Dividing Conflicting Items Fairly

Ayumi Igarashi Pasin Manurangsi The University of Tokyo Google Research igarashi@mist.i.u-tokyo.ac.jp pasin@google.com Hirotaka Yoneda The University of Tokyo yoneda-h@g.ecc.u-tokyo.ac.jp

Abstract

We study the allocation of indivisible goods under conflicting constraints, represented by a graph. In this framework, vertices correspond to goods and edges correspond to conflicts between a pair of goods. Each agent is allocated an independent set in the graph. In a recent work of Kumar et al. [21], it was shown that a maximal EF1 allocation exists for interval graphs and two agents with monotone valuations. We significantly extend this result by establishing that a maximal EF1 allocation exists for any graph when the two agents have monotone valuations. To compute such an allocation, we present a polynomial-time algorithm for additive valuations, as well as a pseudo-polynomial time algorithm for monotone valuations. Moreover, we complement our findings by providing a counterexample demonstrating a maximal EF1 allocation may not exist for three agents with monotone valuations; further, we establish NP-hardness of determining the existence of such allocations for every fixed number $n\geq 3$ of agents. All of our results for goods also apply to the allocation of chores.

1 Introduction

How can we allocate a resource fairly? This problem was first formalized by the pioneering work of Steinhaus [25] and has since been extensively studied in the fields of economics, mathematics, and computer science under the umbrella of fair division. Applications of fair division arise in many real-life situations, including the allocation of courses among students [11], the division of family inheritance among family members [15], and the division of household chores between couples [19].

A central notion of fairness in fair division is envy-freeness, which requires that every agent is allocated their most preferred bundle in the allocation. However, such a fairness guarantee is impossible to achieve when dealing with indivisible resources, such as courses, houses, or tasks. Consequently, recent literature on discrete fair division has focused on approximate fairness, exploring various concepts and algorithmic results [1]. One particular influential relaxation of envy-freeness is, envy-freeness up to one good (EF1), introduced by Budish [10], allowing agents to remove one good from others’ bundle to eliminate envy. This concept has garnered significant attention over the past decade. It is known that for general classes of monotone valuations, an EF1 allocation exists and can be computed efficiently [23].

Most studies on fair division assume that any allocation is feasible. While this assumption may hold in some cases, many practical scenarios involve constraints that restrict the structure of allocations. For instance, consider the allocation of multiple offices among several people over different periods of time. If the time intervals associated with two offices overlap, they cannot be assigned to the same person. Similar constraints arise in job scheduling, where overlapping shifts cannot be allocated to the same employee. Another example is the allocation of players to sports teams. If two players have overlapping areas of expertise, it is preferable not to assign them to the same team. A versatile framework for modeling such constraints, explored in a series of recent papers [12, 18, 21], represents conflicts among indivisible resources using a graph structure, where vertices correspond to resources and edges represent conflicts.

Conflicting constraints introduce new challenges, as standard fairness and efficiency concepts often become unattainable. Notably, canonical efficiency concepts such as completeness and Pareto-optimality are incompatible with EF1 under these constraints. In fact, with conflicting constraints, a complete allocation that assigns all goods may not always exist. Furthermore, even if such an allocation exists, there are simple instances where a complete EF1 allocation is unattainable [18].

Kumar et al. [21] studied chore allocation under conflicting constraints, observing that even on a path, EF1 is incompatible with Pareto-optimality for two agents with identical additive valuations.¹¹1In [21], Pareto-optimality is defined with respect to maximal allocations. To address this limitation, they introduced the notion of maximality. A maximal allocation ensures that no unassigned item can be feasibly allocated to some agent. Kumar et al. showed that for interval graphs—a common structure in job scheduling—a maximal EF1 allocation among two agents always exists and can be efficiently found for monotone valuations. However, the existence of a maximal EF1 allocation for more general graph families remains unresolved, offering a rich avenue for further research.

Our contributions.

In this paper, we study the allocation of indivisible goods under conflicting constraints. Our goal is to identify conditions under which a maximal EF1 allocation exists and can be efficiently computed. Our main contributions are as follows:

1.

Two-Agent-Case: We significantly extend the result of Kumar et al. [21] by establishing that a maximal EF1 allocation exists for any graph when the agents have monotone valuations. Further, we develop efficient algorithms for finding such allocations, including a polynomial-time algorithm for additive valuations and a pseudo-polynomial-time algorithm for monotone valuations. Note that the two-agent case is of particular importance in fair division, with various applications, including inheritance division, house-chore division, and divorce settlements [7, 8, 19].
2.

Three or More Agents: We establish a sharp dichotomy from the two-agent case in terms of both existence and computational complexity. First, we provide an example where a maximal EF1 allocation fails to exist, even for three agents with monotone valuations. While Hummel and Hetland [18] previously identified a counterexample for four agents, no such example was known for three agents. We also prove the NP-hardnesss of determining the existence of a maximal EF1 allocation for a fixed number $n\geq 3$ of agents with monotone valuations.
3.

Chore Allocation: Finally, we consider the problem of chore allocation, where each agent has a monotone non-increasing valuation. We show that the existence of a maximal EF1 allocation under identical valuations directly translates from the goods case, establishing that all our results for goods hold for chores as well.

Related work.

There is a growing body of research on fair division under constraints. For a comprehensive survey on the topic, see Suksompong [26]. Conflicting constraints in the context of fair division were introduced by Chiarelli et al. [12] and have been further explored by [18, 21, 5, 22]. Chiarelli et al. [12] explored different fairness objectives from ours, focusing on partial allocations that maximize the egalitarian social welfare—defined as the value of a bundle received by the worst-off agent. Hummel and Hetland [18] studied complete allocations satisfying fairness criteria such as EF1 and MMS (maximin fair share). Biswas et al. [5] generalized the model of [18], taking into account capacity of resources. Li et al. [22] and Kumar et al. [21] considered an interval scheduling problem, with the goal of achieving fairness concepts such as EF1 and MMS. While Li et al. [22] focused on goods allocation with flexible intervals, Kumar et al. [21] examined chore allocation.

Our problem is closely related to the problem of equitable coloring in graph theory, which corresponds to a complete EF1 allocation when agents have uniform valuations (namely, all goods are equally valued $1$ ). In Appendix B, we make this connection more precise and show that for such valuations, a maximal EF1 allocation exists on trees.

A related type of constraints to conflicting constraints is the connectivity constraints of a graph [4, 6], where each agent receives a connected bundle of a graph. Note that while connectivity constraints imposed by a sparse graph such as a tree allow fewer feasible allocations, in our setting, sparsity implies greater flexibility, as it increases the number of feasible allocations.

2 Preliminaries

For any natural number $s\in\mathbb{N}$ , let $[s]=\{1,2,\ldots,s\}$ .

Problem instance.

We use $M=[m]$ to denote the set of goods and $N=[n]$ to denote the set of agents. Let $G=(M,E)$ denote an undirected graph, where each vertex corresponds to a good and each edge corresponds to a conflict. Each agent $i$ has a valuation function $v_{i}:2^{M}\rightarrow\mathbb{R}_{+}$ ; here, $\mathbb{R}_{+}$ is the set of non-negative reals. We assume that $v_{i}(\emptyset)=0$ . For simplicity, the valuation of a single good $g\in M$ , $v_{i}(\{g\})$ , is denoted by $v_{i}(g)$ . An instance of our problem is given by the tuple $(N,M,\mathcal{V},G)$ where $\mathcal{V}=(v_{1},v_{2},\dots,v_{n})$ denotes a valuation profile. We use $K_{s,t}$ to denote a complete bipartite graph with a bipartition in which one part contains $s$ vertices and another part includes $t$ vertices.

Valuation function.

A valuation function $v_{i}$ is monotone non-decreasing if $v_{i}(S)\leq v_{i}(T)$ for every $S\subseteq T\subseteq M$ . Unless specified otherwise, we refer to such a function simply as monotone. It is additive if $v_{i}(S)=\sum_{g\in S}v_{i}(g)$ holds for every $S\subseteq M$ and $i\in N$ . A valuation profile $\mathcal{V}$ is called identical if every agent $i\in N$ has the same valuation function; in this case, we denote this function by $v$ . Let $T(m)$ denote the time to compute valuation $v_{i}(S)$ for a given $S\subseteq M$ .

Allocation.

An allocation is an ordered subpartition $\mathcal{A}=(A_{1},\ldots,A_{n})$ of $M$ where for every pair of distinct agents $i,j\in N$ , $A_{i}\cap A_{j}=\emptyset$ , $\cup_{i\in N}A_{i}\subseteq M$ , and for each $i\in N$ , $A_{i}$ is an independent set of $G$ , namely, there is no pair of goods in $A_{i}$ that are adjacent to each other. The subset $A_{i}$ is called the bundle of agent $i$ . An allocation is complete if all goods are allocated, i.e., $\cup_{i\in N}A_{i}=M$ .

Fairness and efficiency notions.

An allocation is envy-free if for every pair of agents $i,j\in N$ , we have $v_{i}(A_{i})\geq v_{i}(A_{j})$ [13, 14]. It is called envy-free up to one good (EF1) if for every pair of agents $i,j\in N$ , either $A_{j}=\emptyset$ , or there exists some good $g\in A_{j}$ such that $v_{i}(A_{i})\geq v_{i}(A_{j}\setminus\{g\})$ [10, 23].

As discussed in the Introduction, observe that a complete allocation may not always exist (e.g., consider a complete graph $K_{2}$ with one agent). Further, even when a complete allocation exists, a complete EF1 allocation may not exist: For instance, in a setting with $n$ agents and a star where the center has a value of $0$ and each of the $n+1$ leaves has a value of $1$ , an agent receiving the center cannot receive any other good, while at least one agent receives two or more leaves.

Besides completeness, another commonly used notion of efficiency in fair division is Pareto-optimality. Similar to the chore setting [21], EF1 is incompatible with Pareto-optimality for goods instances. Specifically, an allocation $\mathcal{A}^{\prime}$ Pareto-dominates another allocation $\mathcal{A}$ if $v_{i}(A^{\prime}_{i})\geq v_{i}(A_{i})$ for every $i\in N$ and the inequality is strict for at least one agent. An allocation $\mathcal{A}$ is Pareto-optimal (PO) if there is no other allocation that Pareto-dominates it. Consider two agents and a cycle of four goods with values $1,3,1,3$ in order. In any EF1 allocation, no agent should receive both goods of value $3$ , and no agent can receive two goods of different values due to conflicting constraints. However, such an allocation is Pareto-dominated by assigning the two high-value goods to one agent and the two low-value goods to the other.

As in Kumar et al. [21], we therefore consider a relaxed efficiency notion of maximality. An allocation $\mathcal{A}$ is maximal if for every agent $i\in N$ and every unallocated good $g\in M\setminus\bigcup_{i\in N}A_{i}$ , $g$ is adjacent to some good in $A_{i}$ . Our goal is to achieve an allocation that simultaneously satisfies maximality and EF1.

3 Two Agents

In this section, we consider the case of two agents.

3.1 Cut-and-Choose Protocol

When there are two agents, we can use a well-known strategy called cut-and-choose protocol [9].

Theorem 1.

Suppose that a maximal EF1 allocation always exists and can be computed in $\tau$ time for instances with two agents and identical valuations. Then, a maximal EF1 allocation always exists and can be computed in $\tau+2T(m)$ time for instances with two agents.

Proof.

Let $(A_{1},A_{2})$ a maximal EF1 allocation in a hypothetical scenario that valuations of both agents are $v_{1}$ . Then, agent $2$ chooses a preferred bundle, leaving the reminder for $1$ . The resulting allocation is maximal and EF1. ∎

Now, the question is whether a maximal EF1 allocation exists for identical valuations. In subsections 3.2-3.5, we assume that the valuations are identical and monotone, and valuations for both agents are represented by $v(S)\ (S\subseteq M)$ .

3.2 Proof Strategy of Kumar et al.

To describe our proof strategy, let us (informally) review Kumar et al.’s proof of the existence of maximal EF1 allocation when $G$ is a path. At a high level, the idea of the proof is to construct a chain of maximal allocations $\mathcal{A}^{(0)},\dots,\mathcal{A}^{(m-1)}$ , as illustrated in Figure 1, satisfying the following two properties: (i) adjacent allocations only “differ slightly” and (ii) $(A^{(0)}_{1},A^{(0)}_{2})=(A^{(m-1)}_{2},A^{(m-1)}_{1})$ . The latter implies that the signs of $v(A^{(0)}_{1})-v(A^{(0)}_{2})$ and $v(A^{(m-1)}_{1})-v(A^{(m-1)}_{2})$ are different. Therefore, there exists an $i$ that the signs of $v(A^{(i)}_{1})-v(A^{(i)}_{2})$ and $v(A^{(i+1)}_{1})-v(A^{(i+1)}_{2})$ are different. At this point, they show that at least one of $\mathcal{A}^{(i)}$ or $\mathcal{A}^{(i+1)}$ must be EF1, which shows the existence of a maximal EF1 allocation. This concludes the proof overview of [21] for path graphs. They also extended this method to interval graphs, although the construction of the chain of maximal allocations becomes significantly more involved. Indeed, as explained and formalized below, our main contribution is to give a construction of a chain for any graph.

Refer to caption — Figure 1: An example of chain of allocations for a path graph, $\mathcal{A}^{(0)},\dots,\mathcal{A}^{(5)}$ . The number written in each vertex is a valuation of the corresponding good. Red and blue vertices are those assigned to agent $1$ and $2$ , respectively.

3.3 Useful Definitions and Lemmata

Our construction will require several generalizations of definitions and lemmata from Kumar et al. [21]. We believe that these tools can be useful beyond the context of our work. Firstly, we use the following definition of “adjacency”. Compared to [21], our definition (Definition 2) is more relaxed, in that it does not place any requirement on $|A^{\prime}_{1}\setminus A_{1}|$ and $|A_{2}\setminus A^{\prime}_{2}|$ whereas Kumar et al.’s enforces that these are at most one. Such a relaxation is crucial since our “chain” constructed below violates the aforementioned Kumar et al.’s condition.

Definition 2.

A pair $(\mathcal{A},\mathcal{A}^{\prime})$ of allocations is ordered adjacent if $|A_{1}\setminus A^{\prime}_{1}|\leq 1$ and $|A^{\prime}_{2}\setminus A_{2}|\leq 1$ .

The following is the key lemma that enables to find an EF1 allocation at the point when $v(A_{1})-v(A_{2})$ crosses from positive to negative.

Lemma 3.

Let $(\mathcal{A},\mathcal{A}^{\prime})$ be any ordered adjacent pair of allocations. Further, assume that the following conditions hold:

1.

$v(A_{1})\geq v(A_{2})$
2.

$v(A^{\prime}_{1})\leq v(A^{\prime}_{2})$

Then, at least one of $\mathcal{A}$ or $\mathcal{A}^{\prime}$ must be EF1.

Proof.

Suppose that the allocation $\mathcal{A}$ is not EF1. Since $v(A_{1})\geq v(A_{2})$ , agent $a_{2}$ envies agent $a_{1}$ even if one good is removed from $A_{1}$ . Since $|A_{1}\setminus A^{\prime}_{1}|\leq 1$ , we must have $v(A_{1}\cap A^{\prime}_{1})=v(A_{1}\setminus(A_{1}\setminus A^{\prime}_{1}))\geq v(A_{2})$ . As a result,

$\textstyle v(A^{\prime}_{1})$	$\textstyle\geq v(A_{1}\cap A^{\prime}_{1})$
	$\textstyle\geq v(A_{2})$
	$\textstyle\geq v(A^{\prime}_{2}\cap A_{2})=v(A^{\prime}_{2}\setminus(A^{\prime}_{2}\setminus A_{2})),$	(1)

where the first and third inequalities hold because $v$ is monotone. Now, consider the allocation $\mathcal{A}^{\prime}$ :

•

Since $|A^{\prime}_{2}\setminus A_{2}|\leq 1$ , (1) implies that agent $1$ does not envy agent $2$ after removing a good from $A^{\prime}_{2}$ .
•

Agent $2$ does not envy agent $1$ since $v(A^{\prime}_{1})\leq v(A^{\prime}_{2})$ .

Therefore, this allocation is EF1. ∎

We generalize Kumar et al.’s method by defining a gapless chain, and prove that any gapless chain must contain an EF1 allocation. We stress again that our requirements are weaker than the chain used in Kumar et al.’s²²2Namely, we use a weaker adjacency notion and we only require the sign flip (first two conditions) whereas Kumar et al. require that $(A^{(0)}_{1},A^{(0)}_{2})=(A^{(k)}_{2},A^{(k)}_{1})$ ., but still suffices to ensure EF1 for identical valuations.

Definition 4.

A sequence of allocation $\mathcal{A}^{(0)},\dots,\mathcal{A}^{(k)}$ is a gapless chain if it satisfies the following conditions:

1.

$v(A^{(0)}_{1})\geq v(A^{(0)}_{2})$ .
2.

$v(A^{(k)}_{1})\leq v(A^{(k)}_{2})$ .
3.

$(\mathcal{A}^{(i-1)},\mathcal{A}^{(i)})$ is ordered adjacent for every $i\in[k]$ .

Lemma 5.

If $\mathcal{A}^{(0)},\dots,\mathcal{A}^{(k)}$ is a gapless chain, there exists an $i\in\{0,\dots,k\}$ that $\mathcal{A}^{(i)}$ is EF1.

Proof.

From the first two conditions, there exists $i\in[k]$ that $v(A^{(i-1)}_{1})\geq v(A^{(i-1)}_{2})$ and $v(A^{(i)}_{1})\leq v(A^{(i)}_{2})$ . Lemma 3 then implies that at least one of $\mathcal{A}^{(i-1)}$ or $\mathcal{A}^{(i)}$ is EF1. ∎

Given Lemma 5, our main task is thus to (efficiently) construct a gapless chain of maximal allocations (for any graph $G$ ). We devote the remainder of this section to this task.

3.4 Proof of Existence

Algorithm 1 ChainEF1

(S;G=(M,E),v)

S=\{s_{1},\dots,s_{k}\}

is a maximal independent of

G

2:for

t\in M\setminus S

\Gamma_{t}:=\{i\in[k]\mid\{s_{i},t\}\in E\}

\triangleright

Non-empty since

S

is a maximal independent set

p_{t}=\min_{i\in\Gamma_{t}}i

q_{t}=\max_{i\in\Gamma_{t}}i

6:end for

X_{1}\leftarrow\emptyset

8:for

t\in M\setminus S

in increasing order of

q_{t}

9: if

t

has no neighbor in

X_{1}

then

10:

X_{1}\leftarrow X_{1}\cup\{t\}

11: end if

12:end for

13:

X_{2}\leftarrow\emptyset

14:for

t\in M\setminus S

in decreasing order of

p_{t}

15: if

t

has no neighbor in

X_{2}

then

16:

X_{2}\leftarrow X_{2}\cup\{t\}

17: end if

18:end for

19:for

i=0,\dots,k

20:

A^{(i)}_{1}=\{s_{i+1},\dots,s_{k}\}\cup\{t\in X_{1}\mid q_{t}\leq i\}

21:

A^{(i)}_{2}=\{s_{1},\dots,s_{i}\}\cup\{t\in X_{2}\mid p_{t}>i\}

22: if

\mathcal{A}^{(i)}=(A^{(i)}_{1},A^{(i)}_{2})

is EF1 then

23: return

\mathcal{A}^{(i)}

24: end if

25:end for

26:return NULL

Now, we prove that, a maximal EF1 allocation always exists for two agents, as stated below.

Theorem 6.

For $n=2$ agents, any graph $G$ and any monotone valuation $v$ , there exists a maximal EF1 allocation.

Although we only claim the existence in the above theorem, we will in fact present an algorithm for finding such an allocation, as its running time will be discussed in the next section. Our algorithm is presented in Algorithm 1 where the input $S$ can be any maximal independent set of $G$ . To prove Theorem 6, we will use one that maximizes $v(S)$ . In Figure 2, we provide an example of a chain constructed by the algorithm.

We now prove a couple of crucial lemmata. Starting with the fact that each allocation is valid and maximal:

Lemma 7.

For every $i\in\{0,\dots,k\}$ , $\mathcal{A}^{(i)}$ is a valid maximal allocation.

Proof.

First, we prove that the allocation is valid:

•

No good is assigned to both agents. This is obvious for goods in $S$ . For good $t\in M\setminus S$ , since $q_{t}\geq p_{t}$ , at most one of the conditions $q_{t}\leq i$ or $p_{t}>i$ can hold; thus, it is assigned to at most one agent.
•
No adjacent goods are assigned to the same agent. Consider two goods $g,g^{\prime}\ (g\neq g^{\prime})$ that are assigned to agent $1$ and consider the following cases:
1. 1.
  
  Both goods are in $S$ . They are not adjacent because $S$ is an independent set.
2. 2.
  
  Both goods are in $M\setminus S$ . They are not adjacent because $g,g^{\prime}\in X_{1}$ and $X_{1}$ is an independent set.
3. 3.
  
  One good is from $S$ and the other is from $M\setminus S$ . Assume w.l.o.g. that $g\in S$ and $g^{\prime}\in M\setminus S$ . From this, we must have $g\in\{s_{i+1},\dots,s_{k}\}$ and that $q_{g^{\prime}}\leq i$ . By the definition of $q_{g^{\prime}}$ , this ensures that $g,g^{\prime}$ are not adjacent.

Next, we prove maximality of $\mathcal{A}^{(i)}$ , i.e., that no good in $M\setminus(A_{1}^{(i)}\cup A_{2}^{(i)})$ can be assigned to one of the agents. Let $g\in M\setminus(A_{1}^{(i)}\cup A_{2}^{(i)})$ be any unassigned good. Consider three following cases:

1.

$q_{g}\leq i$ . Since $g$ is adjacent to $s_{q_{g}}\in A^{(i)}_{2}$ , $g$ cannot be assigned to agent $2$ . Moreover, since $g\notin A^{(i)}_{1}$ , it must be that $g\notin X_{1}$ . From how $X_{1}$ is constructed, there exists $t\in X_{1}$ such that $q_{t}\leq q_{g}$ and $t$ is adjacent to $g$ . As $t\in A^{(i)}_{1}$ , $g$ cannot be assigned to agent $1$ .
2.

$p_{g}>i$ . This case follows from an analogous argument to the first case.
3.

$p_{g}\leq i<q_{g}$ . In this case, $g$ is adjacent to $s_{p_{g}}\in A^{(i)}_{2}$ and $s_{q_{g}}\in A^{(i)}_{1}$ . Thus, $t_{j}$ cannot be assigned. ∎

Next, we show that $\mathcal{A}^{(0)},\dots,\mathcal{A}^{(k)}$ is a gapless chain. However, this requires an additional requirement that $v(S)$ is no smaller than $v(X_{1}),v(X_{2})$ .

Lemma 8.

If $v(S)\geq v(X_{1})$ and $v(S)\geq v(X_{2})$ , then $\mathcal{A}^{(0)},\dots,\mathcal{A}^{(k)}$ is a gapless chain.

Proof.

Note that $\mathcal{A}^{(0)}=(S,X_{2})$ and $\mathcal{A}^{(k)}=(X_{1},S)$ . Thus, the first two conditions are satisfied by our assumption $v(S)\geq v(X_{1}),v(S)\geq v(X_{2})$ . Finally, for every $i\in[k]$ , we can see that $A_{1}^{(i-1)}\setminus A_{1}^{(i)}=\left\{s_{i}\right\}$ and $A_{2}^{(i)}\setminus A_{2}^{(i-1)}=\left\{s_{i}\right\}$ . Thus, $(\mathcal{A}^{(i-1)},\mathcal{A}^{(i)})$ is ordered adjacent. ∎

Lemmas 5, 7 and 8 together immediately imply the following:

Lemma 9.

If $v(S)\geq v(X_{1})$ and $v(S)\geq v(X_{2})$ , then Algorithm 1 outputs a maximal EF1 allocation.

Our main theorem of this section (Theorem 6) now then follows easily from Lemma 9 by choosing an appropriate $S$ .

Proof of Theorem 6.

Let $S$ be a maximal independent set of $G$ that maximizes $v(S)$ . Since $X_{1},X_{2}$ are independent set and $v$ is monotone, we have $v(S)\geq v(X_{1})$ and $v(S)\geq v(X_{2})$ . Thus, Lemma 9 ensures that running Algorithm 1 on input $S$ yields a maximal EF1 allocation. ∎

3.5 Algorithm

Recall in the proof of Theorem 6 that we pick $S$ to be a maximal independent set with largest valuation. Doing so trivially would require enumerating through all $2^{m}$ subsets of $M$ . In this section, we give a simple algorithm (Algorithm 2) that significantly improves upon this running time. In particular, it runs in polynomial-time for additive valuations and pseudo-polynomial time for general monotone valuations.

Algorithm 2 SwapEF1

(G=(M,E),v)

g^{*}\leftarrow\operatorname{argmax}_{g\in M}v(g)

S^{i}\leftarrow

any maximal independent set of

G

containing

g^{*}

i\leftarrow 0

4:while True do

5: if

\textsc{ChainEF1}(S^{i};G,v)\neq

NULL then

6: return

\textsc{ChainEF1}(S^{0};G,v)

7: end if

X^{i}_{1},X^{i}_{2}\leftarrow X_{1},X_{2}

\textsc{ChainEF1}(S^{i};G,v)

\ell\leftarrow\operatorname{argmax}_{\ell^{\prime}\in\{1,2\}}v(X^{i}_{\ell^{\prime}})

10:

S^{i+1}\leftarrow

any maximal independent set containing

X^{i}_{\ell}

11:

i\leftarrow i+1

12:end while

Our algorithm running time is stated below in Theorem 10. Note that in the second case $B$ is the number of different values the valuation function can take. This implies that, if the valuations $v(S)$ are all integers, the running time is pseudo-polynomial.

Theorem 10.

When there are $n=2$ agents, there exists an algorithm that can find a maximal EF1 allocation and its running time is as follows:

•

$O(3^{m/3}\cdot(m\cdot T(m)+|E|))$ for monotone valuations,
•

$O(B\cdot(m\cdot T(m)+|E|))$ for monotone valuations such that the number of distinct values of $v(S)$ is at most $B$ (i.e. $|\{v(S)\mid S\subseteq M\}|\leq B$ ), and,
•

$O(m\log m\cdot(m\cdot T(m)+|E|))$ for additive valuations.

Proof.

Observe that the preprocessing time and the running time of each loop is $O(m\cdot T(m)+|E|)$ . Thus, it suffices to bound the number of iterations of the loop in each case.

•

For general monotone valuations, by Lemma 9, each loop either terminates or we must have $v(X^{i}_{\ell})>v(S^{i})$ . From monotonicity, this implies $v(S^{i+1})>v(S^{i})$ . This means that $S^{0},S^{1},\dots$ are all distinct maximal independent sets of $G$ . Since there are at most $3^{m/3}$ maximal independent sets [24], the number of iterations is at most $3^{m/3}$ .
•

Next, suppose that $v(S)$ can take at most $B$ distinct values. From the argument above, $v(S^{0}),v(S^{1}),\dots$ are strictly increasing and are thus distinct. As a result, the number of iterations is at most $B$ .
•

Finally, suppose that $v$ is additive. We claim that the following holds for each loop $i$ that does not terminate:

$\textstyle v(S^{i+1})>\frac{m}{m-1}\cdot v(S^{i}).$ (2)

Before we prove (2), let us first use it to bound the number of iterations. Notice that (2) implies that, after loop $i$ , we must have $v(S^{i+1})>\left(\frac{m}{m-1}\right)^{i+1}\cdot v(S^{0})\geq\left(\frac{m}{m-1}\right)^{i+1}\cdot v(g^{*})$ . Moreover, our choice of $g^{*}$ ensures that $v(S^{i+1})\leq m\cdot v(g^{*})$ . Thus, the number of iterations is at most $\log_{m/(m-1)}m+1=O(m\log m)$ .

To see that (2) holds, first recall from Algorithm 1 that, if Algorithm 1 returns NULL, it has considered either $(S^{i},X^{i}_{\ell})$ or $(X^{i}_{\ell},S^{i})$ already and has determined that this is not EF1. Since $v(X^{i}_{\ell})>v(S^{i})$ , this means that, for any good $g\in X^{i}_{\ell}$ , we must have $v(S)<v(X^{i}_{\ell}\setminus\left\{g\right\})$ . When we pick $g\in X^{i}_{\ell}$ with the largest $v(g)$ , we have

$v(g)\geq\frac{1}{\left|X^{i}_{\ell}\right|}v(X^{i}_{\ell})\geq\frac{1}{m}v(X^{i}_{\ell}).$

Hence, $v(S^{i})<v(X^{i}_{\ell}\setminus\{g\})\leq\frac{m-1}{m}v(X^{i}_{\ell})$ , proving (2). ∎

4 Three or More Agents

4.1 Negative Examples

While a maximal EF1 allocation always exists when there are two agents, it turns out that this is not the case for three agents or more.

{restatable}

theoremExampleThreeAgents For $n=3$ agents, there exists an instance with identical monotone valuations where no maximal EF1 allocation exists.

Proof.

Consider the following instance with $7$ goods. The graph consists of a complete bipartite graph $K_{3,3}$ with bipartition $X=\{1,2,3\}$ and $Y=\{4,5,6\}$ together with two edges $\{1,7\}$ and $\{4,7\}$ . Each of the three agents has an identical monotone valuation $v$ such that

•

$v(\emptyset)=0$ ;
•

$v(S)=1$ if $|S|=1$ and $S\in\{\{1\},\{4\}\}$ ;
•

$v(S)=2$ if $|S|=1$ and $S\not\in\{\{1\},\{4\}\}$ ;
•

$v(S)=3$ if $S$ is $\{2,7\}$ , $\{3,7\}$ , $\{5,7\}$ , or $\{6,7\}$ ;
•

$v(S)=4$ for any other $S\subseteq M$ .

Note that the symmetry on the graph and valuation holds: swapping $2$ and $3$ , swapping $5$ , $6$ , and swapping $\{1,2,3\}$ and $\{4,5,6\}$ all lead to the same instance, and swapping the bundles of agents does not change whether the allocation is EF1 or not because valuations are identical.

Specifically, consider any maximal allocation $\mathcal{A}$ . By maximality, good $7$ must be allocated to some agent since it has degree $2$ . Since agents have identical valuations, without loss of generality, suppose it is allocated to agent $1$ . Thus, agent $1$ can receive neither of good $1$ nor $4$ ; let us assume that if $1$ is allocated, then it is allocated to agent $2$ and if $4$ is allocated, then it is allocated to agent $3$ . Further, agent $1$ can receive a good from at most one part of the bipartition $X$ and $Y$ due to conflicting constraints. Without loss of generality, suppose it is $Y$ . Now consider the following cases.

1.

If none of the goods in $Y$ is allocated to agent $1$ , then $\mathcal{A}$ must allocate $X$ to agent $2$ and $Y$ to agent $3$ .
2.

Suppose that exactly one of the goods in $Y$ , say good $5$ , is allocated to agent $1$ . In this case, observe that $4$ must be allocated. This is because if $4$ is unallocated, this means that both agents $2$ and $3$ receive at least one good from $X$ and $6$ must be allocated to agent $1$ , a contradiction. Thus, the only possible maximal allocations are $(\{5,7\},\{1,2,3\},\{4,6\})$ and $(\{5,7\},\{6\},\{4\})$ . A similar argument holds when $6$ is allocated to agent $1$ due to the symmetry of the graph.
3.

Suppose that two goods, $5$ and $6$ , in $Y$ are allocated to agent $1$ . If $4$ is allocated, then the only possible maximal allocation is $(\{5,6,7\},\{1,2,3\},\{4\})$ . If $4$ is unallocated, this means that both agents $2$ and $3$ receive at least one good from $X$ and all goods in $X$ must be allocated to either of such agents. Thus, the only possible maximal allocations are $(\{5,6,7\},\{1\},\{2,3\})$ , $(\{5,6,7\},\{1,2\},\{3\})$ , together with $(\{5,6,7\},\{1,3\},\{2\})$ .

For this reason, there are only $6$ maximal allocations to consider (refer to Figure 3). All of them are not EF1, because:

•

Allocations #1, #4, #5, #6: one agent takes one good but there is another agent who takes three goods
•

Allocation #2: $v(\{5,7\})=3$ but $v(\{1,2\})=v(\{1,3\})=v(\{2,3\})=4$ .
•

Allocation #3: $v(4)=1$ but $v(5)=v(7)=2$ . ∎

The instance constructed in the proof of Theorem 4.1 uses an identical monotone valuation. It remains an open question whether a counterexample exists for three agents with (even identical) additive valuations.

For every number $n\geq 4$ of agents, Hummel and Hetland [18] presented an instance with identical additive valuations and a complete bipartite graph $K_{n-1,n-1}$ for which a complete EF1 allocation does not exist. In such cases, a complete allocation does exist, as the maximum degree of the graph is less than the number of agents. Thus, this counterexample implies that achieving both EF1 and maximality is impossible even when four agents have identical additive valuations. Here, we provide a smaller example using $K_{3,n-1}$ , which turns out to be smallest since for $m\leq n+1$ , there always exists a maximal EF1 allocation.

{restatable}

propositionExampleFourAgents For every number $n\geq 4$ of agents, there is an instance with identical additive valuations, $m=n+2$ , and $G=K_{3,n-1}$ where no maximal EF1 allocation exists.

Proof.

Consider the following instance with $m=n+2$ goods and a complete bipartite graph $K_{3,n-1}$ with the left part containing three goods and the right part containing $n-1$ goods. Let $a_{1},a_{2},a_{3}$ denote the three goods on the left side and $b_{1},b_{2},\ldots,b_{n-1}$ . Each agent has an identical additive valuation where

v(g)=\begin{cases}2&(g\in\{a_{1},a_{2},a_{3}\})\\ 3&(\text{otherwise}).\end{cases}

Consider an arbitrary maximal allocation $\mathcal{A}$ . Since the maximum degree of the graph is $n-1$ , all goods must be allocated in this allocation. Due to pigeonhole’s principle, there exists an agent X that cannot receive any of goods on the right.

•

Case 1: If there is only one such agent, goods $b_{1},b_{2},\ldots,b_{n-1}$ are allocated to $n-1$ other agents, which means that each of such agents takes one good of value $3$ . The remaining goods $a_{1},a_{2},a_{3}$ must be taken by agent X. Thus, agent X is envied by each of the other agents even after removing one of the goods $a_{1},a_{2},a_{3}$ from X’s bundle.
•

Case 2: If there are two or more such agents, there exists an agent Y who takes two or more goods among goods $b_{1},b_{2},\ldots,b_{n-1}$ ; further, there exists an agent Z who does not receive any of the goods on the right and takes at most one good from the left side. Then, agent Y is envied by agent Z after removing one of the goods from Y’s bundle.

Therefore, in either case, the allocation $\mathcal{A}$ is not EF1. ∎

{restatable}

propositionnPlusOne For $n$ agents with monotone valuations and $m$ goods with $m\leq n+1$ , there exists a maximal EF1 allocation.

Proof.

Suppose $m\leq n+1$ . Consider the following algorithm.

1.

For each agent $i=1,2,\dots,n$ in this order, run one cycle of the round-robin algorithm. Namely, agent $i$ chooses the most preferred good $g^{*}$ among the remaining goods (namely, $g^{*}\in\operatorname{argmax}_{g\in R}v_{i}(g)$ where $R$ is the set of remaining goods). If there is no remaining good, the agent does not receive any good.
2.

After that, at most one good $g$ remains. If $g$ can be feasibly assigned to some agent $i$ (namely, $g$ is not adjacent to the good received by $i$ in the first phase), assign it to any of such agents.

It is obvious that the resulting allocation is maximal. If $m\leq n$ , the resulting allocation is clearly EF1 since each agent receives at most one good. If $m=n+1$ , consider an agent $i$ who gets two goods. After removing the first good she has chosen, each of the other agents does not envy $i$ since the last good $g$ was not selected in the first phase. ∎

4.2 NP-Hardness

Now, we will show that it is NP-hard to decide whether a maximal EF1 allocation exists, as stated below.

Theorem 11.

Given the graph and valuations, determining whether a maximal EF1 allocation exists is NP-hard for:

1.

any fixed $n\geq 4$ , even when restricted to identical and additive valuation, and,
2.

$n=3$ , even when restricted to identical and monotone valuation.

In fact, our proof can transform any negative example into an NP-hardness result, as stated more precisely below.

Lemma 12.

Suppose that there exists an instance $\tilde{I}=([n],\tilde{M},\tilde{v},\tilde{G}=(\tilde{M},\tilde{E}))$ (with identical valuation) where the number $n$ of agents and the number $|\tilde{M}|$ of items are both constants, such that no maximal EF1 allocation exists. Then, it is NP-hard to decide whether a maximal EF1 allocation exists for $n$ agents with identical valuations. Furthermore, if $\tilde{v}$ is additive, then this NP-hardness applies even when the valuations are additive.

Theorem 11 is an immediate corollary of Lemma 12 where $\tilde{I}$ is the instance from Figure 3 or Section 4.1. Note that we state Lemma 12 in this generic form so that, if subsequent work finds such an instance $\tilde{I}$ for additive valuation for $n=3$ , then the NP-hardness would follow as a corollary.

Reduction.

We will reduce from the Independent Set (IS) problem, which is one of Karp’s classic NP-hard problems [20]. In the IS problem, we are given a graph $H=(V_{H},E_{H})$ and a positive integer $t$ , and the goal is to decide whether $H$ contains an IS of size $t$ .

At a high-level, our reduction starts from $\tilde{I}$ and adds to it $n$ copies of the graph $H$ , where each good has the same value $\lambda$ . Roughly speaking, we wish the $i$ -th copy of $H$ (denoted by $X_{i}$ in the proof below) to give “extra goods” to the $i$ -th agent, in case that agent envies some other agent by more than one good. The crux of the reduction is that such an agent can “catch up” (and thus satisfy EF1) iff there is a sufficiently large independent set in $H$ . This is not yet a complete reduction since, $H$ may not have a maximal independent set of a certain prescribed size. To alleviate this, we introduce “dummy goods” with zero value (denoted by $Y_{i}$ below) to ensure that we can pick any desired number of goods from each copy of $H$ . Finally, some additional edges are also added to ensure that each agent selects goods from a single copy of $H$ .

For the proof below, we will use the following notation: for any valuation $v$ and set $S$ of goods, let $v^{-1}(S):=\min_{j\in S}v(S\setminus\{j\})$ denote the value of $S$ after its most valuable good is removed. We use the convention $v^{-1}(\emptyset)=0$ .

Proof of Lemma 12.

Recall $\tilde{I}$ from the lemma statement. Let³³3 $\gamma$ can be computed in $O(1)$ time by bruteforce. $\gamma:=\min_{\tilde{\mathcal{A}}}\max_{i,i^{\prime}\in[n]}\left(\tilde{v}^{-1}(\tilde{A}_{i})-\tilde{v}(\tilde{A}_{i^{\prime}})\right)$ where the outer minimum is over all maximal allocation $\tilde{\mathcal{A}}$ of $\tilde{I}$ . By the assumption on $\tilde{I}$ , we have $\gamma>0$ . Let $\lambda:=\gamma/t$ .

Let $(H=(V_{H},E_{H}),t)$ denote the IS instance. Our reduction constructs the instance $I=([n],M,v,G)$ as follows:

•

Goods: $M=\tilde{M}\cup X_{1}\cdots\cup X_{n}\cup Y_{1}\cup\cdots\cup Y_{n}$ where $X_{i}=\{x_{i,w}\mid w\in V_{H}\}$ and $Y_{i}=\{y_{i,w}\mid w\in V_{H}\}$ are sets (each of size $|V_{H}|$ ) of additional goods.
•
Graph: $G$ contains the following edges:
1. [(i)]
2. 1.
  
  All edges in $\tilde{G}$ ,
3. 2.
  
  $(x_{i,u},x_{i,w})$ for all $i\in[n]$ and $(u,w)\in V_{H}$ ,
4. 3.
  
  $(x_{i,w},y_{i,w})$ for all $i\in[n]$ and $w\in V_{H}$ ,
5. 4.
  
  all pairs of vertices in $(X_{i}\cup Y_{i})\times(X_{i^{\prime}}\cup Y_{i^{\prime}})$ for all distinct $i,i^{\prime}\in[n]$ .
•

Valuation: For all $S\subseteq M$ , let $v(S)=\tilde{v}(S\cap\tilde{M})+\lambda|S\cap X|$ where $X:=X_{1}\cup\cdots\cup X_{n}$ . That is, the valuations on original goods remain the same, the valuations of each good in $X$ is $\lambda$ , and the goods in $Y_{1}\cup\cdots\cup Y_{n}$ have valuations zero.

See Figure 4 for an illustration of the instance $I$ .

It is clear that the reduction runs in polynomial time, and that, if $\tilde{v}$ is additive, then $v$ is also additive.

(YES)

Suppose that $H$ contains an IS of size $t$ . Let $\tilde{\mathcal{A}}^{*}$ be a maximal allocation of $\tilde{I}$ such that $\max_{i,i^{\prime}\in[n]}\left(\tilde{v}^{-1}(\tilde{A}^{*}_{i})-\tilde{v}(\tilde{A}^{*}_{i^{\prime}})\right)=\gamma$ . We may assume w.l.o.g. that $v^{-1}(\tilde{A}^{*}_{1})\geq v^{-1}(\tilde{A}^{*}_{2}),\dots,v^{-1}(\tilde{A}^{*}_{n})$ . For each $i\in[n]$ , we construct $A^{*}_{i}$ as follows:

1.

Let $c_{i}:=\lceil\max\{0,\tilde{v}^{-1}(\tilde{A}^{*}_{1})-\tilde{v}(\tilde{A}_{i}^{*})\}/\lambda\rceil\leq t$ .
2.

Let $S_{i}$ be any (non-necessarily maximal) IS of size $c_{i}$ in $H$ , which exists since $H$ contains an IS of size $t$ .
3.

Let $A^{*}_{i}=\tilde{A}^{*}_{i}\cup\{x_{i,v}\}_{v\in S_{i}}\cup\{y_{i,v}\}_{v\in(V_{H}\setminus S_{i})}$

Observe that each good in $X_{i}\cup Y_{i}$ can only belong to $A^{*}_{i}$ , and it is obvious that there is no edge between goods in $A^{*}_{i}$ . Thus, $\mathcal{A}^{*}=(A^{*}_{1},\dots,A^{*}_{n})$ is a valid allocation. To see that this is maximal, note that the goods from $Y_{i}$ (resp., $X_{i}$ ) together with type-(iii) edges ensure that no other goods in $X_{i}$ (resp., $Y_{i}$ ) can be added to $A^{*}_{i}$ . Since at least one good from $X_{i}\cup Y_{i}$ is picked, type-(iv) edges ensure that no goods in $X_{i^{\prime}}\cup Y_{i^{\prime}}$ for $i^{\prime}\neq i$ can be added to $A^{*}_{i}$ .

Finally, we argue that $\mathcal{A}^{*}$ is EF1. To bound $v^{-1}(A^{*}_{i})$ , note that $v(A^{*}_{i})=\tilde{v}(\tilde{A}^{*}_{i})+c_{i}\lambda$ . Consider two cases based on $c_{i}$ .

•

If $c_{i}=0$ , we have $v^{-1}(A^{*}_{i})=\tilde{v}^{-1}(\tilde{A}^{*}_{i})\leq\tilde{v}^{-1}(\tilde{A}^{*}_{1})$ .
•

If $c_{i}>0$ , by definition of $c_{i}$ , we have $v(A^{*}_{i})<\tilde{v}^{-1}(\tilde{A}^{*}_{1})+\lambda$ . Thus, $v^{-1}(A^{*}_{i})\leq v(A^{*}_{i})-\lambda<\tilde{v}^{-1}(\tilde{A}^{*}_{1})$ .

Thus, in both cases, we have $v^{-1}(A^{*}_{i})\leq\tilde{v}^{-1}(\tilde{A}^{*}_{1})$ .

On the other hand, for any $i^{\prime}\in[n]$ , the definition of $c_{i^{\prime}}$ immediately implies $v(A^{*}_{i^{\prime}})\geq\tilde{v}^{-1}(\tilde{A}^{*}_{1})$ .

By the two previous paragraphs, $\mathcal{A}^{*}$ is EF1.

(NO)

Suppose that $H$ does not contain an IS of size $t$ . Consider any maximal allocation $\mathcal{A}=(A_{1},\dots,A_{n})$ of $I$ . Notice that the allocation $\tilde{\mathcal{A}}=(A_{1}\cap\tilde{M},\dots,A_{n}\cap\tilde{M})$ is maximal w.r.t. $\tilde{I}$ . Thus, there exist $i,i^{\prime}\in[n]$ such that $\tilde{v}^{-1}(\tilde{A}_{i})-\tilde{v}(\tilde{A}_{i^{\prime}})\geq\gamma$ . Due to type-(iv) edges, at most one of $A_{i^{\prime}}\cap X_{1},\dots,A_{i^{\prime}}\cap X_{n}$ can be non-empty. Furthermore, type-(ii) edges imply that the non-empty set must correspond to an independent set in $H$ . From our assumption, this implies $|A_{i^{\prime}}\cap X|<t$ . As a result,

	$\textstyle v^{-1}(A_{i})\geq\tilde{v}^{-1}(A_{i})$	$\textstyle\geq\gamma+\tilde{v}(\tilde{A}_{i^{\prime}})$
		$\textstyle=\gamma+v(A_{i^{\prime}})-\lambda\|A_{i^{\prime}}\cap X\|$
		$\textstyle>\gamma+v(A_{i^{\prime}})-\lambda\cdot t\qquad\overset{(\star)}{\geq}v(A_{i^{\prime}}),$

where $(\star)$ is due to our choice of $\lambda$ . Thus, $\tilde{A}$ is not EF1. ∎

5 Chore Allocation

In this section, we consider the chore version of our problem, where each agent $i$ has a monotone non-increasing valuation function $v_{i}$ , namely, $v_{i}(S)\geq v_{i}(T)$ for every $S\subseteq T\subseteq M$ . An allocation $\mathcal{A}=(A_{1},\dots,A_{n})$ is envy-free up to one chore (EF1 for chores) if for every pair of agents $i,j\in N$ , either $A_{i}=\emptyset$ , or $v_{i}(A_{i}\setminus\{c\})\geq v_{i}(A_{j})$ some $c\in A_{i}$ [2, 3]. For identical valuations, it is easy to see that the existence of a maximal EF1 allocation is equivalent for goods and chores. Specifically, an allocation $\mathcal{A}$ is envy-free up to one chore under valuation $v$ if and only if $\mathcal{A}$ is envy-free up to one good under valuation $-v$ . Thus, a maximal allocation that is envy-free up to one chore exists for $v$ if and only if a maximal allocation that is envy-free up to one good exists for $-v$ .

Consider the case of two agents with monotone non-increasing valuations. Theorem 1 holds in this case, so we can assume without loss of generality that the agents have identical valuations $v$ . By combining the discussion above with the fact that $-v$ is monotone non-decreasing, Theorem 6 guarantees the existence of a maximal EF1 allocation. Moreover, the algorithms presented in Theorems 10, 13 and A remain applicable in this setting.

For three or more agents with monotone non-increasing valuations, there exist instances where a maximal allocation does not exist. In fact, the instances given in Theorem 3 and Proposition 4.1 use an identical valuation $v$ , and the corresponding instance obtained by replacing $v$ with $-v$ yields no maximal allocation that is EF1 for chores. Also, determining whether a maximal EF1 allocation exists or not is NP-hard also for monotone non-increasing valuations. This follows from Theorem 11 since the constructed instance uses an identical monotone non-decreasing valuation.

6 Conclusion

While we give a nearly complete picture of the existence for maximal EF1 allocations for monotone valuations, there are still a few open interesting questions left. First, for $n=2$ , our algorithm (Theorem 10) runs in pseudo-polynomial time for general monotone valuations. Is there a polynomial-time (in $m,T(m)$ ) algorithm for this task?

It might also be worthwhile considering special cases, which can result from restricting either the valuations or the graphs. We list a few intriguing cases below.

•

Additive valuations: The existence of maximal EF1 allocation remains open only for the case $n=3$ since our lower bound in Section 4.1 requires non-additive valuations, but the lower bound for $n\geq 4$ (Figure 3) holds for additive (and identical) valuations.
•

Uniform valuations: All $n\geq 3$ remains open in this case as we are not aware of any lower bound that holds for uniform valuations. We also note that the well-known Hajnal-Szemerédi theorem ([16]) is equivalent to stating that a maximal EF1 allocation exists when the graph has maximum degree at most $n-1$ . Thus, a positive answer to the question for arbitrary graphs will significantly generalize the theorem. We make progress on this question by proving the existence on trees (see Appendix B).
•

Restricted Graph Classes: It might also be interesting to study special graph classes, such as bounded degree graphs (as suggested by the above Hajnal-Szemerédi theorem) or trees. Some other interesting graph classes include interval graphs–which was studied by Kumar et al. [21]–and bipartite graphs.

Acknowledgements

This work was partially supported by JST FOREST Grant Numbers JPMJPR20C1. We thank the anonymous IJCAI 2025 for their valuable comments.

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Appendix

Appendix A Two agents and Special Graph Classes

Bipartite Graphs.

For two agents and special graph classes like bipartite graphs, we can find a maximal EF1 and allocation in polynomial time.

Theorem 13.

When there are $n=2$ agents and $G$ is bipartite, a maximal EF1 allocation can be found in $O\left(\left|E\right|+m\cdot T(m)\right)$ time.

Proof.

Let $M_{1},M_{2}$ the two parts of bipartite graphs that $v(M_{1})\geq v(M_{2})$ ; we may assume w.l.o.g. that $M_{1}$ contains all vertices of degree zero. The algorithm is simply to run Algorithm 1 with $S=M_{1}$ . The running time claim is obvious. Meanwhile, since $X_{1},X_{2}\subseteq M_{2}$ , we have that $v(X_{1}),v(X_{2})\leq v(M_{2})\leq v(M_{1})=v(S)$ . Thus, Lemma 8 guarantees that the algorithm finds an EF1 allocation. ∎

Interval Graphs.

For interval graphs, Kumar et al. [21] proved that a maximal EF1 allocation exists among two agents and can be computed in polynomial time. They extended the idea presented in Figure 1 for path graphs to interval graphs. However, unlike paths, there is an issue that the allocation after the slight change may not be maximal. Kumar et al. addressed this issue by dividing the structure of intervals into five cases, showing that there is a way to amend the allocation to satisfy maximality for each case. However, their proof is technically involved. Here, we provide a simpler proof by utilizing Theorem 6.

{restatable}

theoremIntervalTwo When there are $n=2$ agents with monotone valuations and $G$ is an interval graph, a maximal EF1 allocation can be found in $O(m\log m+m\cdot T(m))$ time.

We consider (generalized) interval scheduling problem as a base.

IntervalScheduling.: Given $n$ intervals $[l_{1},r_{1}),\dots,$ $[l_{n},r_{n})$ and an integer $c\geq 1$ , find a subset of intervals with maximum size such that no point is covered by more than $c$ intervals. For simplicity, we assume that $l_{1},r_{1},\dots,l_{n},r_{n}$ are all distinct.

This can be solved by greedy algorithm: First, we sort intervals so that $r_{1}<\dots<r_{n}$ . Then, for $i=1,\dots,n$ in this order, if $[l_{i},r_{i})$ can be added while satisfying the constraint that each point is covered by at most $c$ intervals, add this to the current solution.

Let’s understand the structure of greedy solution. Let $[l^{\prime}_{1},r^{\prime}_{1}),\dots,[l^{\prime}_{k^{\prime}},r^{\prime}_{k^{\prime}})\ (r^{\prime}_{1}<\dots<r^{\prime}_{k^{\prime}})$ be any valid solution. For convenience, $l^{\prime}_{i},r^{\prime}_{i}$ for $i\geq k^{\prime}+1$ are set to $+\infty$ . Also, let $[l^{*}_{1},r^{*}_{1}),\dots,[l^{*}_{k},r^{*}_{k})\ (r^{*}_{1}<\dots<r^{*}_{k})$ be the greedy solution.

Lemma 14.

$r^{*}_{i}\leq r^{\prime}_{i}$ holds for any $i\in\{1,\dots,k\}$ .

Proof.

We prove by induction of $i$ .

•

Base Case $(i=1)$ : $r^{*}_{1}\leq r^{\prime}_{1}$ holds, because the greedy algorithm first chooses the interval $[l_{i},r_{i})$ with the smallest $r_{i}$ .
•

Induction $(i\geq 2)$ : We assume that $r^{*}_{1}\leq r^{\prime}_{1},\dots,r^{*}_{i-1}\leq r^{\prime}_{i-1}$ holds. Suppose that $r^{*}_{i}>r^{\prime}_{i}$ . Because each point is covered by at most $c$ intervals, $r^{\prime}_{i-c}\leq l^{\prime}_{i}$ must hold. The assumption $r^{*}_{i-c}\leq r^{\prime}_{i-c}$ implies that $r^{*}_{i-c}\leq l^{\prime}_{i}$ , so $[l^{*}_{1},r^{*}_{1}),\dots,[l^{*}_{i-1},r^{*}_{i-1}),[l^{\prime}_{i},r^{\prime}_{i})$ is a valid solution. Because of the construction of the greedy algorithm, $[l^{\prime}_{i},r^{\prime}_{i})$ should have been added before $[l^{*}_{i},r^{*}_{i})$ , a contradiction. Therefore, $r^{*}_{i}\leq r^{\prime}_{i}$ . ∎

The above result also implies that the greedy solution is optimal, i.e. $k\geq k^{\prime}$ .

Lemma 15.

Consider the interval scheduling problem with $c=1$ . Suppose that $[l^{\prime}_{1},r^{\prime}_{1}),\dots,[l^{\prime}_{k},r^{\prime}_{k})$ is an optimal solution. Then, for any $i\in\{0,\dots,k\}$ , $[l^{*}_{1},r^{*}_{1}),\dots,[l^{*}_{i-1},r^{*}_{i-1}),[l^{\prime}_{i},r^{\prime}_{i}),\dots,[l^{\prime}_{k},r^{\prime}_{k})$ is an optimal solution.

Proof.

The inequality that $r^{*}_{i-1}\leq r^{\prime}_{i-1}$ holds due to Lemma 14, and $r^{\prime}_{i-1}<l^{\prime}_{i}$ holds because $[l^{\prime}_{i-1},r^{\prime}_{i-1})$ and $[l^{\prime}_{i},r^{\prime}_{i})$ do not overlap. Therefore, $r^{*}_{i-1}<l^{\prime}_{i}$ holds, so the intervals do not overlap. Also, the number of chosen intervals is $k$ , so it is an optimal solution. ∎

Algorithm 3

\textsc{IntervalEF1}(G=(M,E),v)

[l_{1},r_{1}),\dots,[l_{m},r_{m})\leftarrow

intervals corresponding to each vertex in

G

such that

l_{1},r_{1},\dots,l_{m},r_{m}

are distinct

Z\leftarrow

an optimal solution of IntervalScheduling with

c=2

\triangleright

Let

Z=\{z_{1},\dots,z_{k}\}

to be

r_{z_{1}}<\dots<r_{z_{k}}

Z_{1}\leftarrow\{z_{1},z_{3},z_{5},\dots\}

Z_{2}\leftarrow\{z_{2},z_{4},z_{6},\dots\}

5:if

v(Z_{1})<v(Z_{2})

then swap

Z_{1},Z_{2}

6:end if

7:for

z\in Z_{2}

8: if intervals contained in

Z_{1}\cup\{z\}

do not overlap then

9: Add

z

Z_{1}

and remove

z

from

Z_{2}

10: end if

11:end for

12:

X^{\prime}_{1},X^{\prime}_{2}\leftarrow

a “greedy solution” and a “greedy solution from the opposite direction” for Interval Scheduling of intervals in

M\setminus Z_{1}

with

c=1

13:

\mathcal{C}_{1}\leftarrow

chain of allocations generated by Algorithm 1 setting

S=Z_{1},X_{1}=X^{\prime}_{1},X_{2}=X^{\prime}_{2}

14:

\mathcal{C}_{2}\leftarrow

a chain of allocations which starts with

(Z_{1},Z_{2})

and ends with

(Z_{1},X^{\prime}_{2})

15:

\mathcal{C}_{3}\leftarrow

a chain of allocations which starts with

(X^{\prime}_{1},Z_{1})

and ends with

(Z_{2},Z_{1})

16:

\mathcal{C}\leftarrow

chain of allocations created by concatenating

\mathcal{C}_{2},\mathcal{C}_{1},\mathcal{C}_{3}

in this order

\triangleright

\mathcal{C}

is a gapless chain

17:return any maximal EF1 allocation in

\mathcal{C}

Proof of Theorem A.

We show that IntervalEF1 returns a maximal EF1 allocation. First, the resulting $Z_{1},Z_{2}$ meet the following conditions:

•

$Z_{1}$ is maximal in $M$ . Due to lines 6-8, any $z\in Z_{2}$ cannot be added to $Z_{1}$ . Also, if $z\in Z\setminus(Z_{1}\cup Z_{2})$ can be added, this contradicts the fact that $Z$ is an optimal solution for IntervalScheduling with $c=2$ .
•

$v(Z_{1})\geq v(Z_{2})$ due to line 5 and monotonicity.

Next, we consider $\mathcal{C}_{1}$ . In Algorithm 1, when $S$ is sorted, i.e. $l_{s_{1}}<r_{s_{1}}<\dots<l_{s_{k}}<r_{s_{k}}$ , it suffices to regard $p_{t}$ as $l_{t}$ and $q_{t}$ as $r_{t}$ .⁴⁴4Note that, for example, the actual $p_{t}$ can take the same value even if $l_{t}$ ’s are different, but it does not affect the construction of $X_{2}$ because ties of $p_{t}$ can be resolved in any way. Then, $X_{1}$ becomes a greedy solution, and $X_{2}$ becomes a greedy solution from opposite direction. Hence, the construction of $\mathcal{C}_{1}$ is justified.

Finally, we consider $\mathcal{C}_{3}$ . We use Lemma 15 to construct this chain. We note that both $X^{\prime}_{1}$ and $Z_{2}$ are optimal solutions for IntervalScheduling of intervals in $M\setminus Z_{1}$ .

•

$X^{\prime}_{1}$ is an optimal solution because the greedy solution is optimal (Lemma 14).
•

$Z_{2}$ is an optimal solution because otherwise, this would contradict the optimality of $Z$ (for IntervalScheduling with $c=2$ ).

We use Lemma 15 by setting $[l^{\prime}_{1},r^{\prime}_{1}),\dots,[l^{\prime}_{k},r^{\prime}_{k})$ to intervals of $Z_{2}$ . We create $\mathcal{C}_{3}$ by arranging the following allocation for $i=k,\dots,0$ in this order.

(\{[l^{*}_{1},r^{*}_{1}),\dots,[l^{*}_{i-1},r^{*}_{i-1}),[l^{\prime}_{i},r^{\prime}_{i}),\dots,[l^{\prime}_{k},r^{\prime}_{k})\},Z_{1})

They are maximal due to Lemma 15. By symmetry, we can construct $\mathcal{C}_{2}$ in a similar way.

Combining all these facts, $\mathcal{C}$ is a gapless chain. Note that $\mathcal{C}$ starts with $(Z_{1},Z_{2})$ and ends with $(Z_{2},Z_{1})$ , where $v(Z_{1})\geq v(Z_{2})$ is ensured.

Finally, we analyze the running time of the algorithm.

•

Reconstructing $[l_{1},r_{1}),\dots,[l_{m},r_{m})$ can be done in $O(m)$ time, due to [17].
•

Obtaining an optimal solution of IntervalScheduling can be done in $O(m)$ time by the greedy algorithm.
•

Making $Z_{1}$ maximal (lines 6-8) can be done in $O(m\log m)$ time by using binary search tree.
•

The length of $\mathcal{C}$ is at most $3m+1$ , so it takes $O(m\cdot T(m))$ time to find a maximal EF1 allocation among them.

Thus, the algorithm runs in $O(m\log m+m\cdot T(m))$ time. ∎

Appendix B Uniform Valuations

We consider the case of uniform valuations, i.e., $v_{i}(S)=\left|S\right|$ for all $S\subseteq M$ . For such valuations, our problem is closely related to an equitable coloring of a graph, where the number of vertices assigned to each color differs by at most $1$ . The key difference is that an equitable coloring must be complete—no vertex can remain uncolored—whereas in our problem, we aim to find a maximal equitable (partial) coloring defined as follows.

Given a graph $G=(V,E)$ and $n\in\mathbb{N}$ , a maximal equitable $n$ -coloring is a subpartition $(S_{1},\dots,S_{n})$ of $V$ , where $S_{1},\dots,S_{n}\subseteq V$ , with the following conditions:

•

$S_{1},\dots,S_{n}$ are disjoint.
•

(independence) For any pair of distinct vertices $x,y\in S_{i}$ , the vertices $x,y$ are not adjacent.
•

(maximality) For every $v\in V\setminus(S_{1}\cup\dots\cup S_{n})$ and $i\in\{1,\dots,n\}$ , $v$ has an adjacent vertex in $S_{i}$ .
•

(equitability) $\max_{i\in[n]}\left|S_{i}\right|-\min_{i\in[n]}\left|S_{i}\right|\leq 1$ .

Below, we prove that a maximal equitable $n$ -coloring exists when $G$ is a tree. It remains an open problem whether this can be extended to all graphs.

Theorem 16.

For every tree $T=(V,E)$ and every $n\in\mathbb{N}$ , there exists a maximal equitable $n$ -coloring of $T$ .

Proof.

Consider $T$ as a rooted tree with root $r$ . We prove a stronger fact that there exists a maximal equitable $n$ -coloring of $T$ such that $r$ is colored with a higher color or is uncolored. Here, for a coloring $\mathcal{S}=(S_{1},\dots,S_{n})$ , we define that color $i$ is a higher color if $|S_{i}|=\max_{j\in[n]}|S_{j}|$ .

We will prove this by strong induction on the number of vertices of $T=(V,E)$ . The base case $|V|=1$ is trivial.

For the inductive step, consider any tree $T=(V,E)$ with $|V|>1$ and suppose that the statement holds for all trees with smaller number of vertices. Let $r_{1},\dots,r_{k}$ be the children of $r$ , and let $T_{1},\dots,T_{k}$ be the subtree of $r_{1},\dots,r_{k}$ , respectively. By the inductive hypothesis, for each $i$ , there exists a maximal equitable $n$ -coloring of $T_{i}$ where $r_{i}$ is a higher color or is uncolored; let it be $\mathcal{S}^{(i)}=(S^{(i)}_{1},\dots,S^{(i)}_{n})$ . Without loss of generality, we assume $|S^{(i)}_{1}|\geq\dots\geq|S^{(i)}_{n}|$ . We say that $T_{i}$ is a singular subtree if $\mathcal{S}^{(i)}$ has only one higher color (color $1$ ) and $r_{i}$ has color $1$ . Without loss of generality, for some $c$ , $T_{1},\dots,T_{c}$ are singular subtrees and $T_{c+1},\dots,T_{k}$ are not.

First, we aim to obtain an equitable $n$ -coloring $\mathcal{S}$ of $T_{1}\cup\dots\cup T_{k}$ in the following way:

1.

Initialize $\mathcal{S}=(\emptyset,\dots,\emptyset)$ and $x=0$ .
2.
For $i=1,\dots,k$ , do the following:
1. (a)
  
  $(S_{1},\dots,S_{n})\leftarrow(S_{1}\cup S^{(i)}_{n-x+1},\dots,S_{x}\cup S^{(i)}_{n},S_{x+1}\cup S^{(i)}_{1},\dots,S_{n}\cup S^{(i)}_{n-x})$ where the indices of $\mathcal{S}^{(i)}$ wrap around to 1 after $n$ , and,
2. (b)
  
  $x\leftarrow(x+(\text{number of higher colors in $\mathcal{S}^{(i)}$}))\bmod n$ .

It is not difficult to see that, for every loop just after step 2 (b), colors $1,\dots,x$ are higher colors of $\mathcal{S}$ , and colors $x+1,\dots,n$ are not (or, when $x=0$ , all colors are higher colors). Also, observe that the algorithm works in a way that $r_{1},\dots,r_{\min(c,n)}$ will have colors $1,\cdots,\min(c,n)$ , respectively, as $T_{1},\dots,T_{\min(c,n)}$ are singular subtrees.

The problem is that, in order to obtain a coloring of $T$ , we need to consider vertex $r$ . We divide it into two cases to solve this issue.

•

(Case 1. $c\geq n$ ) $r_{1},\dots,r_{n}$ will have colors $1,\dots,n$ , respectively. Therefore, $r$ can be left uncolored, and $\mathcal{S}$ is already a maximum equitable $n$ -coloring of $T$ .
•

(Case 2. $c<n$ ) We aim to color vertex $r$ with color $n$ . The case when this cannot be done is that some $r_{i}$ already has color $n$ . However, since $r_{1},\dots,r_{c}$ have colors $1,\dots,c\ (<n)$ , we know that $T_{i}$ is not a singular subtree. So, there is another higher color inside $T_{i}$ , say color $x$ (in the context of coloring $\mathcal{S}$ ; $x\neq n$ must hold). Then, we can swap color $x$ and color $n$ inside the entire $T_{i}$ — then $r_{i}$ no longer has color $n$ , and the number of vertices of each color does not change. We repeat this process until no $r_{i}$ has color $n$ . Since color $n$ is used by the least number of vertices, coloring vertex $r$ in color $n$ does not break the equitability condition, and color $n$ becomes a higher color.

Therefore, we obtained the desired coloring of $T$ . ∎