Stability and Welfare in (Dichotomous) Hedonic Diversity Games

Abstract

In a hedonic diversity game (HDG) there are two types of agents (red and blue agents) that need to form disjoint coalitions, i.e., subgroups of agents. Each agent’s preferences over the coalitions depend on the relative number of agents of the same type in her coalition. In the special case of a dichotomous hedonic diversity game (DHDG) each agent distinguishes between approved and disapproved fractions only. We aim at outcomes that are stable against agents’ deviations, and at outcomes that maximize social welfare. In particular, we show that the strict core of a DHDG may be empty even in instances with only three agents, while each HDG with two agents has a non-empty strict core. We also provide several computational complexity results for DHDGs with respect to the number of fractions approved per agent. For instance, we prove that deciding whether a DHDG has a non-empty strict core is $\textsf {NP}$-complete even when each agent approves of at most three fractions. In addition, we show that deciding whether a DHDG admits a Nash stable outcome is $\textsf {NP}$-complete even in restricted settings with only two approved fractions per agent—therewith, improving a result in the literature. For the task of maximizing social welfare, we apply approval scores and Borda scores from voting theory. For DHDGs and approval scores, we draw the sharp separation line between polynomially solvable and $\textsf {NP}$-complete cases with respect to the fixed number of approved fractions per agent. We complement these findings with an $\textsf {NP}$-completeness result for HDGs under Borda scores.

1 Introduction

In a hedonic diversity game (HDG) there are two types of agents (red and blue agents) that need to form disjoint coalitions, i.e., subgroups of agents. Each agent’s preferences over the coalitions depend only on the relative number of agents of the same type in her coalition. An important subclass of hedonic games are so-called Bakers and Millers games (Aziz et al. [3] and Bredereck et al. [13]), where each red agent (baker) wishes to be in a coalition in which the fraction of blue agents (millers) is as large as possible, and each blue agent (miller) wishes to be in a coalition in which the fraction of red agents (bakers) is as large as possible. For another example of a HDG consider the situation in which two academic departments (say R and B) merge into one: while some of the members of R (red agents) are eager to closely collaborate with their new colleagues, i.e., the members of B (blue agents), others might be reluctant to do so and prefer to work on their next project together with other members of R, who they are used to collaborate with; hence, some red agents may prefer a high fraction of blue agents in their coalition, while others prefer a low fraction. Analogously, some of the blue agents prefer a high relative number of red agents in their coalition to a low one, whereas others prefer it the other way round. Similar applications of HDGs appear in student group formation (e.g., among local and exchange students) and international collaboration (see Bredereck et al. [13]).

Table 1 Overview of computational complexity results for different stability notions; the results provided in this paper are in bold. In the table, “in $\textsf{P}$” indicates that a respective outcome always exists and can be found in polynomial time. “$\mathsf NP$-c” means that the corresponding decision problem is $\textsf{NP}$-complete

We are hence concerned with a coalition formation problem, where the aim would be a reasonable outcome, i.e., partition of agents into disjoint coalitions. In this respect we take into account two kinds of solution concepts. On the one hand, we consider concepts that deal with stability against agents’ deviations: here, we focus on the concepts of Nash stability, a famous concept of stability against individual deviations, and strict core stability, which is concerned with stability against group deviations. On the other hand, by adapting scores from voting theory we aim at outcomes that maximize the induced social welfare (i.e., sum of scores). In this work we lay particular focus on the setting of a dichotomous hedonic diversity game (DHDG), the special case of a HDG in which each agent, in a binary way, states her preferences by distinguishing only between “good” fractions, i.e., fractions she approves of, and “bad” fractions, i.e., fractions she disapproves of.

In this paper, we show that deciding whether a DHDG admits a Nash stable outcome is $\textsf{NP}$-complete, even when each agent approves of only two fractions (see also Table 1). Therewith, we improve upon a result in the literature. In addition, we prove that for instances with two agents the strict core of a HDG—and thus of a DHDG—is always non-empty, while for any number $n \ge 3$ there is an instance of a DHDG with n agents in which the strict core is empty, even when each agent approves of only one fraction. Next, we show that the corresponding decision problem whether a DHDG admits a non-empty strict core is $\textsf{NP}$-complete, even when each agent approves of three fractions only. Adapting scores from voting theory, we then aim at outcomes that maximize social welfare (see also Table 2). In that respect, for approval scores we draw the sharp separation line between polynomially solvable and $\textsf{NP}$-complete cases with respect to the fixed number of approved fractions per agent: when each agent approves of exactly one fraction, by a non-trivial reduction to the two-constraint knapsack problem we show that an outcome that maximizes social welfare can be found in polynomial time; as soon as the agents may approve of more fractions, the corresponding decision problem becomes $\textsf{NP}$-complete—more precisely, we prove that $\textsf{NP}$-completeness holds for any fixed number $s\ge 2$ of fractions approved per agent. Finally, we show that maximizing social welfare under Borda scores is computationally difficult.

Table 2 Overview of results for maximizing social welfare under approval and Borda scores; the results provided in this paper are in bold. Here, for the problem that is “in $\textsf{P}$”, a respective outcome can be found in polynomial time. “$\mathsf NP$-c” means that the corresponding decision problem is $\textsf{NP}$-complete, while “pref.” and “p. a.” stand for “preferences” and “per agent” respectively

1.1 Related literature

Hedonic diversity games. The literature on hedonic diversity games is a very recent one. Introduced by Bredereck et al. [13], the main solution concepts considered in HDGs deal with stability against individual or group deviations such as Nash stability, individual stability, and core stability. For the special case of single-peaked preferences, Bredereck et al. [13] show that a Nash stable outcome may fail to exist, whereas an individually stable outcome always exists and can be found in polynomial time. They also show that the core might be empty even when each agent’s preferences are single-peaked, and prove that in general, deciding whether a HDG admits a non-empty core is an $\textsf{NP}$-complete problem. In a follow-up paper, Boehmer and Elkind [9] focus on stability against individual deviations by considering Nash stability and individual stability. They generalize a previous result by Bredereck et al. [13] by showing that an individually stable outcome is guaranteed to exist in any HDG, and that such an outcome can be found in polynomial time. On the negative side, Boehmer and Elkind [9] show that it is $\textsf{NP}$-complete do decide whether a DHDG—and thus a HDG—admits a Nash stable outcome, even when each agent approves of at most 4 fractions. In this work, we sharpen that result and take a step towards a computational complexity dichotomy with respect to the fixed number of approved fractions per agent: we show that the decision problem whether a DHDG admits a Nash stable outcome is $\textsf{NP}$-complete even in a restricted setting in which each agent approves of at most 2 fractions. More generally, we prove that for any fixed number $s\ge 2$ the corresponding decision problem is $\textsf{NP}$-complete even when each agent approves of exactly (or at most) s fractions.

Turning to stability against group deviations, it is known that—in contrast to HDGs—every DHDG has a non-empty core (this observation follows from a more general result for dichotomous hedonic games (see Aziz et al. [2] and also Peters [24])); Boehmer [8] then provides a polynomial time algorithm to find such an outcome. In this paper, we prove that in contrast to the above finding for the core, the strict core of a DHDG might be empty, even in instances with only three agents each of which approving of only one fraction. From a computational complexity perspective, we show that the corresponding decision problem of whether a DHDG admits a non-empty strict core is $\textsf{NP}$-complete even in a restricted setting with exactly (or at most) s approved fractions per agent, for any fixed number $s\ge 3$.

Hedonic games. HDGs are closely related to hedonic games (Drèze and Greenberg [17]), for which stability has been well-studied from different perspectives (see, e.g., Bogomolnaia and Jackson [10], Bloch and Diamantoudi [7], and Karakaya and Klaus [19]). In particular, the subclasses of anonymous hedonic games (see, e.g., Bogomolnaia and Jackson [10]) and fractional hedonic games (see, e.g., Aziz et al. [3]) have a vicinity to HDGs.

A hedonic game is a coalition formation game, in which each agent has preferences over the members of her coalition (for a survey, see, e.g., Aziz and Savani [1]). In anonymous hedonic games, the agents are only concerned with the size of their possible coalitions (and not with the identity of the members in their coalition). For different stability notions such as Nash, individual, and (strict) core stability, stable coalition formation in (anonymous) hedonic games has been well-studied from a computational complexity viewpoint, for instance by Bogomolnaia and Jackson [10], Ballester [4], Olsen [23], and Peters [24]. In fractional hedonic games, each agent links a certain value with each other agent; an agent’s value of her coalition is then given by the average value of the other agents’ values in the coalition. For fractional hedonic games, the computational complexity of finding stable outcomes or deciding whether stable outcomes exist has also been studied in several papers including the ones by Bilò et al. [6], Brandl et al. [12], and Aziz et al. [3]. We point out that a Bakers and Millers game can be understood as a special case of a fractional hedonic game (see also Bredereck et al. [13]); in a Bakers and Millers game, a finest partition in the strict core—which is guaranteed to be non-empty—can be determined in linear time (Aziz et al. [3]). Note that in contrast to the settings of an anonymous hedonic game and a fractional hedonic game, in the problem of a HDG considered in our paper we are concerned with two types of agents who have preferences over the possible fractions of agents of their own type.

Finally, we remark that positional scores from voting theory (see Brams and Fishburn [11] for a survey), originally designed to determine the winner(s) of an election, have been applied to several other settings in order to evaluate outcomes. Most prominently, approval and Borda scores have been used, for instance, in combinatorial optimization problems such as the traveling salesperson problem (Klamler and Pferschy [22]), in fair division problems (see, e.g., Baumeister et al. [5], Darmann and Schauer [16], and Kilgour and Vetschera [21]), and in the group activity selection problem (Darmann [14]).

The structure of this paper is as follows. In Section 2 we formally introduce the model of a (dichotomous) hedonic diversity game and the solution concepts considered. In Sections 3 and 4 we consider dichotomous hedonic diversity games: in Section 3 we present our results for Nash stability and strict core stability, and in Section 4 we focus on outcomes maximizing social welfare measured by means of the total number of approvals. In Section 5, we consider the problem of maximizing social welfare under the use of Borda scores in hedonic diversity games with strict preferences. A preliminary version of this paper appeared as [15].

2 Preliminaries

A hedonic diversity game $G=(R,B,(\succsim _{i})_{i\in R\cup B})$ consists of two disjoint sets R, B of agents¹—the agents in R are called red agents, the agents in B are called blue agents—and we set $N=R\cup B$. Each agent $i\in N$ specifies a weak order $\succsim _{i}$ (with indifference part $\sim _{i}$ and strict preference part $\succ _{i}$) over the set $\Theta$ of all fractions of red agents in some subset of N containing agent i. Hence, for a red agent we have $\Theta =\{\frac{r}{r+b}\mid r\in \{1,\ldots | R | \},b\in \{1,\ldots ,| B | \}\}\cup \{1\}$, and for a blue agent we have $\Theta =\{\frac{r}{r+b}\mid r\in \{1,\ldots |R| \},b\in \{1,\ldots ,| B | \}\}\cup \{0\}$; observe that the cardinality of $\Theta$ is the same for a red and a blue agent.

A subset $C\subseteq N$ is called coalition, and $\mathcal {C}_{i}$ denotes the set of all coalitions containing agent $i\in N$. We interpret $\succsim _{i}$ as the preferences of agent i over all possible fractions of red agents in some coalition containing her. For coalition C, we denote the fraction of red agents in C by $\theta _{R}(C)$. A coalition is mixed if it contains both blue and red agents, otherwise it is pure. A purely red (blue) coalition consists of red (blue) agents only. An outcome $\pi$ is a partition of $R\cup B$ into disjoint coalitions. For outcome $\pi$, let $\pi (i)$ denote the coalition containing agent i; conversely, we write $C\in \pi$ if $\pi (i)=C$ holds for some agent i. Abusing notation, for $C,D\in \mathcal {C}_{i}$ we have $C\succsim _{i}D$ iff $\theta _{R}(C)\succsim _{i}\theta _{R}(D)$ holds; we say that agent i strictly prefers coalition C over coalition D, $C\succ _{i}D$, iff $\theta _{R}(C)\succ _{i}\theta _{R}(D)$ holds.

In a dichotomous hedonic diversity game (DHDG) $G=(R,B,(A_{i})_{i\in R\cup B})$, each agent i specifies a set $A_{i}$ of approved fractions in $\Theta$; agent i is indifferent between all fractions in $A_{i}$ (i.e., for $\theta ,\bar{\theta }\in A_{i}$ we have $\theta \sim _{i}\bar{\theta }$), strictly prefers any $\theta \in A_{i}$ to any $\bar{\theta }\notin A_{i}$, and is indifferent between all fractions not contained in $A_{i}$.

2.1 Solution Concepts

We will consider two kinds of solution concepts: On the one hand, we take into account the game-theoretic notions of stability against individual and group deviations, where we focus on Nash stability and strict core stability; on the other hand, we apply approval scores and Borda scores from voting theory to our setting in order to determine outcomes that maximize (utilitarian) social welfare, i.e, the total sum of scores.

2.1.1 Stability Notions

Nash stable outcomes require that no agent can make herself better off by forming a singleton coalition or by deviating towards some other coalition. Formally, an outcome $\pi$ of a hedonic diversity game is Nash stable, if there is no agent i with $S\cup \{i\}\succ _{i}\pi (i)$ for some $S\in \pi \cup \{\emptyset \}$. In a DHDG, an outcome $\pi$ is hence Nash stable if there is no agent i with $\theta _{R}(\pi (i))\notin A_{i}$ but $\theta _{R}(S\cup \{i\})\in A_{i}$ for some $S\in \pi \cup \{\emptyset \}$.

Strictly core stable outcomes require that there is no group of agents S such that, by forming a deviating coalition, at least one member of S is better off while no member of S changes for the worse. This can be formalized as follows. A coalition $S\subseteq N$ weakly blocks an outcome $\pi$ of N if for every agent $i\in S$ we have $S\succsim _{i}\pi (i)$, and for some $i\in S$ we have $S\succ _{i}\pi (i)$. An outcome $\pi$ is said to be strictly core stable (or in the strict core) if there is no weakly blocking coalition for $\pi$.

2.1.2 Social Welfare

The score of outcome $\pi$ for agent i, $sc_{\pi }(i)$, is a non-negative integer assigned to $\pi (i)$. Under given scores, the social welfare of outcome $\pi$, $SW(\pi )$, is the sum of the scores over all agents: $SW(\pi )=\sum _{i\in N}sc_{\pi }(i)$. We consider the following two kinds of scores.

In a DHDG, the approval score of outcome $\pi$ for agent i is 1 if $\theta _{R}(\pi (i))\in A_{i}$ and 0 otherwise. In a DHDG, using approval scores the social welfare $SW(\pi )$ (or total approval score) of outcome $\pi$ is hence the number of agents $i\in N$ for which $\theta _{R}(\pi (i))\in A_{i}$ holds.

Given a hedonic diversity game $G=(R,B,(\succ _{i})_{i\in R\cup B})$ with strict preferences $\succ _{i}$ over the set $\Theta$, the Borda score of outcome $\pi$ for agent i is given by $sc_{\pi }(i)= | \{ \theta \in \Theta \mid \theta _{R}(\pi (i))\succ _{i}\theta \} |$.

Example 1

In the following instance of a DHDG with $R=\{r_1,r_2,r_3\}$ and $B=\{b_1,b_2,b_3,b_4\}$, each agent approves of exactly one fraction as listed below.

$$\begin{array}{ccccc} r_{1}: &{} \frac{1}{3} &{} ~~~~~ &{} b_{1}: &{} \frac{1}{3}\\ r_{2}: &{} \frac{1}{2} &{} ~~~~~ &{} b_{2}: &{} \frac{1}{3}\\ r_{3}: &{} \frac{3}{7} &{} ~~~~~ &{} b_{3}: &{} \frac{2}{3}\\ &{} &{} &{} b_{4}: &{} 0 \end{array}$$

Let outcome $\pi$ be given by the coalitions $\{r_1,b_1,b_2\}$, $\{r_2,r_3,b_3\}$ and $\{b_4\}$. $\pi$ is not Nash stable, because agent $r_2$ would be better off by joining coalition $\{b_4\}$: $\{r_2,b_4\}$ is of fraction $\frac{1}{1+1}=\frac{1}{2}$ which $r_2$ approves of whereas she does not approve of $\frac{2}{3}$, i.e., the fraction of coalition $\{r_2,r_3,b_3\}$ she is assigned to under $\pi$. On the other hand, $\pi$ is strictly core stable. The only agents that have an incentive to deviate are the agents $r_2$ and $r_3$. Observe that $r_2$ (and $r_3$ respectively) is the only agent approving of $\frac{1}{2}$ (resp. $\frac{3}{7}$), which would require a mixed coalition. However, each blue agent would be worse off in a coalition of fraction $\frac{1}{2}$ (resp. $\frac{3}{7}$).

Next, consider the outcome $\mu$ given by the coalitions $\{r_3,b_1,b_2\}$, $\{r_2,b_3\}$, $\{r_1\},\{b_4\}$. Outcome $\mu$ is Nash stable: each of the agents $r_2,b_1,b_2,b_4$ is in a coalition with fraction she approves of and hence has no incentive to deviate; agent $r_3$ does not approve of $\frac{2}{3}$, 1, or $\frac{1}{2}$, so she does not have an incentive to deviate either; likewise, $r_1$ (and $b_3$ respectively) does not approve of $\frac{1}{2}$ or $\frac{2}{3}$ (resp. $\frac{1}{4}$, $\frac{1}{2}$ or 0) and hence does not have an incentive to deviate. On the other hand, $\mu$ is not strictly core stable because the coalition $S=\{r_1,b_1,b_2\}$ weakly blocks $\mu$ since by forming the deviating coalition S agent $r_1$ is better off while $b_1$ and $b_2$ are not worse off when compared with the coalition assigned under $\mu$.

Finally, observe that the approval score of outcome $\mu$ is 4, while the approval score of $\pi$ is 5. In particular, $\pi$ is an outcome that maximizes the approval score.

3 DHDGs: Nash Stability and the Strict Core

In this section, we consider Nash stability and strict core stability in restricted settings with a fixed number of approvals per agent. In that context, we state that even in instances with a small number of approvals per agent such outcomes may fail to exist, and then turn to the decision problems whether a DHDG admits a Nash stable outcome and a strictly core stable outcome respectively.

3.1 Nash Stability in DHDGs

Nash stable outcomes may fail to exist in a DHDG even in small instances with only two agents as shown in [9]. It is straightforward to generalize their example to any number of agents greater than or equal to two; for the sake of completeness, however, we state this in the proposition below.

Proposition 1

For each $n\ge 2$, there is an instance of a dichotomous hedonic diversity game $G=(R,B,(A_{i})_{i\in R\cup B})$ with $|N|=n$ where each agent approves of exactly one fraction that does not admit a Nash stable outcome.

Proof

Let $R=\{r_1\}$ and $B=\{b_1,b_2,\ldots ,b_{n-1}\}$, where the red agent approves of fraction 1 only and each blue agent approves of fraction $\frac{1}{2}$.² Any outcome that contains a mixed coalition C is not Nash stable, since the red agent $r_1$, who must be a member of C, prefers forming the singleton coalition containing only herself to being engaged in C. However, any outcome with only pure coalitions is not Nash stable, because each of the blue agents would prefer to join the unique purely red coalition $\{r_1\}$. $\square$

Our first computational complexity result states that deciding whether a DHDG admits a Nash stable outcome is computationally intractable, even when restricted to instances with at most two approvals per agent and one type of agents approving of mixed coalitions only.

Theorem 1

The problem of deciding whether a dichotomous hedonic diversity game $G=(R,B,(A_{i})_{i\in R\cup B})$ admits a Nash stable outcome is $\textsf{NP}$-complete, even when (i) each agent approves of at most two fractions and (ii) none of the red agents approves of a purely red coalition.

Proof

We provide a reduction from scExact Cover by 3-Sets (scX3C). An instance of scX3C is a pair $(X,\mathcal{Y})$, where $X=\{1,\ldots ,3q\}$ and $\mathcal{Y}=\{Y_{1},\ldots ,Y_{p}\}$ is a collection of 3-element subsets (3-sets) of X; it is a “yes”-instance iff X can be covered by exactly q sets from $\mathcal{Y}$. We assume that every element of X appears in exactly three sets in $\mathcal{Y}$; scX3C is known to be $\textsf{NP}$-complete even under this restriction [18]. Observe that the restriction implies $p=3q$, which allows us to omit q. Given such a restricted instance of scX3C, we construct an instance $G=(R,B,(A_{i})_{i\in R\cup B})$ of a dichotomous hedonic diversity game as follows. We set $R=\{\hat{r}_{i,j}\mid i\in \{1,\ldots ,3p\}\setminus \{2p\},\text { }1\le j\le 3p+1\}\cup \{r_{k,t}\mid 1\le k\le p,1\le t\le 3\}$ and $B=\{b_{k,t}\mid 1\le k\le p,1\le t\le 3\}$. For $x_{k}\in X$ let $Y_{k_{1}},Y_{k_{2}},Y_{k_{3}}$ denote the three sets of $\mathcal {Y}$ that contain $x_{k}$. We identify $x_{k}\in X$ with the six agents $b_{k,t}$ and $r_{k,t}$, and the set $Y_{k_{t}}$ with the fraction $\frac{4+3k_{t}}{7+3k_{t}}$.

The agents’ approved fractions are as follows. For each k,

• blue agent $b_{k,t}$’s set of approved fractions is $\{0,\frac{4+3k_{t}}{7+3k_{t}}\}$, $t\in \{1,2,3\}$,

• red agent $r_{k,1}$’s set of approved fractions is $\{\frac{5+3k_{2}}{8+3k_{2}},\frac{5+3k_{3}}{8+3k_{3}}\}$,

• red agent $r_{k,2}$’s set of approved fractions is $\{\frac{5+3k_{1}}{8+3k_{1}},\frac{5+3k_{3}}{8+3k_{3}}\}$, and

• red agent $r_{k,3}$’s set of approved fractions is $\{\frac{5+3k_{1}}{8+3k_{1}},\frac{5+3k_{2}}{8+3k_{2}}\}$.

Finally,

• each red agent $\hat{r}_{i,j}$ approves of $\frac{1}{i+1}$ exclusively.

Observe that

• each fraction $\theta =\frac{4+3k_{i}}{7+3k_{i}}$—induced by the 3-set $Y_{k_{i}}$—is approved of by exactly three blue agents, since for each element $x_{k}\in Y_{k_{i}}$ exactly one blue agent approves of $\theta$;

• each fraction $\tilde{\theta }=\frac{5+3k_{i}}{8+3k_{i}}$ is approved of by exactly six red agents, because for each element $x_{k}\in Y_{k_{i}}$ exactly two red agents approve of $\tilde{\theta }$.

We show that $(X,\mathcal{Y})$ admits an exact cover by 3-sets from $\mathcal {Y}$ iff G has a Nash stable outcome.

Assume $(X,\mathcal{Y})$ admits an exact cover, say $Z\subset \mathcal {Y}$, by 3-sets from $\mathcal {Y}$. We derive the following partition of the agents in G:

• For each set $Y_{k}\in Z$ form a coalition made up of the three blue agents approving of $\frac{4+3k}{7+3k}$ together with the six red agents approving of $\frac{5+3k}{8+3k}$ plus exactly $(4+3k)-6$ arbitrarily chosen red agents $\hat{r}_{i,j}$. The remaining 2p blue agents form the purely blue coalition S. The remaining red agents form singleton coalitions each.

Observe that each blue agent approves of its coalition’s fraction, hence no such agent has an incentive to deviate. No red agent approves of 1 or $\frac{1}{2p+1}$, hence no red agent wants to deviate towards a purely red coalition or towards S. In addition, for any choice of $k,\ell \in \mathbb {N}$ we have $\frac{(4+3k)+1}{(7+3k)+1}\not =\frac{1}{\ell }$ because otherwise $\ell =\frac{8+3k}{5+3k}=1+\frac{3}{5+3k}$ in contradiction with $\ell ,k\in \mathbb {N}$. Therefore, no red agent $\hat{r}_{i,j}$ has an incentive to deviate towards a mixed coalition. Finally, for any coalition of fraction $\frac{4+3k}{7+3k}$, by construction the coalition contains all the six agents $r_{k,t}$ approving of $\frac{5+3k}{8+3k}$. Therefore, none of the agents $r_{k,t}$ has an incentive to deviate towards a mixed coalition either. Thus, the partition is Nash stable.

On the other hand, let $\pi$ be a Nash stable outcome. Let C be a mixed coalition in $\pi$. Coalition C must contain exactly the three blue agents approving of its fraction: C cannot contain a blue agent not approving of its fraction since she would otherwise wish to form a singleton coalition instead. Also, each mixed coalition requires at least three blue agents, and in case C contains more than three blue agents at least one of them wishes to form a singleton coalition instead because for each fraction $\theta =\frac{4+3k}{7+3k}$ there are exactly three blue agents approving of $\theta$.

Now, we show that $\pi$ cannot contain a purely blue coalition of size $s\not =2p$, $s\ge 1$. Assume the opposite and let S be such a purely blue coalition of size $s\not =2p$, $s\ge 1$. Observe that there are exactly $(3p+1)$ red agents $\hat{r}_{s,j}$ approving of $\frac{1}{s+1}$ for any choice of $s\not =2p$. Each agent $\hat{r}_{s,j}$ hence prefers $S\cup \{\hat{r}_{s,j}\}$ over its current coalition—and hence has an incentive to deviate—unless she is already in a coalition of fraction $\frac{1}{s+1}$. However, it is impossible that each such agent $\hat{r}_{s,j}$ is in a coalition of fraction $\frac{1}{s+1}$ since this would require $(3p+1)\cdot s>3p$ blue agents. Thus, $\pi$ cannot contain a purely blue coalition of size $s\not =2p$.

Hence, there must be at least p blue agents which are engaged in some mixed coalitions. Since each blue agent needs to approve of the fraction of the mixed coalition C she is part of, C must be of fraction $\theta =\frac{4+3k}{7+3k}$ for some k. Also, recall that fraction $\theta =\frac{4+3k}{7+3k}$ is induced by set $Y_{k}=\{x_{u},x_{v},x_{w}\}$ in $\mathcal {Y}$. Coalition C is thus made up of

• 1. exactly the three blue agents approving of $\theta$, i.e., three agents $b_{u,h_{u}},b_{v,h_{v}},b_{w,h_{w}}$ for some choices of $h_{u},h_{v},h_{w}\in \{1,2,3\}$, and

• 2. exactly $4+3k$ red agents including the six red agents who approve of $\frac{5+3k}{8+3k}$—say $r_{u,t_{u}},r_{u,\bar{t}_{u}}r_{v,t_{v}},r_{v,\bar{t}_{v}}r_{w,t_{w}},r_{w,\bar{t}_{w}}$ for some choices of $t_{u},\bar{t}_{u},t_{v},\bar{t}_{v},t_{w},\bar{t}_{w}$—since otherwise $\pi$ is not Nash stable.

Note that any two mixed coalitions must have different fractions since each mixed coalition must have exactly three blue agents, all of which approving of its fraction, and by construction for each fraction there are exactly three such agents.

In addition, observe that due to Point 2. above, for each u at most one of $b_{u,1},b_{u,2},b_{u,3}$ can be contained in a mixed coalition: the fact that $r_{u,t_{u}}$ and $r_{u,\bar{t}_{u}}$ are contained in mixed coalition C with $\theta =\frac{4+3k}{7+3k}$ implies that $r_{u,\tilde{t}_{u}}$ with $\tilde{t}_{u}\notin \{t_{u},\bar{t}_{u}\}$ who approves of $\frac{5+3\ell }{8+3\ell }$, $\ell \not =k,$cannot be contained in a mixed coalition $D\not =C$ with fraction $\frac{4+3\ell }{7+3\ell }$ because (i) one of $\{r_{u,t_{u}},r_{u,\bar{t}_{u}}\}$ approves of $\frac{5+3\ell }{8+3\ell }$, and (ii) D would need to contain all the six red agents approving of $\frac{5+3\ell }{8+3\ell }$.

Since at least p blue agents need to be engaged in some mixed coalition it follows that for each u exactly one of $b_{u,1},b_{u,2},b_{u,3}$ is contained in some mixed coalition. Due to 1. that coalition C has to contain also the two other blue agents approving of its fraction. As a consequence, the collection Z of sets $Y_{k}$ for which $\pi$ contains a coalition of size $\frac{4+3k}{7+3k}$ forms an exact cover by 3-sets in instance $(X,\mathcal{Y})$. $\square$

As we show below, a corresponding hardness result holds for instances in which each agent approves of exactly s fractions, for any choice of $s\ge 2$.

Proposition 2

For any fixed $s\ge 2$, $s\in \mathbb {N}$, the problem of deciding whether a dichotomous hedonic diversity game $G=(R,B,(A_{i})_{i\in R\cup B})$ admits a Nash stable outcome is $\textsf{NP}$-complete, even when (i) each agent approves of exactly s fractions and (ii) none of the red agents approves of a purely red coalition.

Proof

We assume $|R|,|B|>s$ since s is fixed. We adapt the proof of Theorem 1 by letting

• each agent $b_{k,t}$ also approve of $\frac{h}{h+|B|}$ for $3\le h \le s$,

• each agent $r_{k,t}$ also approve of $\frac{h}{h+|B|}$ for $4\le h \le s+1$,

• each agent $\hat{r}_{i,j}$ also approve of $\frac{1}{1+(|B|-h)}$ for $2\le h \le s$.

The “only if”-part follows analogously to the proof of Theorem 1.

For the “if” part, let $\pi$ be Nash stable. Recall that each blue agent approves of fraction 0. Hence Nash stability implies that each blue agent approves of her coalition’s fraction in $\pi$, because otherwise she would deviate towards forming a singleton coalition instead. Hence, each mixed coalition C in $\pi$ must be either of fraction $\frac{4+3k}{7+3k}$ for some k or of fraction $\frac{h}{h+|B|}$ for some $3\le h \le s$. Observe that the latter is impossible however: by $s<|B|=3p$ at least one of the red agents $r_{k,t}$ must be in a purely red coalition because all the blue agents are in C—hence, such an agent $r_{k,t}$ wants to join C. Recalling that $\frac{4+3k}{7+3k} \not = \frac{1}{\ell }$, for $k,\ell \in \mathbb {N}$, the proof then follows analogously to the proof of Theorem 1. $\square$

3.2 Strict Core Stability in DHDGs

Analogously to the case of Nash stability, in a DHDG also strict core stable outcomes may fail to exist. However, in instances with only two agents the strict core is guaranteed to be non-empty in any HDG and therefore also in any DHDG (Proposition 3). On the other hand, we show that as soon as a third agent (or more agents) emerges, the strict core may be empty (Proposition 4). In addition, we prove that—from a computational viewpoint—it is difficult to decide whether a given instance has a non-empty strict core, even when the number of approved fractions per agent is at most three and one type of agents approves of mixed coalitions only; we conclude the section by showing that hardness also holds for exactly s approved fractions, for any fixed number $s\ge 3$.

Proposition 3

Each instance of a hedonic diversity game $G=(R,B,(\succsim _{i})_{i\in R\cup B})$ with two agents has a non-empty strict core.

Proof

Let $\tilde{r}$ and $\tilde{b}$ denote the red and blue agent respectively. For agent $\tilde{r}$, the set of possible fractions is $\Theta =\{\frac{1}{2},1\}$, and for agent $\tilde{b}$ the set of possible fractions is $\Theta =\{0,\frac{1}{2}\}$. If both agents have $\frac{1}{2}$ top-ranked, then the outcome made up of the grand coalition $\{\tilde{r},\tilde{b}\}$ is strictly core stable. Otherwise, the outcome made up of the singleton coalitions $\{\tilde{r}\}$, $\{\tilde{b}\}$ is strictly core stable, because at least one of the agents strictly prefers her singleton coalition over the grand coalition. $\square$

Remark. Observe that the above proposition implies that for instances with two agents also the core of a HDG is always non-empty.

Proposition 4

For each $n\ge 3$, there is an instance of a dichotomous hedonic diversity game $G=(R,B,(A_{i})_{i\in R\cup B})$ with $|N|=n$ where each agent approves of exactly one fraction such that the strict core is empty.

Proof

Let $R=\{r_1,r_2\}$ and $B=\{b_1,b_2,\ldots ,b_{n-2}\}$, where the agents $r_1, r_2, b_1$ approve of fraction $\frac{1}{2}$, and the remaining agents approve of fraction 0. Consider an outcome $\pi$. Each of the blue agents approving of fraction 0 must be in a purely blue coalition, otherwise $\pi$ is not strictly core stable because a respective agent would prefer to form a singleton coalition instead. In addition, each of $r_1, r_2, b_1$ must be in a coalition of fraction $\frac{1}{2}$: for the sake of contradiction, assume the opposite, and let w.l.o.g. $r_1$ be in a coalition which has some other fraction; then, $S=\{r_1,b_1\}$ (which has fraction $\frac{1}{2}$) weakly blocks $\pi$, because $r_1$ is made better off and $b_1$ is not made worse off. However, each of $r_1, r_2, b_1$ being in a coalition of fraction $\frac{1}{2}$ requires at least one blue agent different from $b_1$, which is ruled out by the fact that each blue agent different from $b_1$ must be in a purely blue coalition. Therewith, $\pi$ is not strictly core stable. $\square$

We now turn to the decision problem whether a DHDG has a non-empty strict core and prove that this problem is computationally hard even in a restricted setting with at most three approved fractions per agent.

Theorem 2

The problem of deciding whether a dichotomous hedonic diversity game $G=(R,B,(A_{i})_{i\in R\cup B})$ admits a strictly core stable outcome is $\textsf{NP}$-complete, even when (i) each agent approves of at most three fractions and (ii) none of the blue agents approves of a purely blue coalition.

Proof

We provide a reduction from the restricted $\textsf{NP}$-complete version of scExact Cover by 3-scSets (scX3C) used in the proof of Theorem 1. Given such an instance $(X,\mathcal{Y})$ of scX3C, where $X=\{1,\ldots ,3q\}$ and $\mathcal{Y}=\{Y_{1},\ldots ,Y_{p}\}$ is a collection of 3-element subsets of X such that every element of X appears in exactly three sets in $\mathcal{Y}$, we construct an instance $G=(R,B,(A_{i})_{i\in R\cup B})$ of a dichotomous hedonic diversity game. Recall that we have $p=3q$. We set $R=\{r_{k}\mid 1\le k\le p\}\cup \{\hat{r}_{k,j}\mid 1\le k\le p,\text { }1\le j\le 3k-2\}$ and $B=\{b_{k}\mid 1\le k\le p\}$. For $x_{k}\in X$ let $Y_{k_{1}},Y_{k_{2}},Y_{k_{3}}$ denote the three sets of $\mathcal {Y}$ that contain $x_{k}$. We identify $x_{k}\in X$ with the agents $b_{k}$ and $r_{k}$, and we associate set $Y_{i}\in \mathcal {Y}$ with the fraction $\frac{1+3i}{4+3i}$. The agents’ approvals are as follows:

• for each k, agent $b_{k}$’s and agent $r_{k}$’s set of approved fractions is $\{\frac{1+3k_{t}}{4+3k_{t}}\mid 1\le t\le 3\}$, and

• for each k and j, agent $\hat{r}_{k,j}$’s set of approved fractions is $\{1,\frac{1+3k}{4+3k}\}$.

Observe that by construction each fraction $\frac{1+3i}{4+3i}$, $1\le i\le p$, is approved of by exactly three blue agents. We now show that $(X,\mathcal{Y})$ admits an exact cover by 3-sets from $\mathcal {Y}$ iff G admits a non-empty strict core.

Assume that in instance $(X,\mathcal{Y})$ there is an exact cover Z by 3-sets. We construct partition $\pi$ of N as follows. For each set $Y_{i}\in Z$ let coalition $C_{i}=\{b_{k},r_{k}\mid x_{k}\in Y_{i}\}\cup \{\hat{r}_{i,j}\mid 1\le j\le 3i-2\}$, and let each $\hat{r}_{k,j}$ with $Y_{k}\notin Z$ form a singleton coalition. Each of the agents in a singleton coalition approves of its fraction. Observe that $C_{i}$ contains exactly three blue agents and $(3+3i-2)$ red agents. The fraction of coalition $C_{i}$ is hence $\frac{1+3i}{4+3i}$ which, due to $x_{k}\in Y_{i}$, is approved of by all of its agents. Note that by the fact that Z is an exact cover each of the agents $r_{k},b_{k}$ is in exactly one mixed coalition. Therefore, each agent is engaged in some coalition and approves of its fraction. Thus, partition $\pi$ is strictly core stable.

On the other hand, assume that there is a strictly core stable outcome $\pi$. For the sake of contradiction, assume that at least one blue agent $b_{k}$ is in a coalition with a fraction she disapproves of. Let $Y_{i}\in \mathcal {Y}$ denote one of the three sets that contain element $x_{k}$. As above, form the coalition $C_{i}=\{b_{\ell },r_{\ell }\mid x_{\ell }\in Y_{i}\}\cup \{\hat{r}_{i,j}\mid 1\le j\le 3i-2\}$ with fraction $\frac{1+3i}{4+3i}$ which is approved of by all members of $C_{i}$. Since $b_{k}\in C_{i}$ holds we can conclude that $C_{i}$ weakly blocks $\pi$, in contradiction with the assumption that $\pi$ is strictly core stable. Therewith, each blue agent must be in a coalition with a fraction $\theta$ she approves of. Note that by construction (for each blue agent, the denominator of each approved fraction exceeds the numerator by three), this requires that all the three agents approving of $\theta$ must be in the same coalition. Thus, the set $Z=\{Y_{k}\in \mathcal {Y}\mid \exists S\in \pi \text { with }\theta _{R}(S)=\frac{1+3k}{4+3k}\}$ forms an exact cover by 3-sets in $(X,\mathcal {Y})$. $\square$

Proposition 5

For any fixed $s \ge 3$, $s \in \mathbb {N}$, the problem of deciding whether a dichotomous hedonic diversity game $G=(R,B,(A_{i})_{i\in R\cup B})$ admits a strictly core stable outcome is $\textsf{NP}$-complete, even when (i) each agent approves of exactly s fractions and (ii) none of the blue agents approves of a purely blue coalition.

Proof

We assume $|R|,|B|>s$ since s is fixed. We adapt the proof of Theorem 2 as follows:

• each agent $b_k$ and each agent $r_k$ also approves of $\frac{|R|}{|R|+4},\frac{|R|}{|R|+5},\ldots ,\frac{|R|}{|R|+s}$,

• each agent $\hat{r}_{k,j}$ also approves of $\frac{1}{1+|B|}$ and of $\frac{1}{1+(|B|-t)}$ for $4\le t \le s$.

The “only if”-part follows analogously to the proof of Theorem 2.

For the “if”-part, analogously to the proof of Theorem 2 it follows that in a strictly core stable outcome, each blue agent must be in a coalition with a fraction $\theta$ she approves of. Observe that an outcome that has a coalition of fraction $\frac{|R|}{|R|+t}$ for some $t\ge 4$ is not strictly core stable, since the agents $\hat{r}_{k,j}$ would deviate towards a purely red coalition. Hence $\theta$ must correspond to some $\frac{1+3k}{4+3k}$ for some $1\le k \le p$. Therewith, the proof follows analogously to the proof of Theorem 2. $\square$

4 Maximizing Social Welfare: a Dichotomy for DHDGs

Apart from stability notions, from a social choice perspective an outcome that maximizes social welfare is of interest. In this section, we consider DHDGs and use approval scores to measure the social welfare induced by an outcome. We first show that an outcome that maximizes social welfare, i.e., total approval score, can be found in polynomial time when each agent approves of exactly one fraction. However, we then prove that the corresponding decision problem turns $\textsf{NP}$-complete already as soon as agents may approve of two fractions. Therewith we draw the sharp separation line between polynomially solvable and $\textsf{NP}$-complete cases with respect to the fixed number of approved fractions per agent.

We introduce some additional notation. For set $N'\subseteq N$ of agents and fraction $\theta$, let $R_{\theta }(N')$ and $B_{\theta }(N')$ denote the set of red and blue agents in $N'$ approving of $\theta$ respectively. Let $\#r(N')$ and $\#b(N')$ denote the number of red and blue agents in set $N'$ respectively.

Theorem 3

In a dichotomous hedonic diversity game $G=(R,B,(A_{i})_{i\in R\cup B})$ with approval scores in which each agent approves of exactly one fraction, an outcome that maximizes social welfare can be found in polynomial time.

Proof

We will reduce a dichotomous hedonic diversity game with a single approval per agent to a two-constraint knapsack problem³. An instance of the two-constraint knapsack problem consists of a set J of items, where each item $j\in J$ is associated with a profit $p_{j}$, a weight $w_{j}$ and a volume $v_{j}$; the goal is to select a subset $J^{*}\subseteq J$ of items of maximum total profit $p^{*}=\sum _{j\in J^{*}}p_{j}$ such that the total weight does not exceed a given weight bound W and the total volume does not exceed a given volume bound V (i.e., $\sum _{j\in J^{*}}w_{j}\le W$ and $\sum _{j\in J^{*}}v_{j}\le V$). By dynamic programming, the maximum profit in an instance of the two-constraint knapsack problem can be determined in $\mathcal {O}(nWV)$ time, determining both the optimal profit and the profit maximizing set of items can be done in $\mathcal {O}(n^{2}WV)$ time (see Ch. 9.3.2 of [20]).

Given a dichotomous hedonic diversity game $G=(R,B,(A_{i})_{i\in R\cup B})$ with exactly one approval per agent, we construct an instance $\mathcal {I}$ of the two-constraint knapsack problem. W.l.o.g., we assume that the fractions in G cannot be reduced anymore, i.e., the numerator and denominator of each fraction $\theta$ in G are coprime.

For each fraction $\theta =\frac{r_{\theta }}{r_{\theta }+b_{\theta }}$ (with $0=\frac{0}{1}$ and $1=\frac{1}{1+0}$) approved of by at least one agent we first partition the set of agents approving of $\theta$ into sets $S_{\theta ^{(i)}}$ and introduce the items for instance $\mathcal {I}$ on basis of these sets.

In order to construct the sets, the idea is that while there are red or blue agents approving of $\theta$, add them to $S_{\theta ^{(1)}}$ as long as it contains less than $r_{\theta }$ red agents ($b_{\theta }$ blue agents); then continue with $S_{\theta ^{(2)}}$, etc. We proceed as follows:

• $q_{\theta }=\max \{c,d\mid c = \lceil * \rceil {\frac{|R_{\theta }(N)|}{r_{\theta }}},d = \lceil * \rceil {\frac{| B_{\theta }(N)|}{b_{\theta }}}\}$;

• for each fraction $\theta$ construct the sets $S_{\theta ^{(i)}}$, $1\le i\le q_{\theta }$, containing agents of $R_{\theta }(N)$ and $B_{\theta }(N)$ exclusively, such that

  • each such set contains at most $r_{\theta }$ red agents and $b_{\theta }$ blue agents, the set $S_{\theta ^{(1)}}$ is non-empty, and

  • for red agent $r\in R_{\theta }(N)$ we have $r\in S_{\theta ^{(i+1)}}$ iff $S_{\theta ^{(i)}}$ contains $r_{\theta }$ agents (for blue agent b of $B_{\theta }(N)$ we have $b\in S_{\theta ^{(i+1)}}$ iff $S_{\theta ^{(i)}}$ contains $b_{\theta }$ agents).

We say that set $S_{\theta ^{(i)}}$ is full, if it contains exactly $r_{\theta }$ red agents and $b_{\theta }$ blue agents.

Observe that each agent of N is contained in exactly one of the sets $S_{\theta ^{(i)}}$ (hence we have at most n such sets), and each agent in set $S_{\theta ^{(i)}}$ approves of $\theta$.

In order to construct instance $\mathcal {I}$ of the two-constraint knapsack problem, for each set $S_{\theta ^{(i)}}$ we introduce an item $\theta ^{(i)}$ with profit $p_{\theta ^{(i)}}=| S_{\theta ^{(i)}} |$, weight $r_{\theta }$ and volume $b_{\theta }$, and set $W=|R|$, $V=|B|$.

To illustrate the proof, we provide the following running example⁴ of a DHDG with exactly one approved fraction per agent. Let $G^{\text {ex}}$ be made up of a set N of 6 red agents $r_i$ and 7 blue agents $b_j$, with approvals as follows:

• agents $r_1,r_2,r_3$ and $b_1,b_2,b_3$ approve of fraction $\frac{2}{3}$;

• agents $r_4$ and $b_4,b_5,b_6,b_7$ approve of $\frac{1}{3}$;

• agent $r_{5}$ approves of $\frac{1}{2}$;

• agent $r_{6}$ approves of $\frac{1}{8}$.

Then, for fraction $\frac{2}{3}$ we get $|R_{\frac{2}{3}}(N)|=3$ and |$B_{\frac{2}{3}}(N)|=3$; with $r_\theta =2$ and $b_\theta =1$ it follows that $q_\theta =\max \{2,3\}=3$. As a consequence, we construct the sets

$$S_{\frac{2}{3}^{(1)}}=\{r_1,r_2,b_1\}, S_{\frac{2}{3}^{(2)}}=\{r_3,b_2\} {\text { and }} S_{\frac{2}{3}^{(3)}}=\{b_3\}.$$

Analogously, we build the sets

$$S_{\frac{1}{3}^{(1)}}=\{r_4,b_4,b_5\}, S_{\frac{1}{3}^{(2)}}=\{b_6,b_7\}, S_{\frac{1}{2}^{(1)}}=\{r_{5}\} {\text { and }} S_{\frac{1}{8}^{(1)}}=\{r_{6}\}.$$

Observe that in our example, the full sets are the sets $S_{\frac{2}{3}^{(1)}}$ and $S_{\frac{1}{3}^{(1)}}$. From $G^{\text {ex}}$ we construct instance $\mathcal {I}^{\text {ex}}$ of the two-constraint knapsack problem by setting $W=|R|=6$, $V=|B|=7$ and introducing the following items:

• item $\frac{2}{3}^{(1)}$ with profit $|S_{\frac{2}{3}^{(1)}}|=3$, weight 2 and volume 1,

• item $\frac{2}{3}^{(2)}$ with profit $|S_{\frac{2}{3}^{(2)}}|=2$, weight 2 and volume 1,

• item $\frac{2}{3}^{(3)}$ with profit $|S_{\frac{2}{3}^{(3)}}|=1$, weight 2 and volume 1,

• item $\frac{1}{3}^{(1)}$ with profit 3, weight 1 and volume 2,

• item $\frac{1}{3}^{(2)}$ with profit 2, weight 1 and volume 2,

• item $\frac{1}{2}^{(1)}$ with profit 1, weight 1 and volume 1,

• item $\frac{1}{8}^{(1)}$ with profit 1, weight 1 and volume 7.

We now proof the theorem by showing that there is solution of $\mathcal {I}$ with profit $\ge p^{*}$ iff there is an outcome $\pi$ for G with $SW\ge p^{*}$.

“$\Rightarrow$”: A solution (= set of items) $J^{*}$ of total profit $\ell$ in $\mathcal {I}$ induces an outcome of social welfare $\ge \ell$ in G as follows. Consider the set $S^{*}=\{S_{\theta ^{(i)}}\mid \theta ^{(i)}\in J^{*}\}$, i.e., $S^{*}$ is the set of sets $S_{\theta ^{(i)}}$ corresponding to the items in $J^{*}$. We construct outcome $\pi$ of our DHDG in two steps.

First, for all sets $S_{\theta ^{(i)}}\in S^{*}$ which are full, define the coalition $C_{\theta ^{(i)}}=S_{\theta ^{(i)}}$. Next, for a non-full set $S_{\theta ^{(i)}}\in S^{*}$, the set contains $| R\cap S_{\theta ^{(i)}}| <r_{\theta }$ red agents or $| B\cap S_{\theta ^{(i)}} | <b_{\theta }$ blue agents; however, the weight and volume of the corresponding item $\theta ^{(i)}$ are $r_{\theta }$ and $b_{\theta }$, respectively. Hence, for each non-full set $S_{\theta ^{(i)}}\in S^{*}$ the weight contribution of $\theta ^{(i)}$ exceeds the number of red agents in $S_{\theta ^{(i)}}$ by $r_{\theta }-| R\cap S_{\theta ^{(i)}} |$ and the volume contribution of $\theta ^{(i)}$ exceeds the number of blue agents in $S_{\theta ^{(i)}}$ by $b_{\theta }-|B\cap S_{\theta ^{(i)}}|$. Together with the choice of $W=|R|$ (and $V=|B|$ respectively) it follows that there must be at least

$$\begin{aligned} \sum _{S_{\theta ^{(i)}}\in S^{*}:\,|S_{\theta ^{(i)}}|<r_{\theta }+b_{\theta }}r_{\theta }-|R\cap S_{\theta ^{(i)}}| \end{aligned}$$

red agents and at least

$$\begin{aligned} \sum _{S_{\theta ^{(i)}}\in S^{*}:\,|S_{\theta ^{(i)}}|<r_{\theta }+b_{\theta }}b_{\theta }-|B\cap S_{\theta ^{(i)}}| \end{aligned}$$

blue agents in N that are not contained in some set of $S^{*}$. Therefore, for all sets $S_{\theta ^{(i)}}\in S^{*}$ which are not full we are able to construct a coalition $C_{\theta ^{(i)}}$ of fraction $\theta$ by “filling up” $S_{\theta ^{(i)}}$ with such red and blue agents—i.e., create $C_{\theta ^{(i)}}$ by adding to $S_{\theta ^{(i)}}$ red and blue agents of $N\setminus \bigcup _{S_{\theta ^{(i)}}\in S^{*}}S_{\theta ^{(i)}}$ until it contains exactly $r_{\theta }$ red and $b_{\theta }$ blue agents.

Now let $\pi$ be the outcome made up of the coalitions $C_{\theta ^{(i)}}$ for $S_{\theta ^{(i)}}\in S^{*}$ plus coalition D containing all remaining agents. Observe that for each $C_{\theta ^{(i)}}\in \pi$ at least $|S_{\theta ^{(i)}}|$ agents in $C_{\theta ^{(i)}}$ approve of its fraction $\theta$. In addition, recall that by definition $p_{\theta ^{(i)}}=|S_{\theta ^{(i)}}|$. Thus, for outcome $\pi$ we have $SW(\pi )\ge \sum _{S_{\theta ^{(i)}}\in S^{*}}|S_{\theta ^{(i)}}|=\sum _{\theta ^{(i)}\in J^{*}}p_{\theta ^{(i)}}=\ell$.

In our running example, let $J^*=\{\frac{2}{3}^{(1)},\frac{2}{3}^{(2)},\frac{1}{3}^{(2)},\frac{1}{2}^{(1)}\}$. Solution $J^*$ meets weight bound $W=6$ and volume bound $V=7$ (total weight and volume of its items are 6 and 5, respectively), and yields a profit of 8. In order to construct outcome $\pi$ of our DHDG we derive $S^{*}=\{S_{\frac{2}{3}^{(1)}},S_{\frac{2}{3}^{(2)}}, S_{\frac{1}{3}^{(2)}},S_{\frac{1}{2}^{(1)}}\}$. Since $S_{\frac{2}{3}^{(1)}}$ is full, we get $C_{\frac{2}{3}^{(1)}}=S_{\frac{2}{3}^{(1)}}=\{r_1,r_2,b_1\}$. The remaining sets of $S^{*}$ are not full; to achieve fraction $\theta$ associated with $S_{\theta ^{(i)}}$, we arbitrarily “fill up” these sets with agents not contained in a set in $S^{*}$—i.e., with agents $r_4,r_6,b_3,b_4,b_5$. E.g., we get $C_{\frac{2}{3}^{(2)}}=\{r_3,r_4,b_2\}$, $C_{\frac{1}{3}^{(2)}}=\{r_6,b_6,b_7\}$, and $C_{\frac{1}{2}^{(1)}}=\{r_5,b_3\}$. Finally, $D=\{b_4,b_5\}$. It follows that $SW(\pi )=3+2+2+1=8$.

“$\Leftarrow$”: Assume there is an outcome $\pi$ of $SW(\pi )=\ell \ge p^{*}$. W.l.o.g. we assume that each coalition in $\pi$ cannot be split into smaller coalitions of the same fraction, i.e., for each coalition C and $\tilde{C}\subset C$ it holds that $\theta _{R}(C)\not =\theta _{R}(\tilde{C})$.

Let $\mathcal {S}$ be the set of coalitions $C\in \pi$ in which all agents approve of its fraction $\theta _{R}(C)$, and let $\mathcal {S}'$ be the set of coalitions $C'$ for which at least one agent disapproves of $\theta _{R}(C')$. Let $N'$ be the set of agents engaged in some coalition $C'\in \mathcal {S}'$, and let $N'_{a}\subseteq N'$ be the set of agents of $N'$ who approve of its coalition fraction, and $N'_{d}\subseteq N'$ be the set of agents of $N'$ who disapprove of its coalition fraction. From $\pi$ we construct a new partition $\pi '$ by regrouping agents of $N'$ as follows:

• for all $\theta '$ approved of by an agent in $N'_{a}$, build the sets $S'_{\theta '^{(i)}}$ from agents in $N'_{a}$ analogously to building the sets $S_{\theta ^{(i)}}$ in the construction of instance $\mathcal {I}$;

• fill up the sets $C'=S'_{\theta '^{(i)}}$ with agents in $N'_{d}$ as follows (i.e., add exactly $(r_{\theta '}-|R_{\theta '}(N')\cap S'_{\theta '^{(i)}}|)$ red and $(b_{\theta '}-|B_{\theta '}(N')\cap S'_{\theta '^{(i)}}|)$ blue agents from $N'_{d}$ to $S'_{\theta '^{(i)}}$):

  • as long as there is a set $C'=S'_{\theta '^{(i)}}$ with $\#r(C')<r_{\theta '}$ (resp. $\#b(C')<b_{\theta '}$), and a red (resp. blue) agent from $N'_{d}$ who approves of $\theta '$, add that agent to the set $S'_{\theta ',j}$ with the smallest index j among such sets;

  • after that, as long as there is a set $C'=S'_{\theta '^{(i)}}$ with $\#r(C')<r_{\theta '}$ (resp. $\#b(C')<b_{\theta '}$), add an arbitrary red (resp. blue) agent from $N'_{d}$ to $C'$;

• the remaining agents of $N_{d}'$ form coalition D.

Observe that compared with $\pi$, for each $\theta '=\theta _{R}(C')$ such that $C'\in \mathcal {S}'$ the number of sets $S'_{\theta '^{(i)}}$ does not exceed the number of coalitions of fraction $\theta '$ in $\pi$. Hence, the number of agents from $N'_{d}$ added to the sets $S'_{\theta '^{(i)}}$ in order to achieve the required fraction $\theta '$ is in fact sufficient since $\pi$ is a feasible partition.

As a consequence, for each $\theta$ such there is coalition C in partition $\pi$ with $\theta _{R}(C)=\theta$, there are at most as many coalitions of fraction $\theta$ in $\pi '$ as in $\pi$. Also, observe that an agent who approves of her coalition’s fraction in $\pi$ also approves of her coalition’s fraction in $\pi '$. Thus, the number of agents engaged in some coalitions in $\pi '\setminus D$ who approve of their coalition’s fraction is at least $\ell$. However, by construction of $\pi '$, there is a one-to-one correspondence between the set of coalitions of $\pi '\setminus D$ and a subset $J'$ of items in $\mathcal {I}$. Therewith, for each coalition $S\in \pi ',\,S\not =D$, there must be an item in instance $\mathcal {I}$ with the profit corresponding to the number of agents approving of $\theta _{R}(S)$. Also, by the choice of the items’ weights and volume, $J'$ is a feasible solution for instance $\mathcal {I}$. Therewith, $\mathcal {I}$ admits solution $J'$ with profit $\ge \ell$.

In the running example, consider outcome $\pi$ given by coalitions $\{r_1,r_2,b_1\}$, $\{r_4,r_5,b_2\}$, $\{r_3,b_4,b_5\}$, $\{r_6,b_6,b_7\}$, $\{b_3\}$ with $SW(\pi )=3+1+2+2=8$. Thus, $\mathcal {S}=\{\{r_1,r_2,b_1\}\}$, and $\mathcal {S}'$ consists of all other coalitions in $\pi$. Hence, the set $N'$ of agents engaged in some coalition in $\mathcal {S}'$ is $N'=\{r_3,r_4,r_5,r_6\}\cup \{b_2,b_3,\ldots ,b_7\}$. In addition, we have $N'_a=\{b_2,b_4,b_5,b_6,b_7\}$ and $N'_d=\{r_3,r_4,r_5,r_6,b_3\}$.

We hence build the sets $S'_{\theta '^{(i)}}$ from agents in $N'_{a}$ for $\theta ' \in \{\frac{2}{3},\frac{1}{3}\}$, as these are the only fractions approved of by some agent in $N'_{a}$. We get $S'_{\frac{2}{3}^{(1)}}=\{b_2\}$, $S'_{\frac{1}{3}^{(1)}}=\{b_4,b_5\}$, and $S'_{\frac{1}{3}^{(2)}}=\{b_6,b_7\}$. In the next step, we “fill up” these sets with agents from $N'_{d}$. First, we add agents who approve of the respective fractions: we add $r_3$ to $S'_{\frac{2}{3}^{(1)}}$ and $r_4$ to $S'_{\frac{1}{3}^{(1)}}$. Then, we add the remaining agents to the sets to achieve the according coalition fractions, i.e., we get $S'_{\frac{2}{3}^{(1)}}=\{r_3,r_5,b_2\}$, $S'_{\frac{1}{3}^{(1)}}=\{r_4,b_4,b_5\}$, and $S'_{\frac{1}{3}^{(2)}}=\{r_6,b_6,b_7\}$. Finally, we build $D=\{b_3\}$. As a consequence, $\pi '$ consists of the coalitions $\{r_1,r_2,b_1\}$, $\{r_3,r_5,b_2\}$, $\{r_4,b_4,b_5\}$, $\{r_6,b_6,b_7\}$ and $D=\{b_3\}$. Observe that $SW(\pi ')=3+2+3+2=10$.

From $\pi '\setminus D$ we derive a solution $J'$ of our instance of the two-constraint knapsack problem: for $\{r_1,r_2,b_1\} \in \pi '\setminus D$, where all of its 3 members approve of its fraction $\frac{2}{3}$, we include in $J'$ item $\frac{2}{3}^{(1)}$ of profit 3; for $\{r_3,r_5,b_2\} \in \pi '\setminus D$, where 2 of its members approve of its fraction $\frac{2}{3}$, we include in $J'$ item $\frac{2}{3}^{(2)}$ of profit 2, etc.; as a result, $J'=\{\frac{2}{3}^{(1)}, \frac{2}{3}^{(2)}, \frac{1}{3}^{(1)}, \frac{1}{3}^{(2)}\}$. Note that $J'$ satisfies the weight and volume constraints and yields a profit of 10.

Finally, observe that the optimal profit in two-constraint knapsack problem can be determined in $\mathcal {O}(nWV)=\mathcal {O}(n^{3})$ time, together with backtracking of the solution this can be done in $\mathcal {O}(n^{4})$ time. $\square$

On the negative side, as soon as agents approve of up to two fractions the problem of deciding whether a DHDG with approval scores admits an outcome with social welfare exceeding some given integer becomes computationally difficult.

Theorem 4

Given integer $\ell$, the problem of deciding whether a dichotomous hedonic diversity game $G=(R,B,(A_{i})_{i\in R\cup B})$ with approval scores admits an outcome with $SW\ge \ell$ is $\textsf{NP}$-complete, even when (i) each agent approves of at most two fractions and (ii) each blue agent approves of only one fraction.

Proof

Again we reduce from the $\textsf{NP}$-complete variant of scExact Cover by 3-Sets (scX3C) restricted to instances $(X,\mathcal{Y})$ with $X=\{1,\ldots ,3q\}$ and $\mathcal{Y}=\{Y_{1},\ldots ,Y_{p}\}$ such that every element of X appears in exactly three sets in $\mathcal{Y}$. Let $\mathcal {I}$ be such a restricted instance of scX3C, and recall that $p=3q$ holds. We construct an instance $G=(R,B,(A_{i})_{i\in R\cup B})$ of a dichotomous hedonic diversity game as follows. We set $R=\{\hat{r}_{i}\mid i\in \{1,\ldots ,\frac{p}{3}(p^{2}+p-6)\}\}\cup \{r_{k,t}\mid 1\le k\le p,\,\,1\le t\le 3\}$, and $B=\{b_{k,t}\mid 1\le k\le p,\,\,1\le t\le 3\}$. For $x_{k}\in X$ the three sets containing $x_{k}$ are denoted by $Y_{k_{1}},Y_{k_{2}},Y_{k_{3}}$. We identify $x_k \in X$ with the six agents $b_{k,t}, r_{k,t}$, and associate set $Y_{k_t} \in \mathcal {Y}$ with the fraction $\frac{p^{2}+k_{t}}{p^{2}+k_{t}+3}$. The agents’ approvals are given by:

• blue agent $b_{k,t}$’s approved fraction is $\frac{p^{2}+k_{t}}{p^{2}+k_{t}+3}$, $t\in \{1,2,3\}$,

• red agent $r_{k,1}$’s set of approved fractions is $\{\frac{p^{2}+k_{2}}{p^{2}+k_{2}+3},\frac{p^{2}+k_{3}}{p^{2}+k_{3}+3}\}$,

• red agent $r_{k,2}$’s set of approved fractions is $\{\frac{p^{2}+k_{1}}{p^{2}+k_{1}+3},\frac{p^{2}+k_{3}}{p^{2}+k_{3}+3}\}$,

• red agent $r_{k,3}$’s set of approved fractions is $\{\frac{p^{2}+k_{1}}{p^{2}+k_{1}+3},\frac{p^{2}+k_{2}}{p^{2}+k_{2}+3}\}$, and

• each red agent $\hat{r}_{i}$ approves of $\frac{1}{1+3p}$ exclusively.

Observe that each fraction $\theta =\frac{p^{2}+k_{t}}{p^{2}+k_{t}+3}$—induced by set $Y_{k_{t}}$—is approved of by exactly three blue and six red agents. We now argue that $(X,\mathcal{Y})$ admits an exact cover by 3-sets from $\mathcal {Y}$ iff G admits an outcome $\pi$ with $SW(\pi )\ge 3p$.

Assume there is an exact cover Z. Recall that Z contains exactly $\frac{p}{3}$ sets of $\mathcal {Y}$. Observe that for any $x_{k}\in X$ exactly one of $Y_{k_{1}},Y_{k_{2},}Y_{k_{3}}$ is in Z. To construct partition $\pi$ in G,

• for each set $Y_{k_{t}}\in Z$, form a coalition C made up of the three blue and six red agents approving of $\theta =\frac{p^{2}+k_{t}}{p^{2}+k_{t}+3}$ together with $(p^{2}+k_{t}-6)$ arbitrarily chosen agents of $\hat{r}_{i}$;

• the remaining agents form singleton coalitions each.

Due to the fact that Z is an exact cover by 3-sets partition $\pi$ is well-defined. Since we are concerned with exactly $\frac{p}{3}$ mixed coalitions, the number of agents of $\hat{r}_{i}$ engaged in a mixed coalition is at most $\frac{p}{3}(p^{2}+p-6)$, and hence partition $\pi$ is feasible. Each mixed coalition contains 9 agents who approve of its fraction, which yields a total social welfare of $SW(\pi )=3p$.

On the other hand, assume there is an outcome $\pi$ with $SW(\pi )\ge 3p$ in G. No agent approves of being in a pure coalition, so only mixed coalitions can contribute a positive value to the social welfare of $\pi$. In order to do so, a mixed coalition C must be either of fraction $\frac{1}{1+3p}$ or of fraction $\frac{p^{2}+j}{p^{2}+j+3}$ for some $1\le j\le p$. In the former case C contains all blue agents and hence must be the only mixed coalition, implying $SW(\pi )\le 1$ in contradiction with our assumption. In the latter case, C requires at least $(p^{2}+j)$ red agents. Given that G contains exactly $\frac{p}{3}(p^{2}+p-6)+3p=\frac{p}{3}(p^{2}+p+3)$ red agents, at most $\frac{p}{3}$ such coalitions can exist, because otherwise at least $(\frac{p}{3}+1)(p^{2}+1)=\frac{p}{3}(p^{2}+3p+1)+1$ red agents would be required. Due to the fact that any fraction $\frac{p^{2}+j}{p^{2}+j+3}$ is approved of by exactly 9 agents, this means that we must have exactly $\frac{p}{3}$ such coalitions and each of them must contain all the 9 agents approving of its fraction. For each k this means that for at most one—and by the fact that we have $\frac{p}{3}$ such coalitions this means for exactly one—$t\in \{1,2,3\}$ there is a coalition of fraction $\frac{p^{2}+k_{t}}{p^{2}+k_{t}+3}$ in $\pi$; hence, for each k exactly one of $b_{k,1},b_{k,2},b_{k,3}$ is engaged in such a coalition, and that coalition contains all the three blue agents approving of its fraction. Therewith, the collection of sets $Z=\{Y_{k_{t}}\mid \exists C\in \pi :\theta _{R}(C)=\frac{p^{2}+k_{t}}{p^{2}+k_{t}+3}\}$ forms an exact cover by 3-sets in $\mathcal {I}$. $\square$

In fact, the above hardness result translates to any fixed number $s\ge 2$ of fractions approved per agent.

Proposition 6

For any fixed $s\ge 2$, $s \in \mathbb {N}$, the problem of deciding whether a dichotomous hedonic diversity game $G=(R,B,(A_{i})_{i\in R\cup B})$ with approval scores admits an outcome with $SW\ge \ell$ for some given integer $\ell$ is $\textsf{NP}$-complete, even when each agent approves of exactly s fractions.

Proof

Note that we can assume that $|R|,|B|,p>s$ hold due to the fact that s is fixed. The proof follows by adapting the proof of Theorem 4 as follows:

• each blue agent also approves of fractions $\frac{|R|}{|R|+1}, \frac{|R|}{|R|+2}, \ldots , \frac{|R|}{|R|+(s-1)}$;

• if $s>2$ each red agent also approves of fractions $\frac{2}{1+|B|},\frac{3}{2+|B|},\ldots ,\frac{s-1}{s-2+|B|}$;

• in addition, agents $\hat{r}_i$ also approve of fraction $\frac{1}{|B|}$.

The “if”-part follows analogously to the proof of Theorem 4. For the “only if”-part, observe the following:

• Any outcome $\mu$ containing a coalition F of fraction $\frac{|R|}{|R|+g}$ for some $1\le g\le s-1$ contains g blue agents and all red agents, and hence any other coalition of $\mu$ must be purely blue. Each of the blue agents in F approves of its fraction, but no red agent does. Thus, such an outcome achieves a social welfare of at most $s-1<3p$ since no agent approves of the fraction of a pure coalition.

• Any outcome $\mu$ containing a coalition D of fraction $\frac{h}{h+|B|-1}$ for some $1\le h\le s-1$ contains h red agents and all but one of the blue agents. Each of the red agents in D may approve of its fraction, but no blue agent does. Besides, each of the agents not in D is either in a purely red coalition or in a coalition E containing exactly one blue agent. There is no agent approving of a pure coalition. As $\mu$ cannot contain a coalition of fraction $\frac{|R|}{|R|+1}$ due to the fact that at least one red agent is engaged in D, there is no agent approving of the fraction of coalition E either. Hence $\mu$ yields a social welfare of at most $s-1<3p$.

With these observations, the “only if”-part follows analogously to the proof of Theorem 4. $\square$

5 Maximizing Social Welfare in HDGs Under Borda Scores

We now leave the setting of DHDGs and consider the case in which each agent’s preferences are given by means of a strict order over the possible coalition fractions. It turns out that in such a scenario under the use of Borda scores maximizing social welfare is computationally hard.

Theorem 5

Given integer $\ell$, the problem of deciding whether a hedonic diversity game $G=(R,B,(\succ _{i})_{i\in R\cup B})$, with strict order $\succ _{i}$ over $\Theta$ for $i\in N$, under Borda scores admits an outcome with $SW\ge \ell$ is $\textsf{NP}$-complete.

Proof

We reduce from the $\textsf{NP}$-complete variant of scExact Cover by 3-Sets (scX3C) restricted to instances $(X,\mathcal{Y})$ with $X=\{1,\ldots ,3q\}$ and $\mathcal{Y}=\{Y_{1},\ldots ,Y_{p}\}$ such that every element of X appears in exactly three sets in $\mathcal{Y}$. Let $\mathcal {I}$ be such a restricted instance of scX3C (recall that $p=3q$ holds). From $\mathcal {I}$ we derive instance $G=(R,B,(\succ _{i})_{i\in R\cup B})$ of a hedonic diversity game as described below. The set of agents is made up of the sets $R=\{r_{i}\mid i\in \{1,\ldots ,p^{5}\}\}$ and $B=\{b_{k}\mid 1\le k\le p\}$. Again, for $x_{k}\in X$ we denote the three sets containing $x_{k}$ by $Y_{k_{1}},Y_{k_{2}},Y_{k_{3}}$. Agent $b_{k}\in B$ represents element $x_{k}\in X$, and we associate fraction $\frac{j+3}{j+6}$ with set $Y_{j}\in \mathcal {Y}$. The agents’ rankings—-up to the respective position where fraction $\frac{1}{|R|+|B|}$ is ranked—are given in Table 3. Let $T=|\Theta |-1$, i.e., T is the maximum possible Borda score for a single agent.

Table 3 Rankings of agents $b_{k}$ and $r_{i}$ up to fraction $\frac{1}{|R|+|B|}$ (used in the proof of Thm. 5)

We claim that $\mathcal {I}$ is a “yes”-instance of scX3C iff G admits an outcome $\pi$ with total Borda score $SW(\pi )\ge \ell =(T-2)p+(T-p)p^{5}$.

“$\Rightarrow$”: Let Z be an exact cover by 3-sets in instance $\mathcal {I}$. Consider partition $\pi$ which

• for each $Y_{j}\in Z$ forms a coalition $C_{j}$ made up of the three blue agents who have $\frac{j+3}{j+6}$ among their top 3 ranked fractions together with $(j+3)$ arbitrarily chosen red agents,

• and assigns the remaining agents (who, by the fact that Z is an exact cover by 3-sets, must all be red agents) to the single coalition D.

By the fact that Z is an exact cover by 3-sets each blue agent is in a coalition with a fraction she ranks first, second, or third. Each red agent is in a coalition of fraction 1 or $\frac{j+3}{j+6}$ for some $1\le j\le p$. Thus, we have $SW(\pi )\ge (T-2)p+(T-p)p^{5} =\ell$.

“$\Leftarrow$”: Let $\pi$ be an outcome with $SW(\pi )\ge \ell$. Note that any outcome in which all blue agents are engaged in the same coalition yields a total Borda score of at most $(T-3)p+(T-p-1)p^{5}<\ell$. Hence, any outcome meeting the desired bound splits the set of blue agents into at least two coalitions.

Assume there is a blue agent $b_{k}$ who is not in a coalition with fraction ranked among her top 3 fractions. Since the blue agents are not in a single coalition, the maximum possible Borda score for that agent is $sc_{\pi }(i)=T-2-|R|-1=T-3-p^{5}$. For the remaining $p-1$ blue agents the maximum Borda score is T. Next, observe that any coalition’s fraction among the first p ranked fractions of agents $r_{i}$ corresponds to $\frac{k+3}{k+6}$ for some $1\le k\le p$, and hence the number of blue agents required in such a coalition is a multiple of 3. Since there are only p blue agents in total, there are less than $\frac{p}{3}(p+3)$ red agents involved in coalitions of fraction $\frac{k+3}{k+6}$ for some k. Thus the largest possible total Borda score contributed by all red agents is bounded by $\frac{p}{3}(p+3)T+(p^{5}-\frac{p}{3}(p+3))(T-p)$. Thus,

$$\begin{aligned} SW(\pi )\le & {} T-3-p^{5}+T(p-1)+\frac{p}{3}(p+3)T+(p^{5}-\frac{p}{3}(p+3))(T-p)\\= & {} T(p+\frac{p}{3}(p+3)+p^{5}-\frac{p}{3}(p+3))-p^{6}+\frac{p^{3}}{3}+p^{2}-3-p^{5}\\= & {} T(p^{5}+p)-p^{6}-p^{5}+\frac{p^{3}}{3}+p^{2}-3\\< & {} T(p^{5}+p)-p^{6}-2p\\= & {} \ell \end{aligned}$$

in contradiction with our assumption.

As a consequence, each blue agent’s coalition must have a fraction ranked among her top 3 fractions. Since each such fraction is among the top 3 fractions of exactly three blue agents, each respective coalition requires exactly 3 blue agents (because such a coalition requires a multiple of 3 agents). Therewith, the collection of sets $Z=\{Y_{j}\mid \exists C\in \pi :\theta _{R}(C)=\frac{j+3}{j+6}\}$ forms an exact cover by 3-sets in $\mathcal {I}$. $\square$

6 Conclusion

A hedonic diversity game can be understood both as a game-theoretic problem and a social choice problem; accordingly, we have considered two kinds of solution concepts: stability notions that originate from game theory on the one hand, and social welfare which stems from social choice theory on the other hand. With respect to the latter, we have taken into account the two most prominent types of scores from voting theory, namely approval scores and Borda scores.

Besides several computational complexity results, we have also shown that—in contrast to the core, which is known to be always non-empty (see [2, 24])—the strict core of a dichotomous hedonic diversity game may be empty even in instances with a small number of agents and only one approved fraction per agent. Some interesting questions, however, remain open. For instance, what is the computational complexity of deciding whether a dichotomous hedonic diversity game admits a Nash stable outcome in the case of exactly one approved fraction per agent? How hard is it to decide whether a dichotomous hedonic diversity game admits a strictly core stable outcome when each agent approves of exactly one fraction or at most two fractions?

An interesting direction for future research would also be to study the computational complexity of the considered problems when, instead of the number of approved fractions, the number of disapproved fractions per agent is fixed. E.g., can we derive a similar dichotomy for the task of maximizing social welfare when each agent approves of all but a fixed number of fractions?

More generally, a possible future research direction could be to study which (additional) plausible domain restrictions allow for positive results related to computing stable outcomes or outcomes that maximize social welfare.