$ with $|S|=j,$ this excludes $(m-1)^{4-j}$ numbers. Since $|u_{i}|\leq k,$ the total number of excluded integers is at most

[TABLE]

Consequently, we can extend our set by an integer $a_{m}\leq 2^{8}k^{4}m^{3}.$ This will automatically be different from from $a_{1},\ldots,a_{m-1},$ since putting $x_{i}=a_{j}$ for all $i\notin S$ in (4.3) we get $a_{m}\neq a_{j}.$ It will also satisfy $a_{m}>a_{m-1}$ by minimal choice of $a_{m-1}.$

We show that the system of equations (4.2) has no nontrivial solutions in the set $\{a_{1},\ldots,a_{m}\}.$ We use induction on $m.$ The statement is obviously true for $m=1.$ We establish it for $m$ assuming for $m-1.$ Suppose that there is a nontrivial solution $(x_{1},x_{2},x_{3},x_{4})$ to (4.2) for some $u_{1},u_{2},u_{3},u_{4}$ with the properties described above. Let $S$ denote the set of those subscripts for which $x_{i}=a_{m}.$ If $\sum_{i\in S}u_{i}\neq 0,$ then this contradicts (4.3). If $\sum_{i\in S}u_{i}=0,$ then by replacing each occurrence of $a_{m}$ by $a_{1},$ we get another nontrivial solution, which contradicts the induction hypothesis. ∎

For more problems and results on Sidon sets and $k$-fold Sidon sets, we refer the interested reader to [11, 14, 4].

We are now ready to prove Theorem 1.5.

Proof of Theorem 1.5..

We start by applying Lemma 4.1 to obtain a Sidon set $M\subset[n^{1/{7}}],$ such that $|M|=\Theta(n^{1/{14}}).$ We then apply Lemma 4.2 with $k=n^{1/7}$ and $N=\frac{1}{4}n^{{11}/{14}},$ to obtain a $k$-fold Sidon set $A\subset[N]$ such that

[TABLE]

where $c$ is defined in Lemma 4.2. Without loss of generality, let us assume $|A|=cn^{1/14}.$

Let $P=\{(i,j)\in\mathbb{Z}^{2}:i\in A,1\leq j\leq n^{13/14}\},$ and let $\mathcal{L}$ be the family of lines in the plane of the form $y=mx+b,$ where $m\in M$ and $b$ is an integer such that $1\leq b\leq n^{13/14}/2.$

Hence, we have

[TABLE]

Notice that each line in $\mathcal{L}$ has exactly $|A|=cn^{1/14}$ points from $P$ since $1\leq b\leq n^{13/14}/2.$ Therefore,

[TABLE]

Claim 4.3**.**

There are no four distinct lines ${\ell}_{1},{\ell}_{2},{\ell}_{3},{\ell}_{4}\in\mathcal{L}$ and four distinct points $p_{1},p_{2},p_{3},p_{4}\in P$ such that ${\ell}_{1}\cap{\ell}_{2}=p_{1},{\ell}_{2}\cap{\ell}_{3}=p_{2},{\ell}_{3}\cap{\ell}_{4}=p_{3},{\ell}_{4}\cap{\ell}_{1}=p_{4}.$

Proof.

For the sake of contradiction, suppose there are four lines $\ell_{1},\ell_{2},\ell_{3},\ell_{4}$ and four points $p_{1},p_{2},p_{3},p_{4}$ with the properties described above. Let $\ell_{i}=m_{i}x+b_{i}$ and let $p_{i}=(x_{i},y_{i}).$ Therefore,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Hence,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

By summing up the four equations above, we get

[TABLE]

By setting $u_{1}=m_{1}-m_{2},u_{2}=m_{2}-m_{3},u_{3}=m_{3}-m_{4},u_{4}=m_{4}-m_{1},$ we get

[TABLE]

where $u_{1}+u_{2}+u_{3}+u_{4}=0$ and $|u_{i}|\leq n^{1/{7}}.$ Since $x_{1},\ldots,x_{4}\in A,$ $(x_{1},x_{2},x_{3},x_{4})$ must be a trivial solution to (4.4). The proof now falls into the following cases, and let us note that no line in $\mathcal{L}$ is vertical.

Case 1. Suppose $x_{1}=x_{2}=x_{3}=x_{4}.$ Then $\ell_{i}$ is vertical and we have a contradiction.

Case 2. Suppose $x_{1}=x_{2}=x_{3}\neq x_{4}$ and $u_{1}+u_{2}+u_{3}=0$ and $u_{4}=0.$ Then $\ell_{1}$ and $\ell_{4}$ have the same slope which is a contradiction. The same argument follows if $x_{1}=x_{2}=x_{4}\neq x_{3},$ $x_{1}=x_{3}=x_{4}\neq x_{2},$ or $x_{2}=x_{3}=x_{4}\neq x_{1}.$

Case 3. Suppose $x_{1}=x_{2}\neq x_{3}=x_{4},$ $u_{1}+u_{2}=0,$ and $u_{3}+u_{4}=0.$ Since $p_{1},p_{2}\in\ell_{2}$ and $x_{1}=x_{2},$ this implies that $\ell_{2}$ is vertical which is a contradiction. A similar argument follows if $x_{1}=x_{4}\neq x_{2}=x_{3},$ $u_{1}+u_{4}=0,$ and $u_{2}+u_{3}=0.$

Case 4. Suppose $x_{1}=x_{3}\neq x_{2}=x_{4},$ $u_{1}+u_{3}=0,$ and $u_{2}+u_{4}=0.$ Then $u_{1}+u_{3}=0$ implies that $m_{1}+m_{3}=m_{2}+m_{4}.$ Since $M$ is a Sidon set, we have either $m_{1}=m_{2}$ and $m_{3}=m_{4}$ or $m_{1}=m_{4}$ and $m_{2}=m_{3}.$ The first case implies that $\ell_{1}$ and $\ell_{2}$ are parallel which is a contradiction, and the second case implies that $\ell_{2}$ and $\ell_{3}$ are parallel, which is again a contradiction. ∎

This completes the proof of Theorem 1.5.∎

5 Concluding Remarks

• •

An old result of Erdős states that every $n$-vertex graph that does not contain a cycle of length $2k,$ has $O_{k}(n^{1+1/k})$ edges. It is known that this bound is tight when $k=2,3,$ and $5$, but it is a long standing open problem in extremal graph theory to decide whether or not this upper bound can be improved for other values of $k.$ Hence, Erdős’s upper bound of $O(n^{5/4})$ when $k=4$ implies Theorem 1.3 when $t=2$ and $m=n.$ It would be interesting to see if one can improve the upper bound in Theorem 1.3 when $t=2.$ For more problems on cycles in graphs, see [18].

• •

The proof of Lemma 3.1 is similar to the proof of the main result in [1]. The main difference is that we use the result of Fox et al. [8] instead of the Ham-Sandwich Theorem. We also note that a similar result was established by Dujmović and Langerman (see Theorem $6$ in [6]).

• •

Recently, Tomon and the second author [17] improved the lower bound in Theorem 1.5 to $n^{9/8+o(1)}$, and more generally, gave a construction of $n$ points and $n$ lines in the plane with no $k\times k$ grid and with at least $n^{4/3-\Theta(1/k)}$ incidences.