Host Community Respecting Refugee Housing

TL;DR

This paper introduces a new model for refugee housing that respects community and refugee preferences, analyzing the existence and computational complexity of stable housing arrangements within a graph-based community topology.

Contribution

It formulates a game-theoretic model for refugee housing, investigates conditions for stable solutions, and analyzes the problem's NP-hardness and parameterized complexity.

Findings

Existence of stable housing arrangements is guaranteed under certain conditions.

The problem is NP-hard even in simple cases.

Parameterized complexity provides insights into computational feasibility.

Abstract

We propose a novel model for refugee housing respecting the preferences of the accepting community and refugees themselves. In particular, we are given a topology representing the local community, a set of inhabitants occupying some vertices of the topology, and a set of refugees that should be housed on the empty vertices of the graph. Both the inhabitants and the refugees have preferences over the structure of their neighborhood. We are specifically interested in the problem of finding housing such that the preferences of every individual are met; using game-theoretical words, we are looking for housing that is stable with respect to some well-defined notion of stability. We investigate conditions under which the existence of equilibria is guaranteed and study the computational complexity of finding such a stable outcome. As the problem is NP-hard even in very simple settings, we employ the parameterized complexity framework to give a finer-grained view of the problem's complexity with respect to natural parameters and structural restrictions of the given topology.

Equations

$$(m\Delta {)}^{\mathcal{O}(m)}\cdot ||b|{|}_{\infty }^2$$

$$\forall i\in I:\sum _{v\in N_G(\iota (i))}{x}_v$$

$$\sum _{v\in V_U}{x}_v$$

$$\alpha (x_i)=\{\begin{array}{ll}1& \text{if }\pi (r_i)=v_i^1\\ 0& \text{otherwise}\end{array}$$

Full text

We are specifically interested in the problem of finding housing such that the preferences of every individual are met; using game-theoretical words, we are looking for housing that is stable with respect to some well-defined notion of stability. We investigate conditions under which the existence of equilibria is guaranteed and study the computational complexity of finding such a stable outcome. As the problem is \NPheven in very simple settings, we employ the parameterised complexity framework to give a finer-grained view of the problem’s complexity with respect to natural parameters and structural restrictions of the given topology.

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1. Introduction

According to the last report of the United Nations High Commissioner for Refugees (UNHCR), there were 89.3 million forcibly displaced persons at the end of 2021 UNHCR (2022a). It is the highest number since the aftermath of World War II and it is for sure that these numbers will even grow. In May 2022, UNHCR announced that a tragic milestone of 100 million displaced persons was reached. They identified the war in Ukraine as the leading cause of the dramatic growth in the last year Nations (2022). Russian aggression not only forced many Ukrainians to leave their homes, but even caused food insecurity and related population movement in many parts of the world, since Ukraine is among the fifth largest wheat exporters in the world Behnassi and El Haiba (2022).

It should be mentioned that political and armed conflicts are not the only causes of forced displacement UNHCR (2022a). One of the most common reasons for fleeing is due to natural disasters. To name just a few, in August 2022, massive floods across Pakistan affected at least two-thirds of the districts and displaced at least 33 million people Mallapaty (2022b, a); the numbers are not yet final. At the same time, a devastating drought in Somalia caused the internal displacement of at least 755,000 people UNHCR (2022a). Furthermore, it is expected that, due to climate change, extremes of the climate will become even more common in the near future Fischer et al. (2021).

Arguably, the best prevention against the phenomenon of forced displacement is not allowing it to appear at all; however, the aforementioned numbers clearly show that these efforts are not very successful. Therefore, in practise, three main solutions are assumed Jallow and Malik (2004). Voluntary repatriation is the most desirable but not very successful option. In many situations, repatriation is not even possible due to ongoing conflicts or a completely devastated environment. Resettlement and integration in the country of origin or abroad are more common. These two solutions require considerable effort from both the newcomers and the host community sides.

The very problematic part of forced displacement is the fact that $38\%$ of all refugees111From the strict sociological point-of-view, not all forcibly displaced persons are classified as refugees. Slightly abusing the notation, we will use the term refugee and displaced person interchangeably. are hosted in only five countries UNHCR (2022a). And these are only the absolute numbers. For example, in Lebanon, every one in four people is a refugee NRC (2022). The redistribution of refugees seems to be a natural solution to this imbalance; however, not all countries are willing to accept all people. One such example can be the Czech Republic, which refuses to accept any Syrian refugees during the 2015 European migrant crisis, currently hosting the largest number of Ukrainian refugees per capita UNHCR (2022b).

Even with working and widely accepted redistributing policies, there is still a need to provide housing in specific cities and communities. From the good examples of such integration strategies OECD (2018) it follows that one of the most important characteristics is that members of the accepting community do not feel threatened by newcomers.

Inspired by this, we propose a novel computational model for refugee housing. Our ultimate goal is to find an assignment of displaced persons into empty houses of a community such that this assignment corresponds to the preferences of the inhabitants about the structure of their neighbourhoods and, at the same time, our model also takes into consideration the preferences of the refugees themselves, as refugees dissatisfied with their neighbourhood have a strong intention to leave the community. More precisely, in our model, we are given a topology of the community, which is an undirected graph, a set of inhabitants together with their assignment to the vertices of the topology and preferences over the shape of their neighbourhood, and a set of refugees with the same requirements on the neighbourhood shape. We want to find a housing of refugees in the empty vertices of the topology such that the housing satisfies a certain criterion, such as stability.

Refugee redistribution has gained the attention of mathematicians and computer scientists very recently. The formal model for capturing refugee resettlement is double-sided matching Delacrétaz and Kominer (2016); Delacrétaz et al. (2019); Aziz et al. (2018); Ahani et al. (2021). That is, on the input we are given a set of locations with multidimensional constraints and a set of refugees with multidimensional features. An example of a constraint can be the number of refugees the locality can accept on one side and the size of family on the refugee side. The question then is whether there exists a matching between localities and refugees respecting all constraints. According to us, this formulation of the refugees resettlement problem more concerns the global perspective of refugee redistributing, not the local housing problem, as we do in our paper. Aziz et al. (2018) study mostly the complexity of finding stable matching with respect to different notions of stability, and it turns out that for most of the stability notions finding a stable matching is computationally intractable (\NPh, in fact). Kuckuck et al. (2019) later refined the model of Aziz et al. (2018) in terms of hedonic games.

Our Contribution

Partly continuing the line of research in refugee resettlement, we introduce a novel model focused on the local housing of new refugees. Previous models Aziz et al. (2018); Ahani et al. (2021); Delacrétaz et al. (2019) can be seen and used as a very effective model on the (inter-) national level to distribute refugees to certain locations, such as states or cities222In fact, the American resettlement agency HIAS use the matching software AnnieTM Moore which was later improved by Ahani et al. (2021).. However, our model can be assumed as the second level of refugee redistribution; once refugees are allocated to some community, we want to house them in a way that respects preferences of both inhabitants and refugees.

In particular, we introduce three variants of refugee housing, each targeting a certain perspective of this problem. Our simplest model, introduced in Section 3, completely eliminates the preferences of refugees and studies only the stability of the housing with respect to the preferences of the inhabitants. We call this variant anonymous housing. Since refugees are assumed to be indistinguishable, inhabitants have preferences over the number of refugees in their neighbourhood.

As stated above, the most successful refugee integration projects have the following property in common; they try to make both inhabitants and refugees as satisfied as possible by various activities to ensure that both groups get to know each other. We believe that our hedonic model, where the preferences of both inhabitants and refugees are based on the identity of particular members of the second group, supports and leads to more stable and acceptable housing. This model is formally defined and studied in Section 4.

The two introduced models have some disadvantages. The first is disrespectful to the refugees’ preferences, while the second is not very realistic, as it is hard to make all inhabitants familiar with all refugees and the other way around. Therefore, our last model can be seen as a compromise between these two extremes. In the diversity setting, introduced in Section 5, all agents (union of inhabitants and refugees) are partitioned into $k$ types, and preferences are over the fractions of agents of each type in the neighbourhood of each agent. Another advantage of this approach is that it nicely captures also the settings where we already have number of integrated refugees and the newcomers want to have some of them in the neighbourhood, or the case of an internally displaced person.

In all the aforementioned variants of the refugee housing problem, agents have dichotomous preferences; that is, they approve some set of alternatives and do not distinguish between them. It can be seen as if the neighbourhood of some agent does not comply with his approval set, he would rather leave the local community, which is very undesired behaviour on both sides.

For all assumed variants, we show that an equilibrium is not guaranteed to exist even in very simple instances. Thus, we study the computational complexity of finding an equilibrium or deciding that no equilibrium exists. To this end, we provide polynomial-time algorithms and complementary NP-hardness results. In order to paint a more comprehensive picture of the computational tractability of the aforementioned problems, we employ a finer-grained framework of parameterised complexity to give tractable algorithms for, e. g., instances where the number of refugees or the number of inhabitants is small, or for certain structural restrictions of the topology. Additionally, we complement many of our algorithmic results with conditional lower-bounds matching the running-time of these algorithms.

Related Work

Our model is influenced by a game-theoretic reformulation of the famous Schelling’s model Schelling (1969, 1971) of residential segregation introduced by Agarwal et al. (2021). Here, we are given a simple undirected graph $G$ and a set of selfish agents partitioned into $k$ types. Every agent wants to maximise the fraction of agents of her own type in her neighbourhood. The goal is then to assign agents to the vertices of $G$ so that no agent can improve her utility by either jumping to an unoccupied vertex or swapping positions with another agent. Follow-up works include those that study the problem from the perspective of computational complexity Kreisel et al. (2022); Echzell et al. (2019) and equilibrium existence guarantees Bilò et al. (2022, 2022); Kanellopoulos et al. (2022).

The second main inspiration for our model is the Hedonic Seat Arrangement problem and its variants recently introduced by Bodlaender et al. (2020). Here, the goal is to find an assignment of agents with preferences to the vertices of the underlying topology. The desired assignment should then meet specific criteria such as different forms of stability, maximising social welfare, or being envy-free. In our model, compared to Hedonic Seat Arrangement of Bodlaender et al. (2020), inhabitants already occupy some vertices of the topology and we have to assign refugees to the remaining (empty) vertices in a desirable way.

Next, the problem of house allocation Abdulkadiroğlu and Sönmez (1999) or housing market Shapley and Scarf (1974) has been extensively studied in the area of mechanism design. Here, each agent owns a house and the objective is to find a socially efficient outcome using reallocations of objects. Later, You et al. (2022) introduced house allocation over social networks that follows current trend in mechanism design initiated by Li et al. (2017), where each individual can only communicate with his neighbours. As stated before, the house allocation is studied mainly from the viewpoint of mechanism design and as such is far from our model.

Finally, hedonic games Drèze and Greenberg (1980); Bogomolnaia and Jackson (2002); Brandt et al. (2016) are a well-studied class of coalition formation games where the goal is to partition agents into coalitions and where the utility of every agent depends on the identity of other agents in his coalition. In anonymous games Banerjee et al. (2001); Bogomolnaia and Jackson (2002), the agents have preferences over the sizes of their coalition. The most recent variants of hedonic games are the so-called hedonic diversity games Bredereck et al. (2019a); Boehmer and Elkind (2020a) where agents are partitioned into $k$ types and preferences are over the ratios of each type in the coalition. The main difference between our model and (all variants of) hedonic games is that in the latter model all coalitions are pairwise disjoint; however, in our case, each agent has his own neighbourhood overlapping with neighbourhoods of other agents.

2. Preliminaries

Let $\mathbb{N}$ denote the set of positive integers. Given two positive integers ${j,j^{\prime}\in\mathbb{N}}$, with $j\leq j^{\prime}$, we call the set $[j,j^{\prime}]=\{j,\ldots,j^{\prime}\}$ an interval, and let $[j]=[1,j]$ and $[j]_{0}=[j]\cup\{0\}$. Let $S$ be a set. By $2^{S}$ we denote the set of all subsets of $S$ and, given $k\leq|S|$, we denote by $\binom{S}{k}$ the set of all subsets of $S$ of size $k$.

Let $R=\{r_{1},\ldots,r_{m}\}$ be a non-empty set of refugees and $I=\{i_{1},\ldots,i_{\ell}\}$ be a set of inhabitants. The set of all agents is defined as $N=R\cup I$. A topology is a simple undirected graph $G=(V,E)$, where $|V|\geq|N|$. For a vertex $v$, we denote by $N(v)$ the set of its neighbours, formally $N(v)=\{u\mid\{u,v\}\in E\}$. The size of the neighbourhood of a vertex $v$ is called its degree and is defined as $\deg(v)=|N(v)|$. The closed neighbourhood of vertex $v$ is defined as $N[v]=N(v)\cup\{v\}$. In this work, we follow the basic graph-theoretical terminology Diestel (2012).

An inhabitants assignment is an injective function $\iota\colon I\to V$.

A set of vertices occupied by the inhabitants is denoted $V_{I}$ and, given an inhabitant $i\in I$, we denote the set of unoccupied vertices in his neighbourhood $U_{i}=N(\iota(i))\setminus V_{I}$. The set of all vertices that are not occupied by inhabitants is denoted $V_{U}=V\setminus V_{I}$.

The goal of every variant of our problem is to find a mapping of refugees to vertices that are not occupied by inhabitants. Formally, housing is an injective mapping $\pi\colon R\to V_{U}$. A set of vertices occupied by refugees with respect to housing $\pi$ is denoted ${V_{\pi}=\{\pi(r)\mid r\in R\}}$. We denote by $\Pi_{G,\iota}$ the set of all possible housings, and we drop the subscript whenever $G$ and $\iota$ are clear from the context.

Parameterised Complexity

We study the problem in the framework of parameterised complexity Cygan et al. (2015); Downey and Fellows (1995); Niedermeier (2006). Here, we investigate the complexity of the problem not only with respect to an input size $n$, but even assuming some additional parameter $k$. The goal is to find a parameter which is small and the “hardness” can be confined to this parameter. The most favourable outcome is an algorithm with running time $f(k)\cdot n^{{\mathcal{O}(1)}}$, where $f$ is any computable function. We call this algorithm fixed-parameter tractable and the complexity class containing all problems that admit algorithms with such running time is called FPT.

Not all combinations of parameters yield to fixed-parameter tractable algorithms. A less favourable outcome is an algorithm running in $n^{f(k)}$ time, where $f$ is any computable function. Parameterised problems admitting such algorithms belong to complexity class XP. To exclude the existence of a fixed-parameter tractable algorithm, one can show that the parameterised problem is W[t]-hard for some $t\geq 1$. This can be done via a parameterised reduction from any problem known to be W[t]-hard.

It could also be the case that a parameterised problem is \NPheven for a fixed value of $k$; we call such problems para-NP-hard and, assuming $\mathsf{P}\not=\textsf{NP}$, such problems do not admit XP algorithms.

Our running-time lower-bounds are based on the well-known Exponential Time Hypothesis (ETH) of Impagliazzo and Paturi (2001); see also Impagliazzo et al. (2001) and the survey of Lokshtanov et al. (2011). This conjecture states that, roughly speaking, there is no algorithm solving 3-SAT in time sub-exponential in the number of variables.

Structural Parameters.

Let $G=(V,E)$ be a graph. A set $C\subseteq V$ is vertex cover of $G$ if $G\setminus C$ is an edgeless graph. The vertex cover number $\operatorname{vc}(G)$ is the minimum size vertex cover in $G$.

The tree-depth of a graph $G=(V,E)$, denoted $\operatorname{td}(G)$, is defined as follows. It is $1$ if $G$ has only a single vertex, $1+\min_{v\in V}\operatorname{td}(G-v)$ if $G$ is connected, and otherwise it is the maximum tree-depth of connected components of $G$.

Definition \thetheorem (Tree decomposition).

A tree decomposition of a graph $G=(V,E)$ is a triple $\mathcal{T}=(T,\beta,r)$, where $T$ is a tree rooted at node $r$ and $\beta\colon V(T)\to 2^{V}$ is a mapping that satisfies:

1. (1)

$\bigcup_{x\in V(T)}\beta(x)=V$; 2. (2)

For every $\{u,v\}\in E$ there exists a node $x\in V(T)$, such that $u,v\in\beta(x)$; 3. (3)

For every $u\in V$ the nodes $\{x\in V(T)\mid u\in\beta(x)\}$ form a connected subtree of $T$.

To distinguish between the vertices of a tree decomposition and the vertices of the underlying graph, we use the term node for the vertices of a given tree decomposition.

The width of a tree decomposition $\mathcal{T}$ is ${\max_{x\in V(T)}|\beta(x)|-1}$. The tree-width of a graph $G$, denoted $\operatorname{tw}(G)$, is the minimum width of a tree decomposition of $G$ over all such decompositions.

3. Anonymous Refugees

In our simplest model of refugee housing, we assume refugees to be non-strategic and concern only the preferences of inhabitants. In this sense, the refugees are, from the viewpoint of inhabitants, anonymous, and the preferences only takes into account the number of refugees in the neighbourhood of each inhabitant. Similar preferences have already been studied in different problems such as anonymous hedonic games Banerjee et al. (2001); Bogomolnaia and Jackson (2002); Ballester (2004).

We formally capture this setting in the computational problem called the Anonymous Refugees Housing problem (ARH for short). A preference of every inhabitant $i\in I$ is a set $A_{i}\subseteq[\deg(\iota(i))]_{0}$ of the approved numbers of refugees in the neighbourhood. Our goal is to decide whether there is a housing $\pi\colon R\to V_{U}$ that respects the preferences of all inhabitants.

Definition \thetheorem.

A housing $\pi\colon R\to V_{U}$ is called inhabitant-respecting if for every $i\in I$ we have $|N(\iota(i))\cap V_{\pi}|\in A_{i}$.

If the approval set $A_{i}$ for an inhabitant $i\in I$ consists of consecutive numbers, we say that the inhabitant $i$ approves an interval.

**Example **\thetheorem

Let the topology be a cycle with four vertices. There are two inhabitants assigned to neighbouring vertices. One of these inhabitants, call her $h_{1}$, has approval set $A_{h_{1}}=\{0,1\}$, and the second one, say $h_{2}$, is not approving any refugees in his neighbourhood, that is, $A_{h_{2}}=\{0\}$. We have $R=\{r\}$. The only valid housing is next to the inhabitant $h_{1}$ as housing $r$ in the neighbourhood of $h_{2}$ clearly does not respect his preference. Also note that in this particular example, all the inhabitants approve intervals.

As our first result, we observe that even in a very simple settings, it is not guaranteed that any inhabitant-respecting refugees housing exists.

**Proposition **\thetheorem

There is an instance of the ARH problem with no inhabitant-respecting refugees housing even if all inhabitants approve intervals.

To prove Section 3, assume an instance with one inhabitant $i$ and two refugees $r_{1}$ and $r_{2}$. Let the topology be $K^{3}$, the inhabitant $i$ be assigned to an arbitrary vertex, and let $A_{i}=\{0\}$. There are exactly two possible refugees housings and in any of them the inhabitant $i$ has two neighbouring refugees; therefore, there is no inhabitant-respecting housing.

In the previous example, we used the fact that the inhabitant $i$ does not approve any refugees in his neighbourhood. We call such inhabitants intolerant. Despite the fact that the instance does not have an inhabitant-respecting housing even if $A_{i}=\{1\}$, we observe that intolerant inhabitants can be safely removed.

**Proposition **\thetheorem

Let $\mathcal{I}=(G,I,R,\iota,(A_{i})_{i\in I})$ be an instance of the ARH problem, $j\in I$ be an inhabitant with $A_{j}=\{0\}$, and $F_{j}=\{\iota(j)\}\cup U_{j}$. $\mathcal{I}$ admits an inhabitant-respecting housing iff the instance $\mathcal{I}^{\prime}=(G\setminus F_{j},I\setminus\{j\},R,\iota,(A_{i})_{i\in I\setminus\{j\}})$ admits an inhabitant-respecting housing.

Proof.

Let $\mathcal{I}$ be a yes-instance, let $j\in I$ be an inhabitant with $A_{j}=\{0\}$, and let $\pi$ be an inhabitant-respecting refugees housing. Since $\pi$ is an inhabitant-respecting housing, there is no refugee in the neighbourhood of $j$, so $\pi$ is a solution even for $\mathcal{I}^{\prime}$.

In the opposite direction, let $\mathcal{I}^{\prime}$ be a yes-instance and $\pi^{\prime}$ be an inhabitant-respecting housing in $\mathcal{I}^{\prime}$. As $\pi^{\prime}$ houses all refugees to $V^{\prime}=V\setminus\{N(\iota(j))\}$ the $\pi^{\prime}$ is also a solution for $\mathcal{I}$. ∎

Due to the definition of approval sets, inhabitants without unoccupied neighbourhood are necessarily assumed intolerant and therefore can be safely removed by Section 3. Hence, we assume only instances without intolerant inhabitants where every inhabitant has at least one unoccupied vertex in her neighbourhood.

**Proposition **\thetheorem

Let $\mathcal{I}=(G,I,R,\iota,(A_{i})_{i\in I})$ be an instance of the ARH problem and $\{u,v\}\in E(G)$ be an edge such that either $u,v\in V_{I}$ or $u,v\in V_{U}$. Then $\mathcal{I}$ admits an inhabitant-respecting housing iff the instance $\mathcal{I}^{\prime}=((V(G),E(G)\setminus\{\{u,v\}\}),I,R,\iota,(A_{i})_{i\in I})$ admits an inhabitant-respecting housing.

Section 3 directly implies that all graphs assumed in this section are naturally bipartite.

{theorem}

Every instance of the ARH problem where the topology is a graph of maximum degree $2$ can be solved in polynomial-time.

Proof.

Our algorithm is based on the dynamic programming approach combined with the gradual elimination of inhabitants’ approval sets and exhaustive application of Section 3. Observe that graph of maximum degree $2$ is a collection of paths and cycles Diestel (2012). We first introduce an algorithm that solves the problem on paths and then show how to improve the algorithm to solve cycles.

Let the topology be a path $P=v_{1}v_{2}\ldots v_{k}$, $k\geq 3$, and suppose that the vertex $v_{1}$ is occupied by an inhabitant $i\in I$. We distinguish two cases based on $A_{i}$ and show how the algorithm proceed. If ${A_{i}=\{1\}}$, then we have to house a refugee on $v_{2}$. However, this adds one refugee in the neighbourhood of the inhabitant $j$ occupying the vertex $v_{3}$. To capture this, we decrease the value of all elements in $A_{j}$. If there are any negative numbers in $A_{j}$ after this operation, we remove all of them from the list. Then we delete $v_{1}$ and $v_{2}$ from $P$, decrease $|R|$ by one, and solve the problem for $P^{\prime}=v_{3}\ldots v_{k}$. If $A_{i}=\{0,1\}$, we have to try both possibilities. That is, we run the algorithm once with $A_{i}=\{0\}$ and once with $A_{i}=\{1\}$. If any run of the algorithm finds a solution, we also have a solution for the original instance.

The described algorithm is exponential in the worst case. To improve the running time, we tabularise the computed partial solutions. Our dynamic programming table $DP$ has three dimensions: the first for an inhabitant, the second for an actual value of $|R|$, and the third for a shape of approval set. The stored value is either yes or no depending on whether the combination of indices yields to a inhabitant-respecting housing. There are ${\mathcal{O}(n)}$ inhabitants, the value of $|R|$ is also in ${\mathcal{O}(n)}$, and there are at most $2$ different approval sets possible for each inhabitant on the path. Therefore, the size of the table is at most ${\mathcal{O}(n^{2})}$, which is also the running time of our algorithm.

If the vertex $v_{1}$ is unoccupied, we can add the dummy vertex $v_{0}$ as a neighbour of $v_{1}$, make it occupied by an inhabitant with approval set $\{0,1\}$ and run the preceding algorithm.

For cycles, we identify, based on approval sets, the best vertex to break this cycle and turning it into a path. In particular, if there is an inhabitant with the approval set $\{2\}$, we delete this vertex together with his two neighbours and obtain a path. In all other cases, we try all (at most three) ways of resolving his neighbourhood. To do so, we run an algorithm similar to the one from the beginning of this proof. ∎

Unfortunately, as the following theorem states, the bounded-degree condition from Section 3 cannot be relaxed any more.

{theorem}

The ARH problem is NP-complete even if the topology is a graph of maximum degree $3$ and all inhabitants approve intervals.

Proof.

Given a housing $\pi$, it is easy to verify in polynomial time whether $\pi$ is inhabitant-respecting by enumerating all inhabitants and comparing their neighbourhood with approval lists. Thus, ARH is indeed in NP.

For NP-hardness, we present a polynomial-time reduction from a variant of the $2$-Balanced $3$-SAT problem which is known to be NP-complete Tovey (1984); Fiala et al. (2005); Berman et al. (2003). In this variant of $3$-SAT, we are given a propositional formula $\varphi$ with $n$ variables $x_{1},\ldots,x_{n}$ and $m$ clauses $C_{1},\ldots,C_{m}$ such that each clause contains at most $3$ literals and every variable appears in at most $4$ clauses – at most twice as a positive literal and at most twice as a negative literal. Later in this paper, we will refer to this reduction as basic reduction.

We construct an equivalent instance $\mathcal{I}$ of ARH as follows. We represent every variable $x_{i}$ by a single variable gadget $X_{i}$ that is a path $t_{i}v_{i}f_{i}$. The vertex $v_{i}$ is occupied by an inhabitant with approval set $\{1\}$. All other vertices are empty and we call the vertex $t_{i}$ the $t$-port and the vertex $f_{i}$ the $f$-port. Every clause $C_{j}$ is represented by a single vertex $c_{j}$ occupied by an inhabitant $h_{j}$ who approves the interval $[1,|C_{j}|]$ and is connected to the $t$-port of the variable gadget $X_{i}$ if the variable $x_{i}$ occurs as a positive literal in $C_{j}$ and to the $f$-port of $X_{i}$ if $x_{i}$ occurs as a negative literal in $C_{j}$, respectively. To complete the reduction, we set $|R|=n$.

For the correctness of the construction, let $\varphi$ be a satisfiable $2$-Balanced $3$-SAT formula and $\alpha$ be a truth assignment. For every variable $x_{i}$, we house a refugee at $v_{i}^{1}$ if $\alpha(x_{i})=1$ and at $v_{i}^{3}$ if ${\alpha(x_{i})=0}$, respectively. This housing is clearly a solution of $\mathcal{I}$ as every inhabitant assigned to the variable gadget is clearly satisfied, every inhabitant $h_{j}$ is also satisfied since $\alpha$ is a satisfying assignment, and we housed $1$ refugee for every variable gadget.

In the opposite direction, observe that there is exactly one refugee assigned to every variable gadget and, thus, in every assignment $\pi$ there is no variable gadget $X_{i}$ such that the $t$-port and the $f$-port are occupied at the same time. Hence, we can set $\alpha(x_{i})$ equal to $1$ if and only if the $t$-port is occupied by a refugee. Clearly, $\alpha$ is a truth assignment as $\pi$ has to satisfy each inhabitant occupying clause vertex.

By definition, every clause contains at most $3$ literals, and thus the degree of every vertex $c_{j}$, $j\in[m]$, is at most $3$. For every variable gadget $X_{i}$, $i\in[n]$, the vertex $v_{i}^{2}$ has degree $2$ and both $t$-port and $f$-port have degree at most $3$ – they are adjacent to $v_{i}^{2}$ and at most two vertices representing clauses. Hence, the bounded-degree condition holds and the construction can be clearly done in polynomial time, finishing the proof. ∎

Since the above results clearly show that the problem is very hard even in simple settings, we turn our attention to the parameterised complexity of the ARH problem. In particular, we study the problem’s complexity from the viewpoint of natural parameters, such as the number of refugees, the number of inhabitants, the number of empty vertices, and various structural parameters restricting the shape of the topology.

{theorem}

The ARH problem is W[2]-hard parameterised by the number of refugees $|R|$ even if all inhabitants approve intervals. Moreover, unless ETH fails, there is no algorithm that solves ARH in $f(|R|)\cdot n^{o(|R|)}$ time for any computable function $f$.

Proof.

We reduce from the Dominating Set problem, which is known to be W[2]-complete Downey and Fellows (1995) and, unless ETH fails, cannot be solved in $f(k)\cdot n^{o(k)}$ time for any computable function $f$ Chen et al. (2005). The instance $\mathcal{I}$ of Dominating Set consists of a simple undirected graph $G=(V,E)$ and an integer $k\in\mathbb{N}$. The goal is to decide whether there is a set $D\subseteq V$ of size at most $k$ such that each vertex $v\in V$ is either in $D$ or at least one of its neighbours is in $D$.

We construct an equivalent instance $\mathcal{I^{\prime}}$ of the ARH problem as follows. We start by defining the topology. For each vertex ${v\in V}$ we add two vertices $\ell_{v}$ and $p_{v}$. The vertex $\ell_{v}$ represents the original vertex and is intended to be free for refugees. The vertex $p_{v}$ is occupied by an inhabitant $h_{v}$ with $A_{h_{v}}=[1,|N[v]|]$. This inhabitant ensures that there is at least one refugee housed in the closed neighbourhood of $p_{v}$. The edge set of the topology $G^{\prime}$ is $\bigcup_{v\in V}\{\{p_{v},\ell_{w}\}\mid w\in N_{G}[v]\}$. To complete the construction, we set $|R|=k$. For an overview of our construction, we refer the reader to Figure 1.

Let $\mathcal{I}$ be a yes-instance and $D$ be a dominating set of size $k$. For every vertex $v\in D$, we house a refugee in the vertex $\ell_{v}$. Since $D$ was a dominating set of size $k$, in the closed neighbourhood of every $v\in V$ in $G$ there is at least one vertex $u\in D$. Therefore, for every inhabitant $h_{v}$, there is at least one refugee in his neighbourhood and $\mathcal{I^{\prime}}$ is indeed a yes-instance.

In the opposite direction, let $\mathcal{I^{\prime}}$ be a yes-instance and $\pi$ be a solution housing. We set $D$ to $\{v\in V\mid\pi^{-1}(\ell_{v})\neq\emptyset\}$. Due to the definition of approved intervals of the inhabitants, it holds for every $v\in V$ either $v$ or at least one of his neighbours is in $D$, as otherwise the inhabitant $h_{v}$ would not be respected.

To complete the proof, we recall that $|R|=k$ and, hence, the presented reduction is indeed a parameterised reduction. Moreover, assume that there is an algorithm $\mathcal{A}$, that solves ARH in $f(|R|)\cdot|\mathcal{I^{\prime}}|^{o(|R|)}$ time. Then we can reduce an instance of the Dominating Set problem to the instance of the ARH problem, solve the reduced instance using algorithm $\mathcal{A}$ and reconstruct a solution of the original instance. This is an algorithm for Dominating Set running in $f(k)\cdot n^{o(k)}$ time, which contradicts ETH. ∎

We complement Section 3 with an algorithm that runs in time matching the lower-bound given in this theorem.

**Proposition **\thetheorem

The ARH problem can be solved in $n^{{\mathcal{O}(|R|)}}$ time, where $n=|V(G)|$. That is, ARH is in XP parameterised by the number of refugees.

Proof.

Our algorithm is a simple brute-force. Let $V_{U}=V(G)\setminus V_{I}$ be the number of empty vertices and let $n=|V|$. Note that $|V_{U}|\leq n$. We try all subsets of $V_{U}$ of size $|R|$ and for each such subset, we check in linear time whether the housing is inhabitant-respecting. This gives us the total running time $n^{\mathcal{O}(|R|)}$. ∎

As the number of refugees is not a parameter promising tractable algorithms even if all inhabitants approve intervals, we turn our attention to the case where the number of inhabitants is small. Our algorithm is based on integer linear programming formulation of the problem and we use the following result of Eisenbrand and Weismantel ([n.d.]).

{theorem}

[(Eisenbrand and Weismantel, [n.d.], Theorem 2.2)] Integer linear programme ${\mathbb{A}x\leq b}$, $x\geq 0$, with $n$ variables and $m$ constraints can be solved in

[TABLE]

time, where $\Delta$ is an upper-bound on all absolute values in $\mathbb{A}$.

{theorem}

If all inhabitants approve intervals, then the ARH problem is fixed-parameter tractable parameterised by the number of inhabitants $|I|$.

Proof.

We solve the ARH problem using an integer linear programming formulation of the problem. We introduce one binary variable $x_{v}$ for every empty vertex $v\in V_{U}$ representing if a refugee is housed on $v$ or not. Next, we add the following constraints.

[TABLE]

where for an inhabitant $i\in I$ the value $\operatorname{low}(i)$ stands for lower-end and $\operatorname{high}(i)$ stands for upper-end of the approved interval by inhabitant $i$, respectively. Equations 1 and 2 ensure that the number of refugees in the neighbourhood of each inhabitant is in its approved interval, while Equation 3 ensures that all refugees are housed somewhere. Using Section 3 we see that the given integer programme can be solved in time $|I|^{\mathcal{O}(|I|)}\cdot n^{\mathcal{O}(1)}$, as $m=2|I|+1$, $\Delta=1$, and $||b||_{\infty}\leq n$. That is, ARH is in FPT parameterised by the number of inhabitants $|I|$. ∎

Note that it would be possible to provide a different ILP formulation of the problem and use the famous theorem of Lenstra Jr. (1983) to show membership in FPT, however, this would yield to an algorithm with much worse (i. e., doubly-exponential) running-time.

The result from Section 3 cannot be easily generalised to the case with inhabitants approving general sets. However, we can show that if the number of intervals in each approval set is bounded, the problem is still fixed-parameter tractable.

{theorem}

The ARH problem is fixed-parameter tractable when parameterised by the combined parameter the number of inhabitants $|I|$ and the maximum number of disjoint intervals $\delta$ in the approval sets.

Proof.

We start our algorithm by guessing an effective interval for every inhabitant $i\in I$. There are at most $\delta^{|I|}$ such guesses and for each guess we run the integer programme from the proof of Section 3. The overall running time of this algorithm is $\delta^{|I|}|I|^{\mathcal{O}(|I|)}\cdot n^{\mathcal{O}(1)}$. ∎

In our next result, we show that the parameter $\delta$ from Section 3 cannot be dropped while keeping the problem tractable.

{theorem}

The ARH problem is W[1]-hard parameterised by the number of inhabitants $|I|$.

Proof.

Our reduction is very similar to the one of Knop et al. (2019, Theorem 3.1). We reduce from the Multicoloured Clique problem where we are given a $k$-partite graph $G=(V_{1}\cup\cdots\cup V_{k},E)$ and the goal is to find a complete subgraph with $k$ vertices such that it contains a vertex from every $V_{i}$, $i\in[k]$. Multicoloured Clique is known to be W[1]-hard with respect to $k$ Fellows et al. (2009). We may assume that every colour class $V_{i}$, $i\in[k]$, is of size $n$. By $E_{i,j}$ we denote the set of edges between colour classes $V_{i}$ and $V_{j}$, that is, ${E_{i,j}=\{\{u,v\}\mid u\in V_{i}\land v\in V_{j}\}}$ and we may assume that for every pair of distinct $i,j\in[k]$ we have $|E_{i,j}|=m$.

We begin the reduction by fixing a bijection $\nu_{i}\colon V_{i}\to[n]$ for every $i\in[k]$ and a bijection $\epsilon_{i,j}\colon E_{i,j}\to[m]$ for every pair of distinct $i,j\in[k]$. Next, for every colour class $V_{i}$ we add a vertex-selection gadget $S_{i}$ consisting of $n$ vertices and leave these vertices empty. The number of refugees housed on the vertices of $S_{i}$ will correspond to a vertex in $V_{i}$ that is part of the clique. Then, for every $E_{i,j}$, $i,j\in[k]$, we add a set $T_{i,j}$ with $m\cdot n^{2}$ empty vertices and connect these to the vertex $M_{i,j}$. The vertex $M_{i,j}$ is occupied by an inhabitant with approval set $\{tn^{2}\mid t\in[m]\}$ and we call $T_{i,j}$ an edge-selection gadget. Similarly to the vertex selection gadget, the number of refugees assigned to $T_{i,j}$ will correspond to the edge selected for the solution. To ensure that the choice performed in the vertex-selection gadgets and the edge-selection gadgets is compatible, we introduce two vertices $G_{i,j}$ and $G_{j,i}$ for every $E_{i,j}$, $i,j\in[k]$ and $i\not=j$. The vertex $G_{i,j}$ is occupied by an inhabitant with approval set $\{\epsilon_{i,j}(e)\cdot n^{2}+\nu_{i}(v)\mid e\in E_{i,j}\land v\in e\land v\in V_{i}\}$ and is adjacent to every vertex in $S_{i}$, and the vertex $G_{j,i}$ is occupied by an inhabitant with approval set $\{\epsilon_{i,j}(e)\cdot n^{2}+\nu_{j}(v)\mid e\in E_{i,j}\land v\in e\land v\in V_{j}\}$ and is adjacent to every vertex in $S_{j}$. To complete the construction, we set $|R|=\binom{k}{2}\cdot m\cdot n^{2}+k\cdot n$ and introduce the same number of auxiliary vertices of degree [math] which are intended for the remaining refugees not assigned to vertex- and edge-selection gadgets. For an overview of our construction, please refer to Figure 2.

For correctness, let $\mathcal{I}=(G,k)$ be a yes-instance and ${v_{i},\ldots,v_{k}}$, where $v_{i}\in V_{i}$, be vertices that form a clique in $G$. For every $i\in[k]$ we assign $\nu_{i}(v_{i})$ refugees to empty vertices of $S_{i}$ and for each pair of distinct $i,j\in[k]$, we house $\epsilon_{i,j}(\{v_{i},v_{k}\})\cdot n^{2}$ refugees on empty vertices of $T_{i,j}$. The remaining refugees are assigned to the auxiliary vertices. The only inhabitants occupy the vertices $M_{i,j}$, $G_{i,j}$, and $G_{j,i}$. The inhabitants of $M_{i,j}$ are easily satisfied since we assign some multiple of $n^{2}$ to every $T_{i,j}$. An inhabitant in $G_{i,j}$ is adjacent to ${\epsilon_{i,j}(\{v_{i},v_{k}\})\cdot n^{2}}$ refugees from $T_{i,j}$ and ${\nu_{i}(v_{i})}$ refugees from $S_{i}$ while $G_{j,i}$ is adjacent to ${\epsilon_{i,j}(\{v_{i},v_{k}\})\cdot n^{2}}$ refugees from $T_{i,j}$ and ${\nu_{j}(v_{j})}$ refugees from $S_{j}$, respectively. This complies with their approval set.

In the opposite direction, let the equivalent ARH instance $\mathcal{I^{\prime}}$ be a yes-instance and $\pi$ be an inhabitant respecting housing in $\mathcal{I^{\prime}}$. Due to the inhabitants of $M_{i,j}$, where $i,j\in[k]$, there is some positive multiple of $n^{2}$ refugees assigned to every $T_{i,j}$ that corresponds to some edge $e\in E_{i,j}$ in the original graph. Moreover, due to inhabitant on $G_{i,j}$, the number of refugees assigned to $S_{i}$ corresponds to the identification of some vertex $v\in V_{i}$ that is necessarily incident to $e$. The same holds for $G_{j,i}$.

It is not difficult to see that the vertices $G_{i,j}$ and $G_{j,i}$ together with $M_{i,j}$, where $i,j\in[k]$ and $i\not=j$, form a vertex cover of the topology $G^{\prime}$ as $S_{i}$ and $T_{i,j}$ for each pair of distinct $i,j\in[k]$ are sets of independent vertices. It follows that ${\operatorname{vc}(G^{\prime})=\binom{k}{2}+k(k-1)={\mathcal{O}(k^{2})}}$ and thus the presented reduction is indeed a parameterised reduction, finishing the proof. ∎

By careful guessing, we can prove that for the combined parameters the number of refugees and the number of inhabitants, we may obtain fixed-parameter tractable algorithm.

**Lemma **\thetheorem

The ARH problem is fixed-parameter tractable when parameterised by the number of refugees $|R|$ and the number of inhabitants $|I|$ combined.

Proof.

We can guess a particular neighbourhood of every refugee, and for every such guess, verify whether it is correct. Let $I_{r_{i}}\subseteq I$ be a neighbourhood guessed for a refugee $r_{i}\in R$. If there is no empty vertex $v$ with $\deg(v)=|I_{r_{i}}|$ at the intersection of neighbourhoods of vertices in $I_{r_{i}}$, we discard this guess. Otherwise, we house the refugee $r_{i}$ on $v$ and continue with $r_{i+1}$. If all refugees are housed, we simply check whether the housing respects the approval sets of all inhabitants.

Overall, there are $(2^{|I|})^{|R|}=2^{|I|\cdot|R|}$ such guesses and every guess can be verified in polynomial-time. Therefore, the running time of our algorithm is $2^{|I|\cdot|R|}\cdot n^{\mathcal{O}(1)}$, that is, ARH is in FPT with respect to $|R|$ and $|I|$ combined. ∎

The last assumed natural parameter is the number of empty vertices the refugees can be assigned to. Note that $|V_{U}|\geq|R|$. This parameterisation yields, in contrast to Section 3, to a simple algorithm running in FPT time which is, although its simplicity, optimal assuming the Exponential Time Hypothesis.

{theorem}

The ARH problem can be solved in $2^{\mathcal{O}(|V_{U}|)}\cdot n^{\mathcal{O}(1)}$ time and, unless ETH fails, there is no algorithm solving ARH in ${2^{o(|V_{U}|)}\cdot n^{\mathcal{O}(1)}}$ time even if all inhabitants approve intervals.

Proof.

First, observe that if $|V_{U}|\leq|R|$, then the instance is trivially no-instance, as it is not possible to house all refugees. Hence, we can enumerate all subsets of $|V_{U}|$ of size $|R|$ and, for every guess, check whether all inhabitants are satisfied with this assignment. This can be done in $2^{\mathcal{O}(|V_{U}|)}\cdot n^{\mathcal{O}(1)}$ time, that is, ARH is in FPT with respect to the number of empty vertices $|V_{U}|$.

For the running-time lower-bound, we recall that $3,4$-SAT cannot be solved in $2^{o(n+m)}$ time unless ETH fails Impagliazzo and Paturi (2001); Impagliazzo et al. (2001). Assume to the contrary, there is an algorithm $\mathcal{A}$ that solves ARH in $2^{o(|V_{U}|)}\cdot n^{\mathcal{O}(1)}$. Then, for every $3,4$-SAT formula $\varphi$ we can use the basic reduction from Section 3, which can be clearly done in polynomial time, to the equivalent instance $\mathcal{I}$ of the ARH problem, solve $\mathcal{I}$ in $2^{o(|V_{U}|)}\cdot n^{\mathcal{O}(1)}$ time and then reconstruct a solution for $\varphi$. Overall, this gives us an algorithm running in $2^{o(|V_{U}|)}\cdot n^{\mathcal{O}(1)}\subseteq 2^{o(n)}\subseteq 2^{o(n+m)}$ time for $3,4$-SAT which contradicts ETH. ∎

In the remainder of this section, we present complexity results concerning various structural restrictions of the topology. Arguably, the most prominent structural parameter is the tree-width of a graph that, informally speaking, expresses its tree-likeness, and which is usually small in real-life networks Maniu et al. (2019). Unfortunately, we can show the following stronger intractability result.

{theorem}

The ARH problem is W[1]-hard parameterised by the tree-depth $\operatorname{td}(G)$ of the topology $G$ and, unless ETH fails, there is no algorithm solving ARH in $f(k)\cdot n^{o(k/\log k)}$ time, where $k=\operatorname{td}(G)$, for any function $f$.

Proof.

We reduce from the Unary Bin Packing problem. Here, we are given a bin capacity $B$, a set of items $A=(a_{1},\ldots,a_{n})$ and number of bins $k$. Our goal is to decide whether there is an assignment $\beta$ of all items to bins with respect to bin capacity. Unary Bin Packing is known to be W[1]-hard with respect to the number of bins and cannot be solved in time $f(k)\cdot n^{o(k/\log k)}$ for any function $f$ Jansen et al. (2013).

Our construction of an equivalent instance $\mathcal{I}^{\prime}$ of the ARH problem is as follows. First, for every $i\in[k]$ we create a bin vertex $B_{i}$ and make it occupied by an inhabitant $b_{i}$ with ${A_{b_{i}}=[B]_{0}}$. Note that these inhabitants approve the whole interval from [math] to $B$. Next, we add items gadgets to map the items to bins. The single item gadget for some $a_{j}\in A$ is a star with $a_{j}$ leaves and centre occupied by an inhabitant $c_{j}$ with $A_{c_{j}}=\{0,a_{j}\}$. This ensures that in the solution either all leaves are occupied by refugees or none of the leaves are. We add an item gadget $X_{j}^{i}$ for every $i\in[k]$ and every $a_{j}\in A$ and add an edge connecting bin vertex $B_{i}$ with all leaves of item gadgets $X_{j}^{i}$, where $i\in[k]$ and $j\in[n]$. Finally, we must ensure that every item is assigned to exactly one bag. This is ensured by a guard vertex $G_{i}$ for every item $a_{i}\in A$. This vertex is occupied by an inhabitant $g_{i}$ with $A_{g_{i}}=\{a_{i}\}$ and is connected to all leaves of item gadgets $X_{i}^{j}$, where $j\in[k]$. The number of refugees in our instance is $|R|=\sum_{a_{i}\in A}a_{i}$. For an illustration of our construction, we refer the reader to Figure 3.

For the correctness, let $\mathcal{I}$ be a yes-instance of the Unary Bin Packing problem and $\beta$ be a solution assignment. For every $a_{i}\in A$, we house $a_{i}$ refugees to the leaves of the item gadget $X_{i}^{\beta(a_{i})}$. Since $\mathcal{I}$ is a yes-instance, every bin vertex has at most $B$ refugees in the neighbourhood. Moreover, each item gadget is either empty or full, in the neighbourhood of every guard inhabitant $g_{i}$ there is exactly $a_{i}$ neighbouring refugees, and, finally, all refugees are housed. Thus, $\mathcal{I^{\prime}}$ is also a yes-instance.

In the opposite direction, let $\mathcal{I}^{\prime}$ be a yes-instance and $\pi$ be a solution housing. Due to definition of the approval sets, every item gadget is either full or empty. In our construction, there are $k$ copies of item gadgets for every item $a_{i}\in A$ and guard vertices secure that exactly one copy is full. Moreover, the bin vertices accept at most $B$ refugees in their neighbourhood. We recall that $|R|=\sum_{a_{i}\in A}a_{i}$. Hence, we define the solution assignment $\beta$ for $\mathcal{I}$ as $\beta(a_{i})=j$, where $j\in[k]$ and $X_{i}^{j}$ is full.

Finally, we discuss the bound on the parameter value. It is not hard to see that the construction have tree-depth ${\mathcal{O}(k)}$; By removing vertices $B_{1},\ldots,B_{k}$, we obtain $n$ connected components $C_{1},\ldots,C_{n}$ and in every $C_{i}$ it is enough to remove the vertex $G_{i}$ to obtain a collection of stars, which have tree-depth at most $2$. Also note that algorithm running in $f(k^{\prime})\cdot|\mathcal{I}^{\prime}|^{o(k^{\prime}/\log k^{\prime})}$ time for ARH would yield, together with our parameterised reduction, to an algorithm running in $f(k)\cdot n^{o(k/\log k)}$ time for Unary Bin Packing, which contradicts ETH. ∎

It is well known, and easy to see, that the tree-width of a graph is at most its tree-depth. Hence, we directly obtain the following result.

**Corollary **\thetheorem

The ARH problem is W[1]-hard parameterised by the tree-width $\operatorname{tw}(G)$ of the topology $G$.

It is well-known, an easy to see, that the tree-width of a graph is at most its vertex-cover number. Hence, due to Section 3, we directly obtain the following result for tree-width.

**Corollary **\thetheorem

The ARH problem is W[1]-hard parameterised by the tree-width $\operatorname{tw}(G)$ of the topology $G$.

Many problems that are computationally hard with respect to tree-width are studied from the viewpoint of less restricted parameters. Vertex cover number is a frequent representative of such parameters Maiti and Dey (2022); Bredereck et al. (2019b); Knop et al. (2022); however, in the ARH problem, not even this restriction of the topology leads to a tractable algorithm.

{theorem}

The ARH problem is W[1]-hard parameterised by the vertex cover number $\operatorname{vc}(G)$ of the topology.

Proof.

Although it was not stated formally, it can be seen that the construction used to prove Section 3 has not only parameter-many inhabitants, but these agents also forms a vertex cover of the topology. Therefore, we directly obtain W[1]-hardness for the setting with bounded vertex cover number. ∎

Nevertheless, if we additionally restrict the approval sets, we obtain the following algorithmic result.

{theorem}

The ARH problem is fixed-parameter tractable parameterised by the vertex cover number $\operatorname{vc}(G)$ if all inhabitants approve intervals.

Proof.

Let $M\subseteq V$ be a minimum size vertex cover of $G$ and let $k=|M|$. It is not hard to see that, due to Section 3, for each connected component $C_{i}$ of $G$ all vertices in $M\cup C_{i}$ are either empty vertices or occupied by inhabitants. By $C_{I}$ we denote components with vertex cover occupied by inhabitants and by $C_{R}$ the components where the vertex cover consists of empty vertices, respectively.

First, we guess the number $k^{\prime}\leq k$ of refugees assigned to components of $C_{R}$. Now, we try all $2^{k^{\prime}}$ subsets and check for each subset whether all refugees are housed and whether the housing is inhabitant-respecting in $C_{R}$. At the same time, we try to find a housing of $|R|-k^{\prime}$ refugees to components in $C_{I}$. To do so, we can use the algorithm from Section 3 as the number of inhabitants is at most $k$. Altogether, we obtain that the ARH problem is in FPT when parameterized by the vertex cover number of the topology if all inhabitants approve intervals. ∎

By the same argumentation used in the proof of Section 3, we obtain the following last result of this section.

**Corollary **\thetheorem

The ARH problem is fixed-parameter tractable when parameterised by the vertex cover number $\operatorname{vc}(G)$ and the maximum number of disjoint intervals $\delta$ combined.

4. Fully Hedonic Preferences

Our second model of refugee housing improves upon the previous model by introducing individual preferences of refugees. Naturally, refugees are no longer anonymous and the identity of every particular refugee matters. The preferences of the inhabitants are again dichotomous, and for every inhabitant $i\in I$ the approval set is a subset of $2^{R}$. Similarly, for a refugee $r\in R$, the approval set $A_{r}$ is a subset of $2^{I}$. Our goal is to find housing that conforms to the preferences of both groups.

Definition \thetheorem.

A housing $\pi\colon R\to V_{U}$ is called respecting if for every $i\in I$ we have $N(\iota(i))\in A_{i}$ and for every $r\in R$ we have $N(\pi(r))\in A_{r}$.

In other words, a housing $\pi$ is respecting if every inhabitant and every refugee approves its neighbourhood. We study the problem of deciding whether there is a respecting housing in the instance with hedonic preferences under the name Hedonic Refugees Housing (HRH for short).

**Example **\thetheorem

Let the topology be a cycle with four vertices. There are two inhabitants $h_{1}$ and $h_{2}$ assigned to neighbouring vertices and two refugees $r_{1}$ and $r_{2}$ to house. The approval set of inhabitant $h_{1}$ is $A_{h_{1}}=\{\{r_{1}\},\{r_{2}\}\}$, that is, $h_{1}$ approves only one refugee in her neighbourhood regardless of the identity. The second inhabitant approves the set $A_{h_{2}}=\{\{r_{2},\},\{r_{1},r_{2}\}\}$. In other words, the inhabitant $h_{2}$ is dissatisfied with having only the refugee $r_{1}$ in the neighbourhood; however, he is fine with neighbouring with both the refugees. For the refugees, we have $A_{r_{1}}=\{\{h_{1}\}\}$ and $A_{r_{2}}=\{\{h_{2}\}\}$. Housing $r_{1}$ in the neighbourhood of $h_{1}$ and $r_{2}$ in the neighbourhood of $h_{2}$ is clearly respecting.

Observe that, since both inhabitants and refugees have preferences only over the other set of individuals, we can remove all edges between empty and occupied vertices, respectively. Hence, all graphs assumed in this section are again bipartite.

Now, we show how the results from Section 3 carry over to the hedonic setting studied in this section. Our first theorem shows that the hedonic setting is also computationally hard on graphs of constant degree. The construction is very similar to the one used to prove Section 3.

{theorem}

The HRH problem is NP-complete even if the topology is a graph of maximum degree $3$.

Proof.

Our reduction is very similar to the reduction used to prove Section 3. We again reduce from the $2$-Balanced $3$-SAT problem where every variable appears at most twice as positive literal and at most twice as negative literal, respectively. However, due to the different nature of the problem, we have to slightly change the variable gadget and approval sets of agents.

For every variable $x_{i}\in\varphi$, we introduce one variable gadget $X_{i}$. The variable gadget again consists of a path $P_{i}=v_{i}^{1}v_{i}^{2}v_{i}^{3}$ that is intended to represent the assignment of a variable $x_{i}$. Together with $P_{i}$, we add a refugee $r_{i}$ and an inhabitant $h_{i}$. Let $C_{x_{i}}=\{C_{j_{1}},\ldots,C_{j_{4}}\}$, where $j_{1},j_{2},j_{3},j_{4}\in[m]$, be a set of clauses in which the variable $x_{i}$ appears. The refugee $r_{i}$ approves the set $\{S\cup\{h_{i}\}\mid S\subseteq 2^{C_{x_{i}}}\}$ and the inhabitant $h_{i}$ approves the set $\{\{r_{i}\}\}$. Moreover, we set $\iota(h_{i})=v_{i}^{2}$.

Every clause $C_{j}\in\varphi$ is represented by a single vertex $c_{j}$. Let $C_{j}$ consist of variables $x_{i_{1}}x_{i_{2}}x_{i_{3}}$. The vertex $c_{j}$ is occupied by an inhabitant $H_{j}$ with approval set $2^{\{x_{i_{1}},x_{i_{2}},x_{i_{3}}\}}\setminus\emptyset$ and for every ${i\in\{i_{1},i_{2},i_{3}\}}$ the vertex $c_{j}$ is connected to $v_{i}^{1}$ if the variable $x_{i}$ occurs as a positive literal in $C_{j}$ and to $v_{i}^{3}$ if the variable $x_{i}$ occurs as a negative literal in $C_{j}$. This finishes the construction.

For the correctness, let $\alpha$ be a satisfying assignment for $\varphi$. For every $x_{i}\in\varphi$, we house the refugee $r_{i}$ to vertex $v_{i}^{1}$ if $\alpha(x_{i})=1$ and to vertex $v_{i}^{3}$ otherwise. Observe that this housing clearly respects each $h_{i}$ and, since $\alpha$ is satisfying, there is at least one refugee in the neighbourhood of each $H_{j}$, as prescribed. In the opposite direction, let $\pi$ be a correct housing for the equivalent HRH instance $\mathcal{I}^{\prime}$. Due to the approval sets of $h_{i}$, $i\in[n]$, every refugee $r_{i}$ is housed to one of the two positions of the variable gadget. Moreover, the approval set of inhabitant $H_{j}$, $j\in[m]$, which represents the clause of the original formula, ensures that there is at least one refugee in his neighbourhood. Hence, the following assignment

[TABLE]

indeed satisfy $\varphi$ and the reduction can be done in polynomial time. ∎

From the proof of Section 4, we can easily distil the following general reduction from the ARH problem to the HRH problem. Let $\mathcal{I}$ be an ARH instance. For every empty vertex $v\in V_{U}$ we add into $R^{\prime}$ one refugee $r_{v}$ with approval set $A_{r_{v}}=N(v)$. Next, for every inhabitant $i\in I$ we add a new inhabitant $h_{i}$ approving the set $A_{h_{i}}=\{\binom{R_{h_{i}}}{a}\mid a\in A_{i}\}$, where $R_{h_{i}}=\{r_{v}\mid v\in N(v)\cap V_{U}\}$. To ensure that only $|R|$ refugees are housed, we extend the construction by a single star with $|V_{E}|-|R|$ leaves and occupy the centre of the star with an inhabitant $G$ approving the set $\binom{R^{\prime}}{|V_{E}|-|R|}$. Moreover, we have to add $\{G\}$ to the approval set of every refugee $r_{v}\in R^{\prime}$. It is not difficult to see that the instances are indeed equivalent; however, the reduction is not polynomial-time, the approval sets can be at worst exponential in the number of empty vertices. Hence, the reduction works only in cases where the number of empty vertices is at most logarithmic in the size of the topology. Unfortunately, this is not the case for most of our polynomial-time reductions, however, we are able to show similar results using different techniques.

{theorem}

The HRH problem is W[1]-hard when parameterised by the combined parameter the vertex cover number $\operatorname{vc}(G)$ of the topology and the number of inhabitants $|I|$.

Proof.

We reduce from the W[1]-hard Multicoloured Clique problem Fellows et al. (2009). We recall that here we are given a $k$-partite graph $G=(V_{1}\cup\cdots\cup V_{k},E)$ and the goal is to find a complete sub-graph with $k$ vertices such that it contains a vertex from every $V_{i}$, $i\in[k]$.

Let $G=(V_{1}\cup\cdots\cup V_{k},E)$ be an instance of the Multicoloured Clique problem such that all colour classes $V_{i}$ are of the same size $n_{G}$. We construct an equivalent instance $\mathcal{I^{\prime}}$ of the HRH problem as follows. For every vertex set $V_{i}$, where $i\in[k]$, we introduce a vertex-selection gadget $S_{i}$ which is a star with $n_{G}$ leaves. We call an arbitrary leaf a selection leaf. This selection leaf serves for a vertex of colour $i$ that should be part of the clique and is the only connection of the vertex-selection gadget with the rest of the topology. Let $\{v_{i}^{1},\ldots,v_{i}^{n_{G}}\}$ be a set of vertices in the colour class $V_{i}$. We introduce one refugee $r_{i}^{p}$ for every $v_{i}^{p}$, $p\in[{n_{G}}]$ and one inhabitant $s_{i}$ which is assigned to the centre $c_{i}$ of $S_{i}$ and approves the set $\{\{r_{i}^{1},\ldots,r_{i}^{n_{G}}\}\}$. Every refugee $r_{i}^{p}$ approves the set $\{\{s_{i}\}\}$.

Then, for every pair of distinct colours $i,j\in[k]$, we introduce one additional guard vertex $g_{i,j}$ securing that there is an edge between vertices selected in incident vertex-selection gadgets. We connect this guard vertex $g_{i,j}$ to selection leaves of vertex-selection gadgets $V_{i}$ and $V_{j}$. Moreover, we introduce an inhabitant $h_{i,j}$ assigned to $g_{i,j}$ with approval set $\{\{r_{i}^{p},r_{j}^{q}\}\mid\{v_{i}^{p},v_{j}^{q}\}\in E\land p,q\in[{n_{G}}]\}$. That is, $h_{i,j}$ approves exactly those pairs of vertices from $V_{i}$ and $V_{j}$ that are connected by an edge.

To be able to house any refugee to selection leaves, we have to extend their approval sets. Thus, for every refugee $r_{i}^{p}$, where $i\in[k]$ and $p\in[{n_{G}}]$, we add to the approval set the set $\{s_{i}\}\cup\{h_{i,j}\mid j\in[k]\setminus\{i\}\}$. This finishes the construction.

For the correctness, let $K=\{v_{1}^{j_{1}},\ldots,v_{k}^{j_{k}}\}$ be a multicoloured clique in $G$. For every vertex-selection gadget $S_{i}$, we house to the selection leaf the refugee $r_{i}^{j_{i}}$ and the refugees $r_{i}^{p}$, where ${p\in[{n_{G}}]\setminus\{j_{i}\}}$ arbitrarily to the remaining leaves of $S_{i}$. As vertices of $K$ form a clique in $G$, there is an edge between each pair of vertices assigned to selection leaves; hence, the housing is correct.

In the opposite direction, suppose that $\mathcal{I^{\prime}}$ is a yes-instance, $\pi$ is a respecting housing and there is a pair of refugees $r_{i}^{p}$ and $r_{j}^{q}$, where $i,j\in[k]$ are distinct and $p,q\in[{n_{G}}]$, assigned to selection leaves such that there is no corresponding $\{v_{i}^{p},v_{j}^{q}\}\in E(G)$. Then $\{r_{i}^{p},r_{j}^{q}\}$ is not in the approval set of the inhabitant $h_{i,j}$, which is a contradiction with $\pi$ being a respecting housing.

Clearly, the removal of the vertex set $\{c_{i}\mid i\in[k]\}$ together with the set $\{g_{i,j}\mid i,j\in[k]\}$ leads to an edgeless graph. As the size of both sets is polynomial in $k$ and all inhabitants are assigned to these vertices, we obtain that the presented reduction is indeed a parameterised reduction and the theorem follows. ∎

As a final result of this section, we prove that the HRH is \NPheven for graphs of tree-width at most $3$.

{theorem}

The HRH problem is para-NP-hard parameterised by the tree-width $\operatorname{tw}(G)$ of the topology.

Proof.

Our reduction is from the Channel Assignment problem which is known to be \NPheven for graphs of tree-width at most $3$ McDiarmid and Reed (2003). Here, we are given an undirected graph $G=(V,E)$, a function $\ell\colon E\to\mathbb{N}$, and a positive integer $\lambda$. The goal is to decide whether there exists a function $f\colon V\to[\lambda]$ such that for every $\{u,v\}\in E$ we have $|f(u)-f(v)|\geq\ell(\{u,v\})$. It should be noted that the problem is \NPheven if $\lambda$ and $\max_{e\in E}\ell(e)$ are constants.

Let $\mathcal{I}=(G,\ell,\lambda)$ be a Channel Assignment instance. The topology of an equivalent HRH instance $\mathcal{I^{\prime}}$ is a graph $G^{\prime}$ which arises from $G$ by subdividing all edges. We call the vertex created by subdivision of the edge $\{u,w\}\in E(G)$ of the original graph $v_{u,w}$. Next, we introduce $\lambda$ refugees $r_{v}^{1},\ldots,r_{v}^{\lambda}$ for every vertex $v\in V(G)$. Then, for every vertex $v_{u,w}$, where $\{u,w\}\in E(G)$, we introduce one inhabitant $h_{u,w}$ and assign it to $v_{u,w}$. The approval set for every refugee $r_{v}^{i}$ is $\{\{h_{u,v}\}\mid v_{u,w}\in N_{G^{\prime}}(v)\}$ and for every inhabitant $h_{u,w}$ it is $\{\{r_{u}^{i},r_{w}^{j}\}\mid|i-j|\geq\ell(\{u,w\})\}$. In this construction, there are $\lambda$ candidates for every empty vertex, hence, we have to make sure that there is a house where to house remaining $\lambda-1$ candidates. To secure this, we introduce for every $v\in V(G)$ star $S_{v}$ with $\lambda-1$ leaves, make the centre of $S_{v}$ occupied by an inhabitant $g_{v}$ approving all subsets of size $\lambda-1$ of the set $\{r_{v}^{1},\ldots,r_{v}^{\lambda}\}$, and we add set ${\{g_{v}\}}$ to the approval set of every refugee $r_{v}^{i}$, $i\in[\lambda]$.

For the correctness, let $f$ be a proper channel assignment for $\mathcal{I}$. For every vertex $v\in V(G)$, we house the refugee $r_{v}^{f(v)}$ to $v$ and we house all the remaining refugees $r_{v}^{i}$, $i\in[\lambda]\setminus\{f(v)\}$ arbitrarily to empty vertices of $S_{v}$. Clearly, all assigned refugees are satisfied with this housing as well as the inhabitant $g_{v}$ assigned to the centre of $S_{v}$. The same is true for inhabitants $h_{u,w}$, as $f$ is a proper channel assignment and $h_{u,w}$ accept only pairs $\{r_{u}^{i},r_{w}^{j}\}$ where $|i-j|\geq\ell(\{u,w\})$.

In the opposite direction, let $\pi$ be a correct housing. The number of refugees to house is equal to the number of empty vertices of $G^{\prime}$, hence, every empty vertex is occupied. The inhabitants occupying vertices $v_{u,w}$ secure that for refugees $r_{u}^{i}$ and $r_{w}^{j}$ assigned to $u$ and $w$, respectively, it holds that $|i-j|\leq\ell(\{u,w\})$, hence, we can set $f(v)=i$, where $r_{v}^{i}=\pi^{-1}(v)$.

We already claimed that the Channel Assignment problem is \NPheven for the graphs of tree-width at most $3$. As the edge subdivision do not increase tree-width, we have that $G^{\prime}$ is also of tree-width at most $3$. Moreover, the reduction can be done in polynomial-time. This finishes the proof. ∎

5. Diversity Preferences

In the anonymous refugee housing, we are not assuming the preferences of individual refugees. Thanks to this property, the model is as simple as possible. The fully hedonic setting from Section 4 precisely captures preferences of both the refugees and the inhabitants. On the other hand, the fully hedonic model is not very realistic, as it is hard to acquaint all inhabitants with all refugees.

Hence, we introduce the third model of refugees housing, where both the inhabitants and the refugees are partitioned into types and agents from both groups have preferences over fractions of agents of each type in their neighbourhood.

Such a diversity goals, where agents are partitioned into types and the preferences of agents are based on the fraction of each type in their neighbourhood or coalition, was successfully used in many scenarios such as school choice Aziz and Sun (2021); Aziz et al. (2019, 2020), public housing Benabbou et al. (2018); Gross-Humbert et al. (2021), hedonic games Bredereck et al. (2019a); Boehmer and Elkind (2020b, a); Darmann (2021); Ganian et al. (2022), multi-attribute matching Ahmadi et al. (2021), or employee hiring Schumann et al. (2019).

Before we formally define the computational problem of our interest, let us introduce further notation. Let $N=I\cup R$ be a set of agents partitioned into $k$ types $T_{1},\ldots,T_{k}$. For a set $S\subseteq N$, we define a palette as a $k$-tuple $\left(\frac{|T_{i}\cap S|}{|S|}\right)_{i\in[k]}$ if $|S|\geq 1$ and $k$-tuple $(0,\ldots,0)$ if $S=\emptyset$. Given an agent $a\in N$, her approval set is a subset of the set $\left\{\left(\frac{|T_{i}\cap S|}{|S|}\right)_{i\in[k]}\mid S\subseteq 2^{N}\right\}$.

Definition \thetheorem.

A housing $\pi\colon R\to V_{U}$ is called diversity respecting if for every inhabitant $i\in I$ the palette for the set $\{h\in I\mid\iota(h)\in N(\iota(i))\}\cup\{r\in R\mid\pi(r)\in N(\iota(i))\}$ is in $A_{i}$, and for every refugee $r\in R$ the palette for the set $\{h\in I\mid\iota(h)\in N(\pi(r))\}\cup\{r^{\prime}\in R\mid\pi(r^{\prime})\in N(\pi(r))\}$ is in $A_{r}$.

The Diversity Refugees Housing problem (DRH for short) then asks whether there is a diversity respecting housing $\pi$. Note that this time, we are not allowed to drop edges between two inhabitants or two refugees and, thus, the graphs assumed in this section are no longer bipartite.

**Example **\thetheorem

Let the topology be a cycle with four vertices. There are two agents of type $T_{1}$. One of these agents is an inhabitant $h_{1}$ approving $\{(1,0),(1/2,1/2)\}$ and the second one is a refugee $r$ approving only agents of his own type, that is, $A_{r}=\{(1,0)\}$. The type $T_{2}$ contains one inhabitant $h_{2}$ approving the set $\{(1,0)\}$. Inhabitants are assigned such that they are neighbours. There are two possible housings for the refugee $r$. She can be either neighbour of $h_{1}$ or $h_{2}$. Since she accepts only agents of her own type in the neighbourhood, the only diversity respecting housing is next to inhabitant $h_{1}$.

Despite that the topology can be much complicated, we can show, using similar construction as in Section 3, that the tractability condition based on the bounded degree cannot be surpassed even in this model.

{theorem}

The DRH problem is NP-complete even if the topology is a graph of maximum degree $3$.

Proof.

We reduce from the $2$-Balanced $3$-SAT where each variable appears at most twice positive and at most twice negative. Let $\varphi$ be a $3,4$-SAT formula. In an equivalent instance $\mathcal{I^{\prime}}$ of DRH, we have $n+1$ types – one for every variable and an extra one for agents representing clauses. Let $T_{n+1}$ be a clause type. For every variable $x_{i}$, we create a variable gadget which is a path $v_{i}^{1}v_{i}^{2}v_{i}^{3}$ on $3$ vertices. We also introduce two agents – one refugee $r_{i}$ and one inhabitant $h_{i}$. Both of them belong to type $T_{i}$. The inhabitant $h_{i}$ is then assigned to the middle vertex of a path and approves (and requires) only agents of his own type in his neighbourhood. This secure that the refugee $r_{i}$ is necessarily assigned to his variable gadget in each housing. The refugee requires in his neighbourhood an agent of his own type and at most $2$ agents of clause type.

For every clause $C_{j}$, we create one vertex $c_{j}$ and occupy it with an inhabitant $g_{j}$ of type $T_{n+1}$. For every variable $x_{i}$ which is part of $C_{j}$, we connect $c_{j}$ to $v_{i}^{1}$ if $x_{i}$ is as positive literal in $C_{j}$, and to $v_{i}^{3}$ if $x_{i}$ appears as a negative literal in $C_{j}$. Let $x_{i_{1}}$, $x_{i_{2}}$, and $x_{i_{3}}$ be three variables forming $C_{j}$. The inhabitant $g_{j}$ then approves to have monotype neighbourhood with types $i_{1}$, $i_{2}$, or $i_{3}$, and neighbourhoods where these types are represented equally – either two of them or all three.

For the correctness, let $\alpha$ be a satisfying assignment for $\varphi$. For every variable $x_{i}$ we house the only refugee of type $T_{i}$ to the vertex $v_{i}^{1}$ if $\alpha(x_{i})=1$, or to the vertex $v_{i}^{2}$ if $\alpha(x_{i})=0$. It is easy to see that this housing is diversity respecting as every $h_{i}$ has only one neighbour of his own type and, since $\alpha$ is satisfying, every $g_{j}$, $j\in[m]$, has at least one neighbouring refugee. In the opposite direction, let $\pi$ be a diversity respecting housing. Then every refugee is housed to the neighbourhood of the inhabitant of their own type. For every $c_{j}$ we have also one inhabitant $g_{j}$ who ensures that there is at least one refugee in the neighbourhood. Hence, we can directly derive the truth assignment $\alpha$ from $\pi$. ∎

In Section 5 we exploit the number of types to ensure that every refugee is housed in the right house. Therefore, the number of types was as large as the number of empty vertices. In the following result, we show that the DRH problem is computationally hard even if the number of types is small.

{theorem}

The DRH problem is NP-complete even if there are only two types of agents.

Proof.

We show the NP-hardness by a reduction from the Set Cover problem which is known to be NP-complete Garey and Johnson (1979). Here, we are given a universe $U=\{u_{1},\ldots,u_{n}\}$, a family $\mathcal{F}$ of subsets of $U$, and an integer $k\in\mathbb{N}$. The goal is to decide whether there is a sub-family $\mathcal{C}\subseteq\mathcal{F}$ of size at most $k$ such that $\bigcup_{C\in\mathcal{C}}C=U$.

Given an instance $\mathcal{I}=(U,\mathcal{F},k)$, we construct an equivalent instance $\mathcal{I^{\prime}}$ of DRH as follows. For every element $u_{i}\in U$ we add one vertex $v_{i}$ and assign to it an inhabitant $h_{i}$. The inhabitant $h_{i}$ is of type $T_{1}$ and his approval set is $\{(0,1)\}$. Next, for every subset $F\in\mathcal{F}$, we create one vertex $v_{F}$ that is adjacent to every $v_{i}$ such that $u_{i}\in F$. To finalise the construction, we add $k$ refugees $r_{1},\ldots,r_{k}$ of type $T_{2}$ approving the set $\{(1,0)\}$.

For the correctness, let $\mathcal{I}$ be a yes-instance and $\mathcal{C}=\{C_{1},\ldots,C_{k}\}$ be a solution for $\mathcal{I}$. For every $i\in[k]$, we house the refugee $r_{i}$ on the vertex corresponding to the set $C_{i}$. In this housing, every refugee is satisfied and, since $\mathcal{C}$ is a set cover, every $h_{i}\in I$ neighbours with at least one refugee. In the opposite direction, let $\pi$ be a diversity respecting housing. We add to $\mathcal{C}$ a set $C_{i}\in\mathcal{C}$ if and only if there is a refugee housed on the corresponding vertex $v_{C_{i}}$. Suppose that there is an element $u\in U$ which is not covered by $\mathcal{C}$. Then there is an inhabitant which is not neighbour of any refugee. However, this could not be the case as $\pi$ is diversity respecting. Hence, the reduction is correct and clearly can be done in polynomial-time. ∎

Note that the NP-hardness proved in Section 5 can be strengthened to a single type of agents; however, we find this situation not very natural in the context of DRH.

Additionally, it is known that the Set Cover problem is W[2]-complete and cannot be solved in $f(k)\cdot n^{o(k)}$ time, unless the ETH fails Cygan et al. (2015). Therefore, from the construction used to prove Section 5, we directly obtain the following corollary.

**Corollary **\thetheorem

The DRH problem is W[2]-hard when parameterised by the number of refugees $|R|$ even if there are only two types of agents and, unless ETH fails, there is no algorithm solving DRH in $f(R)\cdot n^{o(R)}$ time for any computable function $f$.

6. Conclusions

We initiated the study of a novel model of refugee housing. The model mainly targets the situations where refugees need to be accommodated and integrated in the local community. This distinguishes us from the previous settings of refugee resettlement.

Our results identify some tractable and intractable cases of finding stable outcomes from the viewpoint of both the classical computational complexity and the finer-grained framework of parameterised complexity. To this end, we believe that other notions of stability inspired, for example, by the model of Schelling games of Agarwal et al. (2021) or by exchange-stability of Alcalde (1994), should be investigated.

Natural way to tackle the intractability of problems in computational social choice is to restrict the preferences of agents Elkind et al. (2022). One such restriction that should be investigated, especially in the case of anonymous setting, are the single-peaked preferences Arrow (2012) that were successfully used in similar scenarios; see, e.g., Bilò et al. (2022); Yang (2017); Bredereck et al. (2019a) or the survey of Elkind et al. (2022). Beside that, we are interested in the anonymous setting in which every inhabitant $i\in I$ approves an interval $[0,u_{i}]$, where $u_{i}\geq 0$ is an inhabitant-specific upper-bound on the number of refugees in her neighbourhood.

Finally, there are many notions measuring quality of an outcome studied in the literature in both the context of Schelling and hedonic games Elkind et al. (2020); Agarwal et al. (2021); Aziz et al. (2018), and we believe that these notions should be investigated even in the context of refugee housing. In this line, the most prominent notion is the social-welfare of an outcome.

{acks}

The authors acknowledge the support of the Czech Science Foundation Grant No. 22-19557S. Šimon Schierreich was additionally supported by the Grant Agency of the Czech Technical University in Prague, grant No. SGS23/205/OHK3/3T/18.