# On the edge densities of convex mosaics

Máté Kadlicskó  and  Zsolt Lángi  Máté Kadlicskó, Independent researcher,
Budapest, Hungary,
Műegyetem rkp. 3., H-1111 Budapest, Hungary<EMAIL_ADDRESS>Zsolt
Lángi, Department of Algebra and Geometry, Budapest University of Technology
and Economics, and MTA-BME Morphodynamics Research Group,
Műegyetem rkp. 3., H-1111 Budapest, Hungary<EMAIL_ADDRESS>
###### Abstract.

In this paper we investigate the problem of finding the minimum of edge
density in families of convex mosaics with unit volume cells in
$3$-dimensional Euclidean space. We solve this problem for two families. In
particular, we prove that a translative convex mosaic with unit volume cells
has minimal edge density if and only if it is a face-to-face mosaic generated
by unit cubes. We also investigate decomposable mosaics, defined as mosaics
whose cells are Minkowski sums of cells of two lower dimensional convex
mosaics, and show that the edge density of every normal decomposable mosaic
with unit volume cells is at least as large as that of a face-to-face mosaic
with regular triangle-based right prisms as cells. In addition, we use our
method to determine the minimal total edge length of a $3$-dimensional unit
volume parallelohedron.

###### Key words and phrases:

Honeycomb Conjecture, Kelvin’s conjecture, edge density, translative mosaic,
decomposable mosaic, parallelohedron, total edge length, Melzak’s conjecture

###### 2020 Mathematics Subject Classification:

52C22, 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

The aim of this paper is to investigate certain properties of mosaics. A
_mosaic_ or _tiling_ of the Euclidean $n$-space $\mathbb{R}^{n}$ is a
countable family of compact sets $\mathcal{M}$, called cells, with the
property that $\bigcup\mathcal{M}=\mathbb{R}^{n}$, and the interiors of any
two distinct cells are disjoint. A mosaic is _convex_ , if every cell in it is
convex; it is known that in this case every cell is a convex polytope. If, for
a mosaic $\mathcal{M}$, there are universal constants $0<r<R$ such that the
inradius of every cell of $\mathcal{M}$ is at least $r$, and its circumradius
is at most $R$, the mosaic is called _normal_. In this paper we deal only with
normal, convex mosaics. A convex mosaic is called _face-to-face_ (in the plane
also _edge-to-edge_) if every facet of every cell is a facet of exactly one
more cell. Furthermore, a convex mosaic is _translative_ (resp. _congruent_),
if every cell is a translate (resp. a congruent copy) of a given convex
polytope. Clearly, every congruent and translative convex mosaic is normal. A
cell of a translative convex mosaic in $\mathbb{R}^{n}$ is called an _$n$
-dimensional parallelohedron_.

One of the classic problems of geometry, regarding planar mosaics, is the
Honeycomb Conjecture, stating that in a decomposition of the Euclidean plane
into cells of equal area, the regular hexagonal grid has minimal edge density.
This conjecture first appeared in Roman time in a book of Varro about
agriculture [26]. The problem has been the centre of research throughout the
second half of the 20th century [7, 19]; it was solved by Fejes Tóth for
normal, convex mosaics, while the most general version is due to Hales [12],
who dropped the condition of convexity.

A famous analogous problem for mosaics in the $3$-dimensional Euclidean space
$\mathbb{R}^{3}$ was proposed by Lord Kelvin [14] in 1887; he conjectured that
in a tiling of space with cells of unit volume, minimal surface density is
attained by a mosaic composed of slightly modified truncated octahedra. Even
though this conjecture was disproved by Weaire and Phelan in 1994 [28], who
discovered a tiling of space with two slighly curved ‘polyhedra’ of equal
volume and with less surface density than in Lord Kelvin’s mosaic, the
original problem of finding the mosaics with equal volume cells and minimal
surface density has been extensively studied (see, e.g. [15, 9, 21]). On the
other hand, in the author’s knowledge, there is no subfamily of mosaics for
which Kelvin’s problem is solved. Here, in particular, we should recall a yet
unsolved conjecture of Bezdek [3] from 2006, restated in [18] in 2022, that
states that among translative, convex mosaics with unit volume cells, the ones
generated by regular truncated octahedra have minimal surface density.

We intend to investigate a similar problem, namely we search for convex
mosaics with unit volume cells and minimal edge density. In our paper, we
denote the edge density of a mosaic $\mathcal{M}$ by $\rho_{1}(\mathcal{M})$
(for a formal definition of edge density, see Section 1), and the family of
translative convex mosaics of $\mathbb{R}^{3}$ with unit volume cells by
$\mathcal{F}_{tr}$, and the family of unit volume parallelohedra by
$\mathcal{P}_{tr}$.

Our first result is as follows.

######  Theorem 1.

For any $\mathcal{M}\in\mathcal{F}_{tr}$,

$\rho_{1}(\mathcal{M})\geq 3,$

with equality if and only if $\mathcal{M}$ is a face-to-face mosaic with unit
cubes as cells.

To prove Theorem 2, we prove a more general theorem. To state this theorem, we
recall a few properties of $3$-dimensional parallelohedra. First, we note that
every $3$-dimensonal parallelohedron $P$ is the Minkowski sum of at most $6$
segments. Furthermore, every edge of $P$ is a translate of a generating
segment of $P$, and for every generating segment $S$ of $P$ there are $6$ or
$4$ edges of $P$ that are translates of $S$. In these cases we say that $S$
generates a _$6$ -belt_ or a _$4$ -belt_, respectively.

If $P=\sum_{i=1}^{k}S_{i}$ is a $3$-dimensional parallelohedron generated by
the segments $S_{i}$, and $m=(\alpha_{6},\alpha_{4})$ has positive
coordinates, we set

$w_{i}=\left\\{\begin{array}[]{l}\alpha_{6}\hbox{ if }S_{i}\hbox{ generates a
}6-\hbox{belt;}\\\ \alpha_{4}\hbox{ if }S_{i}\hbox{ generates a
}4-\hbox{belt.}\end{array}\right.$

Furthermore, we define the function $w_{m}(P)=\sum_{i=1}^{k}w_{i}L(S_{i})$,
where $L(S_{i})$ denotes the length of $S_{i}$. We prove Theorem 2, in which
$P_{cu}$ denotes a unit cube, $P_{oc}$ denotes a unit volume regular truncated
octahedron, and $P_{pr}$ denotes a regular hexagon-based right prism, where
the edge length of the hexagon is
$\frac{2^{2/3}\alpha_{6}^{1/3}}{3^{5/6}\alpha_{4}^{1/3}}$, and the length of
the lateral edges of $P_{pr}$ is
$\frac{3^{1/6}\alpha_{4}^{2/3}}{2^{1/3}\alpha_{6}^{2/3}}$. We note that the
volume of $P_{pr}$ is also one.

######  Theorem 2.

Let $m=(\alpha_{6},\alpha_{4})$ with $\alpha_{4},\alpha_{6}>0$. Then
$w_{M}(\cdot)$ attains its minimum on $\mathcal{P}_{tr}$. Furthermore, if
$P\in\mathcal{P}_{tr}$ minimizes $w_{m}(\cdot)$ on $\mathcal{P}_{tr}$, then
the following holds.

  * (a)

If $0<\alpha_{4}<\frac{\sqrt{3}}{2}\alpha_{6}$, then $P$ is congruent to
$P_{cu}$.

  * (b)

If $\alpha_{4}=\frac{\sqrt{3}}{2}\alpha_{6}(\approx 0.866025\alpha_{6})$, then
$P$ is congruent to $P_{cu}$ or $P_{pr}$.

  * (c)

If $\frac{\sqrt{3}}{2}\alpha_{6}<\alpha_{4}<\sqrt[4]{\frac{2}{3}}\alpha_{6}$,
then $P$ is congruent to $P_{pr}$.

  * (d)

If $\alpha_{4}=\sqrt[4]{\frac{2}{3}}\alpha_{6}(\approx 0.903202\alpha_{6})$,
then $P$ is congruent to $P_{pr}$ or $P_{oc}$.

  * (e)

If $\sqrt[4]{\frac{2}{3}}\alpha_{6}<\alpha_{4}$, then $P$ is congruent to
$P_{oc}$.

We note that if $m=\left(\frac{1}{2},\frac{1}{2}\right)$, then $w_{m}(P)$
coincides with the mean width of $P$ for any $P\in\mathcal{P}_{tr}$ [18], and
thus, Theorem 2 is a generalization of the main result of [18], stating that
among $3$-dimensional unit volume parallelohedra, regular truncated octahedra
have minimal mean width.

In the paper we also deal with another kind of mosaics.

######  Definition 1.

A convex mosaic $\mathcal{M}$ of the Euclidean $n$-space $\mathbb{R}^{n}$ is
said to be _decomposable_ , if there are linear subspaces $L_{1},L_{2}$ of
dimensions $k$ and $n-k$, respectively, with $0<k<n$ and $L_{1}\cap
L_{2}=\\{o\\}$, and convex mosaics $\mathcal{M}_{1}$ of $L_{1}$ and
$\mathcal{M}_{2}$ of $L_{2}$ such that a set $C$ is a cell of $\mathcal{M}$ if
and only if $C=C_{1}+C_{2}$ for some cells $C_{i}$ of $\mathcal{M}_{i}$, where
$i=1,2$.

In the paper, we denote the family of normal, decomposable mosaics of
$\mathbb{R}^{3}$ with unit volume cells by $\mathcal{F}_{dc}$, and by $P_{pr}$
we denote a right prism whose base is a regular triangle of edge length
$\frac{1}{\sqrt{3}}$, and whose lateral edges have length
$\frac{2}{\sqrt{3}}$.

######  Theorem 3.

For any $\mathcal{M}\in\mathcal{F}_{dc}$,

$\rho_{1}(\mathcal{M})\geq\frac{3\sqrt{3}}{2}.$

Here, equality is attained e.g. if $\mathcal{M}$ is a face-to-face mosaic the
cells of which are congruent to $P_{pr}$.

We note that a simplified version of the proof of Theorem 3 yields a proof of
the Honeycomb Conjecture for normal, convex mosaics. Our results are also
related to a conjecture of Melzak [16] made in 1965, claiming that among
convex polyhedra of unit volume, regular triangle based prisms, with equal-
length edges, have minimal total edge length. This conjecture is still very
far from settled; for the solution of this problem in smaller families of
convex polyhedra and for bounds on this quantity, see e.g. [1, 2, 4, 8, 17].

Corollary 1 is an immediate consequence of Theorem 2, and Corollary 2 can be
proved by a much simplified variant of the proof of Theorem 3. Here, by $L(P)$
we denote the total edge length of the convex polyhedron $P$.

######  Corollary 1.

For any $3$-dimensional parallelohedron $P$ of unit volume, we have

$L(P)\geq 12,$

with equality if and only if $P$ is a unit cube.

######  Corollary 2.

For any prism $P$ of unit volume, we have

$L(P)\geq 2^{2/3}3^{11/6}\approx 11.8962,$

with equality if and only if $P$ is a regular triangle based right prism, and
every edge of $P$ is of length $\frac{2^{2/3}}{3^{1/6}}$.

We make the following conjecture.

######  Conjecture 1.

For any normal, convex mosaic $\mathcal{M}$ with unit volume cells, we have

$\rho_{1}(\mathcal{M})\geq\frac{3\sqrt{3}}{2}.$

The structure of our paper is as follows. In Section 1 we introduce the
concepts used in the paper and collect the necessary tools to prove our
results. In Section 2 we show how Theorem 2 implies Theorem 1. Finally, in
Sections 3 and 4 we prove Theorems 2 and 3, respectively.

##  1\. Preliminaries

Let $\mathcal{M}$ be a convex mosaic in $\mathbb{R}^{3}$. We define the _edge
density_ of $\mathcal{M}$ as the limit

$\rho_{1}(\mathcal{M})=\lim_{R\to\infty}\frac{L(\operatorname{skel}_{1}(\mathcal{M})\cap(R\mathbf{B}^{3}))}{\operatorname{vol}(R\mathbf{B}^{3})},$

if it exists, where $\mathbf{B}^{3}$ is the closed unit ball centered at the
origin $o$, the $1$-skeleton $\operatorname{skel}_{1}(\mathcal{M})$ of
$\mathcal{M}$ is defined as the union of all edges of all cells of
$\mathcal{M}$, $L(\cdot)$ denotes length, and $\operatorname{vol}(\cdot)$
denotes volume. We note that if $\mathcal{M}$ is normal, then
$\rho_{1}(\mathcal{M})$ exists.

It is shown in [20, 27] that an $n$-dimensional convex polytope $P$ is a
parallelohedron if and only if

  * (i)

$P$ and all its facets are centrally symmetric, and

  * (ii)

the projection of $P$ along any of its $(n-2)$-dimensional faces is a
parallelogram or a centrally symmetric hexagon.

It is worth noting that any affine image of a parallelohedron is also a
parallelohedron, and by [5], for any $n$-dimensional parallelohedron $P$ there
is a _face-to-face_ tiling of $\mathbb{R}^{n}$ with translates of $P$ as
cells.

The combinatorial classes of $3$-dimensional parallelohedra were described a
long time ago [6]. More specifically, any $3$-dimensional parallelohedron is
combinatorially isomorphic to one of the following (see Figure 1):

  1. (1)

a cube,

  2. (2)

a hexagonal prism,

  3. (3)

Kepler’s rhombic dodecahedron (which we call a regular rhombic dodecahedron),

  4. (4)

an elongated rhombic dodecahedron,

  5. (5)

a regular truncated octahedron.

We call a parallelohedron combinatorially isomorphic to the polyhedron in (i)
a _type (i) parallelohedron_.

Figure 1. The five combinatorial types of $3$-dimensional parallelohedra. The
type (5) parallelohedron in the picture is the regular truncated octahedron
generated by the six segments connecting the midpoints of opposite edges of a
cube. The type (3) polyhedron is the regular rhombic dodecahedron generated by
the four diagonals of the same cube. The rest of the polyhedra in the picture
are obtained by removing some generating segments from the type (5)
polyhedron.

Every $3$-dimensional parallelohedron is a zonotope; in particular, a type
(1)-(5) parallelohedron is the Minkowski sum of $3,4,4,5,6$ segments,
respectively. A typical parallelohedron is of type (5), as every other
parallelohedron can be obtained by removing some of the generating vectors of
a type (5) parallelohedron, or setting their lengths to zero.

By well-known properties of zonotopes (see e.g. [13, 18, 24]), every edge of a
$3$-dimensional parallelohedron $P$ is a translate of one of the segments
generating $P$. Furthermore, for any generating segment $S$, the faces $F_{i}$
of $P$ that contain a translate of $S$ can be written in a cyclic sequence
$F_{1},F_{2},\ldots,F_{s}=F_{0}$ such that for each $i$, $F_{i}\cap F_{i+1}$
is a translate of $S$; the union of these faces is called the _zone_ of $S$.
By property (ii) of parallelohedra, for any such zone the number of elements
is $4$ or $6$, implying that for any generating segment $S$ of $P$ there are
$4$ or $6$ translates of $S$ among the edges of $P$. In these cases we say
that $S$ generates a _$4$ -belt_ or a _$6$ -belt_, respectively. For
$i=1,2,3,4,5$, a type (i) parallelohedron contains $3,3,0,1,0$ $4$-belts, and
$0,1,4,4,6$ $6$-belts, respectively.

To prove Theorem 2 we adapt the representation of $3$-dimensional
parallelohedra introduced in [18]. More specifically, if $P$ is a
$3$-dimensional parallelohedron, then, up to translation, there are vectors
$v_{1},v_{2},v_{3},v_{4}\in\mathbb{R}^{3}$ such that

(1) $P=\sum_{1\leq i<j\leq 4}[o,\beta_{ij}(v_{i}\times v_{j})]$

for some non-negative real numbers $\beta_{ij}$. Here, if $P$ is a type (5)
parallelohedron, the vectors $v_{i}$ are normal vectors of the pairs of
hexagon faces. As it was shown in [18], we may assume that any three of the
$v_{i}$ are linearly independent, and satisfy $v_{1}+v_{2}+v_{3}+v_{4}=o$. We
may also assume that $V_{123}=1$, where $V_{ijk}$ denotes the value of the
determinant with columns $v_{i},v_{j}v_{k}$. Then, setting
$T=\operatorname{conv}\\{v_{1},v_{2},v_{3},v_{4}\\}$, $T$ is _centered_ , i.e.
its center of mass is $o$, implying that for any pairwise different indices
$i,j,k$, we have $|V_{ijk}|=1$. The sign of $V_{ijk}$ can be determined from
our choice that $V_{123}=1$, the properties of determinants, and from the
identity that for any $\\{i,j,s,t\\}=\\{1,2,3,4\\}$, we have
$V_{ijs}=-V_{ijt}$.

Every $3$-dimensional parallelohedron can be obtained from a type (5)
parallelohedron by setting some of the $\beta_{ij}$s to be zero. In
particular, $P$ is of type (5), or (4), or (3), or (2), or (1) if and only if

  * •

$\beta_{ij}>0$ for all $i\neq j$, or

  * •

exactly one of the $\beta_{ij}$s is zero, or

  * •

exactly two of the $\beta_{ij}$s is zero:
$\beta_{i_{1}j_{1}}=\beta_{i_{2},j_{2}}=0$, and they satisfy
$\\{i_{1},j_{1}\\}\cap\\{i_{2},j_{2}\\}=\emptyset$, or

  * •

exactly two of the $\beta_{ij}$s is zero:
$\beta_{i_{1}j_{1}}=\beta_{i_{2},j_{2}}=0$, and they satisfy
$\\{i_{1},j_{1}\\}\cap\\{i_{2},j_{2}\\}\neq\emptyset$, or

  * •

exactly three of the $\beta_{ij}$s are zero, and there is no
$s\in\\{1,2,3,4\\}$ such that the indices of all nonzero $\beta_{ij}$s contain
$s$, respectively.

In the remaining cases $P$ is planar.

The volume of a parallelohedron $P$ in the above representation is independent
of the $v_{i}$. More specifically,

(2)
$\operatorname{vol}_{3}(P)=f(\beta_{12},\beta_{13},\beta_{14},\beta_{23},\beta_{24},\beta_{34}),$

where the function $f:\mathbb{R}^{6}\to\mathbb{R}$ is defined as

(3)
$f(\tau_{12},\tau_{13},\tau_{14},\tau_{23},\tau_{24},\tau_{34})=\tau_{12}\tau_{13}\tau_{23}+\tau_{12}\tau_{14}\tau_{24}+\tau_{13}\tau_{14}\tau_{34}+\tau_{23}\tau_{24}\tau_{34}+\\\
+(\tau_{12}+\tau_{34})(\tau_{13}\tau_{24}+\tau_{14}\tau_{23})+(\tau_{13}+\tau_{24})(\tau_{12}\tau_{34}+\tau_{14}\tau_{23})+(\tau_{14}+\tau_{23})(\tau_{12}\tau_{34}+\tau_{13}\tau_{24}).$

We remark that $f$ is _not_ the third elementary symmetric function on six
variables, since it has only $16<20=\binom{6}{3}$ members.

Lemma 1 is proved in [18].

######  Lemma 1.

Let $T=\operatorname{conv}\\{p_{1},p_{2},p_{3},p_{4}\\}$ be an arbitrary
centered tetrahedron with volume $V>0$ where the vertices are labelled in such
a way that the determinant with columns $p_{1},p_{2},p_{3}$ is positive. For
any $\\{i,j,s,t\\}=\\{1,2,3,4\\}$, let $\gamma_{ij}=-\langle
p_{s},p_{t}\rangle$ and $\zeta_{ij}=\gamma_{ij}|p_{i}\times p_{j}|^{2}$. Then

  1. (1)

$f(\gamma_{12},\gamma_{13},\gamma_{14},\gamma_{23},\gamma_{24},\gamma_{34})=\frac{9}{4}V^{2}$,

  2. (2)

$\sum_{1\leq i<j\leq 4}\zeta_{ij}=\frac{27}{4}V^{2}$,

where $f$ is the function defined in (3).

To prove our result, we need the following variant of [18, Lemma 5]. Here, for
any $C>0$, we set
$\mathcal{X}_{C}=\left\\{(\tau_{12},\dots,\tau_{34})\in\mathbb{R}^{6}_{\geq
0}:\,\sum_{1\leq i<j\leq 4}\tau_{ij}=C\right\\}$.

######  Lemma 2.

Let $C>0$ and $\lambda\geq 1$. Then for each
$(\tau_{12},\tau_{13},\tau_{14},\tau_{23},\tau_{24},0)\in\mathcal{X}_{C}$ we
have

(4)
$f(\lambda\tau_{12},\tau_{13},\tau_{14},\tau_{23},\tau_{24},0)\leq\frac{16C^{3}\lambda^{3}}{27(4\lambda-1)^{2}},$

with equality if and only if

$\tau_{12}=\frac{4\lambda-3}{2\lambda}\tau_{13}=\ldots=\frac{4\lambda-3}{2\lambda}\tau_{24}=\frac{2\lambda}{12\lambda-3}C.$

###### Proof.

Without loss of generality, we assume that $C=1$ and for simplicity, in the
proof we denote the operator $\frac{\partial}{\partial\tau_{ij}}$ by
$\partial_{ij}$. Given that $\mathcal{X}_{C}$ is compact, the function $f$
attains its maximum on it at some point $(\tau^{*},0)\in\mathcal{X}_{C}$ with
$\tau^{*}=(\tau^{*}_{12},\ldots,\tau_{24}^{*})$.

_Case 1_ : all coordinates of $\tau^{*}$ are positive.
Then, by the Lagrange multiplier method, the gradient of $f$ at $\tau^{*}$ is
parallel to the gradient of the function $\sum_{i\neq j,ij\neq 34}\tau_{ij}-1$
defining the condition, thus, all five partial derivatives of $f$ at
$\tau^{*}$ are equal. From the equations
$(\partial_{13}f)(\tau^{*})=(\partial_{14}f)(\tau^{*})$ and
$(\partial_{23}f)(\tau^{*})=(\partial_{24}f)(\tau^{*})$ and by the assumption
that all coordinates of $\tau^{*}$ are positive, we obtain that
$\tau_{13}^{*}=\tau_{14}^{*}$ and $\tau_{23}^{*}=\tau_{24}^{*}$. Eliminating
$\tau_{14}^{*}$ and $\tau_{24}^{*}$ from the partial derivatives, the equation
$(\partial_{13}f)(\tau^{*})=(\partial_{23}f)(\tau^{*})$, and the fact that all
$\tau_{ij}^{*}$ as well as $\lambda$ are positive, yields that
$\tau_{13}^{*}=\tau_{23}^{*}$. Thus, we need to find the maximum of
$f(\lambda\tau_{12},\tau_{13},\ldots,\tau_{13},0)$ under the condition that
$\tau_{12}+4\tau_{13}=1$. From this, by simple calculus methods, we obtain
that
$f(\lambda\tau_{12},\tau_{13},\ldots,\tau_{13},0)\leq\frac{16\lambda^{3}}{27(4\lambda-1)^{2}}$,
with equality if and only if
$\tau_{12}=\frac{4\lambda-3}{2\lambda}\tau_{13}=\frac{2\lambda}{12\lambda-3}$.

_Case 2_ : $\tau_{12}^{*}=0$ and all other $\tau_{ij}^{*}$ are positive.
Note that
$f(0,\tau_{1}3,\ldots,\tau_{24},0)=(\tau_{13}+\tau_{24})\tau_{14}\tau_{23}+(\tau_{14}+\tau_{23})\tau_{13}\tau_{24}$.
In this case, by the same argument as in Case 1, we obtain that
$\tau_{13}^{*}=\ldots=\tau_{24}^{*}=\frac{1}{4}$, and
$f(\tau^{*})=\frac{1}{16}$.

_Case 3_ : $\tau_{ij}^{*}=0$ for some $ij\neq 12$, and all other
$\tau_{st}^{*}$ are positive.
Without loss of generality, we may assume that $\tau_{24}^{*}=0$. By a method
similar to the one in Case 3, we obtain that if $0<\lambda\leq\frac{1}{2}$,
there is no solution $\tau^{*}$ with positive coordinates, and if
$\lambda>\frac{1}{2}$, then
$\tau_{12}^{*}=\frac{2(2\lambda-1)}{3(4\lambda-1)}$,
$\tau_{13}^{*}=\tau_{14}^{*}=\frac{2\lambda}{3(4\lambda-1)}$, and
$\tau_{23}^{*}=0$, and $f(\tau^{*})=\frac{4\lambda^{2}}{27(4\lambda-1)}$.

_Case 4_ , at least two $\tau_{ij}^{*}$ are zero.
We may assume that exactly two $\tau_{ij}^{*}$ are zero, and that there is no
$s\in\\{1,2,3,4\\}$ such that the indices of all nonzero $\tau_{ij}^{*}$s
contain $s$, as otherwise $f(\tau^{*},0)=0$. Under this condition, depending
on which $\tau_{ij}^{*}$ are not zero, we have that
$f(\tau^{*})=\frac{\lambda}{27}$ or $f(\tau^{*})=\frac{1}{27}$, with equality
if and only if all nonzero $\tau_{ij}^{*}$ are equal to $\frac{1}{3}$.

As a summary, we need to find

$\max\left\\{\frac{16\lambda^{3}}{27(4\lambda-1)^{2}},\frac{1}{16},\frac{4\lambda^{2}}{27(4\lambda-1)},\frac{\lambda}{27},\frac{1}{27}\right\\}.$

Then an elementary computation shows that under the condition $\lambda\geq 1$,
the maximum is $\frac{16\lambda^{3}}{27(4\lambda-1)^{2}}$, and the assertion
follows. ∎

##  2\. The proof of Theorem 1

Theorem 1 readily follows from Theorem 2 by Lemma 3.

######  Lemma 3.

Let $P$ be a unit volume parallelohedron in $\mathbb{R}^{3}$, and let $M$ be a
tiling of $\mathbb{R}^{3}$ with translates of $P$. Set $m=(2,1)$. Then
$\rho_{1}(\mathcal{M})\geq w_{m}(P)$, with equality if $\mathcal{M}$ is face-
to-face.

###### Proof.

First, we assume that $M$ is face-to-face and that $P$ is a cell of $M$. Let
us choose an edge $E$ of $P$. Let $\mathcal{E}$ contain the edges $E^{\prime}$
of $\mathcal{M}$ for which there is a sequence of cells $P_{1},\ldots,P_{k}$
such that

  1. (i)

for any $i=1,2,\ldots,k-1$, $P_{i}\cap P_{i+1}$ is a face containing an edge
that is a translate of $E$;

  2. (ii)

$E$ is an edge of $P_{1}$ and $E^{\prime}$ is an edge of $P_{k}$.

Furthermore, let $\mathcal{P}$ be the family of cells containing edges from
$\mathcal{E}$.

Note that the projection of the members of $\mathcal{P}$ in the direction of
$E$ onto a plane $H$ is a translative, convex, edge-to-edge tiling of $H$. By
the second property of parallelohedra in the list in the beginning of Section
1, the projections of the elements of $\mathcal{P}$ are either parallelograms,
in which every vertex belongs to exactly four parallelograms, or they are
centrally symmetric hexagons, in which every vertex belongs to three hexagons.
Thus, a cell contains $4$ or $6$ translates of any given edge, and in the
first case every edge belongs to exactly $4$ cells, and in the second case
every edge belongs to exactly $3$ cells.

Now, we assign a weight $w_{E}$ to each edge $E$ of $M$; we set
$w(E)=\frac{1}{k}$, where $k$ is the number of cells $E$ belongs to. Let
$N_{r}$ denote the number of cells of $\mathcal{M}$, and $\mathcal{E}_{r}$
denote the family of the edges of the cells contained in $r\mathbf{B}^{3}$. As
$N_{r}=\operatorname{vol}_{3}(r\mathbf{B}^{3})+\mathcal{O}(r^{2})$, we have

$L(\operatorname{skel}_{1}(\mathcal{M})\cap(r\mathbf{B}^{3}))=\sum_{E\in\mathcal{E}_{r}}L(E)+\mathcal{O}(r^{2})=N_{r}\sum_{i=1}^{6}w_{i}|p_{i}|+\mathcal{O}(r^{2}),$

implying Lemma 3 for face-to-face mosaics.

Finally, if $\mathcal{M}$ is a not necessarily face-to-face tiling by
translates of $P$, then we may divide each edge into finitely many pieces in
such a way that relative interior points on each piece belong to the same
cells. It is easy to see that any part of an edge in a $6$-belt belongs to
exactly $3$ cells, and a part of an edge in a $4$-belt belongs to either $4$
or $2$ cells. Thus, repeating the procedure, the weighted sum of the edge
lengths in $r\mathbf{B}^{3}$, defined as in the previous case, yields a value
for $\rho_{1}(M)$ asymptotically not less than that for a face-to-face mosaic
generated by $P$. ∎

##  3\. The proof of Theorem 2

First, for each value of $i$ and $m=(\alpha_{4},\alpha_{6})$, we find the
infimum of $w_{m}(P)$ of the type (i) parallelohedra for each $i$ separately,
which we denote by $w_{i}(\alpha_{4},\alpha_{6})$, then compare these values.

Case 1: $P$ is a type (1) parallelohedron.
Then $P$ is a parallelopiped, and it is easy to see that $w_{m}(P)$ is minimal
if and only if $P$ is a unit cube, and in this case $w_{m}(P)=3\alpha_{4}$.
Thus, $w_{1}(\alpha_{4},\alpha_{6})=3\alpha_{4}$.

Case 2: $P$ is a type (2) parallelohedron.
Then $P$ is a centrally symmetric hexagon-based prism. Clearly, we can assume
that $P$ is a right prism, and by the Discrete Isoperimetric Inequality, we
may assume that its base is a regular hexagon. Let the edge length of the base
be $a$, and the length of the lateral edges be $b$. Then
$\operatorname{vol}_{3}(P)=\frac{3\sqrt{3}}{2}a^{2}b$, and
$w_{m}(P)=3\alpha_{4}a+\alpha_{6}b$. An elementary computation shows that if
$\frac{3\sqrt{3}}{2}a^{2}b=1$, then
$w_{m}(P)\geq\frac{3^{7/6}\alpha_{4}^{2/3}\alpha_{6}^{1/3}}{2^{1/3}}$, with
equality if and only if
$a=\frac{2^{2/3}\alpha_{6}^{1/3}}{3^{5/6}\alpha_{4}^{1/3}}$ and
$b=\frac{3^{1/6}\alpha_{4}^{2/3}}{2^{1/3}\alpha_{6}^{2/3}}$. Hence,
$w_{2}(\alpha_{4},\alpha_{6})=\frac{3^{7/6}\alpha_{4}^{2/3}\alpha_{6}^{1/3}}{2^{1/3}}$.

Case 3: $P$ is a type (3) parallelohedron.
In this case $P$ is generated by $4$ segments, and each segment generates a
$6$-belt. Thus, $w_{m}(P)=\frac{\alpha_{6}}{2}w(P)$, where $w(P)$ denotes the
mean width of $P$. This means that we may apply the proof of [18, Theorem 1]
(see also Table 1 of [18]), and obtain that $w_{m}(P)\geq
3^{1/2}2^{2/3}\alpha_{6}$, with equality if and only if $P$ is a regular
rhombic dodecahedron, implying, in particular, that
$w_{3}(\alpha_{4},\alpha_{6})=3^{1/2}2^{2/3}\alpha_{6}$.

Case 4: $P$ is a type (4) parallelohedron.

First, note that if $\alpha_{4}>\alpha_{6}$, then
$w_{4}(\alpha_{4},\alpha_{6})\geq
w_{4}(\alpha_{6},\alpha_{6},)>w_{5}(\alpha_{6},\alpha_{6})$, where the last
inequality follows from [18, Theorem 1]. In the following we assume that
$\alpha_{4}\leq\alpha_{6}$.

To prove Theorem 2 in this case, we use the notation introduced in Section 1:
we represent $P$ in the form

$P=\sum_{1\leq i<j\leq 4}[o,\beta_{ij}v_{i}\times v_{j}]$

for some $v_{1},v_{2},v_{3},v_{4}\in\mathbb{R}^{3}$ satisfying
$\sum_{i=1}^{4}v_{i}=o$, and for some $\beta_{ij}\geq 0$, with exactly one
$\beta_{ij}$ equal to zero. By our assumptions, $|V_{ijk}|=1$ for any
$\\{i,j,k\\}\subset\\{1,2,3,4\\}$, where $V_{ijk}$ denotes the determinant
with $v_{i},v_{j},v_{k}$ as columns, and we assumed that $V_{123}=1$, which
implies, in particular, that $V_{124}=-1$. Then
$T=\operatorname{conv}\\{v_{1},v_{2},v_{3},v_{4}\\}$ is a centered
tetrahedron, with volume $\operatorname{vol}_{3}(T)=\frac{2}{3}$.

Without loss of generality, we set $\beta_{34}=0$. Then, among the edges of
$P$, the translates of $\beta_{12}v_{1}\times v_{2}$ belong to a $4$-belt, and
the translates of all other edges belong to $6$-belts. Thus, we have

$w_{m}(P)=\alpha_{4}\beta_{12}|v_{1}\times
v_{2}|+\alpha_{6}\sum_{i\in\\{1,2\\},j\in\\{3,4\\}}\beta_{ij}|v_{i}\times
v_{j}|,\hbox{ and}$
$\operatorname{vol}_{3}(P)=f(\beta_{12},\beta_{13},\ldots,\beta_{24},0).$

We follow the two steps of the proof of [18, Theorem 1]. Our first step is to
find the minimum of  $w_{m}(P)$ in the affine class of $P$. For this, we use
Lemma 4, proved by Petty [22] (see also (3.4) in [11] for arbitrary convex
bodies).

######  Lemma 4.

Let $P\subset\mathbb{R}^{n}$ be a convex polytope with outer unit facet
normals $u_{1},\ldots,u_{k}$. Let $F_{i}$ denote the $(n-1)$-dimensional
volume of the $i$th facet of $P$. Then, up to congruence, there is a unique
volume preserving affine transformation $A$ such that the surface area
$\operatorname{surf}(A(P))$ is maximal in the affine class of $P$.
Furthermore, $P$ satisfies this property if and only if its surface area
measure is isotropic, that is, if

(5) $\sum_{i=1}^{k}\frac{nF_{i}}{\operatorname{surf}(P)}u_{i}\otimes
u_{i}=\operatorname{Id}$

where $\operatorname{Id}$ denotes the identity matrix.

Any convex polytope satisfying the conditions in (5) is said to be in _surface
isotropic position_. We note that the volume of the projection body of any
convex polyhedron is invariant under volume preserving linear transformations
(cf. [23]). On the other hand, from Cauchy’s projection formula and the
additivity of the support function (see [10]) it follows that the projection
body of the polytope in Lemma 4 is the zonotope
$\sum_{i_{1}}^{k}[o,F_{i}u_{i}]$. Note that the solution to Minkowski’s
problem [24] states that some unit vectors
$u_{1},\ldots,u_{k}\in\mathbb{R}^{n}$ and positive numbers
$F_{1},\ldots,F_{k}$ are the outer unit normals and volumes of the facets of a
convex polytope if and only if the $u_{i}$s span $\mathbb{R}^{n}$, and
$\sum_{i=1}^{k}F_{i}u_{i}=o$. This yields that there is an $o$-symmetric
convex polyhedron $Q$ whose faces have outer unit normals
$\pm\frac{v_{1}\times v_{2}}{|v_{1}\times v_{2}|}$, and $\pm\frac{v_{i}\times
v_{j}}{|v_{i}\times v_{j}|}$ with $ij=13,23,14,24$, and these faces have area
$\frac{1}{2}\alpha_{4}\beta_{12}|v_{1}\times v_{2}|$ and
$\frac{1}{2}\alpha_{6}\beta_{ij}|v_{i}\times v_{j}|$ with $ij=13,14,23,24$,
respectively. Since $u\otimes u=(-u)\otimes(-u)$ for any $u\in\mathbb{R}^{n}$,
we may apply Lemma 4 for $Q$, and obtain that

(6) $\frac{\alpha_{4}\beta_{12}}{|v_{1}\times v_{2}|}(v_{1}\times
v_{2})\otimes(v_{1}\times
v_{2})+\alpha_{6}\sum_{i\in\\{1,2\\},j\in\\{3,4\\}}\frac{2\beta_{ij}}{|v_{i}\times
v_{j}|}(v_{i}\times v_{j})\otimes(v_{i}\times
v_{j})=\frac{w_{m}(P)}{3}\operatorname{Id},$

which in the following we assume to hold for $P$.

Recall that $x\otimes y=xy^{T}$ and $\langle x,y\rangle=x^{T}y$ for any column
vectors $x,y\in\mathbb{R}^{n}$. Hence, multiplying both sides in (6) by
$v_{3}^{T}$ from the left and $v_{4}$ from the right, it follows that

$\langle v_{3},v_{4}\rangle=\frac{3\alpha_{4}\beta_{12}}{w_{m}(P)|v_{1}\times
v_{2}|}V_{123}V_{124}.$

Since $V_{123}=-V_{124}=1$, we have

$\beta_{12}=-\langle v_{3},v_{4}\rangle|v_{1}\times
v_{2}|\cdot\frac{w_{m}(P)}{3\alpha_{4}}.$

Multiplying both sides in (6) with some other $v_{i}$s, we obtain that for any
$i\in\\{1,2\\}$ and $j\in\\{3,4\\}$, and with the notation
$\\{i,j,s,t\\}=\\{1,2,3,4\\}$, we have

$\beta_{ij}=-\langle v_{s},v_{t}\rangle|v_{i}\times
v_{j}|\cdot\frac{w_{m}(P)}{3\alpha_{6}},$

and obtain similarly that $\langle v_{1},v_{2}\rangle=0$.

Now we set $\zeta_{ij}=-\langle v_{s},v_{t}\rangle|v_{i}\times v_{j}|$ for all
$\\{i,j,s,t\\}=\\{1,2,3,4\\}$, and note that
$\beta_{12}=\frac{w_{m}(P)\zeta_{12}}{3\alpha_{4}}$,
$\beta_{34}=\zeta_{34}=0$, and for any $ij\neq 12,34$,
$\beta_{ij}=\frac{w_{m}(P)\zeta_{ij}}{3\alpha_{6}}$. Substituting back these
quantities into the formulas for $\operatorname{vol}_{3}(P)$ and $w_{m}(P)$
and simplifying, we may rewrite our optimization problem in the following
form: find the maximum of
$\frac{27\operatorname{vol}_{3}(P)}{\left(w_{m}(P)\right)^{3}}=f\left(\frac{\zeta_{12}}{\alpha_{4}},\frac{\zeta_{13}}{\alpha_{6}},\ldots,\frac{\zeta_{34}}{\alpha_{6}},0\right)$,
where $\zeta_{ij}$ is defined as above, in the family of all centered
tetrahedra $T$ with $\operatorname{vol}_{3}(T)=\frac{2}{3}$, under the
condition that $\sum_{1\leq i<j\leq 4}\zeta_{ij}|v_{i}\times v_{j}|=3$. Here,
it is worth noting that the last condition is satisfied for any centered
tetrahedron $T$ with volume $\frac{2}{3}$ by Lemma 1, and thus, it is
redundant. Note that since $f$ is $3$-homogeneous, we have

$f\left(\frac{\zeta_{12}}{\alpha_{4}},\frac{\zeta_{13}}{\alpha_{6}}\ldots,\frac{\zeta_{34}}{\alpha_{4}},0\right)=\frac{1}{\alpha_{6}^{3}}f\left(\mu\zeta_{12},\zeta_{13},\ldots,\zeta_{34},0\right),$

where $\mu=\frac{\alpha_{6}}{\alpha_{4}}\geq 1$.

In the second step we give an upper bound for the values of this function. Set
$\gamma_{ij}=-\langle v_{s},v_{t}\rangle$ and
$\tau_{ij}=\gamma_{ij}|v_{i}\times v_{j}|^{2}$ for all
$\\{i,j,s,t\\}=\\{1,2,3,4\\}$, and note that this yields
$\gamma_{34}=\tau_{34}=0$. To find an upper bound, we apply the Cauchy-
Schwartz Inequality, which states that for any real numbers $x_{i},y_{i}$,
$i=1,2,\ldots,k$, we have
$\sum_{i=1}^{k}x_{i}y_{i}\leq\sqrt{\sum_{i=1}^{k}x_{i}^{2}}\sqrt{\sum_{i=1}^{k}y_{i}^{2}}$,
with equality if and only if the $x_{i}$s and the $y_{i}$s are proportional.
To do this, we write each member of $f$ not containing $\zeta_{12}$ as the
product
$\zeta_{ij}\zeta_{kl}\zeta_{mn}=\sqrt{\gamma_{ij}\gamma_{kl}\gamma_{mn}}\sqrt{\tau_{ij}\tau_{kl}\tau_{mn}}$,
and each member containing $\zeta_{12}$ as
$\mu\zeta_{12}\zeta_{kl}\zeta_{mn}=\sqrt{\gamma_{12}\gamma_{kl}\gamma_{mn}}\sqrt{\mu^{2}\tau_{12}\tau_{kl}\tau_{mn}}$.
Thus, we obtain that

$f(\mu\zeta_{12},\zeta_{13},\ldots,\zeta_{24},0)\leq\sqrt{f(\mu^{2}\tau_{12},\tau_{13},\tau_{14},\tau_{23},\tau_{24},0)}\cdot\sqrt{f(\gamma_{12},\gamma_{13},\gamma_{14},\gamma_{23},\gamma_{24},\gamma_{34})}.$

Here we use the fact that by Lemma 1, $f(\gamma_{12},\ldots,\gamma_{34})=1$
for all centered tetrahedra with volume $\frac{2}{3}$. Furthermore, observe
that by Lemma 1 we have $\sum_{1\leq i\leq j\leq 4}\tau_{ij}=3$ for all such
tetrahedra. Hence, by Lemma 2, we have

$f(\mu^{2}\tau_{12},\tau_{13},\ldots,\tau_{24},0)\leq\frac{16\mu^{6}}{(4\mu^{2}-1)^{2}},$

implying that if $\operatorname{vol}_{3}(P)=1$, then

$w_{m}(P)\geq\frac{3\alpha_{4}^{1/3}(4\alpha_{6}^{2}-\alpha_{4}^{2})^{1/3}}{2^{2/3}}.$

Consequently, we have

(7)
$w_{4}(\alpha_{4},\alpha_{6})\geq\left\\{\begin{array}[]{l}w_{5}(\alpha_{6},\alpha_{6}),\hbox{
if }\alpha_{4}>\alpha_{6};\\\
\frac{3\alpha_{4}^{1/3}(4\alpha_{6}^{2}-\alpha_{4}^{2})^{1/3}}{2^{2/3}},\hbox{
if }\alpha_{4}\leq\alpha_{6},\end{array}\right.$

and equality in the first case does not occur.

Case 5: $P$ is a type (5) parallelohedron. In this case every generating
segment generates a $6$-belt, and we can apply [18, Theorem 1] directly, and
obtain that as $\operatorname{vol}_{3}(P)=1$,

$w_{m}(P)\geq\frac{3}{2^{1/6}}\alpha_{6},$

with equality if and only if $P$ is a regular truncated octahedron.

Table 1 collects our lower bounds for $w_{m}(P)$ for the different types of
parallelohedra. Note that if $i\neq 4$, $w_{i}(\alpha_{4},\alpha_{6})$
coincides with the expression in the $i$th row and second column of Table 1,
whereas $w_{4}(\alpha_{4},\alpha_{6})$ is not less than the quantity in the
$4$h row and the second column.

Type |  $w_{i}(\alpha_{4},\alpha_{6})$ |  Optimal parallelohedra
---|---|---
(1) |  $3\alpha_{4}$ |  unit cube
(2) |  $\frac{3^{7/6}}{2^{1/3}}\alpha_{4}^{2/3}\alpha_{6}^{1/3}$ |  regular hexagon based right prism, with base and lateral edges of lengths $\frac{2^{2/3}\alpha_{6}^{1/3}}{3^{5/6}\alpha_{4}^{1/3}}$ and $\frac{3^{1/6}\alpha_{4}^{2/3}}{2^{1/3}\alpha_{6}^{2/3}}$, respectively
(3) |  $2^{2/3}\cdot 3^{1/2}\alpha_{6}$ |  regular rhombic dodecahedron with edge length $\frac{\sqrt{3}}{2^{4/3}}$
(4) |  $\geq\frac{3\alpha_{4}^{1/3}(4\alpha_{6}^{2}-\alpha_{4}^{2})^{1/3}}{2^{2/3}}$, or $>w_{5}(\alpha_{6},\alpha_{6})$ |  not known
(5) |  $\frac{3}{2^{1/6}}\alpha_{6}$ |  regular truncated octahedron of edge length $\frac{1}{2^{7/6}}$
Table 1. Minima of the weighted mean widths of different types of
parallelohedra

In the final step we determine the minimum of
$\\{w_{i}(\alpha_{4},\alpha_{6}):i=1,2,3,4,5\\}$ as a function of $\alpha_{4}$
and $\alpha_{6}$.

Figure 2. The lower bounds for $w_{i}(\alpha_{4},\alpha_{6})$ from the second
column of Table 1, with $\alpha_{6}=1$, and $0<\alpha_{4}\leq\alpha_{6}$.

An elementary computation shows that
$w_{3}(\alpha_{4},\alpha_{6})>w_{5}(\alpha_{4},\alpha_{6})$ for all
$\alpha_{4},\alpha_{6}>0$, and that
$w_{4}(\alpha_{4},\alpha_{6})\geq\frac{3\alpha_{4}^{1/3}(4\alpha_{6}^{2}-\alpha_{4}^{2})^{1/3}}{2^{2/3}}>\min\\{w_{1}(\alpha_{4},\alpha_{6}),w_{5}(\alpha_{4},\alpha_{6})\\}$
if $0<\alpha_{4}\leq\alpha_{6}$. Furthermore, if $\alpha_{4}>\alpha_{6}$, then
$w_{4}(\alpha_{4},\alpha_{6})>w_{5}(\alpha_{6},\alpha_{6})=w_{5}(\alpha_{4},\alpha_{6})$.
Thus, we need to compare only the values of $w_{1}(\alpha_{4},\alpha_{6})$,
$w_{2}(\alpha_{4},\alpha_{6})$ and $w_{5}(\alpha_{4},\alpha_{6})$, and from
this another elementary computation proves the statement.

##  4\. Proof of Theorem 3

Let $\mathcal{M}\in\mathcal{F}_{dc}$. Then $\mathcal{M}$ can be decomposed
into a planar mosaic $\mathcal{M}_{1}$ in a plane $H$ and a linear mosaic
$\mathcal{M}_{2}$ in a line $L$. Since every cell in $\mathcal{M}$ has volume
one, $L$ is decomposed into congruent segments, and $H$ is tiled with equal
area convex polygons. To minimize the edge density of $\mathcal{M}$, we may
assume that $L$ is orthogonal to $H$. Let the length of the cells in
$\mathcal{M}_{2}$ be denoted by $\ell$. Since $\mathcal{M}$ has unit volume
cells, the area of the cells of $\mathcal{M}_{1}$ is $\frac{1}{\ell}$. Since
$\mathcal{M}$ is normal, it is easy to see that $\mathcal{M}_{1}$ is normal.
We permit the cells of $\mathcal{M}_{1}$ to have straight angles, and in this
way we can regard $\mathcal{M}_{1}$ as an edge-to-edge mosaic.

In the following we assume that $H$ coincides with the $(x,y)$-coordinate
plane, and $L$ coincides with the $z$-axis. Note that by the symmetry of the
mosaic, the edge density of $\mathcal{M}$ is equal to the edge density of the
part of $\mathcal{M}$ in the infinite, half open strip $C=\\{0\leq z<m\\}$.

Consider the half-open right cylinder
$C(R)=\\{(x,y,z)\in\mathbb{R}^{3}:x^{2}+y^{2}\leq R^{2},0\leq z<\ell\\}$ of
volume $R^{2}\pi\ell$. We estimate the total length of the edges of
$\mathcal{M}$ in $C(R)$. Let $k$ and $N$ denote the numbers of cells and nodes
of $\mathcal{M}_{1}$ in $C(R)$, respectively. Let the cells of
$\mathcal{M}_{0}$ in $C(R)$ be $F_{1},F_{2},\ldots,F_{k}$, and let $v_{i}$ be
the number of the edges of $F_{i}$. Let $S$ denote the sum of all interior
angles of all cells of $\mathcal{M}_{1}$ in $C(R)\cap H$. Then, on one hand,

$S=2\pi N+\mathcal{O}(R),$

and on the other hand, it is

$\sum_{i=1}^{k}(v_{i}-2)\pi+\mathcal{O}(R).$

Since $\operatorname{area}(F_{i})=\frac{1}{\ell}$ for all $i$,
$k=R^{2}\pi\ell+\mathcal{O}(R)$. This yields that

$N=\frac{R^{2}\pi\ell}{2}\left(\frac{\sum_{i=1}^{k}v_{i}}{k}-2\right)+\mathcal{O}(R).$

Thus, the total edge length of $\mathcal{M}$ in $C(R)$ is

$L(\operatorname{skel}_{1}(\mathcal{M}_{1})\cap C(R))=\ell
N+\frac{1}{2}\sum_{i=1}^{k}\operatorname{perim}(F_{i})+\mathcal{O}(R),$

where $\operatorname{perim}(\cdot)$ denotes perimeter. Since
$\operatorname{vol}_{3}(C(R))=R^{2}\pi\ell$, we have that

$\rho_{1}(\mathcal{M})=\frac{\ell}{2}(v-2)+\frac{1}{2}\lim_{R\to\infty}\frac{\sum_{i=1}^{k}\operatorname{perim}(F_{i})}{k},$

where $v=\lim_{R\to\infty}\frac{\sum_{i=1}^{k}}{k}$ is the average number of
edges of a cell of $\mathcal{M}$, which exists, since $\mathcal{M}$ is normal.

By the Discrete Isoperimetric Inequality, $\operatorname{perim}(F_{i})\geq
2\sqrt{v_{i}\tan\frac{\pi}{v_{i}}}\cdot\frac{1}{\sqrt{\ell}}$; here the right-
hand side is the perimeter of the regular $v_{i}$-gon of area
$\frac{1}{\ell}$. By the convexity of the function
$x\mapsto\sqrt{x\tan\frac{\pi}{x}}$ on the interval $[3,\infty)$, we have

(8)
$\rho_{1}(\mathcal{M})=\frac{\ell}{2}(v-2)+\sqrt{v\tan\frac{\pi}{v}}\cdot\frac{1}{\sqrt{\ell}}.$

For any $3\leq v\leq 6$, the expression on the right-hand side of (8) is
minimal if $\ell=\frac{\sqrt[3]{v\tan\frac{\pi}{v}}}{(v-2)^{2/3}}$.
Substituting it back into (8), we obtain that

$\rho_{1}(\mathcal{M})\geq\frac{3}{2}\left(v(v-2)\tan\frac{\pi}{v}\right)^{1/3}.$

Then the assertion follows from the facts that the expression on the right-
hand side is strictly increasing on the interval $[3,6]$, and that an average
cell in a planar mosaic has at most $6$ edges.

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Copyright for this paper by its authors. Use permitted under Creative Commons
License Attribution 4.0 International (CC BY 4.0).

COMHUM 2022

[orcid=0000-0002-0877-7063<EMAIL_ADDRESS>[orcid=0000-0002-2619-2835<EMAIL_ADDRESS>[orcid=0000-0003-1203-9108<EMAIL_ADDRESS>
# Data Augmentation for Robust Character Detection in Fantasy Novels

Arthur Amalvy  Vincent Labatut  Richard Dufour  Laboratoire d’Informatique
d’Avignon (LIA), Avignon University, France  Laboratoire des Sciences du
Numérique de Nantes (LS2N), Nantes University, France

(2022)

###### Abstract

Named Entity Recognition (NER) is a low-level task often used as a foundation
for solving higher level NLP problems. In the context of character detection
in novels, NER false negatives can be an issue as they possibly imply missing
certain characters or relationships completely. In this article, we
demonstrate that applying a straightforward data augmentation technique allows
training a model achieving higher recall, at the cost of a certain amount of
precision regarding ambiguous entities. We show that this decrease in
precision can be mitigated by giving the model more local context, which
resolves some of the ambiguities.

###### keywords:

data augmentation named entity recognition character detection

##  1 Introduction

The Character Detection task is concerned with detecting which characters
appear in a text, and where. We decompose it in two subtasks:

  * •

Named Entity Recognition (NER), whose goal is to find character occurrences in
a text.

  * •

Named Entity Disambiguation (NED), which maps character occurrences to their
respective normalized character form.

The result of the character detection task can be used to solve other higher-
level problems. Since its output can affect the performance of other tasks
relying on it, such as character network extraction [1], its degree of success
is of importance.

Dekker et al. [2] perform a study where they evaluate several NER models for
the purpose of extracting character networks from literary texts. According to
their results, performance varies greatly across novels, ranging from great to
pretty low. These authors find that quite a few of the false negatives
produced by the evaluated models stem from two specific types of names: word
names (names that are also regular words, such as Valor or Mercy) and
apostrophed names (such as Rand al’Thor). Replacing them with regular names
allows for better detection, showing that the issue regarding these names is
linked to their form and not to their surrounding context.

We suspect that these shortcomings come from the datasets used to train the
NER systems assessed by Dekker et al. [2]. Indeed, NER systems are usually
trained on the main publicly available datasets (such as Ontonotes [3],
CoNLL-2003 [4] or WikiGold [5]), which come from a few domains (news, web…).
Among these mainstream datasets, none contain texts of the literary
domain111While the Litbank dataset [6] specifically contains only literary
texts, its annotations concern nested NER, an arguably harder task., and
annotating new datasets is costly. This means applying off-the-shelf NER
systems to literary texts suffers from these systems integrating knowledge
specific only to their training data.

More precisely, there may be a mismatch in style between typical person names
from the training dataset and from literary texts. This is particularly the
case in the fantasy genre, where character names can have very unusual styles
depending on the setting.

In order to try and fix this issue, we showcase the application of a specific
data augmentation technique, mention replacement [7], to inject new character
names in the training dataset. Using this technique, we demonstrate
performance improvements for the NER task on a dataset of texts from the
fantasy genre. We release our code and data, that can be used to reproduce our
results, under a free license222https://github.com/CompNet/ddaugNER.

The rest of this article is organized as follows: first, Section 2 provides a
few examples of related data augmentation works. Then, in Section 3, we
present mention replacement and the experiments we perform to assess its
ability to resolve the name mismatch issue we highlighted. We discuss the
results of these experiments in Section 4, and perform further analysis to
shed light on the precision decrease we observe when using mention
replacement. Finally, we conclude with some perspectives regarding NER
performance in literary texts.

##  2 Data Augmentation and Cross-Domain NER

Generally speaking, data augmentation is the generation of new synthetic
training examples. Data augmentation techniques can be used to address data
scarcity problems, or to increase the diversity of the training dataset to
reduce overfitting. While these techniques are ubiquitous in image processing
[8], they are less commonly explored in natural language processing [9], and
even less for NER [7].

A few techniques have been used to try and fix the domain discrepancy between
training and testing datasets. Ding et al. [10] train a language model to
generate new examples by directly including NER tags in the generated text.
Chen et al. [11] train a neural architecture to transform examples from the
training domain to examples closer to the test domain. Very recently, Yang et
al. [12] propose FactMix, a two-step data augmentation process performing
mention replacement followed by masked token replacement. However, as far as
we know, previous works are not specifically concerned with literary texts or
the mismatch in person names between domains.

##  3 Method

In this section, we first describe mention replacement and how we apply it to
texts of the fantasy genre. We then introduce the datasets we use to assess
the performance of this data augmentation technique. Finally, we describe the
specific setting of the experiment we carry to evaluate its performance,
including the model we used and its training parameters.

###  3.1 Mention Replacement

Figure 1: Example of mention replacement. Note that in this case, the sentence
labels have to be slightly modified to account for the replacement entity
higher number of tokens.

Dai and Adel [7] survey a few simple data augmentation schemes for NER,
including mention replacement. It consists in generating new examples by
replacing tokens from an entity mention found in the training dataset by
tokens from another entity of the same type. Figure 2 shows an example of this
process. Based on the same principle, we propose to randomly replace training
entities with ones from a list of fantasy names to increase the training
dataset coverage of this type of names. If the same entity mention is present
multiple times in a training example, we replace every occurrence with the
same fantasy mention when performing augmentation to avoid inconsistencies. We
hope that injecting typical fantasy names into the training dataset using
mention replacement will allow the model to better detect them.

To perform mention replacement, we compose a list of replacement entities that
do not come directly from the evaluation dataset. We scrap the entirety of the
names from The Elder Scrolls series of video games that are mentioned on the
Unofficial Elder Scrolls Pages, a wiki dedicated to The Elder Scrolls
universe333https://en.uesp.net/wiki/Main_Page. In total, we retrieve 22,748
first names, 4,879 last names, 647 suffixes and 37 prefixes. We combine these
to form new mentions: when performing mention replacement, we randomly compose
a new mention by first sampling from a set of valid forms ([first name]+[last
name], [prefix]+[first name]+[last name], …). We weight this sampling by a
rough approximation of each form’s frequency. When a form has been selected,
we uniformly sample a name part for each of the form’s elements.

Since we are interested in knowing the impact of the number of generated
examples, we compare the performance of models trained with different
augmentation rates. We define the augmentation rate as the ratio of examples
generated over the dataset size, so an augmentation rate of 1 would mean
generating as many examples as there are examples in the dataset.

###  3.2 Datasets

#### Training Dataset

We train our model on a modified version of CoNLL-2003 [4], which is one of
the best known and used NER dataset. CoNLL-2003 is composed of 14,041
sentences from news articles. Our modifications consist in including
honorifics as part of entities to be consistent with our evaluation dataset
(see below). The training dataset contains annotations for four classes:
persons (PER), organizations (ORG), locations (LOC) and miscellaneous (MISC).
Despite restricting ourselves to character detection (PER class), we keep
annotations for every class since we observe that training the model with all
classes increase the performance of NER.

#### Evaluation dataset

We use a corrected version of the subset of fantasy novels from the dataset of
Dekker et al. [2], where only persons entities (PER) are annotated. We found
the dataset had to be corrected because we noticed a number of encoding,
tokenization and annotation issues. We fixed the encoding and tokenization
issues manually. In order to consistently correct annotation errors, we
designed an annotation guide and applied a semi-automated correction process.
It consists of 3 steps:

  1. 1.

We apply a number of simple heuristics to identify obvious errors:

     * •

When a span is not annotated as a person occurrence, but it appears in the
list of character names for the book, it might be a false negative.

     * •

When a span is annotated as a person occurrence, but it does not appear in
this list, it might a false positive.

     * •

When a span is annotated as a person occurrence, but the first letter of all
of its tokens is not capitalized, it might be a false positive.

Each time one of these heuristics recommends an annotation change, we manually
check if it is correct before accepting it.

  2. 2.

We then consider the differences between the existing annotations and the
predictions of a BERT model [13] fine-tuned for the NER task, and manually
reconcile them.

  3. 3.

Finally, we manually correct the few remaining errors that we could detect.

| Exact | Partial |
---|---|---|---
Name Set | Match | Match | Unseen
CoNLL-2003 (train) | 9.62% | 76.97% | 13.41%
The Elder Scrolls | 12.25% | 76.38% | 11.37%
Table 1: Overlap of different name sets with the set of character names from
our evaluation dataset. The "exact match" column reports the proportion of PER
labeled tokens from our evaluation dataset that are also in the studied name
set. The "partial match" column is the proportion of PER labeled tokens
containing a wordpiece that exists in the studied name set. Finally, the
"unseen" column indicates the rest of PER labeled tokens from the evaluation
dataset.

The dataset consists in the first chapter of 17 novels. Together, these
chapters are composed of 5,518 sentences and contain 360 unique person names.
Inspired by Taillé et al. [14], we report in Table 1 the name overlap between
our evaluation dataset and:

  * •

The train portion of the CoNLL-2003 [4];

  * •

Our list of names from The Elder Scrolls, used for mention replacement.

###  3.3 Experiment

We fine-tune a BERT model [13] with an added classification head on our
training dataset, and report its performance on our evaluation dataset
depending on the augmentation rate. In order to obtain stable results, we
report the mean of the results from 10 fine-tuning runs for each augmentation
rate that we test. Models without augmentations are trained for two full
epochs, while we adjust the number of training steps for model with
augmentation so that they learn on the same number of examples as models
without augmentation. We use a learning rate of $2\cdot 10^{-5}$ [13], and we
initialize each model with the bert-base-cased checkpoint of the transformers
library [15]. At inference time, each model predicts NER tags separately for
each sentence, and we do not provide it with any additional context. Since the
evaluation dataset only contains annotations for the PER class, we simply
discard predictions made for other classes.

##  4 Results

Vin | Camon | Bilbo | Pug | Arlen
---|---|---|---|---
Galladon | Bran | Selia | Bug | Cob
Raoden | Gandalf | Szeth | Weasel | Chivalry
Logen | Theron | Frodo | Jory | Jezrien
Table 2: Top 20 character names that were better detected by a model trained
with data augmentation (10% augmentation rate) but not by a model trained with
the original dataset only. Figure 2: Precision, recall and F1-score against
augmentation rate. Error bars represent the 95% confidence interval.

Results can be seen in Figure 2. We report performance according to the
CoNLL-2003 guidelines [4] and using the Python seqeval library [16]. We
observe a notable increase in recall using augmentation, while precision
decreases. The recall increase can be attributed to the model picking up on
more unusual character names, as Table 2 shows.

To try to understand the decrease in precision when using data augmentation,
we perform some additional experiments. Based on our manual investigation of
the false positives, we formulate two hypotheses that we want to test:

  * $h_{1}$

Adding more examples with PER entities modifies the class distribution in the
dataset, progressively leading to a situation of class imbalance where these
entities are way more frequent than those of the other classes (LOC, ORG and
MISC). As we observe that removing other entity classes from the training
dataset is detrimental to PER detection performance (possibly because having
more classes means the model is better at distinguishing between them), we
suppose that class imbalance can be detrimental as well.

  * $h_{2}$

Increasing name variety in the training set encourages the model to try and
make PER predictions for tokens whose classes are ambiguous given a specific
sentence. We deem a token “ambiguous” when the context given to the model does
not suffice to infer its correct class (for example, when it is not possible
to distinguish between a PER and a LOC entity using only an input sentence).

###  4.1 Class Imbalance

Figure 3: Precision, recall and F1-score against augmentation rate for
different augmentation methods. Error bars represent the 95% confidence
interval.

To check if the decrease in precision is due to class imbalance, we test two
different methods to inject names in the training dataset that should not
suffer from the class imbalance problem:

  * •

Upsample and balance: After adding newly generated examples to the dataset, we
copy some of the original examples and add them to the dataset to restore the
original class distribution.

  * •

Replace: Instead of adding newly generated examples to the dataset, we
substitute them for the original training samples.

Figure 3 shows the performance of a model trained on a dataset modified with
the above augmentation methods. The upsample and balance strategy still
increases recall, but does not fix the precision issue. Meanwhile, the replace
strategy still makes precision decrease, and also generally decreases
performance for an augmentation rate greater than 20%.

As none of these two methods are able to fix the precision issue, we conclude
that class imbalance is not a plausible cause of precision decreasing.

###  4.2 Name Variety and Ambiguous Occurrences

In order to check $h_{2}$, we analyze the difference in false positives
between a model trained with augmentation and a model trained without it. As
errors can vary between runs, we perform three training runs for each model
and keep the false positives that are consistent between the runs. We then
observe the set of false positives of the model trained with augmentation that
are not present in the set of false positives of the model trained without
augmentation. This allows us to analyze errors that are specific to our
augmentation scheme.

We manually find that for 62% (58/93) of these false positives, the model
attributes a PER label to some tokens even though the context alone is not
sufficient to know the correct label of the entity, such as in the following
example from Elantris:

Raoden stood, and as he did, his eyes fell on Elantris again.

Here, the model correctly predicts that Elantris is a named entity, but gives
it the class PER while Elantris is actually a city (which corresponds to the
LOC class). Of course, predicting the correct class of Elantris in this
example would only be a matter of luck even for a perfect model, as the given
context alone is not sufficient.

Overall, this result shows a bias towards the PER class, that was introduced
by increasing the variety of possible names at training time.

To try and mitigate this effect, we supply the model with broader context, in
hope that this context can help resolving some ambiguities. In our previous
example, the next sentence gives enough context to predict the correct entity
type for Elantris:

Raoden stood, and as he did, his eyes fell on Elantris again. Resting in the
great city’s shadow, Kae seemed like an insignificant village by comparison.

We define a context size of $n$ as giving the model the $n$ sentences that
precede the considered sentence, as well as the $n$ sentences that follow it.

Figure 4: F1-score, precision and recall against different context sizes for a
model trained without augmentation (Noaug) and a model trained with
augmentation (The Elder Scrolls) (60% augmentation ratio). Error bars
representent the 95% confidence interval.

Figure 4 shows the effect of giving more context to models trained with
augmentation (The Elder Scrolls) or without it (Noaug). As can be seen,
context is beneficial to both models, steadily increasing F1-score. Precision
increases a lot for the model trained with augmentation, as previously
ambiguous entities are now correctly labeled. Using our previously described
method for error analysis, we find that only 48% (30/63) of the entities found
in false positives are deemed ambiguous for a context size of 1. Meanwhile,
the recall augmentation for the Noaug model can be attributed to it predicting
PER entities in previously ambiguous cases. While adding more context results
in greater recall for the Noaug model, it still never surpasses the recall of
the The Elder Scrolls model.

##  5 Conclusion

We demonstrated the usage of a simple data augmentation technique to better
detect fantasy names when performing named entity recognition. While this
technique greatly increases model recall, precision decreases as the model
sometimes does not have enough information to predict the correct class of
some entities. We showed that this issue can be mitigated by giving the model
more local information. However, for the same context size, our technique
always result in greater recall but lower precision. In the context of
character detection, we argue that this trade is beneficial: while false
positives can be filtered after performing NER (automatically or manually),
false negatives are not easily recovered. We also note that increasing NER
context has a positive impact on recall in general.

Further work might investigate the role of context in NER. While we showed
that it can be used to disambiguate some mentions, we were only interested in
local context. Nevertheless, necessary disambiguation information is not
always present in the local context, And might exist in other places for each
novel.

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