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Agreeable sets with matroidal constraints

  • Laurent Gourvès
Article

Abstract

This article deals with the challenge of reaching an agreement for a group of agents who have heterogeneous preferences over a set of goods. In a recent work, Suksompong (in: Subbarao (ed) Proceedings of the twenty-fifth international joint conference on artificial intelligence, IJCAI 2016, New York, pp 489–495, 2016) models a problem of this kind as the search of an agreeable subset of a given ground set of goods. A subset is agreeable if it is weakly preferred to its complement by every agent of the group. Under natural assumptions on the agents’ preferences such as monotonicity or responsiveness, an agreeable set of small cardinality is guaranteed to exist, and it can be efficiently computed. This article deals with an extension to subsets which must satisfy extra matroidal constraints. Worst case upper bounds on the size of an agreeable set are shown, and algorithms for computing them are given. For the case of two agents having additive preferences, we show that an agreeable solution can also be approximately optimal (up to a multiplicative constant factor) for both agents.

Keywords

Allocation of indivisible goods Matroids Approximation 

1 Introduction

Social choice theory addresses the recurrent challenge of reaching an agreement for a group of agents who have possibly heterogeneous preferences. The underlying problems often have a combinatorial nature like the fair allocation of indivisible items, or the election of a committee. Beyond its existence, the quick computation of a common solution that several agents consider as being agreeable, or fair, has recently received much attention, see for example two recent textbooks on computational social choice (Brandt et al. 2016; Rothe 2016).

Suppose we are given a set of goods S and a group of agents N. When it is not possible to keep all the goods, a compromise must be found. A possible candidate is an agreeable set \(T \subseteq S\) such that every agent in N finds T at least as good as its complement \(S{\setminus } T\). This approach is particularly relevant when the complement \(S{\setminus }T\) is left in the possession of an opponent of the group of agents N.

As an example, consider a sport league which consists of a set of teams S. A company who sells sporting goods of a given brand, wants to sign a sponsoring contract with a subset T of S. The other teams \((S {\setminus } T)\) receive sporting goods from a rival company whose brand is different. In this context, the persons who work in the board of the first company would accept T if they agree on the fact that it is at least as good as \(S {\setminus } T\). Said differently, they should not envy their rival.

The problem of finding an agreeable set has been introduced by Suksompong (2016). Agreeability is related to envy-freeness (Foley 1967), a well established notion of fairness where one wants to distribute some goods over a group of agents so that everyone finds his share at least as good the share of another agent. Agreeability takes the viewpoint of a group instead of the viewpoint of every single agent. Moreover, agreeability is unilateral as one wishes that the group of agents N does not envy the opponent’s share.

In his work, Suksompong makes a natural monotonicity assumption on every preference relation over the possible subsets of S. A preference relation is monotonic if every \(T'\) is at least as good as T when \(T \subseteq T'\). An obvious agreeable set contains all the elements of S. In the previous example with the sport league, it would correspond to a monopoly, which is unlikely. Thus, it is more realistic to find a small agreeable set. How small can an agreeable set be? Following Suksompong (2016), every instance with monotonic preferences admits an agreeable set of size \(\min \left( \lceil \frac{m+n-1}{2} \rceil , m \right) \) where n and m denote the number of agents in N and the number of goods, respectively. Beyond the existence of an agreeable set, its efficient algorithmic construction is possible when the preference relations are responsive (a special case of monotonicity that can be found in e.g. Barberà et al. 2004; Steven and Brams 2012), and \(|N| \in \{2,3\}\).

In the present work, we intend to extend the problem of finding a small agreeable set to richer cases. Concretely, we introduce an additional constraint on the set of items that the agents, together with their opponent, select. Let us provide two examples of this extension.

Example 1

A university is recruiting q students who will work in one of the two departments: physics and biology. By the law, a certain parity between men and women must be satisfied, i.e., the set of recruited students must contain at most \(\alpha q\) members of each sex, for a given parameter \(\alpha \in (0.5,1]\). The sets of female and male candidates are denoted by \(S_f\) and \(S_m\), respectively. Let T and C be the set of students recruited by the department of physics and biology, respectively. A feasible solution must satisfy \(T \cap C =\emptyset \), \(T\cup C \subseteq S_f \cup S_m\), \(|T\cup C| \le q\), and \(\max ( |(T\cup C) \cap S_f|, |(T\cup C) \cap S_m|) \le \alpha q\).

An instance of this example can be \(S_f=\{f_1,f_2,f_3,f_4,f_5\}\), \(S_m=\{m_1,m_2,m_3,m_4\}\), \(q=5\), and \(\alpha =3/5\). If \(T=\{f_3,m_2,m_3\}\), then C can be any member of \( \mathcal{C} = \{\{f,m\} \in S_f {\setminus } \{f_3\} \times \{m_1, m_4\}\} \cup \{\{f,f'\} \mid f\ne f' \text{ and } f,f' \in S_f {\setminus } \{f_3\}\},\) or a proper subset of some \(\{f,m\} \in \mathcal{C}\) if less than q students are recruited. \(\square \)

Example 2

The main cities of a given country need to be connected with highways. The connection network consists of links between pairs of cities. The objective is to guarantee the existence of a path between every pair of cities. Since the construction and maintenance are expensive, the network should be inclusion-wise minimal. Two rival companies compete for the construction and maintenance of the network. We denote by T and C the links of the first and second company, respectively. Thus, T and C should be disjoint and their union should be a tree spanning all big cities.

For instance, consider the network of Fig. 1. If \(T=\{(a,b),(b,d)\}\) then C can contain at most one edge in \(\{(a,e),(b,e),(d,e)\}\) (dotted edges) and at most one edge in \(\{(a,c),(b,c),(d,c)\}\) (dashed edges). \(\square \)

Fig. 1

Each node is a big city and each edge is a possible highway

In Example 1, we take the point of view of a department, say physics, who would like to choose a best set T of students. The opponent is the department of biology who receives C. In Example 2, we take the point of view of one company, say the first one, who would like a best set T of links. The opponent is the second company who builds and maintains C.

Both examples depart from the framework of Suksompong (2016) because the set of selected elements must satisfy extra feasibility contraints. It is not always true that \(C=S {\setminus } T\). Moreover, when T is fixed, the way to define a complement C of T is not necessarily unique.

Therefore, our model imposes to propose an adapted notion of agreeability. We will do it in two ways in this article. Either every agent finds the current solution at least as good as any complement, or every agent finds the current solution at least as good as at least one complement. We are going to refer to these two notions as strong and weak agreeability, respectively.

The focus of the present work is on a certain type of feasibility constraints for the solutions, namely those defined by a matroid (James 1992). Indeed, the model of Suksompong (2016) and Examples 1 and 2 are based on matroids which are free, laminar, and graphic, respectively (defined in Sect.2.1). A matroid is a well studied combinatorial structure which generalizes the notion of independence. Matroid theory has significantly contributed to the understanding of some important combinatorial optimization problems, such as spanning trees and matchings of a graph (Schrijver 2003; Korte and Vygen 2007). The literature on the allocation of resources comprises applications having matroidal constraints like quotas, see for example Gale and Shapley (1962); Gusfield and Irving (1989); Biró et al. (2010). Matroids have also been studied in the context of allocating goods, see for example Gourvès et al. (2013); Ferraioli et al. (2014); Gourvès et al. (2016).

In order to propose a self-contained work, some basic notions of matroid theory are provided in Sect. 2.1. The type of preference relations under consideration are defined in Sect. 2.2: they are monotonic or responsive. The next section deals with adapted notions of agreeability to the specific context of matroids. Namely, agreeability can be strong or weak. Results for the existence of an agreeable set of bounded size are presented in Sect. 3. These contributions are paired with tight polynomial-time algorithms.

Before concluding, the case of two agents having additive preference relations is studied in Sect. 4. Though approximation results cannot be obtained for every instance, we show that it is possible to compute in polynomial time a set which is, at the same time, agreeable and approximately optimal within a constant multiplicative factor for the class of simple matroids.

2 Fundamental concepts

2.1 Matroids

In this document, [i] denotes \(\{1,2, \ldots ,i\}\). For a set S and an element e, we sometimes write \(S+e\) to denote \(S\cup \{e\}\), and \(S-e\) to denote \(S{\setminus } \{e\}\).

Definition 1

A matroid \(({\mathcal {E}},{\mathcal {I}})\) consists of a finite set \({\mathcal {E}}=\{e_1, \ldots ,e_m\}\) of elements and a collection \({\mathcal {I}}\) of subsets of \({\mathcal {E}}\) such that:
  • \(({\mathsf {M1}})\) \(\emptyset \in {\mathcal {I}}\);

  • \(({\mathsf {M2}})\) if \(I_2\subseteq I_1\) and \(I_1\in {\mathcal {I}}\), then \(I_2\in {\mathcal {I}}\);

  • \(({\mathsf {M3}})\) if \(I_1,\,I_2\in {\mathcal {I}}\) such that \(|I_1|<|I_2|\), then there exists \(e \in I_2\backslash I_1\) such that \(I_1\cup \{e\}\in {\mathcal {I}}\).

The elements of \({\mathcal {I}}\) are called independent sets, and inclusion-wise maximal independent sets are called bases.

The rank of a subset \(S \subseteq {\mathcal {E}}\) is defined as \(\texttt {rank}(S):=\max \{|I| : I \subseteq S, \, I \in {\mathcal {I}}\}\). All the bases of a matroid have the same cardinality \(r:=\texttt {rank}({\mathcal {E}})\), also called the rank of the matroid.

The time complexity of an algorithm that manipulates a matroid depends on the time for testing if a set is independent. This is done with a test called the independence oracle. Here, we assume that the independence oracle runs in polynomial time with respect to \(|{\mathcal {E}}|\). For the ease of presentation, we often write that a matroid is part of the input of an algorithm but concretely, we only require \({\mathcal {E}}\) and the independence oracle. Let us give some well known matroids:
  • With every graph \(({\mathcal V},{\mathcal {E}})\) is associated a graphic matroid \(({\mathcal {E}},{\mathcal {I}})\) such that \(I \in {\mathcal {I}}\) iff it is an acyclic set of edges.

  • For a set of m elements \({\mathcal {E}}\) and a bound \(b \in [m]\), a uniform matroid \(({\mathcal {E}},{\mathcal {I}})\) is such that \({\mathcal {I}}:=\{I \subseteq {\mathcal {E}}: |I| \le b\}\). The rank of a uniform matroid is its bound b. In this article, a uniform matroid with bound b is also called a b-uniform matroid.

  • For a partition of \({\mathcal {E}}\) in k subsets \(({\mathcal {E}}_1, \ldots ,{\mathcal {E}}_k)\) and k bounds \((b_1, \ldots ,\) \(b_k)\), a partitional matroid \(({\mathcal {E}},{\mathcal {I}})\) is such that \({\mathcal {I}}:=\{I \subseteq {\mathcal {E}}: |I \cap {\mathcal {E}}_j| \le b_j, \, \forall j \in [k]\}\).

  • The partitional matroid can be extended to the laminar matroid if for each pair of subsets \(({\mathcal {E}}_i,{\mathcal {E}}_j)\), it holds that \({\mathcal {E}}_i \subseteq {\mathcal {E}}_j\), or \({\mathcal {E}}_j \subseteq {\mathcal {E}}_i\), or \({\mathcal {E}}_i \cap {\mathcal {E}}_j=\emptyset \).

The unconstrained case of allocating indivisible goods, i.e. the one studied in e.g. Suksompong (2016); Manurangsi and Suksompong (2017) and Bouveret et al. (2016), corresponds to a free matroid: a free matroid \(({\mathcal {E}},{\mathcal {I}})\) is such that \({\mathcal {I}}:=2^{\mathcal {E}}\). Example 1 corresponds to a laminar matroid.

Partitional and laminar matroids can capture known constraints like upper quotas (Gale and Shapley 1962; Gusfield and Irving 1989) and common upper quotas (Biró et al. 2010). Other applications can be found with other matroids [e.g. see Gourvès et al. (2015, Example 1) and Gourvès et al. (2016, Example 2) for the transversal matroid].

2.2 Preference relations

Let \(\succeq \) denote a reflexive, complete, and transitive preference relation on \(2^{\mathcal {E}}\), where \({\mathcal {E}}\) is a set of elements. The strict part of \(\succeq \) is denoted by \(\succ \). We sometimes write \(e \succeq e'\) instead of \(\{e\} \succeq \{e'\}\).

Definition 2

\(\succeq \) is monotonic if \(E + e \succeq E\) for all \(E \in 2^{\mathcal {E}}\) and \(e \in {\mathcal {E}}\).

Definition 3

\(\succeq \) is responsive if it satisfies the following two conditions:
  • \(\succeq \) is monotonic;

  • \(E -e' + e \succeq E\) for all \(E \in 2^{\mathcal {E}}\) and \(e,e'\) such that \(e \succeq e'\), \(e \in {\mathcal {E}}{\setminus } E\), and \(e' \in E\).

Let us give three examples of responsive preference relations:
  • (i) In an additive preference relation \(\succeq \), for every \(e \in {\mathcal {E}}\), \(\{e\}\) has a non-negative valuation \(\nu (\{e\}) \in \mathbb {R}\). Moreover, \(\nu (\emptyset )=0\) and for every non-empty \(I \subseteq {\mathcal {E}}\), \(\nu (I) = \sum _{e \in I} \nu (\{e\})\). Then, for \(I,I' \subseteq {\mathcal {E}},\) we have \(I \succ I'\) if \(\nu (I) > \nu (I')\).

A given ordering on \({\mathcal {E}}\) is assumed in the following two definitions. Furthermore, for \(I,I' \subseteq {\mathcal {E}}\) such that \(|I|=|I'|\) and \(I\ne I'\), I lexicographically dominates \(I'\) if the smallest element of \(I {\setminus } I'\) is smaller than the smallest element of \(I' {\setminus } I\), where an ordering on \({\mathcal {E}}\) is assumed.
  • (ii) For \(I,I' \subseteq {\mathcal {E}},\) we have \(I \succ I'\) if one of the following two conditions holds:
    • \(|I| > |I'|\);

    • \(|I| = |I'|\) and I lexicographically dominates \(I'\).

  • (iii) For \(I,I' \subseteq {\mathcal {E}},\) we have \(I \succ I'\) if one of the following three conditions holds (e is a given element of \({\mathcal {E}}\) and \(\bigtriangleup \) denotes the symmetric difference):
    • \(e \in I{\setminus } I'\);

    • \(e\notin I\bigtriangleup I'\) and \(|I| > |I'|\);

    • \(e\notin I\bigtriangleup I'\), and \(|I| = |I'|\) and I lexicographically dominates \(I'\).

Let \(\unrhd \) be a preference relation on \({\mathcal {E}}\). As a notation, we reserve \(\succeq \) and \(\unrhd \) for preference relations on \(2^{\mathcal {E}}\) and \({\mathcal {E}}\), respectively. The strict part of \(\succeq \) and \(\unrhd \) are denoted by \(\succ \) and \(\vartriangleright \), respectively.

Definition 4

We say that \(\succeq \) is consistent with \(\unrhd \) when \(e \unrhd e' \Leftrightarrow e \succeq e'\), \(\forall \, e,e' \in {\mathcal {E}}\).

The next observation provides a sufficient, but not necessary, condition for a set I to be preferred to another set J when \(\succeq \) is unknown but \(\succeq \) is consistent with a given relation \(\unrhd \).

Observation 1

For a responsive preference relation \(\succeq \) consistent with \(\unrhd \) and two sets \(I,J\in 2^{\mathcal {E}}\), we have \(I \succeq J\) if there exists a matching that saturates1 J in the bipartite graph \((I \cup J, \{(e_i,e_j) \in I \times J \mid e_i \unrhd e_j \})\).

Proof

Let \(\mu \) be a matching that saturates J in the bipartite graph. Denote by \(\mu (e)\) the element of I that is matched with \(e \in J\), and let \(\mu (J)=\{\mu (e) : e\in J\}\). By the responsiveness of \(\succeq \) and its consistency with \(\unrhd \), \(\mu (J) \succeq J\). Since \(\mu (J) \subseteq I\), the monotonicity property gives \(I \succeq \mu (J)\) and the result follows by transitivity of \(\succeq \). \(\square \)

This observation, together with the notion of most preferred base, will be used several times in rest of this article. Given a matroid \(({\mathcal {E}},{\mathcal {I}})\) and a preference relation \(\unrhd \) on \({\mathcal {E}}\), the most preferred base problem is to find a base B of \(({\mathcal {E}},{\mathcal {I}})\) such that \(B \succeq B'\) holds for every base \(B'\) and every preference relation \(\succeq \) consistent with \(\unrhd \). Interestingly a most preferred base always exists and the well known greedy algorithm for computing a base of maximum weight (additive preference) also solves the most preferred base problem.

Theorem 1

greedy (Algorithm 1) solves the most preferred base problem for all responsive preference relation \(\succeq \) consistent with \(\unrhd \).

Proof

Let \(B=\{e'_1, \ldots ,e'_r\}\) be the output of greedy, and \(B^*=\{e^*_1, \ldots ,e^*_r\}\) is a most preferred base. Suppose w.l.o.g. that \(e'_i \unrhd e'_j\) and \(e^*_i \unrhd e^*_j\) for all \(1 \le i<j\le r\). If \(e'_i \unrhd e^*_i\) for every \(i \in [r]\), then \(B \succeq B^*\) (Observation 1); B is a most preferred base. Otherwise, let k be the smallest index such that \(e'_k \lhd e^*_k\). By \({\mathsf {M3}}\), there exists \(e \in \{e^*_1\ldots ,e^*_k\} {\setminus } \{e'_1\ldots ,e'_{k-1}\}\) such that \(\{e'_1\ldots ,e'_{k-1}\}+e \in {\mathcal {I}}\). Since \(e \unrhd e^*_k \rhd e'_k\), we get a contradiction with the construction of B. \(\square \)

Unfortunately, greedy is not guaranteed to work if \(\succeq \) is monotonic but not responsive. Indeed, suppose \({\mathcal {E}}=\{a,b,c\}\), \({\mathcal {I}}=\{I \subseteq {\mathcal {E}}: |I| \le 2\}\), \(a \unrhd b \unrhd c\) and \(\{b,c\} \succ \{a,b\} \succeq \{a,c\} \succeq \{a\} \succeq \{b\} \succeq \{c\} \succeq \emptyset \).

2.3 Agreeability: instances and problems

An instance of agreeable problem consists of a matroid \(({\mathcal {E}},{\mathcal {I}})\) of rank r, a set N of n agents, and for every agent \(\ell \in N\), a preference relation \(\unrhd _\ell \) over \({\mathcal {E}}\). Though the preference relation \(\succeq _\ell \) of an agent \(\ell \) over \({\mathcal {I}}\) is not part of the input, we assume that \(\succeq _\ell \) is consistent with \(\unrhd _\ell \). Moreover \(\succeq _\ell \) is always monotonic, and is sometimes responsive. The motivation comes from the fact that it is difficult to have a compact representation of \(\succeq _\ell \), as opposed to \(\unrhd _\ell \).

Definition 5

The complement set of \(I \in {\mathcal {I}}\) is \(\mathsf{{C}}(I):=\{J \subseteq {\mathcal {E}}{\setminus } I : I \cup J \in {\mathcal {I}}\}\). Every \(J \in \mathsf{{C}}(I)\) is a complement of I.

Thus, \(|I|+|J| \le r\) holds for every \(J\in \mathsf{{C}}(I)\). Moreover, all the complements of I that are maximal for inclusion have the same size: \(r-|I|\).

The instances considered in Suksompong (2016) are a special case of the model proposed in this article (\(({\mathcal {E}},{\mathcal {I}})\) is a free matroid). Our generalization comes with an important difference since an independent set does not need to have a unique complement. For this reason, we shall introduce a different notion of agreeability when a set is compared to all its complements (strong agreeability, abbreviated s-agreeability), or only compared to a single complement (weak agreeability, abbreviated w-agreeability).

Definition 6

A set \(I \in {\mathcal {I}}\) is s-agreeable to agent \(\ell \) if \(I \succeq _\ell J\) holds for all \( J \in \mathsf{{C}}(I)\). A set \(I \in {\mathcal {I}}\) is w-agreeable to agent \(\ell \) if there exists \(J \in \mathsf{{C}}(I)\) such that \(I \succeq _\ell J\) and J is maximal for inclusion. For t \(\in \{\)s,w\(\}\), a set \(I \in {\mathcal {I}}\) is t-agreeable if it is t-agreeable to every agent in N.

In this definition, both I and its complement J are independent sets of the same matroid. This constraint captures the fact that the agents in N and their possible opponent evolve in the same environment. Thus, they are bound to build solutions which fit this environment. In Examples 1 and 2, the environments are the global parity constraint imposed to the university by the law, and the inclusion-wise minimality of the entire network, respectively.

Definition 6 extends the notion of agreeability for which there is a unique complement (Suksompong 2016). In case of a free matroid, s-agreeability and w-agreeability coincide.

For every instance, every base of the underlying matroid must be (strong or weak) agreeable since its only complement is the empty set. The difficulty relies on the existence, and efficient construction, of an agreeable set of small size. A challenging aspect of the construction problem comes from the fact that the agents’ preferences over \(2^{\mathcal {E}}\) are unknown, but the preferences over \({\mathcal {E}}\) are given. Indeed, it is difficult to elicit and communicate the agents’ preferences over \(2^{\mathcal {E}}\). However, this is easier for the agents’ preferences over \({\mathcal {E}}\). This motivates the following notion of necessarily agreeable set which generalizes (Suksompong 2016, Definition 5).

Definition 7

Given a preference relation \(\unrhd \) over \({\mathcal {E}}\), a set \(I \in {\mathcal {I}}\) is necessarily s-agreeable (resp., necessarily w-agreeable) with respect to \(\unrhd \) if I is s-agreeable (resp., w-agreeable) for any responsive preference relation \(\succeq \) consistent with \(\unrhd \).

For t \(\in \{\)s,w\(\}\), a set \(I \in {\mathcal {I}}\) is necessarily t-agreeable if it is necessarily t-agreeable with respect to \(\unrhd _\ell \), \(\forall \ell \in N\).

3 Small agreeable sets

We first address the existence of an agreeable set whose size is not trivial (i.e. not equal to the rank) for every instance. This type of existence result comes with a hypothesis of monotonicity on the agents’ preferences. The second problem is the algorithmic construction of a necessarily agreeable set, provided that the agents’ preferences are responsive. The preferences of the agents over \({\mathcal {E}}\) are given but the preferences over \(2^{\mathcal {E}}\) are not part of the input. These two problems are addressed for strong and weak agreeability, respectively.

3.1 Strong agreeability

We begin with an existence result for monotonic preferences followed by a constructive method for responsive preferences.

Proposition 1

For every matroid \(({\mathcal {E}},{\mathcal {I}})\) with rank r and every set \(N=[n]\) of agents with monotonic preferences on \({\mathcal {I}}\), there exists \(I \in {\mathcal {I}}\) such that \(|I| \le \lceil \frac{n \, r}{n+1}\rceil \) and I is s-agreeable to all players.

Proof

Choose any base B of \(({\mathcal {E}},{\mathcal {I}})\) which will be cut in \(n+1\) disjoint subsets \(S_1, \ldots , S_{n+1}\). \(S_i\) is the share of agent \(i \in N\) while \(S_{n+1}\) is a complement. Let x and y be two natural numbers satisfying \(|B|=r=(n+1) x + y\) with \(0\le y \le n\). We impose that \(|S_1|=|S_2|=\cdots =|S_y| =x+1\) and \(|S_{y+1}|=\cdots =|S_{n+1}|=x\).

Modify \(S_1, \ldots , S_{n+1}\) as follows. While there exists \(\ell \in N\) satisfying \(\exists J \in \mathsf{{C}}(\bigcup _{i\in N} S_i) \text{ such } \text{ that } J \succ _\ell S_\ell \) and \(|J|=|S_{n+1}|\) , do \(S_{n+1} \leftarrow S_\ell \) and \(S_\ell \leftarrow J\). The process is finite because the preferences are transitive, and every replacement \(S_\ell \leftarrow J\) induces an improvement in agent \(\ell \)’s preference for \(S_\ell \), without deteriorating the preference of agent k for \(S_k\), \(\forall k \in N {\setminus } \{\ell \}\). We eventually get \(n+1\) disjoint sets \(S_1, \ldots , S_{n+1}\) such that \(\bigcup _{i =1}^{n+1} S_i \in {\mathcal {I}}\) and for all \(\ell \in N\) and \(J \in \mathsf{{C}}(\bigcup _{i \in N} S_i)\), \(S_\ell \succeq _\ell J\). The monotonicity property gives \(\bigcup _{i \in N} S_i \succeq _\ell S_\ell \succeq _\ell J\), implying that \(\bigcup _{i \in N} S_i\) is s-agreeable to all agents. The size of \(\bigcup _{i \in N} S_i\) is \(r - |S_{n+1}|\). Since \(|S_i| \in \{x,x+1\}\) holds for every \(i \in [n+1]\), at any time, we get that \(| \bigcup _{i \in N} S_i | \le r-x =\frac{rn+y}{n+1} = \lceil \frac{r n}{n+1} \rceil \). \(\square \)

Proposition 1 deals with monotonic preferences, which is a proper superset of responsive preferences.

Proposition 2

For every r and n, there exists an instance with rank r, a set N of n agents, and responsive preferences for which the size of every s-agreeable set is at least \(\lceil \frac{n \, r}{n+1} \rceil \).

Proof

Consider the r-uniform matroid \(({\mathcal {E}},{\mathcal {I}})\) such that \({\mathcal {E}}=\{e_{i}^\ell : (i,\ell ) \in [r] \times N\}\). For every \(I,I'\in {\mathcal {I}}\), \(I \succ _\ell I'\) iff \(| I \cap \{e_1^\ell , \ldots , e_r^\ell \}| > | I' \cap \{e_1^\ell , \ldots , e_r^\ell \}|\), \(\forall \ell \in N\). Thus, \(\succeq _\ell \) is responsive for all \(\ell \).

To be s-agreeable to an agent \(\ell \in N\), I must satisfy \(x_\ell :=| I \cap \{e_1^\ell , \ldots , e_r^\ell \} | \ge r - |I| \). We get that \(\sum _{\ell \in N} x_\ell \ge \sum _{\ell \in N} (r-|I|)=n(r-|I|)\). Use \(\sum _{\ell \in N} x_\ell =|I|\) to get that \((n+1) |I| \ge nr\). Therefore, \(|I| \ge \lceil \frac{n \, r}{n+1} \rceil \) because |I| is an integer. \(\square \)

Proposition 2 matches with Proposition 1, meaning that the general bound of \(\lceil \frac{n \, r}{n+1}\rceil \) is tight for strong agreeability. This bound can be informally explained: if the complement has size s, then every agent must “value” s elements of the agreeable set (see the second and third examples of a responsive preference in Sect. 2.2). In the worst case, the agents value disjoint sets. Thus, as done in the proof of Proposition 1, the problem is to identify a base (its size is r) cut in \(n+1\) pieces of (almost) equal size, and only one piece can be discarded.

Now, we investigate the construction of an s-agreeable set I for any number \(n=|N|\) of agents (see Algorithm 2 where \(r=\texttt {rank}({\mathcal {E}})\)) when the preference of every agent over \({\mathcal {E}}\) is given. Let \(next: [n] \rightarrow [n]\) be defined as \(next(t)=t+1\) if \(1\le t < n\), and \(next(n)=1\). The idea of the algorithm is the following. The solution is empty at the beginning (it belongs to \({\mathcal {I}}\)) and the agents add an element (the one they like the most) in a round-robin way so that the solution remains in \({\mathcal {I}}\). The algorithm stops after the insertion of \(\lceil \frac{n \, r}{n+1} \rceil \) elements.

Theorem 2

round-robin returns a necessarily s-agreeable set of size \(\lceil \frac{n \, r}{n+1} \rceil \).

Proof

At the end of the algorithm’s execution, I has size \(\lceil \frac{n \, r}{n+1} \rceil \) whereas any complement \(J \in \mathsf{{C}}(I)\) has size \(\lfloor \frac{r}{n+1} \rfloor \). Each agent has inserted at least as many elements as the size of a complement. The fact that an agent t, during her turn, inserts the element \(e \in {\mathcal {E}}{\setminus } I\) coming first in \(\unrhd _t\), ensures that every element \(e'\) that can be subsequently added to I, and be part of the complement, satisfies \(e \unrhd _t e'\). Therefore, the conditions of Observation 1 are met for every agent,2 meaning that I is s-agreeable to all agents. \(\square \)

Note that the proof of Theorem 2 does not exploit the fact the agents add elements to I in a round-robin way. Other orders would work, provided that each agent inserts at least as many elements as the size of a complement. However, round-robin will be used in Sect. 4 to show some performance guarantees, in addition to agreeability.

3.2 Weak agreeability

Let us first give a characterization of w-agreeable sets.

Proposition 3

An independent set I is w-agreeable to an agent with preference relation \(\succeq \) iff there exists a base B such that \(B \supseteq I\) and \(I \succeq B {\setminus } I\).

Proof

On one hand, there exists by definition \(J \in \mathsf{{C}}(I)\) such that \(I \succeq J\) and J is maximal for inclusion. Thus, \(I \cup J\) is a base which includes I. On the other hand, \(B{\setminus } I\) is a complement to I and it is maximal for inclusion because B is a base. \(\square \)

Let us quote the main theorem of Suksompong (2016) which is used in the proof of our next result.

Theorem 3

Suksompong (2016) Assume that there are n players with monotonic preferences on \({\mathcal {S}}\). There exists a subset \(T \subseteq S\) such that \(|T| \le \min \left( \lceil \frac{m+n-1}{2} \rceil , m \right) \) and T is agreeable to all n players. Moreover, there exist monotonic preferences for which the bound \(\min \left( \lceil \frac{m+n-1}{2} \rceil ,m\right) \) is tight.

In Theorem 3, m is the rank of the free matroid. The following theorem extends this results to any matroid of rank r.

Theorem 4

Assume that there are n agents with monotonic preferences on \(2^{\mathcal {E}}\). There exists a subset \(T \subseteq {\mathcal {E}}\) such that \(|T| \le \min \left( \lceil \frac{r+n-1}{2} \rceil , r \right) \) and T is w-agreeable to all n agents. Moreover, there exist monotonic preferences for which the bound \(\min \left( \lceil \frac{r+n-1}{2} \rceil , r \right) \) is tight.

Proof

Take any base of the matroid, say B. Such a base can be computed with greedy in polynomial time (choose \(\unrhd \) arbitrarily). Use Theorem 3 where \(S=B\), and m is the rank r of the corresponding free matroid. Thus, there exists \(T \subseteq B\) such that \(|T| \le \min \left( \lceil \frac{r+n-1}{2} \rceil , r \right) \) and \(T \succeq _\ell B{\setminus } T\) for every agent \(\ell \). Now, by Proposition 3 with B and T, T is a w-agreeable set. For the tightness of the bound, the examples used for the proof of Theorem 3 remain valid since they are free matroids. \(\square \)

As mentioned in Suksompong (2016), and it extends to the matroidal setting, it is remarkable that w-agreeability can be obtained with a set whose size is roughly half the size of a base, plus half an element per agent. Interestingly, the construction of a necessarily w-agreeable set of size \(\min \left( \lceil \frac{r+n-1}{2} \rceil , r \right) \) is possible when \(n=2\). In this case \(\min \left( \lceil \frac{r+n-1}{2} \rceil , r \right) = \lceil \frac{r+1}{2} \rceil \) because \(r\ge 1\) can be assumed.

Theorem 5

For two agents and their preferences over \({\mathcal {E}}\), there exists a polynomial-time algorithm that returns a necessarily w-agreeable set of size \(\lceil \frac{r+1}{2} \rceil \).

Proof

Use greedy with \(\unrhd _1\) in order to construct a most preferred base \(\{f_1, \cdots , f_r\}\) for agent 1. Afterwards, run Algorithm 33 with input \(\{f_1, \cdots , f_r\}\) and \(\unrhd _2\). We get two sets I and J such that \(I\in {\mathcal {I}}\) because \(I \subset \{f_1, \cdots , f_r\}\) (Property \(\mathsf{M2}\) of matroids), \(J \in \mathsf{{C}}(I)\), and \(I\cup J=\{f_1, \ldots ,f_r\}\). For every player \(\ell \in \{1,2\}\), each element \(f \in J\) can be paired with an element \(f' \in I\) such that \(f' \unrhd _\ell f\). The conditions of Observation 1 are met, meaning that I is w-agreeable. In both cases (\(r=2k+1\) or \(r=2k\)), I has cardinality \(k+1=\lceil \frac{r+1}{2} \rceil \). \(\square \)

4 Approximation for additive preference relations

It is assumed in this section that the agents’ preferences are additive. The valuation function of \(t\in N\) is denoted by \(\nu _t\). Hence, \(\forall I,I' \subseteq {\mathcal {E}},\) \(\nu _t(I) \ge \nu _t(I') \Leftrightarrow I \succeq _t I'\). The representation of \(\succeq _t\) is compact since it suffices to store \(\nu _t(\{e\})\) for every \(e \in {\mathcal {E}}\). For the sake of simplicity, we often write \(\nu _t(e)\) instead of \(\nu _t(\{e\})\).

So far, we have seen that a succinct set can be agreeable to a group of agents if the preferences are monotonic. An interesting challenge is to combine this property with another kind of guarantee on the quality of the solution.

As an illustration, consider the instance of Example 2 depicted on Fig. 2. There are two agents with identical valuations for the edges. The set \(\{(a,b),(b,c)\}\) is valued 2, which is as large as the valuation of the unique complement (cd). Thus, \(\{(a,b),(b,c)\}\) is s-agreeable but its valuation is only \(20 \%\) of the optimum because the maximum valuation for a tree is 10 (achieved by \(\{(a,b),(a,c),(c,d)\}\) or \(\{(b,c),(a,c),(c,d)\}\)). A tree with maximum valuation is the best s-agreeable set (without limiting the cardinality of the set), or what the agents would choose if they were alone. Compared to \(\{(a,b),(b,c)\}\), \(\{(a,b),(a,c)\}\) is an s-agreeable set of size 2 but its valuation is \(80\%\) of the optimum. From the agents’ viewpoint, \(\{(a,b),(a,c)\}\) is more attractive.
Fig. 2

The nodes and the edges are the cities and the roads, respectively. The numbers correspond to the valuations

In addition to the notion of agreeability, we are going to use approximation which is the worst-case ratio between the valuation of a given solution and the largest valuation of a solution.

Definition 8

For \(\rho \in (0,1]\), a set \(I \in {\mathcal {I}}\) is \(\rho \) -approximate for agent t if \(\forall B \in {\mathcal {I}}\), \(\nu _t(I) \ge \rho \, \nu _t(B)\). An algorithm is \(\rho \) -approximate for agent t if its output is \(\rho \)-approximate for every instance.

When the matroid is free, every agreeable set is also 1 / 2-approximate for every agent (Suksompong 2016). We are going to study the question of the approximation ratio for a larger class of matroids.

When the rank of the matroid is strictly smaller than the number of agents, the following example illustrates that there are instances without any \(\rho \)-approximate solution (disregarding the agreeability).

Example 3

Consider the b-uniform matroid such that \({\mathcal {E}}=\{e_1, \ldots ,e_m\}\) and \(1 \le b<n\le m\). For any agent \(\ell \in N\), \(\nu _\ell (e_k)=1\) if \(k=\ell \), and \(\nu _\ell (e_k)=0\) otherwise. Since \(b<n\), every solution T has an agent t such that \(e_t \notin T\), whereas a solution containing \(e_t\) exists.

Even when there are only two agents, the following example illustrates that there are instances with rank larger than 2 but without any \(\rho \)-approximate solution.

Example 4

Consider the partitional matroid such that \({\mathcal {E}}=\{e_1, \ldots ,e_m\}\) and \({\mathcal {I}}=\{I \subset {\mathcal {E}}: |I \cap \{e_1,e_2\}| \le 1\}\). For \(\ell \in \{1,2\}\), \(\nu _\ell (e_k)=1\) if \(k=\ell \), and \(\nu _\ell (e_k)=0\) otherwise.

Examples 3 and 4 illustrate that, as opposed to the case of a free matroid, no algorithm can be \(\rho \)-approximate for every instance. However, we shall see that there is room for interesting approximation results if we consider two agents and simple matroids. A matroid is simple if every pair of elements is independent (James 1992). For example, the graphic matroid associated with a simple graph (there is at most one edge between two nodes) is a simple matroid. A b-uniform matroid is simple if \(b \ge 2\).

We are going to analyze round-robin for s-agreeability and another algorithm for w-agreeability (Algorithm 4), and see that they have approximation ratios which are lower bounded by constants. Matroids of rank 1 are ignored since we restrict ourselves to simple matroids.

4.1 Strong agreeability

We shall see with Theorems 6 and 7 that round-robin combines two nice properties in simple matroids for two agents: it produces a strong agreeable set and this set is a constant approximation for both agents. The approximation ratios are given in Table 1.
Table 1

The approximation ratios of round-robin where \(\lceil r/3 \rceil /r \ge 1/3\)

Rank r of the matroid

2

3

4

\(r\ge 5\)

Approximation ratio for agent 1

1 / 2

1 / 3

1 / 2

\(\lceil r/3\rceil /r\)

Approximation ratio for agent 2

1 / 2

1 / 3

1 / 4

1 / 3

The following lemma will be often used in the following proofs.

Lemma 1

Let \(({\mathcal {E}},{\mathcal {I}})\) be a matroid with rank r, \(\{f_1,\ldots ,f_r\}\) a most preferred base of \(({\mathcal {E}},{\mathcal {I}})\) for some agent i such that \(\nu _i(f_1) \ge \nu _i(f_2) \ge \cdots \ge \nu _i(f_r)\), and \(E \in {\mathcal {I}}\) such that \(|E|<r\). If \(e \in {\mathcal {E}}{\setminus }\ E\) is agent i’s most preferred element such that \(E+e \in {\mathcal {I}}\), then \(\forall j \in \{|E|+1, \cdots ,r\}\), \(\nu _i(e) \ge \nu _i(f_{j})\).

Proof

By property \({\mathsf {M3}}\) of matroids, an element of \(\{f_1, \ldots , f_{|E|+1}\}\) can be added to E. Since \(f_{|E|+1}\) is the element of smallest value in \(\{f_1, \ldots , f_{|E|+1}\}\), we get that \(\nu _i(e) \ge \nu _i(f_{|E|+1})\). Use \(\nu _i(f_{|E|+1}) \ge \nu _i(f_{|E|+2}) \ge \cdots \ge \nu _i(f_r)\) to complete the proof. \(\square \)

Theorem 6

round-robin for two agents is \(\lceil r/3\rceil /r\)-approximate for agent 1.

Proof

In this proof \(F=\{f_1,\ldots ,f_r\}\) denotes a most preferred base for agent 1 where
$$\begin{aligned} \nu _1(f_1) \ge \nu _1(f_2) \ge \cdots \ge \nu _1(f_r) \end{aligned}$$
(1)
The output of round-robin is \(I=\{e_1, \ldots ,e_{\lceil 2 r/3 \rceil }\}\) and we assume that \(e_i\) is inserted before \(e_j\) whenever \(i<j\). We will consider three cases: \(r =3x\), \(r =3x-1\), and \(r =3x+1\).

Suppose \(r =3x\). In this case \(\lceil r/3\rceil /r=1/3\), and I contains \(\lceil 2 r/3 \rceil =2x\) elements. During the execution of round-robin, agent 1 chooses the odd elements \(\{e_{2i+1} : i=0..x-1\}\) where \(e_{2x-1}\) is the last element inserted by agent 1.

Fix some \(i\ge 0\). Agent 1 inserts \(e_{2i+1}\) when the current solution is \(\{e_1, \ldots , e_{2i}\}\). By Lemma 1 we get that
$$\begin{aligned} \nu _1(e_{2i+1}) \ge \nu _1(f_j), \; \forall j \ge 2i+1 \end{aligned}$$
(2)
Associate with each \(e_{2i+1}\) the set \(F_i:=\{ f_{2i+1}, f_{2i+2},f_{2x+i+1}\}\). Note that for two indices i and \(i'\), \(F_i \cap F_{i'}=\emptyset \); moreover \(F=\bigcup _{i =0}^{x-1} F_i\). Use (2) to get that \(\nu _1(e_{2i+1}) \ge \nu _1(f)\) for every \(f \in F_i\). Since \(F_i\) contains at most 3 elements, we deduce that \(3 \nu _1(e_{2i+1}) \ge \nu _1(F_i)\). Thus, \(3 \nu _1(I) \ge 3 \sum _{i =0}^{x-1} \nu _1(e_{2i+1}) \ge \sum _{i =0}^{x-1} \nu _1(F_i)=\nu _1(F)\). In other words, I is 1 / 3-approximate for agent 1.

Suppose \(r =3x-1\). In this case \(\lceil r/3\rceil /r=\frac{x}{3x-1}\), and I contains \(\lceil 2 r/3 \rceil =2x\) elements. Agent 1 chooses the odd elements \(\{e_{2i+1} : i=0..x-1\}\) of I. One can partition F in three sets: \(F_1=\{f_{2i+1} : i=0..x-1\}\), \(F_2=\{f_{2i} : i=1..x\}\), and \(F_3=\{f_{i} : 2x+1 \le i \le 3x-1\}\).

Since \(\nu _1(e_{2i+1}) \ge \nu _1(f_{2i+1})\) for \(i=0..x-1\) by Lemma 1, we deduce that
$$\begin{aligned} \nu _1(\{e_{2i+1} : i=0..x-1\}) \ge \nu _1(F_1) \end{aligned}$$
(3)
Since \(\nu _1(e_{2i+1}) \ge \nu _1(f_{2i+2})\) for \(i=0..x-1\) by Lemma 1, we deduce that
$$\begin{aligned} \nu _1(\{e_{2i+1} : i=0..x-1\}) \ge \nu _1(F_2) \end{aligned}$$
(4)
Since \(\nu _1(e_{2i+1}) \ge \nu _1(f_{j})\) for \(i=0..x-1\) and \(2x+1 \le j \le 3x-1\) by Lemma 1, we deduce that \((x-1) \nu _1(e_{2i+1}) \ge \nu _1(F_3)\) and then
$$\begin{aligned} (x-1) \nu _1(\{e_{2i+1} : i=0..x-1\}) \ge x \cdot \nu _1(F_3) \end{aligned}$$
(5)
Combine (3) multiplied by x, (4) multiplied by x, and (5), to get that \((3x-1) \nu _1(\{e_{2i+1} : i=0..x-1\}) \ge x (\nu _1(F_1)+\nu _1(F_2)+\nu _1(F_3))= x \cdot \nu _1(F)\). In other words, I is \(\frac{x}{3x-1}\)-approximate for agent 1.

Suppose \(r =3x+1\). In this case \(\lceil r/3\rceil /r=\frac{x+1}{3x+1}\), and I contains \(\lceil 2 r/3 \rceil =2x+1\) elements. Agent 1 chooses the odd elements \(\{e_{2i+1} : i=0..x\}\) of I. Let O denote \(\{e_{2i+1} : i=0..x\}\).

One can partition F in three sets: \(F_1=\{f_{2i+1} : i=0..x\}\), \(F_2=\{f_{2i} : i=1..x\}\), and \(F_3=\{f_{i} : 2x+2 \le i \le r\}\). Note that \(|F_1|=|O| =x+1\) and \(|F_2|=|F_3|=x\).

By Lemma 1 we have \(\nu _1(e_{2i+1}) \ge \nu _1(f_{2i+2})\) for \(i=0..x-1\). Summing these inequalities gives
$$\begin{aligned} \nu _1(O) - \nu _1(e_{2x+1}) \ge \nu _1(F_2) \end{aligned}$$
(6)
that we multiply by \(x+1\) to get
$$\begin{aligned} (x+1)\nu _1(O) - (x+1)\nu _1(e_{2x+1}) \ge (x+1)\nu _1(F_2) \end{aligned}$$
(7)
By Lemma 1 we have \(\nu _1(e_{2x+1}) \ge \nu _1(f_{2x+2+i})\) for \(i=0..x-1\). Summing these inequalities gives
$$\begin{aligned} x\nu _1(e_{2x+1}) \ge \nu _1(F_3) \end{aligned}$$
(8)
that we multiply by \(\frac{x+1}{x}\) to get
$$\begin{aligned} (x+1)\nu _1(e_{2x+1}) \ge (1+1/x)\nu _1(F_3) \end{aligned}$$
(9)
We have \(x\nu _1(O) \ge (x+1)\nu _1(F_3)\) because \(\nu _1(e) \ge \nu _1(f)\) for every \((e,f) \in O \times F_3\) by Lemma 1, \(|O|=x+1\), and \(|F_3|=x\). When \(x\ge 2\), multiply \(x\nu _1(O) \ge (x+1)\nu _1(F_3)\) by \(\frac{x-1}{x}\) to get that
$$\begin{aligned} (x-1)\nu _1(O) \ge (x-1/x)\nu _1(F_3) \end{aligned}$$
(10)
By Lemma 1 we have \(\nu _1(e_{2i+1}) \ge \nu _1(f_{2i+1})\) for \(i=0..x-1\). Summing these inequalities gives
$$\begin{aligned} \nu _1(O) \ge \nu _1(F_1) \end{aligned}$$
(11)
If \(x\ge 2\), then we can combine (7), (9), (10), and (11) mutiplied by \(x+1\) to get that \((3x+1)\nu _1(O) \ge (x+1)(\nu _1(F_1)+\nu _1(F_2)+\nu _1(F_3))=(x+1)\nu _1(F)\). Since \(\nu _1(I) \ge \nu _1(O)\), I is \(\frac{x+1}{3x+1}\)-approximate for agent 1. If \(x=1\) then \(\frac{x+1}{3x+1}=\frac{1}{2}\). Combine (6), (8), and (11) to get that \(2 \nu _1(O) \ge \nu _1(F_1)+\nu _1(F_2)+\nu _1(F_3)=\nu _1(F)\). Since \(\nu _1(I) \ge \nu _1(O)\), I is \(\frac{x+1}{3x+1}\)-approximate for agent 1. \(\square \)

Theorem 7

round-robin for two agents and simple matroids is 1 / 3-approximate for agent 2 when \(r\ge 5\), and 1 / r-approximate for agent 2 when \(r\in \{2,3,4\}\).

Proof

Let us begin with the case \(r \ge 5\). We denote by \(G=\{g_1,\ldots ,g_r\}\) a most preferred base for agent 2 where
$$\begin{aligned} \nu _2(g_1) \ge \nu _2(g_2) \ge \cdots \ge \nu _2(g_r) \end{aligned}$$
(12)
The output of round-robin is \(I=\{e_1, \ldots ,e_{\lceil 2 r/3 \rceil }\}\) and we assume that \(e_i\) is inserted before \(e_j\) whenever \(i<j\). We have \(\lceil 2 r/3 \rceil \ge 4\) because \(r\ge 5\). During the execution of round-robin, agent 2 chooses the even elements \(\{e_{2i} : i=1..x\}\) where \(x \ge 2\) because \(\lceil 2 r/3 \rceil \ge 4\).
By Lemma 1 we get that
$$\begin{aligned} \nu _2(e_{2i}) \ge \nu _2(g_j), \; \forall j \ge 2i \end{aligned}$$
(13)
We have \(\nu _2(e_{1})+ \nu _2(e_{2}) \ge \nu _2(g_{1})\) because the matroid is simple. Indeed, either \(e_1=g_1\) and the inequality holds directly, or \(e_1\ne g_1\) and \(\{e_1,g_1\}\) must be independent, which means that \(\nu _2(e_{2}) \ge \nu _2(g_{1})\). Use (12) to get that
$$\begin{aligned} 3 (\nu _2(e_{1})+\nu _2(e_{2})) \ge \nu _2(g_{1})+\nu _2(g_{2})+\nu _2(g_{3}) \end{aligned}$$
(14)
For \(i=2..x-1\), associate the set \(\{ g_{3i-2}, g_{3i-1}, g_{3i}\}\) with \(e_{2i}\). Since \(i\ge 2 \Leftrightarrow 2i\le 3i-2\), use (13) and (12) to get that \(\nu _2(e_{2i})\ge \nu _2(g_{3i-2}) \ge \nu _2(g_{3i-1}) \ge \nu _2(g_{3i})\). It follows that
$$\begin{aligned} 3 \nu _2(e_{2i})\ge \nu _2(g_{3i-2})+\nu _2(g_{3i-1})+\nu _2(g_{3i}) \end{aligned}$$
(15)
For \(i=x\), we associate the set \(\{ g_j : 3x-2 \le j \le r\}\) with \(e_{2x}\). Depending on r, the size of \(\{ g_j : 3x-2 \le j \le r\}\) is between 1 and 3. With the same arguments as above we get that
$$\begin{aligned} 3 \nu _2(e_{2x})\ge \sum _{3x-2 \le j \le r} \nu _2(g_j) \end{aligned}$$
(16)
From (14), (15) where \(i=2..x-1\), and (15) we deduce that \(3\nu _2(I)\ge 3 (\nu _2(e_{1})+\nu _2(e_{2})+\nu _2(e_{4})+\cdots +\nu _2(e_{2x})) \ge \nu _2(G)\), meaning that I is 1 / 3-approximate for agent 2 when \(r\ge 5\).

To conclude, consider the case \(r \in \{2,3,4\}\). round-robin picks \(\lceil 2r/3\rceil \ge 2\) elements. Since the matroid is simple, round-robin must pick the most preferred element for agent 2 (it is either the first or the second element inserted in the solution). As the maximum utility for agent 2 is at most r times the utility for the most preferred element, we deduce that round-robin is 1 / r-approximate for agent 2. \(\square \)

Proposition 4

There exist instances for which the approximation ratios given in Theorems 6 and 7 are reached by round-robin.

Proof

The proof is made on the rank r.

Take the graphic matroid associated with the 3 node graph of Fig. 3 (rank 2). The most preferred bases of agent 1 and 2 are \(\{(a,b),(b,c)\}\) and \(\{(a,c),(b,c)\}\), respectively. They both have value 2 for it. A possible execution of round-robin is to select (ab) and then (ac). This solution contains \(\lceil \frac{n \, r}{n+1} \rceil \) elements (\(r=n=2\)). It is 1 / 2-approximate for both agents.

Take the graphic matroid associated with the 4 node graph of Fig. 3 (rank 3). The most preferred bases of agent 1 and 2 are \(\{(a,b),(c,d),(a,d)\}\) and \(\{(a,c),(c,b),(b,d)\}\), respectively. They both have value 3 for it. A possible execution of round-robin is to select (ab) and then (bc). This solution contains \(\lceil \frac{n \, r}{n+1} \rceil \) elements (\(r=3\) and \(n=2\)). It is 1 / 3-approximate for both agents.

Take the graphic matroid associated with the 5 node graph of Fig. 3 (rank 4). The most preferred base of agent 1 and 2 are \(\{(a,e),(e,b),(b,d),(c,d)\}\) and \(\{(e,b),(a,b),(b,c),(b,d)\}\), respectively. They both have value 4 for it. A possible execution of round-robin is to select (ae), (ab), and then (ac). This solution contains \(\lceil \frac{n \, r}{n+1} \rceil \) elements (\(r=4\) and \(n=2\)). It is 1 / 2-approximate for agent 1 and 1 / 4-approximate for agent 2.

Take the graphic matroid associated with the graph of Fig. 4 (rank \(\ge 5\)). Agent 1 has utility 1 for every edge \(f_i\), and utility 0 for any other edge. Agent 2 has utility 1 for edges \(g_1\), \(g_2\) and \(g_3\), and utility 0 for any other edge. The most preferred base of agent 1 is \(\{f_i : i..r\}\) and \(\nu _1(\{f_i : i..r\})=r\). A most preferred base for agent 2 includes \(\{g_1,g_2,g_3\}\); the maximum utility for agent 2 is 3. A possible execution of round-robin is to select \(f_1\), \(g_1\), \(f_3\), \(g_4\), \(f_5\), and so on,4 until \(\lceil 2 r/3 \rceil \) elements are picked. Agents 1 and 2 have value \(\lceil r/3\rceil \) and 1 for this solution, respectively. Thus, the solution is \(\lceil r/3\rceil /r\)-approximate for agent 1 and 1 / 3-approximate for agent 2. \(\square \)

Fig. 3

Three graphic matroids with rank 2, 3 and 4, respectively. The first and second coordinates are the valuation of the first and second agent for the edges, respectively

Fig. 4

A graphic matroid with rank \(\ge 5\)

4.2 Weak agreeability

The algorithm described in the proof of Theorem 5 cannot be used as it is because it provides a guarantee only for the first agent. For example, take a graphic matroid defined on a graph composed of two disjoint spanning trees \(T_1\) and \(T_2\). Agent \(i \in \{1,2\}\) has valuation 1 for every edge of \(T_i\), and valuation 0 for every edge of \(T_{3-i}\). For this instance, the algorithm described in the proof of Theorem 5 outputs a subset of \(T_1\) for which agent 2 has valuation 0. However agent 2 has a positive valuation for \(T_2\). Nevertheless, approximation results under weak agreeability can be obtained if we restrict ourselves to two agents and simple matroids. These results are based on Algorithm 4.
Note that step 1 of Algorithm 4 corresponds to an algorithm called generalized alt-greedy in Gourvès et al. (2015). Furthermore, the output I of Algorithm 4 is a w-agreeable set of size \(\lceil \frac{r+1}{2} \rceil \) by Theorem 5.
Table 2

Reachable approximation ratios for w-agreeable solution of size \(\lceil \frac{r+1}{2}\rceil \)

Rank r of the matroid

2

3

4

Odd

Even

\(r\ge 5\)

\(r=4q+2\ge 6\) or \(r=4(q+1)\ge 8\)

Approximation ratio for agent 1

1 / 2

1 / 3

1 / 2

1 / 4

\((q+1)/(4q+2)\)

Approximation ratio for agent 2

1 / 2

1 / 3

1 / 4

1 / 5

1 / 5

Theorem 8

For two agents having additive preferences over the independent sets of a simple matroid, Algorithm 4 builds in polynomial time a w-agreeable set of size \(\lceil \frac{r+1}{2}\rceil \) whose approximation ratios are given Table 2.

Proof

Suppose for the moment that \(r\ge 5\). A base \(\{e_1, \ldots ,e_r\}\) is constructed in Algorithm 4 (step 1). Agent 1 and 2 have inserted the elements with odd and even index, respectively. When \(r\ge 5\), there are four cases: \(r\in \{4q+1,4q+2,4q+3,4q+4\}\) where q is a positive integer. The elements of \(\{e_1, \ldots ,e_r\}\) are renamed as \(\{f_1, \ldots ,f_r\}\) such that \(\nu _1(f_1) \ge \nu _1(f_2) \ge \ldots \ge \nu _1(f_r)\). If r is even then \(f_1\) and \(f_2\) are kept, while the remaining elements are paired as follows: \((f_3,f_4)\), \((f_5,f_6)\), and so on. The element with maximum valuation \(\nu _2\) is kept in each of these pairs. If r is odd then \(f_1\) is kept and the remaining elements are paired as follows: \((f_2,f_3)\), \((f_4,f_5)\), and so on. The element with maximum valuation \(\nu _2\) is kept in each of these pairs. The output of Algorithm 4 is I.

From the viewpoint of agent 1, the elements put in I are, in the worst case, \(f_1,f_2,f_4,f_6,\ldots \) when r is even, and \(f_1,f_3,f_5,\ldots \) when r is odd. Observe that \(\nu _1(f_1) \ge \nu _1(e_1)\), and \(\nu _1(f_i) \ge \nu _1(e_{2i-1})\) for \(i \in \{2, \ldots ,\lceil r/2 \rceil \}\) because in each odd round of round-robin, agent 1 selects the element with largest valuation \(\nu _1\) that can be inserted in the current solution. Thus, from the viewpoint of agent 1, the elements that are kept are, in the worst case, \(e_1,e_3,e_7,e_{11},\ldots \) when r is even, and \(e_1,e_5,e_9,\ldots \) when r is odd. More precisely, the sequence is \(e_1,e_3,e_7,e_{11},\ldots ,e_{4q+3}\) when \(r=4(q+1)\), \(e_1,e_3,e_7,e_{11},\ldots ,e_{4q-1}\) when \(r=4q+2\), and \(e_1,e_5,e_9,e_{13},\ldots ,e_{4q+1}\) when \(r\in \{4q+1, 4q+3\}\).

Let \(A^*=\{a_1, \ldots ,a_r\}\) be a most preferred base for agent 1 where \(\nu _1(a_1) \ge \nu _1(a_2) \ge \ldots \ge \nu _1(a_r)\). We know from Lemma 1 that \(\nu _1(e_i) \ge \nu _1(a_i)\) holds when i is odd. It follows that
$$\begin{aligned} \nu _1(I) \ge \nu _1(\{a_1,a_3,a_7,a_{11},\ldots ,a_{4q+3}\})&\text{ when }&r=4(q+1), \end{aligned}$$
(17)
$$\begin{aligned} \nu _1(I) \ge \nu _1(\{a_1,a_3,a_7,a_{11},\ldots ,a_{4q-1}\})&\text{ when }&r=4q+2, \text{ and }\end{aligned}$$
(18)
$$\begin{aligned} \nu _1(I) \ge \nu _1(\{a_1,a_5,a_9,a_{13},\ldots ,a_{4q+1}\})&\text{ when }&r\in \{4q+1, 4q+3\}. \end{aligned}$$
(19)
Suppose \(r=4(q+1)\) for some integer \(q \ge 1\). Consider the set \(A_1=\{a_1\} \cup \{a_{4i+3}: 0 \le i \le q\}\). For \(i=0..q-1\), deduce from \(\nu _1(a_{4i+3}) \ge \nu _1(a_{4i+4}) \ge \nu _1(a_{4i+5}) \ge \nu _1(a_{4i+6})\) that
$$\begin{aligned} (4q+2) \nu _1(a_{4i+3}) \ge (q+1)\nu _1(\{a_{4i+3},a_{4i+4},a_{4i+5}\}) + (q-1)\nu _1(a_{4i+6}) \end{aligned}$$
(20)
For \(i=0..q-1\), use \(\nu _1(a_{1}) \ge \nu _1(a_{4i+6})\) to get that
$$\begin{aligned} 2q \nu _1(a_{1}) \ge 2 \sum _{i=0}^{q-1} \nu _1(a_{4i+6}) \end{aligned}$$
(21)
Use \(\nu _1(a_{4q+3}) \ge \nu _1(a_{4q+4})\) to obtain
$$\begin{aligned} 2(q+1) \nu _1(a_{4q+3}) \ge (q+1) \nu _1(\{a_{4q+3},a_{4q+4}\}) \end{aligned}$$
(22)
Use \(\nu _1(a_{1}) \ge \nu _1(a_{2})\) to obtain
$$\begin{aligned} 2(q+1) \nu _1(a_{1}) \ge (q+1) \nu _1(\{a_{1},a_{2}\}) \end{aligned}$$
(23)
Combine Inequalities (20) to (23) to obtain
$$\begin{aligned} (4q+2) \nu _1(A_1) \ge (q+1)\nu _1(A^*) \end{aligned}$$
(24)
Use Inequalities (17) and (24) to get that \( (4q+2) \nu _1(I) \ge (q+1) \nu _1(A^*)\). In other words, I is \((q+1)/(4q+2)\)-approximate for agent 1 when \(r=4(q+1)\).
Suppose \(r=4q+2\) for some integer \(q \ge 1\). Consider the set \(A_2=\{a_1\} \cup \{a_{4i+3}: 0 \le i \le q-1\}\). In this case, Inequalities (20), (21), and (23) hold. Combine them to obtain
$$\begin{aligned} (4q+2) \nu _1(A_2) \ge (q+1)\nu _1(A^*) \end{aligned}$$
(25)
Use Inequalities (18) and (25) to get that \( (4q+2) \nu _1(I) \ge (q+1) \nu _1(A^*)\). In other words, I is \((q+1)/(4q+2)\)-approximate for agent 1 when \(r=4q+2\).
Suppose \(r\in \{4q+1,4q+3\}\) for some integer \(q \ge 1\). Consider the set \(A_3=\{a_{4i+1}: 0 \le i \le q\}\). For any \(i\in \{0,\ldots ,q\}\), use \(\nu _1(a_{4i+1}) \ge \nu _1(a_{4i+2}) \ge \nu _1(a_{4i+3}) \ge \nu _1(a_{4i+4})\) that
$$\begin{aligned} 4 \nu _1(a_{4i+1}) \ge \nu _1(\{a_{4i+1},a_{4i+2},a_{4i+3},a_{4i+4}\}) \end{aligned}$$
(26)
Sum Inequality (26) for \(i=0..q\) to get
$$\begin{aligned} 4 \nu _1(A_3) \ge \nu _1(A^*) \end{aligned}$$
(27)
Inequalities (19) and (27) give \(4 \nu _1(I) \ge \nu _1(A^*)\), which means that I is 1 / 4-approximate for agent 1 when \(r\in \{4q+1,4q+3\}\). Note that \(r\in \{4q+1,4q+3\}\) is equivalent to r odd and \(r\ge 5\).
Now we take the viewpoint of agent 2 who has inserted the elements with even indexes in the base \(\{e_1, \ldots ,e_r\}\). By construction \(\nu _2(e_2) \ge \nu _2(e_4) \ge \nu _2(e_6)\), and so on. Observe that the element of \(\{e_1, \ldots ,e_r\}\) with maximum valuation \(\nu _2\) belongs to the output of Algorithm 4. Therefore, \(e_2\), or \(e_1\) if \(\nu _2(e_1) > \nu _2(e_2)\), must belong to I. One cannot guarantee that \(e_4\) belongs to I because \(e_4\) can be paired with \(e_2\). In that case only \(e_2\) is kept. However, \(e_4\) and \(e_6\) cannot be both paired with \(e_2\), meaning that I must contain an element valued at least \(\nu _2(e_6)\). Using this argument repeatedly, I must contain an element valued at least \(\nu _2(e_{4i+2})\) for \(i=1..\lfloor \frac{r-2}{4}\rfloor \).
$$\begin{aligned} \nu _2(I) \ge \nu _2(e_2) + \sum _{i=1}^{\lfloor (r-2)/4\rfloor } \nu _2(e_{4i+2}) \end{aligned}$$
(28)
Take a most preferred base \(B^*=\{b_1, \ldots ,b_r\}\) for agent 2 where \(\nu _2(b_1) \ge \nu _2(b_2) \ge \ldots \ge \nu _2(b_r)\). Use Lemma 1 and the fact that the matroid is simple to get that
$$\begin{aligned} \nu _2(e_2) \ge \nu _2(b_1), \text{ and } \nu _2(e_{4i+2}) \ge \nu _2(b_{4i+2}) \text{ for } i \in \{1, \ldots ,\lfloor (r-2)/4\rfloor \} \end{aligned}$$
(29)
Inequalities (28) and (29) give
$$\begin{aligned} \nu _2(I) \ge \nu _2(b_1) + \sum _{i=1}^{\lfloor (r-2)/4\rfloor } \nu _2(b_{4i+2}) \end{aligned}$$
(30)
Consider the set \(B_1=\{b_1\} \cup \{b_{4i+2} : 1 \le i \le \lfloor (r-2)/4\rfloor \}\), subset of \(B^*\). Use \( \nu _2(b_{1}) \ge \nu _2(b_{j})\) for \(j \in \{1,2,3,4,5\}\) to get that
$$\begin{aligned} 5 \nu _2(b_{1}) \ge \nu _2(\{b_{1},b_{2},b_{3},b_{4},b_{5}\}) \end{aligned}$$
(31)
For \(i \in \{1,\ldots ,\lfloor \frac{r-2}{4} \rfloor \}\), deduce from \(\nu _2(b_{4i+2}) \ge \nu _2(b_{j})\) for \(4i+2 \le j \le \max (4i+5, r)\) that
$$\begin{aligned} 4 \nu _2(b_{4i+2}) \ge \sum _{j=4i+2}^{\max (4i+5, r)}\nu _2(b_{j}) \end{aligned}$$
(32)
Combine Inequalities (31) and (32) to get that \(5\nu _2(b_{1})+4 \sum _{i=1}^{\lfloor \frac{r-2}{4} \rfloor } \nu _2(b_{4i+2}) \ge \nu _2(B^*)\). The lefthand part being upper bounded by \(5\nu _2(B_1)\), we obtain
$$\begin{aligned} 5 \nu _2(B_1) \ge \nu _2(B^*) \end{aligned}$$
(33)
Use Inequalities (30) and (33) to obtain \(5 \nu _2(I) \ge \nu _2(B)\), i.e. I is 1 / 5-approximate for agent 2.

Different approximation ratios are reached by Algorithm 4 when \(r\in \{3,4\}\). When \(r=3\), the solution is of size 2 and it contains the most preferred element of each agent. Therefore, it is 1 / 3-approximate for both agents by Lemma 1. When \(r=4\), the solution is of size 3. It contains the first and third preferred element of agent 1, and the most preferred element of agent 2. Therefore, it is 1 / 2-approximate for agent 1 and 1 / 4-approximate for agent 2 by Lemma 1.

The case \(r=2\) corresponds to step 3 of Algorithm 4. Both agents have inserted their most preferred element. Thus, the solution is 1 / 2-approximate for both agents because the matroid is simple. \(\square \)

Proposition 5

There exist instances for which the approximation ratios given in Theorem 8 are reached by Algorithm 4.

Proof

For \(r \in \{2,3,4\}\), use the instances depicted on Fig. 3.

For \(r=5\), take the graphic matroid associated with the graph of Fig. 5. A most preferred base for agent 1 consists of the edges \(\{(a,b),(b,f),(f,e),(b,c),(c,d)\}\) (plain edges). A most preferred base for agent 2 consists of the edges \(\{(a,f),(a,e),(b,e),(e,c),(e,d)\}\) (dotted edges). Agents 1 and 2 value their most preferred base 4 and 5, respectively. Algorithm 4 may take the edges: (ab), (ae), (ef), (ec), (cd). The reordering with respect to \(\nu _1\) can be (ab), (ef), (cd), (ec), (ae). The rank being odd, I consists of (ab), the edge that agent 2 prefers between (ef) and (cd), that is (cd), and one edge out of \(\{(e,c),(a,e)\}\), say (ec). Therefore, \(\nu _1(I)=\nu _2(I)=1+\epsilon \). When \(\epsilon \rightarrow 0\), I is 1 / 4-approximate for agent 1 and 1 / 5-approximate for agent 2.

Examples with \(r >5\) and r odd can be derived from the graph of Fig. 5. Indeed, add two new edges valued (0, 0) between one node, say a, and two new nodes. This operation increases the rank by two units (r remains odd) and it can be done as many times as necessary. For this extended instance, the previous paragraph indicates that Algorithm 4 can output a solution which is 1 / 4-approximate for agent 1 and 1 / 5-approximate for agent 2.

For r even and \(r\ge 6\), take the graphic matroid associated with the graph of Fig. 4. The rank can be singly even (\(r=4q+2\)) or doubly even (\(r=4(q+1)\)), which defines a parameter q is each case. Agent 1 has value 1 for every edge \(f_i\) such that \(1 \le i \le 4q+2\) and her value for the other edges is 0. In particular, when \(r=4(q+1)\), agent 1 has value 0 for the edges \(f_{4q+3}\) and \(f_{4q+4}\). Agent 2 has value 1 for every edge \(g_i\) such that \(1 \le i \le 5\) and her value for the other edges is 0. A most preferred base for agents 1 and 2 consists of the edges \(\{f_1,\ldots ,f_r\}\) (plain edges) and the edges \(\{g_1,\ldots ,g_r\}\) (dotted edges), respectively. Agents 1 and 2 value their most preferred base \(4q+2\) and 5, respectively. Algorithm 4 may take the edges: \(f_1,g_1,f_3,g_4,f_5,g_6,\ldots ,f_{r-1},g_r\). The reordering with respect to \(\nu _1\) can be \(f_1,f_3,f_5,\ldots , f_{r-1},g_r, \ldots , g_4,g_1\). The rank being even, I consists of the first two edges \(\{f_1,f_3\}\), plus the edge that agent 2 prefers in each pair of consecutive edges. The algorithm returns a solution which contains \(q+1\) edges valued 1 by agent 1 and one edge valued 1 by agent 2 which are \(\{f_1, f_3, f_7, \ldots , f_{4q-1}\}\) and \(\{g_1\}\), respectively. Thus, the solution is \(\frac{q+1}{4q+2}\)-approximate for agent 1 and 1 / 5-approximate for agent 2. \(\square \)

Fig. 5

A graphic matroid with rank \(\ge 5\). The first and second coordinates are the valuation of the first and second agent for the edges, respectively. Every dotted edge is valued 0 and 1 by agents 1 and 2, respectively

5 Conclusion and future works

This article is devoted to the existence and computation of succinct agreeable sets under additional matroidal constraints. An s-agreeable set of cardinality \(\lceil \frac{nr}{n+1}\rceil \) always exists if the agents’ preferences are monotonic, and no better (worst case) bound on the cardinality can be found unless it applies to a special case. When the agents have responsive preferences, an s-agreeable set of cardinality \(\lceil \frac{nr}{n+1}\rceil \) can be efficiently computed with round-robin.

A w-agreeable set of cardinality \(\lceil \frac{r+n-1}{2}\rceil \) always exists if the agents preferences are monotonic, and no better (worst case) bound on the cardinality can be found unless it applies to a special case. When there are two agents with responsive preferences, a w-agreeable of cardinality \(\lceil \frac{r+1}{2}\rceil \) can be efficiently computed with Algorithm 4.

Here, \(\lceil \frac{n \, r}{n+1} \rceil \) and \(\lceil \frac{r+n-1}{2}\rceil \) are worst case bounds. There exist instances for which our algorithms fail to return an agreeable set of smallest cardinality. A future direction would be to design algorithms which output (strong or weak) agreeable sets of smallest cardinality. This was recently shown as a computationally difficult task, even when the difficulty coming from the representation of \(\succeq _\ell \), \(\forall \ell \in N\), is put aside (Manurangsi and Suksompong 2017). However, an agreeable set of approximately optimal size5 can be computed efficiently if the matroid is free (Manurangsi and Suksompong 2017). Thus, is it possible to efficiently compute (strong or weak) agreeable sets of approximately minimum size?

Given the preference of every agent over \({\mathcal {E}}\), can we extend Theorem 5 and produce a succinct w-agreeable set for more than two agents? The case of three agents is resolved in Suksompong (2016) where a polynomial number of queries to preference oracle can be made in order to partially elicit the agents’ preferences.

We have shown that for two agents with additive preferences, agreeability and approximation can be combined. An interesting question is whether the approximation ratios of Tables 1 and 2 can be improved. Preliminary answers for s-agreeability can be provided when \(r\in \{2,3,4\}\). For \(r=2\), consider the partitional matroid such that \({\mathcal {E}}=\{e_1,e_2,e_3,e_4\}\), and \({\mathcal {I}}=\{I \subseteq {\mathcal {E}}: |I\cap \{e_1,e_2\}| \le 2 \}\). Suppose agent \(i \in \{1,2\}\) has utility 1 for \(e_i\) and \(e_{i+2}\), and utility 0 for the other elements. The optimal utility is 2 for both agents. No algorithm can outperform round-robin because no feasible solution of cardinality \(\lceil \frac{2 \cdot 2}{3}\rceil =2\) is better than 1 / 2-approximate for both agents. For \(r \in \{3,4\}\), similar arguments can be used with the instances of Fig. 3 (rank 3 and 4).

Another question is whether approximately good agreeable sets for more than two agents can be computed. We believe that there is room for positive results if every group of k elements is independent, where k is an upper bound on the number of agents. Matroids having this property are said to be k-simple (Gourvès et al. 2015).

Footnotes

  1. 1.

    A matching saturates a set if it is incident to all its elements.

  2. 2.

    Note that the definition of a necessarily s-agreeable set applies to responsive preferences.

  3. 3.

    The algorithm, inspired of (Suksompong 2016, Theorem 1), takes pairs of consecutives elements of \(\{f_1, \ldots ,f_r\}\) and keeps the one preferred by the second agent. The objective is that both agents prefer the elements that are kept to the elements that are discarded.

  4. 4.

    Note that if \(g_2\) or \(g_3\) is added after \(f_3\), then a cycle would be created.

  5. 5.

    The approximation ratio is \(O(m \ln \ln m / \ln m)\) where m is the number of items, and no polynomial-time \(o(m/ \ln m)\)-approximate algorithm exists.

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© Springer Science+Business Media, LLC, part of Springer Nature 2018

Authors and Affiliations

  1. 1.CNRS, PSL, LAMSADEUniversité Paris-DauphineParisFrance

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