Efficiency and Envy-Freeness in Fair Division of Indivisible Goods

Home , Fair division

Efﬁciency and envy-freeness in fair division of indivisible goods: logical representation and complexity Sylvain Bouveret Jer´ omeˆ Lang IRIT (UPS-CNRS) and IRIT ONERA Universit´ePaul Sabatier - CNRS 31055 Toulouse Cedex 4, France 31062 Toulouse Cedex, France [email protected] [email protected]

Abstract The above discussion reveals the existence of a gap: compact representation and complexity issues for fair division We study fair division of indivisible goods among have received very little attention until now, apart of the re- agents from the point of view of compact repre- cent work [Lipton et al., 2004] which studies approximation sentation and computational complexity. We iden- schemes for envy-freeness. The need for compact representa- tify the complexity of several problems, including tion arises from the following dilemna, formulated by several that of deciding whether there exists an efficient social choice theorists: either (a) allow agents to express and envy-free allocation when preferences are rep- any possible preference relation on the set of all subsets of resented in a succinct way. We also draw connec- items, and end up with an exponentially large representation tions to nonmonotonic reasoning. (such as in [Herreiner and Puppe, 2002]); or (b) severely re- strict the set of expressible preferences, typically by assuming 1 Introduction additive independence between items, and then design proce- dures where agents express preferences between single items, Allocation of goods among agents has been considered from thus giving up the possibility of expressing preferential de- different perspectives in social choice theory and AI. In this pendencies such as complementarity and substitutability ef- paper we focus on fair division of indivisible goods without fects among items; this is the path followed by [Brams et al., money transfers. Fair division makes a prominent use of fair- 2000] and [Demko and Hill, 1998]. Yet, as we advocate in ness criteria such as equity and envy-freeness, and on this this paper, conciliating conciseness and expressivity is possi- point totally depart from auctions, where only efficiency is ble, by means of compact representation. relevant (and moreover a specific form of efficiency, since the criterion to be maximized is the total revenue of the auc- As in most works on fair allocation of indivisible items tioneer). Envy-freeness is a key concept in the literature on we focus on the joint search for envy-freeness and efficiency. fair division: an allocation is envy-free if and only if each The impossibility to guarantee the existence of an efficient agent likes her share at least as much as the share of any envy-free allocation implies that determining whether there other agent. Ensuring envy-freeness is considered as cru- exists such an allocation is a crucial task, since a positive an- cial; however, envy-freeness alone does not suffice as a cri- swer leads to choose such an allocation whereas a negative terion for finding satisfactory allocations, therefore it has to answer calls for a relaxation of one of the criteria. We con- be paired with some efficiency criterion, such as Pareto opti- sider this problem from the point of view of compact repre- mality. However, it is known that for any reasonable notion of sentation and computational complexity. We focus first in efficiency, there are profiles for which no efficient and envy- the simple case where agents have dichotomous preferences, free allocation exists (see [Brams et al., 2000])1. that is, they simply express a partition between satisfactory Whereas social choice theory has developed an important and unsatisfactory shares. The interest of such a restriction literature on fair division, computational issues have rarely is that in spite of the expressivity loss it imposes, it will be been considered. On the other hand, artificial intelligence shown to be no less complex than the general case, while be- has studied these issues extensively, but until now has fo- ing much simpler to expose. The most natural representa- cused mainly on combinatorial auctions and related prob- tion of a dichotomous preference is by a single propositional lems, investigating issues such as compact representation formula, where variables correspond to goods. Expressing as well as complexity and algorithms. Complexity issues envy-freenessand efficiency within this logical representation for negotiation (where agents exchange goods by means of reveals unexpected connections to nonmonotonic reasoning. deals) have also been studied (e.g. [Dunne et al., 2005; We identify the complexity of the key problem of the existence of an envy-free and Pareto-efficient allocation, which Chevaleyre et al., 2004]). See also [Bouveret et al., 2005] p for a preliminary complexity study of fair division problems. turns out to be Σ2-complete; we also identify the complexity of several other problems obtained either by imposing some 1This is even trivial if every good must be assigned to someone: restrictions on the latter or by replacing Pareto-efficiency by in this case, there are profiles for which not even an envy-free allo- other criteria. In Section 5 we extend this result to the case of cation exists. non-dichotomous, compactly represented preferences. 2 Background 2.3 Computational complexity 2.1 Fair division problems In this paper we will refer to some complexity classes located in the polynomial hierarchy. We assume the reader to be fa- Definition 1 miliar with the classes NP and coNP. BH2(also referred to A fair division problem is a tuple I,X, where P = h Ri as DP) is the class of all languages of the form L1 ∩ L2 p NP • I = {1,...,N} is a set of agents; where L1 is in NP and L2 in coNP. ∆2= P is the class of all languages recognizable by a deterministic Turing machine • X = {x1,...,xp} is a set of indivisible goods; p working in polynomial time using NP oracles. Likewise, Σ2= • R = hR1,...,RN i is a preference profile, where each NP p p p X NP . Θ2 = ∆2[O(log n)] is the subclass of ∆2 of prob- Ri is a reflexive, transitive and complete relation on 2 . lems that only need a logarithmic number of oracles. See for Ri is the preference relation of agent i. ARiB is alterna- instance [Papadimitriou, 1994] for further details. tively denoted by Ri(A, B) or by A i B; we note A ≻i B (strict preference) for A i B and not B i A and A ∼i B 3 Fair division problems with dichotomous (indifference) for A i B and B i A. preferences: logical representation In addition, Ri is said to be monotonous if and only if for all A, B, A ⊆ B ⊆ X implies B A. R = hR ,...,R i We start by considering in full detail the case where prefer- i 1 N ences are dichotomous. is monotonous if and only if Ri is monotonous for every i. Definition 3 Ri is dichotomous if and only if there exists a Definition 2 X subset Goodi of 2 such that for all A, B ⊆ X, A i B if An allocation for I,X, is a mapping π I • P = h Ri : → and only if A ∈ Goodi or B 6∈ Goodi. R = hR1,...,RN i X 2 such that for all i and j 6= i, π(i) ∩ π(j)= ∅. If for is dichotomous if and only if every Ri is dichotomous. every x ∈ X there exists a i such that x ∈ π(i) then π is a complete allocation. There is an obvious way of representing dichotomous preferences compactly, namely by a propositional formula ϕi (for ′ ′ • Let π, π two allocations. π dominates π if and only if each agent i) of the language LX (a propositional symbols for ′ (a) for all i, π(i) i π (i) and (b) there exists an i such each good) such that Mod(ϕ)= Goodi. Formally: ′ that π(i) ≻i π (i). π is (Pareto-) efficient if and only if X ′ ′ Definition 4 Let Ri be a dichotomouspreference on 2 , with there is no π such that π dominates π. X Goodi its associated subset of 2 , and ϕi a propositional • An allocation π is envy-free if and only if π(i) i π(j) formula on the propositional language LX . We say that ϕi holds for all i and all j 6= i. represents Ri if and only if Mod(ϕi)= Goodi. 2.2 Propositional logic Clearly, for any dichotomous preference Ri there is a formula ϕi representing Ri; furthermore, this formula is unique Let V be a finite set of propositional variables. LV is the up to logical equivalence. propositional language generated from V , the usual connectives ¬, ∧ and ∨ and the Boolean constants ⊤ and ⊥ in the Example 1 X = {a,b,c} and Goodi = {{a,b}, {b,c}}. 2 Note that R is not monotonous. Then ϕ = (a ∧ b ∧ ¬c) ∨ usual way . An interpretation M for LV is an element of i i 2V , i.e., a truth assignment to symbols: for all x ∈ V , x ∈ M (¬a ∧ b ∧ c) represents Ri. (resp. x 6∈ M) means that M assigns x to true (resp. to false). An easy but yet useful result (whose proof is omitted): V Mod(ϕ) = {M ∈ 2 | M |= ϕ} is the set of all models of X ϕ (the satisfaction relation |= is defined as usual, as well as Proposition 1 Let Ri be a dichotomous preference on 2 . satisfiability and logical consequence). The following statements are equivalent: A literal is aformulaof LV of the form x oroftheform ¬x, 1. Ri is monotonous; where x ∈ V . A formula ϕ is under negative normal form (or 2. Goodi is upward closed, that is, A ∈ Goodi and B ⊇ A NNF) if and only if any negation symbol in ϕ appears only in imply B ∈ Goodi. literals. Any formula can be turned in polynomial time into an equivalent NNF formula. For instance, a ∧ ¬(b ∧ c) is not 3. Ri is representable by a positive propositional formula. underNNF but is equivalentto the NNF formula a∧(¬b∨¬c). From now on, we assume that allocation problems P are A formula is positive if it contains no occurrence of the represented in propositional form, namely, instead of I, X negation symbol. For instance, a ∧ (b ∨ ¬c) and a ∨ (¬a ∧ b) and R we only specify hϕ1,...,ϕN i. I and X are obviously are not positive, whereas a ∧ (b ∨ c) and (a ∧ c) ∨ (a ∧ b) are. determined from hϕ1,...,ϕN i. ⊤ and ⊥ are considered positive as well. Let P = hϕ1,...,ϕN i be an allocation problem with di- Let ϕ ∈ LV . V ar(ϕ) ⊆ V is the set of propositional chotomous preferences; then for each i ≤ N, we rewrite ϕi ∗ ∗ variables appearing in ϕ. For instance, V ar((a ∧ c) ∨ (a ∧ into ϕi obtained from ϕi by replacing every variable x by b)) = {a,b,c} and V ar(⊤)= ∅. the new symbol xi. For instance, if ϕ1 = a ∧ (b ∨ c) and ∗ ∗ Lastly, if S = {ϕ1,...,ϕn} is a finite set of formulas then ϕ2 = a ∧ d then ϕ1 = a1 ∧ (b1 ∨ c1) and ϕ2 = a2 ∧ d2. S = ϕ1 ∧ . . . ∧ ϕn is the conjunction of all formulas of S. For all i ≤ N, let Xi = {xi, x ∈ X} An allocation for V a standard allocation problem P corresponds to a model of 2 Note that connectives → and ↔ are not allowed; this is impor- V = X1 ∪. . .∪XN satisfying at most one xi for each x ∈ X. tant for the definition of positive formulas (to come). In other terms, there is a bijective mapping between the set of possible allocations and the models of the following formula ΓP = x∈X i6=j ¬(xi ∧ xj ) Example 2 (cont’d) The maximal ΓP -consistent subsets of V V ∗ ∗ ∗ ∗ ∗ ∗ If allocation are required to be complete, then ΓP is replaced Φ are S1 = {ϕ1, ϕ2}, S2 = {ϕ1, ϕ3} and S3 = {ϕ2, ϕ3}. C by Γ =ΓP ∧ (x1 ∨. . .∨xn). The rest in unchanged. S1 ∧ ΓP has only one model: {b1,c1,a2}. S2 ∧ ΓP has P Vx∈X V V Let V = {xi | i = 1,...,N, x ∈ X}. Any interpretation two models: {a1,b3} and {b1,c1,a3}. S3 ∧ ΓP has one model: a ,b . Therefore the four efficientV allocations for M of Mod(ΓP ) is such that it is never the case that xi and xj { 2 3} are simultaneously true for i 6= j, therefore we can map M ∈ P are (bc,a, −), (a, −,b), (bc, −,a) and (−,a,b). None of them is envy-free. Mod(ΓP ) to an allocation F (π) = M simply defined by π(i) = {x | M |= xi}. This mapping is obviously bijective, and we denote F −1(M) the allocation corresponding to an 3.3 Efficient and envy-free allocations interpretation M of Mod(ΓP ). We are now in position of putting things together. Since 3.1 Envy-freeness envy-free allocations corresponds to the models of ΛP and We now show how the search for an envy-free allocation can efficient allocations to the models of maximal ΓP -consistent ∗ subsets of , the existence of an efficient and envy-free be mapped to a satisfiability problem. Let ϕj|i be the formula ΦP ∗ ∗ (EEF) allocation is equivalent to the following condition: obtained from ϕi by substituting every symbol xi in ϕi by ∗ ∗ there exists a maximal Γ -consistent subset S of Φ such xj : for instance, if ϕ = a1 ∧ (b1 ∨ c1) then ϕ = a2 ∧ P P 1 2|1 that S ∧ Γ ∧ Λ is consistent. In this case, the models (b ∨ c ). (Obviously, ϕ∗ = ϕ∗.) P P 2 2 i|i i of theV latter formula are the EEF allocations. Interestingly, this is an instance of a well-known problem in nonmonotonic Proposition 2 Let P = hϕ1,...,ϕN i be an allocation problem with dichotomous preferences under propositional form, reasoning: ∗ 4 and the formulas ϕj|i and mapping F as defined above. Let Definition 6 A supernormal default theory is a pair D = ∗ ∗ hβ, ∆i with ∆ = {α ,...,αm}, where α ,...,αm and β ΛP = ϕ ∨ ¬ϕ 1 1 Vi=1,...,N h i Vj6=i j|ii are propositional formulas. A propositional formula ψ is a Then π is envy-free if and only if F (π) |=ΛP . skeptical consequence of D, denoted by D |∼∀ ψ, if and only The proof is simple, so we omit it. The search for envy-free if for all S ∈ MaxCons(∆,β) we have S ∧ β |= ψ. allocations can thus be reduced to a satisfiability problem: V −1 Proposition 4 Let P = hϕ1,...,ϕN i a fair division prob- {F (M) | M |=ΓP ∧ΛP } is the set of envy-freeallocations lem. Let D = hΓ , Φ i. Then there exists an efficient and for P Note that, importantly, Γ ∧Λ has a polynomial size P P P P P envy-free allocation for P if and only if D 6|∼∀ ¬Λ . (precisely, quadratic) in the size of the input data. P Example 2 This somewhat unexpected connection to nonmonotonic reasoning has several implications. First, EEF allocations ϕ1 = a ∨ (b ∧ c); ϕ2 = a; ϕ3 = a ∨ b. correspond to the models of S ∧ ΓP ∧ ΛP for S ∈ ΛP = ((a1 ∨ (b1 ∧ c1)) ∨ (¬(a2 ∨ (b2 ∧ c2)) ∧ ¬(a3 ∨ (b3 ∧ c3))) MaxCons(ΦP , ΓP ); however, VMaxCons(ΦP , ΓP ) may be ∧ (a2 ∨ (¬a1 ∧ ¬a3)) ∧ ((a3 ∨ b3) ∨ (¬(a1 ∨ b1) ∧ ¬(a2 ∨ b2)); exponentially large, which argues for avoiding to start com- Mod(ΓP ∧ ΛP )= {{c1}, {c1,b3}, {c2,b3}, {c2}, {b3}, {c3}, ∅}. puting efficient allocations and then filtering out those that There are therefore 7 envy-free allocations, namely (c, −, −), are not envy-free, but rather compute EEF allocations in a (c, −,b), (−,c,b), (−, c, −), (−, −,b), (−, −,c) and single step, using default reasoning algorithms thus, fair di- (−, −, −). Note that none of them is complete. vision may benefit from computational work in default logic 3.2 Efficient allocations and connex domains such as belief revision and answer set programming. Moreover, alternative criteria for selecting ex- Definition 5 Let ∆= {α1,...,αm} a set of formulae and β tensions in default reasoning (such as cardinality, weights or a formula. S ⊆ ∆ is a maximal β-consistent subset of ∆ iff ′ priorities) correspond to alternative efficiency criteria in allo- (a) S ∧ β is consistent and (b) there is no S such that S ⊂ cation problems. S′ ⊆V∆ and S′ ∧ β is consistent. Let MaxCons(∆,β) be the set of allV maximal β-consistent subsets of ∆. 4 Allocation problems with dichotomous Proposition 3 Let P = hϕ1,...,ϕN i an allocation prob- ∗ ∗ preferences: complexity lem. Let ΦP = {ϕ1,...,ϕN }. Then π is efficient for P if ∗ ∗ p and only if {ϕi | F (π) |= ϕi } is a maximal ΓP -consistent It is known that skeptical inference is Π2-complete [Gottlob, subset of ΦP . 1992]; now, after Proposition 4, the problem of the existence of an EEF allocation can be reduced to the complement of a This simple result, whose proof is omitted, suggests that ef- skeptical inference problem, which immediately tells that it ficient allocations can be computed from the logical expres- sion Φ of the problem, namely, by computing the maximal allocations). This can be tempered by (a) there are many practical ΓP -consistent subsets of Φ; call them {S1,...,Sq}. Thenfor q cases where the number of maximal consistent subsets is small; (b) each Si, let Mi = Mod( Si ∧ ΓP ) and let M = ∪i=1Mi. it is generally not asked to look for all efficient allocations; if we Then F −1(M) is the set ofV all efficient allocations for Φ3. look for just one, then this can be done by computing one maximal ΓP -consistent subset of Φ. 3 4 Note that there are in general exponentially many maximal ΓP - “Supernormal” defaults are also called “normal defaults without consistent subsets of Φ (and therefore exponentially many efficient prerequisites” (e.g., [Reiter, 1980]). p is in Σ2. Less obviously, we now show that it is complete for 2. 2,...,n envy noone; this class, even if preferences are required to be monotonous. p 3. n +1 can only envy n +2; Let us first note that skeptical inference remains Π2- complete under these two restrictions (to which we refer as 4. n +2 can only envy n +1; RSI, for RESTRICTEDSKEPTICALINFERENCE): (a) ψ = ϕ1; Proof: First, note that for any i, j 6= i, i envies j if and only (b) n ≥ 2. (Here is the justification: ∆ |∼ ψ if and only if π(i) |= ¬ϕi and π(j) |= ϕi. if ∆ ∪ {ψ, ψ} |∼ ψ.) Equivalently, RSI is the problem of 1. Let i = 1 and j ∈ {2,...,n,n +2}. Assume 1 envies 1 deciding whether, given ∆ = hα1,...,αni with n ≥ 2, all j. Then π(j) |= ϕ1. π being regular, x 6∈ π(j), therefore maximal consistent subsets of ∆ contain α1. π(j) |= β1. Now, since π is regular, π(j) does not contain i i any v nor any v¯ ; now, β1 can only be made true by Proposition 5 The problem EEFEXISTENCE of determining i i whether there exists an efficient and envy-free allocation fora variables v or v¯ (which cannot be the case here) unless it is given problem P with monotonous, dichotomous preferences a tautology. Now, if β1 is a tautology, then 1 is satisfied by π under logical form is Σp-complete. and cannot envy j, a contradiction. 2 2. Let i ∈ {2,...,n} and j 6= i. If i envies j then RSI i i We show hardness by the following reduction from π(j) |= βi ∧ x , which is impossible because x 6∈ π(j), due (the complement problem of RSI) to EEFEXISTENCE. Given to the regularity of π. any finite set of propositional formulae, let V V ar ∆ ∆ = (∆) 3. Let i = n +1. Assume n +1 envies 1 then π(1) |= ϕn+1. the set of propositional symbols appearing in ∆, and let Since π(1) |= y is impossible (because π is regular), we have P = H(∆) the following instance of EEFEXISTENCE: n i n i 1 π(1) |= v∈V∆ i=1 v ∨ i=1 v¯ ∧ x , which implies V V nV n (1) I = {1, 2, ..., n +2}; that either π(1) |= vi or π(1) |= v¯i . Vv∈V∆ Vi=1 Vi=1 X = {vi|v ∈ V ,i ∈ 1...n}∪{v¯i|v ∈ V ,i ∈ 1...n} Both are impossible because π is regular and n ≥ 2. The (2) ∆ ∆ case j ∈{2,...,n} is similar. ∪{xi|i ∈ 1...n}∪{y}; 4. Let i = n +2 and j 6= n. If i envies j then π(j) |= y, (3) for each i = 1,...,n, let βi be obtained from αi by the which is impossible because π is regular. following sequence of operations: (i) put αi into NNF form ′ ′ (let αi be the result); (b) for every v ∈ V∆, replace, in αi, Lemma 3 Let π be a regular allocation satisfying n +1 and i each (positive) occurrence of v by v and each occurrence of n +2. Let M(π) be the interpretation on V∆ obtained from i ¬v by v¯ ; let βi be the formula obtained. Then π by: for all v ∈ V∆, M(π) |= v (i.e., v ∈ M(π)) if n +1 1 n 1 receives v¯ ,..., v¯ , and M(π) |= ¬v otherwise, i.e., if n +1 • ϕ1 = β1 ∨ x ; 1 n receives v ,...,v . Then π is envy-free iff M(π) |= α1. i • for i =2,...,n, ϕi = βi ∧ x ; Proof: Let π be a regular allocation satisfying n + 1 and n i n i 1 n + 2. Since π satisfies n + 2, y ∈ π(n + 2). Now, π • ϕn+1 = v ∨ v¯ ∧ x ∨y; Vv∈V ar(∆) Vi=1 Vi=1 satisfies n + 1 without giving him y, therefore, for any v, n +1 receives either all the vi’s or all the v¯i’s. This shows • ϕn+2 = y. that our definition of M(π) is well-founded. Now, since π is regular, it is envy-free if and only if (a) 1 does not envy Lemma 1 An allocation π for P is said to be regular if and n+1, (b) n+1 does not envy n+2 and (c) n+2 does not envy only if for all i 6= n, π(i) ⊆ σ(i), where n+1. Since n +1 and n +2 are satisfied by π, we get that π • for all i 6= n, σ(i)= {vi, v¯i} {xi}; is envy-free if and only if 1 does not envy n+1, that is, if and Sv∈V∆ S i i 1 only if either π(1) |= ϕ1 or π(n + 1) 6|= ϕ1. Now, π(n + 1) • σ(n +1)= {v , v¯ } {x ,y}; 1 Sv∈V∆,i=1,...,n S contains x , therefore π(n + 1) |= ϕ1, which entails that π • σ(n +2)= {y}. is envy-free if and only if π(1) |= ϕ1. This is equivalent to π(1) |= β , because 1 does not get x1 (which is assigned to Let now π defined by π (i)= π(i) ∩ σ . Then 1 R R i n+1), which in turn is equivalentto M(π) by construction. 1. πR is regular; 2. π is efficient if and only if π is is efficient; Lemma 4 For each interpretation M over V∆, let us define R X πM : I → 2 by: 3. if π is envy-free then πR is is envy-free. 1 1 • πM (1) = {v | M |= v}∪{v¯ | M |= ¬v}; i Proof: (1) is obvious. For all i, the goods outside σ(i) do • for each i ∈ 2,...,n, πM (i) = {v | M |= v} ∪ not have any influence on the satisfaction of i (since they {v¯i | M |= ¬v}∪{xi}; do not appear in αi), therefore πR(i) ∼i π(i), from which 1 i • πM (n +1) = {x }∪{v¯ | M |= v,i = 1,...,n} ∪ (2) follows. The formulas αi being positive, the preference {vi | M |= ¬v,i =1,...,n}; relations are monotonous, therefore π j π j holds i ( ) i R( ) • π (n +2)= {y} for all i, j. Now, if π is envy-free then for all i, j we have M π(i) i π(j), therefore πR(i) ∼i π(i) i π(j) i πR(j) Then: and πR is envy-free, from which (3) follows. 1. πM is a well-defined and regular allocation satisfying n +1 and n +2;

Lemma 2 If π is regular then 2. MπM = M (MπM is obtained from πM as in Lemma3).

1. 1 can only envy n +1; 3. for any i ∈ 1,...,n, πM satisfies i iff M |= αi. 4. πM is efficient iff M satisfies a maximal consistent sub- Proof: Let π be an efficient and envy-free allocation. By set of ∆. Lemma 1, πR is regular, efficient and envy-free. By Lemma Proof: 6, πR satisfies n +1 and n +2. Then by Lemma 5, M(πR) satisfies a maximal consistent subset of ∆, and by Lemma 1. πM does not give the same good to more than one individ- ual, therefore it is an allocation. The rest is straightforward. 3, M(πR) |= α1. Therefore Sat(M(πR), ∆) is a maximal consistent subset of ∆ and contains α1. i 2. if M |= v then πM (n + 1) contains {v¯ | i = 1,...,n} and therefore M(πM ) |= v. The case M |= ¬v is similar. Lemma 8 If there exists a maximal consistent subset of ∆ i containing α then there exists an EEF allocation. 3. let i ∈ 2,...,n. Since πM gives x to i, πM satisfies i if 1 i and only if πM (i) |= β , which is equivalent to M |= αi. If Proof: Assume that there exists a maximal consistent subset 1 i = 1 then, since πM does not give x to 1, πM satisfies 1 if S of ∆ containing α1, and let M be a model of S. By point 1 and only if πM (1) |= β , which is equivalent to M |= α1. 4 of Lemma 4, πM is efficient. By point 1 of Lemma 4, πM is regular; then by Lemma 2, 4. from point 3, {i, πM satisfiesi} = {i,M |= αi}. Now, since preferences are dichotomous, an allocation π is πM is envy-free if and only if (i) 1 does not envy n +1, (ii) efficient if and only if the set of individuals it satisfies is n + 1 does not envy n + 2 and (iii) n + 2 does not envy maximal with respect to inclusion. Therefore, π is efficient n +1. By point 1 of Lemma 4, πM satisfies n +1 and n +2, M therefore (ii) and (iii) hold. Lastly, by point 5 of Lemma 4, if and only if M satisfies a maximal consistent subset of ∆. M |= α1 implies that πM satisfies 1, therefore (i) holds as Lemma 5 Let π be a regular and efficient allocation satisfy- well and πM is envy-free. ing n+1 and n+2. Then M(π) satisfies a maximal consistent subset of ∆. Proof of Proposition 5: from Lemmas 7 and 8, the existence of a maximal consistent subset of ∆ containing α1 Proof: π is regular and satisfies n +1 and n +2, therefore and the existence of an efficient and envy-free allocation 1 π(n +2) = {y}, x ∈ π(n + 1), and M(π) is well-defined. for H are equivalent. Clearly, H is computed in ′ P = (∆) Let π obtained from π by polynomial time. Therefore, H is a polynomial reduction • π′(1) = {v1|M(π) |= v}∪{v¯1|M(π) |= ¬v}; from RSI to EEFEXISTENCE, which shows that the latter p p • for each i = 2,...,n: π′(i) = {vi|M(π) |= v} ∪ problem is Σ2-hard, and therefore Σ2-complete. {v¯i|M(π) |= ¬v}∪{xi}; • π′(n+1) = {x1}∪{vi|M(π) |= ¬v}∪{v¯i|M(π) |= v}; p As a corollary, this Σ2-completeness result holds for gen- • π′(n +2)= {y}. eral (not necessarily monotonous) dichotomous preferences. π being regular and satisfying n + 1 and n + 2, we have As a consequenceof this high complexity,it is worth study- π(n +1) = π′(n + 1), π(n +2) = π′(n + 2), and then for ing restrictions and variants of the latter problem for which each i, π(i) ⊆ π′(i): indeed, let j ∈ {2,...,n} (for n +1 complexity may fall down. We start by considering identical and n +2 this inclusion is obviously satisfied); then (a) π dichotomous preference profiles, that is, all agents have the is regular, therefore π(j) ⊆ σ(j); now, all goods of σ(j) same preference, i.e. the same formula ϕ. ′ 1 are either in π (j) or in π(n + 1) (namely: x if j = 1 and Proposition 6 EEFEXISTENCE with N identical dichoto- all the vj such that M(π) |= ¬v and all the v¯j such that mous, monotonous preferences is NP-complete, for any fixed M(π) |= v); therefore, π(j) ⊆ π′(j) ∪ π(n + 1), which, N ≥ 2. together with π(1) ∩ π(n +1) = ∅, implies π(j) ⊆ π′(j). Due to space limitations the proofs of this result and the Since preferences are monotonous, all individuals satisfied 5 by π are satisfied by π′ as well; and since π is efficient, π and following ones are omitted . Note that we have here a hard- π′ satisfy the same set of individuals. Now, we remark that ness result for any fixed number of agents (≥ 2). Things are ′ ′ different with Proposition 5, for which hardness does not hold π = πM(π). By Lemma 4, π is efficient iff M(π) satisfies a maximal consistent subset of ∆, from which we conclude. when N is fixed. Namely, the following holds for N =2: Proposition 7 EEFEXISTENCE for two agents with Lemma 6 Any envy-free and efficient allocation for P sati- monotonous dichotomous preferences is NP-complete. fies n +1 and n +2. Unlike Proposition 5, these results are sensitive to whether Proof: Suppose π doesn’t satisfy n +1; then y 6∈ π(n + 1); preferences are required to be monotonous or not. now, if y ∈ π(n + 2) then n +1 envies n +2; if y 6∈ π(n + 2) Proposition 8 EEFEXISTENCE with N identical dichoto- then π is not efficient because giving y to n +2 would satisfy mous preferences is coBH -complete, for any fixed N ≥ 2. n +2 and thus lead to a better allocation than π. 2 Now, suppose π does not satisfy n +1, i.e., y 6∈ π(n + 2); Proposition 9 EEFEXISTENCE for 2 agents with dichoto- if y ∈ π(n + 1) then n +2 envies n +1; if y 6∈ π(n + 1) mous preferences is coBH2-complete. then again, π is not efficient because giving y to n +2 would Complexity decreases as well if we weaken Pareto-efficiency satisfy n +2 and thus lead to a better allocation than π. by only requiring allocations to be complete:

Lemma 7 If there exists an EEF allocation, then there exists 5They can be found in the long version of the paper, accessible at a maximal consistent subset of ∆ containing α1. http://www.irit.fr/recherches/RPDMP/persos/JeromeLang/papers/eef.pdf. Proposition 10 The existence of a complete envy-free allo- 6 Concluding remarks cation for agents with monotonous, dichotomous preferences We have identified the exact complexity of the key problem is NP-complete, even for 2 agents with identical preferences. of deciding whether there exists an efficient and envy-free Lastly, replacing Pareto-efficiency by an utilitarianistic notion allocation when preferences are represented compactly, in of efficiency results in a complexity decrease as well: several contexts; we have also considered variations of the Proposition 11 The existence of an envy-free allocation sat- problem. We have also drawed connections to a well-studied isfying a maximal number of agents with monotonous di- problem in nonmonotonic reasoning. The next step will p consist in designing and experimenting algorithms for the chotomous preferences is Θ2-complete. search of an EEF allocation (when it exists) and approxima- 5 Non-dichotomous preferences tion notions for defining optimal allocations when there is no EEF allocation (see [Lipton et al., 2004] for approximate We now consider the case where preferences are no longer envy-freeness, although not coupled with efficiency). dichotomous. Again, since an explicit description of preferences is exponentially large, the need for a compact descrip- Acknowledgements: We thank Michel Lemaˆıtre for stimu- tion thereof is clear. Many languages exist for succinct rep- lating discussions about fair division and compact represen- resentation of preference. However, Proposition 5 extends to tation, and Thibault Gajdos for stimulating discussions about any language, provided that (a) it extends propositional logic, envy-freeness and for pointing to us some relevant papers. i.e., it is able to express compactly any dichotomous preference represented by a propositional formula; (b) comparing two sets of goodscan be done in polynomialtime. Conditions References (a) and (b) are met by many languages for succinct represen- [Bouveret et al., 2005] S. Bouveret, H. Fargier, J. Lang, and tation of preference6. Under assumptions (a) and (b): M. Lemaˆıtre. Allocation of indivisible goods: a general model and some complexity results. In Proceed- Corollary 1 EEFEXISTENCE with monotonous preference p ings of AAMAS 05, 2005. Long version available at under logical form is Σ2-complete. http://www.irit.fr/recherches/RPDMP/persos/JeromeLang/papers/aig.pdf. For the latter result preferences do not have to be numerical since Pareto efficiency and envy-freeness are purely ordinal [Brams et al., 2000] S. Brams, P. Edelman, and P. Fishburn. notions. Now, if preferences are numerical, which implies the Fair division of indivisible items. Technical Report RR possibility of intercomparing and aggregating preferences of 2000-15, C.V. Starr Center for Applied Economics, New several agents, then, besides Pareto-efficiency, we may con- York University, 2000. sider efficiency based on social welfare functions. We con- [Chevaleyre et al., 2004] Y. Chevaleyre, U. Endriss, S. Es- sider here only the two most classical way of aggregating a tivie, and N. Maudet. Multiagent resource allocation with collection of utility functions hu1,...,uni into a social wel- k-additive utility functions. In Proc. DIMACS-LAMSADE fare function sw: utilitarianism (sw = ui) and egalitari- Workshop on Computer Science and Decision Theory, vol- Pi anism (sw = mini ui). ume 3 of Annales du LAMSADE, pages 83–100, 2004. Proposition 12 Given a collection of utility functions on 2R [Demko and Hill, 1998] S. Demko and T.P. Hill. Equitable given in compact form: distribution of indivisible items. Mathematical Social Sci- • the problem of the existence of an envy-free allocation ences, 16:145–158, 1998. p maximizing utilitarian social welfare is ∆2-complete, [Dunne et al., 2005] P. Dunne, M. Wooldridge, and M. Lau- even if N =2. rence. The complexity of contract negotiation. Artificial • the problem of the existence of an envy-free allocation Intelligence, 2005. To appear. p maximizing egalitarian social welfare is ∆2-complete, [Gottlob, 1992] G. Gottlob. Complexity results for non- even if N =2. monotonic logics. Journal of Logic and Computation, A last case that has not been considered is the case of addi- 2:397–425, 1992. tive numerical preferences. In the latter case, the utility func- [Herreiner and Puppe, 2002] D. Herreiner and C. Puppe. A tion of agent i ≤ N is simply expressed by the p numbers simple procedure for finding equitable allocations of in- ui({xj }), j = 1,...,p. While the existence of a complete divisible goods. Social Choice and Welfare, 19:415–430, envy-free allocation is easily shown to be NP-complete (see 2002. [Lipton et al., 2004]), things become much harder with EEF [Lang, 2004] J. Lang. Logical preference representation and EXISTENCE: all we know is that this problem is NP-hard and combinatorial vote. Annals of Mathematics and Artificial in p, but its precise complexity remains an open problem. Σ2 Intelligence, 42(1):37–71, 2004. 6For the sake of illustration, we pick here one of the most sim- [Lipton et al., 2004] R. Lipton, E. Markakis, E. Mossel, and ple ones, similar to those used for combinatorial auctions: agents’ A. Saberi. On approximately fair allocations of indivisible preferences are numerical (i.e., utility functions) and are represented goods. In Proceedings of EC’04, 2004. by a set of propositional formulas, each of which is associated with a weight denoting its importance; the utility of a set of goods is [Papadimitriou, 1994] Ch. H. Papadimitriou. Computational the sum of the weights of the formulas satisfied. Preferences are complexity. Addison–Wesley, 1994. monotonous if all formulas are positive and all weights are positive. [Reiter, 1980] R. Reiter. A logic for default reasoning. Arti- See for instance [Lang, 2004] for a survey of logical languages for ficial Intelligence, 13:81–132, 1980. compact preference representation.