Construction of ε-Nets for the Space of Planar Convex Bodies Endowed with the Banach–Mazur Metric

Abstract

In Chuanming Zong’s program to attack Hadwiger’s covering conjecture, which is a long-standing open problem from convex and discrete geometry, the construction of $\epsilon$-nets for the space of convex bodies endowed with the Banach–Mazur metric plays a crucial role. Recently, Gao et al. provided a possible way of constructing $\epsilon$-nets for $\left({\mathcal{K}}^n,d_{BM}\right)$ based on finite subsets of ${\mathbb{Z}}^n$ theoretically. In this work, we present an algorithm to construct $\epsilon$-nets for $\left({\mathcal{K}}^2,d_{BM}\right)$ and a $(1/4)$-net for $\left({\mathcal{C}}^2,d_{BM}\right)$ is constructed. To the best of our knowledge, this is the first concrete $\epsilon$-net for $\left({\mathcal{C}}^2,d_{BM}\right)$ for such a small $\epsilon$.

1. Introduction

A compact convex set $K\subseteq {\mathbb{R}}^n$ having interior points is called a $convex body$, whose $\mathit{interior}$ and $\mathit{boundary}$ are denoted by $intK$ and $bdK$, respectively. Let ${\mathcal{K}}^n$ be the set of convex bodies in ${\mathbb{R}}^n$, and let ${\mathcal{C}}^n$ be the set of convex bodies in ${\mathcal{K}}^n$ that are symmetric about the $origin o$. If A is a set in ${\mathbb{R}}^n$, we denote the $convex hull$ of A as $convA$. Several classical open problems from discrete and convex geometry can be considered via studying certain functionals that are uniformly continuous on $({\mathcal{K}}^n,d_{BM})$. Take Hadwiger’s covering conjecture for example. Let $c\left(K\right)$ be the least number of translates of $intK$ needed to cover K.

Conjecture 1

(Hadwiger’s covering conjecture). For each $K\in {\mathcal{K}}^n$, we have

$$c\left(K\right)\le 2^n;$$

the equality holds if and only if K is a parallelotope.

This conjecture has been completely resolved only for the case of $n=2$ (cf. [1]) and remains open for $n\ge 3$. For further information on this conjecture, we refer to [2,3,4,5,6,7,8,9,10,11,12,13,14].

In fact, $c\left(K\right)$ equals the minimum number of $smaller homothetic copies$ (sets having the form $x+\gamma K$ with $x\in {\mathbb{R}}^n$ and $\gamma \in (0,1)$) of K needed to cover K. Chuanming Zong introduced a quantitative program to tackle this conjecture via estimating

$${\Gamma }_m\left(K\right):=min\left\{\gamma >0\mid \exists \left\{x_i\mid i\in \left[m\right]\right\}\subseteq {\mathbb{R}}^n\mathrm{s}.\mathrm{t}.K\subseteq \bigcup _{i\in \left[m\right]}(x_i+\gamma K)\right\},$$

where $\left[m\right]=\left\{i\in {\mathbb{Z}}^{+}\mid 1\le i\le m\right\}$. It is evident that

$$c\left(K\right)\le m\iff {\Gamma }_m\left(K\right)<1.$$

The map

$$\begin{array}{cc}\hfill {\Gamma }_m(·):{\mathcal{K}}^n& \to [0,1]\hfill \\ \hfill K& \mapsto {\Gamma }_m\left(K\right)\hfill \end{array}$$

is called the covering functional with respect to m. For each $m\in {\mathbb{Z}}^{+}$, ${\Gamma }_m(·)$ is an affine invariant. More precisely,

$${\Gamma }_m\left(K\right)={\Gamma }_m\left(T\left(K\right)\right),\forall T\in {\mathcal{A}}^n,$$

where ${\mathcal{A}}^n$ is the set of non-degenerate affine transformations from ${\mathbb{R}}^n$ to ${\mathbb{R}}^n$. It is natural to use the (multiplicative) Banach–Mazur metric $d_{BM}^M$ to measure the distance between two convex bodies, where, for each pair of $K_1,K_2\in {\mathcal{K}}^n$, $d_{BM}^M(K_1,K_2)$ is defined as

$$d_{BM}^M(K_1,K_2):=min\left\{\gamma \ge 1\mid \exists x\in {\mathbb{R}}^n,T\in {\mathcal{A}}^n\mathrm{s}.\mathrm{t}.K_1\subseteq T\left(K_2\right)\subseteq \gamma K_1+x\right\}.$$

Set

$$d_{BM}(K_1,K_2)=ln{d}_{BM}^M(K_1,K_2).$$

It is well known that $({\mathcal{K}}^n,d_{BM})$ is a compact metric space (convex bodies that are affinely equivalent are regarded as the same element, of course).

Let

$$c_n=max\left\{{\Gamma }_{2^n}\left(K\right)\mid K\in {\mathcal{K}}^n\right\}.$$

It is clear that

$$c\left(K\right)\le 2^n,\forall K\in {\mathcal{K}}^n\iff c_n<1.$$

Since ${\Gamma }_{2^n}\left(K\right)$ is continuous on the compact metric space ${\mathcal{K}}^n$, there exists $\epsilon >0$ such that, for each $K,K^{\prime }\in {\mathcal{K}}^n$ with $d_{BM}(K,K^{\prime })\le \epsilon$, the inequality

$$|{\Gamma }_{2^n}\left(K\right)-{\Gamma }_{2^n}\left(K^{\prime }\right)| \le {\displaystyle \frac{1}{2}}(1-c_n)$$

(1)

holds.

Let $\epsilon$ be a positive number, and let $K_1,K_2,\dots ,K_{l(n,\epsilon )}$ be $l(n,\epsilon )$ convex bodies in ${\mathcal{K}}^n$, where $l(n,\epsilon )$ is an integer depending on n and $\epsilon$. If, for each $K\in {\mathcal{K}}^n$, there exists a corresponding $K_i$ satisfying

$$d_{BM}^M(K,K_i)\le 1+\epsilon ,$$

then, we call $\mathcal{N}=\{K_1,K_2,\dots ,K_{l(n,\epsilon )}\}$ an ε-net for ${\mathcal{K}}^n$. If, for a given $\epsilon >0$, we can construct an ε-net $\mathcal{N}$ and verify that ${\Gamma }_{2^n}\left(K_i\right)\le c_n$ holds for each $K_i\in \mathcal{N}$, then, by (1), we have

$${\Gamma }_{2^n}\left(K\right)\le {\displaystyle \frac{1}{2}}(1+c_n)<1,\forall K\in {\mathcal{K}}^n.$$

Therefore, Zong proposed the following four steps in [15] to attack Hadwiger’s covering conjecture:

(1)

Obtain a good guess ${\widehat{c}}_n$ of $c_n$ by estimating ${\Gamma }_{2^n}\left(K\right)$ for special classes of convex bodies;

(2)

Choose a suitable $\epsilon >0$;

(3)

Construct an ε-net $\mathcal{N}$ of ${\mathcal{K}}^n$;

(4)

For each $K\in \mathcal{N}$, verify that ${\Gamma }_{2^n}\left(K\right)\le {\widehat{c}}_n$.

Clearly, constructing a suitable $\epsilon$-net $\mathcal{N}$ is crucial in this program. Recently, three algorithms have been designed to estimate ${\Gamma }_m\left(K\right)$ [2,3,4].

Zong showed in [15] that

$$l(n,\epsilon )\le {⌊{\displaystyle \frac{7n}{\epsilon }}⌋}^{c·{14}^n·n^{2n+3}·{\epsilon }^{-n}},$$

where c is a universal constant. When n is large and $\epsilon$ is small, the right-hand side is huge. Moreover, one should not expect the cardinality of an $\epsilon$-net to be small, since there are known lower bounds on the cardinality of such nets, see, e.g., [16,17].

In [18], Shenghua Gao et al. provided a possible way of constructing $\epsilon$-nets for $({\mathcal{K}}^n,d_{BM})$ based on finite subsets of ${\mathbb{Z}}^n$ theoretically. However, there is still no practical algorithm for this purpose. In this paper, we propose an algorithmic method for the first time for constructing $\epsilon$-nets with a reasonable cardinality in the two-dimensional case. Furthermore, we construct a $(1/4)$-net for $({\mathcal{C}}^2,d_{BM})$ consisting of 30,306 sets. This is the first concrete $\epsilon$-net for $({\mathcal{C}}^2,d_{BM})$.

2. A General Framework

Definition 1.

Let $K\subseteq {\mathbb{Z}}^n$ and $i\in \left[n\right]$. If

$$K=convK\cap {\mathbb{Z}}^n,$$

then, K is called a convex lattice set. If

$$x,y\in K,x-y\in \mathrm{span}\left\{e_i\right\}\Rightarrow [x,y]\cap {\mathbb{Z}}^n\subseteq K,$$

then K is said to be a lattice set convex in the i-th coordinate.

Clearly, if K is a convex lattice set, then, for each $i\in \left[n\right]$, K is convex at the i-th coordinate.

Proposition 1.

If $K\subseteq {\mathbb{Z}}^n$ is a convex lattice set and H is a closed halfspace of ${\mathbb{R}}^n$, then $K\cap H$ is also a convex lattice set.

Proof. 

We only need to show that $conv(K\cap H)\cap {\mathbb{Z}}^n\subseteq K\cap H$. Actually,

$$\begin{array}{c}\hfill conv(K\cap H)\cap {\mathbb{Z}}^n\subseteq (convK\cap {\mathbb{Z}}^n)\cap H=K\cap H.\end{array}$$

   □

Set

$$Q^n={[-1,1]}^n,\forall n\in {\mathbb{Z}}^{+}.$$

For each $K\in {\mathcal{K}}^n$, let

$$\alpha \left(K\right)=max\left\{\alpha >0\mid \exists T\in {\mathcal{A}}^n\mathrm{s}.\mathrm{t}.\alpha Q^n\subseteq T\left(K\right)\subseteq Q^n\right\}.$$

Theorem 1.

Let $K\in {\mathcal{K}}^n$, $M\in {\mathbb{Z}}^{+}$, and $\alpha \in (0,\alpha (K\left)\right]$. There exists a convex lattice set $G_M$ contained in $M{Q}^n$ such that

$$d_{BM}^M(K,conv(G_M+{\displaystyle \frac{1}{2}}{Q}^n))\le 1+{\displaystyle \frac{1}{M\alpha }}.$$

Moreover, if $K\in {\mathcal{C}}^n$, we may require further $G_M$ to be symmetric with respect to o.

Proof. 

The construction is based on the proof of Theorem 6 in [18]. By applying a suitable affine transformation if necessary, we may assume that

$$\alpha \left(K\right)Q^n\subseteq K\subseteq Q^n.$$

Thus $\alpha Q^n\subseteq K\subseteq Q^n$, which implies that

$$\alpha M{Q}^n\subseteq MK\subseteq M{Q}^n.$$

Set

$$G_M=\left\{z\in {\mathbb{Z}}^n\mid (z+{\displaystyle \frac{1}{2}}{Q}^n)\cap MK\ne \mathrm{\varnothing }\right\}.$$

Then

$$MK\subseteq G_M+{\displaystyle \frac{1}{2}}{Q}^n\subseteq conv(G_M+{\displaystyle \frac{1}{2}}{Q}^n).$$

For each $z\in G_M$, there exists a point $x\in MK$ such that $x-z\in {\displaystyle \frac{1}{2}}{Q}^n$. Thus, $z\in x+{\displaystyle \frac{1}{2}}{Q}^n$. It follows that

$$G_M\subseteq MK+{\displaystyle \frac{1}{2}}{Q}^n\subseteq M{Q}^n+{\displaystyle \frac{1}{2}}{Q}^n,$$

which implies that

$$MK\subseteq conv(MK+Q^n)=MK+Q^n\subseteq MK+{\displaystyle \frac{1}{\alpha }}K=M(1+{\displaystyle \frac{1}{M\alpha }})K,$$

and that (since $G_M\subseteq {\mathbb{Z}}^n$) $G_M\subseteq M{Q}^n$. Therefore,

$$d_{BM}^M(MK,conv(G_M+{\displaystyle \frac{1}{2}}{Q}^n))=d_{BM}^M(K,conv(G_M+{\displaystyle \frac{1}{2}}{Q}^n))\le 1+{\displaystyle \frac{1}{M\alpha }}.$$

Next, we show that $G_M$ is a convex lattice set. Clearly, $K\subseteq convK\cap {\mathbb{Z}}^n$. For each point $z\in convK\cap {\mathbb{Z}}^n$, by Carathéodory’s theorem, there exist $m\in [n+1]$, a set of m points $\{z_1,\dots ,z_m\}\subseteq K$, and a set of numbers $\{{\lambda }_1,\dots ,{\lambda }_m\}\subseteq [0,1]$ such that

$$z=\sum _{i\in \left[m\right]}{\lambda }_i{z}_i \mathrm{and} \sum _{i\in \left[m\right]}{\lambda }_i=1.$$

For each $i\in \left[m\right]$, there exists a point $x_i\in MK$ such that $z_i-x_i\in {\displaystyle \frac{1}{2}}{Q}^n$. It follows that

$$-(z-\sum _{i\in \left[m\right]}{\lambda }_i{x}_i)=-\left(\sum _{i\in \left[m\right]}{\lambda }_i(z_i-x_i)\right)\in {\displaystyle \frac{1}{2}}{Q}^n.$$

Since $MK$ is convex, we have

$$\sum _{i\in \left[m\right]}{\lambda }_i{x}_i\in MK.$$

Therefore,

$$(z+{\displaystyle \frac{1}{2}}{Q}^n)\cap MK\ne \mathrm{\varnothing }.$$

Hence, $z\in G_M$.

In the rest, we consider the case where $K\in {\mathcal{C}}^n$. Suppose that $z\in G_M$. Then, there exists a point $x\in {\displaystyle \frac{1}{2}}{Q}^n$ such that $z+x\in MK$, which implies that $-z-x\in -MK=MK$. Thus,

$$(-z+{\displaystyle \frac{1}{2}}{Q}^n)\cap MK\ne \varnothing ,$$

which shows that $-z\in G_M$. This completes the proof.    □

Put

$${\alpha }_0^n=inf\left\{\alpha \left(K\right)\mid K\in {\mathcal{K}}^n\right\} \mathrm{and} {\beta }_0^n=inf\left\{\alpha \left(K\right)\mid K\in {\mathcal{C}}^n\right\}.$$

Theorem 2

(cf. [19]). Let $K\in {\mathcal{K}}^n$. Then, the John ellipsoid E (the unique ellipsoid of maximum volume contained in K) of K, whose center is c, satisfies

$$E\subseteq K\subseteq c+n(E-c).$$

Let $K\in {\mathcal{K}}^n$; by a suitable affine transformation if necessary, we may assume that the Euclidean unit ball $B_2^n$ is the John ellipsoid of K. From Theorem 2, we have

$$B_2^n\subseteq K\subseteq n{B}_2^n,$$

or, equivalently,

$${\displaystyle \frac{1}{n}}{B}_2^n\subseteq {\displaystyle \frac{1}{n}}K\subseteq B_2^n.$$

Thus,

$${\displaystyle \frac{1}{n\sqrt{n}}}{Q}^n\subseteq {\displaystyle \frac{1}{n}}{B}_2^n\subseteq {\displaystyle \frac{1}{n}}K\subseteq B_2^n\subseteq Q^n,$$

which implies that

$${\alpha }_0^n\ge {\displaystyle \frac{1}{n\sqrt{n}}}.$$

Since $Q^n,C\in {\mathcal{C}}^n$, we have

$$d_{BM}^M(Q^n,C)=min\left\{\gamma \ge 1\mid \exists T\in {\mathcal{T}}^n\mathrm{s}.\mathrm{t}.Q^n\subseteq T\left(C\right)\subseteq \gamma Q^n\right\},$$

where ${\mathcal{T}}^n$ is the set of non-degenerate linear transformations from ${\mathbb{R}}^n$ to ${\mathbb{R}}^n$. Then

$$d_{BM}^M(Q^n,C)=min\left\{\gamma \ge 1\mid \exists T\in {\mathcal{T}}^n\mathrm{s}.\mathrm{t}.{\displaystyle \frac{1}{\gamma }}{Q}^n\subseteq T\left(C\right)\subseteq Q^n\right\}.$$

Therefore,

$${\displaystyle \frac{1}{d_{BM}^M(Q^n,C)}}\le \alpha \left(C\right).$$

Hence,

$$inf\left\{{\displaystyle \frac{1}{d_{BM}^M(Q^n,C)}}\mid C\in {\mathcal{C}}^n\right\}\le inf\left\{\alpha \left(C\right)\mid C\in {\mathcal{C}}^n\right\}.$$

It follows that

$${\beta }_0^n\ge {(max\left\{d_{BM}^M(Q^n,C)\mid C\in {\mathcal{C}}^n\right\})}^{-1}.$$

It is proved in [20] that

$$max\left\{d_{BM}^M(Q^2,C)\mid C\in {\mathcal{C}}^2\right\}\le {\displaystyle \frac{3}{2}}.$$

Thus, ${\beta }_0^2\ge 2/3$.

Corollary 1.

Let $\epsilon >0,K\in {\mathcal{K}}^n$, and $C\in {\mathcal{C}}^n$.

(1) 

There exists a convex lattice set $G_1$ contained in

$$⌈{\displaystyle \frac{1}{{\alpha }_0^n\epsilon }}⌉Q^n$$

and containing

$$⌊{\alpha }_0^n⌈{\displaystyle \frac{1}{{\alpha }_0^n\epsilon }}⌉⌋Q^n\cap {\mathbb{Z}}^n$$

such that

$$d_{BM}^M(K,G_1+{\displaystyle \frac{1}{2}}{Q}^n)\le 1+\epsilon .$$

(2) 

There exists a centrally symmetric convex lattice set $G_2$ contained in

$$⌈{\displaystyle \frac{1}{{\beta }_0^n\epsilon }}⌉Q^n$$

and containing

$$⌊{\beta }_0^n⌈{\displaystyle \frac{1}{{\beta }_0^n\epsilon }}⌉⌋Q^n\cap {\mathbb{Z}}^n$$

such that

$$d_{BM}^M(C,G_2+{\displaystyle \frac{1}{2}}{Q}^n)\le 1+\epsilon .$$

Proof. 

We only need to prove (1), and (2) can be proved in a similar way. Clearly, $\alpha \left(K\right)\ge {\alpha }_0^n$. By applying a suitable affine transformation if necessary, we may assume that

$${\alpha }_0^n{Q}^n\subseteq K\subseteq Q^n.$$

Let M be the least positive integer such that ${\displaystyle \frac{1}{M{\alpha }_0^n}}\le \epsilon$. Then, $M=⌈{\displaystyle \frac{1}{{\alpha }_0^n\epsilon }}⌉$. As in the proof of Theorem 1, there exists $G_M\subseteq M{Q}^n$ such that

$$d_{BM}^M(K,conv(G_M+{\displaystyle \frac{1}{2}}{Q}^n))\le 1+{\displaystyle \frac{1}{M{\alpha }_0^n}}\le 1+\epsilon .$$

Moreover, ${\alpha }_0^{n}M{Q}^n\subseteq MK$, which implies that

$$\left({\alpha }_0^{n}M{Q}^n\right)\cap {\mathbb{Z}}^n\subseteq MK.$$

For each $z\in {\mathbb{Z}}^n$, since $z\in MK$ implies that $z\in G_M$, we have

$$\left(\lfloor {\alpha }_0^{n}M\rfloor Q^n\right)\cap {\mathbb{Z}}^n=\left({\alpha }_0^{n}M{Q}^n\right)\cap {\mathbb{Z}}^n\subseteq MK\cap {\mathbb{Z}}^n\subseteq G_M.$$

$G_1:=G_M$ has the desired properties.    □

Theoretically, to construct an $\epsilon$-net for ${\mathcal{K}}^n$ (or ${\mathcal{C}}^n$), one needs to enumerate all convex lattice sets contained in $M_2{Q}^n$ and containing $M_1{Q}^n\cap {\mathbb{Z}}^n$ for some positive integers $M_1<M_2$. Then, one needs to take the quotient set of the resulting set with respect to the affine equivalence. Each of these steps is not easy. In the following, we focus on the planar case.

3. Construction of $\mathit{\epsilon }$-Nets for $({\mathcal{K}}^{\mathbf{2}},{\mathit{d}}_{\mathit{BM}})$

Let G be the group of transformations whose matrices are

$$\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right),\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right),\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right),\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right),\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right).$$

Proposition 2.

Let $I_1$ and $I_2$ be two subsets of ${\mathbb{Z}}^2$. If there exists $T\in G$ such that $I_2=T\left(I_1\right)$, then

$$T(conv(I_1+{\displaystyle \frac{1}{2}}{Q}^2))=conv(I_2+{\displaystyle \frac{1}{2}}{Q}^2).$$

Proof. 

It is clear that, for each $T\in G$, $T\left({\displaystyle \frac{1}{2}}{Q}^2\right)={\displaystyle \frac{1}{2}}{Q}^2$. From Theorem 3.16 in [5],

$$\begin{array}{cc}\hfill conv(I_2+{\displaystyle \frac{1}{2}}{Q}^2)& =conv(T\left(I_1\right)+T\left({\displaystyle \frac{1}{2}}{Q}^2\right))\hfill \\ \hfill & =conv\left(T(I_1+{\displaystyle \frac{1}{2}}{Q}^2)\right)\hfill \\ \hfill & =T(conv(I_1+{\displaystyle \frac{1}{2}}{Q}^2)).\hfill \end{array}$$

   □

Let $M_1$ and $M_2$ be two positive integers satisfying $M_1<M_2$. Let $H_0^{+}$ and $H_0^{-}$ be the closed halfplanes of ${\mathbb{R}}^2$ bounded by the line span $\left\{e_2\right\}$ and containing $(1,0)$ and $(-1,0)$, respectively.

Let K be a convex lattice set contained in $M_2{Q}^2$ and containing $M_1{Q}^2\cap {\mathbb{Z}}^2$. Then, both $K\cap H_0^{+}$ and $K\cap H_0^{-}$ are convex lattice sets. Moreover, $K\cap H_0^{-}$ can be viewed as $-(K^{\prime }\cap H_0^{+})$, where $K^{\prime }$ is another convex lattice set with the desired properties.

Based on these observations, we can construct $\epsilon$-nets for $({\mathcal{K}}^2,d_{BM})$ with the following steps.

Step 1. Enumerate all subsets G of ${\mathbb{Z}}^2$ that are convex in the 2nd coordinate and satisfy

$$M_1{Q}^2\cap {\mathbb{Z}}^2\cap H_0^{+}\subseteq G\subseteq M_2{Q}^2\cap H_0^{+}.$$

Let $i\in [0,M_2]\cap \mathbb{Z}$. Set

$$C_i=\left\{(i,j)\mid j\in [-M_2,M_2]\cap \mathbb{Z}\right\}.$$

Case 1. $i\in [M_1+1,M_2]$. In this case, the intersection of the vertical line passing through $(i,0)$ and the convex lattice set could be ∅, or a singleton, or a set of the form

$$[(i,{\beta }_1),(i,{\beta }_2)]\cap {\mathbb{Z}}^2,$$

with ${\beta }_1\ne {\beta }_2$. In this case, we set

$${\mathcal{S}}_i=\{\varnothing \}\cup \phi (C_i,1)\cup \phi (C_i,2),$$

where $\phi (A,i)$ stands for the collection of all subsets of A consisting of i elements.

Case 2. $i\in \left[0,M_1\right]$. In this case, the intersection of the vertical line passing through $(i,0)$ and the convex lattice set is a set of the form

$$[(i,{\beta }_1),(i,{\beta }_2)]\cap {\mathbb{Z}}^2,$$

where ${\beta }_1\le -M_1$ and ${\beta }_2\ge M_1$.

Hence, we set

$${\mathcal{S}}_i^1=\{\varnothing \}\cup \phi (\left\{(i,j)\mid M_1<j\le M_2,j\in \mathbb{Z}\right\},1)$$

and

$${\mathcal{S}}_i^2=\{\varnothing \}\cup \phi (\left\{(i,j)\mid -M_2\le j<-M_1,j\in \mathbb{Z}\right\},1).$$

We do not consider the situation $j=\pm M_1$, since we take the union of the resulting set with $M_1{Q}^2\cap {\mathbb{Z}}^2$ later.

The set is

$$\mathcal{P}=\left(\prod _{i=0}^{M_1}({\mathcal{S}}_i^1\times {\mathcal{S}}_i^2)\right)\times \prod _{i=M_1+1}^{M_2}{\mathcal{S}}_i.$$

For each member

$$P=(S_0^1,S_0^2,\dots ,S_{M_1}^1,S_{M_1}^2,S_{M_1+1},\dots ,S_{M_2})\in \mathcal{P},$$

we can recover a lattice set that is convex in the 2nd coordinate in the following way. $♯A$ denotes the cardinality of a set A.

Let $i\in [0,M_2]\cap \mathbb{Z}$.

Case 1. $i\in [M_1+1,M_2]$.

• If $♯S_i\le 1$, then we set $T_i=S_i$.
• If $S_i$ has the form $\{(i,{\beta }_1),(i,{\beta }_2)\}$, then we set $T_i=[(i,{\beta }_1),(i,{\beta }_2)]\cap {\mathbb{Z}}^2$.

Case 2. $i\in \left[0,M_1\right]$. If $S_i^1=\varnothing$, we set ${\beta }_1=M_1$; otherwise, let ${\beta }_1$ be the 2nd coordinate of the point in $S_i^1$. Similarly, if $S_i^2=⌀$, we set ${\beta }_2=-M_1$; otherwise, we let ${\beta }_2$ be the 2nd coordinate of the point in $S_i^2$. Let $T_i=[(i,{\beta }_1),(i,{\beta }_2)]\cap {\mathbb{Z}}^2$.

The set $T\left(P\right)={\bigcup }_{i\in [0,M_2]}{T}_i$ is a lattice set convex in the 2nd coordinate, see Algorithm 1.

Algorithm 1 Convert P to set $T\left(P\right)$.

Require:  $P\in \mathcal{P}$ Ensure:  $T\left(P\right)$, a lattice set convex in the 2nd coordinate generated by P.    1: Initialize a set $T\left(P\right)\leftarrow \varnothing$    2: for$i\in [0,M_1]$do    3:     $S_i^1\leftarrow P\left[2i\right]$ ▹$P\left[i\right]$ denotes the i-th elements of P    4:     $S_i^2\leftarrow P[2i+1]$    5:     if $S_i^1=\varnothing$ then    6:         $ub\leftarrow M_1$    7:     else    8:         $ub\leftarrow {\beta }_1$▹ Second coordinate of the point in $S_i^1$    9:     end if  10:     if $S_i^2=\varnothing$ then  11:         $lb\leftarrow -M_1$  12:     else  13:         $lb\leftarrow {\beta }_2$▹ Second coordinate of the point in $S_i^2$  14:     end if  15:     $T_i\leftarrow \left\{i\right\}\times [lb,ub]$  16:     $T\left(P\right)\leftarrow T\left(P\right)\cup T_i$  17: end for  18: for$i\in [M_1+1,M_2]$ do  19:     $S_i\leftarrow P[i+M_1+1]$  20:     if $♯S_i\le 1$ then  21:         $T\left(P\right)\leftarrow T\left(P\right)\cup S_i$  22:     else if $♯S_i=2$ then  23:         $lb\leftarrow min\left\{x\mid (i,x)\in P[i+M_1+1]\right\}$  24:         $ub\leftarrow max\left\{x\mid (i,x)\in P[i+M_1+1]\right\}$  25:         $T_i\leftarrow \left\{i\right\}\times [lb,ub]$  26:         $T\left(P\right)\leftarrow T\left(P\right)\cup T_i$  27:     end if  28: end for  29: return$T\left(P\right)$

Step 2. Obtain all convex lattice sets satisfying

$$M_1{Q}^2\cap {\mathbb{Z}}^2\subseteq G\subseteq M_2{Q}^2.$$

The case for ${\mathcal{K}}^2$: Let ${\mathcal{L}}_0$ and $\mathcal{L}$ be an empty list. For each $P\in \mathcal{P}$, if $T\left(P\right)$ is a convex lattice set, then we add $T\left(P\right)$ to ${\mathcal{L}}_0$; otherwise, we ignore P. Let $(-T_1,T_2)\in {\mathcal{L}}_0\times {\mathcal{L}}_0$. If $T_1\cup T_2$ is a convex lattice set, then we add $T_1\cup T_2$ to $\mathcal{L}$.

The case for ${\mathcal{C}}^2$: Let $\mathcal{L}$ be a new empty list. For each $P\in \mathcal{P}$, if $T\left(P\right)\cup (-T(P\left)\right)$ is a convex lattice set, we add $T\left(P\right)\cup (-T(P\left)\right)$ to $\mathcal{L}$; otherwise, we ignore P.

For each $P\in \mathcal{P}$, Algorithm 2 is used to check whether $T\left(P\right)$ is a lattice set convex in the 1st coordinate. Thereafter, Algorithm 3 is used to obtain all convex lattice sets in $T\left(\mathcal{P}\right)=\left\{T\left(P\right)\mid P\in \mathcal{P}\right\}$.

Algorithm 2 Checking convexity in the 1st coordinate for a lattice set.

Require: $T\left(P\right)\in T\left(\mathcal{P}\right)$    1:  for$i\in [-M_2,M_2]$do    2:     $Y_i\leftarrow \left\{x\mid (x,i)\in T\left(P\right)\right\}$    3:     if $♯Y_i\ge 2$ then    4:         $lb\leftarrow min\left(Y_i\right)$    5:         $ub\leftarrow max\left(Y_i\right)$    6:         if $♯Y_i<ub-lb+1$ then    7:            return False    8:         end if    9:     end if  10: end for  11: return True

Algorithm 3 Obtain all convex lattice sets in $T\left(\mathcal{P}\right)$.

Require: $\mathcal{P}$    1: ${\mathcal{L}}_0\leftarrow \varnothing$    2: for$P\in \mathcal{P}$do    3:     $S_0\leftarrow \mathrm{to}\_\mathrm{set}\left(P\right)$▹ Algorithm 1    4:     if is_x_convex$\left(S_0\right) \mathbf{then}$▹ Algorithm 2    5:         $S\leftarrow conv\left(S_0\right)$    6:         if $♯S_0=♯(S\cap {\mathbb{Z}}^2)$ then    7:            ${\mathcal{L}}_0\leftarrow {\mathcal{L}}_0\cup S$    8:            continue    9:         end if  10:     end if  11: end for  12: return${\mathcal{L}}_0$

Step 3. Let $\mathcal{L}$ be the list obtained in the last step. For each set $A\in \mathcal{L}$, remove every element $B\in \mathcal{L}\setminus \left\{A\right\}$ for which there exist $T\in G$ and $x\in {\mathbb{R}}^2$ such that

$$T\left(A\right)+x=B,$$

to obtain a new list ${\mathcal{L}}^{\prime }$, see Algorithm 4.

Algorithm 4 Removing redundant members in $\mathcal{L}$.

Require: $\mathcal{L},G$    1: ${\mathcal{L}}^{\prime }\leftarrow \varnothing$    2: $\mathcal{S}\leftarrow \varnothing$    3: for$A\in \mathcal{L}$ do    4:     Compute the centroid $c_A$ of $convA$    5:     $A^{\prime }=convA-c_A$    6:     if $A^{\prime }\in \mathcal{S}$ then    7:         continue    8:     end if    9:     ${\mathcal{L}}^{\prime }\leftarrow {\mathcal{L}}^{\prime }\cup \left\{A\right\}$  10:     Set $G\left(A^{\prime }\right)=\left\{T\left(A^{\prime }\right)\mid T\in G\right\}$  11:     $\mathcal{S}\leftarrow \mathcal{S}\cup G\left(A^{\prime }\right)$  12: end for  13: return$$

Step 4. Constructing the $\epsilon$-net.

• For each set $A\in {\mathcal{L}}^{\prime }$:

  ‒

  Generating the convex body $conv(A+(1/2)Q^2)$. Clearly, we only need to store the set of vertices of each convex polytope.

  ‒

  Temporarily store the results in a new list ${\mathcal{L}}_{\mathrm{temp}}$.

• Update the list ${\mathcal{L}}^{\prime }$ by replacing it with ${\mathcal{L}}_{\mathrm{temp}}$.

The updated list ${\mathcal{L}}^{\prime }$ is a desired $\epsilon$-net.

From [15] (Theorem A), if $d_{BM}^M\left(K,L\right)\le 1+\epsilon$ holds for some positive number $\epsilon$, then

$$|{\Gamma }_m\left(K\right)-{\Gamma }_m\left(L\right)|\le \epsilon .$$

This implies that, if one can construct an $\epsilon$-net $\mathcal{N}$ for $({\mathcal{K}}^n,d_{BM})$ with

$$\epsilon <1-sup\left\{{\Gamma }_{2^n}\left(L\right)\mid L\in \mathcal{N}\right\},$$

then,

$${\Gamma }_{2^n}\left(K\right)\le sup\left\{{\Gamma }_{2^n}\left(L\right)\mid L\in \mathcal{N}\right\}+\epsilon <1,\forall K\in {\mathcal{K}}^n.$$

In [21], M. Lassak proved that

$${\Gamma }_4\left(K\right)\le {\displaystyle \frac{\sqrt{2}}{2}}$$

holds for each planar convex body K, and this bound can be attained.

Therefore, to confirm Hadwiger’s covering conjecture for ${\mathcal{C}}^2$, we need to construct an ε-net of $({\mathcal{C}}^2,d_{BM})$ for some $\epsilon \in (0,1-\sqrt{2}/2)$.

If we take $M_2=6$ and $M_1=4$, then, from Corollary 1, the set of all convex lattice sets contained in $M_2{Q}^2$ and containing $M_1{Q}^2\cap {\mathbb{Z}}^2$ is a $(1/4)$-net for $({\mathcal{C}}^2,d_{BM})$. In the following, we construct such a net.

For $i\in \{5,6\}$, we have

$$\begin{array}{c}\hfill {\mathcal{S}}_i=\{\varnothing \}\cup \phi (\left\{(x,y)\mid x\in \{5,6\},y\in [-6,6]\cap \mathbb{Z}\right\},1)\\ \hfill \cup \phi (\left\{(x,y)\mid x\in \{5,6\},y\in [-6,6]\cap \mathbb{Z}\right\},2).\end{array}$$

For $i\in [0,4]\cap \mathbb{Z}$, we have

$${\mathcal{S}}_i^1=\{\varnothing \}\cup \phi (\{(i,5),(i,6)\},1)$$

and

$${\mathcal{S}}_i^2=\{\varnothing \}\cup \phi (\{(i,-5),(i,-6)\},1).$$

Since

$$♯{\mathcal{S}}_i^1=♯{\mathcal{S}}_i^2=3$$

and $♯{\mathcal{S}}_i=1+13+C_{13}^2=92$, we have

$$♯\mathcal{P}={(3\times 3)}^5\times {(1+13+C_{13}^2)}^2=499,790,736.$$

Thus, $♯T\left(\mathcal{P}\right)=499,790,736$. By Step 2, there are 434,849 members in $T\left(P\right)$ that are convex lattice sets. After applying Algorithm 4, there are 30,306 sets left. Finally, we obtain the convex hull of the Minkowski sum of these convex lattice sets and $(1/2)Q^2$ as in Step 4. The family of these convex lattice sets is a $(1/4)$-net for $({\mathcal{C}}^2,d_{BM})$.

In [4], Han et al. proposed an algorithm based on CUDA for estimating covering functionals of two-dimensional convex polytopes. On a computer equipped with an AMD Ryzen 9 3900X 12-core processor, and the NVIDIA A4000 graphics processor, and taking $B_2^2$ for example, this algorithm takes approximately 0.217 s to obtain an estimation of ${\Gamma }_4\left(B_2^2\right)$ that is no greater than $0.75$. Thus, we can complete the verification of the 30,306 two-dimensional convex bodies in this $(1/4)$-net for $({\mathcal{C}}^2,d_{BM})$ within approximately 2 h and confirm that ${\Gamma }_4\left(K\right)\le {\Gamma }_4(conv(G_M+(1/2)Q^2))+1/4<1$.

Remark 1.

An ε-net $\mathcal{N}$ for $({\mathcal{K}}^n,d_{BM})$ or $({\mathcal{C}}^n,d_{BM})$ is said to be reduced if no two members in $\mathcal{N}$ are affinely equivalent. Most likely, the $(1/4)$-net we obtained is not reduced. Clearly, for each member L of this $(1/4)$-net, $2L$ is the convex hull of a convex lattice set. There are several recent results for deciding whether two convex lattice sets are affinely equivalent; see, e.g., [22]. To the best of our knowledge, there is no very efficient algorithm for this purpose yet.