(Almost Full) EFX Exists for Four Agents (and Beyond)
((Almost Full) EFX Exists for Four Agents (and Beyond)
Ben Berger ∗ Tel Aviv [email protected] Avi Cohen ∗ Tel Aviv [email protected] Michal Feldman ∗ Tel Aviv [email protected] Fiat † Tel Aviv Universityfi[email protected] 2, 2021
Abstract
The existence of EFX allocations is a major open problem in fair division, even for additive valuations.The current state of the art is that no setting where EFX allocations are impossible is known, and EFXis known to exist for ( i ) agents with identical valuations, ( ii ) 2 agents, ( iii ) 3 agents with additivevaluations, ( iv ) agents with one of two additive valuations and ( v ) agents with two valued instances. Itis also known that EFX exists if one can leave n − n is the number of agents.We develop new techniques that allow us to push the boundaries of the enigmatic EFX problembeyond these known results, and, arguably, to simplify proofs of earlier results. Our main results are( i ) every setting with 4 additive agents admits an EFX allocation that leaves at most a single itemunallocated, ( ii ) every setting with n additive valuations has an EFX allocation with at most n − nice cancelable valuations (a newclass, including additive, unit-demand, budget-additive and multiplicative valuations, among others).Furthermore, using our new techniques, we show that previous results for additive valuations extend tonice cancelable valuations. The question of justness, fairness and division of resources and commitments dates back to Aristotle [Chr42].
Distributional justice , the “just” allocation of limited resources, is fundamental in the work of Rawls [Raw99].Some evidence of the great interest in Rawls’ work is that numerous editions of his book have been citedover 100,000 times.The mathematical study of fair division is credited to Hugo Steinhaus, Bronislaw Knaster and StefanBanach. The tale is that they would meet at the Scottish Caf´e in Lvov where they wrote a book of openproblems — the “Scottish book” — preserved by Steinhaus throughout the war and subsequently translatedby Ulam and published in the United States. Subsequent editions of this book were written following theend of the war. In 1944 Hugo Steinhaus proposed the problem of dividing a cake into n pieces so thatevery agent gets at least a 1 /n fraction of her total utility (“proportional division”). Steinhaus was activelyworking despite then living under German occupation in fear for his life. A solution to Proportional divisionof a cake was credited to Banach and Knaster by Steinhaus in 1949 [Ste49]. ∗ Supported by the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovationprogram (grant agreement No. 866132), and by the Israel Science Foundation (grant number 317/17). † Supported by the Israel Science Foundation (grant number 1841/14) and the Blavatnik fund. a r X i v : . [ c s . G T ] F e b stronger notion of fairness is that of an envy free (EF) allocation — introduced by Gamow and Stern[GS58] for cake cutting, and in the context of general resource allocation by Foley [Fol67]. Unfortunately, ifgoods are indivisible, envy free allocations need not exist. Consider the trivial case of one indivisible good— if Alice gets the good, others will be envious. Lipton et al. [LMMS04] and Budish [Bud11] consider arelaxed notion of envy freeness, namely envy freeness up to some item (EF1) — an allocation is EF1 if forevery pair of agents Alice and Bob, there is an item that we can remove from Alice’s allocation such thatBob will not be interested in swapping his allocation with what remains of Alice’s allocation.EF1 allocations always exist but their fairness guarantees are questionable. Consider for example a settingwhere Alice and Bob have identical valuations over 3 items a, b, c with respective values 1 , ,
2. Arguably, afair allocation would assign a, b to one of the players, and c to the other one, giving each a value 2. However,the allocation that assigns a, c to Alice and b to Bob is also EF1.The notion of Envy Freeness up to any item (EFX) was introduced by Caragiannis et al. [CKM + + any item is discarded . I.e. , it suffices to consider removing the itemwith minimal marginal value (to Bob) from Alices’s allocation. Indeed, in the example above, the only EFXallocations are those that allocate a, b to one player and c to the other player.A major open problem is “when do EFX allocations exist?”. The current state of our knowledge issomewhat embarrassing. We do not know how to rule out EFX allocations in any setting, and yet, they areknown to exist only in several restricted cases. In particular, Plaut and Roughgarden, [PR20], prove thatEFX valuations exist for 2 agents with arbitrary valuations, and for any number of agents with identicalvaluations. Even for the simple case of additive valuations (where the value of a bundle of items is simply thesum of values of individual goods), EFX is only known to exist in settings with 3 agents (Chaudhury, Garg,and Mehlhorn [CGM20]), in settings with only one of two types of additive valuations (Mahara [Mah20]),or when the value of every agent to every item can take one of two permissible values (Amanatidis et al. [ABFR + partial EFX allocation? i.e. , an EFX allocation in which only a small amount of items can beunallocated? The idea of partial allocations for EF and EFX allocations has appeared in multiple papers,e.g. [BKK13, CGG13, CGH19]. Caragiannis, Gravin and Huang, [CGH19] show that discarding some itemsgives good EFX allocations for the rest (achieving 1/2 of the maximum Nash Welfare). Chaudhury et. al [CKMS20] show that given n agents with arbitrary valuations, there always exists an EFX allocation with atmost n − In this paper we develop new techniques, based upon ideas that appear in [CGM20, CKMS20]. [CGM20]introduced the notion of champion edges with respect to a single unallocated good, and used it to makeprogress with respect to the lexicographic potential function in order to eventually reach an EFX allocation.We extend the notion of champion edges beyond a single unallocated item, to sets of items, allocated ornot, and derive useful structural properties that allow us to make more aggressive progress within a graphtheoretic framework.Our techniques are powerful enough to allow us to ( i ) push the boundaries of EFX existence beyondknown results, and ( ii ) present substantially simpler proofs for previously known results. Our results aredescribed below, and are summarized in Table 1.Our main result concerns EFX allocation for four agents. Extending EFX existence from three to fouragents is highly non-trivial. Indeed, [CGM +
21] discovered an instance with 4 additive agents in whichthere exists an EFX allocation with one unallocated item such that no progress can be made based on the2exicographic potential function. We show that one unallocated item is the only possible obstacle to EFXexistence in any setting with four agents.
Theorem 1 (Main Result):
Every setting with 4 additive agents admits an EFX allocation with at mosta single unallocated item (which is not envied by any agent).Moreover, this result extends beyond additive valuations to a new class of valuations that we term nicecancelable valuations (including additive, unit-demand, budget additive and more).To prove Theorem 1, we show that for any EFX allocation with at least two unallocated items, it ispossible to reshuffle bundles and reallocate them in such a way that advances the lexicographic potentialfunction and preserves EFX. The proof requires solving a complex puzzle, and exemplifies the extensive useof our new techniques.The immediate open problem is whether one can go the additional mile and allocate the one item thatremains. A natural approach to solving this problem is by using a different potential function. Notably, ournew techniques are orthogonal to the choice of the potential function, and may prove useful in analyzingother potential functions.Our second result makes an additional progress in settings with an arbitrary number of agents.
Theorem 2:
Every setting with n additive agents admits an EFX allocation with at most n − n − n − Beyond additive valuations.
We extend the results of [CGM20] (EFX for 3 additive agents) and [Mah20](EFX for n agents with one of two additive valuations) beyond additive valuations to a class of valuationsthat we term nice cancelable valuations, defined as follows: a valuation v is cancelable if for any two bundles S, T ⊆ M and item g ∈ M \ ( S ∪ T ) we have v ( S ∪ { g } ) > v ( T ∪ { g } ) ⇒ v ( S ) > v ( T ) . It is nice cancelableif there exists a cancelable and non-degenerate valuation ( i.e. , different bundles have different values) thatrespects all strong inequalities of v .Nice cancelable valuations include additive valuations but also valuations such as budget additive ( v ( S ) =min( (cid:80) j ∈ S v ( j ) , B ) for some value B ), unit demand ( v ( S ) = max j ∈ S v ( j )), multiplicative valuations ( v ( S ) = (cid:81) j ∈ S v ( j )), and others.Moreover, one can have different nice cancelable valuations for different agents . We remark that theoriginal proofs, as written, do not directly generalize to this more general class of valuations. It is our newtechniques that allow us to generalize the results to this class. Theorem 3:
EFX existence for 3 agents and EFX existence for two types of valuations (any number ofagents) extends beyond additive valuations to all nice cancelable valuations.
Simplification of proofs for known results.
Our new techniques greatly simplify existing proofs of EFXexistence for 3 agents [CGM20] and for the case of 2 types of additive valuations [Mah20]. Admittedly,simplicity is a matter of subjective judgment, but at least in terms of character count, the proof for the caseof 3 agents with no envy in [CGM20] (some 5 pages) drops to half a page using our new techniques. Similarly,the proof for settings with 2 types of additive valuations in [Mah20] (some 8 pages) drops to one page usingour new techniques. Moreover, in both settings, the simplified proofs apply beyond additive valuations toall nice cancelable valuations. E.g. , an EFX allocation exists for three agents when agent a has a multiplicative valuation, b has a budget additive valuation,and c has a unit demand valuation. Another example, fix any two nice cancelable valuations, some agents have the 1st andothers have the 2nd, an EFX allocation still exists. etting Prior results Our resultsEFX for 3 agents Additive [CGM20] Beyond additive ∗ EFX for n agents, one of 2 valuations Additive [Mah20] Beyond additive ∗ Partial EFX for n agents ≤ n − ≤ n − ≤ ≤ Table 1: All our results hold for (*) nice cancelable valuations, generalizing additive valuationsWe believe that we have only scratched the surface of the power of our new techniques, and hope theywill prove useful in making further progress on the EFX problem.
Our proof techniques lie within a graph theoretic framework. Given an EFX allocation X , we describe agraph M X (see Definition 3.2) where vertices are associated with agents and there are three types of edges:envy edges i j , champion edges i g j , where g is an unallocated item, and generalized champion edges i H | S j , where H is some subset of items (allocated or not) and S is a subset of j ’s allocation in X .The use of such graphs, with envy and champion edges (but no generalized championship edges) haspreviously appeared in the literature and is a key component in the proof of an EFX allocation for 3 additiveagents [CGM20]. The new ingredient introduced in this paper is the notion of generalized champion edges.We show how to find such edges (Section 3.1), and use them to reach a new EFX allocation that advancesthe lexicographic potential function of [CGM20].The key idea in all our results is to reshuffle the existing allocation to obtain a new allocation with higherpotential, while preserving EFX. This follows the same proof template as in Chaudhury et al. — but wehave more options to play with by using the generalized championship edges.An envy edge i j suggests a possible reshuffling where agent i gets j ’s current allocation. A championedge i g j suggests another reshuffling, where agent i gets a subset of agent j ’s current allocation, alongwith the currently unallocated item g . A generalized champion edge i H | S j suggests giving agent i agent j ’s allocation along with some arbitrary set of items H (that may be arbitrarily allocated among otheragents, or be unallocated), while freeing up the set of items S .Our proofs require solving a complex puzzle, where the goal is to find a cycle consisting of envy, champion,and generalized champion edges, such that the union of all sets S freed up along with the currently unallocateditems suffice for the suggested allocation along the cycle.Finding the appropriate generalized championship edges is a major technical component of our techniques(see Section 3.1). We show how to find such edges, based on existing edges. Then, these edges allow us toreshuffle the current allocation and advance the potential.In several of our proofs we consider the case where M X has envy edges separately from the case where ithas not. For our applications, it turns out that if there are no envy edges, one can Pareto improve the EFXallocation (See Section 5.1), in particular this advances the potential function. If there are envy edges itmay no longer be possible to Pareto improve (as pointed out by [CGM20]) but one can nonetheless advancethe potential function itself (see Section 5.2). Varian [Var74] introduced competitive equilibria from equal incomes (CEEI) which assigns equal budgets toall agents and computes prices such that every agent gets a most favored set of (divisible) items, and theresult is envy free.Lipton et al. [LMMS04] give a greedy algorithm for producing an EF1 allocation, this adds items toan agent that is not envied, or — alternatively — switches bundles around an envy cycle, the so-calledenvy-cycles procedure. 4aragiannis et al. [CKM +
19] show that maximizing Nash welfare (maximizing the product of agentutilities) gives an EF1 allocation that is also Pareto optimal. Approximations for maximizing Nash welfarewere presented by Cole and Gkatzelis [CG15] and subsequent papers.Recall that envy free allocations need not exist — but — at least for two agents, there are envy freeallocations that discard few items [BT00, BKK13]. In [BKK13] the resulting allocation is Pareto optimal,envy free, and maximal (no EF allocation allocates more items).There have been several papers on truthful fair division. Mosel and Tamuz [MT10] and Cole, Gkatzelisand Goel [CGG13] deal with proportional fairness. The mechanism in the latter uses no money and providesgood guarantees — the mechanism discards a fraction of the of the resources to achieve truthfulness.In [CGH19] Caragiannis, Gravin, and Huang show that for additive valuations, there is always an EFXallocation of a subset of items with a Nash welfare (product of agent utilities) that is at least half of themaximum possible Nash welfare for the original set of items. Moreover, this is the best possible Nash welfareof all EFX allocations.Other extensions such as group EFX allocations and EF1 for matroid settings were considered respectivelyby [KSV20] and [GMT14].While we concentrate on envy free allocations and relaxations thereof — there are many other possiblenotions of fairness, e.g., equitable division [Jon02] — where the subjective value of every agent for herallocation is the same, and maximin share [Bud11] where an agent gets the most preferred share she couldguarantee herself if she was to suggest a partition and be allocated her least valuable bundle of this partition.
The next section, Section 2, presents the model, introduces “nice cancelable valuations”, gives definitionsfrom [CGM20] as well as new definitions, notation, and proofs. Useful properties of nice cancelable valuationsare presented in Appendix A.Section 3 defines “generalized championship” and establishes the technical framework underlying ourEFX existence results. It may be helpful to consider Figure 1 while reading the section. How to findgeneralized championship edges is described in Section 3.1.Section 4 proves that it suffices to discard at most n − We consider a setting with n agents, and a set M of m items. Each agent has a valuation v i : 2 M → R ≥ ,which is normalized and monotone, i.e. v ( S ) ≤ v ( T ) whenever S ⊆ T and v ( ∅ ) = 0.For two sets of items S, T ⊆ M , we write S < i T if v i ( S ) < v i ( T ). Similarly we define S > i T , S ≤ i T , S ≥ i T , S = i T if v i ( S ) > v i ( T ), v i ( S ) ≤ v i ( T ), v i ( S ) ≥ v i ( T ), v i ( S ) = v i ( T ), respectively.We denote a valuation profile by v = ( v , . . . , v n ). An allocation is a vector X = ( X , . . . , X n ) of disjointbundles, where X i is the bundle allocated to agent i . Given an allocation X , We say that agent i envies aset of items S if X i < i S . We say that agent i envies agent j , denoted i j , if i envies X j . We say thatagent i strongly envies a set of items S if there exists some h ∈ S such that i envies S \ { h } . Likewise wesay that agent i strongly envies agent j if i strongly envies X j . X is called envy-free (EF) if no agent enviesanother. X is called envy-free up to any good (EFX) if no agent strongly envies another.5 .1 Nice Cancelable Valuations We consider a class of valuation functions that can be viewed as a generalization of additive valuations.These include additive, unit-demand and budget-additive valuations, among others.
Definition 2.1.
A valuation v is cancelable if for any two bundles S, T ⊆ M and an item g ∈ M \ ( S ∪ T ), v ( S ∪ { g } ) > v ( T ∪ { g } ) ⇒ v ( S ) > v ( T ) . A valuation v is non-degenerate if v ( S ) (cid:54) = v ( T ) for any two different bundles S, T . A valuation v (cid:48) is saidto respect another valuation v if for every two bundles S, T ⊆ M such that v ( S ) > v ( T ) it also holds that v (cid:48) ( S ) > v (cid:48) ( T ). Definition 2.2.
A cancelable valuation v is nice if there is a non-degenerate cancelable valuation v (cid:48) thatrespects v . In particular, any non-degenerate cancelable valuation is a nice valuation (by setting v (cid:48) ≡ v ).We can show that in order to prove the existence of an EFX allocation for a given valuation profile v = ( v , . . . , v n ) of nice cancelable valuations, it is without loss of generality to assume that all of thevaluations are non-degenerate (Lemma A.1 in Appendix A). Thus, for the remainder of this paper weassume that all valuations are cancelable and non-degenerate. Under this assumption it is easy to verifythat for any valuation v and bundles S, T, R such that R ⊆ M \ ( S ∪ T ) we have v ( S ∪ R ) > v ( T ∪ R ) ⇔ v ( S ) > v ( T ) . Other useful claims on cancelable valuations are deferred to the appendix.
All our EFX existence results follow the same paradigm: given an arbitrary EFX allocation X with k unallocated goods, construct a new partial EFX allocation that advances some fixed potential function.Since there are finitely many allocations, there must exist an EFX allocation with at most k − X , Y , we say that Y Pareto dominates X if Y i ≥ i X i for every i ∈ [ n ], and there exists some i for which the inequality is strict.Chaudhury et al. [CGM20] have shown that there need not exist a Pareto-dominating EFX allocation when n = 3 and k ≥
1. To overcome this obstacle they introduce an alternative “lexicographic” progress measurewhich we shall also use:
Definition 2.3 ([CGM20]) . Fix some arbitrary ordering of the agents a , . . . , a n . The allocation Y domi-nates X if for some k ∈ [ n ], we have that Y a j = a j X a j for all 1 ≤ j < k , and Y a k > a k X a k .Note that Pareto-domination implies domination but not vice versa. Lemma 2.4.
If for every partial EFX allocation X with k unallocated items, there exists a partial EFXallocation Y that dominates X , then there exists a partial EFX allocation with at most k − k − Proof.
The first part of the lemma follows directly from the fact that the number of possible allocations isfinite and domination is a total order relation.We now show that if in a given partial EFX allocation X some agent envies the set of unallocated itemsthen there exists a partial EFX allocation Y that Pareto dominates X . Analogous to the above, this provesthe second part of the lemma.If some agent envies U , then there exists a subset T of U that some agent i envies and T is a smallestsubset of U that some agent envies. We obtain Y by replacing X i with T . Y is EFX, since i did not stronglyenvy anyone before and now she is better off, and no one strongly envies i by minimality of T .In fact, in our results we almost always progress via Pareto-domination. In the few cases we do not, wefind an allocation in which a (the most important agent in the ordering) is strictly better off. Hereinafterwe denote this agent a vip . 6 .3 Most Envious Agents Hereinafter, we fix some partial EFX allocation X and some unallocated good g . We denote by U the setof goods that are unallocated in X (thus g ∈ U ). The following are variants of definitions from [CGM20],[CKMS20].We say that i is most envious of a set of items S , if there exists a subset T ⊆ S , such that i envies T andno agent strongly envies T . When more than one such T exists, we choose one of them arbitrarily unlessstated otherwise. The set S \ T is referred to as the corresponding discard set . Example 2.5.
Consider the valuation profile depicted in Table 2, describing a setting with 7 items M = { a, b, c, d, e, f, g } and 3 agents with additive valuations. Let X be the allocation where X = { a, b, c } , X = { d } , X = { e, f } . Consider the set S = { c, d, e } . Agent 1 is most envious of S since X < { d } and noagent may strongly envy a singleton. The corresponding discard set is S \ { d } = { c, e } . Agent 2 is alsomost envious of S since X < { c, e } , and no agent strongly envies { c, e } . The corresoponding discard set is S \ { c, e } = { d } . Definition 2.6 ([CGM20]) . We say that i champions j with respect to g , denoted i g j , if i is most enviousof X j ∪ { g } . The corresponding discard set is denoted D gi,j . Note that i envies the set ( X j ∪ { g } ) \ D gi,j , butno agent strongly envies it.An important case considered frequently in the paper is where g / ∈ D gi,j . In this case X j = ( X j \ D gi,j ) ·∪ D gi,j . Definition 2.7 ([CGM20]) . If i g j and g / ∈ D gi,j , then we say that i g -decomposes j into top and bottom half-bundles ( X j \ D gi,j ) and D gi,j , respectively (in short, i g -decomposes j ). If there is no concern ofambiguity, then we denote the top and bottom half-bundles by T j and B j , respectively (note that different g -decomposers of j may induce different top and bottom half-bundles). Under this notation, we have( X j ∪ { g } ) \ D gi,j = T j ∪ { g } .In the following observations from [CGM20], i g j and i (cid:54) j are the respective negations of i g j , i j . Observation 2.8.
Every agent i has a champion with respect to g . Proof.
Since valuations are non-degenerate and monotone, X i < i X i ∪ { g } . Since i envies X i ∪ { g } , this sethas a most envious agent, and by definition that agent is a champion of i with respect to g . Observation 2.9. If i g j and i (cid:54) j , then g / ∈ D gi,j , i.e. i g -decomposes j . Proof.
By definition of D gi,j , agent i envies ( X j ∪{ g } ) \ D gi,j . If g ∈ D gi,j then ( X j ∪{ g } ) \ D gi,j is a subset of X j ,implying that agent i envies a subset of X j and thus the set X j as well. Hence, i envies j , in contradictionto our assumption. Observation 2.10. If i g j and j is g -decomposed into X j = T j ·∪ B j , then X i > i T j ∪ { g } . Proof.
Assume that j is g -decomposed by some agent k into X j = T j ·∪ B j . By definition of D gk,j , no agentstrongly envies ( X j ∪ { g } ) \ D gk,j = T j ∪ { g } . Therefore, any agent that envies this set is a most enviousagent of X j ∪ { g } , and thus a g -champion of j . Hence, agent i does not envy that set. Observation 2.11. If i g -decomposes j , i g k and k is g -decomposed, then T k < i T j . Proof.
We have, T k ∪ { g } < i X i < i T j ∪ { g } , where the first inequality is by Observation 2.10 since i g k and the second inequality is by definition of basic championship (since i g -decomposes j ).7 b c d e f g M X corresponding to thesetting described in Ta-ble 2 and the allocation X = { a, b, c } , X = { d } , X = { e, f } . Onlya subset of the edges isshown. A crucial component in our techniques is the extension of Definition 2.6 to an arbitrary set of items H . Itwill be useful to have a notation that contains some information regarding the discarded items. Definition 3.1. i champions j with respect to ( H | S ) , denoted i H | S j , where H ⊆ M \ X j and S ⊆ X j ,if i is most envious of ( X j \ S ) ∪ H . The corresponding discard set is denoted D H | Si,j .As opposed to basic championship, not every agent j has an ( H | S )-champion (consider an extremeexample where H = ∅ , S = X j ). If i H | S j , then giving i the desired bundle implied by the championshipreleases S to be reallocated to other agents. For example, if we also know that k S | S (cid:48) (cid:96) , then these twochampion relations can be “used” simultaneously in a transition to a new EFX allocation.We say that a set of items T is released by i H | S j if T ⊆ S ∪ D H | Si,j . We denote the negation of i H | S j by i H | S j . Definition 3.2.
The champion graph M X = ([ n ] , E ) with respect to X is a labeled directed multi-graph.The vertices correspond to the agents, and E consists of the following 3 types of edges:1. Envy edges: i j iff i envies j .2. Champion edges: i g j iff i champions j with respect to g , where g is an unallocated good.3. Generalized champion edges: i H | S j iff i champions j with respect to H | S .We refer to envy and champion edges as basic edges . Hereinafter, the edge notations above will sometimesrefer to the edges of the champion graph and sometimes refer to the statements they convey. For example,we will sometimes refer to “ i g j ” as an edge in M X and sometimes as shorthand that i is a g -championof j , and the meaning will be clear from the context. Futhermore, it is not hard to verify that i g j iff i { g } | ∅ j and that i j iff i ∅ | ∅ j . Thus, we can treat every basic edge in M X as a generalizedchampion edge. Example 3.3.
Consider the instance given in Table 2, and let X be the partial EFX allocation where X = { a, b, c } , X = { d } , X = { e, f } . Figure 1 depicts the graph M X . We haven’t drawn all edges; rather,we chose a subset of the edges that illustrate the different types of edges. Item g is unallocated in X , thus U = { g } . Since { a, b, c } < { d } , 1 2. Moreover, combined with the fact that no one strongly envies { d } ,it also means that 1 g
2. Since { d } < { b, g } and no one strongly envies { b, g } , 2 g
1. Similarly, since { d } < { f, g } and no one strongly envies { f, g } , 2 g
3, and since { e, f } < { c, g } and no one stronglyenvies { c, g } , 3 g
1. Finally, it holds that 2 { a, b } | { e } { d } < { a, b, f } and no one stronglyenvies { a, b, f } .Given a cycle C = a → a → · · · → a k → a of edges and an agent a i in the cycle, succ ( a i ) and pred ( a i )denote, respectively, the successor and predecessor of a i along the cycle.8 efinition 3.4. A cycle C = a H | S a H | S · · · H k − | S k − a k H k | S k a in M X is called Pareto improvable (PI) if for every i, j ∈ [ k ] we have H i ∩ H j = ∅ , and either H i ⊆ U or H i is released bysome edge a (cid:96) H (cid:96) | S (cid:96) a succ ( (cid:96) ) , for (cid:96) ∈ [ k ]. A PI cycle which is composed entirely of basic edges is called a basic PI cycle .By definition, every agent a i along a PI cycle envies some subset A i ⊆ X succ ( i ) that no agent stronglyenvies. The following simple but very (!) useful lemma asserts that reallocating A i to agent a i for every i produces a Pareto-dominating EFX allocation. Lemma 3.5. If M X contains a Pareto improvable cycle, then there exists a (partial) EFX allocation Y thatPareto dominates X . Furthermore, every agent i along the cycle satisfies Y i > i X i . Proof.
Let C be a Pareto-improvable cycle in M x , and assume w.l.o.g. that C = 1 H | S H | S · · · H k − | S k − k H k | S k
1. Define the allocation Y as follows: for every agent i , Y i = (cid:40)(cid:0)(cid:0) X succ ( i ) \ S i (cid:1) ∪ H i (cid:1) \ D H i | S i i, succ ( i ) i ∈ [ k ] ( i.e. , i on cycle C ) X i otherwiseFirst, note that the assumptions on the sets H i , S i ensure that the sets Y i are disjoint. That is, Y isindeed an allocation. For every i ∈ [ k ], i is the ( H i | S i )-champion of succ ( i ). Thus X i < i Y i , by definition ofgeneralized championship. Since the bundles did not change for agents outside the cycle, we conclude that Y Pareto-dominates X .It remains to show that Y is EFX. Since no agent becomes worse off in the transition from X to Y , itsuffices to show that in allocation X no agent strongly envies the set Y i for all i ∈ [ n ]. Indeed, if i is anagent outside the cycle, by the fact that X is EFX, no agent strongly envies Y i = X i in X . If i is an agentin the cycle, no agent strongly envies Y i = (cid:0)(cid:0) X succ ( i ) \ S i (cid:1) ∪ H i (cid:1) \ D H i | S i i, succ ( i ) in X , by definition of generalizedchampionship. Corollary 3.6 (Following [CKMS20]) . If M X contains an envy-cycle, A self- g -champion (an agent i satis-fying i g i ) or a cycle composed of envy edges and at most one h -champion edge for every h ∈ U , thenthere exists a (partial) EFX allocation Y that Pareto dominates X . Note that these are exactly the basicPI cycles . Remark 3.7.
Lemma 3.5 can be generalized to handle disjoint cycles. The fact that C is a cycle is usedin the proof of the lemma only to ensure that every agent whose bundle is reallocated, is also given analternative bundle in the new allocation. The same is true if C is a set of vertex-disjoint cycles rather thana single cycle. We may then define C as an edge set { a i H i | S i succ ( a i ) } i ∈ [ k ] , and if the conditions in thedefinition of a Pareto-improvable cycle are satisfied, then Lemma 3.5 still applies. In this case we refer to C as a Pareto-improvable edge set . In this section we describe a way to discover new generalized champion edges in M X . These will almostalways be of the form k S | B j j where B j ⊆ X j is some bottom half-bundle induced by a g -decomposer of j (see discussion below Definition 2.6). Therefore, to facilitate readability we use the following convention: Convention 3.8.
For any two agents k, j and any set S disjoint from X j , we write k S | ◦ j (resp. ( S | ◦ )-champion) as shorthand for k S | B j j (resp. ( S | B j )-champion), where the half-bundle B j will be clearfrom the context. We could have defined a PI cycle more generally, e.g. to to allow the set H i to be a combination of unallocated goods anditems released from several edges. The proposed definition is hopefully easier to digest and suffices for our purposes. Envy cycles, the simplest form of basic PI cycles, were considered in [LMMS04]. Basic PI-cycles were considered in[CKMS20] using different terminology — championship was only defined in [CGM20]. Our definition of a PI-cycle captures andgeneralizes these notions.
Definition 3.9.
A cycle C = a g a g · · · g a k g a with at least two g -champion edges in M X is called a good g -cycle if:1. All agents along the cycle are different.2. There are no parallel envy edges, i.e., a i (cid:54) succ ( a i ) for all i .3. There are no internal g -champion edges, i.e. , for every i, j ∈ [ k ], a i g a j iff a j = succ ( a i ). Observation 3.10.
Agents j on a good g -cycle are g -decomposed by pred ( j ) into X j = T j ·∪ B j . Proof.
This holds by Observation 2.9 since pred ( j ) g j and pred ( j ) (cid:54) j by definition of a good g -cycle.We next show how to discover new generalized champion edges in the presence of a good g -cycle. Thefollowing two observations are useful: Observation 3.11. If i (cid:54) j then i B j | ◦ j . Proof.
Recall that a generalized championship relation i H | S j is required to satisfy H ∩ S = ∅ . Here H = S = B j . Thus, if B j (cid:54) = ∅ then the requirement clearly does not hold. Otherwise B j = ∅ and i ∅ | ∅ j implies that i j , which we assume does not hold. In any case, the relation i B j | ◦ j cannot hold. Observation 3.12.
For any two agents i, j along a good g -cycle, pred ( i ) B i | ◦ j . Proof. If i = j the statement holds by Observation 3.11. Assume otherwise. By definition of a good g -cycle, pred ( i ) (cid:54) i and pred ( i ) g i . Therefore, X pred ( i ) > pred ( i ) T i ∪ B i > pred ( i ) T j ∪ B i , where the second inequalityis by Observation 2.11 and cancelability. Thus, pred ( i ) does not envy T j ∪ B i and the claim follows. Lemma 3.13.
Let C be a good g -cycle. For any agent i along the cycle, there exists an agent a such that a B i | ◦ succ ( i ). Proof. C is a good cycle, hence i g -decomposes succ ( i ) into X succ ( i ) = T succ ( i ) ∪ B succ ( i ) . Furthermore, i g i and i is g -decomposed into X i = T i ∪ B i . Thus by Observation 2.11 (with j = succ ( i ) and k = i in theObservation statement) we have T i < i T succ ( i ) . Hence, by cancelability, X i = T i ∪ B i < i T succ ( i ) ∪ B i . Sincethe set T succ ( i ) ∪ B i is envied by some agent, it must have a most envious agent and the claim follows. Lemma 3.14.
Let C be a good g -cycle. Let i, j, k be agents along the cycle. If k B i | ◦ j , then thereexists an agent a (not necessarily in the cycle) such that a B i | ◦ succ ( k ). Proof. If k = pred ( j ), then j = succ ( k ) and we are done (take a = k ). Assume otherwise. C is a good cycle,hence k g -decomposes succ ( k ) into X succ ( k ) = T succ ( k ) ∪ B succ ( k ) . Furthermore, k g j since k (cid:54) = pred ( j ),and j is g -decomposed into X j = T j ∪ B j . Thus by Observation 2.11 we have T j < i T succ ( k ) . Hence, X k < k T j ∪ B i < k T succ ( k ) ∪ B i , where the first inequality holds by k B i | ◦ j and the second by cancelability.Since the set T succ ( k ) ∪ B i is envied by some agent, it must have a most envious agent and the claim follows.For every bottom half-bundle B i along a good g -cycle C , applying Lemma 3.13 provides an initial B i -champion edge. If this edge is internal to the cycle, i.e., the source of the edge is in the cycle, then we canapply Lemma 3.14 to discover a new B i -champion edge. Once again, if the new edge is internal to the cycle,then we can reapply Lemma 3.14. We can repeat this process to discover more and more B i -champion edges,until either the new edge has already been previously discovered, or it is external (i.e., its source is outsidethe cycle).There are two particular types of internal B i -champions edges that are useful to us. Definition 3.15.
Let C be a good g -cycle. Let i, j, k be three agents along C . If i B k | ◦ j and k ison the path from j to i in C , then we say that the edge i B k | ◦ j is a good edge (or good B k -edge). If (cid:96) B k | ◦ j for some agent (cid:96) outside the cycle C , then we say that the edge (cid:96) B k | ◦ j is an external edge (or external B k -edge). 10he figure on the right illustrates Definition 3.15. The red edges forma good g -cycle C among 4 agents, C = 1 g g g g
1. Theedge 2 B | B good edge , since 1 is on the path from 4 to 2 in C .On the other hand, 3 B | B C . 5 B | B Theorem 3.16.
Let C be a good g -cycle. For every agent j along the cycle, there exists either a good B j -edge in C , or an external B j -edge in C . Proof.
Assume without loss of generality that C = 1 g g · · · g k g j = 1, i.e., we tryto find B -champion edges. By Lemma 3.13 there exists an edge (cid:96) B | ◦ (cid:96) . If this isan external B -edge we are done. Otherwise, (cid:96) is an agent along C , and thus by Lemma 3.14 there existsan edge (cid:96) B | ◦ succ ( (cid:96) ). As long as the result of Lemma 3.14 is not an external edge we may apply thelemma repeatedly. Hence, if no such iteration results in an external edge, we obtain a sequence of agents( (cid:96) i ) ∞ i =1 such that (cid:96) i +1 B | ◦ succ ( (cid:96) i ) for every i ≥ i ≥
1, we have (cid:96) i +1 ≤ (cid:96) i then the edge (cid:96) i +1 B | ◦ succ ( (cid:96) i ) is a good edge (since the pathfrom succ ( (cid:96) i ) to (cid:96) i +1 includes 1). Hence, if none of these edges are good, then (cid:96) i < (cid:96) i +1 for every i ≥
1, incontradiction to C being of finite length. Thus, one of these edges must be good, hence we are done.The following observation and its corollary allow us to narrow down the possible configurations of B j -edges obtained from Theorem 3.16. Observation 3.17. If i B j | ◦ k and i (cid:54) k then B k < i B j . Proof.
Since i (cid:54) k and i B j | ◦ k we have T k ∪ B k = X k ≤ i X i < i ( X k \ B k ) ∪ B j = T k ∪ B j , and bycancelability this implies B k < i B j . Corollary 3.18.
Let C be a good g -cycle. Consider the set of B j -edges guaranteed by Theorem 3.16 forevery agent j along the cycle. If all these edges are external, then they cannot all share the same sourceagent, unless that agent envies some agent along the cycle (the figure below demonstrates an impossibleconfiguration). Proof.
Assume towards contradiction that there is some agent a which is the source of all B i -edges given byTheorem 3.16, and a does not envy any agent along the cycle. Let j = arg min i { v a ( B i ) | i is an agent along C } .By assumption, there exists some j (cid:48) along C such that a B j | ◦ j (cid:48) , and a (cid:54) j (cid:48) . By Observation 3.17, B j > a B j (cid:48) . Hence, v a ( B j ) > v a ( B j (cid:48) ), in contradiction to the definition of j . n − unallocated goods In this section we prove the following:
Theorem 4.1.
For every profile of n additive valuations (and more generally, nice cancelable valuations),there exists an EFX allocation X with at most n − X no agent envies theset of unallocated items.The following lemma shows that the basic edges of the graph M X follow a very particular structure. Thislemma is used in the proof of Theorem 4.1. 11igure 2: An illustration of the proof of Lemma 4.2 for n = 4, k = 3. Figures ( a ) − ( d ) represent the 4iterations in the proof. Red, blue and green edges are champion edges w.r.t. unallocated goods g , g , g ,respectively. Lemma 4.2.
Let X be a partial EFX allocation with at least n − G be M X restricted to basic edges, i.e. , envy edges and basic champion edges as per Definition 2.6.If G does not admit a basic PI cycle, then the number of unallocated goods is exactly n −
1, and G is a union of n − g -champion edges for someunallocated item g (see Figure 2 (d) for the case of n = 4). Proof.
Recall that U denotes the set of unallocated items, and let U = { g , . . . , g k } for some k ≥ n −
1. Let e be an arbitrary incoming g -champion edge of agent 1 (such an edge exists by Observation 2.8).If the source of this edge is agent 1 we are done (we have a self champion). Assume w.l.o.g. that thesource of e is agent 2, and consider its incoming g -champion edge, denoted e . If the source of e is agent 2or agent 1, we are done (in the first case we have a self g -champion, and in the second case we have a basicPI cycle of size 2). Assume w.l.o.g. that the source of e is agent 3. We can continue this way to concludethat w.l.o.g we have a directed path n g n − n − g n − · · · g k ≥ n , thenconsider the incoming g n -champion edge of agent n . No matter what the source of this edge is, it must closea basic PI cycle.Ergo, k = n −
1. Consider the incoming g -champion edge of agent n , denoted e n . The source of thisedge must be agent 1 (every other option closes a basic PI cycle). Now consider the incoming g -championedge of agent 1, e . Similarly, the source of this edge must be agent 2. We can again continue this way untilwe get to the incoming g n − -champion edge of agent n −
2, denoted e n − n − , and conclude that its source isagent n − g -champion edge of agent n −
1, and continuewith the same reasoning, to finally conclude that the source of the incoming g n − -champion edge of agent n is agent 1 (see Figure 2 (d)).At this point there is a Hamiltonian cycle n g i n − g i · · · g i g i n for every i . Any otherbasic champion or envy edge must close a basic PI cycle and thus does not exist by assumption. Thisconcludes the second part of the lemma.We are now ready to prove Theorem 4.1. Proof.
By Lemma 2.4, it suffices to prove that if X has at least n − Y . By Lemma 3.5, it suffices to find a PI cycle in M X .By Lemma 4.2, there are exactly n − G is a union of n − n = 4. The Pareto-improvable cycle ishighlighted in yellow.cycles, one for each unallocated good g , . . . , g n − :1 g g · · · · · · n g g g · · · · · · n g g n − g n − · · · · · · n g n − k and every unallocated good g there exists a good D gk − ,k -championedge, from some agent j to some agent j (cid:48) . (Indeed, since G is a union of parallel Hamiltonian cycles, noexternal edges exists.) Choose some arbitrary such agent k and unallocated good g and corresponding agents j, j (cid:48) , and let Z = D gk − ,k . We show that j Z | ◦ j (cid:48) closes a Pareto improvable cycle in M X .By definition of a good edge, there is a unique path P consisting of g -champion edges from agent j (cid:48) to j ,passing through agent k . By Lemma 4.2, for every q ∈ U , every edge in P has a parallel q -champion edge.Note that P has at most n − j is agent | P | , agent j (cid:48) is agent 1, let k (cid:48) be the new index for agent k ,and let P = 1 , , . . . , k (cid:48) − , k (cid:48) , . . . | P | . Note that Z is now the discard set of the champion edge k (cid:48) − g k (cid:48) . Let U (cid:48) = U \ { g } ( | U (cid:48) | = n − U (cid:48) where U (cid:48) = { r , r , . . . , r k (cid:48) − , r k (cid:48) , . . . , r n − } .We now describe a Pareto-improvable cycle:1 r r r · · · r k (cid:48)− k (cid:48) − g k (cid:48) r k (cid:48) k (cid:48) + 1 r k (cid:48) +1 · · ·· · · r | P |− | P | Z | ◦ Z is the discard set of the champion edge k (cid:48) − g k (cid:48) , it is discarded along the cycle. All other edgesin the cycle are with respect to distinct unallocated goods. Therefore, this is a Pareto-improvable cycle(seeFigure 3 for an illustration) . In this section we prove our main result, namely that every setting with 4 additive agents admits an EFXallocation with at most one unallocated good. We prove this theorem for the class of nice cancelablevaluations which contains additive. By Lemma 2.4 it suffices to prove:
Theorem 5.1.
Let X be an EFX allocation on 4 agents with nice cancelable valuations, with at least twounallocated items. Then, there exists an EFX allocation Y that dominates X .13igure 4: High-level roadmap of the proof of Theorem 5.1. | C g | and | C h | are the lengths of some good cyclesof items g and h , respectively. ( ∗ ) The g and h champions of agent 4 can be either agent 1 or 2.The proof involves a rigorous case analysis, which exemplifies the extensive use of our new techniques.We have attempted to make the proofs as accessible as possible through the use of extensive aids such asfigures and colors.By assumption, there exist two unallocated goods which we denote g , h . The proof distinguishes betweentwo main cases, namely whether X is envy-free (Section 5.1) or not (Section 5.2). When X is envy-free, weshow that a Pareto improvable (PI) cycle always exists. This is shown via a case analysis that depends onthe lengths of the good g - and h -cycles which must exist in the champion graph M X .When X has envy, we argue that the only interesting case is where the basic edges follow some specificstructure, modulo permuting the agent identities (otherwise, a PI cycle exists). Then, we show that thereis an EFX allocation in which agent a vip (per the lexicographic potential) is better off relative to X . Since a vip could be any one of the agents (due to the identity permutation), the proof splits to cases accordingly,where the case a vip = 2 is treated separately from the case where a vip is one of the other three agents. Ourapproach here is heavily inspired by and follows a similar high-level structure to that of [CGM20] in theiranalysis of the envy case in their 3 agent result. Our proof structure is depicted in Figure 4. In this section we show that if X is envy-free, then we can always find a PI cycle or edge set in M X (seeRemark 3.7), implying (by Lemma 3.5) the existence of a Pareto-dominating EFX allocation Y .Recall that every agent i has an incoming g -champion edge and an incoming h -champion edge (Observa-tion 2.8), and thus M X contains a g -cycle and an h -cycle. If there is a self g or h champion we are done byCorollary 3.6. Thus, these cycles are of size at least 2, and contain no envy edges, and are therefore goodcycles. Denote them by C g , C h , respectively.Note that if one of these has size 1 (that is, there is a self g or h -champion) then we are done by Corollary3.6, thus we assume that | C g | , | C h | ∈ { , , } .By Observation 3.10, C g (resp. C h ) induces a g (resp. h )-decomposition of X j for any agent j in thecycle. In the following we denote the g (resp. h ) -decomposition by X j = T gj ·∪ B gj (resp. X j = T hj ·∪ B hj ).We shall make repeated use of Observations 3.11 and 3.12. For concise presentation, we write here theimplications that will be repeatedly used in this section: For every two agents i, j we have: • i B gj | ◦ j • If i and j reside on the same good g -cycle, then pred ( i ) B gi | ◦ j Analogous claims hold for h . We remind the reader that i B gk | ◦ j , i B hk | ◦ j are shorthand for i B gk | B gj j , i B hk | B hj j , respectively. 14ecall that each of C g , C h can be of size 2, 3, or 4. Assume w.l.o.g. that | C g | ≤ | C h | . Thus, there aresix cases to consider. Case 1: | C g | = | C h | = 2. Assume w.l.o.g. that C g = 1 g g
1. By Theorem 3.16, there existseither a good or external B g -edge going into agent 2, and there exists either a good or external B g -edgegoing into agent 1. If one of these is good we are done: for example, the only possible good B g -edge is1 B g | ◦ B g | ◦ g g B g ).Thus both edges have to be external, i.e. , their sources are agents 3 or 4. Assume w.l.o.g. that 3 B g | ◦
2. We cannot also have 3 B g | ◦ B g | ◦
1, and we have thefollowing structure:Consider C h . If C h = 1 h h g h C h = 3 h h
3, then following the analogous reasoning for C g we can assume thatwe have external B h and B h edges, in which case we have one of the following two structures (highlightededges are part of PI cycles or PI edge sets):In the left graph, the two cycles 1 B h | ◦ B g | ◦
1, 2 B h | ◦ B g | ◦ B h | ◦ B g | ◦ B h | ◦ B g | ◦ M X contains the good g -cycle 3 g g
3, thensimilar reasoning also shows that we have a PI cycle (or PI edge set). In other words, if M X contains twodisjoint g -cycles of size 2 or two disjoint h -cycles of size 2, then we are done.It remains to consider the case where C h intersects C g at exactly one agent. If C h = 1 h h h B g | ◦ g
1. The case C h = 2 h h C h = 2 h h C h = 1 h h
1, which are also symmetric. Thus, w.l.o.g.we assume C h = 2 h h B h and B h edges (coming out of agents 1 and 4). If 1 B h | ◦ B h | ◦
2, then we get the PI cycle 1 B h | ◦ h g
1. Thus 1 B h | ◦ B h | ◦ h -champion of agent 4 (such exists by Observation 2.8). If 3 h
4, we are done viathe PI edge set 1 B h | ◦ g
1, 3 h B h | ◦
3. If 2 h
4, we are done: 1 g h B g | ◦
1. Thus, assume 1 h i.e. , 4 h h -champion of agent 1. If 4 h h -cycles, a situation we have already dealt with at the start of Case 1. If 2 h g h
1. Therefore, 3 h g -champion of 3. If 1 g g
3, we have a size 2 cycle with g and h edgesand we are done. Thus, 4 g g -champion of 4. If 3 g g -cycles of size 2 and weare done. If 2 g
4, then we are done: 2 g B h | ◦ h
2. Thus assume 1 g
4, and we have thestructure:In this case we are done via the PI edge set 1 g B g | ◦
1, 2 h B g | ◦ Case 2: | C g | = 2 , | C h | = 3. Assume w.l.o.g. that C g = 1 g g
1. If C h passes through both agents1 and 2 then we are done since we are guaranteed to have a size 2 cycle with g and h edges. Thus C h passesthrough exactly one of them, and we can assume w.l.o.g. that C h = 1 h h h B g and B g edges guaranteed by Theorem 3.16 are external.If we have 3 B g | ◦ B g | ◦
2, we are done: 1 h B g | ◦ g
1. Thus assume we have theedges 3 B g | ◦ B g | ◦ g -champion of 3. If 1 g
3, we are done: 1 g h
1. If 2 g
3, we are done:1 h B g
1, 2 g B g
2. Thus 4 g
3, and we have the structure:16onsider C h . We ask which agent i satisfies i B h | ◦ succ (4) in C h ). We cannot have 1 B h | ◦ pred (4) in C h . If 4 B h | ◦
3, we are done: 1 h B h | ◦ B g | ◦ g
1. If 2 B h | ◦ h B g | ◦
1, 2 B h | ◦ B g | ◦
2. Thus, wemust have 3 B h | ◦ We now ask which agent i satisfies i B h | ◦ B h | ◦ succ (3) in C h ). We cannot have 1 B h | ◦
1, since 1 = pred (4) in C h . If 3 B h | ◦
1, we are done:1 h g B h | ◦
1. If 4 B h | ◦
1, we are done: 1 h B h | ◦
1. Thus 2 B h | ◦
1, and wehave the structure:Finally, we ask which agent i satisfies i B h | ◦ B h | ◦
4, as otherwise, together with 2 B h | ◦ B h < B h < B h ,contradiction. We cannot have 3 B h | ◦
4, as 3 = pred (1) in C h . Thus, we must have 4 B h | ◦
4, and weare done: 1 g B h | ◦
1, 4 B h | ◦ Case 3: | C g | = 2 , | C h | = 4. Assume (again) w.l.o.g. that C g = 1 g g
1. As in Case 1 we canassume w.l.o.g. that we have the edges 3 B g | ◦
2, 4 B g | ◦ C h . If it contains either the edge 1 h h g and h edges. Thus, we have one of the following two structures:In the left graph we have the PI edge set 1 h B g | ◦
1, 2 h B g | ◦
2, and in the right graph wehave the PI cycle 1 h B g | ◦ g
1. Thus we are done.
Case 4: | C g | = 3 , | C h | = 3. Assume w.l.o.g. that C g = 1 g g g
1. Assume first that C h is parallel to C g , i.e., C h = 1 h h h
1. Consider C g . By Theorem 3.16 there exists a good orexternal B gj -edge for every j ∈ { , , } . By Corollary 3.18, not all of these can come out of agent 3 (i.e.,be external). Thus, w.l.o.g., there is a good B g -edge, which can be either one of 1 B g | ◦ , B g | ◦ , B g | ◦
4, and it is easy to verify that in each of the cases we obtain a PI cycle by carefully choosingcomplementary g and h edges from C g and C h , respectively. Thus we assume w.l.o.g. that C h = 1 h h h g h g -champion of 3 and an h -champion of 4. If agent 1 is one of them, then weimmediately get a 2-cycle with g and h edges. The same is true if both 3 h g
3. Thus either Note that as opposed to a self g -loop, 3 B h | ◦ B h is not released within the cycle. g h
4. Since both cases are symmetric we may assume w.l.o.g. that 2 h
4, and we have thestructureConsider C h . By Theorem 3.16, M X contains a good or external B h -edge. Assume first that we have a good B h -edge. The only possible good B h -edges are 3 B h | ◦
1, 3 B h | ◦ B h | ◦
2. In the first twocases we are done: 1 g h B h | ◦ h B h | ◦
2, respectively. Thus we assume thatwe have 1 B h | ◦
2, and we get the structure:By Theorem 3.16, M X contains a good or external B h -edge. If it is a good B h -edge, then it is one of2 B h | ◦
3, 2 B h | ◦ B h | ◦
1, and in all cases we get a PI cycle:1 B h | ◦ B h | ◦ h , h B h | ◦ B h | ◦ h B h | ◦ , respectively. If it is an external B h -edge, then it is one of 4 B h | ◦ B h | ◦ B h -edge going into 2), and these cases admit the respective PI cycles:1 h g B h | ◦ B h | ◦ g B h | ◦ h . Thus, we are done with the case that M X contains a good B h -edge.Assume now that M X contains an external B h -edge. If M X contains an external B h -edge, then it mustbe 4 B h | ◦
3, as the other possibility 4 B h | ◦ h g B h | ◦ B h -edge is not drawn):By Corollary 3.18, since there are external B h and B h edges, there cannot be an external B h -edge. Thus,the B h -edge guaranteed by Theorem 3.16 must be a good edge, which can be one of 1 B h | ◦
3, 1 B h | ◦ B h | ◦
3. These cases admit the respective PI cycles:1 B h | ◦ h , B h | ◦ g B h | ◦ h g B h | ◦ h . Thus, we may assume that M X does not contain an external B h -edge.We ask which agent i satisfies i B h | ◦ B h | ◦ pred (2) in C h . We cannot have 4 B h | ◦ B h -edges.Thus 2 B h | ◦ B h | ◦
3. If 3 B h | ◦
3, then we get the following structure (again, the assumedexternal B h -edge is not drawn): 18he assumed external B h -edge can be either 4 B h B h
2, and in both cases we are done viaPI edge sets: 1 h g B h | ◦ , B h | ◦ g B h | ◦ , B h | ◦ , respectively. Thus 2 B h | ◦
3, and we have the following structure (again, the assumed external B h -edgeis not drawn):It is easy to verify that if agent 2 has other g or h champions except 1, then either we get an immediate PIcycle or we get a size 2 good h -cycle, which along with the size 3 g -cycle that we already have is a case wealready solved (Case 2). Thus 1 is the unique champion of 2 (w.r.t to both g , h ), and recall that 1 (cid:54) Claim 5.2.
Let i, j be agents such that i (cid:54) j , and i is the unique g and h champion of j . If h < i g , thenthere exists a choice of D hi,j and D gi,j such that D hi,j ⊆ D gi,j .Assume that h < g . By Claim 5.2 there exists a choice of D h , and D g , such that D h , ⊆ D g , . Note that B h and B g are arbitrary bottom half-bundles ( i.e. , discard sets that do not contain h and g , respectively)whose specific identity had no impact on the proof thus far. Thus we can redefine B h and B g to be thischoice of D h , and D g , , respectively (note that these do not contain h and g by Observation 2.9), andhave B h ⊆ B g . Thus, the cycle 1 g B h | ◦ h g -edge releases B g whichcontains B h .We can therefore assume that g < h , which leads by symmetric reasoning to the assumption B g ⊆ B h .Consider C g . By Theorem 3.16, M X contains a good or external B g -edge. If it is a good B g -edge, then it isone of 1 B g | ◦
2, 1 B g | ◦
4, 2 B g | ◦
4, and in all cases we get a PI cycle:1 B g | ◦ h g , B g | ◦ g , h B g | ◦ g B g -edge.By Theorem 3.16, M X contains a good or external B g -edge. If it is a good B g -edge, then it is one of2 B g | ◦
4, 2 B g | ◦ B g | ◦
1, and in all cases we get a PI cycle:1 h B g | ◦ g , g B g | ◦ , and 1 g h B g | ◦ , respectively, where the first cycle is indeed PI since the h -edge releases B h which contains B g . Thus, we canassume that there is an external B g -edge.By Corollary 3.18, since there are external B g and B g edges, there cannot be an external B g -edge. Thus,the B g -edge guaranteed by Theorem 3.16 must be good, i.e. it is one of 4 B g | ◦
1, 4 B g | ◦
2, 1 B g | ◦
2. In the first two cases we immediately get a PI cycle:1 h g B g | ◦ , g B g | ◦ B g | ◦
2, and we have the following structure (the assumed B g , B g and B h externaledges are not drawn): 19onsider the external B g edge, which can be one of 3 B g | ◦ B g | ◦
4. If it is 3 B g | ◦
4, then we are done: 1 B g | ◦ h B g | ◦ g
1. Thus we have the edge 3 B g | ◦
2, and thestructure isFinally, we consider the external B g -edge. It cannot be 3 B g | ◦
1, as otherwise, together with3 B g | ◦ B g < B g < B g , contradiction. Hence it must be 3 B g | ◦
4, and we are done: 1 B g | ◦ h B g | ◦ g Case 5: | C g | = 3 , | C h | = 4. Assume w.l.o.g. that C g = 1 g g g C h = 1 h h h h C h is in the opposite direction we immediately get a PI cycle). We havethe structure:Consider C g . In what follows we reason about possible B g , B g and B g edges, starting with B g . ByTheorem 3.16, M X contains a good or external B g -edge. If it is a good B g -edge, then it is one of 3 B g | ◦
4, 3 B g | ◦
1, 4 B g | ◦
1, and in all cases we get a PI cycle:1 g B g | ◦ h , g B g | ◦ , g h B g | ◦ B g -edge, which can be either 2 B g | ◦ B g | ◦
4, and thus the structure is one of the following:We now ask which agent i satisfies i B g | ◦ succ (1) in C g ). Wecannot have 4 B g | ◦
3, since 4 = pred (1) in C g . If 1 B g | ◦
3, we are done: 1 B g | ◦ h g
1. If 3 B g | ◦
3, we are done regardless of whether 2 B g | ◦ B g | ◦
4; in the first case we havethe PI edge set 1 h B g | ◦
1, 3 B g | ◦
3, and in the second case we have the PI edge set 1 h B g | ◦ g
1, 3 B g | ◦
3. Thus, we must have 2 B g | ◦ B g -edge cannot be 2 B g | ◦
1, as otherwise by Observation 3.17 we get B g < B g < B g , contradiction. Hence we have 2 B g | ◦
4, and we obtain the following structure:20e now ask which agent i satisfies i B g | ◦ succ (4) in C g ).We cannot have 3 B g | ◦
1, since 3 = pred (4) in C g . We cannot have 2 B g | ◦
1, as otherwise we get acontradiction to Corollary 3.18, since there are already external B g and B g edges. If 4 B g | ◦
1, we aredone: 1 h B g | ◦ g B g | ◦
1. Thus we must have 1 B g | ◦ B g | ◦ i that satisfies i B g | ◦ succ (1) in C g ). We cannot have 2 B g | ◦
3, as again we would get a contradiction to Corollary 3.18. We cannot have3 B g | ◦ pred (4) in C g . If 1 B g | ◦ B g | ◦ g h
1. Thus we musthave 4 B g | ◦
3, and we are done: 3 g B g | ◦ Case 6: | C g | = 4 , | C h | = 4. It is easy to verify that if C g and C h are not parallel to each other thenthere must exist a 2-cycle with g and h edges. Thus we may assume w.l.o.g. that C g = 1 g g g g C h = 1 h h h h
1, i.e., we have the following structure:Consider C g . By Theorem 3.16, M X contains a good B g -edge (note that there are no external edges asboth cycles pass through all agents), which can either be parallel to one of the edges in C g (1 B g | ◦ B g | ◦
3, or 3 B g | ◦ B g | ◦ B g | ◦
4) or going back (1 B g | ◦ g -edge and maybealso an h -edge.Since all half-bundles are symmetric, we may thus assume that all good edges with respect to any bottomhalf-bundle are parallel to C g and C h . Assume w.l.o.g. that one of these is 1 B gi | ◦ i , and notethat i (cid:54) = 2. Considering the 3 possibilities for the good B g -edge gives us one of the following 3 structures:and in any case we can form a PI cycle by taking the B gi and B g edges, and among the two options forcompleting with one g and one h edge, choose the option so that the h edge does not go into i .21 .2 X is Not Envy-Free We first consider the structure of the envy edges and champion edges w.r.t. g and h . We show that there isonly one non-trivial such structure in the following lemma, whose proof is deferred to Appendix B. Lemma 5.3.
Assuming M X contains no PI cycle, and up to renaming of the agents, the possible g and h champions of 4 are only agents 1 and 2, and all other basic edges in M X are exactly as depicted in thefollowing figure:The above lemma and Observation 2.9 imply: Remark 5.4. agents 1,2 and 3 are g and h decomposed by their unique champion, 2,3 and 4, respectively.In the following we denote the g (resp., h ) -decomposition by X j = T gj ·∪ B gj (resp. X j = T hj ·∪ B hj ), for j ∈ { , , } . Without loss of generality, we assume that g < h (otherwise, switch the names of h and g andthe champion graph described above remains the same). By Claim 5.2, we may redefine B g and B h suchthat they also satisfy B g ⊆ B h . (1)From this point on we don’t always find a PI cycle. Instead, we show that there is a dominating EFXallocation in which the agent a vip strictly improves (while other agents may become worse-off), unless wefind a PI cycle in X . Since we renamed the agents depending on the graph structure, agent a vip can be anyone of the agents. Hence we split into cases according to the identity of a vip . a vip (cid:54) = 2 As a first attempt consider the following allocation, denoted X (cid:48) : X (cid:48) = X (1) T g B g (2) T g g (3) T h h (4) All agents are better off in X (cid:48) : Agent 1 is better off, since X < X . Agent 2 is better off, since X = T g ∪ B g < T g ∪ B g , by Observation 2.11 (since 2 g g
2) and cancelability. Agent 3 isbetter off, since X < T g ∪ { g } (since 3 g X < T h ∪ { h } (since 4 h The only strong envy in X (cid:48) is from 1 to 2:
No agent strongly envies the sets X (cid:48) = X , X (cid:48) = T g ∪ { g } , X (cid:48) = T h ∪ { h } as all agents are better off and no agent strongly envied any of them in X (bydefinition of basic championship).It remains to show that agents 3 and 4 do not strongly envy X (cid:48) . Agent 3 does not envy X (cid:48) , since X (cid:48) > X > X = T g ∪ B g > T g ∪ B g = X (cid:48) , where the first inequality is due to 3 (cid:54) X (cid:48) , then X < X (cid:48) < T g ∪ B g < T g ∪ B g , where the third inequality is by Observa-tion 2.11 and cancelability. The following claim shows that there must now exist a PI-cycle. The proof ofthe claim is deferred to Appendix B. Claim 5.5. If X < T g ∪ B g , then there exists a PI cycle in M X .22y Claim 5.5, we may assume that agent 4 does not strongly envy X (cid:48) . It remains to consider agent 1.If she does not strongly envy X (cid:48) then X (cid:48) is an EFX allocation that Pareto-dominates X . Thus we assumethat the only strong envy in X (cid:48) is from 1 to 2 as we wanted to show.Now, notice that T g ∪ B g > max (cid:8) X , T g ∪ g, T h ∪ h, X (cid:9) in X (cid:48) , since agent 2 doesn’t envy agents 3and 4 in both X and X (cid:48) , and agent 2 is better off in X (cid:48) . Let Z ⊆ T g ∪ B g be a set of minimal size satisfying Z > max (cid:8) X , T g ∪ g, T h ∪ h, X (cid:9) , i.e. , the inequality no longer holds after removing any item from Z .If X > Z then it is easy to find a dominating allocation as shown in the following lemma. The proof ofthe lemma is deferred to Appendix B. Lemma 5.6. If X > Z then there is an EFX allocation Y that dominates X .By Lemma 5.6 we may assume for the rest of the proof that X < Z . Consider the allocation X (cid:48)(cid:48) definedas follows: X (cid:48)(cid:48) = Z (1) max { X , T g ∪ g, T h ∪ h, X } (2) T g ∪ g or ∗ X (3) X or † T h ∪ h (4) ∗ X (cid:48)(cid:48) = T g ∪ { g } unless X (cid:48)(cid:48) = T g ∪ { g } , in which case X (cid:48)(cid:48) = X ; † X (cid:48)(cid:48) = X unless X (cid:48)(cid:48) = X , in which case X (cid:48)(cid:48) = T h ∪ { h } . Claim 5.7. X (cid:48)(cid:48) is EFX.
Proof.
Agent 1 envies no one since
Z > X and she did not envy X , T g ∪ { g } and T h ∪ { h } in previouslyconsidered allocations when she was worse off.Agent i , for i ∈ { , } , does not strongly envy another agent: Notice that X (cid:48)(cid:48) i ≥ i X i . We have shown X i > i T g ∪ B g ≥ i Z in our analysis of allocation X (cid:48) . Thus, X (cid:48)(cid:48) i > i X (cid:48)(cid:48) . Moreover, in allocation X no agentstrongly envies T g ∪ g and T h ∪ h , by definition of champion.Agent 2 does not strongly envy 1 by definition of Z , and does not strongly envy agents 3 or 4 since X (cid:48)(cid:48) = max (cid:8) X , T g ∪ g, T h ∪ h, X (cid:9) ≥ max (cid:8) X (cid:48)(cid:48) , X (cid:48)(cid:48) (cid:9) .Agent 1 is strictly better off in allocation X (cid:48)(cid:48) . Thus, if agent a vip is agent 1, then X (cid:48)(cid:48) dominates X , andwe are done. Hence, it remains to show that agents 3 and 4 can be made better off. If X (cid:48)(cid:48) = X , then both3 and 4 are better off and we are done. We split to cases according to the rest of the possibilities for theidentity of X (cid:48)(cid:48) : Case A : X (cid:48)(cid:48) = T g ∪ { g } , Case B : X (cid:48)(cid:48) = T h ∪ { h } , Case C : X (cid:48)(cid:48) = X We shall need the following claim that holds in all three cases (the proof is deferred to Appendix B):
Claim 5.8.
Agent 4 is most envious of T h ∪ { h } in X (cid:48)(cid:48) (in cases A , B , C ). Case A
In this case X (cid:48)(cid:48) is X (cid:48)(cid:48) = Z (1) T g g (2) X (3) X (4)Notice that h and the set B g remain unallocated. Let b ∈ B g be an arbitrary good in B g (note that B g (cid:54) = ∅ since otherwise, as valuations are non-degenerate, we have X < X ∪ { g } = T g ∪ { g } , in contradiction toObservation 2.10 since 1 g X ).Consider the champion graph M X (cid:48)(cid:48) restricted to envy edges and champion edges with respect to h and b . Agent 2 envies agent 1 by definition of Z , and agent 3 envies agent 2 since 3 g X . Thuswe have the structure: 23 ubcase: a vip = 3. By Observation 2.8, agent 3 has a b -champion in M X (cid:48)(cid:48) . If this b -champion is agent 1,2or 3, we obtain a PI-cycle: 1 b b b
3, respectively. By Lemma 3.5, sincethis PI-cycle always includes agent 3, it follows that there exists a partial EFX allocation X (cid:48)(cid:48)(cid:48) that Paretodominates X (cid:48)(cid:48) , in which agent 3 is better off. Hence, X (cid:48)(cid:48)(cid:48) dominates X in case agent a vip is agent 3, asdesired. Thus, assume agents 1,2 and 3 are not b -champions of agent 3. This leaves agent 4 as the unique b -champion of agent 3 in allocation X (cid:48)(cid:48) , and we have the structureWe next show that in the original allocation X , either 4 b b X (cid:48)(cid:48) ,4 is the most envious agent of X ∪ { b } while agents 2 and 3 are not. In the allocation X the bundles of3 and 4 were the same, and agent 2 was better off relative to X (cid:48)(cid:48) . Therefore, neither agent 2 nor 3 can bemost envious agents of X ∪ { b } in allocation X . It follows that either 1 or 4 were most envious agents of X ∪ { b } , i.e. , 4 b b X .We conclude that there exists a PI-cycle in X in both cases, which we could have applied in hindsight:4 b h g b h g
1, respectively. Both are indeed PI-cycles, since the edge2 g B g of which the item b is a member. Hence, we are done by Lemma 3.5. Subcase: a vip = 4. Agent 4 is the most envious agent of T h ∪ { h } by Claim 5.8, implying that 4 h X (cid:48)(cid:48) (since X (cid:48)(cid:48) = X ). Thus we have the structure:By Observation 2.8, there exists a b -champion of agent 4. Regardless of who that champion is we obtaina PI cycle that includes agent 4:1 b h , b h , b h b . By Lemma 3.5, since the resulting PI-cycle includes agent 4, it follows that there exists a partial EFXallocation X (cid:48)(cid:48)(cid:48) that Pareto dominates X (cid:48)(cid:48) , in which agent 4 is better off. Hence, X (cid:48)(cid:48)(cid:48) dominates X in caseagent a vip is agent 4, as desired. Case B or C
In these cases the allocation X (cid:48)(cid:48) isCase B Case C X (cid:48)(cid:48) = Z (1) T h h (2) T g g (3) X (4) X (cid:48)(cid:48) = Z (1) X (2) T g g (3) X (4)24n both cases agent 3 is better off ( X < T g ∪ { g } since 3 g X ). Thus we may assume a vip = 4.We have that 2 1 in allocation X (cid:48)(cid:48) , by definition of Z . In Case B we have 4 2 by Claim 5.8, since X (cid:48)(cid:48) = T h ∪ { h } . In Case C we have 4 h X (cid:48)(cid:48) = X . Therefore we have the followingstructure in the allocation X (cid:48)(cid:48) :In both cases, B g remains unallocated. Let b ∈ B g be an arbitrary good in B g . Recall that B g is nonempty, since 1 g X . By Observation 2.8, there is some agent i ∈ [4] such that i b X (cid:48)(cid:48) .If i is 1,2 or 4, then we obtain a PI-cycle in both Cases B and C . In Case B these cycles are1 b , b , b . In Case C these cycles are 1 b h , b h , b . Agent 4 is along each of the PI-cycles described above, hence by Lemma 3.5 there exists an EFX allocationin which agent 4 is better off, and we are done. Thus, we assume agent 3 is the unique b -champion of agent4 in allocation X (cid:48)(cid:48) . We obtain the following structure:We split the rest of our analysis of Cases B and C into three subcases:I X > T g ∪ B g ; II 2 g X ; III 2 g X . Subcase
I : X > T g ∪ B g . In this case, in any allocation where agent 1 gets X or Z (which is worthmore than X to agent 1), we can give T g ∪ B g to another agent knowing that agent 1 will not envy it.(This is somewhat reminiscent of our approach in constructing allocation X (cid:48) . There, the allocation hadbeen EFX if X > T g ∪ B g .) Note that agent 3 envies T g ∪ B g in X since X = T g ∪ B g < T g ∪ B g , byObservation 2.11 and cancelability. Thus there exists an agent i which is most envious agent of that set, i.e. , i B g | ◦
2. We ask who i is, ignoring agent 1 (that is, the most envious agent of T g ∪ B g restricted toagents 2,3,4). Ignoring agent 1 is fine as long as agent 1 receives X or Z in the allocations we construct. i cannot be agent 4. We show this by essentially using Observation 3.12 since 4 = “ pred (3)” in M X .While we do not have a good cycle in M X and cannot really use that observation, its proof still applies: wehave X > X = T g ∪ B g > T g ∪ B g where the first inequality is by 4 (cid:54) B g | ◦
2. By the next claim, if i = 3 we aredone. Claim 5.9. If i = 3 then M X contains a PI cycle. 25he proof of Claim 5.9 has been deferred to Appendix B.We are left with the case i = 2. In the transition to X (cid:48)(cid:48) from X , agent 2 became worse off while therest of the agents did not. Thus, ignoring agent 1, agent 2 is also most envious of T g ∪ B g in X (cid:48)(cid:48) . Since X (cid:48)(cid:48) = Z > X > T g ∪ B g , agent 1 cannot be the real most envious agent in X (cid:48)(cid:48) and thus agent 2 is thereal one. Since X (cid:48)(cid:48) = T g ∪ { g } , this means that 2 B g | g X (cid:48)(cid:48) and we obtain a PI cycle in both casesB , C :Note that B g can indeed be allocated to agent 2, as B g ⊆ B h (by Equation (1)) and B h is available inboth cases (unallocated in Case B and released by the edge 4 h X (cid:48)(cid:48)(cid:48) that Paretodominates X (cid:48)(cid:48) , in which agent 4 is better off. Hence, X (cid:48)(cid:48)(cid:48) dominates X in case agent a vip is agent 4, asdesired. Subcase
II : g . In this case, M X restricted to basic edges has the structure:We start with a proof sketch: Notice that 2 g g g g -cycle in M X , hence 2 g -decomposes 4 into X = T g ·∪ B g . We first show that X > T g ∪ B g (see Lemma 5.10). This allows us toapply a similar approach to the previous case: If agent 1 gets Z (which is better for her than X ), then wecan safely give T g ∪ B g to another agent knowing that agent 1 will not envy.As in the previous case, we need to show that there is an agent other than 1 that envies the set T g ∪ B g .(In the previous case the fact that 3 envies T g ∪ B g was immediate.) Here, the assumption that we are notin Subcase I allows us to reach this conclusion (see Lemma 5.11).Having shown that, we then find a PI-cycle in the sub-allocation X − = (cid:104) X , X , X (cid:105) , the application ofwhich leaves Z unallocated (see Lemma 5.12). Finally, we can then allocate Z to agent 1 to obtain an EFXallocation that Pareto-dominates X (see Corollary 5.13). Lemma 5.10. X > T g ∪ B g . Lemma 5.11.
Agent 3 envies T g ∪ B g in X .The proofs of Lemmata 5.10 and 5.11 have been deferred to Appendix B.Consider the 3-agent suballocation X − = (cid:104) X , X , X (cid:105) . Lemma 5.12.
There is an EFX allocation Y − over agents 2,3,4 that Pareto-dominates X − , in which Z remains unallocated. Proof.
By Lemma 5.11, agent 3 envies T g ∪ B g . In particular, there exists an agent i ∈ { , , } such that i B g | ◦ X − . i cannot be agent 2, because 2 = pred (4) in the good g -cycle (Observation 3.12). If i isagent 3, we obtain the PI-cycle 3 B g | ◦ g h M X − . If i is agent 4, we obtain the PI-cycle4 B g | ◦ g M X − . 26n both cases, by Lemma 3.5 there exists an EFX allocation Y − over agents 2,3 and 4 that Paretodominates the allocation X − . Moreover, in both cases the set Z ( ⊆ T g ∪ B g ) is unallocated in Y − and weare done. Corollary 5.13.
The allocation Y obtained from Y − by allocating Z to agent 1 is EFX and Pareto-dominates X . Proof.
First, Y Pareto-dominates X since Y − Pareto-dominates X − and Y = Z > X > X . We nowshow that Y is EFX. The allocation Y − is EFX by construction, thus it suffices to show that agent 1 doesnot strongly envy any other agent, and that no agent strongly envies agent 1. Agent 1 does not strongly envy anyone:
Recall that
Z > X > X . The possible bundles in Y − are T g ∪ { g } , T h ∪ { h } , X and a subset of T g ∪ B g . Agent 1 did not strongly envy the first two bundles inthe original allocation X , by definition of champion, and did not envy X . Since she is now better off shedoes not strongly envy those bundles in Y . Finally, agent 1 does not envy a subset of T g ∪ B g , since Z > X > T g ∪ B g , where the last inequality is due to Lemma 5.10. Therefore, agent 1 does not strongly envy any other agentin Y . No agent strongly envies agent 1:
Agent 2 does not strongly envy 1 since she did not strongly envythe bundle Z in allocation X (cid:48)(cid:48) , where she was worse off. We have already shown that X > T g ∪ B g ( ⊇ Z )and X > T g ∪ B g ( ⊇ Z ) when we constructed the allocation X (cid:48) . Thus, since neither agent 3 nor 4 becameworse off in the transition from allocation X to Y , we conclude that agents 3 and 4 do no envy agent 1. Subcase
III : g . In this case we must have 1 g
4, and M X restricted tobasic edges has the structure:We now briefly sketch our strategy to find a PI cycle in M X (cid:48)(cid:48) that includes agent 4. Since agent 1is not only a g -champion of agent 4 in X , but also envies her, it might be the case that this is a trivialchampionship, i.e. , D g , = { g } . Such a championship relationship is not very useful, since no items arereleased from this edge that can used in other generalized championship edges. Our first step is to show thatthis is not necessarily the case: we show that agent 1 envies a strict subset of X together with g , that is,we show that agent 1 g -decomposes 4 into top and bottom half-bundles T g and B g (unless there exists a PIcycle in X ). This is established in Lemma 5.14.Given this assumption, our next step is Lemma 5.16 in which we show that in the allocation X (cid:48)(cid:48) there isan agent i (cid:54) = 4 such that i B g | B g
4. When i ∈ { , } , we immediately a PI cycle that includes agent 4 inboth cases B and C , as depicted below: 27hese are indeed PI cycles since B h is unallocated in Case B and released by the 4 h B g ⊆ B h by Equation (1). The problematic case is when i = 3, giving us the structureTo solve this case we show, in Lemma 5.17, that 2 B g | g X (cid:48)(cid:48) . Since B g is released by the3 B g | B g Lemma 5.14.
Agent 1 g -decomposes agent 4 in the allocation X . Proof.
Consider the allocation X (cid:48)(cid:48) . Recall that in the allocation X (cid:48)(cid:48) agent 3 is the unique b -champion ofagent 4. In particular we have 4 b
4, implying D b , (cid:54) = ∅ . Furthermore 3 b
4, and together with thefact that 3 (cid:54) X (cid:48)(cid:48) = T g ∪ { g } > X > X = X (cid:48)(cid:48) ) this implies b / ∈ D b , by Observation 2.9. Weconclude that there exists some item d ∈ D b , which is not b , i.e. , d ∈ X (cid:48)(cid:48) = X . We obtain X ≤ X (cid:48)(cid:48) < ( X ∪ { b } ) \ D b , ≤ ( X ∪ { b } ) \ { d } = ( X \ { d } ) ∪ { b } , (2)Where the first inequality holds since agent 3 is better off in allocation X (cid:48)(cid:48) , the second holds by definitionof basic championship, the third holds since valuations are monotone and the equality holds since b (cid:54) = d .By the following claim we may assume b < g . The proof of the claim has been deferred to Appendix B. Claim 5.15. b < g , unless there exists a PI cycle in M X .28y Claim 5.15 and Equation (2) we obtain X < ( X \ { d } ) ∪ { b } < ( X \ { d } ) ∪ { g } , where the secondinequality holds by cancelability. Since agent 3 envies ( X \ { d } ) ∪ { g } , there exists a most envious agentof this set. Let i ∈ [4] be that agent. Clearly, i g
4. Since we are in Subcase III , i (cid:54) = 2. If i = 3 or i = 4 we obtain a PI-cycle: 3 g h
3, 4 g
4, respectively. Hence, i = 1. Therefore, there exists achoice of D g , that contains d . Since the set D g , has not been used in this scenario of our proof thus far,we may define it such that d ∈ D g , , rather than arbitrarily. Finally, we have that g / ∈ D g , , as otherwise X < ( X ∪ { g } ) \ { d, g } = X \ { d } , in contradiction to allocation X being EFX. This shows that agent 1 g -decomposes agent 4, as claimed.By Lemma 5.14 agent 1 g -decomposes 4 into top and bottom half-bundles denoted T g and B g . The nextlemma reveals a generalized champion edge in M X (cid:48)(cid:48) that in most cases immediately closes a PI-cycle thatincludes agent 4. The proof of the lemma has been deferred to Appendix B. Lemma 5.16.
There exists i ∈ [4] such that i B g | B g M X (cid:48)(cid:48) .Consider the agent i that satisfies i B g | B g M X (cid:48)(cid:48) which is guaranteed by Lemma 5.16. First, i (cid:54) = 4since X (cid:48)(cid:48) = X > X = T g ∪ B g > T g ∪ B g , where the first inequality holds by 4 (cid:54) M X and the second holds by Observation 2.11 and cancelability.If i = 1 or i = 2 then we obtain a PI-cycle that includes agent 4. To see that the cycles we propose areindeed PI-cycles, recall that B g ⊆ B h by Equation (1) and in Case B the set B h remains unallocated whilein Case C the set B h is released by the edge 4 h B g | B g M X (cid:48)(cid:48) , we obtain the PI-cycle 1 B g | B g B g | B g h B g | B g M X (cid:48)(cid:48) , we obtain the PI-cycle 2 B g | B g B g | B g h i = 3, giving us the structure29o solve this case we next establish (see Lemma 5.17) the existence of the edge 2 B g | g X (cid:48)(cid:48) = T g ∪ { g } ), giving us the PI-cycle 3 B g | B g B g | g B g | B g h B g | g X (cid:48)(cid:48)(cid:48) in which agent 4 is strictly better off. Since we assumed that agent a vip is 4, it follows that X (cid:48)(cid:48)(cid:48) dominates X , as desired. The following lemma finishes this case. Lemma 5.17. B g | g M X (cid:48)(cid:48) . Proof.
First we prove the following auxiliary claim: (its proof is deferred to Appendix B)
Claim 5.18. B g | ◦ M X , unless there exists a PI-cycle in M X .Now, since X (cid:48)(cid:48) = T g ∪ { g } , we need to show that agent 2 is the most envious agent of T g ∪ B g . In thetransition from allocation X to X (cid:48)(cid:48) , agent 2 is worse off while all other agents are not, and thus it is enoughto show that agent 2 was most envious of T g ∪ B g also in X .We first show that there is some agent who envies T g ∪ B g . Note that if B g < B g then X = T g ∪ B g < T g ∪ B g , i.e. , agent 2 envies T g ∪ B g . The following claim helps us deal with the case B g < B g . Its proofhas been deferred to Appendix B. Claim 5.19. If B g < B g then 3 B g | ◦ M X , unless there exists a PI-cycle in M X .Assume B g < B g . Claim 5.19 implies that X < T g ∪ B g < T g ∪ B g , where the second inequality isby Observation 2.11 and cancelability. Hence, the set T g ∪ B g is envied by agent 3, and in particular thereis a most envious agent of T g ∪ B g in the allocation X . To complete the proof of Lemma 5.17 we show thatthis agent must be 2.Assume towards contradiction that 1 is the most envious agent of T g ∪ B g , then T g ∪ B g = X < X < T g ∪ B g (the first inequality follows by 1 (cid:54) B g < B g by cancelability. Moreover, byObservation 2.11, T g < T g . Therefore, by cancelability Z ⊆ T g ∪ B g < T g ∪ B g < T g ∪ B g = X , in contradiction to the assumption Z > X .If agent 3 (resp., 4) is the most envious agent of T g ∪ B g , then 3 B g | ◦ B g | ◦
2) closesthe PI-cycle 3 B g | ◦ h B g | ◦ g B g | ◦ h B g | ◦
4) in M X . Recall that1 B g | ◦ T g ∪ B g and the lemma now follows. a vip = 2 Recall that the structure of M X (restricted to envy, g and h edges) is as follows:30ith only the g and h edges going into agent 4 missing (where the source of either of these can be either 1or 2). Observation 5.20.
For i ∈ { , } we have X > i max i { X , T h ∪ { h } , T g ∪ { g } , X } Proof.
We have X > X , T h ∪ { h } , T g ∪ { g } , X since 1 (cid:54)
2, by Observation 2.10 (since 1 h g (cid:54)
3, respectively. The symmetric claim for agent 4 holds for thesame reasons.By the above observation, we can define for i ∈ { , } the bundle Z i to be the subset of X obtained byiterative removal of the item of least marginal value to agent i as long as the leftover bundle is still betterthan max i { X , T h ∪ { h } , T g ∪ { g } , X } for i .By Lemma A.4, Z i is a smallest size subset of X s.t. Z i > i max i { X , T h ∪{ h } , T g ∪{ g } , X } . Note thatthe fact that Z i is of minimal cardinality holds trivially under additive valuations. Lemma A.4 generalizesthis fact to all nice cancelable valuations.Denote w = 1 , (cid:96) = 4 if | Z | < | Z | and w = 4 , (cid:96) = 1 if | Z | ≥ | Z | . Define the allocation X (cid:48) as: X (cid:48) = X ( w ) max (cid:96) { X , T h ∪ h, T g ∪ g, X } ( (cid:96) ) T g ∪ g or ∗ X (2) X or † T h ∪ h (3) ∗ X (cid:48) = T g ∪ { g } unless X (cid:48) (cid:96) = T g ∪ { g } , in which case X (cid:48) = X ; † X (cid:48) = X unless X (cid:48) (cid:96) = X , in which case X (cid:48) = T h ∪ { h } . The Only Possible Strong Envy in X (cid:48) is From (cid:96) to w : For i ∈ { , , w } , agent i does not stronglyenvy any of the sets X , T g ∪ { g } , T h ∪ { h } , X , X , as agent i is not worse off and she did not strongly envyany of those before (by definition of basic championship, and the fact that X is EFX). Since this list includesall possible sets in X (cid:48) , it follows that agents 2 , w envy no other agent in X (cid:48) . Finally, (cid:96) does not envyagents 2 and 3 since she receives her most valued bundle out of 4 options, (some of) the leftovers of whichgo to agents 2 , If X (cid:48) = T g ∪ { g } : In this case agent 2 is better off relative to X . Update X (cid:48) by replacing X (cid:48) w with Z w . Since | Z w | ≤ | Z (cid:96) | , and since Z (cid:96) is a smallest size bundle such that Z (cid:96) > (cid:96) max (cid:96) { X , T h ∪ { h } , T g ∪ { g } , X } = X (cid:48) (cid:96) it follows immediately that Z w \ { h } < (cid:96) X (cid:48) (cid:96) for any h ∈ Z w . Thus (cid:96) does not strongly envy w and we obtainan EFX allocation where agent 2 is better off and we are done. If X (cid:48) = X : In this case the allocation X (cid:48) is X (cid:48) = X ( w ) T g g ( (cid:96) ) X (2) X (3)and observe that h and all items of B g are unallocated. Take some arbitrary item b ∈ B g . Recall that B g isnon empty, as otherwise 1 would have been a self g -champion in the original allocation X . We split to casesaccording to whether X (cid:48) is EFX or not, i.e. , whether (cid:96) strongly envies w or does not. Note that even if X (cid:48) is EFX we are not done yet since agent 2 is not better off relative to X . Case: X (cid:48) is EFX. In X (cid:48) we have 2 (cid:96) since 2 g X . We also have (cid:96) w by Observation 5.20. Besides that thereis no more envy in X (cid:48) .Consider a b -champion of agent 2 in X (cid:48) (such exists by Observation 2.8). If 2, (cid:96) or w are such, then weget the following respective PI cycles that include agent 2:2 b , (cid:96) b , (cid:96) w b Y in which agent 2 is better off relative to X , and we aredone (recall that X (cid:48) = X ). Thus, we can assume that 3 is the unique b -champion of 2, and M X (cid:48) currentlyhas the structure: 31 laim 5.21. Restricted to agents 2,3,4, the most envious agent of X ∪ { b } is agent 3. Proof.
Note that in the transition from X (cid:48) to X , agents 2 and 3 stay the same, and agent 4 is either betteroff or stays the same as well (depending on whether (cid:96) = 4 or w = 4). Thus, since 3 is most envious of X ∪ { b } in X (cid:48) , we get that 3 is also most envious of that set in X (restricted to agents 2,3,4).Consider the cycle 1 4 h b g M X : (note that 2 g b ∈ B g )This cycle might not really exist, since it could be the case that 1 is the real b -champion of agent 2.Assume for now that this is not the case, i.e. , 3 b M X . Then this is a PI cycle, and byLemma 3.5, there is an allocation Y that Pareto-dominates X . Note that in Y agent 1 receives X . Wetherefore claim that Y is EFX even if 1 had been the most envious agent of X ∪ { b } , and this finishes theproof. Since 3 is the most envious agent in X restricted to agents 2,3,4, the only issue that could prevent Y from being EFX is if 1 strongly envies 3 in Y . But this is not the case since Y = X ≥ X (cid:48) > ( X ∪ { b } ) \ D b , = Y where the second inequality holds since 1 b X (cid:48) . Case: X (cid:48) is Not EFX.
Recall that in this case (cid:96) strongly envies w . Define the set Z ⊆ X as the subset of X obtained byiterative removal of the item of least marginal value to agent w , until agent (cid:96) does not strongly envy w anymore. We obtain the allocation X (cid:48)(cid:48) defined as follows: X (cid:48)(cid:48) = Z ( w ) T g g ( (cid:96) ) X (2) X (3)Clearly, (cid:96) does not strongly envy w in X (cid:48)(cid:48) . Note that by definition of Z (cid:96) and Z , it must be the case that | Z (cid:96) | ≤ | Z | and consequently | Z w | ≤ | Z (cid:96) | ≤ | Z | . Since both Z and Z w are obtained from X by iterativeremoval of the item with least marginal contribution, we have Z w ⊆ Z , and we conclude that w does notenvy any other agent in X (cid:48)(cid:48) by monotonicity of v w and the definition of Z w . Thus X (cid:48)(cid:48) is an EFX allocation.Note however that (cid:96) might still envy w . If this is indeed the case, then from here we can proceed byessentially repeating the previous case with X (cid:48)(cid:48) instead of X (cid:48) . We thus assume that (cid:96) (cid:54) w . Let c ∈ X \ Z be the last item to be removed from X to obtain Z . By definition of Z , (cid:96) strongly envies Z ∪ { c } . Recallhowever that c is the least valued item in Z ∪ { c } according to w . Thus, for any item x ∈ Z ∪ { c } we have Z = ( Z ∪ { c } ) \ { c } > w ( Z ∪ { c } ) \ { x } establishing that w does not strongly envy Z ∪ { c } in X (cid:48)(cid:48) . We conclude that (cid:96) is the most envious agent of X (cid:48)(cid:48) w ∪ { c } (recall that 2 , X which is a superset), i.e., (cid:96) c w .From here we can, as above, essentially repeat the proof of the previous section with X (cid:48)(cid:48) instead of X (cid:48) and the champion edge (cid:96) c w instead of the envy edge (cid:96) w .32 Simplification and Extension of Known Results
In this section we simplify the proofs of full EFX existence for 3 additive agents [CGM20] and for n agentswith one of two fixed additive valuations [Mah20]. Moreover, our proofs extend beyond additive to all nicecancelable valuations. Our proofs demonstrate the versatility of our techniques. By Lemma 2.4 it is sufficient to prove the following theorem.
Theorem 6.1.
Let X be an EFX allocation for 3 agents with nice cancelable valuations, with at least oneunallocated item. Then, there exists an EFX allocation Y that dominates X .By assumption there is an item g that is unallocated in X . The original proof in [CGM20] distinguishesbetween two main cases according to whether X is envy-free or not. In the case where X is not envy-freethe original proof extends almost immediately to nice cancelable valuations (with one exception, see below).The property of an additive valuation v that is applied there is that for any bundles S, T, R such that R is disjoint from both S and T , v ( S ∪ R ) > v ( T ∪ R ) ⇔ v ( S ) > v ( T ), and this property also holds for nicecancelable valuations as pointed out in Section 2.The one exception is in Section 4.2 in [CGM20] (right after Observation 16), in which their proof requiresa subtle adjustment, as follows. In their sub-case “ a = 2” they define the set Z i , for i = 1 ,
3, to be a smallestsubset of X such that Z i > i max i ( X \ G ∪ g, X ). In our case, the set Z i needs to be defined as the subsetof X obtained by iteratively removing the item of least marginal value to agent i , as long as the leftoverbundle is still better than max i ( X \ G ∪ g, X ) from agent i ’s point of view. Lemma A.4 then guaranteesthat the new Z i is indeed a smallest-size subset of X as in the original proof. The rest of the argumentthen follows as in the original proof.We now turn to the envy-free case. In the original proof the following property of an additive valuation v is used to prove the sub-case where M X contains a g -cycle of size 2: if v ( S ) > v ( T ) and v ( S ) > v ( R ), where R and T are disjoint, then 2 · v ( S ) > v ( T ∪ R ). This property does not hold in general for nice cancelablevaluations ( e.g. , v is multiplicative and v ( S ) = 5 , v ( R ) = v ( T ) = 4), thus their proof does not extend to nicecancelable valuations.In what follows we present a proof, based on our new techniques, that applies to all nice cancelablevaluations, and is also substantially simpler than the original proof. Lemma 6.2.
Let X be a partial envy-free allocation on 3 agents. Then there exists an EFX allocation Y that Pareto dominates it. Proof.
By Lemma 3.5, it suffices to find a Pareto-improvable (PI) cycle in the champion graph M X . Sinceevery agent has an incoming g -champion edge (Observation 2.8), M X has a cycle of g edges. Let C be such acycle of minimal length. If C is a self loop, then we are done by Corollary 3.6. Hence, it remains to considerthe cases where C has length two or three. Since X is envy-free, C is a good g -cycle, and as such it inducesa g -decomposition of X j into top and bottom half-bundles X j = T j ∪ B j for every agent j along the cycle. Case 1: | C | = 2. Assume w.l.o.g. that C = 1 g g
1. By Theorem 3.16, M X contains a good orexternal B -edge and a good or external B -edge. By Corollary 3.18 they cannot both be external (sincetheir source would be the same, agent 3). Thus, w.l.o.g. M X contains a good B -edge, which can only be1 B | ◦
2, and we obtain the PI cycle 1 B | ◦ g Case 2: | C | = 3. W.l.o.g. let C = 1 g g g
1. As C contains all 3 agents, there are noexternal edges going into C . Thus by Theorem 3.16, M X contains a good B i -edge, for each i ∈ [3].If for some i ∈ [3] the good B i -edge is i B i | ◦ pred ( i ), then we get the PI cycle i B i | ◦ pred ( i ) g i , and we are done. Thus we can assume that all good edges are parallel to the edges of C , i.e., from j to succ ( j ). W.l.o.g., assume there is a good edge from agent 1 to agent 2. This good edge cannot be the good B -edge, because 1 B | ◦ B i | ◦
2, for some i ∈ { , } ,and the good B -edge must be either 2 B | ◦ B | ◦
1. The former case admits the PI cycle1 B i | ◦ B | ◦ g
1; the latter admits the PI cycle 1 B i | ◦ g B | ◦ .2 EFX for agents with one of two valuations Consider a setting with n agents, where any agent has one of two valuations v a , v b . Let a , . . . , a t denotethe agents with valuation v a , and b , . . . , b (cid:96) denote the agents with valuation v b , ordered such that X a ≤ a X a ≤ a . . . ≤ a X a t and X b ≤ b X b ≤ b . . . ≤ b X b (cid:96) . The following theorem shows that if v a and v b are nice cancelable valuations, then given any partial EFXallocation, there exists an EFX allocation that Pareto dominates it. This implies (by Lemmata 2.4, A.1)that every instance in this setting admits a full EFX allocation. Theorem 6.3.
In every setting with two nice cancelable valuations, given any partial EFX allocation, thereexists an EFX allocation that Pareto dominates it.Before presenting the proof, we present a useful observation. We say that envy (resp., most envious) propagates backward within the valuation class a if whenever some agent a i envies a set S (resp., is mostenvious of a set S ), then for every j < i , agent a j envies S (resp., is most envious of S ) as well. We say thatchampionship propagates backward within the valuation class a in an analogous way. We define backwardpropagation within the valuation class b analogously. One can easily verify that envy propagates backward.The following observation shows that so does championship. Observation 6.4.
Championship propagates backward within the same valuation class.
Proof.
We prove the claim for valuation class a . The proof for valuation class b is analogous. We show thatthe relation most envious propagates backward; by extension, championship propagates backward as well.Suppose that for some i ∈ { , . . . , (cid:96) } , agent a i is most envious of a set S , and let D be the discard set of S with respect to a i . This means that a i envies S \ D . Since envy propagates backward, so does agent a j . Bydefinition of a discard set, no agent strongly envies S \ D . Therefore, a j is most envious of S .We are now ready to prove Theorem 6.3. Proof.
Fix a partial EFX allocation, and let g be an unallocated item. By Corollary 3.6 we may assumethat no agent is a g -self champion. We first claim that a g b and b g a . (3)Indeed, if b j g b for some j , then b is a g -self champion by backward propagation. Thus, since b must have a g -champion (Observation 2.8), then a j g b for some j . Since championship propagatesbackward, a g b . By symmetry, b g a .We may also assume that no agent envies a (and similarly, b ). Clearly, no a j envies a . It remains toshow that no b j envies a . Indeed, if some agent b j envies a , then b envies a , and together with the factthat a g b , we have a Pareto-improvable cycle, so we are done by Lemma 3.5. Similarly, if any agentenvies b , we have a Pareto-improvable cycle.By Equation (3) and the assumption that no agent envies a or b , a g b g a is a good g -cycle,thus by Observation 2.9 the bundles of a and b decompose into top and bottom half-bundles. Let T a and B a (resp., T b and B b ) be the top and bottom half-bundles of a (resp., b ), respectively.We next argue that a B a | ◦ b . Since a g b g a is a good cycle, by Theorem 3.16 there existsa good or external B a -edge that goes into b . If this is a good B a -edge then it can only be a B a | ◦ b . If this is an external B a -edge, it can’t be b j B a | ◦ b since b B a | ◦ b (by Observation 3.12)and championship propagates backward. Hence, the external B a -edge must be a j B a | ◦ b for some j ∈ { , . . . , t } . Again, since championship propagates backward, a B a | ◦ b .It follows that we have a Pareto-improvable cycle consisting of a B a | ◦ b and b g a . We maynow apply Lemma 3.5 to conclude the proof. 34 cknowledgment We are very grateful to Karin Egri for her great help in producing the figures used in this paper. We aredeeply in debt to Bhaskar Ray Chaudhury, Jugal Garg, Kurt Mehlhorn, Ruta Mehta and Misra Pranabendufor their wonderful papers and for sharing unpublished ongoing results.
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Proofs and Claims from Section 2
Lemma A.1.
To prove the existence of an EFX allocation for a given valuation profile v = ( v , . . . , v n )of nice cancelable valuations, it is without loss of generality to assume that all of the valuations are non-degenerate. Proof.
For each i ∈ [ n ], v i is a nice cancelable valuation. Hence, for every i ∈ [ n ] there exists a non-degeneratecancelable valuation v (cid:48) i that respects v i . Let v (cid:48) = ( v (cid:48) , . . . , v (cid:48) n ) be the corresponding valuation profile.We show that any allocation which is EFX for the profile v (cid:48) is also EFX for the profile v . Let X =( X , . . . , X n ) be an EFX allocation for the profile v (cid:48) and assume towards contradiction that X is not anEFX allocation for the profile v . Hence, under the profile v there exists an agent i that strongly envies anotheragent j , i.e. , v i ( X i ) < v i ( X j \{ h } ) for some h ∈ X j . Since v (cid:48) i respects v i , it follows that v (cid:48) i ( X i ) < v (cid:48) i ( X j \{ h } ),in contradiction to X being EFX over the profile v (cid:48) .The family of nice cancelable valuations contains some well-known classes of valuations. Additive valu-ations are clearly cancelable and are shown to be nice in [CGM20]. The following lemma shows that thisclass contains many other classes of valuations, including unit-demand, budget-additive and multiplicative. Lemma A.2.
Unit-demand, budget-additive and multiplicative valuations are nice cancelable.
Proof.
Budget-Additive:
Let v be a budget-additive valuation, i.e. , for every S ⊆ M , v ( S ) = min (cid:88) g ∈ S v ( g ) , B , for some B >
0. We start by proving v is cancelable. Consider S, T ⊆ M and g ∈ M \ ( S ∪ T ) such that v ( S ∪ { g } ) > v ( T ∪ { g } ) . First, v ( S ∪ { g } ) ≤ B , by definition of v . Therefore, v ( T ∪ { g } ) < B , so v is additive over T ∪ { g } .It follows that v ( T ) = v ( T ∪ { g } ) − v ( { g } ). Second, since budget-additive valuations are sub-additive, v ( S ) ≥ v ( S ∪ { g } ) − v ( { g } ). Combining these two observations we get v ( S ) ≥ v ( S ∪ { g } ) − v ( { g } ) > v ( T ∪ { g } ) − v ( { g } ) = v ( T ) . This proves that v is cancelable.We next prove that v is nice. Define the valuation v (cid:48) : 2 M → R ≥ as the underlying additive valuationof v , i.e. , for every S ⊆ M , v (cid:48) ( S ) = (cid:88) g ∈ S v ( g ) . We now show that v (cid:48) respects v .Suppose v ( S ) > v ( T ) for some S, T ⊆ M . Since v ( S ) ≤ B , it follows that v ( T ) < B , and thus v isadditive over T . By definition of v (cid:48) , this implies that v ( T ) = v (cid:48) ( T ). Furthermore, notice that v (cid:48) ( S ) ≥ v ( S ).Thus, v (cid:48) ( S ) ≥ v ( S ) > v ( T ) = v (cid:48) ( T ) . This proves that v (cid:48) respects v .Finally, since v (cid:48) is an additive valuation it is nice and cancelable (as shown in [CGM20]). Therefore,there exists a non-degenerate cancelable valuation v (cid:48)(cid:48) that respects v (cid:48) . Because v (cid:48) respects v , it follows bytransitivity that v (cid:48)(cid:48) respects v as well. Since v (cid:48)(cid:48) is non-degenerate and cancelable, this proves that v is nice. Multiplicative : Let v be a multiplicative valuation, i.e. , for every S ⊆ M , v ( S ) = (cid:89) g ∈ S v ( g ) . Unit-Demand:
Let v be a unit-demand valuation, i.e. , for every S ⊆ M , v ( S ) = max g ∈ S v ( g ) . We first show v is cancelable. Consider S, T ⊆ M and g ∈ M \ ( S ∪ T ) such that v ( S ∪ { g } ) > v ( T ∪ { g } ) . Clearly, g is not the maximal element in S ∪ { g } , otherwise we would have v ( S ∪ { g } ) = v ( { g } ) ≤ v ( T ∪ { g } ).Therefore, v ( S ∪ { g } ) = v ( S ). We get v ( S ) = v ( S ∪ { g } ) > v ( T ∪ { g } ) ≥ v ( T ) . This proves that v is cancelable.We next prove that v is nice. Define δ = min S,T ⊆ M, v ( S ) (cid:54) = v ( T ) | v ( S ) − v ( T ) | . That is, δ is the minimal difference between the value of any two non-equal valued sets of items. Let g , . . . , g m − be the items in M ordered in non-decreasing value, ties broken arbitrarily. Let ε = 2 − ( m +1) δ .Define the valuation v (cid:48) : 2 M → R ≥ as follows: v (cid:48) ( S ) = v ( S ) + ε (cid:88) i : g i ∈ S i . To complete the proof we need to show that v (cid:48) is a non-degenerate cancelable valuation and that v (cid:48) respects v . We begin with the latter. Suppose v ( S ) > v ( T ) for some S, T ⊆ M . This implies that v ( S ) > v ( T ) + δ/ δ . Moreover, since (cid:80) m − i =0 i < m , we have v (cid:48) ( T ) = v ( T ) + ε (cid:88) i : g i ∈ T i < v ( T ) + ε · m = v ( T ) + δ . We get v (cid:48) ( S ) > v ( S ) > v ( T ) + δ > v (cid:48) ( T ) , as required.We next prove v (cid:48) is non-degenerate. For all S, T ⊆ M such that v ( S ) (cid:54) = v ( T ), we have shown abovethat v (cid:48) ( S ) (cid:54) = v (cid:48) ( T ). So it remains to show that v (cid:48) ( S ) (cid:54) = v (cid:48) ( T ) whenever v ( S ) = v ( T ) and S (cid:54) = T . Since v ( S ) = v ( T ), to prove that v (cid:48) ( S ) (cid:54) = v (cid:48) ( T ) it suffices to show that ε (cid:88) i : g i ∈ S i (cid:54) = ε (cid:88) i : g i ∈ T i , which clearly holds for every S (cid:54) = T .Finally, we prove v (cid:48) is cancelable. Consider S, T ⊆ M and g j ∈ M \ ( S ∪ T ) such that v (cid:48) ( S ∪ { g j } ) > v (cid:48) ( T ∪ { g j } ) . It is impossible that v ( S ∪ { g j } ) < v ( T ∪ { g j } ), since v (cid:48) respects v . If v ( S ∪ { g j } ) > v ( T ∪ { g j } ), then v ( S ) > v ( T ) since v is cancelable. Since v (cid:48) respects v , this implies v (cid:48) ( S ) > v (cid:48) ( T ), so we are done. We areleft with the case where v ( S ∪ { g j } ) = v ( T ∪ { g j } ). Since v (cid:48) ( S ∪ { g j } ) > v (cid:48) ( T ∪ { g j } ), this implies ε · (cid:88) i : g i ∈ S ∪{ g j } i > ε · (cid:88) i : g i ∈ T ∪{ g j } i . ε · j from both sides, ε (cid:88) i : g i ∈ S i > ε (cid:88) i : g i ∈ T i . (4)If v ( S ) ≥ v ( T ) then Equation (4) implies v (cid:48) ( S ) > v (cid:48) ( T ) and we are done. We complete the proof by showingthat the case v ( S ) < v ( T ) is impossible. Assume towards contradiction that v ( S ) < v ( T ). Let g s and g t bethe highest valued items in S and T , respectively (breaking ties according to the ordering we defined on theitems). Since v ( S ) < v ( T ) and v is unit-demand, v ( g s ) < v ( g t ). Due to our ordering, it follows that s < t .Therefore, (cid:88) i : g i ∈ S i ≤ s (cid:88) i =0 i < s +1 ≤ t ≤ (cid:88) i : g i ∈ T i , in contradiction to Equation (4). This shows that v (cid:48) is cancelable.On the other hand, not all cancelable valuations are nice, as shown in Proposition A.3. Proposition A.3.
Not all cancelable valuations are nice. Even when restricted to submodular cancelablevaluations it need not be nice.
Proof.
We define a valuation v over the set of items M = { a, b, c, d, e, f } and show that it is cancelable andsubmodular but not nice. The following table defines the value of v over each singleton. a b c d e fv
101 102 102 103 103 104Notice that v ( a ) < v ( b ) = v ( c ) < v ( d ) = v ( e ) < v ( f ), i.e., the values are non-decreasing from left toright.The next table defines the value of v over each pair of items. For convenience, we depict this table as amatrix, where the coordinate ( x, y ) contains the value v ( { x, y } ) (the matrix is symmetric). a b c d e fa - 152 152 153 153 154 b
152 - 152 155 155 156 c
152 152 - 155 155 156 d
153 155 155 - 155 156 e
153 155 155 155 - 156 f
156 156 156 156 156 -Finally, for any set S ⊆ M containing three or more items we define v ( S ) = 200, and we set v ( ∅ ) = 0.This completes the definition of v over all subsets of M . Cancelability:
Consider a pair of sets
S, T ⊆ M and an item g ∈ M \ ( S ∪ T ) such that v ( S ∪ { g } ) >v ( T ∪ { g } ). First consider the cases where | S | = | T | = 1. Therefore, S ∪ { g } and T ∪ { g } are both pairsof items whose values are located in row g of the matrix above. It is not hard to verify that within eachrow of the matrix the values are non-decreasing left to right with equality only between the columns ofequal valued items (the columns b and c and the columns d and e contain equal values). Therefore, the factthat v ( S ∪ { g } ) > v ( T ∪ { g } ) implies that v ( S ∪ { g } ) is located to the right of v ( T ∪ { g } ) in row g of thematrix. Therefore, v ( S ) ≥ v ( T ). The case v ( S ) = v ( T ) implies v ( S ∪ { g } ) = v ( T ∪ { g } ) since the columnsof equal valued singletons are identical in the matrix of pairs. Thus, v ( S ) > v ( T ). This completes the case | S | = | T | = 1.If | T | >
2, then v ( T ∪ { g } ) = 200 and there exists no set S such that v ( S ∪ { g } ) > | T | ≤
1. The case | T | = 0 is easy. In this case, v ( S ∪ { g } ) > v ( T ∪ { g } ) implies that S isnon-empty. Since T = ∅ and S (cid:54) = ∅ , we obtain v ( S ) > v ( T ), so cancelability is maintained.We are left with the case | T | = 1. The case | S | = 0 is not possible since no singleton has a higher valuethan a pair of items. The case | S | = | T | = 1 has been handled above. So it is left to consider | S | > T | = 1. In this case, notice that v ( S ) >
150 while v ( T ) < v ( S ) > v ( T ), as desired. Thisproves that v is cancelable. Submodularity:
It suffices to show that for every item g and every pair of sets S and T such that S ⊆ T ⊆ M \ { g } we have that v ( S ∪ { g } ) − v ( S ) ≥ v ( T ∪ { g } ) − v ( T ). This condition clearly holds if S = T ,so assume S (cid:40) T . Therefore, | S | < | T | . Notice that the following holds for any set Z ⊆ M and any item x ∈ M \ Z : if | Z | = 0, 100 Assume towards contradiction that there exists a non-degenerate cancelable valuation v (cid:48) thatrespects v . Notice that the following inequalities hold:154 = v ( { a, f } ) < v ( { b, e } ) = 155 , (5)155 = v ( { d, e } ) < v ( { c, f } ) = 156 , (6)152 = v ( { b, c } ) < v ( { a, d } ) = 153 . (7)Therefore, the analogous inequalities must hold for v (cid:48) . Consider the comparison between { a, d, f } and { b, d, e } in v (cid:48) . v (cid:48) is non-degenerate, so v (cid:48) ( { a, d, f } ) (cid:54) = v (cid:48) ( { b, d, e } ). If v (cid:48) ( { a, d, f } ) > v (cid:48) ( { b, d, e } ), then bycancelability v (cid:48) ( { a, f } ) > v (cid:48) ( { b, e } ), which contradicts Equation (5), so v (cid:48) ( { a, d, f } ) < v (cid:48) ( { b, d, e } ) . (8)Similarly, due to Equation (6) we obtain v (cid:48) ( { b, d, e } ) < v (cid:48) ( { b, c, f } ) , (9)and from Equation (7) we get v (cid:48) ( { b, c, f } ) < v (cid:48) ( { a, d, f } ) . (10)However, combining Equations (8) and (9) we get v (cid:48) ( { a, d, f } ) < v (cid:48) ( { b, c, f } ), in contradiction to Equa-tion (10). This proves that there exists no non-degenerate cancelable valuation v (cid:48) such that v (cid:48) respects v ,and thus v is not nice.If an additive agent strongly envies some bundle S , then, iterative removal of the least valued item untilstrong envy is eliminated results in a smallest size subset of S that the agent envies. The next lemma showsthat this property extends to cancelable valuations. Lemma A.4. Let v be a cancelable valuation. Let T be some bundle, and let S a subset of T . Let Z be thesubset obtained from T by iteratively removing the item with least marginal contribution until the leftoverbundle has size | S | . Then v ( Z ) ≥ v ( S ). Proof. Define T = T , and for j ≥ T j +1 = T j \ { c } , where c ∈ T j is the item with least marginalcontribution to T j . It suffices to prove that for every 0 ≤ j ≤ | T | we have T j = arg max S ⊆ T : | S | = | T |− j v ( S ) . We prove by induction on j . For j = 0 the claim is immediate. Assume that the claim is true for j and weprove for j + 1. Let c ∈ T j be the item with least marginal contribution to T j , hence by definition we have40 j +1 = T j \ { c } . Let S ⊆ T such that | S | = | T j +1 | . We need to show that v ( S ) ≤ v ( T j +1 ). If S = T j +1 , thenthis is immediate. Therefore, assume S (cid:54) = T j +1 . Since S and T j +1 have the same size, this means that thereis some item b ∈ T j +1 \ S , and thus S ∪ { b } and T j have the same size. By the induction hypothesis we get v ( S ∪ { b } ) ≤ v ( T j ) = v ( T j \ { b } ∪ { b } ) , implying v ( S ) ≤ v ( T j \ { b } ) ≤ v ( T j \ { c } ) = v ( T j +1 ) where the first inequality holds by cancelability, andthe second by definition of c . The claim follows. B Proofs for Section 5 Proof of Claim 5.2. Choose some arbitrary discard set D hi,j . By Observation 2.9, h / ∈ D hi,j and thus X i < i ( X j ∪ { h } ) \ D hi,j = ( X j \ D hi,j ) ∪ { h } < i ( X j \ D hi,j ) ∪ { g } , where the first inequality holds since 1 h 2, the equality holds since h / ∈ D hi,j and the second inequality isby cancelability since h < i g . Therefore, since i is the unique g -champion of j , it follows that i is the mostenvious agent of ( X j \ D hi,j ) ∪ { g } (otherwise that subset has another most envious agent and consequently X j ∪ { g } has other most envious agents except i ). Denote the corresponding minimally envied subset by S .Thus, we can choose D gi,j to be X j \ S . Clearly D hi,j ⊆ D gi,j . Proof of Lemma 5.3. By assumption, there is an envy edge in M X . Rename the agents such that 1 4.By Observation 2.8, agent 1 has a g -champion. If 4 g 1, then 1 4 g g -champion of 1. Rename that agent such that 2 g 1. Agent 2 has an h champion.If 4 h h h g h g 1, respectively. Thus3 h 2. We obtain the following structure:Agent 3 has a g -champion. If 2 g g h 2, thus 1 g g h h -champion of 1 closes a PI cycle.Consequently we must have 3 g h h g 2, hence we may also assume that1 h h g g Claim B.1. If 1 g h M X contains a PI cycle. Proof. We prove for 1 g h g g g g -cycle, which, by Observation 3.10, induces a g -decomposition X j = T gj ·∪ B gj for j ∈ { , , } . 41y Theorem 3.16 M X contains a good or external B g -edge. If it is external then it is one of 4 B g | ◦ B g | ◦ B g | ◦ B g | ◦ g B g | ◦ h g B g -edge whichis either one of 1 B g | ◦ 2, 1 B g | ◦ B g | ◦ 2. In the first two cases we obtain a PI cycle:1 B g | ◦ g B g | ◦ h g 1, respectively. Thus we may assume 3 B g | ◦ h -champion of 3 which can be either 1,2 or 4. In all these cases weget a PI cycle:1 h B g | ◦ g , h g h B g | ◦ g , respectively. We have shown a PI cycle in every case and thus we are done.By Claim B.1 we may assume 4 is the unique g and h champion of 3. We thus obtain the followingstructure:It is not hard to verify that any additional envy edge closes a PI cycle. Moreover, any additional h or g champion of 1,2 or 3 closes a PI cycle. This fact is not hard to verify for agents 1 and 2, and we have shownit explicitly for agent 3 in Claim B.1. As for agent 4, the possible g and h champions of 4 are agents 1 and2 (3 g Proof of Claim 5.5. Since 4 envies T g ∪ B g , there exists a most envious agent of this bundle. That is, thereexists an agent i such that i B g | ◦ i cannot be 3, since X > X = T g ∪ B g > T g ∪ B g , where thefirst inequality is by 3 (cid:54) i = 1, i = 2, i = 4, we obtain the respective PI cycles:1 B g | ◦ g h , B g | ◦ g , B g | ◦ g h . Proof of Lemma 5.6. Consider the allocation Y obtained from X (cid:48) by replacing X (cid:48) with Z : Y = X (1) Z (2) T g g (3) T h h (4)We claim this allocation is EFX. No agent envies agent 2: Agents 3 and 4 do not envy agent 2 since they did not envy her in X (cid:48) andwe only removed items from X (cid:48) in the transition to Y . Agent 1 does not envy agent 2, since we assume X > Z . Agent 2 envies no other agent: This follows from the definition of Z .42ince only the bundle of agent 2 has been changed in the transition from X (cid:48) to Y we conclude that thereis no strong envy that does not include agent 2 (as there wasn’t any in X (cid:48) ), and consequently Y is EFX.Moreover, agents 1,3, and 4 are better off in allocation Y relative to X . Since we assumed that agent a vip isnot agent 2, it follows that Y dominates X and we are done. Proof of Claim 5.8. It suffices to show that only agent 4 envies T h ∪ h in X (cid:48)(cid:48) . Agent 4 envies T h ∪ h , since X (cid:48)(cid:48) = X in all three cases and 4 h X . Agent i , for i ∈ { , } , does not envy T h ∪ h ,since X (cid:48)(cid:48) i ≥ i X i and i h X (Observation 2.10). Agent 2 does not envy T h ∪ h in all threecases since X (cid:48)(cid:48) = max (cid:8) T g ∪ g, T h ∪ h, X (cid:9) ≥ T h ∪ h . Proof of Claim 5.9. Consider the cycle 1 4 g B g | ◦ h M X :This cycle might not really exist, since it could be the case that 1 is the real ( B g | ◦ )-champion of agent 2.Assume for now that this is not the case, i.e. , 3 B g | ◦ M X . Then this is PI cycle, and byLemma 3.5, there is an allocation Y that Pareto-dominates X . Note that in Y agent 1 receives X . Wetherefore claim that Y is EFX even if 1 was the most envious agent of T g ∪ B g in X . Since 3 was the mostenvious agent ignoring agent 1, the only thing that could prevent Y from being EFX is if 1 strongly envies3 in Y . But this is not the case since Y = X > T g ∪ B g ≥ Y where the last inequality holds since Y ⊆ T g ∪ B g . The claim follows. Proof of Lemma 5.10. Consider the good g -cycle 2 g g g X . By Lemma 3.13, there existsan agent i such that i B g | ◦ i cannot be agent 3, because 3 = pred (2) in this cycle (Observation 3.12). If i = 2, we obtain the PI-cycle 2 B g | ◦ h g X , so we are done. If i = 4,then X < T g ∪ B g < T g ∪ B g , where the second inequality is due to Observation 2.11 and cancelability.Therefore, we are done by Claim 5.5.Thus, we may assume that i = 1, i.e. , 1 B g | ◦ 4. It follows that X < T g ∪ B g . Therefore, since1 (cid:54) T g ∪ B g = X < X < T g ∪ B g , and we conclude that T g < T g by cancelability. Finally we use cancelability again to conclude X = T g ∪ B g > T g ∪ B g , as desired. Proof of Lemma 5.11. Consider the good g -cycle 2 g g g X . We first ask which agent i satisfies i B g | ◦ i cannot be agent 2, because 2 = pred (4)in the good g -cycle (Observation 3.12). If i is agent 4, we obtain the PI-cycle 2 g B g | ◦ h M X , hence we are done.If i = 1, then in particular we have T g ∪ B g = X < X < T g ∪ B g , where the first inequality is due to 1 (cid:54) B g | ◦ 3. By cancelability we get B g < B g , and together with Lemma 5.10 we conclude that X = T g ∪ B g > T g ∪ B g > T g ∪ B g , i.e. , weare in Case I which we already solved.Thus, we may assume that i is agent 3, i.e., 3 B g | ◦ 3. It follows that X < T g ∪ B g < T g ∪ B g , where the second inequality is by Observation 2.11 and cancelability. The lemma follows.43 roof of Claim 5.15. Assuming that g < b , We get X < T g ∪ { g } < T g ∪ { b } , where the first inequalityholds since 3 g X and the second by cancelability. In particular there exists a most envious agent of T g ∪ { b } in X , i.e. , there exists an agent i such that i b i cannot be 2, since X > X = T g ∪ B g > T g ∪ B g > T g ∪ { b } , where the first inequality holds by 2 (cid:54) 1, the second by Observation 2.11 andcancelability, and the third by monotonicity. Thus i = 1 , M X :1 b g , b g h b g , respectively. Note that these are indeed PI cycles since the g -champion edge releases B g which contains b . Proof of Lemma 5.16. Recalling that X (cid:48)(cid:48) = X , we need to show that there is a most envious agent of T g ∪ B g in X (cid:48)(cid:48) . We do this by showing X (cid:48)(cid:48) < T g ∪ B g . If B g > B g , then X > X > X = T g ∪ B g > T g ∪ B g , where the first and second inequalities are by 1 4 and 1 (cid:54) 2, respectively, in M X , and the final inequalityfollows from cancelability. This implies that we are in Case I that has been solved earlier.Thus we may assume B g < B g . Moreover T g < T g by Observation 2.11. Therefore, by cancelability T g ∪ B g < T g ∪ B g < T g ∪ B g . Seeing as X (cid:48)(cid:48) = Z ⊆ T g ∪ B g and valuations are monotone, the above equation implies that agent 1envies T g ∪ B g in allocation X (cid:48)(cid:48) , as desired. Proof of Claim 5.18. Recall that X < X < Z ⊆ T g ∪ B g . We thus have X < T g ∪ B g < T g ∪ B g , byObservation 2.11 and cancelability. Since agent 1 envies T g ∪ B g , there exists a most envious agent of thisset. That is, there exists j ∈ [4], such that j B g | ◦ B g | ◦ 4, then we obtain the PI-cycle 2 B g | ◦ h g 2. Note that 3 B g | ◦ 4, since X > X = T g ∪ B g > T g ∪ B g , where the first inequality is by 3 (cid:54) B g | ◦ 4, since X > T g ∪ B g (otherwise we are done by Claim 5.5), and therefore, byObservation 2.11 and cancelability, X > T g ∪ B g > T g ∪ B g . The claim now follows. Proof of Claim 5.19. By Claim 5.18 we have the following structure in M X :Now, we have X = T g ∪ B g < T g ∪ B g , by Observation 2.11 and cancelability. Since agent 4 envies T g ∪ B g , there exists a most envious agent of this set. That is, there exists i ∈ [4], such that i B g | ◦ 3. If i = 1, then T g ∪ B g > X > X = T g ∪ B g , where the first inequality holds by definition of i andthe second inequality follows from 1 (cid:54) 3. By cancelability, we conclude that B g > B g , and together with T g > T g (Observation 2.11) we get X = T g ∪ B g > T g ∪ B g > T g ∪ B g . by cancelability. Thus we are in Subcase I which has been solved earlier.Suppose next that i = 4. By Claim 5.18, 1 B g | ◦ 4, thus we obtain the PI-cycle 1 B g | ◦ B g | ◦ g h B g is released by 3 g B g is releasedby the edge 1 B g | ◦ i = 2. Since we assume that B g < B g , we have X < T g ∪ B g < T g ∪ B g , where the first inequality follows from 2 B g | ◦ T g ∪ B g and thus there exists an agent k ∈ [4] such that k B g | ◦ 3. Note that 3 B g | ◦ X > X = T g ∪ B g > T g ∪ B g , where the first inequality is by 3 (cid:54) B g | ◦ 3, 2 B g | ◦ B g | ◦ 3, in each of these cases we obtain a PI-cycle:1 B g | ◦ g h , B g | ◦ g B g | ◦ g h , respectively. We are left with the case ii