Connections between Fairness Criteria and Efficiency for Allocating Indivisible Chores
aa r X i v : . [ c s . G T ] J a n Connections between Fairness Criteria and Efficiency forAllocating Indivisible Chores
Ankang Sun ∗ Bo Chen † Xuan Vinh Doan ‡ Abstract
We study several fairness notions in allocating indivisible chores (i.e., items with non-positive values): envy-freeness and its relaxations. For allocations under each fairness cri-terion, we establish their approximation guarantee for other fairness criteria. Under thesetting of additive cost functions, our results show strong connections between these fair-ness criteria and, at the same time, reveal intrinsic differences between goods allocationand chores allocation. Furthermore, we investigate the efficiency loss under these fairnessconstraints and establish their prices of fairness . Fair division is a central matter of concern in economics, multiagent systems, and artificialintelligence (Brams and Taylor, 1996; Bouveret et al., 2016; Aziz et al., 2015). Over the years,there emerges a tremendous demand for fair division when a set of indivisible resources, such asclassrooms, tasks, and properties, are divided among a group of 𝑛 agents. This field has attractedthe attention of researchers and most results are established when resources are considered asgoods that bring positive utility to agents. However, in real-life division problems, the resourcesto be allocated can also be chores which, instead of positive utility, bring non-positive utilityor cost to agents. For example, one might need to assign tasks among workers, teaching loadamong teachers, sharing noxious facilities among communities, and so on. Compared to goods,fairly dividing chores is relatively under-developed. At first glance, dividing chores is similar todividing goods. However, in general, chores allocation is not covered by goods allocation andresults established on goods do not necessarily hold on chores. Studies in Bogomolnaia et al.(2019, 2017); Brânzei and Sandomirskiy (2019) and Freeman et al. (2019, 2020) have alreadypointed out this difference in the context of envy-freeness and equitability, respectively. As anexample (Freeman et al., 2019), when allocating goods a leximin allocation is Pareto optimal ∗ Warwick Business School, University of Warwick. [email protected] † Warwick Business School, University of Warwick. [email protected] ‡ Warwick Business School, University of Warwick. [email protected] A leximin solution selects the allocation that maximizes the utility of the least well-off agent, subject tomaximizing the utility of the second least, and so on. equitable up to any item , however, a leximin solution does not guarantee equitability upto any item in chores allocation.Among the variety of fairness notions introduced in the literature, envy-freeness (EF) isone of the most compelling ones, which has drawn research attention over the past few decadesFoley (1967); Brams and Taylor (1995); Caragiannis et al. (2019a). In an envy-free allocation,no agent envies another agent. Unfortunately, the existence of an envy-free allocation cannot beguaranteed in general when the items to be assigned are indivisible. A canonical example is thatone needs to assign one chore to two agents and the chore has a positive cost for either agent.Clearly, the agent who receives the chore will envy the other. In addition, deciding the existenceof an EF allocation is computationally intractable, even for two agents with identical preferenceLipton et al. (2004). Given this predicament, recent studies mainly devote to relaxations of envy-freeness. One direct relaxation is known as envy-free up to one item (EF1) Lipton et al. (2004);Budish (2011). In an EF1 allocation, one agent may be jealous of another, but by removingone chore from the bundle of the envious agent, envy can be eliminated. A similar but stricternotion is envy-free up to any item (EFX) Caragiannis et al. (2019b). In such an allocation, envycan be eliminated by removing any positive-cost chore from the envious agent’s bundle. Anotherfairness notion, maximin share (MMS) Budish (2011); Amanatidis et al. (2017), generalizes theidea of “cut-and-choose” protocol in cake cutting. The maximin share is obtained by minimizingthe maximum cost of a bundle of an allocation over all allocations. The last fairness notion weconsider is called pairwise maximin share (PMMS) Caragiannis et al. (2019b), which is similarto maximin share but different from MMS in that each agent partitions the combined bundleof himself and any other agent into two bundles and then receives the one with the larger cost.The existing research on envy-freeness and its relaxations concentrates on algorithmic fea-tures of fairness criteria, such as their existence and (approximation) algorithms for findingthem. Relatively little research studies the connections between these fairness criteria them-selves, or the trade-off between these fairness criteria and the system efficiency, known as theprice of fairness . When allocating goods, Amanatidis et al. (2018) compare the four aforemen-tioned relaxations of envy-freeness and provides results on the approximation guarantee of oneto another. However, these connections are unclear in allocating chores. On the price of fair-ness, Bei et al. (2019) study allocating indivisible goods and focuses on the notions for whichcorresponding allocations are guaranteed to exist, such as EF1, maximin Nash welfare , andleximin. Caragiannis et al. (2012) study the price of fairness for both chores and goods, andfocuses on the classical fairness notions, namely, EF, proportionality and equitability. Whenallocating chores, it provides a tight upper bound for the price of proportionality and also showsthat the price of both envy-freeness and equitability are infinite (although such an allocationmay not exist at all). However, in allocating chores, the price of fairness is still unknown forany of the aforementioned four relaxations of envy-freeness. Equitability requires that any pair of agents are equally happy with their bundles. In equitability up to anyitem allocations, the violation of equitability can be eliminated by removing any single item from the happier (ingoods allocation)/ less happy agent (in chores allocation). Nash welfare is the product of agents’ utilities. An allocation of goods (resp. chores) is proportional if the value (resp. cost) of every agent’s bundle is atleast (resp. at most) one 𝑛 -th fraction of his value (resp. cost) for all items.
2n this paper, we fill these gaps by investigating the four relaxations of envy-freeness on twoaspects. On the one hand, we study the connections between these criteria and, in particular, weconsider the following questions:
Does one fairness criterion implies another? To what extentcan one criterion guarantee for another?
On the other hand, we study the trade-off betweenfairness and efficiency (or social cost defined as the sum of costs of the individual agents).Specifically, for each fairness criterion, we investigate its price of fairness , which is defined asthe supremum ratio of the minimum social cost of a fair allocation to the minimum social costof any allocation.
On the connections between fairness criteria, we summarize our main results in Figure 1 onthe approximation guarantee of one fairness criterion for another when the cost functions areadditive, where 𝛼 -Z (formally defined in Section 2) refers to 𝛼 -approximation for fairness undernotion Z. While some of our results show similarity to those in goods allocation Amanatidis et al.(2018), others also reveal the difference between allocating goods and chores. 𝛼 -EFX 𝛼 -PMMS 𝛼 -MMS 𝛼 -EF1LB = UB = 𝛼 𝛼 + (P3.6) 𝛼 =
1: UB=1 (P4.1), 𝛼 >
1: LB = ∞ (P4.3) 𝛼 = : U B = ( P . ) , 𝛼 > : L B = ∞ ( P . ) L B = U B = 𝛼 + 𝛼 + ( P . ) LB = UB = 𝑛𝛼 + 𝑛 − 𝑛 − + 𝛼 (P3.4)LB = ∞ (P4.10) L B = ∞ ( P . ) L B : m a x { 𝑛 𝛼 𝛼 + 𝑛 − , 𝑛𝑛 + } , U B : m i n { 𝑛 𝛼 𝑛 − + 𝛼 , 𝑛 𝛼 + 𝑛 − 𝑛 − + 𝛼 } ( P . ) 𝑛 ≥ : L B = ( P . ) 𝛼 < : L B = 𝑛 𝛼𝛼 + ( 𝑛 − )( − 𝛼 ) , U B = 𝑛 𝛼𝛼 + ( 𝑛 − )( − 𝛼 ) ( P . ) Note: LB and UB stand for lower and upper bound, respectively. For example, the directed edge from 𝛼 -EFX to 𝛼 -PMMS with label LB = UB = 𝛼 𝛼 + means that 𝛼 -EFX implies 𝛼 𝛼 + -PMMS, and this resultis tight. P 𝑥.𝑦 points to Proposition 𝑥 . 𝑦 Figure 1:
Connections between fairness criteria
After comparing each pair of fairness notions, we compare the efficiency of fair allocations3ith the optimal one. To quantify the efficiency loss, we apply the idea of the price of fairnessand our results are summarized in Table 1.EFX PMMS EF1 2-MMS -PMMS 𝑛 = (P5.4) (P5.4) (P5.1) (P5.2) (P5.3) 𝑛 ≥ ∞ ∞ ∞ Θ ( 𝑛 ) ∞ (P5.5) (P5.5) (P5.5) (P5.8) (P5.6) Table 1:
Prices of fairness, where P 𝑥.𝑦 points to Proposition 𝑥 . 𝑦 The fair division problem has been studied for both indivisible goods Lipton et al. (2004);BezákováIvona and DaniVarsha (2005); Caragiannis et al. (2019b) and indivisible chores Aziz et al.(2017, 2019b); Freeman et al. (2020). Among various fairness notions, a prominent one is EFproposed in Foley (1967). But an EF allocation may not exist and even worse, checking theexistence of an EF allocation is NP-complete Aziz et al. (2015). For the relaxations of envy-freeness, Lipton et al. (2004) originate the notion of EF1 and provides an efficient algorithm forEF1 allocations of goods when agents have monotone utility functions. For allocating chores,EF1 is achievable by allocating chores in a round-robin fashion if agents have additive costfunctions Aziz et al. (2019a). Another fairness notion that has been a subject of much interestin the last few years is MMS, proposed by Budish (2011). However, existence of an MMS allo-cation is not guaranteed either for goods Kurokawa et al. (2018) or for chores Aziz et al. (2017),even with additive functions. Consequently, more efforts are on approximation of MMS, withAmanatidis et al. (2017); Ghodsi et al. (2018); Garg and Taki (2020) on goods allocation andAziz et al. (2017); Huang and Lu (2020) on chores allocation. The notions of EFX and PMMSare introduced by Caragiannis et al. (2019b). They consider goods allocation and establish thata PMMS allocation is also EFX when the valuation functions are additive. Beyond the simplecase of 𝑛 =
2, the existence of an EFX allocation has not been settled in general. However, sig-nificant results have been achieved for some special cases. When 𝑛 =
3, the existence of an EFXallocation of goods is proved in Chaudhury et al. (2020). Based on a modified version of leximinsolutions, Plaut and Roughgarden (2020) show that an EFX allocation is guaranteed to existwhen all agents have identical valuations. The work most related to ours is Amanatidis et al.(2018), which is on goods allocation, and provides connections between the four EF relaxations.As for the price of fairness, Caragiannis et al. (2012) show that, in the case of divisible goods,the price of proportionality is Θ (√ 𝑛 ) and the price of equitability is Θ ( 𝑛 ) . Bertsimas et al. (2011)extend the study to other fairness notions, maximin fairness and proportional fairness, andprovides a tight bound on the price of fairness for a broad family of problems. Bei et al. (2019)focus on indivisible goods and concentrates on the fairness notions that are guaranteed to exist. It maximizes the lowest utility level among all the agents. Θ ( 𝑛 ) on the price of maximum Nashwelfare Cole and Gkatzelis (2015), maximum egalitarian welfare Brams and Taylor (1996) andleximin. They also consider the price of EF1 but leave a gap between the upper bound 𝑂 ( 𝑛 ) and lower bound Ω (√ 𝑛 ) . This gap is later closed by Barman et al. (2020) with the results that,for both EF1 and -MMS, the price of fairness is 𝑂 (√ 𝑛 ) . In addition, the price of fairness hasbeen studied in other topics of multi-agent systems, such as machine scheduling Agnetis et al.(2019) and kidney exchange Dickerson et al. (2014). In a fair division problem on indivisible chores, we have a set 𝑁 = { , , . . . , 𝑛 } of 𝑛 agents anda set 𝐸 = { 𝑒 , . . . , 𝑒 𝑚 } of 𝑚 indivisible chores. As chores are the items with non-positive values,each agent 𝑖 ∈ 𝑁 is associated with a cost function 𝑐 𝑖 : 2 𝐸 → 𝑅 ≥ , which maps any subsets of 𝐸 into a non-negative real number. In this paper, we assume 𝑐 𝑖 (∅) = 𝑐 𝑖 is monotone, thatis, 𝑐 𝑖 ( 𝑆 ) ≤ 𝑐 𝑖 ( 𝑇 ) for any 𝑆 ⊆ 𝑇 ⊆ 𝐸 . We say a (set) function 𝑐 (·) is additive if 𝑐 ( 𝑆 ) = Í 𝑒 ∈ 𝑆 𝑐 ( 𝑒 ) for any 𝑆 ⊆ 𝐸 . In the remainder of this paper, we assume all cost functions are additive. Forsimplicity, instead of 𝑐 𝑖 ({ 𝑒 𝑗 }) , we use 𝑐 𝑖 ( 𝑒 𝑗 ) to represent the cost of chore 𝑒 𝑗 for agent 𝑖 .An allocation A : = ( 𝐴 , . . . , 𝐴 𝑛 ) is an 𝑛 -partition of 𝐸 among agents in 𝑁 , i.e., 𝐴 𝑖 ∩ 𝐴 𝑗 = ∅ for any 𝑖 ≠ 𝑗 and ∪ 𝑖 ∈ 𝑁 𝐴 𝑖 = 𝐸 . Each subset 𝑆 ⊆ 𝐸 also refers to a bundle of chores. For anybundle 𝑆 and 𝑘 ∈ N + , we denote by Π 𝑘 ( 𝑆 ) the set of all 𝑘 -partition of 𝑆 , and | 𝑆 | the number ofchores in 𝑆 . We study envy-freeness and its relaxations and are concerned with both exact and approximateversions of these fairness notions.
Definition 2.1.
For any 𝛼 ≥
1, an allocation A = ( 𝐴 , . . . , 𝐴 𝑛 ) is 𝛼 -EF if for any 𝑖, 𝑗 ∈ 𝑁 , 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝛼 · 𝑐 𝑖 ( 𝐴 𝑗 ) . In particular, 1-EF is simply called EF. Definition 2.2.
For any 𝛼 ≥
1, an allocation A = ( 𝐴 , . . . , 𝐴 𝑛 ) is 𝛼 -EF1 if for any 𝑖, 𝑗 ∈ 𝑁 ,there exists 𝑒 ∈ 𝐴 𝑖 such that 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 }) ≤ 𝛼 · 𝑐 𝑖 ( 𝐴 𝑗 ) . In particular, 1-EF1 is simply called EF1. Definition 2.3.
For any 𝛼 ≥
1, an allocation A = ( 𝐴 , . . . , 𝐴 𝑛 ) is 𝛼 -EFX if for any 𝑖, 𝑗 ∈ 𝑁 , 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 }) ≤ 𝛼 · 𝑐 𝑖 ( 𝐴 𝑗 ) for any 𝑒 ∈ 𝐴 𝑖 with 𝑐 𝑖 ( 𝑒 ) >
0. In particular, 1-EFX is simply calledEFX.Clearly, EFX is stricter than EF1. Next, we formally introduce the notion of maximinshare. For any 𝑘 ∈ [ 𝑛 ] = { , . . . , 𝑛 } and bundle 𝑆 ⊆ 𝐸 , the maximin share of agent 𝑖 on 𝑆 among 𝑘 agents is MMS 𝑖 ( 𝑘, 𝑆 ) = min 𝐴 ∈ Π 𝑘 ( 𝑆 ) max 𝑗 ∈[ 𝑘 ] 𝑐 𝑖 ( 𝐴 𝑗 ) . Note Plaut and Roughgarden (2020) consider a stronger version of EFX by dropping the condition 𝑐 𝑖 ( 𝑒 ) >
5e are interested in the allocation in which each agent receives cost no more than his maximinshare.
Definition 2.4.
For any 𝛼 ≥
1, an allocation A = ( 𝐴 , . . . , 𝐴 𝑛 ) is 𝛼 -MMS if for any 𝑖 ∈ 𝑁 , 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝛼 · MMS 𝑖 ( 𝑛, 𝐸 ) . In particular, 1-MMS is called MMS. Definition 2.5.
For any 𝛼 ≥
1, an allocation A = ( 𝐴 , . . . , 𝐴 𝑛 ) is 𝛼 -PMMS if for any 𝑖, 𝑗 ∈ 𝑁 , 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝛼 · min B ∈ Π ( 𝐴 𝑖 ∪ 𝐴 𝑗 ) max { 𝑐 𝑖 ( 𝐵 ) , 𝑐 𝑖 ( 𝐵 )} . In particular, 1-PMMS is called PMMS.Note that the right-hand side of the above inequality is equivalent to 𝛼 · MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) . Example 2.1.
Let us consider an example with three agents and a set 𝐸 = { 𝑒 , . . . , 𝑒 } of sevenchores. The cost function of agents is shown as follows. 𝑒 𝑒 𝑒 𝑒 𝑒 𝑒 𝑒 Agent 1 2 3 3 0 4 2 1Agent 2 3 1 3 2 5 0 5Agent 3 1 5 10 2 3 1 3
Regarding this example, it is not hard to verify that
MMS ( , 𝐸 ) = , MMS ( , 𝐸 ) = , MMS ( , 𝐸 ) = . For instance, agent 2 can partition 𝐸 into three bundles: { 𝑒 , 𝑒 } , { 𝑒 , 𝑒 } , { 𝑒 , 𝑒 , 𝑒 } , so that the maximum cost of any single bundle for her is 7. Moreover, there is noother partitions that can guarantee a better worst-case cost.We now examine allocation A with 𝐴 = { 𝑒 , 𝑒 , 𝑒 } , 𝐴 = { 𝑒 , 𝑒 , 𝑒 } , 𝐴 = { 𝑒 } . It is nothard to see that A is an EF allocation, and accordingly, it is also an EFX, EF1, MMS andPMMS allocation. For another allocation B with 𝐵 = { 𝑒 , 𝑒 , 𝑒 } , 𝐵 = { 𝑒 , 𝑒 , 𝑒 } , 𝐵 = { 𝑒 } ,agent 1 would still envy agent 2 even if chore 𝑒 is eliminated from her bundle, and hence,allocation B is neither exact EF nor EFX. One can verify that B is indeed -EF and -EFX.Moreover, B is an EF1 allocation because agent 1 would not envy others if chore 𝑒 is eliminatedfrom her bundle and agent 3 would not envy others if chore 𝑒 is eliminated from her bundle.As for the approximation guarantee on the notions of MMS and PMMS, it is not hard to verifythat allocation B is -MMS and -PMMS. Let 𝐼 = h 𝑁 , 𝐸, ( 𝑐 𝑖 ) 𝑖 ∈ 𝑁 i be an instance of the problem for allocating indivisible chores and let I be the set of all such instances. The social cost of an allocation A = ( 𝐴 , . . . , 𝐴 𝑛 ) is defined asSC ( A ) = Í 𝑖 ∈ 𝑁 𝑐 𝑖 ( 𝐴 𝑖 ) . The optimal social cost for an instance 𝐼 , denoted by OPT ( 𝐼 ) , is the mini-mum social cost over all allocations for this instance. Following previous work Caragiannis et al.(2012); Bei et al. (2019), when study the price of fairness, we assume that agents cost functionsare normalized to one, i.e., 𝑐 𝑖 ( 𝐸 ) = 𝑖 ∈ 𝑁 .6he price of fairness is the supremum ratio over all instances between the social cost ofthe “best” fair allocation and the optimal social cost, where “best” refers to the one with theminimum cost. Since we consider several fairness criteria, let 𝐹 be any given fairness criterionand define by 𝐹 ( 𝐼 ) as the set (possibly empty) of all allocations for instance 𝐼 that satisfyfairness criterion 𝐹 . Definition 2.6.
For any given fairness property 𝐹 , the price of fairness with respect to 𝐹 isdefined as PoF = sup 𝐼 ∈I min A ∈ 𝐹 ( 𝐼 ) SC ( A ) OPT ( 𝐼 ) , where PoF is equal to +∞ if 𝐹 ( 𝐼 ) = ∅ . We begin with some initial results, which reveal some intrinsic difference in allocating goodsand allocating chores as far as approximation guarantee is concerned. Our proof of any resultin this paper either immediately follows the statement of the result, or can be found in theAppendix clearly indicated. First, we state a simple lemma concerning lower bounds of themaximin share.
Lemma 2.1.
For any agent 𝑖 ∈ 𝑁 and bundle 𝑆 ⊆ 𝐸 , • MMS 𝑖 ( 𝑘, 𝑆 ) ≥ 𝑘 𝑐 𝑖 ( 𝑆 ) , ∀ 𝑘 ∈ [ 𝑛 ] ; • MMS 𝑖 ( 𝑘, 𝑆 ) ≥ 𝑐 𝑖 ( 𝑒 ) , ∀ 𝑒 ∈ 𝑆, ∀ 𝑘 ∈ [ 𝑛 ] .Proof. Let T = ( 𝑇 , . . . , 𝑇 𝑘 ) be the 𝑘 -partition of S defining MMS 𝑖 ( 𝑘, 𝑆 ) ; that is max 𝑇 𝑗 𝑐 𝑖 ( 𝑇 𝑗 ) = MMS 𝑖 ( 𝑘, 𝑆 ) . We start from the first one point. Without loss of generality, assume 𝑐 𝑖 ( 𝑇 ) ≥ 𝑐 𝑖 ( 𝑇 ) ≥ · · · ≥ 𝑐 𝑖 ( 𝑇 𝑘 ) and as a result, we have 𝑐 𝑖 ( 𝑇 ) = MMS 𝑖 ( 𝑘, 𝑆 ) . Then, the following holds 𝑘𝑐 𝑖 ( 𝑇 ) ≥ 𝑘 Õ 𝑗 = 𝑐 𝑖 ( 𝑇 𝑗 ) ≥ 𝑐 𝑖 ( 𝑘 Ø 𝑗 = 𝑇 𝑗 ) = 𝑐 𝑖 ( 𝑆 ) , where the second transition is due to additivity. Due to 𝑐 𝑖 ( 𝑇 ) = MMS 𝑖 ( 𝑘, 𝑆 ) , we have MMS 𝑖 ( 𝑘, 𝑆 )≥ 𝑘 𝑐 𝑖 ( 𝑆 ) . As for the second lower bound, for any given chore 𝑒 ∈ 𝑆 , there must exist a bundle 𝑇 𝑗 ′ containing 𝑒 , i.e., 𝑒 ∈ 𝑇 𝑗 ′ . Due to the monotonicity of cost function, we have 𝑐 𝑖 ( 𝑇 𝑗 ′ ) ≥ 𝑐 𝑖 ( 𝑒 ) ,which combines MMS 𝑖 ( 𝑘, 𝑆 ) = 𝑐 ( 𝑇 ) ≥ 𝑐 ( 𝑇 𝑗 ′ ) , implying MMS 𝑖 ( 𝑘, 𝑆 ) ≥ 𝑐 𝑖 ( 𝑒 ) . (cid:3) Based on the lower bounds in Lemma 2.1, we provide a trivial approximation guarantee forPMMS and MMS.
Lemma 2.2.
Any allocation is 2-
PMMS and 𝑛 - MMS .Proof.
Let A = ( 𝐴 , . . . , 𝐴 𝑛 ) be an arbitrary allocation without any specified properties. Wefirst show it’s already an 𝑛 -MMS allocation. By the first point of Lemma 2.1, for each agent 𝑖 ,we have 𝑐 𝑖 ( 𝐸 ) ≤ 𝑛 · MMS 𝑖 ( 𝑛, 𝐸 ) . Then, due to the monotonicity of the cost function, 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝑐 𝑖 ( 𝐸 ) ≤ 𝑛 · MMS 𝑖 ( 𝑛, 𝐸 ) holds. 7ext, by a similar argument, we prove the result about 2-PMMS. By the first point of Lemma2.1, 𝑐 𝑖 ( 𝐴 𝑖 ∪ 𝐴 𝑗 ) ≤ 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) holds for any 𝑖, 𝑗 ∈ 𝑁 . Again, due to the monotonicityof the cost function, we have 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝑐 𝑖 ( 𝐴 𝑖 ∪ 𝐴 𝑗 ) that implies 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) .Therefore, allocation A is also 2-PMMS, completing the proof. (cid:3) As can be seen from the proof of Lemma 2.2, in allocating chores, if one assigns all choresto one agent, then the allocation still has a bounded approximation for PMMS and MMS.However, when allocating goods, if an agent receives nothing but his maximin share is positive,then clearly the corresponding allocation has an infinite approximation guarantee for PMMSand MMS.
Let us start with EF. According to the definitions, for any 𝛼 ≥ 𝛼 -EF is stronger than 𝛼 -EFXand 𝛼 -EF1. The following proposition presents an approximation guarantee of 𝛼 -EF for MMSand PMMS. Proposition 3.1.
For any 𝛼 ≥ , an 𝛼 - EF allocation is also 𝑛𝛼𝑛 − + 𝛼 - MMS , and this result istight.Proof.
We first prove the upper bound and focus on agent 𝑖 . Let A = ( 𝐴 , . . . , 𝐴 𝑛 ) be an 𝛼 -EFallocation, then according to its definition, 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝛼 · 𝑐 𝑖 ( 𝐴 𝑗 ) holds for any 𝑗 ∈ 𝑁 . By summingup 𝑗 over 𝑁 \ { 𝑖 } , we have ( 𝑛 − ) 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝛼 · Í 𝑗 ∈ 𝑁 \{ 𝑖 } 𝑐 𝑖 ( 𝐴 𝑗 ) and as a result, ( 𝑛 − + 𝛼 ) 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝛼 · Í 𝑗 ∈ 𝑁 𝑐 𝑖 ( 𝐴 𝑗 ) = 𝛼 · 𝑐 𝑖 ( 𝐸 ) where the last transition owing to the additivity of cost functions. Onthe other hand, from the first point of Lemma 2.1, it holds that MMS 𝑖 ( 𝑛, 𝐸 ) ≥ 𝑛 𝑐 𝑖 ( 𝐸 ) , implyingthe ratio 𝑐 𝑖 ( 𝐴 𝑖 ) MMS 𝑖 ( 𝑛, 𝐸 ) ≤ 𝑛𝛼𝑛 − + 𝛼 . Regarding tightness, consider an instance with 𝑛 agents and 𝑛 chores denoted as { 𝑒 , . . . , 𝑒 𝑛 } .Agents have identical cost profile. The cost function of agent 1 is as follows: 𝑐 (cid:0) 𝑒 𝑗 (cid:1) = ( 𝛼, if 𝑗 = , , . . . , 𝑛, , if 𝑗 ≥ 𝑛 + . Consider allocation B = ( 𝐵 , . . . , 𝐵 𝑛 ) with 𝐵 𝑖 = { 𝑒 ( 𝑖 − ) 𝑛 + , . . . , 𝑒 𝑖𝑛 } for any 𝑖 ∈ 𝑁 . It is not hardto verify that allocation B is 𝛼 -EF. As for MMS ( 𝑛, 𝐸 ) , since in total we have 𝑛 chores witheach cost 𝛼 and ( 𝑛 − ) 𝑛 chores with each cost 1, then in the partition defining MMS ( 𝑛, 𝐸 ) ,each bundle contains exactly one chore with cost 𝛼 and 𝑛 − ( 𝑛, 𝐸 ) = 𝑛 − + 𝛼 and the ratio 𝑐 ( 𝐵 ) MMS ( 𝑛,𝐸 ) = 𝑛𝛼𝑛 − + 𝛼 . (cid:3) Proposition 3.2.
For any 𝛼 ≥ , an 𝛼 - EF allocation is also 𝛼 + 𝛼 - PMMS , and this result istight.Proof.
We first prove the upper bound. Let A = ( 𝐴 , 𝐴 , . . . , 𝐴 𝑛 ) be an 𝛼 -EF allocation, thenaccording to the definition, for any two agents 𝑖, 𝑗 ∈ 𝑁 , 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝛼 · 𝑐 𝑖 ( 𝐴 𝑗 ) holds. By additivity, we8ave 𝑐 𝑖 ( 𝐴 𝑖 ∪ 𝐴 𝑗 ) = 𝑐 𝑖 ( 𝐴 𝑖 ) + 𝑐 𝑖 ( 𝐴 𝑗 ) ≥ ( + 𝛼 ) · 𝑐 𝑖 ( 𝐴 𝑖 ) , and consequently, 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝛼𝛼 + · 𝑐 𝑖 ( 𝐴 𝑖 ∪ 𝐴 𝑗 ) holds. On the other hand, from the first point of Lemma 2.1, we know 𝑐 𝑖 ( 𝐴 𝑖 ∪ 𝐴 𝑗 ) ≤ · MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) , and therefore the following holds 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝛼𝛼 + · MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) . As for tightness, consider an instance with 𝑛 agents and 2 𝑛 chores denoted as { 𝑒 , 𝑒 , . . . , 𝑒 𝑛 } .Agents have identical cost profile. The cost function of agent 1 is as follows: 𝑐 (cid:0) 𝑒 𝑗 (cid:1) = ( 𝛼, if 𝑗 = , , , if 𝑗 ≥ . Now, consider an allocation B = ( 𝐵 , . . . , 𝐵 𝑛 ) where 𝐵 𝑖 = { 𝑒 𝑖 − , 𝑒 𝑖 } for any 𝑖 ∈ 𝑁 . It is nothard to verify that allocation B is 𝛼 -EF and except for agent 1, no one else will violate thecondition of PMMS. For any 𝑗 ≥
2, one can calculate MMS ( , 𝐵 ∪ 𝐵 𝑗 ) = + 𝛼 , yielding theratio 𝑐 ( 𝐵 ) MMS ( ,𝐵 ∪ 𝐵 𝑗 ) = 𝛼 + 𝛼 , as required. (cid:3) Proposition 3.2 indicates that the approximation guarantee of 𝛼 -EF for PMMS is indepen-dent of the number of agents. However, according to Proposition 3.1, its approximation guar-antee for MMS is affected by the number of agents. Moreover, this guarantee ratio convergesto 𝛼 as 𝑛 goes to infinity.We remark that none of EFX, EF1, PMMS and MMS has a bounded guarantee for EF. Weshow this by a simple example. Consider an instance of two agents and one chore, and thechore has a positive cost for both agents. Assigning the chore to an arbitrary agent results in anallocation that satisfies EFX, EF1, PMMS and MMS, simultaneously. However, since one agenthas a positive cost on his own bundle and zero cost on other agents’ bundle, such an allocationhas an infinite approximation guarantee for EF.Next, we consider approximation of EFX and EF1. Proposition 3.3. An 𝛼 - EFX allocation is 𝛼 - EF1 for any 𝛼 ≥ . On the other hand, an EF1 allocation is not 𝛽 - EFX for any 𝛽 ≥ .Proof. We first show the positive part. Let A = ( 𝐴 , 𝐴 , . . . , 𝐴 𝑛 ) be an 𝛼 -EFX allocation, thenaccording to its definition, ∀ 𝑖, 𝑗 ∈ 𝑁 , ∀ 𝑒 ∈ 𝐴 𝑖 , 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 }) ≤ 𝛼 · 𝑐 𝑖 ( 𝐴 𝑗 ) holds. This implies A isalso 𝛼 -EF1.For the negative part, consider an instance with 𝑛 agents and 2 𝑛 chores denoted as { 𝑒 , 𝑒 , . . . ,𝑒 𝑛 } . Agents have identical cost profile. The cost function of agent 1 is: 𝑐 ( 𝑒 ) = 𝑝, 𝑐 ( 𝑒 𝑗 ) = , ∀ 𝑗 ≥ 𝑝 ≫
1. Now consider an allocation B = ( 𝐵 , . . . , 𝐵 𝑛 ) with 𝐵 𝑖 = { 𝑒 𝑖 − , 𝑒 𝑖 } , ∀ 𝑖 ∈ 𝑁 .It is not hard to see allocation B is EF1 and except for agent 1, no one else will envy the bundleof others. Thus, we only need to concern agent 1 in calculating the approximation ratio onEFX. By removing chore 𝑒 from bundle 𝐵 , 𝑐 ( 𝐵 \{ 𝑒 }) 𝑐 ( 𝐵 𝑗 ) = 𝑝 holds for any agent 𝑗 ∈ 𝑁 \ { } ,and the ratio 𝑝 → ∞ as 𝑝 → ∞ . (cid:3) Next, we consider the approximation guarantee of EF1 for MMS. In allocating goods,9manatidis et al. (2018) present a tight result that an 𝛼 -EF1 allocation is 𝑂 ( 𝑛 ) -MMS. In con-trast, in allocating chores, 𝛼 -EF1 can have a much better guarantee for MMS. Proposition 3.4.
For any 𝛼 ≥ and 𝑛 ≥ , an 𝛼 - EF1 allocation is also 𝑛𝛼 + 𝑛 − 𝑛 − + 𝛼 - MMS , and thisresult is tight.Proof.
We first prove the upper bound. Let A = ( 𝐴 , . . . , 𝐴 𝑛 ) be an 𝛼 -EF1 allocation and wefocus on agent 𝑖 . If 𝐴 𝑖 = ∅ or 𝐴 𝑖 only contains chores with zero cost for agent 𝑖 , then 𝑐 𝑖 ( 𝐴 𝑖 ) = 𝛼 -EF1 nor MMS. Thus, without loss of generality, we assume 𝐴 𝑖 ≠ ∅ and 𝐴 𝑖 contains chores with strictly positive cost for agent 𝑖 . Let ¯ 𝑒 be the chore with largestcost for agent 𝑖 in bundle 𝐴 𝑖 , i.e., ¯ 𝑒 ∈ arg max 𝑒 ∈ 𝐴 𝑖 𝑐 𝑖 ( 𝑒 ) .By the definition of 𝛼 -EF1, for any 𝑗 ∈ 𝑁 \ { 𝑖 } , 𝑐 𝑖 ( 𝐴 𝑖 \ { ¯ 𝑒 }) ≤ 𝛼 · 𝑐 𝑖 ( 𝐴 𝑗 ) holds. Then, bysumming up 𝑗 over 𝑁 \ { 𝑖 } and adding a term 𝛼𝑐 𝑖 ( 𝐴 𝑖 ) on both sides, the following holds, 𝛼 · Õ 𝑗 ∈ 𝑁 𝑐 𝑖 ( 𝐴 𝑗 ) ≥ ( 𝑛 − + 𝛼 ) 𝑐 𝑖 ( 𝐴 𝑖 ) − ( 𝑛 − ) 𝑐 𝑖 ( ¯ 𝑒 ) . (1)From Lemma 2.1, we have MMS 𝑖 ( 𝑛, 𝐸 ) ≥ max { 𝑛 𝑐 𝑖 ( 𝐸 ) , 𝑐 𝑖 ( ¯ 𝑒 )} , and by additivity, it holds that 𝑛𝛼 MMS 𝑖 ( 𝑛, 𝐸 ) ≥ ( 𝑛 − + 𝛼 ) 𝑐 𝑖 ( 𝐴 𝑖 ) − ( 𝑛 − ) MMS 𝑖 ( 𝑛, 𝐸 ) . (2)Inequality (2) is equivalent to 𝑐 𝑖 ( 𝐴 𝑖 ) MMS 𝑖 ( 𝑛,𝑀 ) ≤ 𝑛𝛼 + 𝑛 − 𝑛 − + 𝛼 , as required.As for tightness, consider an instance with 𝑛 agents and a set 𝐸 = { 𝑒 , . . . , 𝑒 𝑛 − 𝑛 + } of 𝑛 − 𝑛 + 𝑐 (cid:0) 𝑒 𝑗 (cid:1) = 𝛼 + 𝑛 − , 𝑗 = ,𝛼, ≤ 𝑗 ≤ 𝑛, , 𝑗 ≥ 𝑛 + . Now, consider an allocation B = { 𝐵 , . . . , 𝐵 𝑛 } with 𝐵 = { 𝑒 , . . . , 𝑒 𝑛 } and 𝐵 𝑗 = { 𝑒 𝑛 +( 𝑛 − ) ( 𝑗 − )+ , . . . ,𝑒 𝑛 +( 𝑛 − ) ( 𝑗 − ) } for any 𝑗 ≥
2. Since agents have identical cost profile, for any agent 𝑖 and bundle 𝐵 𝑗 with 𝑗 ≥ 𝑐 𝑖 ( 𝐵 𝑗 ) = 𝑐 ( 𝐵 𝑗 ) = 𝑛 −
1, smaller than the cost of bundle 𝐵 . Accordingly, exceptfor agent 1, no one else will violate the condition of 𝛼 -EF1 and MMS. As for agent 1, since 𝑐 ( 𝐵 \ { 𝑒 }) = ( 𝑛 − ) 𝛼 = 𝛼𝑐 ( 𝐵 𝑗 ) , ∀ 𝑗 ≥
2, then we can claim that allocation B is 𝛼 -EF1. To cal-culate MMS ( 𝑛, 𝐸 ) , consider an allocation T = ( 𝑇 , . . . , 𝑇 𝑛 ) with 𝑇 = { 𝑒 } and 𝑇 𝑗 = { 𝐵 𝑗 ∪ (cid:8) 𝑒 𝑗 (cid:9) } for any 2 ≤ 𝑗 ≤ 𝑛 . It is not hard to verify that 𝑐 ( 𝑇 𝑗 ) = 𝛼 + 𝑛 − 𝑗 ∈ 𝑁 . Therefore, wehave MMS ( 𝑛, 𝐸 ) = 𝛼 + 𝑛 − 𝑐 ( 𝐵 ) MMS ( 𝑛,𝐸 ) = 𝑛𝛼 + 𝑛 − 𝑛 − + 𝛼 , completing the proof. (cid:3) We now study 𝛼 -EFX in terms of its approximation guarantee for MMS and provide upperand lower bounds for general 𝛼 ≥ 𝑛 ≥ Proposition 3.5.
When agents have additive cost functions, for any 𝛼 ≥ and 𝑛 ≥ , an 𝛼 - EFX allocation is min (cid:8) 𝑛𝛼𝑛 − + 𝛼 , 𝑛𝛼 + 𝑛 − 𝑛 − + 𝛼 (cid:9) - MMS , while it is not 𝛽 - MMS for any 𝛽 < max (cid:8) 𝑛𝛼 𝛼 + 𝑛 − , 𝑛𝑛 + (cid:9) .Proof. We first prove the upper bound. Let A = ( 𝐴 , . . . , 𝐴 𝑛 ) be an 𝛼 -EFX allocation with 𝛼 ≥ 𝑖 . The upper bound 𝑛𝛼 + 𝑛 − 𝑛 − + 𝛼 directly follows from Proposition 3.310nd 3.4. In what follows, we prove the upper bound 𝑛𝛼𝑛 − + 𝛼 . If 𝐴 𝑖 = ∅ or 𝐴 𝑖 only contains choreswith zero cost for agent 𝑖 , then 𝑐 𝑖 ( 𝐴 𝑖 ) = 𝛼 -EFX. Thus, without loss of generality, we assume that 𝐴 𝑖 ≠ ∅ and meanwhile contains choreswith a strictly positive cost for agent 𝑖 . Let 𝑒 ∗ be the chore in bundle 𝐴 𝑖 having the minimumcost for agent 𝑖 , i.e., 𝑒 ∗ ∈ arg min 𝑒 ∈ 𝐴 𝑖 𝑐 𝑖 ( 𝑒 ) . Next, we divide the proof into two cases. Case 1 : | 𝐴 𝑖 | =
1. Then 𝑒 ∗ is the unique element in 𝐴 𝑖 , and thus 𝑐 𝑖 ( 𝐴 𝑖 ) = 𝑐 𝑖 ( 𝑒 ∗ ) . By thesecond point of Lemma 2.1, 𝑐 𝑖 ( 𝑒 ∗ ) ≤ MMS 𝑖 ( 𝑛, 𝐸 ) holds, and thus, 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ MMS 𝑖 ( 𝑛, 𝐸 ) . Case 2 : | 𝐴 𝑖 | ≥
2. By the definition of 𝛼 -EFX, for any agent 𝑗 ∈ 𝑁 \ { 𝑖 } , 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 ∗ }) ≤ 𝛼 · 𝑐 𝑖 ( 𝐴 𝑗 ) . Since 𝑒 ∗ ∈ arg min 𝑒 ∈ 𝐴 𝑖 𝑐 𝑖 ( 𝑒 ) and | 𝐴 𝑖 | ≥
2, we have 𝑐 𝑖 ( 𝑒 ∗ ) ≤ 𝑐 𝑖 ( 𝐴 𝑖 ) . Then, thefollowing holds, 𝛼𝑐 𝑖 ( 𝐴 𝑗 ) ≥ 𝑐 𝑖 ( 𝐴 𝑖 ) − 𝑐 𝑖 ( 𝑒 ∗ ) ≥ 𝑐 𝑖 ( 𝐴 𝑖 ) , ∀ 𝑗 ∈ 𝑁 \ { 𝑖 } . (3)By summing up 𝑗 over 𝑁 \ { 𝑖 } and adding a term 𝛼𝑐 𝑖 ( 𝐴 𝑖 ) on both sides of inequality (3), thefollowing holds 𝛼𝑐 𝑖 ( 𝐸 ) = 𝛼 Õ 𝑗 ∈ 𝑁 \{ 𝑖 } 𝑐 𝑖 ( 𝐴 𝑗 ) + 𝛼𝑐 𝑖 ( 𝐴 𝑖 ) ≥ 𝑛 − + 𝛼 𝑐 𝑖 ( 𝐴 𝑖 ) . (4)On the other hand, from the first point of Lemma 2.1, we know MMS 𝑖 ( 𝑛, 𝐸 ) ≥ 𝑛 𝑐 𝑖 ( 𝐸 ) , whichcombines inequality (4) yielding the ratio 𝑐 𝑖 ( 𝐴 𝑖 ) MMS 𝑖 ( 𝑛, 𝑀 ) ≤ 𝑛𝛼𝑛 − + 𝛼 . Regarding the lower bound 𝑛𝑛 + , consider an instance with 𝑛 agents and a set 𝐸 = { 𝑒 , 𝑒 , ..., 𝑒 𝑛 } of 2 𝑛 chores. Agents have identical cost profile. The cost function of agent 1 is 𝑐 ( 𝑒 𝑗 ) = ⌈ 𝑗 ⌉ forany 𝑗 ≥
1. It is easy to see MMS 𝑖 ( 𝑛, 𝐸 ) = 𝑛 + 𝑖 . Then, consider an allocation B = ( 𝐵 , ..., 𝐵 𝑛 ) with 𝐵 = { 𝑒 𝑛 − , 𝑒 𝑛 } and 𝐵 𝑖 = ( 𝑒 𝑖 − , 𝑒 𝑛 − 𝑖 ) for any 𝑖 ≥
2. Since agents haveidentical profile, for any agent 𝑖 and bundle 𝐵 𝑗 with 𝑗 ≥
2, we have 𝑐 𝑖 ( 𝐵 𝑗 ) = 𝑐 ( 𝐵 𝑗 ) = 𝑛 . Thus,except for agent 1, no one else will violate the condition of MMS and EFX. As for agent 1, envycan be eliminated by removing any single chore since 𝑐 ( 𝐵 \{ 𝑒 𝑛 }) = 𝑐 ( 𝐵 \{ 𝑒 𝑛 − }) = 𝑛 . Hence,the allocation B is EFX and its approximation guarantee on MMS equals to 𝑐 ( 𝐵 ) MMS ( 𝑛,𝐸 ) = 𝑛𝑛 + ,as required.Next, for lower bound 𝑛𝛼 𝛼 + 𝑛 − , let us consider an instance with 𝑛 agents and a set 𝐸 = { 𝑒 , ..., 𝑒 𝑛 − 𝑛 } of 2 𝑛 − 𝑛 chores. We focus on agent 1 and his cost function is 𝑐 ( 𝑒 𝑗 ) = 𝛼 for 𝑗 ≤ 𝑛 and 𝑐 ( 𝑒 𝑗 ) = 𝑗 ≥ 𝑛 +
1. Now, consider an allocation B = ( 𝐵 , ..., 𝐵 𝑛 ) with 𝐵 = { 𝑒 , ..., 𝑒 𝑛 } , 𝐵 = { 𝑒 𝑛 + , ..., 𝑒 𝑛 − } and 𝐵 𝑗 = { 𝑒 𝑛 − +( 𝑗 − ) ( 𝑛 − ) , ..., 𝑒 𝑛 − +( 𝑗 − ) ( 𝑛 − ) } for any 𝑗 ≥
3. Accordingly, bundle 𝐵 contains 2 𝑛 − 𝐵 𝑗 contains 2 𝑛 − 𝑗 ≥ 𝑖 ≥
2, every agent 𝑖 has cost 0 < 𝛿 < 𝜖 on each single chore in bundle 𝐵 𝑖 with 𝛿 arbitrarilysmall, while his cost on other chores are one. Consequently, except for agent 1, no one else willviolate the condition of MMS and 𝛼 -EFX. As for agent 1, his cost on 𝐵 is the smallest overall bundles and 𝑐 ( 𝐵 \ { 𝑒 }) = 𝛼 ( 𝑛 − ) = 𝛼𝑐 ( 𝐵 ) , as a result, the allocation B is 𝛼 -EFX. ForMMS ( 𝑛, 𝐸 ) , it happens that 𝐸 can be evenly divided into 𝑛 bundles of the same cost (for agent1), so we have MMS ( 𝑛, 𝐸 ) = 𝛼 + 𝑛 − 𝑐 ( 𝐵 ) MMS ( 𝑛,𝐸 ) = 𝑛𝛼 𝛼 + 𝑛 − , completing11he proof. (cid:3) The upper bound in Proposition 3.5 is almost tight since 𝑛𝛼 + 𝑛 − 𝑛 − + 𝛼 − 𝑛𝛼 𝛼 + 𝑛 − < 𝑛 − 𝑛 − + 𝛼 <
1. Inaddition, we highlight that the upper and lower bounds provided in Proposition 3.5 are tightin two interesting cases: (i) 𝛼 = 𝑛 = Proposition 3.6.
For any 𝛼 ≥ , an 𝛼 - EFX allocation is also 𝛼 𝛼 + - PMMS , and this guaranteeis tight.Proof.
We first prove the upper bound and focus on agent 𝑖 . Let A = ( 𝐴 , 𝐴 , . . . , 𝐴 𝑛 ) be an 𝛼 -EFX allocation. If 𝐴 𝑖 = ∅ or 𝐴 𝑖 only contains chores with zero cost for agent 𝑖 , then 𝑐 𝑖 ( 𝐴 𝑖 ) = 𝐴 𝑖 is not empty and contains chores with strictly positive cost for agent 𝑖 . Let 𝑒 ∗ be the chore in 𝐴 𝑖 having the minimum cost for agent 𝑖 , i.e., 𝑒 ∗ ∈ arg min 𝑒 ∈ 𝐴 𝑖 𝑐 𝑖 ( 𝑒 ) . Then, we divide the proofinto two cases. Case 1 : | 𝐴 𝑖 | =
1. Then chore 𝑒 ∗ is the unique element in 𝐴 𝑖 , and thus 𝑐 𝑖 ( 𝑒 ∗ ) = 𝑐 𝑖 ( 𝐴 𝑖 ) . Bythe second point of Lemma 2.1, 𝑐 𝑖 ( 𝑒 ∗ ) ≤ MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) holds for any 𝑗 ∈ 𝑁 \ { 𝑖 } . As a result,we have 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) , ∀ 𝑗 ∈ 𝑁 \ { 𝑖 } . Case 2 : | 𝐴 𝑖 | ≥
2. Since 𝑒 ∗ ∈ arg min 𝑒 ∈ 𝐴 𝑖 𝑐 𝑖 ( 𝑒 ) and | 𝐴 𝑖 | ≥
2, we have 𝑐 𝑖 ( 𝑒 ∗ ) ≤ 𝑐 𝑖 ( 𝐴 𝑖 ) , andequivalently, 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 ∗ }) = 𝑐 𝑖 ( 𝐴 𝑖 ) − 𝑐 𝑖 ( 𝑒 ∗ ) ≥ 𝑐 𝑖 ( 𝐴 𝑖 ) . Then, based on the definition of 𝛼 -EFXallocation,for any 𝑗 ∈ 𝑁 \ { 𝑖 } , the following holds 𝛼 · 𝑐 𝑖 ( 𝐴 𝑗 ) ≥ 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 ∗ }) ≥ · 𝑐 𝑖 ( 𝐴 𝑖 ) . (5)Combining the first point of Lemma 2.1 and Inequality (5), for any 𝑗 ∈ 𝑁 \ { 𝑖 } , we haveMMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) ≥ ( 𝑐 𝑖 ( 𝐴 𝑖 ) + 𝑐 𝑖 ( 𝐴 𝑗 )) ≥ 𝛼 + 𝛼 𝑐 𝑖 ( 𝐴 𝑖 ) . Therefore, for any 𝑗 ∈ 𝑁 \ { 𝑖 } , 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝛼 𝛼 + · MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) holds, as required.As for the tightness, consider an instance with 𝑛 agents and a set 𝐸 = { 𝑒 , . . . , 𝑒 𝑛 } of2 𝑛 chores. Agents have identical cost profile. The cost function of agent 1 is defined as 𝑐 ( 𝑒 ) = 𝑐 ( 𝑒 ) = 𝛼, 𝑐 ( 𝑒 𝑗 ) = , ∀ 𝑗 ≥
3. Now, consider allocation B = ( 𝐵 , . . . , 𝐵 𝑛 ) with 𝐵 𝑖 = { 𝑒 𝑖 − , 𝑒 𝑖 } , ∀ 𝑖 ∈ 𝑁 . It is not hard to verify that, except for agent 1, no one else will violatethe condition of EFX and PMMS. For agent 1, by removing any single chore from his bundle,the remaining cost is 𝛼 times of the cost on others’ bundle. Thus, allocation B is 𝛼 -EFX. Noticethat for any 𝑗 ≥
2, bundle 𝐵 ∪ 𝐵 𝑗 contains exactly two chores with cost 2 𝛼 and two chores withcost 1, then MMS ( , 𝐵 ∪ 𝐵 𝑗 ) = 𝛼 + 𝑐 ( 𝐵 ) MMS ( ,𝐵 ∪ 𝐵 𝑗 ) = 𝛼 𝛼 + , completing theproof. (cid:3) Proposition 3.7.
For any 𝛼 ≥ , an 𝛼 - EF1 allocation is also 𝛼 + 𝛼 + - PMMS , and this guaranteeis tight.Proof.
We first prove the upper bound part. Let A = ( 𝐴 , . . . , 𝐴 𝑛 ) be an 𝛼 -EF1 allocationand we focus on agent 𝑖 . If 𝐴 𝑖 = ∅ or 𝐴 𝑖 only contains chores with zero cost for agent 𝑖 , then12 𝑖 ( 𝐴 𝑖 ) = 𝐴 𝑖 ≠ ∅ and meanwhile 𝐴 𝑖 contains chores with strictly positive costs for agent 𝑖 . To study PMMS, we fix another agent 𝑗 ∈ 𝑁 \ { 𝑖 } , and let 𝑒 ∗ ∈ 𝐴 𝑖 be the chore such that 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 ∗ }) ≤ 𝛼 · 𝑐 𝑖 ( 𝐴 𝑗 ) . We divide our proof into two cases. Case 1 : 𝑐 𝑖 ( 𝑒 ∗ ) > 𝑐 𝑖 ( 𝐴 𝑖 ∪ 𝐴 𝑗 \ { 𝑒 ∗ }) . Consider (cid:8) { 𝑒 ∗ } , 𝐴 𝑖 ∪ 𝐴 𝑗 \ { 𝑒 ∗ } (cid:9) , a 2-partition of 𝐴 𝑖 ∪ 𝐴 𝑗 .Since 𝑐 𝑖 ( 𝑒 ∗ ) > 𝑐 𝑖 ( 𝐴 𝑖 ∪ 𝐴 𝑗 \ { 𝑒 ∗ }) , we can claim that this partition defining MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) ,and accordingly, MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) = 𝑐 𝑖 ( 𝑒 ∗ ) holds. From the first point of Lemma 2.1 and thedefinition of 𝛼 -EF1, the following holds 𝑐 𝑖 ( 𝑒 ∗ ) ≥ ( 𝑐 𝑖 ( 𝐴 𝑖 ) + 𝑐 𝑖 ( 𝐴 𝑗 )) ≥ 𝑐 𝑖 ( 𝐴 𝑖 ) + 𝛼 · 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 ∗ }) (6)Then, based on Inequality (6) and the fact MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) = 𝑐 𝑖 ( 𝑒 ∗ ) , we have 𝑐 𝑖 ( 𝐴 𝑖 ) MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) ≤ 𝛼 + 𝛼 + . Case 2 : 𝑐 𝑖 ( 𝑒 ∗ ) ≤ 𝑐 𝑖 ( 𝐴 𝑖 ∪ 𝐴 𝑗 \ { 𝑒 ∗ }) . By the definition of 𝛼 -EF1, we have 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 ∗ }) ≤ 𝛼 · 𝑐 𝑖 ( 𝐴 𝑗 ) . As a consequence, 𝑐 𝑖 ( 𝐴 𝑖 ) = 𝑐 𝑖 ( 𝑒 ∗ ) + 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 ∗ }) ≤ 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 ∗ }) + 𝑐 𝑖 ( 𝐴 𝑗 )≤ ( 𝛼 + ) · 𝑐 𝑖 ( 𝐴 𝑗 ) , (7)where the first inequality transition is due to 𝑐 𝑖 ( 𝑒 ∗ ) ≤ 𝑐 𝑖 ( 𝐴 𝑖 ∪ 𝐴 𝑗 \ { 𝑒 ∗ }) . Using Inequality (7)and additivity of cost function, we have 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝛼 + 𝛼 + · 𝑐 𝑖 ( 𝐴 𝑖 ∪ 𝐴 𝑗 ) . By the first point of Lemma2.1, we have MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) ≥ 𝑐 𝑖 ( 𝐴 𝑖 ∪ 𝐴 𝑗 ) and then, the following holds, 𝑐 𝑖 ( 𝐴 𝑖 ) MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) ≤ 𝛼 + 𝛼 + . As for tightness, consider an instance of 𝑛 agents and a set 𝐸 = { 𝑒 , . . . , 𝑒 𝑛 + } of 𝑛 + 𝑐 ( 𝑒 ) = 𝛼 + , 𝑐 ( 𝑒 ) = 𝛼 and 𝑐 ( 𝑒 𝑗 ) = , ∀ 𝑗 ≥
3. Then, consider an allocation B = ( 𝐵 , . . . , 𝐵 𝑛 ) with 𝐵 = { 𝑒 , 𝑒 } and 𝐵 𝑗 = { 𝑒 𝑗 + } , ∀ 𝑗 ≥
2. It is not hard to see allocation B satisfying 𝛼 -EF1. As for PMMS, weonly need to consider agent 1 since he is the only one envying other agents. Notice that forany 𝑗 ≥
2, the combined bundle 𝐵 ∪ 𝐵 𝑗 contains three chores with cost 𝛼 + , 𝛼,
1, respectively.Thus, for any 𝑗 ≥
2, we have MMS ( , 𝐵 ∪ 𝐵 𝑗 ) = 𝛼 + 𝑐 ( 𝐵 ) MMS ( ,𝐵 ∪ 𝐵 𝑗 ) = 𝛼 + 𝛼 + ,completing the proof. (cid:3) In addition to the approximation guarantee for PMMS, Proposition 3.7 also has a directimplication in approximating PMMS algorithmically. It is known that an EF1 allocation canbe found efficiently by allocating chores in a round-robin fashion — agents in turn pick theirmost preferred chores from the remaining (Aziz et al., 2019a). Therefore, Proposition 3.7 with 𝛼 = leads to the following corollary, which is the only algorithmic result for PMMS (in choresallocation), to the best of our knowledge. 13 orollary 3.1. The round-robin algorithm outputs a - PMMS allocation in polynomial time.
Note that PMMS implies EFX in goods allocation according to Caragiannis et al. (2019b). Thisimplication also holds in allocating chores as stated in our proposition below.
Proposition 4.1. A PMMS allocation is also
EFX .Proof.
Let A = ( 𝐴 , . . . , 𝐴 𝑛 ) be a PMMS allocation. For sake of contradiction, assume A is notEFX and agent 𝑖 violate the condition of EFX. If 𝐴 𝑖 = ∅ or 𝐴 𝑖 only contains chores with zerocost for agent 𝑖 , then 𝑐 𝑖 ( 𝐴 𝑖 ) = which would not violate the condition of EFX. Thus, withoutloss of generality, we can assume 𝐴 𝑖 ≠ ∅ and 𝐴 𝑖 contains chores with positive cost for agent 𝑖 .As agent 𝑖 violates the condition of EFX, there must exist an agent 𝑗 ∈ 𝑁 and 𝑒 ∗ ∈ 𝐴 𝑖 with 𝑐 𝑖 ( 𝑒 ∗ ) > such that 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 ∗ }) > 𝑐 𝑖 ( 𝐴 𝑗 ) . Note chore 𝑒 ∗ is well-defined owing to 𝑐 𝑖 ( 𝐴 𝑖 ) > .Now, consider the -partition (cid:8) 𝐴 𝑖 \ { 𝑒 ∗ } , 𝐴 𝑗 ∪ { 𝑒 ∗ } (cid:9) ∈ Π ( 𝐴 𝑖 ∪ 𝐴 𝑗 ) . By 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 ∗ }) > 𝑐 𝑖 ( 𝐴 𝑗 ) ,the following holds: 𝑐 𝑖 ( 𝐴 𝑖 ) > max (cid:8) 𝑐 𝑖 ( 𝐴 𝑖 \ { 𝑒 ∗ }) , 𝑐 𝑖 ( 𝐴 𝑗 ∪ { 𝑒 ∗ }) (cid:9) ≥ min B ∈ Π ( 𝐴 𝑖 ∪ 𝐴 𝑗 ) max { 𝑐 𝑖 ( 𝐵 ) , 𝑐 𝑖 ( 𝐵 )} ≥ 𝑐 𝑖 ( 𝐴 𝑖 ) , (8)where the last transition is by the definition of PMMS. Inequality (8) is a contradiction, andtherefore, A must be an EFX allocation. (cid:3) Since EFX implies EF1, Proposition 4.1 directly leads to the following corollary.
Corollary 4.1. A PMMS allocation is also
EF1 . For approximate version of PMMS, when allocating goods it is shown in Amanatidis et al.(2018) that for any 𝛼 , 𝛼 -PMMS can imply 𝛼 − 𝛼 -EF1. However, in the case of chores, our resultsindicate that 𝛼 -PMMS has no bounded guarantee for EF1. Proposition 4.2.
For 𝑛 ≥ , an 𝛼 - PMMS allocation with 𝛼 > is not necessarily 𝛽 - EF1 forany 𝛽 ≥ .Proof. First note according to Lemma 2.2 that we can assume without loss of generality that < 𝛼 < . Consider an instance with 𝑛 agents and 𝑛 + chores 𝑒 . . . , 𝑒 𝑛 + . Agents haveidentical cost profile. For any agent 𝑖 , the cost function is as follow: 𝑐 𝑖 ( 𝑒 ) = 𝛼 − , 𝑐 𝑖 ( 𝑒 ) = and 𝑐 𝑖 ( 𝑒 𝑗 ) = 𝜖 , ∀ 𝑗 ≥ where 𝜖 takes any arbitrarily small positive value. Then, consider anallocation B = ( 𝐵 , . . . , 𝐵 𝑛 ) with 𝐵 = { 𝑒 , 𝑒 } and 𝐵 𝑗 = { 𝑒 𝑗 + } , ∀ 𝑗 ≥ . Consequently, except foragent 1, no one else will violate the condition of EF1 and 𝛼 -PMMS. As for agent 1, notice that 𝛼 − > + 𝜖 and thus, for any 𝑗 ≥ , the combined bundle 𝐵 ∪ 𝐵 𝑗 admits MMS ( , 𝐵 ∪ 𝐵 𝑗 ) = 𝛼 − that implies 𝑐 ( 𝐵 ) MMS ( ,𝐵 ∪ 𝐵 𝑗 ) = 𝛼 . Thus, allocation B is 𝛼 -PMMS. For the guarantee on EF1, as 𝑐 ( 𝐵 𝑗 ) = 𝜖 for any 𝑗 ≥ , then removing the chore with the largest cost from 𝐵 still yields theratio 𝑐 ( 𝐵 \{ 𝑒 }) 𝑐 ( 𝐵 𝑗 ) = 𝜖 → ∞ as 𝜖 → , completing the proof. (cid:3) 𝛼 ≥ , 𝛼 -EFX is stricter than 𝛼 -EF1, the impossibility result on EF1 inProposition 4.2 is also true for EFX. Proposition 4.3.
For 𝑛 ≥ , an 𝛼 - PMMS allocation with 𝛼 > is not necessarily a 𝛽 - EFX allocation for any 𝛽 ≥ . We now study the approximation guarantee of PMMS for MMS. Since these two notionscoincide when there are only two agents, we assume there are at least three agents. We firstprovide a tight bound for 𝑛 = and then give an almost tight bound for general 𝑛 . Proposition 4.4.
For 𝑛 = , a PMMS allocation is also - MMS , and moreover, this bound istight.Proof.
See Appendix A.1. (cid:3)
For general 𝑛 , we use the connections between PMMS, EFX and MMS to find the approxi-mation guarantee of PMMS for MMS. According to Proposition 4.1, a PMMS allocation is alsoEFX, and by Proposition 3.5, EFX implies 𝑛𝑛 + -MMS. As a result, we can claim that PMMSalso implies 𝑛𝑛 + -MMS. With the following proposition we show that this guarantee is almosttight. Proposition 4.5.
For 𝑛 ≥ , a PMMS allocation is 𝑛𝑛 + - MMS but not necessarily ( 𝑛 + 𝑛 + − 𝜖 ) - MMS for any 𝜖 > .Proof. The positive part directly follows from Propositions 4.1 and 3.5. As for the lower bound,consider an instance with 𝑛 (odd) agents and a set 𝐸 = { 𝑒 , . . . , 𝑒 𝑛 } of 𝑛 chores. We focus onagent 1 and his costs 𝑐 (cid:0) 𝑒 𝑗 (cid:1) = ( 𝑛 + , ≤ 𝑗 ≤ 𝑛, , 𝑛 + ≤ 𝑗 ≤ 𝑛. Now, consider an allocation B = ( 𝐵 , . . . , 𝐵 𝑛 ) with 𝐵 = { 𝑒 , 𝑒 } , 𝐵 𝑛 = { 𝑒 𝑛 + , . . . , 𝑒 𝑛 } and 𝐵 𝑗 = { 𝑒 𝑗 + } for any 𝑗 = , . . . , 𝑛 − . For agents 𝑖 ≥ , we let their cost on each of their receivedchore be < 𝛿 < 𝜖 and their cost on each of the other chores be one. As a result, exceptfor agent 1, no one else will violate the condition of MMS and PMMS. For agent 1, since forany 𝑗 ≥ , 𝑐 ( 𝐴 ) ≤ MMS ( , 𝐴 ∪ 𝐴 𝑗 ) holds, then allocation B is PMMS. For MMS ( 𝑛, 𝐸 ) , ithappens that 𝐸 can be evenly divided into 𝑛 bundles of the same cost (for agent 1), so we haveMMS ( 𝑛, 𝐸 ) = 𝑛 + yielding the ratio 𝑐 ( 𝐵 ) MMS ( 𝑛,𝑀 ) = 𝑛 + 𝑛 + , completing the proof. (cid:3) Next, we investigate the approximation guarantee of approximate PMMS for MMS. Let usstart with an example of six chores 𝐸 = { 𝑒 , . . . , 𝑒 } and three agents. We focus on agent 1 andthe cost function of agent 1 is 𝑐 ( 𝑒 𝑗 ) = for 𝑗 = , , and 𝑐 ( 𝑒 𝑗 ) = for 𝑗 = , , , thus clearly,MMS ( , 𝐸 ) = . Consider an allocation A = ( 𝐴 , 𝐴 , 𝐴 ) with 𝐴 = { 𝑒 , 𝑒 , 𝑒 } . It is not hardto verify that allocation A is a -PMMS allocation and also a 3-MMS allocation. Combiningthe result in Lemma 2.2, we observe that allocation A only has a trivial guarantee on the notionof MMS. Motivated by this example, we focus on 𝛼 -PMMS allocations with 𝛼 < .15 roposition 4.6. For any 𝑛 ≥ and < 𝛼 < , an 𝛼 - PMMS allocation is 𝑛𝛼𝛼 +( 𝑛 − ) ( − 𝛼 ) - MMS ,but not necessarily ( 𝑛𝛼𝛼 +( 𝑛 − ) ( − 𝛼 ) − 𝜖 ) - MMS for any 𝜖 > . Before we can prove the above proposition, we need the following two lemmas.
Lemma 4.7.
For any 𝑖 ∈ 𝑁 and bundle 𝑆 ⊆ 𝐸 , suppose MMS 𝑖 ( , 𝑆 ) is defined by a 2-partition T = ( 𝑇 , 𝑇 ) with 𝑐 𝑖 ( 𝑇 ) = MMS 𝑖 ( , 𝑆 ) . If the number of chores in 𝑇 is at least two, then 𝑐 𝑖 ( 𝑆 ) MMS 𝑖 ( ,𝑆 ) ≥ .Proof. For the sake of contradiction, we assume 𝑐 𝑖 ( 𝑆 ) MMS 𝑖 ( ,𝑆 ) < . Since 𝑐 𝑖 ( 𝑇 ) = MMS 𝑖 ( , 𝑆 ) , wehave 𝑐 𝑖 ( 𝑇 ) > 𝑐 𝑖 ( 𝑆 ) , and accordingly, 𝑐 𝑖 ( 𝑇 ) < 𝑐 𝑖 ( 𝑆 ) . Thus, 𝑐 𝑖 ( 𝑇 ) − 𝑐 𝑖 ( 𝑇 ) > 𝑐 𝑖 ( 𝑆 ) holds, andwe claim that each single chore in 𝑇 has cost strictly larger than 𝑐 𝑖 ( 𝑆 ) for agent 𝑖 ; otherwise,by moving the chore with the smallest cost in 𝑇 to 𝑇 , one can find a 2-partition in whichthe cost of larger bundle is smaller than 𝑐 𝑖 ( 𝑇 ) , contradiction. Based on our claim, we have | 𝑇 | = . Notice that for any 𝑒 ∈ 𝑇 , 𝑐 𝑖 ( 𝑒 ) > 𝑐 𝑖 ( 𝑇 ) holds. As a result, moving one chore from 𝑇 to 𝑇 results in a 2-partition, in which the cost of larger bundle is strictly smaller than 𝑐 𝑖 ( 𝑇 ) ,contradicting to the construction of allocation T . (cid:3) Lemma 4.8.
For any 𝑖 ∈ 𝑁 and bundles 𝑆 , 𝑆 ⊆ 𝐸 , if MMS 𝑖 ( , 𝑆 ∪ 𝑆 ) > MMS 𝑖 ( , 𝑆 ) , then MMS 𝑖 ( , 𝑆 ∪ 𝑆 ) ≤ 𝑐 𝑖 ( 𝑆 ) + 𝑐 𝑖 ( 𝑆 ) .Proof. Suppose MMS 𝑖 ( , 𝑆 ) is defined by partition ( 𝑇 , 𝑇 ) and we have MMS 𝑖 ( , 𝑆 ) = 𝑐 𝑖 ( 𝑇 ) .We distinguish two different cases according to the value of 𝑐 𝑖 ( 𝑇 ) . If 𝑐 𝑖 ( 𝑇 ) = 𝑐 𝑖 ( 𝑆 ) , thenconsider ( 𝑇 ∪ 𝑆 , 𝑇 ) , a 2-partition of 𝑆 ∪ 𝑆 . Clearly, MMS 𝑖 ( , 𝑆 ∪ 𝑆 ) ≤ 𝑐 𝑖 ( 𝑇 ∪ 𝑆 ) = 𝑐 𝑖 ( 𝑆 ) + 𝑐 𝑖 ( 𝑆 ) . If 𝑐 𝑖 ( 𝑇 ) > 𝑐 𝑖 ( 𝑆 ) , since MMS 𝑖 ( , 𝑆 ∪ 𝑆 ) > MMS 𝑖 ( , 𝑆 ) , we can claim that 𝑐 𝑖 ( 𝑇 ) − 𝑐 𝑖 ( 𝑇 ) < 𝑐 𝑖 ( 𝑆 ) ; otherwise, the 2-partition ( 𝑇 , 𝑇 ∪ 𝑆 ) defines MMS 𝑖 ( , 𝑆 ∪ 𝑆 ) andMMS 𝑖 ( , 𝑆 ∪ 𝑆 ) = 𝑐 𝑖 ( 𝑇 ) = MMS 𝑖 ( , 𝑆 ) , contradiction. Then, let us consider ( 𝑇 ∪ 𝑆 , 𝑇 ) ,another 2-partition of 𝑆 ∪ 𝑆 . According to our claim, we have 𝑐 𝑖 ( 𝑇 ∪ 𝑆 ) > 𝑐 𝑖 ( 𝑇 ) , andthus, MMS 𝑖 ( , 𝑆 ∪ 𝑆 ) ≤ 𝑐 𝑖 ( 𝑇 ∪ 𝑆 ) < 𝑐 𝑖 ( 𝑆 ) + 𝑐 𝑖 ( 𝑆 ) where the last inequality is due to 𝑐 𝑖 ( 𝑇 ) = 𝑐 𝑖 ( 𝑆 ) − 𝑐 𝑖 ( 𝑇 ) < 𝑐 𝑖 ( 𝑆 ) . (cid:3) We now are ready for providing the proof of Proposition 4.6.
Proof of Proposition 4.6.
We first prove the upper bound. Let A = ( 𝐴 , ..., 𝐴 𝑛 ) be an 𝛼 -PMMSallocation and we focus our analysis on agent 𝑖 . Let 𝛼 ( 𝑖 ) = max 𝑗 ≠ 𝑖 𝑐 𝑖 ( 𝐴 𝑖 ) MMS 𝑖 ( ,𝐴 𝑖 ∪ 𝐴 𝑗 ) and 𝑗 ( 𝑖 ) be theindex such that MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ( 𝑖 ) ) ≤ MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) for any 𝑗 ∈ 𝑁 (tie breaks arbitrarily).By these constructions, clearly, 𝛼 = max 𝑖 ∈ 𝑁 𝛼 ( 𝑖 ) and 𝑐 𝑖 ( 𝐴 𝑖 ) = 𝛼 ( 𝑖 ) · MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ( 𝑖 ) ) . Then,we split our proof into two different cases. Case 1: ∃ 𝑗 ≠ 𝑖 such that MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) = MMS 𝑖 ( , 𝐴 𝑖 ) . Then 𝛼 ( 𝑖 ) = 𝑐 𝑖 ( 𝐴 𝑖 ) MMS 𝑖 ( ,𝐴 𝑖 ) holds.Suppose MMS 𝑖 ( , 𝐴 𝑖 ) is defined by the 2-partition ( 𝑇 , 𝑇 ) with 𝑐 𝑖 ( 𝑇 ) = MMS 𝑖 ( , 𝐴 𝑖 ) . If | 𝑇 | ≥ ,by Lemma 4.7, we have 𝛼 ( 𝑖 ) = 𝑐 𝑖 ( 𝐴 𝑖 ) MMS 𝑖 ( ,𝐴 𝑖 ) ≥ , contradicting to 𝛼 ( 𝑖 ) ≤ 𝛼 < . As a result, wecan further assume | 𝑇 | = . By the first point of Lemma 2.1, we have MMS 𝑖 ( 𝑛, 𝐸 ) ≥ 𝑐 𝑖 ( 𝑇 ) andaccordingly, 𝑐 𝑖 ( 𝐴 𝑖 ) MMS 𝑖 ( 𝑛,𝐸 ) ≤ 𝑐 𝑖 ( 𝐴 𝑖 ) 𝑐 𝑖 ( 𝑇 ) = 𝛼 ( 𝑖 ) ≤ 𝛼 . For < 𝛼 < and 𝑛 ≥ , it is not hard to verifythat 𝛼 ≤ 𝑛𝛼𝛼 +( 𝑛 − ) ( − 𝛼 ) , completing the proof for this case.16 ase 2: ∀ 𝑗 ≠ 𝑖 , MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) > MMS 𝑖 ( , 𝐴 𝑖 ) holds. According to Lemma 4.8, for any 𝑗 ≠ 𝑖 , the following holds MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) ≤ 𝑐 𝑖 ( 𝐴 𝑖 ) + 𝑐 𝑖 ( 𝐴 𝑗 ) . (9)Due to the construction of 𝛼 ( 𝑖 ) , for any 𝑗 ≠ 𝑖 , we have 𝑐 𝑖 ( 𝐴 𝑖 ) ≤ 𝛼 ( 𝑖 ) · MMS 𝑖 ( , 𝐴 𝑖 ∪ 𝐴 𝑗 ) . CombiningInequality (9), we have 𝑐 𝑖 ( 𝐴 𝑗 ) ≥ − 𝛼 ( 𝑖 ) 𝛼 ( 𝑖 ) 𝑐 𝑖 ( 𝐴 𝑖 ) for any 𝑗 ≠ 𝑖 . Thus, the following holds, 𝑐 𝑖 ( 𝐴 𝑖 ) MMS 𝑖 ( 𝑛, 𝐸 ) ≤ 𝑛𝑐 𝑖 ( 𝐴 𝑖 ) 𝑐 𝑖 ( 𝐸 ) ≤ 𝑛𝑐 𝑖 ( 𝐴 𝑖 ) 𝑐 𝑖 ( 𝐴 𝑖 ) + ( 𝑛 − ) − 𝛼 ( 𝑖 ) 𝛼 ( 𝑖 ) 𝑐 𝑖 ( 𝐴 𝑖 ) . (10)The last expression in (10) is monotonically increasing in 𝛼 ( 𝑖 ) , and accordingly, we have 𝑐 𝑖 ( 𝐴 𝑖 ) MMS 𝑖 ( 𝑛, 𝐸 ) ≤ 𝑛𝛼𝛼 + ( 𝑛 − ) ( − 𝛼 ) . As for the lower bound, consider an instance of 𝑛 agents with 𝑛 ∈ N + and a set 𝐸 = { 𝑒 , ..., 𝑒 𝑛 } of 𝑛 chores. Agents have identical cost functions. The cost function of agent 1 isas follows: 𝑐 ( 𝑒 𝑗 ) = 𝛼 for 𝑗 = , ..., 𝑛 and 𝑐 ( 𝑒 𝑗 ) = − 𝛼 for 𝑗 = 𝑛 + , ..., 𝑛 . Now, consider anallocation B = ( 𝐵 , ..., 𝐵 𝑛 ) with 𝐵 𝑖 = { 𝑒 ( 𝑛 − ) 𝑖 + , ..., 𝑒 𝑛𝑖 } for 𝑖 = , ..., 𝑛 . Since 𝛼 > , it is easyto see that, except for agent 1, no one else will violate the condition of PMMS, and moreover,the approximation guarantee on MMS is determined by agent 1. For agent 1, since 𝑛 ∈ N + ,MMS ( , 𝐵 ∪ 𝐵 𝑗 ) = 𝑛 holds for any 𝑗 ≥ , and due to 𝑐 ( 𝐵 ) = 𝑛𝛼 , we can claim that theallocation B is 𝛼 -PMMS. It is not hard to verify that MMS ( 𝑛, 𝐸 ) = 𝛼 + ( 𝑛 − ) ( − 𝛼 ) , yieldingthe ratio 𝑛𝛼𝛼 +( 𝑛 − ) ( − 𝛼 ) , completing the proof. (cid:3) The motivating example right before Proposition 4.6, unfortunately, only works for the caseof 𝑛 = . When 𝑛 becomes larger, an 𝛼 -PMMS allocation with 𝛼 ≥ is still possible to providea non-trivial approximation guarantee on the notion of MMS.We remain to consider the approximation guarantee of MMS for other fairness criteria.Notice that all of EFX, EF1 and PMMS can have non-trivial guarantee for MMS (i.e., betterthan 𝑛 -MMS). However, the converse is not true and even the exact MMS does not provide anysubstantial guarantee for the other three criteria. Proposition 4.9.
For any 𝑛 ≥ , there exists an MMS allocation that is only 2-
PMMS .Proof.
Consider an instance with 𝑛 agents and 𝑝 + 𝑛 − chores denoted as { 𝑒 , . . . , 𝑒 𝑛 + 𝑝 − } where 𝑝 is a nature number taking arbitrarily large value. We focus on agent 1 and his costs 𝑐 (cid:0) 𝑒 𝑗 (cid:1) = ( , ≤ 𝑗 ≤ 𝑛 + 𝑝,𝑝, 𝑛 + 𝑝 + ≤ 𝑗 ≤ 𝑛 + 𝑝 − . Consider allocation B = ( 𝐵 , . . . , 𝐵 𝑛 ) with 𝐵 = (cid:8) 𝑒 , . . . , 𝑒 𝑝 + (cid:9) , 𝐵 𝑖 = (cid:8) 𝑒 𝑝 + 𝑖 (cid:9) , ∀ 𝑖 = , . . . , 𝑛 − , 𝐵 𝑛 − = { 𝑒 𝑛 + 𝑝 − , 𝑒 𝑛 + 𝑝 } and 𝐵 𝑛 = { 𝑒 𝑛 + 𝑝 + , . . . , 𝑒 𝑛 + 𝑝 − } . For agent 𝑖 ≥ , we let their cost on eachof their received chore be 𝑝 and their cost on each of the rest chores be one. Consequently,17xcept for agent 1, no one else will violate the condition of MMS and PMMS. For MMS ( 𝑛, 𝐸 ) ,it happens that 𝐸 can be evenly divided into n bundles of the same cost (for agent 1), so we haveMMS ( 𝑛, 𝐸 ) = 𝑝 + . Accordingly, 𝑐 ( 𝐵 ) = MMS ( 𝑛, 𝐸 ) holds and thus, allocation B is MMS.In terms of approximation guarantee on PMMS, consider the combined bundle 𝐵 ∪ 𝐵 and it isnot hard to verify that MMS ( , 𝐵 ∪ 𝐵 ) = ⌈ 𝑝 + ⌉ yielding the ratio 𝑐 ( 𝐵 ) MMS ( ,𝐵 ∪ 𝐵 ) = 𝑝 + ⌈ 𝑝 + ⌉ → as 𝑝 → ∞ . (cid:3) Proposition 4.10. An MMS allocation is not necessarily 𝛽 - EF1 or 𝛽 - EFX for any 𝛽 ≥ .Proof. By Proposition 3.3, the notion 𝛽 -EFX is stricter than 𝛽 -EF1 for any 𝛽 ≥ , and thus,we only need to show the unbounded guarantee on EF1. Again, we consider the instance givenin the proof of Proposition 4.9. As stated in that proof, B is an MMS allocation, and exceptfor agent 1, no one else will violate the condition of PMMS. Note that PMMS is stricter thanEF1, then no one else will violate the condition of EF1. As for agent 1, each chore in 𝐵 hasthe same cost for him, so we can remove any single chore in 𝐵 and check its performance interms of EF1. Comparing with another bundle 𝐵 , we have the ratio 𝑐 ( 𝐵 \{ 𝑒 }) 𝑐 ( 𝐵 ) = 𝑝 → ∞ as 𝑝 → ∞ . (cid:3) After having compared the fairness criteria between themselves, in this section we study the effi-ciency of these fairness criteria in terms of the price of fairness with respect to social optimalityof an allocation.
We start with the case of two players. Our first result concerns EF1.
Proposition 5.1.
The price of
EF1 is / when there are two agents.Proof. For the upper bound part, we analyze the allocation returned by algorithm
𝐴𝐿𝐺 , whosedetailed description is in Appendix A.2. We first show that 𝐴𝐿𝐺 is well-defined and can alwaysoutput an EF1 allocation. Note that O is the optimal allocation for the underlying instance dueto the order of chores. We consider the possible value of index 𝑠 . Because of the normalized costfunction, trivially, 𝑠 < 𝑚 holds. If 𝑠 = , 𝐴𝐿𝐺 outputs the allocation returned by round-robin(line 6) and clearly, it’s EF1. If the optimal allocation O is EF1 (line 9), we are done. For thiscase, we claim that if 𝑠 = 𝑚 − , then O must be EF1. The reason is that for agent 1, his cost 𝑐 ( 𝑂 ) ≤ 𝑐 ( 𝑂 ) ≤ 𝑐 ( 𝑂 ) where the first transition due to line 1 of 𝐴𝐿𝐺 , and thus he doesnot envy agent 2. For agent 2, since he only receives a single chore in optimal allocation due to 𝑠 = 𝑚 − , clearly, he does not violate the condition of EF1, either. Thus, allocation O is EF1 inthe case of 𝑠 = 𝑚 − . Next, we focus on the case where O is not EF1 and this can only happenwhen ≤ 𝑠 ≤ 𝑚 − . We first show that the index 𝑓 is well-defined. Notice that 𝑐 ( 𝑅 ( 𝑓 + )) is monotonically decreasing on 𝑓 and 𝑐 ( 𝐿 ( 𝑓 )) is monotonically increasing on 𝑓 , it suffices toshow 𝑐 ( 𝑅 ( 𝑠 + )) > 𝑐 ( 𝐿 ( 𝑠 )) . For the sake of contradiction, assume 𝑐 ( 𝑅 ( 𝑠 + )) ≤ 𝑐 ( 𝐿 ( 𝑠 )) .18his is equivalent to 𝑐 ( 𝑂 \ { 𝑒 𝑠 + }) ≤ 𝑐 ( 𝑂 ) , which means agent 2 satisfying EF1 in allocation O . Due to the assumption (line 1), 𝑐 ( 𝑂 ) ≤ 𝑐 ( 𝑂 ) ≤ 𝑐 ( 𝑂 ) holds, and thus, agent 1 is EFunder the allocation O . Consequently, the allocation O is EF1, contradiction. Then, we proveallocation A (line 13) is EF1. According to the order of chores, it holds that 𝑐 ( 𝐿 ( 𝑓 )) 𝑐 ( 𝐿 ( 𝑓 )) ≤ 𝑐 ( 𝑅 ( 𝑓 + )) 𝑐 ( 𝑅 ( 𝑓 + )) . Since 𝑐 ( 𝑅 ( 𝑓 + )) > 𝑐 ( 𝐿 ( 𝑓 )) ≥ , this implies, 𝑐 ( 𝐿 ( 𝑓 )) 𝑐 ( 𝑅 ( 𝑓 + )) ≤ 𝑐 ( 𝐿 ( 𝑓 )) 𝑐 ( 𝑅 ( 𝑓 + )) . By the definition of index 𝑓 , we have 𝑐 ( 𝑅 ( 𝑓 + )) > 𝑐 ( 𝐿 ( 𝑓 )) and therefore 𝑐 ( 𝐿 ( 𝑓 )) < 𝑐 ( 𝑅 ( 𝑓 + )) which is equivalent to 𝑐 ( 𝐴 \ { 𝑒 𝑓 + }) < 𝑐 ( 𝐴 ) . Thus, agent 1 is EF1 under allocation A .As for agent 2, if 𝑓 = 𝑚 − , then 𝐴 = and clearly, agent 2 does not violate the condition ofEF1. We can further assume 𝑓 ≤ 𝑚 − . Since 𝑓 is the maximum index satisfying 𝑓 ≥ 𝑠 and 𝑐 ( 𝑅 ( 𝑓 + )) > 𝑐 ( 𝐿 ( 𝑓 )) , it must hold that 𝑐 ( 𝑅 ( 𝑓 + )) ≤ 𝑐 ( 𝐿 ( 𝑓 + )) , which is equivalent to 𝑐 ( 𝐴 \ { 𝑒 𝑓 + }) ≤ 𝑐 ( 𝐴 ) and so agent 2 is also EF1 under allocation A .Next, we show the social cost of the allocation returned by 𝐴𝐿𝐺 is at most 1.25 times of theoptimal social cost. If 𝑠 = , both agents have the same cost profile, then any allocations havethe optimal social cost and we are done in this case. If allocation O is EF1, then clearly, we aredone. The remaining is the case where allocation O is not EF1. Since 𝑐 ( 𝑂 ) ≤ 𝑐 ( 𝑂 ) ≤ 𝑐 ( 𝑂 ) ,we have 𝑐 ( 𝑂 ) ≤ . Notice that O is not EF1, then 𝑐 ( 𝑂 ) > must hold; otherwise, 𝑐 ( 𝑂 ) ≤ 𝑐 ( 𝑂 ) and allocation O is EF, contradiction. Therefore, under the case where allocation O isnot EF1, we must have 𝑐 ( 𝑂 ) ≤ and 𝑐 ( 𝑂 ) > . Due to 𝑓 + ≥ 𝑠 + and the order of chores,it holds that 𝑐 ( 𝑅 ( 𝑓 + )) 𝑐 ( 𝑅 ( 𝑓 + )) ≥ 𝑐 ( 𝑂 ) 𝑐 ( 𝑂 ) . This implies 𝑐 ( 𝑅 ( 𝑓 + )) ≥ 𝑐 ( 𝑂 ) 𝑐 ( 𝑂 ) 𝑐 ( 𝑅 ( 𝑓 + )) , and equivalently, 𝑐 ( 𝐴 ) = 𝑐 ( 𝐿 ( 𝑓 + )) ≤ − 𝑐 ( 𝑂 ) 𝑐 ( 𝑂 ) 𝑐 ( 𝑅 ( 𝑓 + )) . Again, by the construction of 𝑓 , we have 𝑐 ( 𝐴 ) = 𝑐 ( 𝑅 ( 𝑓 + )) > 𝑐 ( 𝐿 ( 𝑓 )) ≥ 𝑐 ( 𝐿 ( 𝑠 )) = 𝑐 ( 𝑂 ) . 𝑐 ( 𝐴 ) + 𝑐 ( 𝐴 ) ≤ − ( 𝑐 ( 𝑂 ) 𝑐 ( 𝑂 ) − ) 𝑐 ( 𝐴 )≤ − ( 𝑐 ( 𝑂 ) 𝑐 ( 𝑂 ) − ) 𝑐 ( 𝑂 ) = − ( − 𝑐 ( 𝑂 ) 𝑐 ( 𝑂 ) − ) ( − 𝑐 ( 𝑂 )) , (11)where the second inequality is due to 𝑐 ( 𝑂 ) 𝑐 ( 𝑂 ) ≥ and 𝑐 ( 𝐴 ) ≥ 𝑐 ( 𝑂 ) . Based on (11), we havean upper bound on the price of EF1 as followsPrice of EF1 ≤ − ( − 𝑐 ( 𝑂 ) 𝑐 ( 𝑂 ) − ) ( − 𝑐 ( 𝑂 )) 𝑐 ( 𝑂 ) + 𝑐 ( 𝑂 ) . (12)Recall ≤ 𝑐 ( 𝑂 ) ≤ < 𝑐 ( 𝑂 ) ≤ . The partial derivatives of the faction in Inequality (12)with respect to 𝑐 ( 𝑂 ) equals to the following ( 𝑐 ( 𝑂 ) + 𝑐 ( 𝑂 )) ( 𝑐 ( 𝑂 ) − ) . It is not hard to see this derivatives has negative value for all < 𝑐 ( 𝑂 ) ≤ . Thus, the fractionin Inequality (12) takes maximum value only when 𝑐 ( 𝑂 ) = and hence,Price of EF1 ≤ − 𝑐 ( 𝑂 ) 𝑐 ( 𝑂 ) − . Similarly, by taking the derivative with respect to 𝑐 ( 𝑂 ) , the maximum value of this expressionhappens only when 𝑐 ( 𝑂 ) = , then one can easily compute the maximum value of RHS ofInequality (12) is . . Therefore, the price of EF1 ≤ . As for the lower bound, consider an instance with a set 𝐸 = { 𝑒 , 𝑒 , 𝑒 } of three chores.The cost function of agent 1 is 𝑐 ( 𝑒 ) = and 𝑐 ( 𝑒 ) = 𝑐 ( 𝑒 ) = . For agent 2, his cost is 𝑐 ( 𝑒 ) = − 𝜖 and 𝑐 ( 𝑒 ) = 𝑐 ( 𝑒 ) = + 𝜖 where 𝜖 > takes arbitrarily small value. An optimalallocation assigns chore 𝑒 to agent 1 and the rest chores to agent 2, which yields the optimalsocial cost + 𝜖 . However, this allocation is not EF1 since agent 2 envies agent 1 even removingone chore from his bundle. To achieve EF1, agent 2 can not receive both of chores 𝑒 and 𝑒 ,and so, agent 1 must receive one of chore 𝑒 and 𝑒 . Therefore, the best EF1 allocation can beassigning chore 𝑒 and 𝑒 to agent 1 and chore 𝑒 to agent 2 resulting in the social cost + 𝜖 .Thus, the price of EF1 is at least + 𝜖 + 𝜖 → as 𝜖 → , completing the proof. (cid:3) According to Propositions 3.4 and 3.6, EF1 implies 2-MMS and -PMMS. The followingtwo propositions confirm an intuition — if one relaxes the fairness condition, then less efficiencywill be sacrificed. Proposition 5.2.
The price of 2-
MMS is 1 when there are two agents.
Proposition 5.3.
The price of - PMMS is / when there are two agents.Proof. We first prove the upper bound. Given an instance 𝐼 , let O = ( 𝑂 , 𝑂 ) be an optimalallocation of 𝐼 . If the allocation O is already -PMMS, we are done. For the sake of contradiction,we assume that agent 1 violates the condition of -PMMS in allocation O , i.e., 𝑐 ( 𝑂 ) > MMS ( , 𝐸 ) . Suppose 𝑂 = { 𝑒 , . . . , 𝑒 ℎ } and the index satisfies the following rule; 𝑐 ( 𝑒 ) 𝑐 ( 𝑒 ) ≥ 𝑐 ( 𝑒 ) 𝑐 ( 𝑒 ) ≥ · · · ≥ 𝑐 ( 𝑒 ℎ ) 𝑐 ( 𝑒 ℎ ) . In this proof, for simplicity, we write 𝐿 ( 𝑘 ) : = { 𝑒 , ..., 𝑒 𝑘 } for any ≤ 𝑘 ≤ ℎ and 𝐿 ( ) = ∅ . Then, let 𝑠 be the index such that 𝑐 ( 𝑂 \ 𝐿 ( 𝑠 )) ≤ MMS ( , 𝐸 ) and 𝑐 ( 𝑂 \ 𝐿 ( 𝑠 − )) > MMS ( , 𝐸 ) . In the following, we divide our proof into two cases. Case 1: 𝑐 ( 𝐿 ( 𝑠 )) ≤ 𝑐 ( 𝑂 ) . Consider allocation A = ( 𝐴 , 𝐴 ) with 𝐴 = 𝑂 \ 𝐿 ( 𝑠 ) and 𝐴 = 𝑂 ∪ 𝐿 ( 𝑠 ) . We first show allocation A is -PMMS. For agent 1, due to the construction of index 𝑠 ,he does not violate the condition of -PMMS. As for agent 2, we claim that 𝑐 ( 𝐴 ) = − 𝑐 ( 𝑂 \ 𝐿 ( 𝑠 − )) + 𝑐 ( 𝑒 𝑠 ) < + 𝑐 ( 𝑒 𝑠 ) because 𝑐 ( 𝑂 \ 𝐿 ( 𝑠 − )) ≥ 𝑐 ( 𝑂 \ 𝐿 ( 𝑠 − )) > MMS ( , 𝐸 ) ≥ where the first inequality transition is due to the fact that 𝑂 is the bundle assigned to agent 1in the optimal allocation. If 𝑐 ( 𝑒 𝑠 ) < , then clearly, 𝑐 ( 𝐴 ) < ≤ MMS ( , 𝐸 ) . If 𝑐 ( 𝑒 𝑠 ) ≥ ,then 𝑐 ( 𝑒 𝑠 ) = MMS ( , 𝐸 ) and accordingly, it is not hard to verify that 𝑐 ( 𝐴 ) ≤ MMS ( , 𝐸 ) .Thus, A is a -PMMS allocation.Next, based on allocation A , we derive an upper bound on the price of -PMMS. First, by theorder of index, 𝑐 ( 𝐿 ( 𝑠 )) 𝑐 ( 𝐿 ( 𝑠 )) ≥ 𝑐 ( 𝑂 ) 𝑐 ( 𝑂 ) holds, implying 𝑐 ( 𝐿 ( 𝑠 )) ≤ 𝑐 ( 𝑂 ) 𝑐 ( 𝑂 ) 𝑐 ( 𝐿 ( 𝑠 )) . Since 𝐴 = 𝑂 \ 𝐿 ( 𝑠 ) and 𝐴 = 𝑂 ∪ 𝐿 ( 𝑠 ) , we have the following:Price of -PMMS ≤ + 𝑐 ( 𝐿 ( 𝑠 )) − 𝑐 ( 𝐿 ( 𝑠 )) 𝑐 ( 𝑂 ) + 𝑐 ( 𝑂 )≤ + 𝑐 ( 𝐿 ( 𝑠 )) ( 𝑐 ( 𝑂 ) 𝑐 ( 𝑂 ) − ) 𝑐 ( 𝑂 ) + 𝑐 ( 𝑂 ) = + 𝑐 ( 𝐿 ( 𝑠 )) 𝑐 ( 𝑂 ) ( − 𝑐 ( 𝑂 ) − 𝑐 ( 𝑂 )) 𝑐 ( 𝑂 ) + 𝑐 ( 𝑂 )≤ + − ( 𝑐 ( 𝑂 ) + 𝑐 ( 𝑂 )) 𝑐 ( 𝑂 ) + 𝑐 ( 𝑂 )≤ − + × = , where the second inequality due to 𝑐 ( 𝐿 ( 𝑠 )) ≤ 𝑐 ( 𝑂 ) 𝑐 ( 𝑂 ) 𝑐 ( 𝐿 ( 𝑠 )) ; the third inequality due to thecondition of Case 1 ; and the last inequality is because 𝑐 ( 𝑂 ) > MMS ( , 𝐸 ) ≥ . Case 2: 𝑐 ( 𝐿 ( 𝑠 )) > 𝑐 ( 𝑂 ) . We first derive a lower bound on 𝑐 ( 𝑒 𝑠 ) . Since 𝑐 ( 𝑒 𝑠 ) = 𝑐 ( 𝑂 \ 𝐿 ( 𝑠 − )) + 𝑐 ( 𝐿 𝑠 ) − 𝑐 ( 𝑂 ) , combine which with the condition of Case 2 implying 𝑐 ( 𝑒 𝑠 ) >𝑐 ( 𝑂 \ 𝐿 ( 𝑠 − )) − 𝑐 ( 𝑂 ) , and consequently we have 𝑐 ( 𝑒 𝑠 ) > MMS ( , 𝐸 ) − 𝑐 ( 𝑂 ) ≥ where the last transition is due to MMS ( , 𝐸 ) ≥ and 𝑐 ( 𝑂 ) ≤ . Then, we consider twosubcases.If ≤ 𝑐 ( 𝑒 𝑠 ) − 𝑐 ( 𝑒 𝑠 ) ≤ , consider an allocation A = ( 𝐴 , 𝐴 ) with 𝐴 = 𝑂 \ { 𝑒 𝑠 } and 𝐴 = 𝑂 ∪ { 𝑒 𝑠 } . We first show the allocation A is -PMMS. For agent 1, since 𝑐 ( 𝑒 𝑠 ) > ,21 ( 𝐴 ) = 𝑐 ( 𝑂 ) − 𝑐 ( 𝑒 𝑠 ) < ≤ MMS ( , 𝐸 ) . As for agent 2, 𝑐 ( 𝐴 ) = 𝑐 ( 𝑂 ) + 𝑐 ( 𝑒 𝑠 ) ≤ − 𝑐 ( 𝑂 ) + 𝑐 ( 𝑒 𝑠 ) < + 𝑐 ( 𝑒 𝑠 ) . If 𝑐 ( 𝑒 𝑠 ) < , then clearly, 𝑐 ( 𝐴 ) ≤ < MMS ( , 𝐸 ) holds.If 𝑐 ( 𝑒 𝑠 ) ≥ , we have 𝑐 ( 𝑒 𝑠 ) = MMS ( , 𝐸 ) and accordingly, it is not hard to verify that 𝑐 ( 𝐴 ) ≤ MMS ( , 𝐸 ) . Thus, the allocation A is -PMMS. Next, based on the allocation A ,we derive an upper bound regarding the price of -PMMS,Price of -PMMS ≤ 𝑐 ( 𝑂 ) − 𝑐 ( 𝑒 𝑠 ) + 𝑐 ( 𝑂 ) + 𝑐 ( 𝑒 𝑠 ) 𝑐 ( 𝑂 ) + 𝑐 ( 𝑂 )≤ + × = , where the second inequality due to ≤ 𝑐 ( 𝑒 𝑠 ) − 𝑐 ( 𝑒 𝑠 ) ≤ and 𝑐 ( 𝑂 ) > .If 𝑐 ( 𝑒 𝑠 ) − 𝑐 ( 𝑒 𝑠 ) > , consider an allocation A ′ = ( 𝐴 ′ , 𝐴 ′ ) with 𝐴 ′ = { 𝑒 𝑠 } and 𝐴 ′ = 𝐸 \ { 𝑒 𝑠 } .We first show that the allocation A ′ is -PMMS. For agent 1, due to Lemma 2.1, 𝑐 ( 𝑒 𝑠 ) ≤ MMS ( , 𝐸 ) holds. As for agent 2, since 𝑐 ( 𝑒 𝑠 ) ≥ 𝑐 ( 𝑒 𝑠 ) > , we have 𝑐 ( 𝐴 ′ ) = 𝑐 ( 𝐸 ) − 𝑐 ( 𝑒 𝑠 ) < ≤ MMS ( , 𝐸 ) . Thus, the allocation A ′ is -PMMS. In the following, we first derive anupper bound for 𝑐 ( 𝑂 \ { 𝑒 𝑠 }) − 𝑐 ( 𝑂 \ { 𝑒 𝑠 }) , then based on the bound, we provide the targetupper bound for the price of fairness. Since 𝑐 ( 𝑂 ) > and 𝑐 ( 𝑒 𝑠 ) − 𝑐 ( 𝑒 𝑠 ) > , we have 𝑐 ( 𝑂 \ { 𝑒 𝑠 }) − 𝑐 ( 𝑂 \ { 𝑒 𝑠 }) = 𝑐 ( 𝑂 ) − 𝑐 ( 𝑂 ) − ( 𝑐 ( 𝑒 𝑠 ) − 𝑐 ( 𝑒 𝑠 )) < , and then, the followingholds, Price of -PMMS ≤ + 𝑐 ( 𝑂 \ { 𝑒 𝑠 }) − 𝑐 ( 𝑂 \ { 𝑒 𝑠 }) 𝑐 ( 𝑂 ) + 𝑐 ( 𝑂 )≤ + × = . Up to here, we complete the proof of upper bound.Regarding lower bound, consider an instance 𝐼 with two agents and a set 𝐸 = { 𝑒 , 𝑒 , 𝑒 , 𝑒 } offour chores. The cost function for agent 1 is: 𝑐 ( 𝑒 ) = , 𝑐 ( 𝑒 ) = + 𝜖 , 𝑐 ( 𝑒 ) = − 𝜖 , 𝑐 ( 𝑒 ) = where 𝜖 > takes arbitrarily small value. For agent 2, here cost function is: 𝑐 ( 𝑒 ) = 𝑐 ( 𝑒 ) = , 𝑐 ( 𝑒 ) = 𝑐 ( 𝑒 ) = . It is not hard to verify that MMS 𝑖 ( , 𝐸 ) = for any 𝑖 = , . Inthe optimal allocation, the assignment is; 𝑒 , 𝑒 to agent 1 and 𝑒 , 𝑒 to agent 2, resulting inOPT ( 𝐼 ) = + 𝜖 . Observe that to satisfy -PMMS, agent 1 cannot receive both chores 𝑒 , 𝑒 ,and accordingly, the minimum social cost of a -PMMS allocation is by assigning 𝑒 to agent1 and the rest chores to agent 2. Based on this instance, when 𝑛 = , the price of -PMMS isat least + 𝜖 → as 𝜖 → . (cid:3) We remark that if we have an oracle for the maximin share, then our constructive proof ofProposition 5.3 can be transformed into an efficient algorithm for finding a / -PMMS allocationwhose cost is at most times the optimal social cost. Moving to other fairness criteria, we havethe following uniform bound. Proposition 5.4.
The price of
PMMS, MMS , and
EFX are all 2 when there are two agents.Proof.
We first show results on the upper bound. When 𝑛 = , PMMS is identical with MMSand can imply EFX, so it suffices to show that the price of PMMS is at most 2. Given an instance22 , let allocation O = ( 𝑂 , 𝑂 ) be its optimal allocation and w.l.o.g, we assume 𝑐 ( 𝑂 ) ≤ 𝑐 ( 𝑂 ) .If 𝑐 ( 𝑂 ) ≤ , then we have 𝑐 ( 𝑂 ) ≤ − 𝑐 ( 𝑂 ) = 𝑐 ( 𝑂 ) and 𝑐 ( 𝑂 ) ≤ − 𝑐 ( 𝑂 ) = 𝑐 ( 𝑂 ) .So allocation O is an EF and accordingly is PMMS, which yields the price of PMMS equals toone. Thus, we can further assume 𝑐 ( 𝑂 ) > and hence the optimal social cost is larger than . We next show that there exist a PMMS allocation whose social cost is at most 1. W.l.o.g,we assume MMS ( , 𝐸 ) ≤ MMS ( , 𝐸 ) (the other case is symmetric). Let T = ( 𝑇 , 𝑇 ) be theallocation defining MMS ( , 𝐸 ) and 𝑐 ( 𝑇 ) ≤ 𝑐 ( 𝑇 ) = MMS ( , 𝐸 ) . If 𝑐 ( 𝑇 ) ≤ 𝑐 ( 𝑇 ) , thenallocation T is EF (also PMMS), and thus it hold that 𝑐 ( 𝑇 ) ≤ and 𝑐 ( 𝑇 ) ≤ . Therefore,social cost of allocation T is no more than one, which implies the price of PMMS is at mosttwo. If 𝑐 ( 𝑇 ) > 𝑐 ( 𝑇 ) , then consider the allocation T ′ = ( 𝑇 , 𝑇 ) . Since 𝑐 ( 𝑇 ) = MMS ( , 𝐸 ) and 𝑐 ( 𝑇 ) < 𝑐 ( 𝑇 ) , then T ′ is a PMMS allocation. Owing to MMS ( , 𝐸 ) ≤ MMS ( , 𝐸 ) , weclaim that 𝑐 ( 𝑇 ) ≤ 𝑐 ( 𝑇 ) ; otherwise, we have MMS ( , 𝐸 ) = 𝑐 ( 𝑇 ) > 𝑐 ( 𝑇 ) > 𝑐 ( 𝑇 ) , andequivalently, allocation T ′ is a 2-partition where the cost of both bundles for agent 2 is strictlysmaller than MMS ( , 𝐸 ) , contradicting to MMS ( , 𝐸 ) ≤ MMS ( , 𝐸 ) . By 𝑐 ( 𝑇 ) ≤ 𝑐 ( 𝑇 ) ,the social cost of allocation T ′ satisfies 𝑐 ( 𝑇 ) + 𝑐 ( 𝑇 ) ≤ and so the price of PMMS is at mosttwo.Regarding the tightness, consider an instance 𝐼 two agents and a set 𝐸 = { 𝑒 , 𝑒 , 𝑒 } of threechores. The cost function of agent 1 is : 𝑐 ( 𝑒 ) = , 𝑐 ( 𝑒 ) = − 𝜖 and 𝑐 ( 𝑒 ) = 𝜖 where 𝜖 > takes arbitrarily small value. For agent 2, his cost is 𝑐 ( 𝑒 ) = , 𝑐 ( 𝑒 ) = 𝜖 and 𝑐 ( 𝑒 ) = − 𝜖 .An optimal allocation assigns chores 𝑒 , 𝑒 to agent 2, and 𝑒 to agent 1, and consequently, theoptimal social cost equals to + 𝜖 . We first concern the tightness on the notion of PMMS(or MMS, these two are identical when 𝑛 = ). In any PMMS allocations, it must be the casethat an agent receives chore 𝑒 and the other one receives chores 𝑒 , 𝑒 , and thus the socialcost of PMMS allocations is one. Therefore, the price of PMMS and MMS is at least + 𝜖 → as 𝜖 → . As for EFX, similarly, it must be the case that in any EFX allocations, the agentreceiving chore 𝑒 can not receive any other chores. Thus, it not hard to verify that the socialcost of EFX allocations is also one and the price of EFX is at least + 𝜖 → as 𝜖 → . (cid:3) Note that the existence of an MMS allocation is not guaranteed in general Kurokawa et al.(2018); Aziz et al. (2017) and the existence of PMMS or EFX allocation is still open when 𝑛 ≥ . Nonetheless, we are still interested in the prices of fairness in case such a fair allocationdoes exist. Observe that when the number of chore 𝑚 ≤ , the price of EF1, EFX, PMMS istrivially 1. If 𝑚 = , assigning the unique chore to any agent satisfies all these three fairnesscriteria, so does the optimal allocation. If 𝑚 = , in an optimal allocation, it never happensthat both of the two chores are assigned to the same agent. The reason is that if an agent hasthe smallest cost on one chore, then his cost on another chore is higher than someone else dueto the normalized cost function. In the following, we settle down the case of 𝑚 ≥ . Proposition 5.5.
For 𝑛 ≥ and 𝑚 ≥ , the price of EF1, EFX and
PMMS are all infinite. roof. In this proof, 𝜖 always takes arbitrarily small positive value. Consider an instance with 𝑛 agents and 𝑚 chores. For agent 𝑖 ≥ , his cost function is: 𝑐 𝑖 ( 𝑒 𝑗 ) = 𝑚 for any 𝑗 = , . . . , 𝑚 .For agent 1, his cost is 𝑐 ( 𝑒 ) = − 𝜖 ; 𝑐 ( 𝑒 𝑗 ) = for 𝑗 = , . . . , 𝑚 − ; 𝑐 ( 𝑒 𝑚 − ) = 𝑐 ( 𝑒 𝑚 ) = 𝜖 .For agent 2, his cost is 𝑐 ( 𝑒 ) = − 𝑚 ; 𝑐 ( 𝑒 𝑗 ) = for 𝑗 = , . . . , 𝑚 − ; 𝑐 ( 𝑒 𝑚 − ) = 𝑐 ( 𝑒 𝑚 ) = 𝑚 .Last, for agent 3, his cost is 𝑐 ( 𝑒 ) = 𝜖 ; 𝑐 ( 𝑒 𝑗 ) = 𝑚 for 𝑗 = , . . . , 𝑚 − ; and 𝑐 ( 𝑒 𝑚 ) = 𝑚 − 𝜖 . Anoptimal allocation assigns 𝑒 𝑚 − , 𝑒 𝑚 to agent 1 and 𝑒 to agent 3. For each of rest chore, it isassigned to the agents with zero cost on it. Accordingly, the optimal social cost is 𝜖 . However,in any optimal allocations, agent 1 violates the condition of EF1 since he would still envy agent2 even removing one chore from his bundle. Observe that this instance has EF1, EFX andPMMS allocation. For any optimal allocation, if we reassign chore 𝑒 𝑚 to agent 2, then theresulting allocation would satisfy all three fairness notions. However, it is not hard to verifythat the allocation satisfies any one or more fairness notions in question has social cost morethan 𝑚 , yielding an infinite ratio when 𝜖 → . (cid:3) In the context of goods allocation, Barman et al. (2020) present an asymptotically tightprice of EF1, 𝑂 (√ 𝑛 ) . However, as shown by Proposition 5.5, when allocating chores, the priceof EF1 is infinite, which shows a sharp contrast between goods and chores allocation.By using a similar construction to the one in the proof of Proposition 5.5, we can establishthe following proposition. Proposition 5.6.
For 𝑛 ≥ , the price of - PMMS is infinite.Proof.
In this proof, 𝜖 always takes arbitrarily small positive value. Consider an instance with 𝑛 agents and 𝑚 ≥ chores. The cost function of agent 1 is 𝑐 ( 𝑒 ) = − 𝜖 ; 𝑐 ( 𝑒 𝑗 ) = for 𝑗 = , . . . , 𝑚 − ; 𝑐 ( 𝑒 𝑗 ) = 𝜖 for 𝑗 ≥ 𝑚 − . For agent 2, his cost is 𝑐 ( 𝑒 ) = − 𝑚 ; 𝑐 ( 𝑒 𝑗 ) = for 𝑗 = , . . . , 𝑚 − ; 𝑐 ( 𝑒 𝑗 ) = 𝑚 for 𝑗 ≥ 𝑚 − . The cost function of agent 3 is as follows; 𝑐 ( 𝑒 ) = 𝜖 ; 𝑐 ( 𝑒 𝑗 ) = 𝑚 for 𝑗 = , . . . , 𝑚 − and 𝑐 ( 𝑒 𝑚 ) = 𝑚 − 𝜖 . For any agent 𝑖 ≥ , his costis 𝑐 𝑖 ( 𝑒 𝑗 ) = 𝑚 for any ≤ 𝑗 ≤ 𝑚 . An optimal allocation assigns 𝑒 𝑚 − , 𝑒 𝑚 − , 𝑒 𝑚 − , 𝑒 𝑚 to agent1 and 𝑒 to agent 3. For each of rest chore, it is assigned to the agents with zero cost on it.Accordingly, the optimal social cost is 𝜖 . However, in any optimal allocation, compared withthe bundle received by agent 2, agent 1 violates the condition of -PMMS. And to achieve -PMMS, at least one of 𝑒 𝑚 − , 𝑒 𝑚 − , 𝑒 𝑚 − , 𝑒 𝑚 has to be assigned to someone other than agent1, then accordingly, the social cost of a -PMMS allocation is larger than 𝑚 , resulting in theinfinite ratio when 𝜖 → . (cid:3) We are now left with MMS fairness. Let us first provide upper and lower bounds on theprice of MMS.
Proposition 5.7.
For 𝑛 ≥ , the price of MMS is at most 𝑛 and at least 𝑛 .Proof. We first prove the upper bound part. For any instance 𝐼 , if OPT ( 𝐼 ) ≤ 𝑛 , then by firstpoint of Lemma 2.2, any optimal allocations is an MMS allocation. Thus, we can further assumeOPT ( 𝐼 ) > 𝑛 . Notice that the maximum social cost of an allocation is 𝑛 and thus we triviallyhave the upper bound as 𝑛 . 24or the lower bound, consider an instance 𝐼 with 𝑛 agents and 𝑛 + chores 𝐸 = { 𝑒 , . . . , 𝑒 𝑛 + } .For agent 𝑖 = , . . . , 𝑛 , 𝑐 𝑖 ( 𝑒 ) = 𝑐 𝑖 ( 𝑒 ) = and 𝑐 𝑖 ( 𝑒 𝑗 ) = for any 𝑗 ≥ . As for agent 1, 𝑐 ( 𝑒 ) = 𝑛 , 𝑐 ( 𝑒 ) = 𝜖 , 𝑐 ( 𝑒 ) = 𝑛 − 𝜖 and 𝑐 ( 𝑒 𝑗 ) = 𝑛 for any 𝑗 ≥ where 𝜖 > takes arbitrarilysmall value. It is not hard to verify that MMS ( 𝑛, 𝐸 ) = 𝑛 and for any 𝑖 ≥ , MMS 𝑖 ( 𝑛, 𝐸 ) = .In any optimal allocation O = ( 𝑂 , . . . , 𝑂 𝑛 ) , the first two chores are assigned to agent 1 andall rest chores are assigned to agents have cost zero on them. Thus, we have OPT ( 𝐼 ) = 𝑛 .However, in any optimal allocation, agent 1 receives cost larger than his MMS. To achieve anMMS allocation, agent 1 can not receive both chores 𝑒 , 𝑒 , and so at least one of them have tobe assigned to the agent other than agent 1. As a result, the social cost of an MMS allocationis at least + 𝜖 , yielding the price of MMS is at least 𝑛 as 𝜖 → . (cid:3) As mentioned earlier, the existence of MMS allocation is not guaranteed. So we also providean asymptotically tight price of 2-MMS.
Proposition 5.8.
For 𝑛 ≥ , the price of 2- MMS is Θ ( 𝑛 ) Proof.
We first prove the upper bound. By Proposition 3.4, we know any EF1 allocations are 𝑛 − 𝑛 -MMS (also 2-MMS). As we mentioned earlier, round-robin algorithm always output EF1allocations. Consequently, given any instance 𝐼 , the allocation returned by round-robin is also2-MMS. In the following, we incorporate the idea of expectation in probability theory and showthat there exists an order of round-robin such that the output allocation has social cost at most1. Let 𝜎 be a uniformly random permutation of { , . . . , 𝑛 } and A ( 𝜎 ) = ( 𝐴 ( 𝜎 ) , . . . , 𝐴 𝑛 ( 𝜎 )) bethe allocation returned by round-robin based on the order 𝜎 . Clearly, each element 𝐴 𝑖 ( 𝜎 ) is arandom variable. Since 𝜎 is chosen uniformly random, the probability of agent 𝑖 on 𝑗 -th positionis 𝑛 . Fix an agent 𝑖 , we assume 𝑐 𝑖 ( 𝑒 ) ≤ 𝑐 𝑖 ( 𝑒 ) ≤ · · · ≤ 𝑐 𝑖 ( 𝑒 𝑚 ) . If agent 𝑖 is in 𝑗 -th position ofthe order, then his cost is at most 𝑐 𝑖 ( 𝑒 𝑗 ) + 𝑐 𝑖 ( 𝑒 𝑛 + 𝑗 ) + · · · + 𝑐 𝑖 ( 𝑒 ⌊ 𝑚 − 𝑗𝑛 ⌋ 𝑛 + 𝑗 ) . Accordingly, his expectedcost is at most Í 𝑛𝑗 = 𝑛 Í ⌊ 𝑚 − 𝑗𝑛 ⌋ 𝑙 = 𝑐 𝑖 ( 𝑒 𝑙𝑛 + 𝑗 ) . Thus, we have an upper bound of the expected socialcost. E [ 𝑆𝐶 ( A ( 𝜎 ))] ≤ 𝑛 Õ 𝑖 = 𝑛 Õ 𝑗 = 𝑛 ⌊ 𝑚 − 𝑗𝑛 ⌋ Õ 𝑙 = 𝑐 𝑖 ( 𝑒 𝑙𝑛 + 𝑗 ) = 𝑛 𝑛 Õ 𝑖 = 𝑐 𝑖 ( 𝐸 ) = . Therefore, there exists an order such that the social cost of the output is at most 1. Notice thatfor any instance 𝐼 , if OPT ( 𝐼 ) ≤ 𝑛 , then any optimal allocations are also MMS. Thus, we canfurther assume OPT ( 𝐼 ) > 𝑛 , and accordingly, the price of 2-MMS is at most 𝑛 .For the lower bound, consider an instance 𝐼 with 𝑛 agents and a set 𝐸 = { 𝑒 , . . . , 𝑒 𝑛 + } of 𝑛 + chores. The cost function of agent 1 is: 𝑐 ( 𝑒 ) = 𝑐 ( 𝑒 ) = 𝑛 − 𝜖 , 𝑐 ( 𝑒 ) = 𝜖 , 𝑐 ( 𝑒 ) = 𝑐 ( 𝑒 ) = 𝜖 , 𝑐 ( 𝑒 ) = 𝑛 − 𝜖 where 𝜖 > takes arbitrarily small value, and 𝑐 ( 𝑒 𝑗 ) = 𝑛 for any < 𝑗 ≤ 𝑛 + .For agent 𝑖 = , . . . , 𝑛 , his cost is: 𝑐 𝑖 ( 𝑒 𝑗 ) = for any 𝑗 = , , and 𝑐 𝑖 ( 𝑒 𝑗 ) = for 𝑗 ≥ . It isnot hard to verify that MMS ( 𝑛, 𝐸 ) = 𝑛 and for any 𝑖 ≥ , MMS 𝑖 ( 𝑛, 𝐸 ) = . In any optimal25llocation O = ( 𝑂 , . . . , 𝑂 𝑛 ) , the first three chores are assigned to agent 1 and all rest choresare assigned to agents having cost zero on them. Thus, we have OPT ( 𝐼 ) = 𝑛 + 𝜖 . However, inany optimal allocations, the cost received by agent 1 is larger than MMS ( 𝑛, 𝐸 ) . To achievea 2-MMS allocation, agent 1 can not receive all first three chores, and so at least one of themhave to be assigned to the agent other than agent 1. As a result, the social cost of an 2-MMSallocation is at least + 𝑛 + 𝜖 , yielding that the price of 2-MMS is at least 𝑛 + . Combinglower and upper bound, the price of 2-MMS is Θ ( 𝑛 ) (cid:3) In this paper, we are concerned with fair allocations of indivisible chores among agents underthe setting that the cost functions are additive. First we have established pairwise connectionsbetween several relaxations of the envy-free fairness in allocating, which look at how an alloca-tion under one fairness criterion provides an approximation guarantee for fairness under anothercriterion. Some of our results have shown a sharp contrast to what is known in allocating in-divisible goods, reflecting the difference between goods and chores allocation. Then we havestudied the trade-off between fairness and efficiency, for which we have established the price offairness for all these fairness notions. We hope our results have provided an almost completepicture for the connections between these chores fairness criteria together with their individualefficiencies relative to social optimum.
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We first prove the upper bound. Let A = ( 𝐴 , 𝐴 , 𝐴 ) be a PMMS allocation and we focus onagent 1. For the sake of contradiction, we assume 𝑐 ( 𝐴 ) > MMS ( , 𝐸 ) . We can also assumebundles 𝐴 , 𝐴 , 𝐴 do not contain chore with zero cost for agent 1 since the existence of suchchores do not affect approximation ratio of allocation 𝐴 on PMMS or MMS. To this end, we let 𝑐 ( 𝐴 ) ≤ 𝑐 ( 𝐴 ) (the other case is symmetric).We first show that 𝐴 must be the bundle yielding the largest cost for agent 1. Otherwise,if 𝑐 ( 𝐴 ) ≤ 𝑐 ( 𝐴 ) ≤ 𝑐 ( 𝐴 ) , then by additivity 𝑐 ( 𝐴 ) ≤ 𝑐 ( 𝐸 ) ≤ MMS ( , 𝐸 ) , contradictingto 𝑐 ( 𝐴 ) > MMS ( , 𝐸 ) . Or if 𝑐 ( 𝐴 ) < 𝑐 ( 𝐴 ) ≤ 𝑐 ( 𝐴 ) , since 𝐴 and 𝐴 is a 2-partitionof 𝐴 ∪ 𝐴 , then 𝑐 ( 𝐴 ) is at least MMS ( , 𝐴 ∪ 𝐴 ) . On the other hand, since A is a PMMSallocation, we know 𝑐 ( 𝐴 ) ≤ MMS ( , 𝐴 ∪ 𝐴 ) , and thus, 𝑐 ( 𝐴 ) = MMS ( , 𝐴 ∪ 𝐴 ) holds.Based on assumption 𝑐 ( 𝐴 ) > MMS ( , 𝐸 ) and Lemma 2.1, we have 𝑐 ( 𝐴 ) > MMS ( , 𝐸 ) ≥ 𝑐 ( 𝐸 ) , then 𝑐 ( 𝐴 ) ≥ 𝑐 ( 𝐴 ) > 𝑐 ( 𝐸 ) which yields 𝑐 ( 𝐴 ) < 𝑐 ( 𝐸 ) owning to the additivity.As a result, the difference between 𝑐 ( 𝐴 ) and 𝑐 ( 𝐴 ) is lower bounded 𝑐 ( 𝐴 ) − 𝑐 ( 𝐴 ) > 𝑐 ( 𝐸 ) .Due to 𝑐 ( 𝐴 ) = MMS ( , 𝐴 ∪ 𝐴 ) , we can claim that every single chore in 𝐴 has cost strictlygreater than 𝑐 ( 𝐸 ) ; otherwise, ∃ 𝑒 ∈ 𝐴 with 𝑐 ( 𝑒 ) ≤ 𝑐 ( 𝐸 ) , then reassigning chore 𝑒 to 𝐴 yields a 2-partition { 𝐴 \ { 𝑒 } , 𝐴 ∪ { 𝑒 }} with max { 𝑐 ( 𝐴 \ { 𝑒 }) , 𝑐 ( 𝐴 ∪ { 𝑒 })} < 𝑐 ( 𝐴 ) = MMS ( , 𝐴 ∪ 𝐴 ) , contradicting to the definition of maximin share. Since every single chorein 𝐴 has cost strictly greater than 𝑐 ( 𝐸 ) , then 𝐴 can only contain a single chore; otherwise, 𝑐 ( 𝐴 ) ≥ 𝑐 ( 𝐴 ) ≥ | 𝐴 | 𝑐 ( 𝐸 ) ≥ 𝑐 ( 𝐸 ) , implying 𝑐 ( 𝐴 ∪ 𝐴 ) ≥ 𝑐 ( 𝐸 ) , contradiction. However,if | 𝐴 | = , according to the second point of Lemma 2.1, 𝑐 ( 𝐴 ) > MMS ( , 𝐸 ) can neverhold. Therefore, it must hold that 𝑐 ( 𝐴 ) ≥ 𝑐 ( 𝐴 ) ≥ 𝑐 ( 𝐴 ) , which then implies 𝑐 ( 𝐴 ) = MMS ( , 𝐴 ∪ 𝐴 ) = MMS ( , 𝐴 ∪ 𝐴 ) as a consequence of PMMS.Next, we prove our statement by carefully checking the possibilities of | 𝐴 | . According toLemma 2.1, if | 𝐴 | = , then 𝑐 ( 𝐴 ) ≤ MMS ( , 𝐸 ) . Thus, we can further assume | 𝐴 | ≥ . Wefirst consider the case | 𝐴 | ≥ . Since 𝑐 ( 𝐴 ) > MMS ( , 𝐸 ) ≥ 𝑐 ( 𝐸 ) , by additivity, we have 𝑐 ( 𝐴 ) + 𝑐 ( 𝐴 ) < 𝑐 ( 𝐸 ) and moreover, 𝑐 ( 𝐴 ) < 𝑐 ( 𝐸 ) due to 𝑐 ( 𝐴 ) ≤ 𝑐 ( 𝐴 ) . Then thecost difference between bundle 𝐴 and 𝐴 satisfies 𝑐 ( 𝐴 ) − 𝑐 ( 𝐴 ) > 𝑐 ( 𝐸 ) . This allow us toclaim that every single chore in 𝐴 has cost strictly greater than 𝑐 ( 𝐸 ) ; otherwise, reassigninga chore with cost no larger than 𝑐 ( 𝐸 ) to 𝐴 yields another 2-partition of 𝐴 ∪ 𝐴 in which thecost of larger bundle is strictly smaller than MMS ( , 𝐴 ∪ 𝐴 ) , contradiction. In addition, since 𝑐 ( 𝐴 ) = MMS ( , 𝐴 ∪ 𝐴 ) , we claim 𝑐 ( 𝐴 ) ≥ 𝑐 ( 𝐴 \ { 𝑒 }) , ∀ 𝑒 ∈ 𝐴 ; otherwise, ∃ 𝑒 ′ ∈ 𝐴 suchthat 𝑐 ( 𝐴 ) < 𝑐 ( 𝐴 \ { 𝑒 ′ }) , then reassigning 𝑒 ′ to 𝐴 yields another 2-partition of 𝐴 ∪ 𝐴 ofwhich both two bundles’ cost are strictly smaller than MMS ( , 𝐴 ∪ 𝐴 ) , contradiction. Thus,for any 𝑒 ∈ 𝐴 , we have 𝑐 ( 𝐴 ) ≥ 𝑐 ( 𝐴 \ { 𝑒 }) ≥ 𝑐 ( 𝐸 ) · | 𝐴 \ { 𝑒 }| ≥ 𝑐 ( 𝐸 ) , where the lasttransition is due to | 𝐴 | ≥ . However, the cost of bundle 𝐴 is 𝑐 ( 𝐴 ) < 𝑐 ( 𝐸 ) , contradiction.The remaining work is to rule out the possibility of | 𝐴 | = . Let 𝐴 = { 𝑒 , 𝑒 } with 𝑐 ( 𝑒 ) ≤ 𝑐 ( 𝑒 ) (the other case is symmetric). Since 𝑐 ( 𝐴 ) > MMS ( , 𝐸 ) ≥ 𝑐 ( 𝐸 ) , then 𝑐 ( 𝑒 ) > 𝑐 ( 𝐸 ) . Let 𝑆 ∗ ∈ arg max 𝑆 ⊆ 𝐴 { 𝑐 ( 𝑆 ) : 𝑐 ( 𝑆 ) < 𝑐 ( 𝑒 )} (can be empty set) be the29argest subset of 𝐴 with cost strictly smaller than 𝑐 ( 𝑒 ) . Due to 𝑐 ( 𝐴 ) = MMS ( , 𝐴 ∪ 𝐴 ) ,then swapping 𝑆 ∗ and 𝑒 would not produce a 2-partition in which the cost of both bundles arestrictly smaller than 𝑐 ( 𝐴 ) , and thus 𝑐 ( 𝐴 \ 𝑆 ∗ ∪ { 𝑒 }) ≥ 𝑐 ( 𝐴 ) , equivalent to 𝑐 ( 𝐴 \ 𝑆 ∗ ) ≥ 𝑐 ( 𝑒 ) > 𝑐 ( 𝐸 ) . (A-1)Then, by 𝑐 ( 𝐴 ) − 𝑐 ( 𝐴 ) > 𝑐 ( 𝐸 ) and 𝑐 ( 𝐴 \ 𝑆 ∗ ) ≥ 𝑐 ( 𝑒 ) , we have 𝑐 ( 𝑒 ) − 𝑐 ( 𝑆 ∗ ) > 𝑐 ( 𝐸 ) ,which allows us to claim that every single chore in 𝐴 \ 𝑆 ∗ has cost strictly greater than 𝑐 ( 𝐸 ) ;otherwise, we can find another subset of 𝐴 whose cost is strictly smaller than 𝑒 but larger than 𝑐 ( 𝑆 ∗ ) , contradicting to the definition of 𝑆 ∗ . As a result, bundle 𝐴 \ 𝑆 ∗ must contain a singlechore; if not, 𝑐 ( 𝐴 ) > 𝑐 ( 𝐸 ) · | 𝐴 \ 𝑆 ∗ | ≥ 𝑐 ( 𝐸 ) , which implies 𝑐 ( 𝐴 ∪ 𝐴 ∪ 𝐴 ) > 𝑐 ( 𝐸 ) dueto 𝑐 ( 𝐴 ) > 𝑐 ( 𝐸 ) and 𝑐 ( 𝐴 ) ≥ 𝑐 ( 𝐴 ) > 𝑐 ( 𝐸 ) . Thus, bundle 𝐴 \ 𝑆 ∗ only contains one chore,denoted by 𝑒 . So we can decompose 𝐴 as 𝐴 = { 𝑒 } ∪ 𝑆 ∗ where 𝑐 ( 𝑒 ) ≥ 𝑐 ( 𝑒 ) > 𝑐 ( 𝐸 ) .Next, we analyse the possible composition of bundle 𝐴 . To have an explicit discussion, weintroduce two more notions Δ , Δ as follows 𝑐 ( 𝐴 ) = 𝑐 ( 𝐸 ) + Δ ,𝑐 ( 𝐴 ) = 𝑐 ( 𝐸 ) + 𝑐 ( 𝑆 ∗ ) + Δ . (A-2)Recall 𝑐 ( 𝐴 ) > 𝑐 ( 𝐸 ) and 𝑐 ( 𝑒 ) ≥ 𝑐 ( 𝑒 ) > 𝑐 ( 𝐸 ) , so both Δ , Δ > . Similarly, let 𝑆 ∗ ∈ arg min 𝑆 ⊆ 𝐴 { 𝑐 ( 𝑆 ) : 𝑐 ( 𝑆 ) < 𝑐 ( 𝑒 )} , then we claim 𝑐 ( 𝐴 \ 𝑆 ∗ ) ≥ 𝑐 ( 𝑒 ) ; otherwise,swapping 𝑆 ∗ and 𝑒 yields a 2-partition of 𝐴 ∪ 𝐴 in which the cost of both bundles are strictlysmaller than 𝑐 ( 𝐴 ) = MMS ( , 𝐴 ∪ 𝐴 ) , contradicting to the definition of maximin share. Byadditivity of cost functions and Equation (A-2), we have 𝑐 ( 𝐴 ) = 𝑐 ( 𝐸 ) − 𝑐 ( 𝑆 ∗ ) − Δ − Δ ,and accordingly 𝑐 ( 𝐴 ) − 𝑐 ( 𝐴 ) = 𝑐 ( 𝐸 ) + 𝑐 ( 𝑆 ∗ ) + Δ + Δ . This combing 𝑐 ( 𝐴 \ 𝑆 ∗ ) ≥ 𝑐 ( 𝑒 ) yields 𝑐 ( 𝑒 ) − 𝑐 ( 𝑆 ∗ ) ≥ 𝑐 ( 𝐸 ) + 𝑐 ( 𝑆 ∗ ) + Δ + Δ . (A-3)Based on Inequality (A-3), we can claim that every single chore in 𝐴 \ 𝑆 ∗ has cost at least 𝑐 ( 𝐸 ) + 𝑐 ( 𝑆 ∗ ) + Δ + Δ ; otherwise, contradicting to the definition of 𝑆 ∗ . Recall 𝑐 ( 𝐴 ) = 𝑐 ( 𝐸 ) − 𝑐 ( 𝑆 ∗ ) − Δ − Δ , then due to the constraint on the cost of single chore in 𝐴 \ 𝑆 ∗ , wehave | 𝐴 \ 𝑆 ∗ | ≤ . Meanwhile, 𝑐 ( 𝐴 \ 𝑆 ∗ ) ≥ 𝑐 ( 𝑒 ) implying that bundle 𝐴 \ 𝑆 ∗ can not beempty. In the following, we separate our proof by discussing two possible cases: | 𝐴 \ 𝑆 ∗ | = and | 𝐴 \ 𝑆 ∗ | = . Case 1: | 𝐴 \ 𝑆 ∗ | = . Let 𝐴 \ 𝑆 ∗ = { 𝑒 } . Therefore, the whole set 𝐸 is composed byfour single chores and two subsets 𝑆 ∗ , 𝑆 ∗ , i.e., 𝐸 = { 𝑒 , 𝑒 , 𝑒 , 𝑆 ∗ , 𝑒 , 𝑆 ∗ } . Then, we let T = ( 𝑇 , 𝑇 , 𝑇 ) be the allocation defining MMS ( , 𝐸 ) and without loss of generality, let 𝑐 ( 𝑇 ) = MMS ( , 𝐸 ) . Next, to find contradictions, we analyse bounds on both MMS ( , 𝐸 ) and 𝑐 ( 𝐴 ) .Since min { 𝑐 ( 𝑒 ) , 𝑐 ( 𝑒 )} ≥ 𝑐 ( 𝑒 ) ≥ 𝑐 ( 𝐴 ) , we claim that 𝑐 ( 𝐴 ) ≤ 𝑐 ( 𝐸 ) ; otherwise 𝑐 ( 𝐴 ) + 𝑐 ( 𝑒 ) + 𝑐 ( 𝑒 ) > 𝑐 ( 𝐸 ) . Notice that 𝐸 contains three chores with the cost at least 𝑐 ( 𝐸 ) each, if any two of them are in the same bundle under T , then MMS ( , 𝐸 ) > 𝑐 ( 𝐸 ) 𝑐 ( 𝐴 ) MMS ( ,𝐸 ) < , contradiction. Or if each of { 𝑒 , 𝑒 , 𝑒 } is contained in adistinct bundle, then the bundle also containing chore 𝑒 has cost at least 𝑐 ( 𝐴 ) as a result of min { 𝑐 ( 𝑒 ) , 𝑐 ( 𝑒 )} ≥ 𝑐 ( 𝑒 ) and 𝐴 = { 𝑒 , 𝑒 } . Thus, MMS ( , 𝐸 ) ≥ 𝑐 ( 𝐴 ) holds, contradictingto 𝑐 ( 𝐴 ) > MMS ( , 𝐸 ) . Case 2: | 𝐴 \ 𝑆 ∗ | = . Let 𝐴 \ 𝑆 ∗ = { 𝑒 , 𝑒 } and accordingly, the whole set can be decomposedas 𝐸 = { 𝑒 , 𝑒 , 𝑒 , 𝑆 ∗ , 𝑒 , 𝑒 , 𝑆 ∗ } . Note the upper bound 𝑐 ( 𝐴 ) ≤ 𝑐 ( 𝐸 ) still holds since min { 𝑐 ( 𝐴 \ 𝑆 ∗ ) , 𝑐 ( 𝑒 )} ≥ 𝑐 ( 𝑒 ) . Then, we analyse the possible lower bound of MMS ( , 𝐸 ) . Ifchores 𝑒 , 𝑒 are in the same bundle of T , then MMS ( , 𝐸 ) > 𝑐 ( 𝐸 ) holds and so 𝑐 ( 𝐴 ) MMS ( ,𝐸 ) < ,contradiction. Thus, chores 𝑒 , 𝑒 are in different bundles in T . Then, if both chores 𝑒 , 𝑒 are inthe bundle containing 𝑒 or 𝑒 , then we also have MMS ( , 𝐸 ) > 𝑐 ( 𝐸 ) implying 𝑐 ( 𝐴 ) MMS ( ,𝐸 ) < ,contradiction. Therefore, only two possible cases; that is, both 𝑒 , 𝑒 are in the bundle differentfrom that containing 𝑒 or 𝑒 ; or the bundle having 𝑒 or 𝑒 contains at most one of 𝑒 , 𝑒 . Subcase 1: both 𝑒 , 𝑒 are in the bundle different from that containing 𝑒 or 𝑒 ; Recall 𝑐 ( 𝑒 ) > 𝑐 ( 𝐸 ) + 𝑐 ( 𝑆 ∗ ) and the fact min { 𝑐 ( 𝑒 ) , 𝑐 ( 𝑒 ) , 𝑐 ( 𝑒 ∪ 𝑒 )} > 𝑐 ( 𝐸 ) , the bundlealso containing 𝑒 has cost strictly greater than 𝑐 ( 𝐸 ) . Thus, MMS ( , 𝐸 ) > 𝑐 ( 𝐸 ) , whichcombines 𝑐 ( 𝐴 ) ≤ 𝑐 ( 𝐸 ) implying 𝑐 ( 𝐴 ) MMS ( ,𝐸 ) < < , contradiction. Subcase 2: bundle having 𝑒 or 𝑒 contains at most one of 𝑒 , 𝑒 . Recall 𝑐 ( 𝑒 ) ≥ 𝑐 ( 𝑒 ) and min { 𝑐 ( 𝑒 ) , 𝑐 ( 𝑒 )} ≥ 𝑐 ( 𝐸 ) + 𝑐 ( 𝑆 ∗ ) + Δ + Δ , thus in allocation T there always exist abundle with cost at least 𝑐 ( 𝐸 ) + 𝑐 ( 𝑆 ∗ ) + Δ + Δ + 𝑐 ( 𝑒 ) and results in the ratio 𝑐 ( 𝐴 ) MMS ( , 𝐸 ) ≤ 𝑐 ( 𝑒 ) + 𝑐 ( 𝑒 ) 𝑐 ( 𝐸 ) + 𝑐 ( 𝑆 ∗ ) + Δ + Δ + 𝑐 ( 𝑒 ) . (A-4)In order to satisfying our assumption 𝑐 ( 𝐴 ) MMS ( ,𝐸 ) > , the RHS of Inequality (A-4) must bestrictly greater than , which implies the following 𝑐 ( 𝑒 ) > 𝑐 ( 𝐸 ) + 𝑐 ( 𝑆 ∗ ) + Δ + Δ . (A-5)However, based on the first equation of (A-2) and 𝑐 ( 𝑒 ) ≤ 𝑐 ( 𝑒 ) , we have 𝑐 ( 𝑒 ) ≤ 𝑐 ( 𝐸 ) + Δ < 𝑐 ( 𝐸 ) + 𝑐 ( 𝑆 ∗ ) + Δ + Δ due to Δ , Δ > . This contradicts to Inequality (A-5).Therefore, 𝑐 ( 𝐴 ) MMS ( ,𝐸 ) > can never hold under Case 2 . Up to here, we complete the proof ofthe upper bound.Next, as for tightness, consider an instance with three agents and a set 𝐸 = { 𝑒 , ..., 𝑒 } ofsix chores. Agents have identical cost functions. The cost function of agent 1 is as follows: 𝑐 ( 𝑒 𝑗 ) = , ∀ 𝑗 = , , and 𝑐 ( 𝑒 𝑗 ) = , ∀ 𝑗 = , , . It is easy to see that MMS ( , 𝐸 ) = . Then,consider an allocation B = { 𝐵 , 𝐵 , 𝐵 } with 𝐵 = { 𝑒 , 𝑒 } , 𝐵 = { 𝑒 } and 𝐵 = { 𝑒 , 𝑒 , 𝑒 } . Itis not hard to verify that allocation B is PMMS and due to 𝑐 ( 𝐵 ) = , we have the ratio 𝑐 ( 𝐵 ) MMS ( ,𝐸 ) = . A.2 Algorithm 1
The following efficient algorithm, which we call
𝐴𝐿𝐺 , outputs an EF1 allocation with a cost at31ost times the optimal social cost under the case of 𝑛 = . In the algorithm, we use notations: 𝐿 ( 𝑘 ) : = { 𝑒 , . . . , 𝑒 𝑘 } and 𝑅 ( 𝑘 ) : = { 𝑒 𝑘 , . . . , 𝑒 𝑚 } for any ≤ 𝑘 ≤ 𝑚 . Algorithm 1Input:
An instance 𝐼 with two agents. Output: an EF1 allocation of instance 𝐼 Check which agent receives less cost under optimal allocations (we assume agent 1 receivesless cost under optimal allocations). Order chores such that 𝑐 ( 𝑒 ) 𝑐 ( 𝑒 ) ≤ 𝑐 ( 𝑒 ) 𝑐 ( 𝑒 ) ≤ · · · ≤ 𝑐 ( 𝑒 𝑚 ) 𝑐 ( 𝑒 𝑚 ) , tie breaks arbitrarily. For chore 𝑒 with 𝑐 ( 𝑒 ) = , put it at the front and chore 𝑒 with 𝑐 ( 𝑒 ) = at back Find index 𝑠 such that 𝑐 ( 𝑒 𝑠 ) < 𝑐 ( 𝑒 𝑠 ) and 𝑐 ( 𝑒 𝑠 + ) ≥ 𝑐 ( 𝑒 𝑠 + ) . if 𝑠 = then Run a round-robin on instance 𝐼 return the output else Let O be the allocation with 𝑂 = 𝐿 ( 𝑠 ) and 𝑂 = 𝑅 ( 𝑠 + ) if allocation O is EF1 then return allocation O else find the maximum index 𝑓 ≥ 𝑠 such that 𝑐 ( 𝑅 ( 𝑓 + )) > 𝑐 ( 𝐿 ( 𝑓 )) return allocation A with 𝐴 = 𝐿 ( 𝑓 + ) and 𝐴 = 𝑅 ( 𝑓 + ) end if end ifend if