Property Testing for Point Sets on the Plane

Abstract

A configuration is a point set on the plane, with no three points collinear. Given three non-collinear points p, q and $r\in \mathbb {R}^2$, let $\chi (p,q,r)\in \{-1,1\}$, with $\chi (p,q,r)=1$ if and only if, when we traverse the circle defined by those points in the counterclockwise direction, we encounter the points in the cyclic order $p,q,r,p,q,r,\dots$. For simplicity, extend $\chi$ by setting $\chi (p,q,r)=0$ if p, q and r are not pairwise distinct. Two configurations A and $B\subset \mathbb {R}^2$ are said to have the same order type if there is a bijection $\iota :A\rightarrow B$ such that $\chi (p,q,r)=\chi (\iota (p),\iota (q),\iota (r))$ for all $(p,q,r)\in A^3$. We say that a configuration C contains a copy of a configuration A if there is $B\subset C$ with A and B of the same order type. Given a configuration F, let $\mathrm{Forb}(F)$ be the set of configurations that do not contain a copy of F. The distance between two configurations A and B with $|A|=|B|=n$ is given by

$$\begin{aligned} \mathrm{dist}(A,B) = \min _\iota \frac{1}{2n^3}\sum _{(p,q,r)\in A^3} |\chi (p,q,r)-\chi (\iota (p),\iota (q),\iota (r))|, \end{aligned}$$

where the minimum is taken over all bijections $\iota :A\rightarrow B$. Roughly speaking, we prove the following property testing result: being free of a given configuration is efficiently testable. Our result also holds in the general case of hereditary properties $\mathcal {P}=\mathrm{Forb}(\mathcal {F})$, defined by possibly infinite families $\mathcal {F}$ of forbidden configurations. Our results complement previous results by Czumaj, Sohler and Ziegler and others, in that we use a different notion of distance between configurations. Our proofs are heavily inspired on recent work of Fox and Wei on testing permutations and also make use of the regularity lemma for semi-algebraic hypergraphs of Fox, Pach and Suk. An extremal function involving order types, which may be of independent interest, plays an important rôle in our arguments.

A Proof Sketch for Lemma 1

We write by l(x, y) as the line defined by two points x, y in the plane. Let l be a line and B be a set of points. Then we say l crosses B if there exist $b_1, b_2\in B$ such that they are at the different sides of l, or, equivalently, the line segment defined by $b_1, b_2$ intersects l. Our key claim is the following.

Claim

Let $b\in B_i$, then any line $l\ni b$ crosses at most one other block $B_j$.

Proof

Assume that $l\ni b$ crosses both $B_j$ and $B_{j'}$. Pick the points $u, u'\in B_j$, $v, v'\in B_{j'}$ such that u and v are at the same side of l, and $u'$, $v'$ are at the other side of l. Consider two lines l(u, b) and $l(u',b)$. These two lines partition the plane into four regions. By the definition of the blow-up, we know that $\chi (b,u,v)=\chi (b,u,v')$ and $\chi (b,u',v)=\chi (b,u',v')$. Thus, v and $v'$ must be in the same region. Moreover, since v and $v'$ must be at the different sides of l, they must be both in one of the two regions that intersect l. Then note that $\chi (b,u,v)\ne \chi (b,u',v')$ holds, which is a contradiction.

Now we start the sketch of the proof. For any $i\in [m]$, we find a largest subset $C_i\subset B_i$ in convex position. By the definition of $ES_F$, each $C_i$ has size at least $f\mathtt{poly}(m)$. Let $C_i=\{v^{(i)}_1,\dots , v^{(i)}_{p}\}$ where the points are ordered along the convex polygon under clockwise order. We now color each line segment $v^{(i)}_j v^{(i)}_{j+1}$ as follows:

• if $l(v^{(i)}_j, v^{(i)}_{j+1})$ crosses $C_p$ for some $c\in [m]$, $c\ne i$, then give $v^{(i)}_j v^{(i)}_{j+1}$ color p;

• otherwise, we rotate the line $l(v^{(i)}_j, v^{(i)}_{j+1})$ around $v^{(i)}_{j+1}$ clock-wise. Give $v^{(i)}_j v^{(i)}_{j+1}$ color p if $C_p$ is the first block that the line crosses.

Note that the coloring is well-defined because of the claim. In total this defines a coloring with m colors, and note that each color class consists of two (possibly empty) sets of consecutive line segments of the convex polygon $C_i$. Thus by the Pigeonhole principle, there is a subset $C_i'=\{u^{(i)}_1,\dots ,u^{(i)}_{q}\}\subset C_i$ such that $|C_i'|\ge f\mathtt{poly}(m)$, and all $(u^{(i)}_r, u^{(i)}_{r+1})$ for $1\le r<q$ have the same color. We repeat this process for all $i\in [m]$.

Next we consider $C_1'$ and all lines $(u^{(i)}_{q/2}, u^{(i)}_{q/2 +1})$ for every index i such that $C'_i$ received color 1. These lines divide $C'_1$ in at most $\mathtt{poly}(m)$ regions. Hence, by averaging, we can find a subset $D'_1\subset C'_1$ of 2f points such that for every such i, either $\{u^{(i)}_1,\dots ,u^{(i)}_{q/2}\}$ or $\{u^{(i)}_{q/2+1},\dots ,u^{(i)}_q\}$ are entirely on the same side of $D_1'$. Thus, abusing the notation, we pick either $\{u^{(i)}_1,\dots ,u^{(i)}_{q/2}\}$ or $\{u^{(i)}_{q/2+1},\dots ,u^{(i)}_q\}$ as our new $C_i'$, and we still have $|C_i'|\ge f\,\,\mathtt{poly}(m)$.

We can continue the operation for every $C'_r$ ($1\le r\le m$) (but considering only the points survived from previous iterations). This gives sets $A_1,\dots ,A_m$ each of size f satisfying Lemma 1 (a) and (b), because during the whole process, a set $C'_r$ is shrunk at most twice: by a factor of $\mathtt{poly}(m)$ when it plays the role of $C'_j$, or by a factor of 2 when it plays the role of $C'_i$.    $\square$