# Many Touchings Force Many Crossings

• János Pach
• Géza Tóth
Conference paper
Part of the Lecture Notes in Computer Science book series (LNCS, volume 10692)

## Abstract

Given n continuous open curves in the plane, we say that a pair is touching if they have only one interior point in common and at this point the first curve does not get from one side of the second curve to its other side. Otherwise, if the two curves intersect, they are said to form a crossing pair. Let t and c denote the number of touching pairs and crossing pairs, respectively. We prove that $$c \ge {1\over 10^5}{t^2\over n^2}$$, provided that $$t\ge 10n$$. Apart from the values of the constants, this result is best possible.

## Keywords

Planar curves Touching Crossing

## 1 Introduction

In the context of the theory of topological graphs and graph drawing, many interesting questions have been raised concerning the adjacency structure of a family of curves in the plane or in another surface [5]. In particular, during the past four decades, various important properties of string graphs (i.e., intersection graphs of curves in the plane) have been discovered, and the study of different crossing numbers of graphs and their relations to one another has become a vast area of research. A useful tool in these investigations is the so-called crossing lemma of Ajtai, Chvátal, Newborn, Szemerédi and Leighton [1, 7]. It states the following: Given a graph of n vertices and $$e>4n$$ edges, no matter how we draw it in the plane by not necessarily straight-line edges, there are at least constant times $$e^3/n^2$$ crossing pairs of edges.

This lemma has inspired a number of results establishing the existence of many crossing subconfigurations of a given type in sufficiently rich geometric or topological structures [2, 6, 10, 11, 12].

In this note, we will be concerned with families of curves in the plane. By a curve, we mean a non-selfintersecting continuous arc in the plane, that is, a homeomorphic image of the open interval (0, 1). Two curves are said to touch each other if they have precisely one interior point in common and at this point the first curve does not pass from one side of the second curve to the other. Any other pair of curves with nonempty intersection is called crossing. A family of curves is in general position if any two of them intersect in a finite number of points and no three pass through the same point.

Let n be even, t be a multiple of n, and suppose that $$n\le t<{n^2\over 4}$$. Consider a collection A of $$n-{2t\over n}>{n\over 2}$$ pairwise disjoint curves, and another collection B of $${2t\over n}$$ curves such that
1. (i)

$$A\cup B$$ is in general position,

2. (ii)

each element of B touches precisely $${n\over 2}$$ elements of A, and

3. (iii)

no two elements of B touch each other.

The family $$A\cup B$$ consists of n curves such that the number of touching pairs among them is t. The only pairs of curves that may cross each other belong to B. Thus, the number of crossing pairs is at most $${2t/n \atopwithdelims ()2}\le {2t^2\over n^2}$$. See Fig. 1.

The aim of the present note is to prove that this construction is optimal up to a constant factor, that is, any family of n curves and t touchings has at least constant times $${t^2\over n^2}$$ crossing pairs.

### Theorem 1

Consider a family of n curves in general position in the plane which determines t touching pairs and c crossing pairs.

If $$t\ge 10n$$ , then we have $$c \ge {1\over 10^5}{t^2\over n^2}$$. This bound is best possible up to a constant factor.

We make no attempt to optimize the constants in the theorem.

Pach et al. [8] established a similar relationship between t, the number of touching pairs, and C, the number of crossing points between the curves. They proved that $$C \ge t(\log \log (t/n))^{\delta }$$, for an absolute constant $$\delta >0$$. Obviously, we have $$C\ge c$$. There is an arrangement of n red curves and n blue curves in the plane such that every red curve touches every blue curve, and the total number of crossing points is $$C=\varTheta (n^2\log n)$$; cf. [4]. Of course, the number of crossing pairs, c, can never exceed $${n\atopwithdelims ()2}$$.

Between n arbitrary curves, the number of touchings t can be as large as $$(\frac{3}{4} + o(1)){n\atopwithdelims ()2}$$; cf. [9]. However, if we restrict our attention to algebraic plane curves of bounded degree, then we have $$t=O(n^{3/2})$$, where the constant hidden in the notation depends on the degree [3].

## 2 Proof of Theorem

### Lemma

Given a family of $$n\ge 3$$ curves in general position in the plane, no two of which cross, the number of touchings, t, cannot exceed $$3n-6$$.

### Proof

Pick a different point on each curve. Whenever two curves touch each other at a point p, connect them by an edge (arc) passing through p. In the resulting drawing, any two edges that do not share an endpoint are represented by disjoint arcs. According to the Hanani-Tutte theorem [13], this means that the underlying graph is planar, so that its number of edges, t, satisfies $$t\le 3n-6$$.    $$\Box$$

### Proof of Theorem

We proceed by induction on n. For $$n\le 20$$, the statement is void. Suppose that $$n>20$$ and that the statement has already been proved for all values smaller than n.

We distinguish two cases.

CASE A: $$t \le 10n^{3/2}$$.

In this case, we want to establish the stronger statement
$$c\ge {1\over 10^4}{t^2\over n^2}.$$
By the assumption, we have
\begin{aligned} {1\over 10^4}{t^2\over n^2}\le {n\over 100}. \end{aligned}
(1)
Let $$G_t$$ (resp., $$G_c$$) denote the touching graph (resp., crossing graph) associated with the curves. That is, the vertices of both graphs correspond to the curves, and two vertices are connected by an edge if and only if the corresponding curves are touching (resp., crossing).
Let T be a minimal vertex cover in $$G_c$$, that is, a smallest set of vertices of $$G_c$$ such that every edge of $$G_c$$ has at least one endpoint in T. Let $$\tau =|T|$$. Let U denote the complement of T. Obviously, U is an independent set in $$G_c$$. According to the Lemma, the number of edges in $$G_t[U]$$, the touching graph induced by U, satisfies
\begin{aligned} |E(G_t[U])|< 3|U|\le 3n. \end{aligned}
(2)
By the minimality of T, $$G_c$$ has at least $$|T|=\tau$$ edges. That is, we have $$c \ge \tau$$, so we are done if $$\tau \ge {1\over 10^4}{t^2\over n^2}$$.
From now on, we can and shall assume that $$\tau < {1\over 10^4}{t^2\over n^2}.$$ By (1), we have $${1\over 10^4}{t^2\over n^2}\le {n\over 100}$$. Hence, $$|T|\le {n\over 100}$$ and
\begin{aligned} |U|=n-|T| \ge {99n\over 100}. \end{aligned}
(3)
Let $$U'\subseteq U$$ denote the set of all vertices in U that are not isolated in the graph $$G_c$$. By the definition of T, all neighbors of a vertex $$v\in U$$ in $$G_c$$ belong to T. If $$|U'|\ge {1\over 10^4}{t^2\over n^2}$$, then we are done, because $$c\ge |U'|.$$
Therefore, we can assume that
\begin{aligned} |U'|< {1\over 10^4}{t^2\over n^2}\le {n\over 100}, \end{aligned}
(4)
where the second inequality follows again by (1) (Fig. 2).

Letting $$U_0=U\setminus U'$$, by (3) and (4) we obtain $$|U_0|=|U|-|U'|\ge {98n\over 100}$$. Clearly, all vertices in $$U_0$$ are isolated in $$G_c$$.

Suppose that $$G_t[T\cup U']$$ has at least $${t\over 10}$$ edges. Consider the set of curves $$T\cup U'$$. We have $$n_0=|T\cup U'|\le {2n\over 100}$$ and, the number of touchings, $$t_0=|E(G_t[T\cup U'])|\ge {t\over 10}$$. Therefore, by the induction hypothesis, for the number of crossings we have $$c_0=|E(G_c[T\cup U'])|\ge {1\over 10^5}{t_0^2\over n_0^2}\ge {1\over 10^4}{t^2\over n^2}$$ and we are done. Hence, we assume in the sequel that $$G_t[T\cup U']$$ has fewer than $${t\over 10}$$ edges.

Consequently, for the number of edges in $$G_t$$ running between T and $$U_0$$, we have
\begin{aligned} |E(G_t[T,U_0])|\ge t-|E(G_t[T\cup U'])|-|E(G_t[U_0\cup U'])|\ge t-{t\over 10}-3n>{t\over 2}. \end{aligned}
(5)
Here we used the assumption that $$t\ge 10n$$.

Let $$\chi =\chi (G_c[T])$$ denote the chromatic number of $$G_c[T]$$. In any coloring of a graph with the smallest possible number of colors, there is at least one edge between any two color classes. Hence, $$G_c[T]$$ has at least $${\chi \atopwithdelims ()2}\ge {1\over 10^4}{t^2\over n^2}$$ edges, and we are done, provided that $$\chi >{1\over 70}\cdot {t\over n}$$.

Thus, we can suppose that
\begin{aligned} \chi =\chi (G_c[T])\le {1\over 70}\cdot {t\over n}. \end{aligned}
(6)
Consider a coloring of $$G_c[T]$$ with $$\chi$$ colors, and denote the color classes by $$I_1, I_2, \ldots , I_{\chi }$$. Obviously, for every j, $$I_j\cup U_0$$ is an independent set in $$G_c$$. Therefore, by the Lemma, $$G_t[I_j\cup U_0]$$ has at most 3n edges. Summing up for all j and taking (6) into account, we obtain
$$|E(G_t[T,U_0])|\le \sum _{j=1}^{\chi }|E(G_t[I_j\cup U_0])| \le {1\over 70}\cdot {t\over n}3n\le {t\over 20},$$
contradicting (5). This completes the proof in CASE A.
CASE B: $$t\ge 10n^{3/2}$$. Set $$p={10n^3\over t^2}\le {1 \over 10}$$. Select each curve independently with probability p. Let $$\mathbf{n'}$$, $$\mathbf{t'}$$, and $$\mathbf{c'}$$ denote the number of selected curves, the number of touching pairs, and the number of crossing pairs between them, respectively. Clearly,
\begin{aligned} E[\mathbf{n'}]=pn, \;\; E[\mathbf{t'}]=p^2t, \;\; E[\mathbf{c'}]=p^2c. \end{aligned}
(7)
The number of selected curves, $$\mathbf{n'}$$, has binomial distribution, therefore,
\begin{aligned} \mathrm{Prob}[|\mathbf{n'}-pn|>{1\over 4}pn]<{1\over 3}. \end{aligned}
(8)
By Markov’s inequality,
\begin{aligned} \mathrm{Prob}[\mathbf{c'}>3p^2c]<{1\over 3}. \end{aligned}
(9)
Consider the touching graph $$G_t$$. Let $$d_1, \ldots , d_n$$ denote the degrees of the vertices of $$G_t$$, and let $$e_1, \ldots , e_t$$ denote its edges, listed in any order. We say that an edge $$e_i$$ is selected (or belongs to the random sample) if both of its endpoints were selected. Let $$X_i$$ be the indicator variable for $$e_i$$, that is,
$$X_i=\left\{ \begin{array}{ll} 1\;\;\; \text{ if } e_i \text{ was } \text{ selected, }\\ 0\;\;\; \text{ otherwise. }\\ \end{array} \right.$$
We have $$E[X_i]=p^2$$. Let $$\mathbf{t'}=\sum _{i=1}^tX_i$$. It follows by straightforward computation that for every i,
$$\mathrm{var}[X_i]=E[(X_i-E[X_i])^2]=p^2-p^4,$$
If $$e_i$$ and $$e_j$$ have a common endpoint for some $$i\ne j$$, then
$$\mathrm{cov}[X_i, X_j]=E[X_iX_j]-E[X_i]E[X_j]=p^3-p^4.$$
If $$e_i$$ and $$e_j$$ do not have a common vertex, then $$X_i$$ and $$X_j$$ are independent random variables and $$\mathrm{cov}[X_i, X_j]=0$$. Therefore, we obtain
$$\sigma ^2=\mathrm{var}[\mathbf{t'}] =\sum _{i=1}^t\mathrm{var}[X_i]+\sum _{1\le i\ne j\le t}\mathrm{cov}[X_i, X_j]$$
$$=(p^2-p^4)t+(p^3-p^4)\sum _{i=1}^nd_i(d_i-1)<p^2t+2p^3nt.$$
From here, we get $$\sigma<\sqrt{p^2t}+\sqrt{2p^3nt}<p^2t=E[\mathbf{t'}]$$. By Chebyshev’s inequality,
$$\mathrm{Prob}[|\mathbf{t'}-p^2t|\ge \lambda \sigma ]\le {1\over \lambda ^2}.$$
Setting $$\lambda ={1\over 4}$$,
\begin{aligned} \mathrm{Prob}[|\mathbf{t'}-p^2t|\ge {p^2t\over 4}]\le {1\over 4^2}<{1\over 3}. \end{aligned}
(10)
It follows from (8), (9), and (10) that, with positive probability, we have
\begin{aligned} |\mathbf{n'}-pn|\le {1\over 4}pn,\;\;\;\; \mathbf{c'}\le 3p^2c,\;\;\;\;|\mathbf{t'}-p^2t|\le {1\over 4}p^2t. \end{aligned}
(11)
Consider a fixed selection of $$n'$$ curves with $$t'$$ touching pairs and $$c'$$ crossing pairs for which the above three inequalities are satisfied. Then we have
$$t'\ge {3\over 4}p^2t={300\over 4}\cdot {n^6\over t^3},$$
$$n'\le {5\over 4}pn={50\over 4}\cdot {n^4\over t^2},$$
and, hence,
\begin{aligned} t'\ge {6n^2\over t}n'\ge 10n'. \end{aligned}
(12)
On the other hand,
$$t'\le {5\over 4}p^2t={500\over 4}\cdot {n^6\over t^3},$$
$$n'\ge {3\over 4}pn={30\over 4}\cdot {n^4\over t^2},$$
so that
\begin{aligned} 10(n')^{3/2}\ge 10\cdot {30^{3/2}\over 4^{3/2}}\cdot {n^6\over t^3}>t'. \end{aligned}
(13)
According to (12) and (13), the selected family meets the requirements of the Theorem in CASE A. Thus, we can apply the Theorem in this case to obtain that $$c'\ge {1\over 10^4}{t'^2\over n'^2}$$. In view of (11), we have
$$3p^2c\ge c',\;\;\;\; t'\ge {3\over 4}p^2t,\;\;\;\; n'\le {5\over 4}pn.$$
Thus,
$$3p^2c\ge c'\ge {1\over 10^4}{t'^2\over n'^2}\ge {1\over 10^4}{(3p^2t/4)^2\over (5pn/4)^2} ={1\over 10^4}\left( {3\over 5}\right) ^2{p^2t^2\over n^2}.$$
Comparing the left-hand side and the right-hand side, we conclude that
$$c\ge {1\over 10^5}{t^2\over n^2},$$
as required. This completes the proof of the Theorem.   $$\Box$$

## Notes

### Acknowledgment

The work of János Pach was partially supported by Swiss National Science Foundation Grants 200021-165977 and 200020-162884. Géza Tóth’s work was partially supported by the Hungarian National Research, Development and Innovation Office, NKFIH, Grant K-111827.

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