[affiliation=1]AntonioAlmudévar [affiliation=2,3]ThéoMariotte
[affiliation=1]AlfonsoOrtega [affiliation=2]MarieTahon
[affiliation=1]LuisVicente [affiliation=1]AntonioMiguel
[affiliation=1]EduardoLleida

# Predefined Prototypes for Intra-Class Separation and Disentanglement

###### Abstract

Prototypical Learning is based on the idea that there is a point (which we
call prototype) around which the embeddings of a class are clustered. It has
shown promising results in scenarios with little labeled data or to design
explainable models. Typically, prototypes are either defined as the average of
the embeddings of a class or are designed to be trainable. In this work, we
propose to predefine prototypes following human-specified criteria, which
simplify the training pipeline and brings different advantages. Specifically,
in this work we explore two of these advantages: increasing the inter-class
separability of embeddings and disentangling embeddings with respect to
different variance factors, which can translate into the possibility of having
explainable predictions. Finally, we propose different experiments that help
to understand our proposal and demonstrate empirically the mentioned
advantages.

###### keywords:

prototypical learning, predefined prototypes, intra-class separability,
disentanglement, explainability

##  1 Introduction

Prototype theory is a concept from cognitive science and linguistics that
suggests that humans categorize objects and ideas by comparing them to a
mental representation of a prototype of each category or class [1]. These
prototypes encapsulate the most central and representative features of each
class. On the other hand, many machine learning systems base their operation
on extracting representations [2, 3] (from now on we will refer to them
interchangeably as embeddings), which are typically vectors that somehow
encode the most representative features of the inputs. An approach that has
gained popularity due to its good performance in a multitude of applications
consists of introducing the idea of prototyping in machine learning systems
[4]. These systems allow, in addition to extracting a representation of each
input, to obtain a representation with the most representative features of the
inputs of a class, which is called prototype. Moreover, these systems are
typically trained so that all representations of the same class are close
together in space. Therefore, as a result of these approximations, we end up
having several sets of representations, each one in an area of space and
corresponding to a class. Each of these sets of representations is around the
prototype of that class. However, there are two properties that in general are
not de facto fulfilled in this type of systems and that may be desirable for
different tasks and applications. These desirable properties are:

  * •

The embeddings of different classes are separated in space. This better use of
all available space and usually results in a better performance in tasks such
as classification, anomaly detection, object detection or biometric
recognition [5, 6, 7].

  * •

It is possible to associate some concrete dimensions of these representations
with concrete human-understandable features so that a change of a feature
produces changes in only a few dimensions of the space. This is has some
advantages such as  (i) having more control over data creation in generative
models [8], or  (ii) providing the ability to explain and interpret model
predictions [9].

In this paper we propose a modification on the prototypical systems that
preserves their default advantages and, in addition, allows solving the two
problems presented. This modification consists in having the human predefine
the prototypes before the training of the system, so that one of the
objectives is that all the representations of a class are in an area of the
space that has been decided with a human criterion and therefore can be
explained. Typically, prototypes are trainable or simply computed as the
average of the representations of a class and, to the best of our knowledge,
the approach of imposing concrete conditions on them prior to training is an
approach not explored in the literature. This may have several advantages, but
in this paper we focus on its ability to achieve the two properties explained
in the previous paragraph and whose are immediate to understand. First, if we
predefine the prototypes before training, we can impose that they are far
apart, by defining them as an orthogonal set of vectors, for example. On the
other hand, we can define the prototypes so that some of their dimensions
correspond to concrete human-interpretable factors.

To demonstrate with examples the two previous points, we propose two types of
experiments demonstrating with them the correct operation of our proposal. In
the first one, we solve a typical audio classification problem where we impose
that representations of different classes are orthogonal to each other,
showing that an implication of this is a better accuracy. On the other hand,
we solve an emotion classification task where concrete dimensions of the
representations correspond to acoustic parameters, allowing an explanation of
how these parameters relate to the different emotions to be classified. More
details on these experiments can be found at
https://github.com/not/available/in/the/version/for/review.

##  2 Related Work

Prototypical Learning. Its use was initially proposed for few-shot
classification and class prototypes were calculated as the average of the few
embeddings available for each class. During training, it is imposed that all
embeddings are close to the prototype of their class, which implies that they
are all close in space. Subsequently, different applications and variations of
the prototypical systems have been proposed. For example, they have been used
for unsupervised domain adaptation [10] or to build explainable systems [11,
12].

Intra-Class Separation. Due to the reasons explained in section1, different
loss functions have been proposed that aim to separate embeddings of different
classes in space. Among these loss functions, the following stand out: Center
Loss [5], Focal Loss [6], Orthogonal Projection Loss [7] or that of the
Variational Classifier [13]. The proposal of the present work allows defining
the prototypes in such a way that they are far from each other and,
consequently, the embeddings of the different classes are also far from each
other.

Disentanglement. Although there is no consensus on its meaning, what is widely
accepted is the intuition that it is the separation of the representations in
the different variation factors of the data [2, 14]. That is, changing a
variation factor in the data should change only a part of the representation.
In [15] it is shown that it is not possible to achieve unsupervised
disentanglement without inductive biases in the data or in the model.
Therefore, multiple works have also been proposed focused on supervised
disentanglement [16, 17, 18].

##  3 Proposed Method

###  3.1 General System Description

Let $\mathcal{D}=\\{\\{x^{(i)},y^{(i)},\alpha^{(i)}\\}\\}_{i=1}^{n}$ be a
dataset where $x^{(i)}\in\mathbb{R}^{p}$ are each of the inputs,
$y^{(i)}\in\mathbb{R}^{C}$ is the vector containing the class to which
$x^{(i)}$ belongs and $\alpha^{(i)}\in\mathcal{A}$ is an abstract containing
the factors of interest over $x^{(i)}$. We note that $\alpha^{(i)}$ can be
extracted by an additional system and are the factors over which we intend to
disentangle our representations. Deep Learning classifier systems are usually
divided into two parts: $F_{\theta}:\mathbb{R}^{p}\rightarrow\mathbb{R}^{k}$,
which we call embeddings extractor, and
$G_{\phi}:\mathbb{R}^{k}\rightarrow\mathbb{R}^{C}$, which is the classifier
network and is typically a linear layer plus a Softmax. On the other hand, in
our system, we have
$P:\mathbb{R}^{C}\times\mathcal{A}\rightarrow\mathbb{R}^{k}$, which we call
the prototype extractor. We note that $P$ does not depend on any trainable
parameter, that is, it is not modified throughout the training, which is the
main novelty of our work. From these three components we can compute
embeddings as $z^{(i)}=F_{\theta}(x^{(i)})$, predictions as
$\tilde{y}^{(i)}=G_{\phi}(z^{(i)})$ and prototypes as
$p^{(i)}=P(y^{(i)},\alpha^{(i)})$. The two objectives of this system are:  (i)
that $y^{(i)}$ and $\tilde{y}^{(i)}$ are similar to have a good classification
performance, and  (ii) that $z^{(i)}$ and $p^{(i)}$ are similar for the
embedding extractor $F_{\theta}$ to behave similarly to $P$.  This results in
the following loss function:

$\mathcal{L}=CE\left(y^{(i)},\tilde{y}^{(i)}\right)+\lambda_{p}||z^{(i)}-p^{(i)}||_{2}^{2}$
(1)

where $CE$ is the cross-entropy loss and $\lambda_{p}$ is a hyperparameter
that we set to $1/k$ in the experiments. In algorithm 1 we summarize the
training procedure.

Algorithm 1 Training Algorithm for the Predefined Prototypes System

Dataset $\mathcal{D}$, Prototypes Extractor $P$, $\lambda_{p}$

$\bm{\theta}$, $\bm{\phi}\leftarrow$ Initialize parameters

repeat

$\mathcal{D}^{N}\leftarrow\\{(x^{(i)},y^{(i)},\alpha^{(i)}\\}\\}_{i=1}^{N}$
(Minibatch from $\mathcal{D}$)

for $i=1$ to $N$ do

$z^{(i)}\leftarrow F_{\theta}(x^{(i)})$

$\tilde{y}^{(i)}\leftarrow G_{\phi}(z^{(i)})$

$p^{(i)}\leftarrow P(y^{(i)},\alpha^{(i)})$

end for

$\mathcal{L}\leftarrow\frac{1}{N}\sum_{i=1}^{N}CE\left(y^{(i)},\tilde{y}^{(i)}\right)+\lambda_{p}||z^{(i)}-p^{(i)}||_{2}^{2}$

$\bm{\theta},\bm{\phi}\leftarrow$ Update using gradients of $\mathcal{L}$

until convergence of $\bm{\theta}$ and $\bm{\phi}$

Finally, we propose that $P$ is a multilinear map, i.e.:

  * •

$P(ay^{(i_{a})}+by^{(i_{b})},\alpha^{(i)})=aP(y^{(i_{a})},\alpha^{(i)})+bP(y^{(i_{b})},\alpha^{(i)})$,
which allows working with soft labels (which implies, among other things, the
ability to use regularization techniques such as mixup [19], widely used in
audio classification).

  * •

$P(y^{(i)},a\alpha^{(i_{a})}+b\alpha^{(i_{b})})=aP(y^{(i)},\alpha^{(i_{a})})+bP(y^{(i)},\alpha^{(i_{b})})$,
which provides ease in managing continuous variation factors.

###  3.2 Examples of Prototypes Extractors

The main novelty of our system with respect to others is the proposal that $P$
is not modified during training and can also be defined by the human. This,
despite being an unexplored idea to the best of our knowledge, may have a
large number of applications. Specifically, in this paper we give two formulas
that allow us to solve two very popular problems. These are:  (i) maximize the
distance between embeddings of different types and  (ii) disentangle subsets
of embeddings variables with respect to specified factors of variation.  In
the following, we give the formulas we propose to define $P$ that allow to
solve these problems.

####  3.2.1 Maximize Intra-Class Separability

In this first example, we have that
$P(y^{(i)},\alpha^{(i)})=P(y^{(i)}):\mathbb{R}^{C}\rightarrow\mathbb{R}^{k}$,
since our only goal is to separate embeddings of different classes
independently of the input factors. Subsequently, we define (using Singular
Value Decomposition, for example) an orthogonal basis
$V=\\{v^{(j)}\\}_{j=1}^{C}$ such that $v^{(j)}\in\mathbb{R}^{k}$, so that
$P(j)=v^{(j)}$. In this way, the prototypes of the different classes are
orthogonal to each other and, consequently, the embeddings of elements of
different classes will tend to be quasi-orthogonal, thus being far apart in
space. Although typically the dimension of the embedding is higher than the
number of classes, i.e., $k\geq C$, there may be scenarios in which this is
not the case, which would prevent one from being able to define an orthogonal
$V$ basis as described. In the case where $k<C$, we propose to define an
orthogonal basis $W=\\{w^{(j)}\\}_{j=1}^{C}$ such that
$w^{(j)}\in\mathbb{R}^{C}$, subsequently define a Johnson Lindenstrauss
Transform (JLT) $T:\mathbb{R}^{C}\rightarrow\mathbb{R}^{k}$, so that finally
$v^{(j)}=T(w^{(j)})\in\mathbb{R}^{k}$ for $j=1,2,\dots,C$. JLTs are nearly
distance-preserving transformations, so embeddings of different classes will
end up being maximally separated in space [20]. Fast algorithms exist to
compute these JLTs [21].

####  3.2.2 Embeddings Disentanglement with respect to Features

In this second example we define our prototype extractor as
$P(y^{(i)},\alpha^{(i)})=P_{\alpha}(\alpha^{(i)}):\mathcal{A}\rightarrow\mathbb{R}^{k}$,
i.e., it depends on the factors but not on the labels. The way in which this
$P_{\alpha}(\alpha^{(i)})$ is defined is highly dependent on the scenario and
application. However, one way we have found to obtain a good classification
performance and disentangled embeddings is to set
$P_{\alpha}(\alpha^{(i)})=\left(P^{\prime}_{\alpha}(\alpha^{(i)}),\bm{0}\right)$,
where $(\cdot,\cdot)$ means concatenation,
$P^{\prime}_{\alpha}(\alpha^{(i)}):\mathcal{A}\rightarrow\mathbb{R}^{k_{f}}$
and $\bm{0}$ is the all-zero vector of length $k-k_{f}$. Thus, the goal is
that we can clearly separate our embeddings into two parts:  (i) one that
depends on the factors and is disentangled with respect to them, and  (ii) one
whose value does not depend directly on the factors and whose elements are
close to zero.  The purpose of this second part of the embedding is to give
the model a greater degree of freedom in organizing its embeddings and to
capture all the factors of variation that influence the predictions
$\tilde{y}^{(i)}$ but are not captured in $\alpha^{(i)}$. Without this second
part (or, equivalently, if $k=k_{f}$), predictions would be made from only
$\alpha^{(i)}$, which would generally result in lower accuracy, since there
are more factors that can affect $\tilde{y}^{(i)}$ and are not in
$\alpha^{(i)}$. In fact, by comparing the two parts of the embeddings, we can
obtain information the proportion the predictions determined by the known
factors $\alpha^{(i)}$ and by unknown factors. Although one might think that
defining the second part of the prototypes as $\bm{0}$ might lead to its
elements tending to always be worth 0, this is not necessarily the case. As is
the case in the Variational Autoencoder [22], the loss function to be
optimized defined in the equation (1), is defined by two terms. The fact that
cross-entropy must be minimized will lead our embeddings to differ across
classes if this helps to minimize cross-entropy. If, on the other hand, our
predictions can be made very accurately only from $\alpha^{(i)}$ (which is
highly unlikely), the elements of the second part of the embeddings will tend
to take values close to 0, which is positive for two reasons:  (i) it
indicates that there are no factors other than those in $\alpha^{(i)}$ that
affect the predictions, which gives us new knowledge about our dataset; and
(ii) it tells us that there are elements of the embeddings that are
unnecessary, thus reducing the size of the embeddings and, consequently, the
complexity of the model.  The way to define $P^{\prime}_{\alpha}$ is very
scenario-dependent, so we do not give a general formula to define it. In
section 4.3.2 we describe an emotion recognizer in which embeddings are
disentangled with respect to a set of acoustic parameters and follow the
previous formulation.

Table 1: Accuracy for the different Datasets, Embeddings Extractors and Loss Functions | ESC-50 | KS2
---|---|---
| AST | BEATs | AST | BEATs
Cross-entropy | $93.97\pm 0.21$ | $91.05\pm 0.41$ | $\mathbf{92.05\pm 0.04}$ | $88.94\pm 0.13$
Focal Loss | $94.40\pm 0.36$ | $91.10\pm 0.49$ | - | -
OPL | $94.11\pm 0.37$ | $91.50\pm 0.20$ | - | -
Predefined Prototypes | $\mathbf{94.52\pm 0.02}$ | $\mathbf{91.72\pm 0.30}$ | $91.45\pm 0.06$ | $\mathbf{89.42\pm 0.09}$

##  4 Experiments

###  4.1 Datasets

Environmental Sound Classification (ESC-50) [23] includes 2,000 ambient sound
recordings categorized into 5 classes, each lasting 5 seconds. Throughout our
experiments, we adhere to the standard 5-fold cross-validation approach.

Speech Commands V2 (KS2) [24] comprises 105,829 one-second clips of spoken
keywords, each annotated with one of 35 word classes. Officially, it is
segmented into 84,843 training clips, 9,981 test clips, and 11,005 validation
clips.

IEMOCAP (ER) [25] consists of approximately 12 hours of speech showcasing four
distinct emotions. Our evaluation uses the standard 5-fold cross-validation
method proposed in [26].

###  4.2 Embeddings Extractors

ECAPA-TDNN [27] integrates attention mechanisms and parallel processing for
efficient sequential data analysis, particularly suited for tasks like speech
recognition.

Audio Spectrogram Transformer (AST) [28] renowned as the pioneer in using
Transformer architectures for audio, has set a benchmark due to its
exceptional performance. We initialize the model with pre-trained weights from
Imagenet [29] and Audioset [30] and fine-tune it for each specific scenario.

BEATs [31] is an audio pre-training framework for learning representations
from Audio Transformers, in which an acoustic tokenizer and a self-supervised
audio model are optimized. We also start from the pre-trained model with
Audioset.

###  4.3 Results

In all the next experiments, we use a sampling frequency of 16kHz. The inputs
to every system are 128 mel-filterbank calculated in 25 ms windows every 10
ms. We normalize the mean and standard deviation to 0 and 0.5, respectively.

####  4.3.1 Audio Classification with Intra-Class Separation

To compare the performance of our proposal with others in the literature, we
evaluate them on ESC-50 and KS2 using AST and BEATs as embedding extractors.
The hyperparameters used to obtain the inputs and to train each of them are
based on [28, 31]. To evaluate our proposal, we compare it with identical
systems where the only differentiating factor is the loss function.
Specifically, we compare our loss function with cross-entropy, Focal Loss and
OPL. The latter two, like our proposal, aim at increasing the distance between
embeddings of different classes. They give excellent performances in different
classification scenarios. We note that we have not been able to train them for
KS2, since mixup is used, which results in having soft labels, and these two
proposals do not allow working with soft labels. Conversely, our proposal
does, since the prototype extractor is a multilinear map. In our proposal, we
define the prototypes according to the procedure explained in 3.2.1. We have
run all the described experiments three times and we show the mean and
standard deviation of the accuracy in table 1. We see that in all cases except
one, our system outperforms the rest on average. This seems to indicate that
it is effective in separating embeddings of different classes and that this
translates into better accuracy.

Figure 1: Joint probabilities of each emotion and acoustic parameter (by
levels) in the training dataset.

####  4.3.2 Disentangled Emotion Recognition

Next we give an illustrative example of how to design a classifier whose
embeddings are disentangled with respect to given factors of variation. We
propose to use IEMOCAP as dataset to classify the emotions {neutral, happy,
angry, sad} and choose {pitch median, pitch std, loudness} as factors with
respect to which to disentangle our embeddings. We select only female voices,
to minimize pitch differences between speakers. The reason for choosing these
three acoustic parameters as factors is that they have been found to be
related to emotions [32, 33, 34]. To extract pitch median and pitch std, we
use [35]; and to obtain loudness we follow [36]. On the other hand, although
$P^{\prime}_{\alpha}$ can take continuous values as inputs and can be any type
of function, here we discretize the values of the factors to three levels
according to their value with respect to the $1/3$ and $2/3$ quantiles
calculated in the training set. This facilitates interpretability so that,
from now on, we will say that each factor is either low, medium or high. The
joint probability of each emotion and level of these factors is shown in
Figure 1. Subsequently, we define a coding function $C(\alpha^{(i)}_{k})$ as:

$C(\alpha^{(i)}_{k})=\left\\{\begin{array}[]{ll}(1,0,0)&\text{if}\,\alpha^{(i)}_{k}\
\text{is}\ \textit{low}\\\ (0,1,0)&\text{if}\,\alpha^{(i)}_{k}\ \text{is}\
\textit{medium}\\\ (0,0,1)&\text{if}\,\alpha^{(i)}_{k}\ \text{is}\
\textit{high}\end{array}\right.$ (2)

where $\alpha^{(i)}_{1}$, $\alpha^{(i)}_{2}$ and $\alpha^{(i)}_{3}$ are the
pitch median, pitch std and loudness of the input $x^{(i)}$, respectively and
$\alpha^{(i)}=\left\\{\alpha^{(i)}_{1},\alpha^{(i)}_{2},\alpha^{(i)}_{3}\right\\}$.
Finally, we define
$P^{\prime}_{\alpha}(\alpha^{(i)})=\left(C\left(\alpha^{(i)}_{1}\right),C\left(\alpha^{(i)}_{2}\right),C\left(\alpha^{(i)}_{3}\right)\right)$.
Thus, an audio with low pitch median, medium pitch std and high loudness would
correspond to a prototype $\left((1,0,0,0,1,0,0,0,1),\bm{0}\right)$. As
embedding extractor $F_{\theta}$ we use an ECAPA-TDNN with embedding size
$k=16$ (thus, length of $\bm{0}$ is 7). As classification net $G_{\phi}$ we
use a linear layer without bias, i.e. $G_{\phi}\equiv
W_{\phi}=(w^{\phi}_{jc})\in\mathbb{R}^{k\times C}$ and the prediction is
calculated as $\tilde{y}^{(i)}=W_{\phi}^{T}\cdot z^{(i)}$. Equivalently, we
define the relevance matrix as:

$\Gamma^{(i)}=(\gamma^{(i)}_{jc})\ \text{such that}\
\gamma^{(i)}_{jc}=w^{\phi}_{jc}\cdot z^{(i)}_{j}$ (3)

where $\gamma^{(i)}_{jc}$ contains information on how the $j$ component of
embedding $z^{(i)}$ influences the probability of class $c$. Next we analyze
these relevance matrices of different embeddings shown in Figure 2 to
illustrate this better.

  * •

In Figure 2(a) we see a case where neutral emotion has been predicted with a
probability of 0.68. We can see that having low pitch std and loudness added
probabilities to both sad and neutral, but the fact that the pitch median is
medium makes the model to decrease the probability of sad. As we can see in
Figure 1, it is uncommon to have medium pitch median given the emotion sad.

  * •

In Figure 2(b) we can suspect that the loudness is between medium and high,
since the activated part of high adds probability to angry, but the activated
part of medium subtracts to angry and adds to happy, which is the emotion with
the highest probability. As we can see, the probability of having a high
loudness when the emotion is angry is very high.

  * •

In Figure 2(c) we see clearly that having a high loudness and medium pitch std
has a great influence in predicting angry.

  * •

In Figure 2(d) we can see that having low pitch std and low loudness increases
the probability of neutral and sad. However, component 10 has a great
influence on ultimately increasing the probability of sad.

It is interesting to analyze component 10, since it seems to have similar
behavior for neutral and happy and for angry and sad in all figures. This
raises the suspicion that this component holds a factor related to the
positivity or negativity of the emotion, but which we have not tried to
disentangle explicitly. This is an example of the importance of giving freedom
to a part of the embeddings. This part ends up learning factors that we have
not explicitly tried to disentangle but seem to have influence on predictions
and interpretable meaning. In addition, we see how the components associated
with other factors are far from being 0 in most cases, as explained in section
3.2.2. Finally, we mention that the accuracy of the described system is 55.53
and that of one with the same characteristics but with cross-entropy loss
54.86, i.e. we obtain a similar performance in both cases.

(a) $p_{neu}=0.68$, $p_{hap}=0.15$, $p_{ang}=0.05$, $p_{sad}=0.12$

(b) $p_{neu}=0.02$, $p_{hap}=0.87$, $p_{ang}=0.10$, $p_{sad}=0.01$

(c) $p_{neu}=0.01$, $p_{hap}=0.04$, $p_{ang}=0.91$, $p_{sad}=0.04$

(d) $p_{neu}=0.20$, $p_{hap}=0.10$, $p_{ang}=0.08$, $p_{sad}=0.62$

Figure 2: Relevance matrices $\Gamma^{(i)}$ for different embeddings $z^{(i)}$
and their corresponding predictions $\tilde{y}^{(i)}$

##  5 Conclusions

In this paper, we have proposed the possibility of defining the prototypes of
a system before training and with a human criterion. We have argued some of
the advantages and applications that this may have. Specifically, we have
argued how this idea can be used to  (i) increase the distance between
embeddings of different classes and, consequently, improve the accuracy of a
classifier; and  (ii) disentangle embeddings with respect to given variation
factors, being able to have more control over them and providing the
possibility to explain predictions.  However, the ultimate goal of the paper
is to give the idea that it may be convenient in some cases to let the human
define the prototypes instead of letting the system learn them by itself. The
two proposed applications are intended to illustrate use cases in which
applying this idea is both convenient and advantageous.

##  6 Acknowledgements

Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aenean commodo
ligula eget dolor. Aenean massa. Cum sociis natoque penatibus et magnis dis
parturient montes, nascetur ridiculus mus. Donec quam felis, ultricies nec,
pellentesque eu, pretium quis, sem. Nulla consequat massa quis enim. Donec
pede justo, fringilla vel, aliquet nec, vulputa.

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# On monohedral tilings of a regular polygon

Bushra Basit  and  Zsolt Lángi  Department of Geometry, Budapest University of
Technology, Egry József utca 1., Budapest, Hungary, 1111
<EMAIL_ADDRESS>MTA-BME Morphodynamics Research Group and Department
of Geometry, Budapest University of Technology, Egry József utca 1., Budapest,
Hungary, 1111<EMAIL_ADDRESS>
###### Abstract.

A tiling of a topological disc by topological discs is called monohedral if
all tiles are congruent. Maltby (J. Combin. Theory Ser. A 66: 40-52, 1994)
characterized the monohedral tilings of a square by three topological discs.
Kurusa, Lángi and Vígh (Mediterr. J. Math. 17: article number 156, 2020)
characterized the monohedral tilings of a circular disc by three topological
discs. The aim of this note is to connect these two results by characterizing
the monohedral tilings of any regular $n$-gon with at most three tiles for any
$n\geq 5$.

###### Key words and phrases:

dissection, monohedral tiling, topological disc, Jordan region

###### 2010 Mathematics Subject Classification:

52C20, 52B45, 52A10

Partially supported by the BME Water Sciences & Disaster Prevention TKP2020
Institution Excellence Subprogram, grant no. TKP2020 BME-IKA-VIZ, the NKFIH
grant K134199, the János Bolyai Research Scholarship of the Hungarian Academy
of Sciences, and the ÚNKP-20-5 New National Excellence Program by the Ministry
of Innovation and Technology.

##  1\. Introduction

Subsets of the Euclidean plane $\mathbb{R}^{2}$ homeomorphic to the Euclidean
closed circular unit disc $\mathbf{B}^{2}$ centered at the origin $o$ are
usually called _topological discs_ or _Jordan regions_. A family of
topological discs $\\{D_{1},D_{2},\ldots,D_{k}\\}$ whose union is a
topological disc $D$ and whose elements are mutually nonoverlapping (i.e.
their interiors are mutually disjoint), is called a _tiling_ , _decomposition_
, or _dissection_ of $D$, and the elements of the family are called _tiles_. A
tiling is called _monohedral_ , if all tiles are congruent to a given
topological disc, which is often called _prototile_ [3].

The history of the investigation of tilings goes back to ancient times and
well beyond the boundary of mathematics (see e.g. [6, 18]). The aim of this
paper is to examine one such problem. A result of Maltby [14] in 1994 states
that a square cannot be dissected into three non-rectangular congruent
topological discs. Along the same line, Yuan, C. Zamfirescu and T. Zamfirescu
[19] proved, answering a question of Danzer, that in any monohedral tiling of
a square by five _convex_ tiles, the prototile is a rectangle, and conjectured
that the same holds if the number of tiles is an odd prime. This question has
been recently answered in [16] in the special case that the prototile is a
$q$-gon with $q\geq 6$ or it is a right-angled trapezoid, and a computer-
assisted proof has been given in [15] for seven or nine tiles.

We intend to investigate a similar question, also based on the result of
Maltby in [14]. To state our main result, we call a monohedral tiling of a
regular $n$-gon $P$, centered at the origin $o$, by tiles
$D_{1},D_{2},\ldots,D_{k}$ _rotationally generated_ if the rotation around $o$
and with angle $\frac{2\pi}{k}$ leaves $P$ invariant, and permutes the tiles
(cf. Figure 1).

######  Theorem 1.

Let $P$ be a regular $n$-gon with $n\geq 5$, and let $\mathcal{F}$ be a
monohedral tiling of $P$ by $k$ topological discs, where $2\leq k\leq 3$. Then
either $k=2$, $n$ is odd and $\mathcal{F}$ contains the two halves of $P$
dissected by a line of symmetry of $P$, or $n$ is divisible by $k$ and
$\mathcal{F}$ is rotationally generated.

Figure 1. A rotationally generated tiling of a regular $9$-gon $P$ with three
tiles

We note that the same theorem with the Euclidean circular disc in place of $P$
was proved in [13]. Theorem 1 can be regarded as a result connecting the one
in [14] for squares and the one in [13] for circular discs. The proof of
Theorem 1 is based on (geometric, combinatorial and topological) tools from
both [13] and [14], and also on some new ideas.

Finally, we remark that in the past few years a ‘dual version’ of this
problem, namely the investigation of dissecting the Euclidean plane into
mutually _incongruent_ tiles with equal area under various constraints has
also gained a significant interest. For related results the interested reader
is referred to [2, 3, 4, 5, 10, 11]. The number of dissections of a square
into equal area rectangles was estimated in [1, 7]. For the investigation of
monohedral dissections of geometric figures using a different notion of
dissection, see e.g. [8, 9].

The structure of our paper is as follows. In Section 2 we introduce the
necessary notation and tools to prove our main result. In Section 3 we present
the proof of Theorem 1. Finally, in Section 4 we collect some additional
remarks.

##  2\. Preliminaries

In the paper, for any set $X\subset\mathbb{R}^{2}$, we denote the interior,
the boundary, the closure and the convex hull of $X$ by
$\operatorname{int}(X),\partial X$, $\operatorname{cl}(X)$ and
$\operatorname{conv}(X)$, respectively. Furthermore, if $X$ is bounded and
nonempty, then $\operatorname{diam}(X)$ denotes its diameter. For any
$x,y\in\mathbb{R}^{2}$, by $[x,y]$ we denote the closed segment with endpoints
$x,y$. By a simple curve we mean a continuous curve which does not cross
itself, and a simple, closed curve is a simple curve whose two endpoints
coincide. With a little abuse of notation, we call the points of a simple, not
closed curve, different from its endpoints, interior points of the curve.
Finally, for brevity, we call a topological disc simply a _disc_.

In the proof, $P$ denotes a regular $n$-gon with unit side-length centered at
$o$, and vertices $p_{1},p_{2},\ldots,p_{n}$ in counterclockwise order. We set
$\mathcal{F}=\\{D_{1},D_{2},\ldots,D_{k}\\}$, where all $D_{i}$s are congruent
to a disc $D$, and for $i=1,2,\ldots,k$ let $S_{i}=D_{i}\cap\partial P$. For
any value of $i\neq 1$, we choose an isometry
$g_{1i}:\mathbb{R}^{2}\to\mathbb{R}^{2}$ satisfying $g_{1i}(D_{1})=D_{i}$, and
set $g_{i1}=g_{1i}^{-1}$, and $g_{ij}=g_{1i}^{-1}\circ g_{1j}$ for all $i,j$.

We remark that every disc is compact, and thus, it is Lebesgue measurable. On
the other hand, the boundary of a disc is not necessarily rectifiable; as an
example we may choose e.g. the Koch snowflake (for more ‘esoteric’ examples,
see [17]). In the proof for any disc $D$ we use the notation
$\operatorname{area}(D)$ and $\operatorname{perim}(D)$ for the area and the
perimeter of $D$, respectively, and we use the latter one only if $\partial D$
is clearly rectifiable. If $\Gamma$ is a rectifiable curve, then by
$l(\Gamma)$ we mean the length of $\Gamma$. In particular, this yields that if
$\partial D$ is rectifiable for some disc $D$, then we have $l(\partial
D)=\operatorname{perim}(D)$.

We start with some preliminary lemmas and remarks.

######  Remark 1.

Since any $D_{i}$ is a disc, any two points of $D_{i}$ can be connected by a
continuous curve which contains only interior points of $D_{i}$, apart from
its endpoints. In the paper, we call such a curve an _in-curve_ of $D_{i}$.

######  Remark 2.

We note that if some isometry $g_{ij}$ is a reflection about a line $L$, then
$L$ separates $D_{i}$ and $D_{j}$. Indeed, suppose for contradiction that
there are points $x,y\in D_{i}$ in different open half planes bounded by $L$,
and let $\Gamma$ be an in-curve of $D_{i}$ connecting $x$ and $y$. Then there
is a point $z$ of $\Gamma$ on $L$. Thus, $g_{ij}(z)=z$, implying that
$\operatorname{int}(D_{i})\cap\operatorname{int}(D_{j})\neq\emptyset$; a
contradiction.

######  Lemma 1.

If $\operatorname{diam}(D)=\operatorname{diam}(P)$, then $k=2$. Furthermore,
either $n$ is odd and $\mathcal{F}$ contains the two halves of $P$ dissected
by a line of symmetry of $P$, or $n$ is divisible by $2$ and $\mathcal{F}$ is
rotationally generated.

###### Proof.

Under our conditions, each $D_{i}$ contains a diametrically opposite pair of
points of $P$, or in other words, the two endpoints of a longest diagonal of
$P$. First, observe that if $p_{i_{1}},p_{i_{2}}\in D_{i}$ and
$p_{j_{1}},p_{j_{2}}\in D_{j}$ are mutually distinct diametrically opposite
points of $P$ where $p_{i_{1}},p_{j_{1}},p_{i_{2}},p_{j_{2}}$ are in this
cyclic order in $\partial P$, then any in-curve of $D_{i}$ connecting
$p_{i_{1}}$ and $p_{i_{2}}$ would cross any in-curve of $D_{j}$ connecting
$p_{j_{1}}$ and $p_{j_{2}}$, leading to a contradiction. Thus, there is a
vertex of $P$ contained in any diameter of any $D_{i}$. Without loss of
generality, let us assume that $p_{1}$ is such a vertex.

First, consider the case that $n=2m$ for some integer $m\geq 3$. Since
$[p_{1},p_{m+1}]$ is the unique diameter of $P$ containing $p_{1}$, it follows
that $p_{m+1}\in D_{i}$ for all values of $i$. Furthermore, any isometry
mapping a longest diagonal of $P$ into a longest diagonal of $P$ is a symmetry
of $P$, implying that $g_{ij}$ is a symmetry of $P$ for all $i,j$. Thus,
$g_{ij}$ is the reflection about the line $L$ through $[p_{1},p_{m+1}]$, or
the bisector $L^{\prime}$ of $[p_{1},p_{m+1}]$, or $o$. On the other hand, if
$g_{ij}$ is the reflection about $L^{\prime}$, the fact that $p_{1}$ and
$p_{m+1}$ are in different open half planes bounded by $L^{\prime}$
contradicts Remark 2. Thus, $g_{ij}$ is the reflection about $L$ or about $o$
for any $i\neq j$. Since $g_{il}=g_{ij}\circ g_{jl}$ for all $i,j,l$ by
definition, the fact that there are only two possible isometries as $g_{ij}$
implies that $k\leq 2$. If $g_{12}$ is the reflection about $o$, then we are
done. If $g_{12}$ is the reflection about $L$, then from Remark 2 it follows
that $L$ separates $D_{1}$ and $D_{2}$, and the tiling is rotationally
generated.

Finally, consider the case that $n=2m+1$ for some integer $m\geq 2$, and let
$L$ denote the line through $[o,p_{1}]$. By our conditions, any tile $D_{i}$
contains $p_{m}$ or $p_{m+1}$. Suppose for contradiction that a tile contains
both $p_{m}$ and $p_{m+1}$. Then any tile contains either $p_{1}$, $p_{m}$ and
$p_{m+1}$, or $p_{1},p_{2}$ and $p_{m+1}$, or $p_{1},p_{n}$ and $p_{m}$. In
other words, for any $i\neq j$, $g_{ij}$ is the reflection about either $L$,
or the bisector of $[p_{1},p_{m}]$, or the bisector of $[p_{1},p_{m+1}]$. On
the other hand, any of these cases contradicts Remark 2. Thus, any tile
contains either $p_{m}$ or $p_{m+1}$.

Assume that there are at least two tiles containing one of them, say $p_{m}\in
D_{1},D_{2}$. Then $g_{12}$ is the reflection about either the line
$L^{\prime}$ through $[p_{1},p_{m}]$, or the bisector of $[p_{1},p_{m}]$, or
the midpoint of $[p_{1},p_{m}]$. Here the second case contradicts Remark 2. In
the first and the third cases we have that $D_{1},D_{2}\subset P\cap
P^{\prime}$, where $P^{\prime}=g_{12}(P)$ (cf. Figure 2). Thus,
$P^{\prime}\cap P\subsetneq P$ yields that $k=3$, and $P\setminus
P^{\prime}\subseteq D_{3}$. Then the compactness of $D_{3}$ implies that
$p_{m},p_{m+1}\in D_{3}$; a contradiction.

Figure 2. An illustration for the proof of Lemma 1

We have shown that $k=2$, and there is a unique tile containing $p_{m}$ and a
unique tile containing $p_{m+1}$. Let these tiles be $D_{1}$ and $D_{2}$,
respectively. Let $q$ be the midpoint of $[p_{m},p_{m+1}]$ and assume, without
loss of generality, that $q\in D_{1}$. Then the only congruent copy of $P$
containing $p_{1},p_{m},q$ is $P$, implying that $g_{12}(P)=P$. Since we also
have $g_{12}([p_{1},p_{m}])=[p_{1},p_{m+1}]$, this yields that $g_{12}$ is the
reflection about the line $L$ through $[o,p_{1}]$, from which the assertion
readily follows. ∎

Next, we recall Lemma 2.3 from [13].

######  Lemma 2.

Let $\\{D_{1},D_{2},D_{3}\\}$ be a tiling of the disc $D$ where for $i=1,2,3$
$D_{i}$ is a disc such that $S_{i}=D_{i}\cap\partial D$ is a nondegenerate
simple continuous curve. Then $D_{1}\cap D_{2}\cap D_{3}$ is a singleton
$\\{q\\}$, and for any $i\neq j,D_{i}\cap D_{j}$ is a simple continuous curve
connecting $q$ and a point in $\partial D$.

Our next lemma is a generalization of Lemma 2.

######  Lemma 3.

Let the topological disc $D$ be decomposed into three topological discs
$D_{1}$, $D_{2}$ and $D_{3}$. For $i=1,2,3$, set $S_{i}=D_{i}\cap\partial D$.
Then, with a suitable choice of indices, exactly one of the following holds
(cf. Figure 3).

  1. (1)

$S_{1}$ contains at most two points, and $S_{2}$ and $S_{3}$ are connected
arcs whose union covers $\partial D$.

  2. (2)

$S_{1}$ is the union of two disjoint, connected, nondegenerate arcs, the sets
$S_{2},S_{3},D_{1}\cap D_{2}$, $D_{1}\cap D_{3}$ are connected arcs, and
$D_{2}\cap D_{3}=\emptyset$.

  3. (3)

$S_{2}$, $S_{3}$, $D_{1}\cap D_{2}$, $D_{1}\cap D_{3}$, $D_{2}\cap D_{3}$ are
connected arcs, $D_{1}\cap D_{2}\cap D_{3}$ is a singleton $\\{q\\}$, and
$S_{1}$ is either a connected arc, or the union of a connected arc and
$\\{q\\}$.

Figure 3. The three topological types described in Lemma 3 with a Euclidean
disc as $D$. We note that the points $q_{1}$ and $q_{2}$ in the left panel,
and $q$ in the right panel may lie on $\partial D$.

###### Proof.

Assume that one of the $S_{i}$s, say $S_{1}$, has more than one component, and
let $q_{1},q_{2},\ldots,q_{m}$ be points of $S_{1}$, in this cyclic order,
contained in different components of $S_{1}$. Let $r_{1},r_{2},\ldots,r_{m}$
be points in ${\partial D}\setminus S_{1}$ such that
$q_{1},r_{1},q_{2},\ldots,q_{m},r_{m}$ are in this cyclic order in ${\partial
D}$. Note that every $r_{j}$ belongs to $S_{2}$ or $S_{3}$, and no two of them
belongs to the same set. Indeed, if, say, $r_{j_{1}}$ and $r_{j_{2}}$ belongs
to $S_{2}$, where $j_{1}\neq j_{2}$, then any in-curve in $D_{2}$ connecting
them, and any in-curve in $S_{1}$ connecting $q_{j_{1}}$ and $q_{j_{2}}$ would
cross, which is a contradiction. Thus, we have $m=2$, which also yields, by
the same argument, that $S_{2}$ and $S_{3}$ are connected. If neither
component of $S_{1}$ is a singleton, then the closure of $(\partial
D_{1})\setminus S_{1}$ contains two disjoint, simple curves which, apart from
their endpoints, are contained in $\operatorname{int}(D)$. Since in this case
$D_{2}$ and $D_{3}$ can be separated by an in-curve of $D_{1}$ disjoint from
$D_{2}\cup D_{3}$, it follows by compactness that the two components of
$\operatorname{cl}((\partial D_{1})\setminus S_{1})$ coincide with $D_{1}\cap
D_{2}$ and $D_{1}\cap D_{3}$, implying (2). If exactly one component of
$S_{1}$ is a singleton, a similar argument can be applied, implying (3).
Finally, if both components of $S_{1}$ are singletons, the conditions in (1)
are satisfied.

Assume that all $S_{i}$s are connected. Since $S_{1}\cup S_{2}\cup
S_{3}=\partial D$ and every $S_{i}$ is a simple connected arc properly
contained in $\partial D_{i}$, we have that at least two of the $S_{i}$s
contain more than one point. If one of them, say $S_{1}$, contains at most one
point, then (1) follows. If $S_{1}$, $S_{2}$ and $S_{3}$ are nondegenerate,
simple arcs, then the conditions of Lemma 2 are satisfied, implying (3). ∎

######  Definition 1.

Let $\\{D_{1},D_{2},D_{3}\\}$ be a tiling of the topological disc $D$. If the
discs $D_{1},D_{2},D_{3}$ satisfy the conditions in ($i$) of Lemma 3 with
$i=1$, $i=2$ or $i=3$, we say that the tiling is a _Type $i$ decomposition_ of
$D$.

######  Remark 3.

Assume that $D$ is decomposed into two topological discs $D_{1},D_{2}$, and
for $i=1,2$, set $S_{i}=D_{i}\cap\partial D$. Note that since the number of
tiles is more than one, we have that no $S_{i}$ coincides with $D$.
Furthermore, by the argument in the proof of Lemma 3 we also have that $S_{1}$
and $S_{2}$ are connected. Motivated by this property, we call any tiling of
$D$ with two topological discs a Type 1 decomposition of $P$.

Lemmas 4-6 and Definitions 2-3 are from [13].

######  Lemma 4.

Let $G$ and $C$ be simple curves. Then $G$ contains finitely many congruent
copies of $C$ which are mutually disjoint, apart from possibly their
endpoints.

######  Definition 2.

A _multicurve_ (see also [12]) is a finite family of simple curves, called the
_members of the multicurve_ , which are parameterized on nondegenerate closed
finite intervals, and any point of the plane belongs to at most one member, or
it is the endpoint of exactly two members. If $\mathcal{F}$ and $\mathcal{G}$
are multicurves, $\bigcup\mathcal{F}=\bigcup\mathcal{G}$, and every member of
$\mathcal{F}$ is the union of some members of $\mathcal{G}$, we say that
$\mathcal{G}$ is a _partition_ of $\mathcal{F}$.

######  Definition 3.

Let $\mathcal{F}$ and $\mathcal{G}$ be multicurves. If there are partitions
$\mathcal{F}^{\prime}$ and $\mathcal{G}^{\prime}$ of $\mathcal{F}$ and
$\mathcal{G}$, respectively, and a bijection
$f\colon\mathcal{F}^{\prime}\to\mathcal{G^{\prime}}$ such that $f(C)$ is
congruent to $C$ for all $C\in\mathcal{F}^{\prime}$, we say that $\mathcal{F}$
and $\mathcal{G}$ are _equidecomposable_.

######  Lemma 5.

If $\mathcal{F}$ and $\mathcal{G}$ are multicurves with
$\bigcup\mathcal{F}=\bigcup\mathcal{G}$, then $\mathcal{F}$ and $\mathcal{G}$
are equidecomposable.

######  Lemma 6.

If $\mathcal{F}$ and $\mathcal{G}$ are equidecomposable, and their subfamilies
$\mathcal{F}^{\prime}\subseteq\mathcal{F}$ and
$\mathcal{G}^{\prime}\subseteq\mathcal{G}$ are equidecomposable, then
$\mathcal{F}\setminus\mathcal{F}^{\prime}$ and
$\mathcal{G}\setminus\mathcal{G}^{\prime}$ are equidecomposable.

We finish with a remark and a definition.

######  Remark 4.

Let $P$ be a regular polygon, and let $\\{D_{1},D_{2},D_{3}\\}$ be a
monohedral tiling of $P$. For $i=1,2,3$, set $S_{i}=D_{i}\cap\partial P$. Note
that for any $i\neq j$,
$g_{ij}(S_{i})\subset(\partial\operatorname{conv}(D_{j}))\cap(\partial
D_{j})$. Furthermore, if there is a vertex $p_{t}$ of $P$ in the interior of
$S_{j}$ with $p_{t}\in g_{ij}(S_{i})$, then $S_{j}=g_{ij}(S_{i})$. Thus,
$S_{j}$ and $g_{ij}(S_{i})$ are either disjoint apart from their endpoints, or
their intersection contains subsets of the sides adjacent to the endpoints, or
they coincide. In the first case we say that $S_{j}$ and $g_{ij}(S_{i})$ are
_nonoverlapping_ , and in the second case that they are _slightly overlapping_
(cf. Figure 4). We observe that $S_{i}$ and $S_{j}=g_{ij}(S_{i})$ are
nonoverlapping, slightly overlapping and equal if and only if $g_{il}(S_{i})$
and $g_{jl}(S_{j})$ are nonoverlapping, slightly overlapping and equal,
respectively, for an arbitrary value of $l$.

Figure 4. An illustration for Remark 4. In the left panel $S_{2}$ and
$g_{12}(S_{1})$ do not overlap; in the right panel $S_{2}$ and $g_{12}(S_{1})$
slightly overlap.

##  3\. Proof of Theorem 1

By Lemma 3 and Remark 3, it is sufficient to prove the theorem for Type 1,
Type 2 and Type 3 decompositions of $P$. In the following, we present the
proof for each type in a separate subsection. Throughout the proof, we assume
that no $D_{i}$ contains diametrically opposite points of $P$, as otherwise
the assertion readily follows from Lemma 1.

###  3.1. Proof for Type $1$ decompositions

We choose our notation in such a way that $S_{1}$ and $S_{2}$ are
nondegenerate connected arcs in $\partial P$ whose union is $P$, and $S_{3}$,
if it exists, contains at most two points. Observe that in this case $n$ is
even, as otherwise $S_{1}$ or $S_{2}$ contains at least $\frac{n+1}{2}$
vertices of $P$, inculding a pair of diametrically opposite points of $P$.
Consider the sets $S_{1}$ and $S_{1}^{\prime}=g_{21}(S_{2})$. By Remark 4, we
distinguish three cases.

_Case 1_ , $S_{1}$ and $S_{1}^{\prime}$ do not overlap.
In this case they are nonoverlapping arcs in the boundary of
$\operatorname{conv}(D_{1})$ whose total length is $\operatorname{perim}(P)$,
which implies that $\operatorname{conv}(D_{1})=P$ and
$\partial(\operatorname{conv}D_{1})=S_{1}\cup S_{1}^{\prime}$. On the other
hand, since in this case $S_{1}\cup S_{1}^{\prime}$ is a simple, closed curve,
we have $D_{1}=\operatorname{conv}(D_{1})=P$, which contradicts the assumption
that $k>1$.

_Case 2_ , $S_{1}$ and $S_{1}^{\prime}$ slightly overlap.
Let $L$ be a sideline of $P$ containing an interior point of $S_{1}^{\prime}$,
and let $L^{\prime}$ be the supporting line of $P$ parallel to $L$. Let
$G_{1}$ and $G_{2}$ denote the two components of $\operatorname{cl}(\partial
P\setminus(L\cup L^{\prime}))$. Clearly, since $L$ contains a common endpoint
of $S_{1}$ and $S_{2}$, at least one of $G_{1}$ and $G_{2}$ contains no
endpoint of $S_{1}$ and $S_{2}$ in its interior, and hence, we may assume that
e.g. $G_{2}\subset S_{2}$. Then the facts that $S_{1}^{\prime}$ and $S_{1}$
are slightly overlapping and $S_{1}^{\prime}\subset P$ yield that $L^{\prime}$
also contains a point of $S_{1}^{\prime}$, and $G_{1}\subset S_{1}$.
Furthermore, since in this case $S_{1}\cup S_{1}^{\prime}$ is a simple closed
curve in $\partial D_{1}$, we have that $D_{1}=\operatorname{conv}(S_{1}\cup
S_{1}^{\prime})$. Let $q$ and $q^{\prime}$ be the midpoints of the sides of
$P$ on $L$ and $L^{\prime}$, respectively, and observe that the translate
$G_{2}^{\prime}$ of $G_{2}$ whose endpoints are $q$ and $q^{\prime}$ is
contained in $D_{1}$. Thus, the area of $D_{1}$ is greater than or equal to
the area of $\operatorname{conv}(G_{1}\cup G_{2}^{\prime})$, and the area of
the latter region is strictly greater than $\frac{\operatorname{area}(P)}{2}$.
This contradicts the fact that the examined tiling of $P$ is monohedral.

_Case 3_ , $S_{1}=S_{1}^{\prime}$.
In this case $g_{21}$ is either the reflection about the line $L$ through the
two endpoints of $S_{1}$ and $S_{2}$, or it is the reflection about $o$. This
implies Theorem 1 for $k=2$. Furthermore, if $k=3$, then
$D_{3}=\operatorname{cl}(P\setminus(D_{1}\cup D_{2}))$ is symmetric to $L$ or
$o$, respectively, and $P$ has an even number of sides.

Assume that $k=3$ and $g_{21}$ is the reflection about $o$. Then, since
$D_{1}$, $D_{2}$ and $D_{3}$ are all congruent, it follows that both $D_{1}$
and $D_{2}$ are centrally symmetric. As $D_{1},D_{2}\subset P$ we also have
that the centers of $D_{1}$ and $D_{2}$ are contained on the line through $o$
parallel to the two sides of $P$ containing the common endpoints of $S_{1}$
and $S_{2}$. Let these two sides of $P$ be $E$ and $E^{\prime}$, and let the
centers of symmetry of $D_{1}$ and $D_{2}$ be $c_{1}$ and $c_{2}$,
respectively. From the properties of central symmetry and the fact that
$S_{1}=S_{1}^{\prime}$, we have that for $i=1,2$, $E\cap S_{i}$ and
$E^{\prime}\cap S_{i}$ are segments of length $1/2$. Furthermore, for $i=1,2$
the union of $S_{i}$ and and its reflection about $c_{i}$ is a simple closed
convex curve in $\partial D_{i}$, implying that its convex hull is $D_{i}$.
Thus, $D_{1}$ and $D_{2}$ overlap; a contradiction.

Finally, assume that $k=3$ and $g_{21}$ is the reflection about the line $L$
passing through the common endpoints of $S_{1}$ and $S_{2}$. Since $L$ is a
symmetry line of $P$ and no $D_{i}$ contains diametrically opposite points of
$P$, it follows that $L$ passes through the midpoints of two opposite sides
$E,E^{\prime}$ of $P$. Furthermore, since $D_{3}$ is symmetric to the line
$L$, we obtain that $D_{1}$ and $D_{2}$ have lines of symmetry, which we
denote by $L_{1}$ and $L_{2}$, respectively. Since both discs are contained in
the infinite strip bounded by the two sidelines of $P$ through $E$ and
$E^{\prime}$, $L_{1}$ and $L_{2}$ are parallel to $L$, or coincide with the
line $L^{*}$ through $o$ perpendicular to $L$.

Assume that one of $L_{1}$ and $L_{2}$, say $L_{1}$, is parallel to $L$, and
let $S^{\prime}$ denote the reflection of $S_{1}$ about $L_{1}$. Since
$S_{1}\cup S^{\prime}$ is a simple, closed curve in $\partial D_{1}$, we have
$S_{1}\cup S^{\prime}=\partial D_{1}$, which yields that
$D_{1}=\operatorname{conv}(S_{1}\cup S^{\prime})$. On the other hand, as the
endpoints of $S_{1}$ are midpoints of two opposite sides of $P$, from this an
elementary computation shows that
$\operatorname{area}(D_{1})>\frac{\operatorname{area}(P)}{3}$, a
contradiction. Thus, we have $L_{1}=L^{*}$, and we remark that our argument
shows that _any_ line of symmetry of $D_{1}$ coincides with $L^{*}$, and,
applying this argument, we obtain the same statement for $D_{2}$. On the other
hand, if both $D_{1}$ and $D_{2}$ are symmetric to $L^{*}$, then the same
holds for $D_{3}$. Hence, $D_{3}$ is symmetric to both $L$ and $L^{*}$, which
yields that $D_{1}$ (resp., $D_{2}$) has a line of symmetry different from
$L^{*}$, which contradicts our previous observation.

###  3.2. Proof for Type $2$ decompositions

We apply the notation of Lemma 3. Furthermore, following Figure 3, we denote
the two components of $S_{1}$ by $S_{1}^{\prime}$ and $S_{1}^{\prime\prime}$.
We distinguish three cases.

_Case 1_ , both $S_{1}^{\prime}$ and $S_{1}^{\prime\prime}$ contain vertices
of $P$.
Without loss of generality, we may assume that the vertices of $P$ in
$S_{1}^{\prime}$ are $p_{1},p_{2},\ldots,p_{m}$ for some $1\leq m\leq n-1$.
First, we show that $n$ is even. Suppose for contradiction that $n=2t+1$ for
some $t\geq 2$. Then, since $S_{1}$ contains no diametrically opposite points
of $P$, we have that $p_{t+1},p_{t+2},\ldots,p_{t+m+1}$ belong to the same set
$S_{i}\neq S_{1}$. Without loss of generality, we may assume that they, and
also $p_{m+1}$, belong to $S_{2}$. This yields that
$p_{m+1},p_{m+2},\ldots,p_{m+t+1}$ belong to $S_{2}$, and thus, $S_{2}$
contains at least $t+1$ vertices of $P$, which contradicts our assumption that
$D_{2}$ does not contain diametrically opposite points of $P$. Thus, we have
that $n$ is even.

Let $n=2t$ for some $t\geq 3$. Similarly like in the previous paragraph, we
have that $p_{t+1},\ldots,p_{t+m}$ are not points of $S_{1}^{\prime\prime}$,
and thus, they all belong to $S_{2}$ or all belong to $S_{3}$. Without loss of
generality, assume that they belong to $S_{2}$. Thus,
$p_{m+1},\ldots,p_{m+t}\in S_{2}$. Since no $S_{i}$ contains diametrically
opposite points of $P$, we also have $m\leq t$ and $p_{m+t+1}\in
S_{1}^{\prime\prime}$. This implies that $l(S_{3})<l(S_{2})$, and that $S_{3}$
contains strictly less vertices of $P$ than $S_{2}$. Since neither $S_{1}$ nor
$S_{2}$ contains diametrically opposite vertices of $P$, we also have that the
endpoints of $S_{2}$ are interior points of two sides of $P$. The facts that
$S_{2}$ contains exactly $t$ vertices of $P$ and $S_{1}$ is disconnected yield
also that $S_{2}\neq g_{12}(S_{1})$ and $S_{2}\neq g_{32}(S_{3})$.
Furthermore, $g_{32}(S_{3})$ and $S_{2}$ are not slightly overlapping, since
otherwise $S_{3}$ and $g_{23}(S_{2})$ are slightly overlapping (cf. Remark 4),
which contradicts the fact that in this case the side of $P$ opposite of the
overlap is contained in $S_{2}$. Thus, we have that either $S_{1}$,
$g_{21}(S_{2})$ and $g_{31}(S_{3})$ are mutually nonoverlapping, or $S_{1}$
and $g_{21}(S_{2})$ slightly overlap.

If $S_{1}$, $g_{21}(S_{2})$ and $g_{31}(S_{3})$ are mutually nonoverlapping,
then their total length is equal to $\operatorname{perim}(P)$, implying that
$\operatorname{perim}(\operatorname{conv}(D_{1}))\geq\operatorname{perim}(P)$.
This yields that $D_{1}=\operatorname{conv}(D_{1})=P$, which contradicts our
assumption that $k=3$ for any Type 2 decomposition of $P$. Thus, the only
possibility left is that $S_{1}$ and $g_{21}(S_{2})$ slightly overlap. Since
the endpoints of $S_{2}$ are interior points of two opposite sides of $P$,
this yields that $g_{21}(S_{2})$ contains at least one endpoint of $S_{2}$. On
the other hand, since apart from its endpoints, no point of $S_{1}$ may belong
to $D_{2}$, we also have that $g_{21}(S_{2})$ contains both endpoints of
$S_{2}$, and also that it is a translate of $S_{2}$. Let the endpoints of
$S_{2}$ on $[p_{m},p_{m+1}]$ and $[p_{m+t},p_{m+t+1}]$ be $q$ and
$q^{\prime}$, respectively. Then $g_{21}([p_{m+1},q])$ is either $[p_{m},q]$
or $[p_{m+t+1},q^{\prime}]$, which yields that $q$ is the midpoint of
$[p_{m},p_{m+1}]$ and $q^{\prime}$ is the midpoint of $[p_{m},p_{m+1}]$ (cf.
Figure 5). On the other hand, since $g_{21}(S_{2})\subset\partial D_{1}$, it
separates $D_{2}$ from $D_{1}\cup D_{3}$. In other words, $D_{2}$ is the
region bounded by the union of $S_{2}$ and the part of $g_{21}(S_{2})$
connecting $q$ and $q^{\prime}$. But this and the fact $S_{3}$ is not empty
yields that $D_{1}$ is the translate of $D_{2}$ by the vector $q-p_{m+1}$, and
hence, $D_{3}=\operatorname{cl}(P\setminus(D_{1}\cup D_{2}))$ is not congruent
to $D_{1}$ and $D_{2}$; a contradiction.

Figure 5. An illustration for Case 1 in Subsection 3.2, where $t=4$ and $m=1$.
In the picture $S_{1}^{\prime}$ and $S_{1}^{\prime\prime}$ are drawn with
solid, $S_{2}$ and $S_{3}$ with dashed, and $g_{21}(S_{2})$ with dotted lines.

_Case 2_ , exactly one of $S_{1}^{\prime}$ and $S_{1}^{\prime\prime}$ contains
a vertex of $P$.
Without loss of generality, assume that the vertices of $P$ in
$S_{1}^{\prime}$ are $p_{1},p_{2},\ldots,p_{m}$ for some $1\leq m\leq n$.

First, we consider the case that $n$ is odd, namely that $n=2t+1$ for some
integer $t\geq 2$. Observe that since the diameter graph of the vertex set of
$P$ is an odd cycle, and hence, it cannot be colored with two colors, every
$S_{i}$ contains a vertex of $P$ in its interior. We show that the side of $P$
containing $S_{1}^{\prime\prime}$ is opposite of a vertex in $S_{1}^{\prime}$.
Indeed, suppose for contradiction that
$[p_{t+1},p_{t+2}],\ldots,[p_{t+m},p_{t+m+1}]$ are disjoint from
$S_{1}^{\prime\prime}$. Then they all belong to the same $S_{i}$, and thus, we
may assume that $p_{m+1},\ldots,p_{t+m+1}$ belong to $S_{2}$. But then $S_{2}$
contains diametrically opposite vertices of $P$, which contradicts our
assumption. Hence, we have that for some $1\leq i\leq m$,
$S_{i}^{\prime\prime}\subset[p_{i+t},p_{i+t+1}]$.

Note that the fact that $S_{1}$ is disconnected implies that $g_{1i}(S_{1})$
does not coincide with $S_{i}$. Assume that, say, $g_{12}(S_{1})$ slightly
overlaps $S_{2}$. Then $g_{12}$ is the composition of a symmetry of $P$ and a
(nondenegerate) translation parallel to a side of $P$, which contradicts the
fact that $g_{12}(S_{1})\subset P$. Thus, we have that $S_{1}$ does not
overlap $g_{21}(S_{2})$ and $g_{31}(S_{3})$. Without loss of generality, we
may assume that $l(S_{2})\geq l(S_{3})$. For $i=2,3$, let $T_{i}$ denote the
segment connecting the endpoints of $S_{i}$,
$K_{i}=\operatorname{conv}(S_{i})$ and
$C_{i}=\operatorname{cl}(\operatorname{conv}(D_{1})\cap K_{i})$ (cf. Figure
6). Recall that $g_{21}(S_{2})$ and $g_{31}(S_{3})$ are contained in
$\partial(\operatorname{conv}(D_{1}))$ and they do not overlap $S_{1}$. Since
$S_{2}$ and $S_{3}$ contain vertices of $P$, we also have that they do not
overlap $T_{2}$ and $T_{3}$. Thus, in particular, $C_{2}$ or $C_{3}$ is a
plane convex body with perimeter at least $l(T_{2})+l(S_{2})$ or
$l(T_{3})+l(S_{2})$, respectively. As
$\operatorname{perim}(K_{i})=l(S_{i})+l(T_{i})$ for $i=2,3$, this yields that
$g_{21}(S_{2})$ coincides with $S_{2}$ or $S_{3}$, a contradiction.

Figure 6. An illustration for Case 2 in Subsection 3.2, where $t=4$ and $m=1$.
In the picture $S_{1}^{\prime}$ and $S_{1}^{\prime\prime}$ are drawn with
dashed, and $S_{2}$ and $S_{3}$ are drawn with solid lines.

In the remaining part of Case 2, we assume that $n$ is even, i.e. $n=2t$ for
some $t\geq 3$. Assume that $S_{1}^{\prime},S_{2},S_{1}^{\prime\prime},S_{3}$
are in counterclockwise direction on $\partial P$. Note that at least one of
$S_{2}$ and $S_{3}$ contains vertices of $P$. Furthermore, similarly like for
$n$ odd, the fact that neither $S_{2}$ nor $S_{3}$ contains diametrically
opposite points yields that there are points $q^{\prime}\in S_{1}^{\prime}$
and $q^{\prime\prime}\in S_{1}^{\prime\prime}$ on opposite sides of $P$.

Clearly, $g_{12}(S_{1})\neq S_{2}$ as $S_{1}$ is disconnected. Assume that
they are slightly overlapping. Then $g_{12}(q^{\prime})$ and
$g_{12}(q^{\prime\prime})$ lie on opposite sides of $P$, and hence, they
belong to $S_{2}$. Thus, $S_{2}$ contains $t$ vertices of $P$, and
$q^{\prime},q^{\prime\prime}$ and their images under $g_{12}$ are on the same
two sides of $P$. This yields that $g_{12}$ is either a translation parallel
to these sides, or the composition of such a translation with a reflection to
a line parallel or perpendicular to these sides, or the origin. Let these two
sides be $E^{\prime}$ and $E^{\prime\prime}$ with $q^{\prime}\in E^{\prime}$
$q^{\prime\prime}\in E^{\prime\prime}$. Then $g_{12}((E^{\prime}\cup
E^{\prime\prime})\cap S_{1})=(E^{\prime}\cup E^{\prime\prime})\cap S_{2}$.
Since $E^{\prime}\cap S_{1}$, $E^{\prime}\cap S_{2}$, $E^{\prime\prime}\cap
S_{2}$ and $E^{\prime\prime}\cap S_{1}$ are in counterclockwise order on
$\partial P$, we have that $g_{12}$ is a translation parallel to $E^{\prime}$,
or its composition with the reflection about the line through $o$ and parallel
to $E^{\prime}$. But both cases contradict the fact that
$E^{\prime\prime}\cap(S_{1}\cup S_{2})$ is strictly shorter than
$E^{\prime}\cap(S_{1}\cup S_{2})=E^{\prime}$.

We have obtained that $g_{12}(S_{1})$ and $S_{2}$ do not overlap. It follows
similarly that $g_{13}(S_{1})$ and $S_{3}$ do not overlap, or equivalently,
that $g_{21}(S_{2})$ and $g_{31}(S_{3})$ do not overlap $S_{1}$. But in this
case we may apply the argument used for $n$ odd.

_Case 3_ , neither $S^{\prime}_{1}$ nor $S^{\prime\prime}_{1}$ contains a
vertex of $P$.
Recall that by the definition of Type 2 configuration (cf. Lemma 3), both
$S_{1}^{\prime}$ and $S_{1}^{\prime\prime}$ are nondegenerate segments.

As we remarked in Case 2, if $n$ is odd, then the diameter graph of the vertex
set of $P$ is an odd cycle, which is not $2$-colorable. But this contradicts
the assumptions that none of $S_{1}$, $S_{2}$, $S_{3}$ contains diametrically
opposite vertices of $P$, and $S_{1}$ does not contain a vertex of $P$. Thus,
the condition of Case 3 is satisfied only if $n$ is even, and $S_{1}^{\prime}$
and $S_{1}^{\prime\prime}$ lie on opposite sides of $P$. Let these sides of
$P$ be $E^{\prime}$ and $E^{\prime\prime}$ with $S_{1}^{\prime}\subset
E^{\prime}$.

Clearly, we have $g_{21}(S_{2})\neq S_{1}$. Consider the case that
$g_{21}(S_{2})$ slightly overlaps $S_{1}$. Then the endpoints of
$g_{21}(S_{2})$ lie on $E^{\prime}$ and $E^{\prime\prime}$, which yields that
either $D_{2}$ is the region bounded by
$S_{2}\cup\left(g_{21}(S_{2})\setminus(E^{\prime}\cup
E^{\prime\prime})\right)$, or $D_{3}$ is the region bounded by
$S_{3}\cup\left(g_{21}(S_{2})\setminus g_{21}(E^{\prime}\cup
E^{\prime\prime})\right)$. From these two cases we obtain
$\operatorname{area}(D_{2})<\frac{1}{3}\operatorname{area}(P)$ and
$\operatorname{area}(D_{3})<\frac{1}{3}\operatorname{area}(P)$, respectively,
which contradicts the fact that the tiling is monohedral. Thus, we are left
with the case that $g_{21}(S_{2})$ and $S_{1}$ do not overlap. Similarly, we
obtain that $g_{31}(S_{3})$ and $S_{1}$ do not overlap. But then $S_{1}$,
$g_{21}(S_{2})$ and $g_{31}(S_{3})$ are mutually nonoverlapping arcs in
$\partial(\operatorname{conv}(D_{1}))$ with total length equal to
$\operatorname{perim}(P)$, implying that $D_{1}=\operatorname{conv}(D_{1})=P$;
a contradiction.

###  3.3. Proof for Type $3$ decompositions

The proof presented in this subsection roughly follows the structure of the
proof in [13] with some of the arguments borrowed from there; in particular,
depending on the number of coinciding arcs among $S_{1},g_{21}(S_{2})$ and
$g_{31}(S_{3})$, we distinguish three cases.

_Case 1_ , no two of $S_{1},g_{21}(S_{2})$ and $g_{31}(S_{3})$ coincide.
If no two of these arcs overlap, then the fact that their total length is
equal to $\operatorname{perim}(P)$ yields that
$D_{1}=\operatorname{conv}(D_{1})=P$; a contradiction. Thus, we have that at
least one pair among them overlaps. Before proceeding further, we use this
observation to show that all of $S_{1}$, $S_{2}$ and $S_{3}$ contain at least
one vertex of $P$ different from their endpoints. Indeed, suppose for
contradiction that one of them, say $S_{1}$, contains no vertex of $P$ in its
interior. Then, since $D_{1},D_{2}$ and $D_{3}$ contain no diametrically
opposite points of $P$, we have that $n$ is even, and that both $S_{2}$ and
$S_{3}$ contain interior points on opposite edges of $P$. Thus, if
$g_{21}(S_{2})$ and $S_{1}$ slightly overlap, then $S_{1}$ would intersect a
pair of opposite edges of $P$ in nondegenerate segments and the configuration
is not Type $3$; a contradiction. The cases that $g_{31}(S_{3})$ slightly
overlaps $S_{1}$ or $g_{21}(S_{2})$ can be eliminated by a similar argument.
Thus, we obtain that $S_{1}$, $g_{21}(S_{2})$ and $g_{31}(S_{3})$ do not
overlap; a contradiction. Hence, in the remaining part of Case 1 we assume
that all of $S_{1}$, $S_{2}$ and $S_{3}$ contain vertices of $P$ different
from their endpoints.

Consider the case that there are at least two overlapping pairs among
$S_{1},g_{21}(S_{2})$ and $g_{31}(S_{3})$; without loss of generality, we may
assume that $g_{21}(S_{2})$ and $g_{31}(S_{3})$ slightly overlap $S_{1}$.
Observe that for $i=2,3$, $D_{i}\subseteq P$, which yields that
$D_{1}=g_{i1}(D_{i})\subseteq g_{i1}(P)$. Thus, $D_{1}\subseteq Q=P\cap
g_{21}(P)\cap g_{31}(P)$, and $Q$ contains $S_{1}$, $g_{21}(S_{2})$ and
$g_{31}(S_{3})$ on its boundary. The total number of full sides of $P$ in the
arcs $S_{1}$, $S_{2}$ and $S_{3}$ is at least $(n-3)$, and, by the definition
of slight overlap, congruent copies of these sides appear on the boundary of
$Q$. Hence, $\partial Q$ contains at least $(n-3)$ unit segments. On the other
hand, $Q$ is the intersection of three translates of $P$, such that for both
$g_{21}(P)$ and $g_{31}(P)$ the translation vector is parallel to a side of
$P$, which implies that $Q$ contains at most $(n-4)$ unit segments on its
boundary; a contradiction.

We are left with the case that exactly one pair of $S_{1},g_{21}(S_{2})$ and
$g_{31}(S_{3})$ overlaps. Without loss of generality, we may assume that
$S_{1}$ and $g_{21}(S_{2})$ overlap. Let us choose our notation such that the
vertices of $P$ on $S_{1}$ are $p_{1},p_{2},\ldots,p_{m}$, and
$S_{1},S_{2},S_{3}$ are in counterclockwise order around $P$. For any $i\neq
j$, let $q_{ij}$ be the common endpoint of $S_{i}$ and $S_{j}$. We assume that
$q_{23}\in[p_{l},p_{l+1}]$ with $q_{23}\neq p_{l+1}$. By our assumption, we
have that $g_{21}(q_{12})$ or $g_{21}(q_{23})$ lies in the interior of
$S_{1}$. Depending on which of $q_{12}$ and $q_{23}$ lies on which side of
$S_{1}$, we distinguish four cases.

If $g_{21}(q_{12})$ lies in the interior of $S_{1}$ and
$g_{21}(q_{12})\in[p_{k},p_{k+1}]$. Then $q_{12}$ is the midpoint of
$[p_{m},p_{m+1}]$ and $g_{21}([q_{12},p_{m+1}])=[q_{12},p_{m}]$. This implies
that $\partial D_{1}$ and $\partial D_{2}$ cross at $q_{12}$; a contradiction.

If $g_{21}(q_{12})$ lies in the interior of $S_{1}$ and
$g_{21}(q_{12})\in[p_{n},p_{1}]$. Then $g_{21}(q_{12})=p_{1}$,
$g_{21}(p_{m+1})=q_{13}$ (or equivalently, $g_{12}(p_{1})=q_{12}$ and
$g_{12}(q_{13})=p_{m+1}$), and $q_{13}$ and $q_{12}$ are interior points of
$[p_{m},p_{m+1}]$ and $[p_{n},p_{1}]$, respectively. Note that
$\operatorname{conv}(D_{2})\subset P$ implies that
$\operatorname{conv}(D_{1})\subset g_{21}(P)$. On the other hand, $g_{21}(P)$
is the translate of $P$ by the vector $q_{12}-p_{n}$. Let
$P_{0}=P\cap(q_{12}-p_{n}+P)$ (cf. Figure 7). Then $P_{0}$ is a convex polygon
whose every side is parallel to some side of $P$. Let $C=S_{1}\cup
g_{21}(S_{2})$, and observe that $C$ and $g_{31}(S_{3})$ are nonoverlapping
curves in $\partial(\operatorname{conv}(D_{1}))$. If $q_{23}$ is not a vertex
of $P$, then the total turning angle along the curves $C$ and $g_{32}(S_{3})$
is $2\pi$, which implies that $\partial(\operatorname{conv}(D_{2}))$ is the
union of $C$ and $g_{31}(S_{3})$ and possibly two segments such that the lines
through them contain segments from both $C$ and $g_{32}(S_{3})$; this
contradicts the fact that $\operatorname{conv}(D_{1})$ is contained in
$P_{0}$. If $q_{23}$ is a vertex of $P$, we may apply the same argument after
observing that $\operatorname{conv}(C\cup g_{31}(S_{3}))$ is a convex polygon,
and the fact that $\operatorname{conv}(C\cup
g_{31}(S_{3}))\subset\operatorname{conv}(D_{1})\subset P_{0}$ implies that the
turning angle of $\operatorname{conv}(C\cup g_{31}(S_{3}))$ at
$g_{21}(q_{23})$ is at least $\frac{2\pi}{n}$.

Figure 7. An illustration for Case 1 in Subsection 3.3

Finally, in the remaining cases, if $g_{21}(q_{23})$ lies in the interior of
$S_{1}$ with $g_{21}(q_{23})\in[p_{m},p_{m+1}]$ or
$g_{21}(q_{23})\in[p_{n},p_{1}]$, we can apply the argument in the previous
case.

_Case 2_ , exactly two of $S_{1},g_{21}(S_{2})$ and $g_{31}(S_{3})$ coincide.
Without loss of generality, we may assume that $S_{1}=g_{21}(S_{2})$. In the
consideration in this case, for brevity, for any $i\neq j$ we let $q_{ij}$
denote the intersection point of $S_{i}$ and $S_{j}$, let $q$ denote the
unique point in $D_{1}\cap D_{2}\cap D_{3}$, and set $C_{ij}=D_{i}\cap D_{j}$.
By Lemma 3, we have that $C_{ij}$ is a simple (possibly degenerate) curve
connecting $q_{ij}$ and $q$.

Note that since $S_{3}$ cannot contain more than $\frac{n}{2}$ vertices of
$P$, we have that each of $S_{1}$ and $S_{2}$ contains at least $\frac{n}{4}$
and at most $\frac{n}{2}$ vertices of $P$. This and $n\geq 5$ implies that
$g_{21}(P)=P$; that is, $g_{21}$ is an isometry of $P$. In particular,
$g_{21}$ (and consequently $g_{12}$) is either a reflection about a symmetry
line of $P$, or a rotation around $o$ with angle $\alpha=\frac{2m\pi}{n}$ for
some integer $1\leq m\leq n$. Depending on the type of $g_{21}$, we
distinguish two subcases.

_Subcase 2.1_ , $g_{21}$ is a rotation around $o$. Then the angle of rotation
is $\alpha=\frac{2m\pi}{n}$ for some integer $\frac{n}{4}\leq m<\frac{n}{2}$,
which implies, in particular, that $l(S_{1})=l(S_{2})=m$, and $l(S_{3})=n-2m$.

Observe that since $o$ is a fixed point of $g_{21}$, we have either $o\in
D_{1}\cap D_{2}$ or $o\in\operatorname{int}(D_{3})$. First, assume that $o\in
D_{1}\cap D_{2}$. If $o=q$, then $g_{21}(q_{12})=q_{13}$,
$g_{21}(q_{23})=q_{12}$, and $S_{1}$ and $S_{3}$ are congruent, yielding that
the tiling is rotationally generated. If $o\neq q$, then $o$ has a closed
circular neighborhood $B$ disjoint from $D_{3}$. Let $t\mapsto C(t)$ be a
continuous parametrization of the curve $C_{12}$ with $C(0)=o$, and let
$t^{+}=\sup\\{t:C([0,t])\subset B\\}$, and $t^{-}=\inf\\{t:C([t,0])\subset
B\\}$. Then $g_{21}(C(t^{\pm}))=C(t^{\mp})$, implying that $g_{21}$ is a
reflection about $o$, which contradicts the condition that the configuration
is Type 3. Thus, we have $o\in\operatorname{int}(D_{3})$.

Let $q_{2}=g_{12}(q)$ and $q_{1}=g_{21}(q)$. Then $q_{2}\in\partial D_{2}$,
$q_{1}\in\partial D_{1}$ and $q_{1},q_{2}\notin\partial(P)$. Let $P_{0}\subset
D_{3}$ be the homothetic copy of $P$ of maximum homothety ratio centered at
$o$. Then $P_{0}$ touches at least one of the curves $C_{13}$ and $C_{23}$,
say, at a point $x_{2}\in C_{23}$. Let $x_{1}=g_{21}(x_{2})$. By the
definition of $P_{0}$, we have $x_{1}\in D_{3}$, and by $x_{2}\in D_{2}$, we
have $x_{1}\in D_{1}$. From this it follows that $x_{1}\in C_{13}$, implying
that $C_{13}\cap g_{21}(C_{23})\neq\emptyset$. As $C_{12}\cup C_{13}$ is a
connected curve from $q_{12}$ to $q_{13}$, and $g_{21}(C_{23})$ is a connected
curve in $C_{12}\cup C_{13}$ from $q_{12}$ to $q_{1}$, the relation
$C_{13}\cap g_{21}(C_{23})\neq\emptyset$ implies that $q_{1}$ is an interior
point of $C_{13}$, from which $q_{2}\in C_{23}$ also follows. For $i=1,2$, let
$C_{i3}^{s}$ and $C_{i3}^{q}$ denote the closed arcs of $C_{i3}$ from $q_{i3}$
to $q_{i}$, and from $q_{i}$ to $q$, respectively (cf. Figure 8). Then
$g_{21}(C_{23}^{s})=C_{12}=g_{12}(C_{13}^{s})$ and
$g_{21}(C_{23}^{q})=C_{13}^{q}$ yield that the corresponding arcs are
congruent. Thus, by Lemma 6, as $\partial D_{1}=S_{1}\cup C_{13}^{q}\cup
C_{13}^{s}\cup C_{12}$ and $\partial D_{3}=S_{3}\cup C_{13}^{q}\cup
C_{13}^{s}\cup C_{23}^{q}\cup C_{23}^{s}$, the equidecomposability of
$\partial D_{1}$ and $\partial D_{3}$ yields that $S_{3}\cup C_{13}^{q}$ and
$S_{1}$ are equidecomposable, implying that $C_{13}^{q}$ (and also
$C_{23}^{q}$) is a polygonal curve of length $l(S_{1})-l(S_{3})$. This yields,
in particular, that $l(S_{1})>l(S_{3})$,
$\alpha=\frac{2m\pi}{n}>\frac{2\pi}{3}$, $m>\frac{n}{3}$. As a counterpoint to
this inequality, we remark that from the facts that no $S_{i}$ contains
diametrically opposite vertices of $P$ and $\alpha<\pi$, it follows that
$S_{3}$ contains at least two vertices of $P$ as interior points and
$l(S_{3})\geq 2$. This also yields that if the endpoints of $S_{3}$ are
vertices of $P$, then $l(S_{3})\geq 4$.

Figure 8. An illustration for Subcase 2.1 in Subsection 3.3 with $n=11$ and
$m=4$.

Note that as $g_{21}$ is a rotation around $o$, either all $q_{ij}$s are
interior points of some edges of $P$, or all are vertices. We consider only
the case that all the $q_{ij}$ are interior points, as in the other case a
similar argument proves the assertion.

Let us call a _copy_ of $S_{i}$ in $\partial D_{j}$ a subset $S$ of $\partial
D_{j}$ congruent to $S_{i}$ such $S$ is not a proper subset of some connected
curve $S^{\prime}\subset\partial D_{j}$ with the property that the unique
congruent copy of $\partial P$ containing $S$ contains also $S^{\prime}$, and
observe that any two copies of $S_{i}$ in $\partial D_{j}$ are either
nonoverlapping or slightly overlapping. Recall that by Lemma 4, for any values
of $i$ and $j$, $\partial D_{j}$ contains finitely many copies of $S_{i}$.

If $q$ is an interior point of a copy $S$ of $S_{1}$ in $\partial D_{3}$,
then, denoting the part of $S$ in $C_{23}^{s}$ by $S^{\prime}$ and the part in
$C_{23}^{q}$ by $S^{\prime\prime}$, from the fact that
$l(c_{13}^{q})=l(S_{1})-l(S_{3})\leq m-2=l(S_{1})-2$, it follows that
$S^{\prime}$ and $g_{21}(S^{\prime\prime})$ have a common vertex in their
interiors. Thus, they belong to the boundary of the same congruent copy
$P^{\prime}$ of $P$. On the other hand, from this we obtain that $g_{21}$ is a
symmetry of $P^{\prime}$, and hence, $P^{\prime}$ is centered at $o$, a
contradiction.

Similarly, if both $q_{2}$ and $q$ are interior points of the same copy $S$ of
$S_{1}$ in $\partial D_{1}$, then its rotated copy $S^{\prime}$ around $o$ by
$\alpha$ is a copy of $S_{1}$ in $\partial D_{1}$ containing both $q$ and
$q_{1}$ in its interior. Thus, $S$ and $S^{\prime}$ overlap. If the
overlapping arc contains a vertex of both, then there is a unique congruent
copy $P^{\prime}$ of $P$ containing them. But then the rotation around $o$ by
$\alpha$ is a symmetry of $P^{\prime}$, which, combined with the fact that
$S\cup S^{\prime}\subset P\cap\partial P^{\prime}$ yields that $P=P^{\prime}$;
a contradiction. On the other hand, if $S\cap S^{\prime}$ contains no vertex
of both, then its length is at most $1/2$, implying that $l(C_{23}^{q})\geq
l(S_{1})-1$, a contradiction.

It follows from $l(C_{23}^{q})\leq l(S_{1})-2$ that no copy of $S_{1}$ is
contained in $C_{23}^{q}$. Hence, any copy $S$ of $S_{1}$ in $\partial D_{i}$
for some $i=1,2,3$ satisfies at least one of the following:

  1. (a)

$S=S_{1}$ or $S=S_{2}$.

  2. (b)

$S$ contains one of the $q_{ij}$s as an interior point, i.e. it slightly
overlaps one of the $S_{i}$s.

  3. (c)

$S$ belongs to exactly one of $C_{23}^{q}$, $C_{13}^{q}$ or $C_{12}$.

  4. (d)

$S$ contains exactly one of $q_{1}$, $q_{2}$ or $q$ in its interior.

For $\mathrm{x}\in\\{\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\\}$, we call
a copy satisfying (x) a _Type (x)_ copy. We note that unless a copy satisfies
(b) and (d) simultaneously, it satisfies exactly one property in the above
list, and if there is a copy satisfying (b) and (d), then any type (b) copy is
also type (d) and vice versa. Furthermore, observe that at most one copy of
$S_{1}$ contains $q_{1}$ or $q_{2}$ in its interior. We show that for any
$\mathrm{x}\in\\{\mathrm{b},\mathrm{c},\mathrm{d}\\}$, $\partial D_{2}$ and
$\partial D_{3}$ contain the same number of type (x) copies. For
$\mathrm{x}=\mathrm{b},\mathrm{c}$, this statement follows from the
observation that $C_{12}$ and $C_{13}^{s}$ are rotated copies of $C_{23}^{s}$,
and the same rotations carry the edge containing $q_{23}$ into the edges
containing $q_{12}$ and $q_{13}$, respectively. For $\mathrm{x}=\mathrm{d}$,
we observe that no copy of $S_{1}$ in $\partial D_{2}$ contains $q_{2}$ in its
interior, and a copy in $\partial D_{2}$ contains $q_{2}$ or $q$ in its
interior if and only if a copy in $\partial D_{3}$ contains $q$ or $q_{1}$ in
its interior, respectively. On the other hand, $\partial D_{2}$ contains
exactly one type (a) copy, namely $S_{2}$, while $\partial D_{3}$ contains no
such copy, yielding that $\partial D_{2}$ contains one more copy of $S_{1}$
than $\partial D_{3}$, contradicting the fact that $D_{1}$ and $D_{3}$ are
congruent.

_Subcase 2.2_ , $g_{21}$ is a reflection about a symmetry line $L$ of $P$.
Without loss of generality, we assume that $L$ is the $y$-axis, and the common
point of $S_{1}$ and $S_{2}$ lies on the positive half of $L$. Note that
$g_{12}=g_{21}$, and $D_{3}=\operatorname{cl}(P\setminus(D_{1}\cup D_{2}))$ is
symmetric to $L$.

Clearly, by Remark 2, $L$ separates $D_{1}$ and $D_{2}$, and from this it
readily follows that $D_{1}\cap D_{2}=[q_{12},q]$. Furthermore, we have that
$D_{1}$ and $D_{2}$ are the closures of the subsets of $P\setminus D_{3}$
contained in the two closed half planes bounded by $L$.

Let $S=g_{13}(S_{1})$. Assume that $S$ slightly overlaps $S_{3}$. Then $S$
crosses $L$. By the symmetry of $D_{3}$, the reflected copy $S^{\prime}$ of
$S$ about $L$ also belongs to $\partial D_{3}$ and it also crosses $L$. Thus,
$S$ and $S^{\prime}$ overlap, and $S\cup S^{\prime}$ intersects $L$ at a right
angle. From this we have that $D_{3}=\operatorname{conv}D_{3}$ is the convex
region bounded by $S_{3}\cup S\cup S^{\prime}$, and an elementary computation
yields that $\operatorname{area}(D_{3})>\frac{\operatorname{area}(P)}{3}$; a
contradiction. Thus, we have that $S$ does not overlap $S_{3}$.

Let $S^{\prime}$ be the reflected copy of $S$ about $L$. If $S$ does not
contain $q$ in its interior, then the facts that
$l(S)+l(S^{\prime})+l(S_{3})=l(P)$ and that $S,S^{\prime},S_{3}$ are subsets
of $\partial(\operatorname{conv}(D_{3}))$ yield that
$D_{3}=\operatorname{conv}(D_{3})=P$, which contradicts our assumptions. Thus,
$S$ and $S^{\prime}$ overlap. If $S\cap S^{\prime}$ contains a vertex in its
interior, then $S\cup S^{\prime}$ belongs to the boundary of a regular
$n$-gon, implying that $g_{31}(S\cup S^{\prime})\subset\partial P$. Hence, $S$
and $S^{\prime}$ either slightly overlap, or they coincide.

Consider the case that $S$ and $S^{\prime}$ slightly overlap, and let $E=S\cap
S^{\prime}$. Then $g_{31}(E)\subset g_{31}(S)=S_{1}$ lies on the edge
containing $q_{12}$, or the edge containing $q_{13}$. Since in the first case
$\partial D_{1}$ and $\partial D_{2}$ crosses, we have that $g_{31}(E)$ lies
on the edge containing $q_{13}$; we remark that the property that $S_{1}$ and
$g_{31}(S^{\prime})$ slightly overlaps with $q_{13}\in S_{1}\cap
g_{31}(S^{\prime})$ implies also that $q_{13}$ is not a vertex of $P$.

Let $L_{1}$ be the supporting line of $P$ parallel to $L$ such that the
infinite strip bounded by $L$ and $L_{1}$ contains $S_{1}$. If $L_{1}\cap P$
is disjoint from $S_{1}$, then $D_{3}$ contains diametrically opposite points
of $P$, contradicting our assumptions. Thus, either $q_{13}\in L_{1}\cap P$ or
$L_{1}\cap P$ belongs to the interior of $S_{1}$. If $q_{13}\in L_{1}\cap P$,
then $L_{1}\cap P$ is a side of $P$, and both endpoints of $g_{31}(S\cup
S^{\prime})$ lie on $L$, implying that
$D_{1}=\operatorname{conv}(D_{1})=\operatorname{conv}(g_{31}(S\cup
S^{\prime}))$, and thus, $\partial D_{1}$ does not contain a part congruent to
$S_{3}$; a contradiction. Hence, we are left with the case that $L_{1}\cap P$
belongs to the interior of $S_{1}$. Let $q^{\prime}=g_{31}(q)$ and let
$L^{\prime}$ be the line intersecting $\partial P$ orthogonally at
$q^{\prime}$ (cf. Figure 9). Then $q_{12}$ is a unique point in $D_{1}$
farthest from $q^{\prime}$. On the other hand, by symmetry, the distance of
the two endpoints of $g_{31}(S\cup S^{\prime})$ from $q^{\prime}$ are the
same. Since one of these endpoints is $q_{12}$, it follows that the two
endpoints coincide, implying that $g_{31}(S\cup S^{\prime})$ is a simple,
closed, convex curve in $\partial D_{1}$. Thus,
$D_{1}=\operatorname{conv}(g_{31}(S\cup S^{\prime}))$, or equivalently,
$D_{3}=\operatorname{conv}(S\cup S^{\prime})$, which contradicts the
assumption that $S_{3}\subset\partial D_{3}$.

Figure 9. An illustration for Subcase 2.2 in Subsection 3.3.

Finally, we consider the case that $S=S^{\prime}$. Then $S$ is symmetric to
$L$, yielding that $S_{1}$ is symmetric to the line $L^{\prime}=g_{31}(L)$.
Thus, $L^{\prime}$ is the bisector of either an edge or an angle of $P$, which
implies that $o\in L^{\prime}$. Since $D_{3}$ is symmetric to $L$, we also
have that $D_{1}$ is symmetric to $L^{\prime}$. Let $g$ be the reflection
about $L^{\prime}$, and set $g^{\prime}_{12}=g_{12}\circ g$;
$g^{\prime}_{13}=g_{13}$ and $g^{\prime}_{23}=g^{\prime}_{13}\circ
g^{\prime-1}_{12}$. Clearly, the transformation $g^{\prime}_{ij}$ is an
isometry mapping $D_{i}$ into $D_{j}$, and $g^{\prime}_{12}$ is a rotation
around $o$. Thus, in this case we can apply the consideration in Subcase 2.1.

_Case 3_ , all of $S_{1},g_{21}(S_{2})$ and $g_{31}(S_{3})$ coincide.

Since this yields that all of $S_{1},g_{21}(S_{2})$ and $g_{31}(S_{3})$
contain vertices of $P$, it follows that $g_{12}$ and $g_{13}$ are symmetries
of $P$. This implies that $o$ is a fixed point of both $g_{21}$ and $g_{31}$,
and thus, the unique common point of $D_{1}$, $D_{2}$ and $D_{3}$ is $o$. If
both $g_{21}$ and $g_{31}$ are rotations about $o$, then the tiling is clearly
rotationally generated, and we are done. Assume that, e.g. $g_{21}$ is a
reflection about a symmetry line $L$ of $P$. Then Remark 2 implies that $L$
separates $D_{1}$ and $D_{2}$, from which we obtain that the curve $D_{1}\cap
D_{2}$ is a straight line segment connecting the common point of $S_{1}$ and
$S_{2}$ to $o$. By the properties of rotations, from this we also have that
for any $i\neq j$, $D_{i}\cap D_{j}$ is a straight line segment connecting the
common point of $S_{i}$ and $S_{j}$ to $o$. This, combined with the fact that
$l(S_{1})=l(S_{2})=l(S_{3})$, readily yields that the tiling is rotationally
generated.

##  4\. Concluding remarks and open questions

######  Remark 5.

A simplified version of the proof of Theorem 1 can be applied to prove the
same statement for monohedral tilings of a regular triangle with at most three
tiles.

The authors have found no results about monohedral tilings of convex polygons
in spherical or hyperbolic plane. This is our motivation to state the
following problem. Before doing it, we note that the symmetry group of an
equiangular convex quadrilateral in spherical or hyperbolic plane contains the
symmetry group of a Euclidean rectangle as a subgroup.

######  Problem 1.

Let $\mathbb{M}^{2}$ denote the spherical plane $\mathbb{S}^{2}$ or the
hyperbolic plane $\mathbb{H}^{2}$, and let $P\subset\mathbb{M}^{2}$.
Characterize the monohedral tilings of $P$ with at most three topological
discs if $P$ is a

  * (i)

circular disc;

  * (ii)

a regular polygon;

  * (iii)

an equiangular convex quadrilateral.

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