# The Honeycomb Conjecture in normed planes and an alpha-convex variant of a
theorem of Dowker

Zsolt Lángi  and  Shanshan Wang  Zsolt Lángi, Department of Algebra and
Geometry, Budapest University of Technology and Economics,
Műegyetem rkp. 3., H-1111 Budapest, Hungary, and
Alfréd Rényi Institute of Mathematics,
Reáltanoda utca 13-15, H-1053, Budapest, Hungary<EMAIL_ADDRESS>Shanshan
Wang, Department of Algebra and Geometry, Budapest University of Technology
and Economics,
Műegyetem rkp. 3., H-1111 Budapest, Hungary<EMAIL_ADDRESS>
###### Abstract.

The Honeycomb Conjecture states that among tilings with unit area cells in the
Euclidean plane, the average perimeter of a cell is minimal for a regular
hexagonal tiling. This conjecture was proved by L. Fejes Tóth for convex
tilings, and by Hales for not necessarily convex tilings. In this paper we
investigate the same question for tilings of a given normed plane, and show
that among normal, convex tilings in a normed plane, the average squared
perimeter of a cell is minimal for a tiling whose cells are translates of a
centrally symmetric hexagon. We also show that the question whether the same
statement is true for the average perimeter of a cell is closely related to an
$\alpha$-convex variant of a theorem of Dowker on the area of polygons
circumscribed about a convex disk. Exploring this connection we find families
of norms in which the average perimeter of a cell of a tiling is minimal for a
hexagonal tiling, and prove some additonal related results. Finally, we apply
our method to give a partial answer to a problem of Steinhaus about the
isoperimetric ratios of cells of certain tilings in the Euclidean plane,
appeared in an open problem book of Croft, Falconer and Guy.

###### Key words and phrases:

Honeycomb Conjecture, normed plane, isoperimetrix, L’huilier’s theorem, normal
tiling, convex tiling, Steinhaus’s problem

###### 2020 Mathematics Subject Classification:

52C20, 52A40, 52A38

Partially supported by the National Research, Development and Innovation
Office, NKFI, K-147544 grant, and BME Water Sciences & Disaster Prevention
TKP2020 Institution Excellence Subprogram, grant no. TKP2020 BME-IKA-VIZ

##  1\. Introduction

The aim of this paper is to investigate certain properties of mosaics. A
_mosaic_ or _tiling_ of the Euclidean $d$-space $\mathbb{R}^{d}$ is a
countable family of compact sets $\mathcal{T}$, called cells, with the
property that $\bigcup\mathcal{T}=\mathbb{R}^{d}$, and the interiors of any
two distinct cells are disjoint. Here, for convenience, it is usually assumed
that every cell has nonempty interior. A tiling is _convex_ , if every cell in
it is convex; it is known that in this case every cell is a convex polytope
[23, Theorem 1]. If, for a convex tiling $\mathcal{T}$, there are universal
constants $0<\hat{r}<\hat{R}$ such that every cell of $\mathcal{T}$ contains a
ball of radius $\hat{r}$ and is contained in a ball of radius $\hat{R}$, the
tiling is called _normal_. In this paper we deal only with normal, convex
tilings.

One of the classic problems of geometry, regarding planar mosaics, is the
Honeycomb Conjecture, which states, roughly speaking, that in a decomposition
of the Euclidean plane into cells of equal area, the average perimeter of the
cells is minimal for the regular hexagonal grid; i.e. a tiling in which every
cell is a unit area regular hexagon. This conjecture first appeared in Roman
time in a book of Varro about agriculture [26]. Despite its ancient origin, it
took a surprisingly long time to find a satisfactory proof for this conjecture
[18]: it was solved by L. Fejes Tóth for normal, convex mosaics [10] in the
1940s , while the most general version is due to Hales [15] in 2001, who
dropped the condition of convexity. Despite its solution, this problem still
seems to be interesting for mathematicians [3, 4], engineers [28] and even for
philosophers [19].

Clearly, it is a natural question to ask if a similar statement holds for
tilings in any normed plane. The aim of this paper to investigate this
problem. To do it, we recall a few well-known facts about normed planes. In
particular, we recall that every origin-symmetric convex disk $M$ (i.e. every
compact, convex set $M$ with nonempty interior and satisfying $M=-M$) is the
unit disk of a normed plane, and the unit disk of a normed plane is an origin-
symmetric convex disk. In the paper, for a normed plane $\mathcal{M}$, we
denote the unit disk of $\mathcal{M}$ by $M$, and the norm of a point
$p\in\mathcal{M}$ by $||p||_{M}$. For convenience, we imagine that
$\mathcal{M}$ also have an underlying Euclidean structure, and define the
Euclidean norm of $p$ by $||p||$.

We also note that any finite dimensional real normed space can be equipped
with a Haar measure, and that this measure is unique up to multiplication of
the standard Lebesgue measure by a scalar (cf. e.g. [25]). This scalar does
not play a role in our investigation and in the paper
$\operatorname{area}(\cdot)$ denotes $2$-dimensional Lebesgue measure.
Finally, as usual, we define the perimeter of any convex disk $K$ in
$\mathcal{M}$, also called the _$M$ -perimeter_ of $K$ and denoted by
$\operatorname{perim}_{M}(K)$, as the supremum of the sums of the $M$-lengths
of the sides of any convex polygon inscribed in $K$ (see e.g. [22]).

We ask the following question, where by a hexagonal tiling we mean a tiling
whose cells are translates of a given centrally symmetric hexagon.

######  Question 1.1.

Is it true that in any normal, convex tiling in any normed pane $\mathcal{M}$,
with unit area cells, the average perimeter of a cell, if it is exists, is
minimal for a hexagonal tiling?

Before stating our first result, we note that there are optimization problems
in discrete geometry whose solution for tilings in the Euclidean plane is a
regular hexagonal tiling, while in any normed plane it is a (not necessarily
regular) hexagonal tiling. As examples, we mention the _densest circle packing
problem_ (see [9, 11, 20, 21] for its solution in the Euclidean plane and in
any normed plane) as well as the _simultaneous packing and covering problem_
[29], which can be regarded as isoperimetric problems involving the area and
inradius of cells, and the circumradius and inradius of cells of a tiling,
respectively.

In the paper we need the following definition, where by $\mathbf{B}^{2}$ we
denote the Euclidean closed unit disk centered at the origin $o$, and
$\operatorname{card}(\cdot)$ denotes the cardinality of a set.

######  Definition 1.2.

Let $\mathcal{T}$ be a convex, normal tiling in the normed plane
$\mathcal{M}$. For any $R>0$, let $\mathcal{T}(R)=$ denote the family of cells
of $\mathcal{T}$ contained in $R\mathbf{B}^{2}$. Let $\alpha>0$. We define the
_lower average $\alpha$th powered perimeter_ of a cell of $\mathcal{T}$ as the
quantity

(1.1)
$\underline{P}_{\alpha}(\mathcal{T})=\liminf_{R\to\infty}\frac{\sum_{C\in\mathcal{T}(R)}\left(\operatorname{perim}_{M}(C)\right)^{\alpha}}{\operatorname{card}(\mathcal{T}(R))}.$

We define the _upper average $\alpha$th powered perimeter_ of a cell of
$\mathcal{T}$, denoted by $\overline{P}_{\alpha}(\mathcal{T})$, in the same
way, replacing $\liminf$ by $\limsup$. If
$\underline{P}_{\alpha}(\mathcal{T})=\overline{P}_{\alpha}(\mathcal{T})$, we
call this quantity the average $\alpha$th powered perimeter of a cell of
$\mathcal{T}$, and denote it by $P_{\alpha}(\mathcal{T})$. We define the
quantities
$\underline{P}_{\log}(\mathcal{T}),\overline{P}_{\log}(\mathcal{T})$ and
$P_{\log}(\mathcal{T})$ similarly, replacing
$\left(\operatorname{perim}_{M}(C)\right)^{\alpha}$ by
$\log\left(\operatorname{perim}_{M}(C)\right)$ in the above definitions.

We note that, as usual in convex geometry, we may denote the quantities
$\underline{P}_{\log}(\mathcal{T})$, $\overline{P}_{\log}(\mathcal{T})$ and
$P_{\log}(\mathcal{T})$ also by
$\underline{P}_{0}(\mathcal{T}),\overline{P}_{0}(\mathcal{T})$ and
$P_{0}(\mathcal{T})$, respectively. We also remark that whereas the above
procedure is the standard way to define the ‘average’ of a certain geometric
quantity of infinitely many objects in the plane, the Euclidean disk
$\mathbf{B}^{2}$ in the definition of $\mathcal{T}(R)$ is often replaced in
the literature by other figures, for instance squares or hexagons (see e.g.
[11, 12]).

######  Remark 1.3.

For any $\alpha,\beta\in[0,\infty)$ with $\alpha<\beta$ and any normal, convex
tiling in $\mathcal{M}$, we have
$\underline{P}_{\alpha}(\mathcal{T})\leq\underline{P}_{\beta}(\mathcal{T})$
and $\overline{P}_{\alpha}(\mathcal{T})\leq\overline{P}_{\beta}(\mathcal{T})$.
Furthermore, if $\mathcal{T}$ is a hexagonal tiling, we have equality in all
the previous inequalities.

Our first result is the following.

######  Theorem 1.4.

For any normed plane $\mathcal{M}$ there is a hexagonal tiling
$\mathcal{T}_{hex}$ of $\mathcal{M}$ such that for any convex, normal tiling
$\mathcal{T}$ of $\mathcal{M}$, we have

$\underline{P}_{2}(\mathcal{T})\geq P_{2}(\mathcal{T}_{hex}).$

We will see that the proof of Theorem 1.4 relies on the following classical
result of Dowker [8] on the area of convex polygons circumscribed about a
convex disk.

######  Theorem 1.5 (Dowker).

For any convex disk $K$ in $\mathbb{R}^{2}$, let

$A_{K}(n)=\inf\\{\operatorname{area}(P):P\hbox{ is a convex }n\hbox{-gon
circumscribed about }K\\}.$

Then the sequence $\\{A_{K}(n)\\}$ is convex. In other words, for any $n\geq
4$, we have

$A_{K}(n-1)+A_{K}(n+1)\geq 2A_{K}(n).$

Based on this result, we define two properties.

######  Definition 1.6.

Let $\alpha\in(0,\infty)$. We say that the normed plane $\mathcal{M}$
satisfies the _$\alpha$ -honeycomb property_, if there is a hexagonal tiling
$\mathcal{T}_{hex}$ of $\mathcal{M}$ such that for any convex, normal tiling
$\mathcal{T}$ of $\mathcal{M}$, we have

$\underline{P}_{\alpha}(\mathcal{T})\geq P_{\alpha}(\mathcal{T}_{hex}).$

Similarly, we say that it satisfies the _log-honeycomb_ (or _$0$ -honeycomb_)
_property_ if the same holds for the lower average log-perimeter of a cell of
$\mathcal{T}$.

Observe that by Theorem 1.4, any norm with unit disk $M$ satisfies the
$2$-honeycomb property.

######  Definition 1.7.

Let $\alpha\in(0,\infty)$. We say that a convex disk $K$ satisfies the
_$\alpha$ -Dowker property_ if the sequence $\\{A_{K}^{\alpha}(n)\\}$ is
convex. Furthermore, we say that it satisfies the _log-Dowker_ (or _$0$
-Dowker_) _property_ if the sequence $\\{\log A_{K}(n)\\}$ is convex.

In our proof we only need the following, weaker variant of the $\alpha$-Dowker
property.

######  Definition 1.8.

Let $\alpha\in(0,\infty)$. We say that a convex disk $K$ satisfies the _weak
$\alpha$-Dowker property_ if

(1.2) $\frac{n-6}{n-m}A_{K}^{\alpha}(m)+\frac{6-m}{n-m}A_{K}^{\alpha}(n)\geq
2A_{K}^{\alpha}(6)$

holds for any $3\leq m<6<n$. Similarly, we say that $K$ satisfies the _weak
log-Dowker_ (or _weak $0$-Dowker_) _property_ if

(1.3) $\frac{n-6}{n-m}\log A_{K}^{(}m)+\frac{6-m}{n-m}\log A_{K}(n)\geq 2\log
A_{K}(6)$

holds for any $3\leq m<6<n$.

For brevity, if $\alpha=1$ in any of the above definitions, we may omit it
from the notation. In our next result, we recall the notion of _isoperimetrix_
of a normed plane $\mathcal{M}$ with unit disk $M$, defined as the rotated
copy of the Euclidean polar of $M$, about the origin $o$, by $\frac{\pi}{2}$.
We denote the isoperimetrix of $\mathcal{M}$ by $M_{iso}$.

######  Theorem 1.9.

Let $M$ be an $o$-symmetric convex disk. For any $\alpha\in[0,\infty)$, if the
isoperimetrix $M_{iso}$ of $\mathcal{M}$ satisfies the weak $\alpha$-Dowker
property, then the normed plane $\mathcal{M}$ satisfies the
$(2\alpha)$-honeycomb property.

It is an elementary exercise to check
$A_{\mathbf{B}^{2}}(n)=n\tan\frac{\pi}{n}$, implying that $\mathbf{B}^{2}$
satisfies the log-Dowker property. Thus, Theorem 1.9 immediately yields
Corollary 1.10.

######  Corollary 1.10.

The Euclidean plane satisfies the log-honeycomb property.

The structure of the paper is as follows.

In Section 2, we prove Theorems 1.4 and 1.9. In Sections 3 and 4 we
investigate the $\alpha$-Dowker and $\alpha$-convex properties of convex disks
and normed planes, respectively. More specifically, in Section 3 we consider
convex polygons and polygonal norms, while in Section 4 we examine properties
of convex disks and norms which are not necessarily polygonal. In Section 5 we
consider a problem of Steinhaus, appeared in the problem book [7] of Croft,
Falconer and Guy, asking if it is true that among tilings of the Euclidean
plane with tiles whose diameters are at least some universal constant $D>0$,
the maximum isoperimetric ratio of the cells is minimal for a regular
hexagonal tiling. Using the approach of the proof of Theorem 1.4, we show that
for convex, normal tilings, a stronger statement holds in any normed plane,
and also investigate possible generalizations of our method for this problem.
We finish the paper with an additional remark in Section 6.

In our investigation, we denote by $\int(X)$, $\operatorname{bd}(X)$ and
$\operatorname{conv}(X)$ the _interior, boundary_ and the _convex hull_ of a
set $X$, and for any two points $p,q$, we denote by $[p,q]$ the closed
straight line segment connecting them. Finally, we denote the Euclidean unit
circle $\operatorname{bd}\mathbf{B}^{2}$ by $\mathbb{S}^{1}$.

##  2\. Proof of Theorems 1.4 and 1.9

First, we observe that Theorem 1.9, combined with Theorem 1.5 readily implies
Theorem 1.4. Thus, we need only prove Theorem 1.9. Throughout this section, we
deal with a given convex, normal tiling $\mathcal{T}$ in a normed plane
$\mathcal{M}$. We permit the angles of cells to be straight angles; due to
this, we can regard $\mathcal{T}$ as an _edge-to-edge_ tiling, in which every
edge of every cell belongs to exactly one more cell. During our investigation,
$\mathcal{T}(R)$ denotes the family of cells of $\mathcal{T}$ contained in
$R\mathbf{B}^{2}$, and for every $C\in\mathcal{T}$, $v(C)$ denotes the number
of sides of $C$. We start with collecting some elementary observation and
known facts about tilings and normed planes.

######  Lemma 2.1.

Let $0<\hat{r}<\hat{R}$ be given such that every cell of $\mathcal{T}$
contains a Euclidean disk of radius $\hat{r}$ and is contained in a Euclidean
disk of radius $\hat{R}$. Then every cell $C\in\mathcal{T}$ has at most
$\frac{9\hat{R}^{2}}{\hat{r}^{2}}-1$ neighbors.

###### Proof.

For every cell $C\in\mathcal{T}$, there is a Euclidean disk of radius $3R$
that contains $C$ as well as all its neighbors. The area of every cell is at
least $r^{2}\pi$. Thus, the assertion follows from a simple area estimate. ∎

######  Lemma 2.2.

Let $\mathcal{T}_{bd}(R)$ denote the family of cells of $\mathcal{F}$ that
intersect the circle $R\mathbb{S}^{1}$. Then
$\operatorname{card}(\mathcal{T}_{bd}(R))=\Theta(R)$.

###### Proof.

Since the diameter of every cell $C\in\mathcal{T}$ is at most $2\hat{R}$,
every cell is contained in the closure of the set
$(R+2\hat{R})\mathbf{B}^{2}\setminus(R-2\hat{R})\mathbf{B}^{2}$. Thus,
$\operatorname{card}(\mathcal{T}_{bd}(R))\leq\frac{(R+2\hat{R})^{2}\pi-(R-2\hat{R})^{2}\pi}{\hat{r}^{2}\pi}=8\hat{R}R$,
implying that $\operatorname{card}(\mathcal{T}_{bd}(R))=\mathcal{O}(R)$. On
the other hand, the union of these cells covers $R\mathbb{S}^{2}$, and the
diameter of each cell is at most $2\hat{R}$. Thus,
$\operatorname{card}(\mathcal{T}_{bd}(R))\geq\frac{2\pi}{2\arcsin\frac{\hat{R}}{R}}$,
implying that $\operatorname{card}(\mathcal{T}_{bd}(R))=\Omega(R)$. ∎

######  Definition 2.3.

We define the _lower average number of sides_ of a cell of $\mathcal{T}$ as

$\underline{v}(\mathcal{T})=\liminf_{R\to\infty}\frac{\sum_{C\in\mathcal{T}(R)}v(C)}{\operatorname{card}(\mathcal{T}(R))}.$

We define the _upper average number of sides_ $\overline{v}(\mathcal{T})$ of a
cell of $\mathcal{T}$ in the same way, replacing $\liminf$ by $\limsup$. If
$\underline{v}(\mathcal{T})=\overline{v}(\mathcal{T})$, we call this quantity
the _average number of sides_ of a cell of $\mathcal{T}$.

######  Lemma 2.4.

For any $R>2\bar{R}$, set
$v_{R}(\mathcal{T})=\frac{\sum_{C\in\mathcal{T}(R)}v(C)}{\operatorname{card}(\mathcal{T}(R))}$.
Then, $v_{R}(\mathcal{T})\leq 6+\mathcal{O}\left(\frac{1}{R}\right)$, and
$\overline{v}(\mathcal{T})\leq 6$.

###### Proof.

Consider the graph $G$ whose vertices are the vertices of the cells of
$\mathcal{T}(R)$, and two vertices are connected by an edge if they are the
endpoints of an edge of a cell in $\mathcal{T}(R)$. Then $G$ is a planar
graph. Let $V(G)$, $E(G)$ and $F(G)$ denote the numbers of vertices, edges and
faces of $G$. Furthermore, let $\mathcal{E}_{bd}(G)$ denote the family of the
edges of $G$ in the boundary of $\bigcup\mathcal{T}(R)$. We note that every
edge in $\mathcal{E}_{bd}(G)$ belongs to a cell of $\mathcal{T}$ intersecting
$R\mathbb{S}^{1}$, and thus, it follows from Lemmas 2.1 and 2.2 that
$E_{bd}(G):=\operatorname{card}\operatorname{card}(\mathcal{E}_{bd}(G))=\Theta(R)$.
These lemmas, an area estimate and the observation that
$\bigcup\mathcal{T}(R)$ contains $(R-2\bar{R})\mathbf{B}^{2}$ and is contained
in $(R+2\bar{R})\mathbf{B}^{2}$ shows that $V(G)$, $E(G)$ and $F(G)$ are all
of order of magnitude $\Theta(R^{2})$.

We observe that there might be vertices of $G$ whose degree is less than $3$.
If $p$ is such a vertex, then its degree is $2$, it lies in
$\operatorname{bd}\left(\bigcup\mathcal{T}(R)\right)$, and it belongs to two
edges in $\mathcal{E}_{bd}(G)$. Thus, the number of such edges is at most
$E_{bd}(G)=\mathcal{O}(R)$. We apply series reduction on these edges, and
obtain another planar graph $G^{\prime}$ in which the degree of every vertex
is at least $3$. If $V(G^{\prime}),E(G^{\prime})$ and $F(G^{\prime})$ denotes
that numbers of vertices, edges and faces of $G^{\prime}$, then by the
construction of $G^{\prime}$ we have $V(G^{\prime})\leq V(G)$ and
$E(G^{\prime})\leq E(G)$ with
$V(G)-V(G^{\prime})=E(G)-E(G^{\prime})=\mathcal{O}(R)$, and
$F(G^{\prime})=F(G)$.

Applying Euler’s formula for $G^{\prime}$ and using the fact that the degree
of every vertex of $G^{\prime}$ is $3$ yields that $E(G^{\prime})\leq
3F(G^{\prime})-6$, implying that $E(G)\leq 3F(G)+\mathcal{O}(R)$. On the other
hand, $v_{R}(\mathcal{T})\leq\frac{2E(G)}{F(G)}$, and as $E(G)$ and $F(G)$ are
of $\Theta(R^{2})$, it follows that $v_{R}(\mathcal{T})\leq
6+\mathcal{O}\left(\frac{1}{R}\right)$, which readily yields the assertion. ∎

The original isoperimetric inequality, observed by Zenodorus in ancient
Greece, states that among regions of a given perimeter in the Euclidean plane,
Euclidean disks have maximum area. This result was generalized for any normed
plane by Busemann [5] in the following form.

######  Theorem 2.5 (Busemann).

Let $\mathcal{M}$ be a normed plane. The area enclosed by a simple, closed
curve $\Gamma$ of a given $M$-length is maximized if $\Gamma$ is the boundary
of a plane convex body $K$ homothetic to the so-called _isoperimetrix_
$M_{iso}$ of $\mathcal{M}$, obtained as the polar of the rotated copy of the
unit disk $M$ of $\mathcal{M}$ by $\frac{\pi}{2}$ (see Figure 1).

Figure 1. The isoperimetrix $M_{iso}$ of a norm with unit disk $M$. The dotted
circle is the Euclidean unit disk $\mathbf{B}^{2}$ centered at $o$. The left-
hand side panel shows $M$ and its polar $M^{\circ}$, the isoperimetrix in the
righ-hand side panel is a rotated copy of $M^{\circ}$ rotated by
$\frac{\pi}{2}$.

We note that by Theorem 2.5, the isoperimetrix of the Euclidean plane is a
Euclidean circular disk.

A variant of this inequality, in Euclidean plane, states that among convex
polygons with given perimeter and fixed directions of its sides, the ones
circumscribed about a circular disk have maximal area. This statement is
called L’huilier’s inequality, which is generalized for arbitrary normed
planes by Chakerian [6] in the following way.

######  Theorem 2.6 (Chakerian).

Let $\mathcal{M}$ be a normed plane with unit disk $M$. Let the isoperimetrix
of the plane be $M_{iso}$. Let $K$ be an arbitrary convex $n$-gon in
$\mathcal{M}$, and let $K^{*}$ be the convex $n$-gon circumscribed about
$M_{iso}$ whose sides have the same outer unit normals as the sides of $K$.
Let the $M$-perimeter of $K$ be $L$, the area of $K$ be $F$, and the area of
$K^{*}$ be $f$. Then

$L^{2}-4fF\geq 0,$

with equality if and only if $K$ is homothetic to $K^{*}$.

Now we are ready to prove Theorem 1.9. We simplicity, we assume that $M_{iso}$
satisfies the $\alpha$-Dowker property for some $\alpha\in(0,\infty)$; for the
case $\alpha=0$ a straightforward modification of our argument can be applied.
From now on we also assume that every cell of $\mathcal{T}$ has unit area. We
need to consider the following quantity.

$\underline{P}_{2\alpha}(\mathcal{T})=\liminf_{R\to\infty}\frac{\sum_{C\in\mathcal{T}(R)}\left(\operatorname{perim}_{M}(C)\right)^{\alpha}}{\operatorname{card}(\mathcal{T}(R))}.$

For any $C\in\mathcal{T}(R)$, let $C^{*}$ denote the convex polygon
circumscribed about $M_{iso}$ such that the sides of $C$ and $C^{*}$ have the
same outer unit normals, and let $v(C)$ denote the number of sides of $C$.
Then, by Theorem 2.6, and since $\operatorname{area}(C)=1$ for any
$C\in\mathcal{T}(R)$, we have

$\underline{P}_{2\alpha}(\mathcal{T})\geq
4\liminf_{R\to\infty}\frac{\sum_{C\in\mathcal{T}(R)}\left(\operatorname{area}(C^{*})\right)^{\alpha}}{\operatorname{card}\mathcal{T}(R)}\geq
4\lim_{R\to\infty}\frac{\sum_{C\in\mathcal{T}(R)}\left(A_{M_{iso}}(v(C))\right)^{\alpha}}{\operatorname{card}\mathcal{T}(R)}.$

where $A_{M_{iso}}(v(C))$ denotes the area of a minimum area polygon with
$v(C)$ sides, circumscribed about $M_{iso}$ (see Theorem 1.5). By our
assumptions, the sequence $\\{\left(A_{M_{iso}}(n)\right)^{\alpha}\\}$ is
convex, and by Lemma 2.4 $\overline{v}(\mathcal{T})\leq 6$. Thus,

$\underline{P}_{2\alpha}(\mathcal{T})\geq
4\left(A_{M_{iso}}(6)\right)^{\alpha}.$

Now the only thing we need to show that for some hexagonal tling
$\mathcal{T}_{hex}$ of $\mathcal{M}$, we have
$P_{2\alpha}(\mathcal{T}_{hex})=4\left(A_{M_{iso}}(6)\right)^{\alpha}$. To do
it we recall another theorem of Dowker from [8].

######  Theorem 2.7 (Dowker).

Let $K$ by an $o$-symmetric plane convex body. Then, for every $m\geq 2$,
there is a centrally symmetric convex $(2m)$-gon $P$ circumscribed about $K$
with $\operatorname{area}(P)=A_{K}(2m)$.

By Theorem 2.7, there is an $o$-symmetric hexagon $H$ circumscribed about
$M_{iso}$ of area $A_{M_{iso}}(6)$. Let $\mathcal{T}_{hex}$ be a tiling whose
cells are translates of $H$. Then the desired equality follows from the
equality part of Theorem 2.6.

##  3\. Results about polygonal norms

The goal of this section to investigate the $\alpha$-honeycomb properties of
poygonal norms, and the $\alpha$-Dowker properties of polygons.

###  3.1. An algorithm to check the $\alpha$-honeycomb property of a polygonal
norm

In this subsection we give a simple algorithm that check if the isoperimetrix
of a polygonal norm satisfies the $\alpha$-Dowker, or weak $\alpha$-Dowker
property for some given $\alpha\in[0,\infty)$. We assume that $\alpha>0$, and
note that our argument can be modified in a natural way for the log-Dowker or
weak log-Dowker property.

######  Remark 3.1.

If the unit disk $M$ of a norm is a $(2k)$-gon, then $M_{iso}$ is also a
$(2k)$-gon, and for any $n\geq 2k$, we clearly have
$A_{M}(n)=\operatorname{area}(M_{iso})$. Thus, to check if $M^{*}$ satisfies
the $\alpha$-Dowker property, it is sufficient to check if the inequality

(3.1) $A_{M_{iso}}^{\alpha}(n-1)+A_{M_{iso}}^{\alpha}(n+1)\geq
2A_{M_{iso}}^{\alpha}(n)$

for values $4\leq n\leq 2k$. Similarly, to check if $M_{iso}$ satisfies the
weak $\alpha$-Dowker property, it is enough to check if the inequality in
(1.2) is satisfied for $m=3,4,5$ and $6<n\leq 2k$.

We also recall that there is an algorithm, given by Aggarwal, Chang and Yap
[1] that computes, for any given convex polygon $P$ with $s$ sides, the area
of a minimum area convex $t$-gon for any given $3\leq t<s$, in $O(s^{2}\log
s\log t)$ steps.

Thus, given an $o$-symmetric $(2k)$-gon $M$, we can provide the following
algorithm to check if, say, the weak $\alpha$-Dowker property is satisfied for
$M_{iso}$.

  1. Step 1

: We compute $M_{iso}$. Since $M_{iso}$ is the rotated copy of the polar of
$M$ by $\frac{\pi}{2}$, and since the polar of an $o$-symmetric convex
$(2k)$-gon can be computed in $\mathcal{O}(k)$ steps, this step requires
$\mathcal{O}(k)$ steps.

  2. Step 2

: Using the algorithm in [1], we compute the quantities $A_{M{iso}}(n)$ for
all $3\leq n\leq 2k$. This step requires $\mathcal{O}(k^{3}\log^{2}k)$ steps.

  3. Step 3

: We can check inequality in (1.2) for $m=3,4,5$ and all $6<n\leq 2k$, in
$\mathcal{O}(k)$ steps.

To check the $\alpha$-Dowker property of $M_{iso}$, in Step 3 we check the
inequality in (3.1) in $\mathcal{O}(k)$ steps.

###  3.2. Norms where the unit disk is a regular polygon

In this subsection we investigate which normed planes, with an $o$-symmetric
regular polygon as their unit disk, satisfy the honeycomb property. Our main
result is as follows.

######  Theorem 3.2.

If the unit disk of a normed plane $\mathcal{M}$ is a regular $(2k)$-gon with
$k\neq 4,5,7$, then $\mathcal{M}$ satisfies the honeycomb property.

Note that if the unit disk $M$ of $\mathcal{M}$ is a regular $(2k)$-gon,
then$M_{iso}$ is a also a regular $(2k)$-gon. Thus, Theorem 3.2 readily
follows from combining Theorem 3.3 and Theorem 1.9.

######  Theorem 3.3.

A regular $(2k)$-gon $P_{k}$, with $k\geq 2$, satisfies the weak
$\frac{1}{2}$-Dowker property if and only if $k\neq 4,5,7$.

######  Remark 3.4.

During the proof we show (see Lemma 3.7) that if $k\geq 4$, then
$A_{2k-2}(P_{k})$ is a convex combination of $A_{2k-1}(P_{k})$ and
$A_{2k-3}(P_{k})$, implying that in this case $P_{k}$ does not satisfy the
$\alpha$-Dowker property for any $\alpha<1$.

To prove Theorem 3.3, we first find the values of $A_{n}(P_{k})$ of the
regular $(2k)$-gon $P_{k}$, with $k\geq 2$ and $3\leq n\leq 2k$. We note also
that for any $n\geq 2k$, we have $A_{n}(P_{k})=\operatorname{area}(P_{k})$. In
the following we assume that the regular polygon $P_{k}$ is circumscribed
about the unit disk $\mathbf{B}^{2}$. We denote the quantity $A_{n}(P_{k})$ by
$A(k,n)$. For our investigation, we need Lemma 3.5, which is a slightly
stronger version of Lemma 1 from [1].

######  Lemma 3.5.

Let $3\leq n\leq m$. Let $P$ be a convex $m$-gon, and let $Q$ denote a minimum
area convex $n$-gon circumscribed about $P$. Then the midpoint of every side
of $Q$ belongs to $P$. Furthermore, if $n\geq 4$, then there is a minimum area
convex $m$-gon $Q^{\prime}$ such that there is at most one side of
$Q^{\prime}$ that does not contain a side of $P_{n}$, and if $S$ is such a
side of $Q^{\prime}$, then the sum of the angles of $Q^{\prime}$ on $S$ is
strictly less than $\pi$.

###### Proof.

Clearly, every side of $Q$ intersects $P$. Suppose that there is a side $S$ of
$Q$ such that the midpoint of $S$ does not belong to $P$. Let $q$ be the point
of $P\cap S$ closest to the midpoint of $S$. Then slightly rotating the
sideline of $S$ around $q$ in a suitable direction yields a circumscribed
polygon whose area is strictly less than $A_{n}(P)$, a contradiction.

Now, let $Q^{\prime}$ be a minimum area convex $m$-gon circumscribed about $P$
with the property that it has the smallest number of sides that contain only a
vertex of $P$. Let $S$ be such a side of $Q^{\prime}$. Then, by the previous
paragraph, the vertex $p$ of $P$ on $S$ is the midpoint of $S$. Let $\alpha$
and $\beta$ denote the two angles of $Q^{\prime}$ on $S$. If
$\alpha+\beta>\pi$, then slightly rotating the sideline of $Q^{\prime}$
through $S$ about $p$ decreases the area of $Q^{\prime}$, contradicting the
assumption that $Q^{\prime}$ is a minimum area circumscribed $m$-gon. If
$\alpha+\beta=\pi$, then the same rotation does not change the area of
$Q^{\prime}$, and thus, rotating this line until it first reaches another
vertex of $S$ we obtain a minimum area circumscribed polygon having strictly
less sides that contain only a vertex of $P$. Thus, we have
$\alpha+\beta<\pi$. But any convex $n$-gon with $n\geq 4$ has at most one side
satisfying this property. ∎

######  Remark 3.6.

We note that if $P$ is $o$-symmetric and $n$ is even, then the method of the
proof of Lemma 3.5, combined with Theorem 2.7, yields that there is a minimum
area convex $m$-gon $Q$ circumscribed about $P$ such that $Q$ is
$o$-symmetric, and every side of $Q$ contains a side of $P$.

The main lemma to prove Theorem 3.3 is Lemma 3.7.

######  Lemma 3.7.

Let $k\geq 2$ and $3\leq n\leq 2k$ be positive integers, and let $2k=nq+r$,
where $q,r$ are nonnegative integers, and $r<n$. If $m\geq 4$, then

(3.2) $A(k,n)=(n-r)\tan\frac{\pi q}{2k}+r\tan\frac{\pi(q+1)}{2k}.$

Furthermore,

(3.3) $A(k,3)=\begin{cases}3\tan\frac{\pi}{3},&\text{if}~{}2k=3q,\\\
\frac{2}{\cos^{2}\left(\frac{\pi}{2k}\right)}\left(2\sin\left(\frac{q\pi}{k}\right)+\sin\left(\frac{(q+1)\pi}{k}\right)\right),&\text{if}~{}n=3q+1,\\\
\frac{2}{\cos^{2}\left(\frac{\pi}{2k}\right)}\left(\sin\left(\frac{q\pi}{k}\right)+2\sin\left(\frac{(q+1)\pi}{k}\right)\right),&\text{if}~{}n=3q+2.\end{cases}$

###### Proof.

Let $Q$ be a minimum area $m$-gon circumscribed about $P_{k}$ with the
smallest number of sides that contain only a vertex of $P_{k}$. By Lemma 3.5
and Remark 3.6, we can assume that if $m$ is even, then every side of $Q$
contains a side of $P$, and otherwise at most one side of $Q$ contains only a
vertex of $P$. Let the sides of $P_{k}$ be denoted as
$S_{1},S_{2},\ldots,S_{2k}$ in counterclockwise order in
$\operatorname{bd}(P_{k})$.

We distinguish two cases.

_Case 1_ , $n\geq 4$.

Set $B(k,n)=(n-r)\tan\frac{\pi q}{2k}+r\tan\frac{\pi(q+1)}{2k}$. We intend to
show that $A(k,n)=B(k,n)$ for every choice of $k,n$.

_Subcase 1.1_ : every side of $Q$ contains a side of $P_{k}$. Without loss of
generality, we may assume that the indices of the sides of $P_{k}$ contained
in a side of $Q$ are of the form
$s_{1},s_{1}+s_{2},\ldots,s_{1}+k_{s}+\ldots+s_{n}=2k$ for some positive
integers $s_{i}$, where $s_{i}<k$ for every value of $i$. Since $P_{k}$ is
circumscribed about $\mathbf{B}^{2}$, we have

$\operatorname{area}(Q)=\sum_{i=1}^{n}\tan\frac{s_{i}\pi}{2k}.$

Note that as the function $x\mapsto\tan x$ is strictly convex on the domain
$\left(0,\frac{\pi}{2}\right)$, it follows that $\operatorname{area}(Q)\geq
B(k,n)$, and equality is attained with a suitable choice of the $s_{i}$.

_Subcase 1.2_ , there is a unique side of $Q$ that contains only a vertex of
$P_{k}$. Observe that by our assumptions, in this case $n\geq 5$ is odd. From
this, by Lemma 3.5, it follows that $k\geq n\geq 5$. Let the sides of $Q$ be
denoted by $E_{1},E_{2},\ldots,E_{n}$ in counterclockwise order. For
$i=2,\ldots,n$, let $2\alpha_{i}$ denote the turning angle of $Q$ at the
common vertex of $E_{i-1}$ and $E_{i}$, and let $2\alpha_{1}$ denote the
turning angle at the common vertex of $E_{n}$ and $E_{1}$. Without loss of
generality, let $E_{1}$ be the side of $Q$ that contains only a vertex of
$P_{k}$. Then, for any $i$ with $3\leq i\leq m$, $\alpha_{i}$ is an integer
multiple of $\frac{\pi}{2k}$, and $\sum_{i=1}^{n}\alpha_{i}=\pi$. Since
$P_{k}$ is circumscribed about $\mathbf{B}^{2}$, the distance of the line of
$E_{1}$ from $o$ is strictly greater than $1$. Let us translate the line of
$E_{1}$ towards $o$ until its distance is $1$, and let $Q^{\prime}$ denote the
convex $m$-gon obtained in this way (see Figure 2). Then

$\operatorname{area}(Q)>\operatorname{area}(Q^{\prime})=\sum_{i=_{1}}^{n}\tan\alpha_{i}.$

By Lemma 3.5, we have that $\alpha_{1}+\alpha_{2}>\frac{\pi}{2}$. Since this
quantity is an integer multiple of $\frac{\pi}{2k}$, it follows that
$\alpha_{1}+\alpha_{2}\geq\frac{\pi}{2}+\frac{s\pi}{2k}$ for some positive
integer $s$. Without loss of generality, we assume that
$\alpha_{1}\leq\alpha_{2}$.

Figure 2. An illustration for Case 2 of the proof of Lemma 3.7 with $n=k=5$.

We intend to construct a convex polygon circumscribed about $P_{k}$ with area
smaller than $\operatorname{area}(Q^{\prime})$. To do it, we assume that
$s=1$; for the case $s\geq 2$ a straightforward modification of our argument
can be given. First, note that as $E_{1}$ contains no side of $P_{n}$, we have
that $\alpha_{2}-\alpha_{1}\neq\frac{\pi}{2k}$. Consider the case that
$\alpha_{2}-\alpha_{1}>\frac{\pi}{2k}$, then there are integer multiples
$\alpha_{2}^{\prime}\geq\alpha_{1}^{\prime}$ of $\frac{\pi}{2k}$ such that
$\alpha_{1}^{\prime}+\alpha_{2}^{\prime}=\alpha_{1}+\alpha_{2}$. Then, by the
convexity of the $x\mapsto\tan x$ function the expression
$\tan\alpha_{1}^{\prime}+\tan\alpha_{2}^{\prime}+\sum_{i=_{3}}^{n}\tan\alpha_{i}$
is the area of a convex $n$-gon $Q^{\prime\prime}$ circumscribed about
$P_{k}$, and
$\operatorname{area}(Q^{\prime\prime})<\operatorname{area}(Q^{\prime})<\operatorname{area}(Q)$,
contradicting our assumption the minimality of the area of $Q$.

Assume that $\alpha_{2}-\alpha_{1}<\frac{\pi}{2k}$. Then
$\frac{\pi}{4}<\alpha_{1}\leq\frac{\pi}{4}+\frac{\pi}{4k}\leq\alpha_{2}<\frac{\pi}{4}+\frac{\pi}{2k}$.
Since $\sum_{i=_{3}}^{n}\tan\alpha_{i}=\frac{\pi}{2}-\frac{\pi}{2k}$ and
$5\leq n\leq k$, for at least one value of $i$ we have

$\alpha_{i}\geq\frac{1}{n-2}\left(\frac{\pi}{2}-\frac{\pi}{2k}\right)=\frac{(k-1)\pi}{2k(n-2)}>\frac{\pi}{2k}.$

Without loss of generality, let $\alpha_{3}>\frac{\pi}{2k}$. Since
$\alpha_{3}$ is an integer multiple of $\frac{\pi}{2k}$, this yields that
$\alpha_{3}\geq\frac{\pi}{k}>\alpha_{1}$. Set
$\alpha_{2}^{\prime}=\alpha_{2}+\frac{\pi}{2k}$,
$\alpha_{3}^{\prime}=\alpha_{3}-\frac{\pi}{2k}$. Then, by the convexity of the
${}^{x}\mapsto\tan x$ function, we have that
$\sum_{i=1}^{n}\tan\alpha_{i}>\alpha_{1}+\alpha_{2}^{\prime}+\alpha_{3}^{\prime}+\sum_{i=3}^{n}\tan\alpha_{i}$.
Here $\alpha_{2}^{\prime}-\alpha_{1}>\frac{\pi}{2k}$, and applying the
argument in the previous case obtain a convex $n$-gon $Q^{\prime\prime}$
circumscribed about $P_{k}$ such that
$\operatorname{area}(Q^{\prime\prime})<\operatorname{area}(Q^{\prime})$, a
contradiction.

This shows that $A(k,n)=B(k,n)$ for any $k\geq 2$ and $4\leq n\leq 2k$.

_Case 2_ , $n=3$.

Let $Q$ be a minimum area triangle circumscribed about $P_{k}$. By Lemma 3.5,
the midpoint of every side of $Q$ belongs to $P$. We denote the vertices of
$P_{k}$ by $p_{1},p_{2},\ldots,p_{2k}$ in counterclockwise direction.
Depending on the number of sides of $Q$ that contain only a vertex of $P_{k}$,
we distinguish four subcases.

_Subcase 2.1_ , every side of $Q$ contains only a vertex of $P_{k}$. Let these
vertices of $P_{k}$ be $p_{i},p_{j},p_{l}$, where we can assume that $1\leq
i<j<l\leq 2k$. Since these vertices are the midpoints of the corresponding
sides of $Q$, $\operatorname{conv}\\{p_{i},p_{j},p_{k}\\}$ is the midtriangle
of $Q$ and hence, the sides of this triangle are parallel to the sides of $Q$.
Since $Q$ is circumscribed about $P_{k}$, it follows that $j-i,k-j$ and
$i+n-k$ differ by at most one. Thus, by elementary computation we obtain that
if $Q$ is a minimal area triangle circumscribed about $P_{k}$, and
$\operatorname{bd}(Q)$ contains no side of $T$, then, with the notation
$2k=3q+r$, where $q,r$ are nonnegative integers and $r\leq 2$,

(3.4)
$\operatorname{area}(Q)=\frac{2}{\cos^{2}\left(\frac{\pi}{2k}\right)}\left((3-r)\sin\left(\frac{q\pi}{k}\right)+r\sin\left(\frac{(q+1)\pi}{k}\right)\right).$

_Subcase 2.2_ , exactly one side of $Q$ contains a side of $P_{k}$. Let the
sides of $Q_{3}$ be $E_{1},E_{2},E_{3}$, and assume that $[p_{n},p_{1}]\subset
E_{1}$, and $p_{i}\in E_{2}$ and $p_{j}\in E_{3}$. Since the latter two
vertices are midpoints of $E_{2}$ and $E_{3}$, respectively, we have that
$[p_{i},p_{j}]$ is parallel to $[p_{n},p_{1}]$, implying that $j=n+1-i$. Let
$q$ denote the midpoint of $E_{1}$, and recall that $q\in[p_{n},p_{1}]$. Note
that $\operatorname{conv}\\{q,p_{i},p_{j}\\}$ is the midtriangle of $Q$, and
thus, the area of $Q$ does not depend on the position of $q$. This yields that
if $\operatorname{conv}\\{q,p_{i},p_{j}\\}$ is the midtriangle of a triangle
circumscribed about $P_{k}$, then so is
$\operatorname{conv}\\{p_{1},p_{i},p_{j}\\}$ or
$\operatorname{conv}\\{p_{n},p_{i},p_{j}\\}$. Thus, we may apply the argument
from Subcase 2.1 and obtain the same quantities for $\operatorname{area}(Q)$
as in (3.4).

_Subcase 2.3_ , exactly two sides of $Q$ contain a side of $P_{k}$. We apply
the approach of the previous cases, and lt $\operatorname{conv}\\{p,q,r\\}$ be
the midtriangle of $Q$, where $p$ is a vertex of $P_{k}$. Then, like in
Subcase 2.2, we can observe that both $[p,q]$ and $[p,r]$ are parallel to some
sides of $P_{k}$. But this yields that both $q,r$ are vertices of $P_{k}$, and
we obtain the estimate in (3.4).

_Subcase 2.4_ , every side of $Q$ contains a side of $P_{k}$. Let
$T=\operatorname{conv}\\{p,q,r\\}$ denote the midtriangle of $Q$. Since every
side of $T$ is parallel to a side of $P_{k}$, the axial symmetry of $P_{k}$
with respect to the bisector of its every side and a simple geometric
observation shows that $p,q,r$ are the midpoints of some sides of $P_{k}$. But
then $T$ is axially symmetric with repect to the bisector of its every side,
implying that $T$ is regular. But from this we immediately obtain that $2k$ is
divisible by $3$, and that $\operatorname{area}(Q)=3\tan\frac{\pi}{3}$.

We need to compare this quantity to the one in the case $r=0$ of (3.4). Since

$3\tan\frac{\pi}{3}=6\sin\left(\frac{2\pi}{3}\right)<\frac{6\sin\left(\frac{2\pi}{3}\right)}{\cos^{2}\left(\frac{\pi}{2k}\right)}$

for all $k\geq 2$, the second part of Lemma 3.7 immediately follows. ∎

Now we prove Theorem 3.3.

###### Proof of Theorem 3.3.

We need to determine which are the values of $k\geq 2$ satisfy the inequality

(3.5) $\frac{n-6}{n-m}\sqrt{A(k,m)}+\frac{6-m}{n-m}\sqrt{A(k,n)}\geq
2\sqrt{A(k,6)}$

for all $3\leq m<6<n$.

Recall that by definition, $\mathbf{B}^{2}\subset P_{k}$ for all values of
$k$, implying that $A_{n}(BB^{2})\leq A_{n}(P_{k})$ for all values of $n$. An
elementary computation shows that $A_{n}(\mathbf{B}^{2})=n\tan\frac{\pi}{n}$
for all $n$. Thus, we have

$\frac{n-6}{n-m}\sqrt{A(k,m)}+\frac{6-m}{n-m}\sqrt{A(k,n)}\geq\frac{n-6}{n-m}\sqrt{m\tan\frac{\pi}{m}}+\frac{6-m}{n-m}\sqrt{n\tan\frac{\pi}{n}}\geq\\\
\geq\frac{1}{2}\sqrt{5\tan\frac{\pi}{5}}+\frac{1}{2}\sqrt{7\tan\frac{\pi}{7}}.$

Furthermore, for any $2k=6q+r$, where $q\geq 1$ and $0\leq r<6$,

$A(k,6)=(6-r)\tan\frac{\pi q}{2k}+r\tan\frac{\pi(q+1)}{2k}.$

Since $(6-r)\frac{\pi q}{2k}+r\frac{\pi(q+1)}{2k}=\pi$ and by the convexity of
the function $x\mapsto\tan x$ on $(0,\frac{\pi}{2})$, we have that for any
fixed value of $r$, $A(k,6)$ is a decreasing sequence of $q$. Using this, an
elementary computation shows that

$\sqrt{A(2k,6)}\leq\frac{1}{2}\sqrt{5\tan\frac{\pi}{5}}+\frac{1}{2}\sqrt{7\tan\frac{\pi}{7}}$

for any $q\geq 3$ and $0\leq r<6$. Thus, (3.5) holds for any $k\geq 9$. We
also observe that since $A(k,n)=\operatorname{area}(P_{k})$ for any $n\geq
2k$, (3.5) clearly holds if $k=2$ or $k=3$. Thus, we are left with the cases
$k=4,5,6,7,8$, for which a direct computation of the wo sides of (3.5) for any
$3\leq m<6<n\leq 2k$, using a Maple 18.00 software, yields the assertion. ∎

##  4\. Results about not necessarily polygonal norms

In this section we collect our results about the $\alpha$-honeycomb properties
of general norms, and the $\alpha$-Dowker properties of general convex disks.
Our first result shows sufficiently smooth convex disks ‘asymptotically’
satisfy the log-Dowker property.

######  Theorem 4.1.

If $K$ is a convex disk in $\mathbb{R}^{2}$ with $C^{4}$-class boundary and
strictly positive curvature everywhere, then there is some value
$n(K)\in\mathbb{R}$ such that for any $n\geq n(K)$, we have

$\log A_{K}(n-1)+\log A_{K}(n+1)\geq 2\log A_{K}(n).$

In the proof we intend to use the following result of Ludwig [17].

######  Lemma 4.2 (Ludwig).

Let $K$ be a convex disk in $\mathbb{R}^{2}$ with $C^{4}$-class boundary and
strictly positive curvature everywhere. Then there are quantities $A,B>0$ and
$C\in\mathbb{R}$, depending only on $K$, such that

$A_{K}(n)=A+\frac{B}{n^{2}}+\frac{C}{n^{4}}+o\left(\frac{1}{n^{4}}\right).$

###### Proof of Theorem 4.1.

Let us define $f(n)=A_{K}(n-1)A_{K}(n+1)-A^{2}_{K}(n)$. Then

$f(n)=\left(A+\frac{B}{(n-1)^{2}}+\frac{C}{(n-1)^{4}}+o\left(\frac{1}{n^{4}}\right)\right)\cdot\\\
\cdot\left(A+\frac{B}{(n+1)^{2}}+\frac{C}{(n+1)^{4}}+o\left(\frac{1}{n^{4}}\right)\right)-\left(A+\frac{B}{n^{2}}+\frac{C}{n^{4}}+o\left(\frac{1}{n^{4}}\right)\right)^{2}.$

Thus, an elementary computation yields that

$f(n)=AB\left(\frac{1}{(n-1)^{2}}+\frac{1}{(n+1)^{2}}-\frac{2}{n^{2}}\right)+AC\left(\frac{1}{(n-1)^{4}}+\frac{1}{(n+1)^{4}}-\frac{2}{n^{4}}\right)+\\\
+B^{2}\left(\frac{1}{(n-1)^{2}(n+1)^{2}}-\frac{1}{n^{4}}\right)+o\left(\frac{1}{n^{4}}\right).$

On the other hand, it can be checked that the orders of magnitude of
$\frac{1}{(n-1)^{4}}+\frac{1}{(n+1)^{4}}-\frac{2}{n^{4}}$ and
$\frac{1}{(n-1)^{2}(n+1)^{2}}-\frac{1}{n^{4}}$ are both
$o\left(\frac{1}{n^{4}}\right)$. Thus, after simplification, we obtain that

$f(n)=\frac{2AB(n^{4}+1)}{(n^{2}-1)^{2}n^{2}}+o\left(\frac{1}{n^{4}}\right),$

which is positive if $n$ is sufficiently large. ∎

Recall that $A_{\mathbf{B}^{2}}(n)=n\tan\frac{\pi}{n}$. In the following
theorem, we let

$\varepsilon_{0}=\frac{\sqrt{A_{\mathbf{B}^{2}}(5)}+\sqrt{A_{\mathbf{B}^{2}}(7)}-2\sqrt{A_{\mathbf{B}^{2}}(6)}}{\sqrt{A_{\mathbf{B}^{2}}(5)}+\sqrt{A_{\mathbf{B}^{2}}(7)}+2\sqrt{A_{\mathbf{B}^{2}}(6)}}=0.002623\ldots,$

and denote the Hausdorff distance of the convex bodies $K,L$ by $d_{H}(K,L)$.

######  Theorem 4.3.

Let $\mathcal{M}$ be a normed plane with unit disk $M$, and assume that
$d_{H}(M,\mathbf{B}^{2})\leq\varepsilon_{0}$. Then $\mathcal{M}$ satisfies the
honeycomb property.

###### Proof.

Observe that $d_{H}(M,\mathbf{B}^{2})\leq\varepsilon$ implies that
$(1-\varepsilon)\mathbf{B}^{2}\subseteq
M\subseteq(1+\varepsilon)\mathbf{B}^{2}$. On the other hand, since the
isoperimetrix $M_{iso}$ of $\mathcal{M}$ is a rotated copy of the polar of
$M$, this yields that $\frac{1}{1+\varepsilon}\mathbf{B}^{2}\subseteq
M_{iso}\subseteq\frac{1}{1-\varepsilon}\mathbf{B}^{2}$.

By Theorem 1.9, it is sufficient to prove that $M_{iso}$ satisfies the weak
$\frac{1}{2}$-Dowker property. Since for any $n\geq 3$ and $K\subseteq L$, we
have $A_{K}(n)\leq A_{L}(n)$, it follows that
$\frac{n\tan\frac{\pi}{n}}{(1+\varepsilon)^{2}}\leq
A_{M_{iso}}(n)\leq\frac{n\tan\frac{\pi}{n}}{(1-\varepsilon)^{2}}$. This yields
that it is sufficient to prove that for any $3\leq m<6<n$ and
$\varepsilon\leq\varepsilon_{0}$, we have

$\frac{1}{1+\varepsilon}\left(\frac{n-6}{n-m}\sqrt{m\tan\frac{\pi}{m}}+\frac{6-m}{n-m}\sqrt{n\tan\frac{\pi}{n}}\right)\geq\frac{2}{1-\varepsilon}\sqrt{6\tan\frac{\pi}{6}}.$

By the properties of $A_{\mathbf{B}^{2}}(n)=n\tan\frac{\pi}{n}$, the left-hand
side is minimal if $m=5$ and $n=7$. Thus, it is sufficient to show that

$\frac{1}{1+\varepsilon}\left(\sqrt{5\tan\frac{\pi}{5}}+\sqrt{7\tan\frac{\pi}{7}}\right)\geq\frac{2}{1-\varepsilon}\sqrt{6\tan\frac{\pi}{6}}$

holds for any $\varepsilon\leq\varepsilon_{0}$. But this readily follows from
the definition of $\varepsilon_{0}$. ∎

in Section 3 we have seen that there are polygons which do not satisfy the
$\alpha$-Dowker property for any $\alpha<1$. Our next theorem can be regarded
as a counterpoint of this observation.

######  Theorem 4.4.

Let $K$ be smooth and strictly convex. Then there is some value $\alpha<1$
such that $K$ satisfies the weak $\alpha$-Dowker property.

###### Proof.

First, observe that by our conditions, any supporting line of $K$ intersects
$K$ in a unique point, and for any point $p$ of $\operatorname{bd}(K)$ there
is a unique supporting line of $K$ that contains $p$. This yields, in
particular, that for any $n\geq 3$ and any minimum area convex $n$-gon $Q_{n}$
circumscribed about $K$, $Q_{n}$ has exactly $n$ sides, and the tangent points
on those sides are distinct. For any point $p\in\operatorname{bd}(K)$, we
denote the unique closed half plane supporting $K$ at $p$ by $H(p)$, and the
boundary of $H(p)$ by $L(p)$. We call boundary points of $K$ with opposite
unit normal vectors _opposite_ , or _antipodal_ points. Note that if
$p,q\in\operatorname{bd}(K)$ are not antipodal points of $K$, there there is a
unique closed arc of $\operatorname{bd}(K)$ connecting them that does not
contain antipodal points. We denote this arc by $\widehat{pq}$, and denote by
$A(\widehat{pq})$ the area of the bounded, connected region $R(\widehat{pq})$
whose boundary consists of $\widehat{pq}$ and the segments of $L(p)$ and
$L(q)$ connecting $p,q$ to the intersection point of the two lines. We observe
that the problem of finding $A_{K}(n)$ coincides with the problem of finding a
tiling of $\operatorname{bd}(K)$ by $n$ arcs $\widehat{x_{i}x_{i+1}}$,
$i=1,2,n$, $x_{n+1}=x_{i}$ such that
$\sum_{i=1}^{n}A(\widehat{x_{i},x_{i+1}})$ is minimal.

First, we fix an arbitrary $n\geq 4$, and show that
$A_{K}(n-1)+A_{K}(n+1)>2A_{K}(n)$. To do it, we follow the proof of the
original theorem of Dowker [8, Theorem 1]

Let $Q_{n-1}$ and $Q_{n+1}$ denote a minimum area convex $(n-1)$-gon and
$(n+1)$-gon, respectively, circumscribed about $K$. Let the tangent points of
$K$ on the sides of $Q_{n-1}$ be $p_{1},p_{2},\ldots,p_{n-1}$ in
counterclockwise order, and we denote the tangent points of $Q_{n+1}$ in
counterclockwise order on $\operatorname{bd}(K)$ by
$q_{1},q_{2},\ldots,q_{n+1}$, where the indices are understood mod $(n-1)$ and
$(n+1)$, respectively. Note that then $Q_{n-1}=\bigcap_{i=1}^{n-1}H(p_{i})$
and $Q_{n+1}=\bigcap_{j=1}^{n+1}H(q_{j})$. Since the areas of these regions
are finite, for any two consecutive tangent points $p_{i}$, $p_{i+1}$, no arc
$\widehat{p_{i}p_{i+1}}$ contains antipodal points of $K$, and the same is
true for the points $q_{j}$.

Due to the possible existence of coinciding points in the above two sequences,
we unite these sequences as a single sequence $v_{1},v_{2},\ldots,v_{2n}$ in
such a way that the points are in this counterclockwise order in
$\operatorname{bd}(K)$, $v_{1}=x_{1}$, and removing all the $p_{i}$ (resp. all
the $q_{j}$) from this sequence we obtain the sequence $q_{1},\ldots,q_{n+1}$
(resp. $p_{1},\ldots,p_{n-1}$). In the proof we regard this sequence as a
cyclic sequence, where the indices are determined mod $2n$, and, with a little
abuse of notation, we say that $\widehat{v_{i}v_{j}}$ _covers_
$\widehat{v_{k}v_{l}}$ if $\widehat{v_{k}v_{l}}\subseteq\widehat{v_{i}v_{j}}$
and $i<k<l<j<i+2n$.

Observe that the family of arcs $\widehat{p_{i}p_{i+1}}$,
$\widehat{q_{j}q_{j+1}}$ is a $2$-tiling $\mathcal{T}_{0}$ of
$\operatorname{bd}(K)$; that is every point of $\operatorname{bd}(K)$ belongs
to at least two such arcs, and no point belongs to the interior of more than
$2$ such arcs. Our main goal will be to modify the $2$-tiling
$\mathcal{T}_{0}$ in such a way that the value of $f$ does not decrease but
the number of covering pairs strictly decreases.

Note that since $\mathcal{T}_{0}$ is the union of two tilings consisting of
$(n-1)$ and $(n+1)$ arcs, respectively, $\mathcal{T}_{0}$ contains covering
pairs. Assume that $\widehat{v_{i}v_{j}}$ covers $\widehat{v_{k}v_{l}}$. Then
let $\mathcal{T}_{1}$ denote the $2$-tiling of $\S^{1}$ in which
$\widehat{v_{i}v_{j}}$ and $\widehat{v_{k}v_{l}}$ are replaced by
$\widehat{v_{i}v_{l}}$ and $\widehat{v_{k}v_{j}}$. Our main observation, as in
the proof of [8, Theorem 1] is that in this case

(4.1) $A(\widehat{v_{i}v_{j}})+A(\widehat{v_{k}v_{l}})\geq
A(\widehat{v_{i}v_{l}})+A(\widehat{v_{k}v_{j}}),$

and if the four points are pairwise distinct, then here we have strict
inequality.

According to our conditions,
$\sum_{S\in\mathcal{T}_{0}}A(S)\leq\sum_{S\in\mathcal{T}_{1}}A(S)$.
Furthermore, as $\mathcal{T}_{0}$ and $\mathcal{T}_{1}$ are $2$-tilings, and
$\widehat{v_{k}v_{l}}$ is already covered twice by the arcs considered in the
modification, the number of covering pairs in $\mathcal{T}_{1}$ is strictly
less than in $\mathcal{T}_{0}$. Repeating this procedure we obtain a
$2$-tiling $\mathcal{T}_{m}$ of $\operatorname{bd}(K)$ for which
$\sum_{S\in\mathcal{T}_{0}}A(S)\geq\sum_{S\in\mathcal{T}_{,}}A(S)$ and which
does not contain covering pairs. Then, $A_{t}$ decomposes into two tilings
$\\{\widehat{v_{1},v_{3}},\widehat{v_{3}v_{5}},\ldots,\widehat{v_{2n-1}v_{1}}\\}$
and
$\\{\widehat{v_{2},v_{4}},\widehat{v_{4}v_{6}},\ldots,\widehat{v_{2n}v_{2}}\\}$,
each of which contains exactly $n$ arcs.

Let $V_{1}$ be defined as the circumscribed convex $n$-gon touching $K$ at the
vertices $v_{i}$ with odd indices, and $V_{2}$ be the circumscribed convex
$n$-gon touching $K$ at the vertices $v_{i}$ with even indices. Then

(4.2)
$\operatorname{area}(Q_{n-1})+\operatorname{area}(Q_{n+1})\geq\operatorname{area}(V_{1})+\operatorname{area}(V_{2}).$

Since $V_{1}$ and $V_{2}$ are convex $n$-gons circumscribed about $K$, if in
(4.2) we have strict inequality, or if $\operatorname{area}(V_{i})>A_{K}(n)$
for $i=1$ or $i=2$, we are done. Thus, in the following we assume for
contradiction that
$\operatorname{area}(Q_{n-1})+\operatorname{area}(Q_{n+1})=\operatorname{area}(V_{1})+\operatorname{area}(V_{2})$,
and $\operatorname{area}(V_{1})=\operatorname{area}(V_{2})=A_{K}(n)$. Then, by
the remark right after (4.1) about the equality case in the inequality, there
is no arc $\widehat{p_{i}p_{i+1}}$ that contains points $q_{j},q_{j+1}$ in its
interior, and vice versa. In particular, there are at least two vertices of
$P_{n-1}$ that coincide with some vertices of $P_{n+1}$.

Without loss of generality, let $p_{2}=q_{2}$ corresponding to $v_{3}$ and
$v_{4}$, respectively, in the united sequence of vertices. Then
$\\{p_{1},q_{1}\\}=\\{v_{1},v_{2}\\}$ and
$\\{p_{3},q_{3}\\}=\\{v_{5},v_{6}\\}$. For $i=1,3$, let $p_{i}^{\prime}$
denote the intersection point of $L(p_{i})$ and $L(p_{2})$, and let
$q_{i}^{\prime}$ denote the intersection point of $L(q_{i})$ and
$L(q_{2})=L(p_{2})$. Note that since $Q_{n-1}$ is a minimum area convex
$(n-1)$-gon circumscribed about $K$, the tangent point on every side is the
midpoint of the side. Thus, $p_{2}$ is the midpoint of
$[p_{1}^{\prime},p_{3}^{\prime}]$, and we obtain similarly that $q_{2}=p_{2}$
is the midpoint of $[q_{1}^{\prime},q_{3}^{\prime}]$.

_Case 1_ , $p_{1}=v_{1}$, $p_{3}=v_{6}$. Then, since $V_{1},V_{2}$ are minimum
area circumscribed $n$-gons, we have that $p_{2}=q_{2}$ is the midpoint of
$[p_{1}^{\prime},q_{3}^{\prime}]$ as well as that of
$[q_{1}^{\prime},p_{3}^{\prime}]$. But then
$||p_{3}^{\prime}-p_{2}||=||q_{3}^{\prime}-p_{2}||=||p_{1}^{\prime}-p_{2}||=||q_{3}^{\prime}-p_{2}||$,
implying that $p_{1}=q_{1}$ and $p_{3}=q_{3}$.

_Case 2_ , $p_{1}=v_{2}$, $p_{3}=v_{5}$. Then by a similar argument, we obtain
that $p_{1}=q_{1}$ and $p_{3}=q_{3}$.

_Case 3_ , $p_{1}=v_{1}$ and $p_{3}=v_{5}$. Then we have that either
$p_{1}=q_{1}$ and $p_{3}=q_{3}$, or $p_{1}\neq q_{1}$ and $p_{3}\neq q_{3}$.

_Case 4_ , $p_{1}=v_{2}$ and $p_{3}=v_{6}$. Then we have that either
$p_{1}=q_{1}$ and $p_{3}=q_{3}$, or $p_{1}\neq q_{1}$ and $p_{3}\neq q_{3}$.

Now, let us call a point $q_{j}$ (resp. $p_{i}$) a _double vertex_ , if it
coincides with some $p_{i}$ (resp. $q_{j}$). If for some $j$, both $q_{j}$ and
$q_{j+1}$ are double vertices, than $q_{j}$ is a double vertex for every $j$
by the previous consideration, which contradicts the fact that $Q_{n+1}$ has
strictly more vertices than $Q_{n-1}$. If there is no value of $j$ such that
both $q_{j}$ and $q_{j+1}$ are double vertices, then by the previous
consideration any arc of $\operatorname{bd}(K)$ connecting two consecutive
double vertices contain the same number of vertices from $Q_{n-1}$ and
$Q_{n+1}$. But this also contradicts the facet that $Q_{n+1}$ has strictly
more vertices than $Q_{n-1}$.

Until now, we have shown that for any $n\geq 4$, we have
$A_{K}(n-1)+A_{K}(n+1)>2A_{K}(n)$. Thus, for any $n\geq 4$, either
$A^{\bar{\alpha}}_{K}(n-1)+A^{\bar{\alpha}}_{K}(n+1)>2A^{\bar{\alpha}}_{K}(n)$
is satisfied for all $\bar{\alpha}>0$, or there is a value $0<\alpha_{n}<1$
such that $A^{\bar{\alpha}}_{K}(n-1)+A^{\bar{\alpha}}_{K}(n+1)\geq
2A^{\bar{\alpha}}_{K}(n)$ is satisfied if and only if
$\bar{\alpha}\geq\alpha_{n}$.

Note that since $K$ is not a polygon, the sequence $\\{A_{K}(n)\\}$ is
strictly decreasing, and every element is strictly greater than
$\operatorname{area}(K)$. Thus, there is some $\alpha_{0}<1$ and a positive
integer $n(\alpha_{0})>6$ such that for any $n>n(\alpha_{0})$ and
$m\in\\{3,4,5\\}$, the inequality (1.2) is satisfied for any
$\alpha\geq\alpha_{0}$. Now, the quantity
$\alpha=\max\\{\alpha_{0},\alpha_{1},\ldots,\alpha_{n(\alpha_{0})}\\}<1$
satisfies the requirements in Theorem 4.4. ∎

##  5\. A conjecture of Steinhaus

We recall the following conjecture, appearing as Problem C15 in [7].

######  Conjecture 1 (Steinhaus).

For any tiling $\mathcal{T}$ in the Euclidean plane with tiles whose diameters
are at least $D$ for some fixed $D>0$, the maximum isoperimetric ratio
$\frac{\operatorname{perim}(C)^{2}}{\operatorname{area}(C)}$ of the cells $C$
of $\mathcal{T}$ is minimal if $\mathcal{T}$ is a regular hexagonal tiling.

To state our result regarding this conjecture, we need Definition 5.1.

######  Definition 5.1.

Let $\mathcal{T}$ be a tiling of a normed plane $\mathcal{M}$. Let
$\mathcal{T}(R)$ denote the family of cells of $\mathcal{T}$ in
$R\mathbf{B}^{2}$. Then the _lower average isoperimetric ratio_ of a cell of
$\mathcal{T}$ is defined as

$\underline{I}(\mathcal{T})=\liminf_{R\to\infty}\frac{\sum_{C\in\mathcal{T}(R)}\frac{\operatorname{perim}_{M}(C)^{2}}{\operatorname{area}(C)}}{\operatorname{card}(\mathcal{T}(C))}.$

If we replace the $\liminf$ in the above definition by $\limsup$, we obtain
the _upper average isoperimetric ratio_ $\overline{I}(\mathcal{T})$ of a cell.
If these quantities are equal, the common value is called the _average
isoperimetric ratio_ of a cell, denoted by $I(\mathcal{T})$.

Our result is the following.

######  Theorem 5.2.

For any normed plane $\mathcal{M}$ there is a hexagonal tiling
$\mathcal{T}_{hex}$ of $\mathcal{M}$ such that for any convex, normal tiling
$\mathcal{T}$ of $\mathcal{M}$, we have

$\underline{I}(\mathcal{T})\geq I(\mathcal{T}_{hex}).$

Furthermore, if $\mathcal{M}$ is a Euclidean plane, that $\mathcal{T}_{hex}$
is a regular hexagonal tiling.

###### Proof.

The first part of Theorem 5.2 follows from a straightforward modification of
the proof of Theorem 1.9. For the second statement we remark that, according
to his proof, the cells of $\mathcal{T}_{hex}$ are homothetic copies of a
minimum area hexagon circumscribed about the isoperimetrix of the norm. Thus,
if the norm is Euclidean, then these hexagons are regular. ∎

Theorem 5.2 can be stated also in the following, slightly more general form.

######  Theorem 5.3.

For any normed plane $\mathcal{M}$ there is a hexagonal tiling
$\mathcal{T}_{hex}$ of $\mathcal{M}$ such that for any convex tiling
$\mathcal{T}$ of $\mathcal{M}$ satisfying $\overline{v}(\mathcal{T})\leq 6$
(resp. $\underline{v}(\mathcal{T})\leq 6$), we have

$\underline{I}(\mathcal{T})\geq I(\mathcal{T}_{hex})\quad(\hbox{resp.
}\overline{I}(\mathcal{T})\geq I(\mathcal{T}_{hex})).$

There is a construction in [14] showing that it is possible to tile the plane
with convex heptagons. In this tiling, as we move away from the origin, the
cells get longer and thinner, showing that it is not normal. On the other
hand, an elementary geometric consideration proves Remark 5.4, where for any
convex disk $K$ in $\mathbb{R}^{2}$, we denote by $I(K)$,
$\operatorname{ir}(K)$ and $\operatorname{cr}(K)$ the isoperimetric ratio
$\frac{\operatorname{perim}^{2}(K)}{\operatorname{area}(K)}$, the inradius and
the circumradius of $K$ respectively; the latter two quantities are defined as
the radii of a largest Euclidean disk contained in $K$ and of the smallest
Euclidean disk containing $K$, respectively.

######  Remark 5.4.

For any $\lambda>0$ there is some $\mu>0$ such that if $I(K)\leq\lambda$, then
$\frac{\operatorname{cr}(K)}{\operatorname{ir}(K)}\leq\mu$.

###### Proof.

Let $\mathcal{F}$ denote the family of convex disks with unit perimeter and
containing $o$. Then for any $K\in\mathcal{F}$,
$\operatorname{area}(K)\geq\frac{1}{\lambda}$ and $K\subset\mathbf{B}^{2}$,
and thus, $\mathcal{F}$ is universally bounded. Since
$\operatorname{cr}(K)\leq 1$ for every $K\in\mathcal{F}$, we need to show that
there is some $\tau>0$ such that $\operatorname{ir}(K)\geq\tau$ for all
$K\in\mathcal{F}$. Suppose for contradiction that this is not so. Then, there
is a sequence $\\{K_{n}\\}$ of elements of $\mathcal{F}$ such that
$\operatorname{ir}(K_{n})\to 0$. It is known (see e.g. p. 215, Ex. 6.2 of
[27]) that for any convex disk we have
$\operatorname{ir}(K)\geq\frac{w(K)}{3}$, where $w(K)$ is the minimal width of
$K$. Thus, we have $w(K_{n})\to 0$. But this, combined with
$K_{n}\subset\mathbf{B}^{2}$, clearly contradicts our assumption that
$\operatorname{area}(K_{n})\geq\frac{1}{\lambda}$ for all values of $n$. ∎

In light of Theorem 5.3, Remark 5.4 and the construction in [14], it seems
meaningful to ask the following.

######  Question 5.5.

Let $\mathcal{T}$ be a convex tiling of $\mathbb{R}^{2}$ such that there are
universal constants $\lambda,\mu>0$ such that for every cell $C$ of
$\mathcal{T}$, we have $\operatorname{ir}(C)\geq\lambda$ and
$\frac{\operatorname{cr}(C)}{\operatorname{ir}(C)}\leq\mu$. Is it true that
$\overline{v}(\mathcal{T})\leq 6$ (resp. $\underline{v}(\mathcal{T}))\leq 6$?

In the following example we show that if we use a square instead of
$\mathbf{B}^{2}$ in the definition of average, the answer for upper average is
negative. To do it, let $\mathbf{S}$ be the square with vertex set
$\\{-1,1\\}^{2}$, and for any convex tiling $\mathcal{T}$, with the notation
$\mathcal{T}_{\square}(R)=\\{C\in\mathcal{T}:C\subseteq R\mathbf{S}\\}$, we
set

(5.1)
$\overline{v}_{\square}(\mathcal{T})=\limsup_{R\to\infty}\frac{\sum_{C\in\mathcal{T}_{\square}(R)}v(C)}{\operatorname{card}(\mathcal{T}_{\square}(R))},$

where $v(C)$ denotes the number of sides of $C$.

######  Example 5.6.

We define a tiling $\mathcal{T}$ of $\mathbb{R}^{2}$ in consecutive steps.
More specifically, we start with the family
$\mathcal{T}_{0}=\\{\mathbf{S}\\}$, and for every positive integer $k$, in the
$k$th step step we add some cells to the family $\mathcal{F}_{k-1}$ of the
already defined cells to tet the family $\mathcal{F}_{k}$. We will do it in
such a way that after each step, $\bigcup\mathcal{F}_{k}=N_{k}\mathbf{S}$ for
some positive integer $N$. Each time we use one of Step A and Step B,
described below:

_Step A_ : Let $\bigcup\mathcal{F}_{k-1}=N_{k-1}\mathbf{S}$, and let the sides
of this square be $E_{1},E_{2},E_{3},E_{4}$. Consider a rectangle $R$ of side
lengths $N_{k-1}$ and $2N_{k-1}$. We set
$\mathcal{T}_{k}=\mathcal{T}_{k-1}\cup\\{R_{1},R_{2},R_{3},R_{4}\\}$, where
each $R_{i}$ is congruent to $R$, and one half of a longer side of $R_{i}$ is
$E_{i}$. We do it in such a way that the rectangles and
$\bigcup\mathcal{T}_{k-1}$ are pairwise nonoverlapping (see Figure 3). _Step
B_ : Let $\bigcup\mathcal{F}_{k-1}=N_{k-1}\mathbf{S}$. We fill the region
obtained as the closure of $(N_{k-1}+2)\mathbf{S}\setminus(N_{k-1}\mathbf{S})$
by $4N_{k-1}+4$ translates of $\mathbf{S}$.

Figure 3. An illustration for the construction in Example 5.6. The four
rectangles surrounding the central square are obtained by applying Step A. The
squares in the outermost layer are obtained by applying Step B.

In the following part we describe a suitable infinite sequence of $A$s and
$B$s, where the k$th$ element is $A$ or $B$, if $\mathcal{T}_{k}$ is obtained
from $\mathcal{T}_{k-1}$ by applying Step A or Step B, respectively. Then we
set $\mathcal{T}=\bigcup_{k=1}^{\infty}\mathcal{T}_{k}$. Before describing
this sequence, we remark that independently of its definition, if an edge
$E_{1}$ of a cell $C_{1}\in\mathcal{T}$ intersects an edge $E_{2}$ of another
cell $C_{2}$ in a nondegenerate segment, then $E_{1}\subseteq E_{2}$ or
$E_{2}\subseteq E_{1}$. Thus, we can apply an arbitrarily small deformation to
the cells of $\mathcal{T}$ to obtain an edge-to-edge convex tiling
$\mathcal{T}^{\prime}$ of $\mathbb{R}^{2}$ with the same combinatorial
properties; this deformation can be done subsequently after each step. As a
consequence, for simplicity, if a side of a polygon, appearing in one of the
two steps above, is dissected into smaller segments by the vertices of the
neighbors of the cell, we regard each smaller segment as a separate edge of
the cell. Using this interpretetion, in the following, we call the edges of
$\mathcal{T}$ in $\operatorname{bd}\left(\bigcup\mathcal{T}_{k}\right)$
_boundary edges_ of $\mathcal{T}_{k}$, and the edges of $\mathcal{T}$ in
$\operatorname{int}\left(\bigcup\mathcal{T}_{k}\right)$ _interior edges_ of
$\mathcal{T}_{k}$.

Fix any positive number $\nu>0$. First, consider the case that we apply Step A
$k_{1}$ times, and after that we apply Step B once. Then
$N_{k_{1}}=3^{k_{1}}$, and thus, $\mathcal{T}_{k_{1}}$ has $43^{k_{1}}$
boundary edges. In addition, $\mathcal{T}_{k_{1}}$ consists of $4k_{1}+1$
cells. We count the number of interior edges: Note that $\mathcal{T}_{0}$ has
$4$ edges. Then in each step before the $k_{1}$th one we add $12$ edges, which
are all interior edges of $\mathcal{T}_{k_{1}}$. Finally, in the $k_{1}$th
step we add four interior edges of $\mathcal{T}_{k_{1}}$. Thus, in total,
$\mathcal{T}_{k_{1}}$ has $12k_{1}-4$ interior edges. Now, if we set
$R_{1}=N_{k_{1}}+\frac{1}{2}$, then

$v_{\square}(R_{1})=\frac{\sum_{C\in\mathcal{T}_{\square}(R_{1})}v(C)}{\operatorname{card}(\mathcal{T}_{\square}(R_{1}))}=\frac{43^{k_{1}}+2(12k_{1}-4)}{4k_{1}+1}.$

Observe that if $k_{1}$ is sufficiently large, then $v_{\square}(R_{1})>\nu$.
We choose $k_{1}$ to satisfy this inequality, and fix this value in the
following.

Now we apply Step A $k_{2}-k_{1}-1$ times and after that we apply Step B once.
Furthermore, we set $R_{2}=N_{k_{2}}+\frac{1}{2}$. Note that if $k_{2}$ is
sufficiently large, then the contribution of $\mathcal{T}_{k_{1}}$ to both the
numerator and the denominator of the fraction defining $v_{\square}(R_{2})$ is
negligible, and thus, by the argument in the previus paragraph, we can choose
some sufficiently large integer $k_{2}$ such that $v_{\square}(R_{2})>\nu$.

Continuing this process, we obtain a tiling $\mathcal{T}$, and a sequence
$\\{R_{n}\\}$ with $R_{n}\to\infty$, such that $v_{\square}(R_{n})>\nu$ for
all values of $n$. This yields that
$\overline{v}_{\square}(\mathcal{T})\geq\mu$.

##  6\. An additional remark

Estimates regarding the numbers of sides of tilings has been a longstanding
problem of geometry. The classical book of Grünbaum and Shephard [14] shows
that it is possible to tile the Euclidean plane with convex hexagons, but that
any normal tiling contains infinitely many cells with at most six sides. Here
we recall that a (not necessarily convex) tiling $\mathcal{T}$ of
$\mathbb{R}^{2}$ is called _normal_ if

  * (1)

every cell is homeomorphic to $\mathbf{B}^{2}$;

  * (2)

the intersection of every pair of cells is connected;

  * (3)

the cells are _universally bounded_ , i.e. there are universal constants
$0<\hat{r}<\hat{R}$ such that every cell contains a ball of radius $\hat{r}$
and is contained in a ball of radius $\hat{R}$.

The statement of Grünbaum and Shephard was reproved in a simpler way by
Kazanci and Vince in [16]. For convex, normal mosaics, more is known. More
specifically, the second part of Lemma 2.4, stating that the upper average
number of sides of a cell of such a tiling is at most six is proved in the
book [12] of L. Fejes Tóth. The fact that if all cells have at least six
sides, then the number of sides with more than six is finite is due to
Stehling [24]. Akopyan [2] gave a quantitative estimate for the number of
these sides, while Frettlöh et al. [13] showed that the order of magnitude
given in [2] is sharp.

We remark that the proofs of the Lemmas 2.1-2.4 (as well as Definition 2.3)
can be generalized for all normal mosaics in a straightforward way. Thus, we
gave a short proof of the following statement, generalizing the result of
Fejes Tóth for any normal mosaic.

######  Theorem 6.1.

For any normal mosaic $\mathcal{T}$ in $\mathbb{R}^{2}$, we have
$\overline{v}(\mathcal{T})\leq 6$.

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# Pre-training Transformers for Molecular Property Prediction Using Reaction
Prediction

Johan Broberg  Maria Bånkestad  Erik Ylipää

###### Abstract

Molecular property prediction is essential in chemistry, especially for drug
discovery applications. However, available molecular property data is often
limited, encouraging the transfer of information from related data. Transfer
learning has had a tremendous impact in fields like Computer Vision and
Natural Language Processing signaling for its potential in molecular property
prediction. We present a pre-training procedure for molecular representation
learning using reaction data and use it to pre-train a SMILES Transformer. We
fine-tune and evaluate the pre-trained model on 12 molecular property
prediction tasks from MoleculeNet within physical chemistry, biophysics, and
physiology and show a statistically significant positive effect on $5$ of the
$12$ tasks compared to a non-pre-trained baseline model.

Machine Learning, ICML, Transformer, Molecular Property Prediction, Transfer
Learning, Pre-training

##  1 Introduction

Molecular property prediction has long been used to quickly screen new
molecule leads in drug development. The accuracy of these methods is crucial,
since false negatives incur high costs when a lead is taken to an experimental
phase. Lately, machine learning has become one of the standard tools for
molecular property prediction. However, the main challenge is the limited
amount of available data to train models on. One solution to this problem is
to curate larger datasets using domain expertise. This can be a costly and
time consuming approach but has the advantage that larger parts of the
relevant molecular domain can be covered (Thawani et al., 2020). Another
approach to solve the data scarcity problem is that of _transfer learning_
(Zhuang et al., 2019), where knowledge in one task is used to improve
performance in another. This is an active area of research in the chemistry
domain, mainly using Graph Neural Networks (GNNs) (Kipf & Welling, 2016) and
Transformer models (Vaswani et al., 2017). While transfer learning has proven
to be very successful in domains such as Natural Language Processing (Howard &
Ruder, 2018; Devlin et al., 2018; Liu et al., 2019) and Computer Vision
(Girshick et al., 2013), the same clear success is yet to happen in chemistry.

This work is a continuation of the master’s thesis (Broberg, 2022) and
explores a pre-training strategy for molecular representation learning based
on chemical reaction prediction. We use it to pre-train a transformer encoder
and compare its performance to a randomly initialized one on a wide range of
molecular property prediction tasks. We show statistically significant
improvements on $5$ of the $12$ datasets using a significance level
$\alpha=0.05$ with Bonferroni correction (Bonferroni, 1935, 1936; Noble,
2009).

Figure 1: Illustration of our pre-training architecture. Note that each
molecule fragment is encoded independently of the others.

##  2 Related Work

Most recent work on pre-training deep models for molecular property prediction
uses either GNNs or Transformers.

With GNNs, it is common to use multiple learning objectives that aims to
improve representation on different levels (node/edge/graph) (Hu et al., 2020;
Rong et al., 2020; Liu et al., 2021). Node and edge level pre-training tasks
generally aim to capture graph structural regularities of molecules. Examples
of such tasks are prediction of masked node or edge attributes or using node
embeddings to predict information about the neighborhood structure. Graph
level tasks may also be based on graph structural information but there are
also approaches that more explicitly utilize information from the chemistry
domain. For example, 3D molecular structure data (Stärk et al., 2021; Fang et
al., 2022; Liu et al., 2021) and graph motifs with their connections to
functional groups (Rong et al., 2020; Zhang et al., 2021) have been used for
graph level pre-training.

For Transformers applied to molecular property prediction using SMILES
(Weininger, 1988), a common pre-training approach is to randomly mask parts of
the input string. ChemBERTa (Chithrananda et al., 2020), MolBERT (Fabian et
al., 2020), and SMILES-BERT (Wang et al., 2019) are only some of the works
that explore this method. Though proven beneficial, this approach has not led
to the huge improvements seen in NLP by, e.g., the BERT model (Devlin et al.,
2018) using the very related masked-language-modeling (MLM) pre-training task.
Chithrananda et al. (2020) show a diminishing performance gain on three tasks
(BBBP, ClinTox (CT_TOX), and Tox21 (SR-p53)) using a masked token prediction
pre-training strategy where an increase in dataset size from $10^{6}$ to
$10^{7}$ only lead to a $\approx 3\%$ mean ROC-AUC increase and a $\approx
2\%$ mean PRC-AUC increase. This indicates that pre-training by recovering
masked tokens alone might not scale well enough to train powerful property
prediction models.

Another recent line of research tries to adapt the Transformer architecture to
take molecular graphs as input (Maziarka et al., 2020; Yoo et al., 2020; Ying
et al., 2021). Such modifications allow Transformers explicit access to the
graph structure, which otherwise must be learned implicitly from string
representations. Furthermore, it can also allow information such as node
features and edge features to be included in the model input. These models
have achieved remarkable results and have in some cases been pre-trained using
structure-based tasks (Maziarka et al., 2020; Yoo et al., 2020), but the main
contribution to their predictive power seem to stem from their architecture
rather than their choice of pre-training task.

A relatively unexplored source of information for pre-training in the
chemistry domain is reaction data. To our knowledge, the only published work
using such data for pre-training is by Wang et al. (2021). They base their
approach on GNNs and a contrastive learning task that teaches the model to
encode molecules such that two aggregated sets of molecular encoding vectors
lie near each other in the encoding space if they make up the left- and right-
hand side of a chemical reaction respectively, and far away if they do not.

Figure 2: Example reaction with molecules represented as their structural
formulas and as SMILES.

Our approach differs from that of Wang et al. (2021) in that we model the pre-
training step as a generative reaction prediction task instead of a
contrastive learning task. Furthermore, we use a transformer architecture and
SMILES representations of molecules while they use a GNN architecture and
graph representations.

##  3 Background

###  3.1 Chemical Reactions

A chemical reaction is a transformation of one set of molecules into another.
The molecules present before the reaction are called reactants, while the
molecules created through the reaction are called products. Molecules might be
part of the reaction but not contribute themselves with any atoms to the
product molecules created. These are called reagents. In reaction prediction
one tries to predict the product molecules of a reaction given the reactants
and reagents. Normally, a reaction produces multiple product molecules. In our
work, we have limited the scope by using data with only a single product.

###  3.2 SMILES

SMILES, Simplified-Molecular-Input-Line-Entry-System is a linearization of the
molecular graph. A molecule has many possible SMILES depending on where the
linearization starts and what branches to take. There are certain rules for
producing canonical SMILES of molecules, where the same molecule will always
encode to the same SMILES. A commonly used strategy when using SMILES as
inputs to neural networks is to randomize the SMILES, by essentially starting
the linearization at a random place and take random branches when traversing
the molecular graph. This has shown to act as a powerful data augmentation
method when working with SMILES representations of molecules (Bjerrum, 2017;
Arús-Pous et al., 2019).

###  3.3 Transformer

The Transformer is an architecture operating on mathematical sets and was
introduced by Vaswani et al. (2017) in the context of neural machine
translation. It has been widely used for sequential data such as natural
language. A positional encoding is then added to the input data to provide the
sequential structure to the model. The original (full) Transformer consists of
an encoder and a decoder and is typically used for translation tasks. On its
own, the Transformer encoder can be used for sequence
classification/regression/representation tasks (Devlin et al., 2018). The key
component of the architecture is the multi-head attention mechanism which
enables the model to attend to all elements of its input at once. For a
detailed description of the full Transformer model and for Transformer encoder
models such as BERT we refer to (Vaswani et al., 2017) and (Devlin et al.,
2018) respectively.

##  4 Method

Our approach is based on the Molecular Transformer by Schwaller et al. (2019),
in which reaction prediction is modeled as a sequence translation problem for
which a full Transformer model is used. Like Schwaller et al. (2019), we
represent molecules as SMILES. By using canonicalized SMILES for our product
molecules we obtain a fixed target sequence which makes the generative process
easier compared to generating graphs, where an order needs to be induced on
the edge set (Vinyals et al., 2016).

###  4.1 Pre-Training Architecture

In the Molecular Transformer, the encoder is applied to SMILES strings
containing all reactants and reagents of the corresponding reaction. In our
pre-training phase, the Transformer encoder is applied to each reactant and
reagent independently. That is, the encoder can only attend to the tokens in
the same SMILES fragment, not across fragments. For the set of reactants and
reagents $R=\\{r_{1},r_{2},...\\}$ in a given reaction, each SMILES fragment
$r_{i}$ produces an encoding $\mathbf{h}_{i}\in\mathbb{R}^{L\times d}$ where
$L$ is the maximum sequence length and $d$ is the token embedding dimension.
The set of such encodings $H=\\{\mathbf{h}_{1},\mathbf{h}_{2},...\\}$ are then
aggregated into a single vector $h_{R}\in\mathbb{R}^{d}$ representing the
entire reaction by first applying element-wise addition across the encodings
and then averaging across the sequence axis according to

$h_{R}=\text{Aggregate}(H)=\text{Mean}(\text{Sum}(H)).$ (1)

The aggregated reaction vector $h_{R}$ is then passed to the decoders
“encoder-decoder attention” layers as a memory key to all tokens in the
product (target) SMILES.

We average across the sequence axis because, in the downstream fine-tuning
tasks, we will use encoded molecules together with a Multi-Layered Perceptron
(MLP). This mean that we will also need to aggregate the encoded molecules
into fixed sized vectors. We choose to include this aggregation in the pre-
training step, so that the single vector representation will be forced to
contain all information needed for the decoding.

###  4.2 Fine-Tuning Architecture

In the fine-tuning phase, we only use the encoder component from the pre-
training phase. Encodings of molecules are aggregated across the sequence axis
using the mean. A 2-layered MLP with ReLU activations is used to map the
aggregated molecular encodings to the target values. All parameters (encoder
and MLP) are tuned in this phase.

Figure 3: Illustration of the fine-tuning architecture for a binary
classification task.

###  4.3 Evaluation

Table 1: Performance of the model with and without pre-training showing the
mean and standard deviation over a 10-fold cross-validation.
Data set | Metric |  Model Without pre-training |  Model With pre-training |  Wilcoxon Signed-Rank Test |  Rank-Biserial Correlation
---|---|---|---|---|---
ESOL |  RMSE $\downarrow$ | $0.656\pm 0.082$ | $\mathbf{0.428}\pm 0.077$ | $\mathbf{0.001}$ | $1.000$
FreeSolv |  RMSE $\downarrow$ | $2.057\pm 0.477$ | $\mathbf{1.484}\pm 0.413$ | $\mathbf{0.002}$ | $0.982$
Lipophilicity |  RMSE $\downarrow$ | $1.012\pm 0.038$ | $\mathbf{0.700}\pm 0.035$ | $\mathbf{0.001}$ | $1.000$
PCBA |  PRC-AUC $\uparrow$ | $0.178\pm 0.010$ | $0.175\pm 0.006$ | $0.862$ | $0.309$
MUV |  PRC-AUC $\uparrow$ | $0.023\pm 0.026$ | $0.025\pm 0.017$ | $0.539$ | $0.491$
HIV |  ROC-AUC $\uparrow$ | $0.704\pm 0.050$ | $\mathbf{0.757}\pm 0.052$ | $\mathbf{0.002}$ | $0.982$
BACE |  ROC-AUC $\uparrow$ | $0.726\pm 0.088$ | $\mathbf{0.817}\pm 0.106$ | $\mathbf{0.001}$ | $1.000$
BBBP |  ROC-AUC $\uparrow$ | $0.902\pm 0.104$ | $0.921\pm 0.101$ | $0.019$ | $0.873$
Tox21 |  ROC-AUC $\uparrow$ | $0.799\pm 0.012$ | $0.792\pm 0.013$ | $0.981$ | $0.145$
ToxCast |  ROC-AUC $\uparrow$ | $0.693\pm 0.021$ | $0.701\pm 0.013$ | $0.348$ | $0.582$
SIDER |  ROC-AUC $\uparrow$ | $0.597\pm 0.016$ | $0.578\pm 0.039$ | $0.920$ | $0.255$
ClinTox |  ROC-AUC $\uparrow$ | $0.956\pm 0.0044$ | $0.959\pm 0.038$ | $0.652$ | $0.436$

We fine-tune separately on 12 datasets in MoleculeNet (Wu et al., 2017) using
10-folded cross-validation. Each fold ($10\%$) is used once for evaluation of
hyperparameter tuning, once for validation, once for testing and otherwise for
training. For regression tasks and multi-label classification tasks we use
random splits while for single-label classification tasks we use stratified
splits. On multi-label prediction benchmarks (PCBA, MUV, TOX21, ToxCast,
SIDER, ClinTox) we report the average performance across all tasks as
suggested by Wu et al. (2017).

To evaluate the effect of our transfer learning approach, we compare the pre-
trained model to a randomly initialized one which we train directly on each
molecular property dataset. Evaluation is based on the best performing
checkpoints with respect to the validation set, for each model run. Our null
hypothesis is that reaction prediction pre-training has no effect on molecular
property prediction across all chemical space. The null hypothesis is then
rejected with 95% level of confidence.

We pair the performances on each test fold and use the Wilcoxon signed-rank
test (Wilcoxon, 1945) to determine statistically significant differences
between our pre-trained model and the randomly initialized one on each of the
12 datasets. The Wilcoxon signed-rank test is a non-parametric version of the
Student’s t-test, which does not assume normally distributed data. Since we
evaluate using multiple tests, we also make a Bonferroni correction to
counteract the multiple comparison problem (Noble, 2009). Practically this
means that, to evaluate our null hypothesis with a 95% level of confidence
(significance level $\alpha=0.05$) we use a significance level of
$\alpha_{1}=...=\alpha_{m}=\alpha/m$ when we test for an effect on each
individual dataset. Here $m=12$ and denote the number of datasets we test on.
This means that we for each dataset we use a significance level of
$\alpha_{1}=...=\alpha_{12}=0.05/12=0.00417$.

##  5 Experiment

###  5.1 Data and Pre-Processing

The dataset used in the pre-training phase is from the USPTO database (Lowe,
2012) and consists of $902,581$ samples used for training and $50,131$ samples
used for validation, based on pre-processing and data splits provided by
Schwaller et al. (2017).

For the reactants and reagents we use randomized SMILES while the product
molecules, our targets, were kept in canonicalized form. Due to the memory
complexity of the Transformer we truncate SMILES in the training data to a
maximum sequence length of 157. Of the reactions in the pre-training data,
$99.9\%$ contain reactants, reagents and products whose SMILES are all shorter
than 157, so this truncation threshold only affects $0.1\%$ of the reactions.

###  5.2 Experimental Setup

For our pre-training model we used a four-layered encoder and decoder, with
eight attention heads and a layer width of 256. We pre-trained our model for
150 epochs, using cross-entropy loss and AdamW optimizer with a batch size of
$4096$ and a cosine cyclic learning rate scheduler with base learning rate of
$10^{-5}$ and maximum learning rate of $5\cdot 10^{-4}$.

In the fine-tuning phase we tuned the learning rate for all models on each
fold in the cross validation. We did this in order to fairly compare the
performance. Each tuning was based on 20 runs with learning rates sampled
geometrically in the interval $\left[10^{-6},10^{-3}\right]$. During learning
rate tuning we trained each run for $50$ epochs for all datasets except PCBA
and MUV which were only tuned for one respectively ten epochs due to the large
number of samples in these datasets. The batch size was set to $64$. For the
number of epochs, we used early stopping with $40$ update steps of patience.
Throughout this work we tokenized SMILES by converting each character to their
corresponding ASCII-value.

###  5.3 Results

The results from our experiment are shown in table 1. For each dataset we
present the mean and standard deviation over the 10-folded cross-validation
along with the $p$-value of the Wilcoxon signed-rank test and the rank-
biserial correlation. On 5 of the 12 downstream property prediction tasks our
pre-training strategy show a statistically significant positive effect given
the Bonferroni-corrected significance level $\alpha=0.00417$. We therefore
reject our null hypothesis and conclude that our pre-training approach using
reaction prediction has an effect on molecule property prediction

##  6 Limitations

This pre-training strategy has two limitations that we would like to point
out. We use reactions that only have one major product molecule, but most
reactions contain more than one product. This is an advantage of the approach
proposed by Wang et al. (2021). Another limitation is that we base our
statistical analysis on cross-validation of the downstream tasks. A more
robust statistical analysis would have been based on several different runs of
pre-training.

##  7 Conclusions

In this paper we have presented a pre-training strategy for transformer models
using reaction prediction. We demonstrate a statistically significant effect
on 5 out of 12 datasets from MoleculeNet and conclude that reaction prediction
pre-training has an effect on molecular property prediction. Our results show,
in line with Wang et al. (2021), that chemical reactions can be used to
successfully pre-train models for downstream molecular property prediction
tasks.

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