Pareto Optimal Allocation under Uncertain Preferences
aa r X i v : . [ c s . G T ] O c t Pareto Optimal Allocation under Uncertain Preferences
Haris Aziz
Data61, CSIRO and UNSW Australia
Ronald de Haan
Technische Universit¨at Wien, Vienna, Austria.
Baharak Rastegari
School of Computing Science, University of Glasgow, Glasgow, UK.
Abstract
The assignment problem is one of the most well-studied settings in social choice,matching, and discrete allocation. We consider the problem with the additionalfeature that agents’ preferences involve uncertainty. The setting with uncer-tainty leads to a number of interesting questions including the following ones.How to compute an assignment with the highest probability of being Paretooptimal? What is the complexity of computing the probability that a givenassignment is Pareto optimal? Does there exist an assignment that is Paretooptimal with probability one? We consider these problems under two naturaluncertainty models: (1) the lottery model in which each agent has an inde-pendent probability distribution over linear orders and (2) the joint probabilitymodel that involves a joint probability distribution over preference profiles. Forboth of the models, we present a number of algorithmic and complexity resultshighlighting the difference and similarities in the complexity of the two models.
Keywords:
Assignment Problem, Resource allocation, Pareto optimality,Uncertain Preferences.
JEL : C62, C63, and C78
1. Introduction
When preferences of agents are aggregated to identify a desirable social out-come, Pareto optimality is a minimal requirement. Pareto optimality stipulatesthat there should not be another outcome that is at least as good for all agentsand better for at least one agent. We take Pareto optimality as a central concernand consider a richer version of the classic assignment problem where the twist
Email addresses: [email protected] (Haris Aziz), [email protected] (Ronald de Haan), [email protected] (Baharak Rastegari)
1s that agents may express uncertainty in their preferences. The assignmentproblem is a fundamental setting in which n agents express preferences over n items and each agent is to be allocated one item. The setting is a classicalone in discrete allocation. Its axiomatic and computational aspects have beenwell-studied [2, 6, 3, 9, 10, 15, 21, 22]. Our motivation for studying assignmentwith uncertain preferences is that agents’ preferences may not be completelyknown because of a lack of information or communication.Our work is inspired by the recent work of Aziz et al. [5] who examined thestable marriage problem under uncertain preferences. Uncertainty in prefer-ences has already been studied in voting [16]. Similarly, in auction theory, it isstandard to examine Bayesian settings in which there is probability distributionover the types of the agents. Although computational aspects of Pareto optimaloutcomes have been intensely studied in various settings such as assignment,matching, housing markets, and committee voting [3, 8, 9, 7, 12, 17, 18, 19],there has not been much work on Pareto optimal under uncertain preferences.When agents have uncertain preferences, one can relax the goal of computing aPareto optimal outcome and focus on computing outcomes that have the highestprobability of being Pareto optimal. We will abbreviate Pareto optimal as PO.If an assignment is Pareto optimal with probability one, we will call it certainlyPO.We consider the following uncertainty models: • Lottery Model : For each agent, we are given a probability distributionover linear preferences. • Joint Probability Model : A probability distribution over linear prefer-ence profiles is specified.Note that both the lottery model and the joint probability model represen-tation can be exponential in the number of agents but if the support of theprobability distributions is small, then the representation is compact. Also notethat the product of the independent uncertain preferences in the lottery modelresults in a probability distribution over preference profiles and hence can berepresented in the joint probability model. However, the change in representa-tion can result in a blowup. Thus whereas the joint probability model is moregeneral than the lottery model, it is not as compact. In view of this, complexityresults for one model do not directly carry over to results for the other model.The most natural computational problems that we will consider are as fol-lows. • PO-Probability : what is the probability that a given assignment is PO? • AssignmentWithHighestPO-Probability : compute an assignmentwith the highest probability of being PO.We also consider simpler problems than
PO-Probability : • IsPO-ProbabilityNon-Zero : for a given assignment, is the probabilityof being PO non-zero? 2
IsPO-ProbabilityOne : for a given assignment, is the probability ofbeing PO one?We also consider a problem connected to
AssignmentWithHighestPO-Probability : ExistsCertainlyPO-Assignment asks whether there ex-ists an assignment that has probability one for being PO. Note that
ExistsPossiblyPO-Assignment —the problem of checking whether there ex-ists some PO assignment with non-zero probability—is trivial for all uncertaintymodels in which the induced ‘certainly preferred’ relation is acyclic. The reasonwhy it is trivial is because the certainly preferred relation can be completed in away so that it is transitive and then for the completed deterministic preferences,there exists at least one PO assignment.We say that a given uncertainty model is independent if any uncertain pref-erence profile L under the model can be written as a product of uncertainpreferences L a for all agents a , where all L a ’s are independent [5]. Note thatthe lottery model is independent but the joint probability model is not. Results.
We show that for both the lottery model and the joint probabilitymodel,
ExistsCertainlyPO-Assignment is NP-complete. We also prove that
AssignmentWithHighestPO-Probability is NP-hard for both models. Inview of the results, we see that as we move from deterministic preferences touncertain preferences, the complexity of computing Pareto optimal assignmentsjumps significantly. On the other hand, we show that for a general class of un-certainty models called independent uncertainty models, both problems
IsPO-ProbabilityNon-Zero and
IsPO-ProbabilityOne can be solved in lineartime. Whereas
PO-Probability is polynomial-time solvable for the joint prob-ability model, we prove that the problem
Lottery Joint ProbabilityModel ModelProblems
PO-Probability
IsPO-ProbabilityNon-Zero in P in P
IsPO-ProbabilityOne in P in P
ExistsPossiblyPO-Assignment in P in P(trivially exists) (trivially exists)
ExistsCertainlyPO-Assignment
NP-complete NP-complete
AssignmentWithHighestPO-Prob
NP-hard NP-hard
Table 1: Summary of results. . Preliminaries The setting we consider is the assignment problem which is a triple (
N, O, ≻ )where N is the set of n agents { , . . . , n } , O = { o , . . . , o n } is the set of items,and ≻ = ( ≻ , . . . , ≻ n ) specifies complete, asymmetric, and transitive preferences ≻ i of each agent i over O . We will denote by R ( O ) as the set of all completeand transitive relations over the set of items O . We will denote by ≻ S as thepreference profile of agents from set S ⊂ N .An assignment is an allocation of items to agents, represented as an n × n matrix [ p ( i )( o j )] ≤ i ≤ n, ≤ j ≤ n such that for all i ∈ N , and o j ∈ O , p ( i )( o j ) ∈{ , } ; and for all j ∈ { , . . . , n } , P i ∈ N p ( i )( o j ) = 1. An agent i gets item o j ifand only if p ( i )( o j ) = 1. Each row p ( i ) = ( p ( i )( o ) , . . . , p ( i )( o m )) represents the allocation of agent i .An assignment p is Pareto optimal if there does not exist another assignment q such that q ( i ) % i p ( i ) for all i ∈ N and q ( i ) ≻ i p ( i ) for some i ∈ N .We first note a couple of well-known characterisations of Paretooptimal assignments. An assignment p admits a trading cycle o , i , o , i , . . . , o k − , i k − , o in which p ( i j )( o j ) > j ∈ { , . . . , k − } , o j +1 mod k ≻ j o j mod k for all j ∈ { , . . . , k − } . Fact 1 (Folklore) . An assignment is Pareto optimal if and only if it does notadmit a trading cycle.
We will also use the following characterization of Pareto optimal discreteassignments [1] that is defined with respect to outcomes of serial dictatorship.Serial dictatorship is an assignment mechanism that is specified with respect toa permutation π over N : agents in the permutation are given the most preferreditem that is still not allocated. We will denote by SD ( N, O, ≻ , π ) the outcomeof applying serial dictatorship with respect to permutation π over assignmentproblem ( N, O, ≻ ). Fact 2 (Abdulkadiro˘glu and S¨onmez [1]) . An assignment is Pareto optimal ifand only if it is an outcome of serial dictatorship.
Fact 2 also follows from Proposition 1 by Brams and King [11]. The factsabove show that when preferences are deterministic, a Pareto optimal assign-ment can be computed or verified easily. We will now focus on similar problemsbut with the feature that agents have uncertain preferences.
Example 1. a, b, c (0 . b, a, c (0 . b, a, c c, b, a Consider the assignment abc in which gets a , gets b , and gets c . Theprobability of the assignment being Pareto optimal is 1. On the other hand, theassignment bac has 0.4 probability of being Pareto optimal. . Joint Probability Model We first observe that the
PO-Probability can be solved easily for the jointprobability model.
Theorem 1.
For the joint probability model,
PO-Probability can be solvedin polynomial time.Proof.
The probability that a given assignment is PO is equivalent to the prob-ability weight of the preference profiles for which the assignment is PO. Thiscan be checked as follows. We check the preference profiles for which the givenassignment is PO (for one profile, this can be checked in linear time). Then weadd the probabilities of those profiles for which the assignment is PO. The sumof the probabilities is the probability that the assignment is PO.
Corollary 1.
For the joint probability model,
IsPO-ProbabilityNon-Zero and
IsPO-ProbabilityOne can be solved in polynomial time.
What about
ExistsCertainlyPO-Assignment ? This problem is equiva-lent to checking whether the sets of PO assignments have a non-empty inter-section. We show that this problem is NP-complete even when the probabilitydistribution is over two linear preference profiles.We reduce from the NP-complete problem
SerialDictatorship-Feasibility —check whether there exists a permutation of agents for whichserial dictatorship gives a particular item o to an agent i [20]. SerialDictatorshipFeasibility
Input: (
N, O, ≻ , i ∈ N, o ∈ O )Question: Does there exist a permutation of agents for which serialdictatorship gives a particular item o to an agent i ?For linear preference profiles, the set of Pareto optimal allocations are char-acterized by those that can be achieved via some serial dictatorship. Thus itfollows that the following problem is also NP-complete: check whether thereexists a Pareto optimal allocation in which a specified agent i gets a specifieditem o . Theorem 2.
For the joint probability model,
ExistsCertainlyPO-Assignment is NP-complete even when the probability distribution is over twolinear preference profiles.Proof.
The problem
ExistsCertainlyPO-Assignment is in NP because itcan be checked in polynomial time whether a given assignment is certainly POor not (Theorem 1).To prove NP-hardness, we reduce from the NP-complete problem :
Serial-DictatorshipFeasibility — given (
N, O, ≻ ), check whether there exists apermutation of agents for which serial dictatorship gives a particular item o toan agent i [20]. 5e construct a joint probability over two preference profiles. One of theprofiles is the same as ≻ . In the other preference profile ≻ ′ , agent i has o as themost preferred item and has the same order of preference over all other itemsas in ≻ i . Each agent j ∈ N \ { i } has o as the least preferred item. As for theother items, each j ∈ N \ { i } has the same preferences over the items in O \ { o } as in ≻ j .Our first observation is that an assignment is PO under profile ≻ ′ only if i gets o in it. Claim 1.
An assignment is PO under profile ≻ ′ only if i gets o in it.Proof. The argument is as follows. If i does not get o , then an agent j = i gets it. However both i and j get a more preferred item under profile ≻ ′ byexchanging their items.We now prove that we have a yes instance of SerialDictatorshipFeasi-bility if and only if there exists a certainly PO assignment.Assume that there exists a certainly PO assignment. Then, it must be POunder ≻ ′ implying that, by our claim above, i gets o in this assignment. Thesame assignment must also be PO under profile ≻ which implies that thereexists an assignment that is PO under profile ≻ in which i gets o . In lightof Fact 2, this implies that there exists a serial dictatorship the outcome ofwhich under profile ≻ is the same assignment. Hence, we have a yes instance of SerialDictatorshipFeasibility .Now consider the case when we have a yes instance of
SerialDictator-shipFeasibility . This means that there is a permutation π under which i gets o when serial dictatorship is run. Let us call this assignment by p . Due toFact 2, p is PO under preference profile ≻ . We want to prove that p is POunder each possible preference profile. We already know that it is PO under ≻ so it remains to show that it is PO under ≻ ′ . Due to Fact 2, it is sufficientto prove that for profile ≻ ′ , there exists a corresponding permutation of agentsunder which the outcome of serial dictatorship is p .In fact, we show that for SD ( N, O, ≻ ′ , π ) = p —i.e., the outcome of applyingserial dictatorship with permutation π is p even if the preference profile is ≻ ′ instead of ≻ . In order to prove the statement we prove the following claim. Claim 2.
The following are the same at each round, when applying serial dic-tatorship to profiles ≻ and ≻ ′ , in both cases with respect to permutation π . • the order in which items are allocated. • the allocation of each agent. • set of remaining items. roof. The claim can be proved via induction on the number of rounds of serialdictatorship. For the base case, let us consider agent π (1). If π (1) = i , then π (1) picks up o under both preference profiles. This is because, by construction(1) π is a permutation under which i gets o when serial dictatorship is appliedon ≻ , and (2) i has o as his most preferred item under ≻ ′ . If π (1) = i , then π (1) picks up some item o ′ = o in p = SD ( N, O, ≻ , π ). Note that for π (1), hismost preferred item in both profiles must be o ′ . Hence by the end of the firstround, the same item has been given to the same agent in both ≻ and ≻ ′ .For the induction, let us assume that k rounds have taken place and theorder in which items are allocated, the allocation of each agent in the first k round and the set of unallocated items T is the same under both profiles ≻ and ≻ ′ . Now consider agent π ( k + 1). If π ( k + 1) = i , then i picks up item o under ≻ , implying that o ∈ T , which in turn implies that i must pick up o under ≻ ′ since o is his most preferred item in O under preference ≻ ′ i and hence his mostpreferred item in T . It remains to show what happens when π ( k + 1) = i . Inthat case π ( k +1) picks up some item o ′ = o in SD ( N, O, ≻ , π ). This means that o ′ is the most preferred item of agent π ( k + 1) in set T ⊂ O under preferenceprofile ≻ , implying that o ′ is the most preferred item of agent π ( k + 1) in set T under preference profile ≻ ′ as well. This completes the proof of the claim.We have thus proved that the outcome of applying serial dictatorship withrespect to permutation π is p under both preference profiles ≻ and ≻ ′ . Thus p is PO under both possibly realizable preference profiles. To conclude, we haveproved there exists a certainly PO assignment if and only if we have a yes in-stance of SerialDictatorshipFeasibility . Since
SerialDictatorshipFea-sibility is NP-complete, it follows that
ExistsCertainlyPO-Assignment isNP-complete.
Corollary 2.
For the joint probability model,
AssignmentWithHighestPO-Prob is NP-hard.Proof.
Assume to the contrary that there exists a polynomial-time algorithm tosolve
AssignmentWithHighestPO-Prob . In that case, we can compute suchan assignment p . By Corollary 1, it can be checked in polynomial time whether p is PO with probability one or not. If p is PO with probability one, then weknow that we have a yes instance of ExistsCertainlyPO-Assignment . Oth-erwise, we have a no instance of
ExistsCertainlyPO-Assignment . Hence
ExistsCertainlyPO-Assignment is polynomial-time solvable, a contradic-tion.Before dealing with the lottery model, we present some general algorithmicresults that apply not just to the lottery model but a class of uncertainty modelsthat includes the lottery model.
4. Independent Uncertainty Models
We first present a couple of general results that apply to a large class ofuncertainty models that satisfy independence. Recall that a given uncertainty7odel is independent if any uncertain preference profile L under the model canbe written as a product of uncertain preferences L a for all agents a , where all L a ’s are independent.We first define the certainly preferred relation ≻ certaini for agent i . We write b ≻ certaini c if and only if agent i prefers b over c with probability 1. Theorem 3.
For any independent uncertainty model in which the certainlypreferred relation can be computed in polynomial given, given an assignment itcan be checked in polynomial-time whether another assignment Pareto dominatesit with probability one.Proof.
Given an assignment ω , we create a trading cycle graph G in which eachagent i points to any item o such that o ≻ certaini ω ( i ). We now claim thatthere exists a cycle in G if and only if the assignment ω is Pareto optimal withprobability zero.If there exists a cycle in G , then another assignment Pareto dominates ω with probability one. The reason is that each agent prefers the item he pointsto with probability one. Hence, if we implement the trade in the cycle, eachagent in the cycle gets a certainly more preferred item. Therefore the assignmentis Pareto dominated with probability one.Now suppose that there is an assignment that Pareto dominates ω withprobability one. Equivalently, there exists another assignment in which eachagent with a different allocation gets a certainly strictly more preferred item.But this means that there exists a cycle in G . Theorem 4.
For any independent uncertainty model,
IsPO-ProbabilityOne can be solved in polynomial time.Proof.
Given an assignment ω , we create a trading cycle graph G in which eachagent i points to any item o such that ω ( i ) certaini o . We claim that ω is Paretooptimal with probability one if and only if G does not contain a cycle.We first show that if there exists a cycle, then it is not the case that ω is POwith probability one. Existence of a cycle implies that each agent in the cycleprefers another item to what he has received with non-zero probability, whichin turn implies that if we implement the cycle then each of these agents willreceive a more preferred item with non-zero probability. Therefore ω is Paretodominated with non-zero probability.If it is not the case that ω is Pareto optimal with probability one, thenit must be that that another assignment Pareto dominates it with non-zeroprobability. Equivalently, there exists another assignment in which each agentwith a different allocation gets a different item that is more preferred with non-zero probability. But this means that there exists a cycle in G .
5. Lottery Model
We now focus on the lottery model. Since the lottery model is a independentuncertainty model, Theorems 3 and 4 apply to it.8 heorem 5.
For the lottery uncertainty model,
IsPO-ProbabilityNon-Zero can be solved in polynomial time.Proof.
Consider an assignment p that we want check whether it is PO withnon-zero probability. We use the following algorithm that can be consideredas building a permutation of agents that is consistent with serial dictatorshipproducing the assignment p .Initialize the set of remaining items to O , the remaining agents to N , and the permutation of the agents π to an empty list. Checkif there exists some agent i such that p ( i ) is an available item thatis the most preferred for i in at least one of his preference lists. Ifno such agent exists, return no. If such an agent exists, give theitem to him, append i to the permutation π , remove i from theset of remaining agents, and remove p ( i ) from the set of availableitems. Also select the preference of agent i that had p ( i ) as the mostpreferred remaining item, denoting it by ≻ i . Repeat until no moreitems are left.If the algorithm builds the whole permutation and does not return no, thenwe claim p is Pareto optimal with non-zero probability. When an agent i picksthe item p ( i ) in his turn, it means that the agent has at least one possiblepreference, ≻ i , in which p ( i ) is the most preferred remaining item. Hence whenapplying serial dictatorship to the selected preference profile ≻ with respect to π , each agent i picks p ( i ) when his turn comes, resulting in p as the outcome ofserial dictatorship, hence implying that p is PO with respect to ≻ and thereforePO with non-zero probability.If the algorithm returns no, we argue that p is PO with zero probability.Consider the first point in the algorithm where no agent i has p ( i ) as an availableitem that is the most preferred for i in at least one of his preference lists. Thismeans that no remaining agent gets his most preferred item (for any preferencelist) among the available items. Therefore, for each realisation of the preferencesprofiles, each of the remaining agents is interested in and points to another itemheld by another agent among the remaining agents. This implies the existenceof a trading cycle for each realisation of the preference profiles, where someremaining agents can exchange items among themselves to get a more preferreditem than in p . Thus p is PO with zero probability.We now prove that the problem of checking whether there exists an assign-ment that is PO with probability one is NP-complete. Although the proof issimilar to the proof of Theorem 2, we give a complete argument since a com-plexity result for the joint probability model does not directly imply a similarresult for the lottery model. Theorem 6.
For the lottery model,
ExistsCertainlyPO-Assignment is NP-complete. roof. The problem
ExistsCertainlyPO-Assignment is in NP because itcan be checked in polynomial time whether a given assignment is certainly POor not (Theorem 4). To prove NP-hardness, we use an argument similar to thatused in the proof of Theorem 2.We reduce from the NP-complete problem :
SerialDictatorshipFeasi-bility — given an assignment setting (
N, O, ≻ ), check whether there exists apermutation of agents for which serial dictatorship gives a particular item o toan agent i [20].We construct preferences in which each agent j ∈ N has two preference listswhere one of them is ≻ j . For agent i , we add another preference list ≻ ′ i in which i ’s most preferred item is o and the rest of the items are in the same order asin ≻ i . For each other agent j ∈ N \ { i } , we add a preference list ≻ ′ j which isidentical to ≻ j except that o is moved to the end of the list.Our first observation is that an assignment is PO under profile ≻ ′ only if i gets o in it. If i does not get o , and agent j = i gets it, then both i and j get amore preferred item under profile ≻ ′ by exchanging their items. Hence if thereis any assignment that is certainly PO then it must give o to i .We prove that there exists a certainly PO assignment if and only if we havea yes instance of SerialDictatorshipFeasibility .If we have a no instance of
SerialDictatorshipFeasibility , then in noassignment that is PO under ≻ agent i gets o . On the other hand, an assignmentis PO under ≻ ′ only if i receives o . Therefore, there does not exist any certainlyPO assignment.Now consider the case when we have a yes instance of SerialDictator-shipFeasibility . This means that there is a permutation π under which i gets o when serial dictatorship is run. Let us call this assignment p . Due to Fact 2, p is PO under preference profile ≻ . We want to prove that p is PO under eachpossible preference profile. Due to Fact 2 it is sufficient to prove that for eachpossible realizable preference profile, there exists a corresponding permutationof agents under which the outcome of serial dictatorship is p .In fact, we show that for each possible preference profile ≻ ′′ , SD ( N, O, ≻ ′′ , π ) = p i.e., the outcome of applying serial dictatorship with permutation π is p . In order to prove the statement we prove the following claim. (Note that p is PO under ≻ with respect to π .) Claim 3.
The following are the same at each round, when applying serial dicta-torship to ≻ and any of the realizable preference profiles ≻ ′′ , in both cases withrespect to permutation π • the order in which items are allocated. • the allocation of each agent. • set of remaining items. roof. The claim can be proved via induction on the number of rounds of serialdicatorship. For the base case, let us consider agent π (1). If π (1) = i , then π (1)picks up o in all his possible preferences. This is because, by construction (1) π is a permutation under which i gets o when serial dictatorship is applied on ≻ ,so it must be that i ranks o at the top of his list under ≻ i and (2) i has o as hismost preferred item under ≻ ′ i by construction. If π (1) = i , then π (1) picks upsome item o ′ = o in p = SD ( N, O, ≻ , π ). Note that for π (1), his most preferreditem is the same in all possible profiles. Hence by the end of the first round, thesame item has been given to the same agent in all the realizable preferences.For the induction, let us assume that k rounds have taken place and the orderin which items are allocated, the allocation of each agent in the first k turns andthe set of unallocated items T is the same all the realizable preferences. Nowconsider the agent π ( k + 1). If π ( k + 1) = i , then i picks up item o under ≻ ,implying that o ∈ T , which in turn implies that i must pick o under ≻ ′ i since o ishis most preferred item in O under ≻ ′ i and hence his most preferred item in T .It remains to show what happens when π ( k + 1) = i . In that case π ( k + 1) pickssome item o ′ = o in SD ( N, O, ≻ , π ). This means that o ′ is the most preferreditem of agent π ( k + 1)in set T ⊂ O of agent π ( k + 1) under preference list ≻ π ( k +1) , implying that o ′ is the most preferred item of agent π ( k + 1) in set T under preference ≻ ′ π ( k +1) as well. This completes the proof of the claim.We have thus proved that the outcome of applying serial dictatorship withrespect to permutation π is p under all possible preference profiles. Thus p isPO under each possibly realizable preference profile when we have a yes instanceof SerialDictatorshipFeasibility .To conclude, we have proved there exists a certainly PO assignment if andonly if we have a yes instance of
SerialDictatorshipFeasibility . Since
SerialDictatorshipFeasibility is NP-hard and
ExistsCertainlyPO-Assignment is in NP, it follows that
ExistsCertainlyPO-Assignment isNP-complete.
Corollary 3.
For the lottery model,
AssignmentWithHighestPO-Prob isNP-hard.Proof.
Assume to the contrary that there exists a polynomial-time algorithm tosolve
AssignmentWithHighestPO-Prob . In that case, we can compute suchan assignment p . By Theorem 4, it can be checked in polynomial time whether p is PO with probability one or not. If p is PO with probability one, then weknow that we have a yes instance of ExistsCertainlyPO-Assignment . Oth-erwise, we have a no instance of
ExistsCertainlyPO-Assignment . Hence
ExistsCertainlyPO-Assignment is polynomial-time solvable, a contradic-tion.In light of Theorem 5 and Theorem 4, we know that for the lottery model,it can be checked in polynomial time whether the PO probability of a given as-signment is zero or one, respectively. We now turn to the problem of computing11he probability that a given assignment is PO. We first present a polynomial-time solution for a restricted setting, and then show that
PO-Probability is
Theorem 7.
For the lottery model, if the number of uncertain agents in con-stant, then
PO-Probability is polynomial-time solvable.Proof.
Let ω be a given assignment. Let constant k denote the number ofuncertain agents, and let the maximum number of preferences for any uncertainagent be ℓ . Therefore, the maximum number of preference profiles that arerealizable is ℓ k which is still polynomial in the input since k = O (1). For eachpossible preference profile ≻ , it is easy to compute the probability of ω beingstable under ≻ by simply computing the product of the probabilities of thepreferences chosen of the uncertain agents. Hence, we have reduced the problemto the problem PO-Probability for the joint probability model which can besolved in polynomial time (Theorem 1).
Theorem 8.
For the lottery model,
PO-Probability is
We show
Monotone-
Input: A 2CNF formula that contains no negation.Question: Count the number of satisfying assignments.Let ϕ be a monotone 2CNF formula with clauses c , . . . , c m and variables x , . . . , x n . We construct an instance of PO-Probability as follows. Consideragents 1 , . . . , n and items o , . . . , o n , and take the assignment σ that gives eachagent i item o i .We construct the preferences of the agents as follows. Take an arbitraryagent i . Consider the set { j , . . . , j u } of indices j such that the clause ( x i ∨ x j )occurs in ϕ . (Without loss of generality, this set { j , . . . , j u } is non-empty.)Suppose that j < j < · · · < j u , in order to fix an (arbitrary) order over theseindices. With probability , agent i has o i at the top of his preference list,followed by the rest of the items in arbitrary order. With probability , agent i has the following preference: o j ≻ i · · · ≻ i o j u ≻ i o i ≻ i · · · , where the remainingitems appear in arbitrary order after o i .This way, the possible preference profiles correspond one-to-one to the pos-sible truth assignments over x , . . . , x n . Namely, taking the preference o i ≻ i · · · for agent i corresponds to setting x i to 1, and taking the other preference foragent i corresponds to setting x i to 0. Moreover, each possible preference profileoccurs with probability n .We show that the number of satisfying assignments for ϕ is equal to thenumber of preference profiles under which σ is Pareto optimal. In particular,12e show that σ is PO under a preference profile if and only if the correspondingtruth assignment T satisfies ϕ .( = ⇒ ) Take a possible preference profile ≻ under which σ is PO and suppose,for a contradiction, that the corresponding truth assignment T does not satisfy ϕ . That is, there is some clause c = ( x i ∨ x j ) that is not satisfied, implying thatin T both x i and x j are set to 0. Then we know that agent i prefers o j to o i and agent j prefers o i to o j , hence they are willing to swap their assigned items.Therefore σ is not Pareto optimal under ≻ , a contradiction.( ⇐ =) Take a possible preference profile ≻ and suppose that the correspond-ing truth assignment T satisfies ϕ . We show that we cannot find a Paretoimprovement of σ , implying that σ is PO. Take an arbitrary agent i . First sup-pose that T sets x i to 1. This means that agent i prefers o i to all other items,and so he is not willing to exchange it with another item. Now, suppose that Tsets x i to 0. Take the set { j , . . . , j u } of indices such that the clause ( x i ∨ x j )occurs in ϕ . As x i is set to 0, this means that i prefers o j , . . . , o j u to o i and iswilling to exchange o i with either of these items (but no other item). BecauseT satisfies ϕ , we know that T sets x j , . . . , x j u to 1, and consequently, agents j , . . . , j u prefer items o j , . . . , o j u over all other items (respectively). So neitherof these agents is willing to exchange their assigned item with o i . Therefore, asno Pareto improvement exists, σ is Pareto optimal.The number of satisfying truth assignments of ϕ is then exactly equal to2 n times the probability that assignment σ is Pareto optimal. Thus, PO-Probability is
PO-Probability is in f : Σ ∗ → N . Wecan consider PO-Probability as such a function producing natural numbersin the following way. Without loss of generality, suppose that the probabilitiesin the input are all given as rational numbers with the same denominator d .(We can transform the input in polynomial time to an equivalent input thatsatisfies this property.) Then the probability that the given assignment is Paretooptimal is zd n for some positive integer z . We then consider the problem PO-Probability as the function that returns z , rather than the rational zd n .We argue membership in M that has the property that for each input, the number of accepting pathsof M for this input equals the number z that corresponds to the probability thatthe given matching is Pareto optimal. The existence of such a Turing machineimplies membership in M operates as follows. For eachagent a i , it uses nondeterminism to generate d different (partial) computationpaths. These partial computation paths are concatenated, resulting in d n to-tal computation paths. Suppose that the input specifies ℓ possible preferenceorders for agent a i , occurring with probabilities u d , . . . , u ℓ d , respectively. Thenthe first u partial computation paths generated for a i correspond to the firstpreference order, the next u correspond to the second order, and so on. As aresult, each total computation path corresponds to some preference profile. Atthe end of each computation path, the machine M checks (in deterministic poly-13omial time) whether the assignment is Pareto optimal for the correspondingpreference profile, and accepts if and only if this is the case. It is straightforwardto verify that the number of accepting computation paths of M is exactly thenumber z such that the probability that the assignment is Pareto optimal is zd n .Therefore, we know that PO-Probability is in k is the number of uncertain agents, and ℓ is themaximum number of possible preference orders for these uncertain agents, therunning time of the algorithm outlined in the proof of Theorem 7 is Ω( ℓ k ). Weimprove on this result by showing that there exists a fixed-parameter tractablealgorithm that computes the PO probability for the lottery model—that is, analgorithm running in time f ( k ) n c for some computable function f and somefixed constant c independent of k , where n denotes the input size. In otherwords, we show that the parameterized problem k - PO-Probability , wherethe parameter is the number of uncertain agents, is fixed-parameter tractablefor the lottery model.
Theorem 9.
For the lottery model, k - PO-Probability can be solved in fixed-parameter tractable time.Proof.
Take an arbitrary instance of the problem k - PO-Probability , con-sisting of agents 1 , . . . , n , objects o , . . . , o n , and an assignment σ . Withoutloss of generality, assume that the assignment gives each agent i the object o i ,and that the uncertain agents are agents 1 , . . . , k . For each uncertain agent i ,let ≻ i, , . . . , ≻ i,u i denote the different possible preferences for agent i .Additionally, assume without loss of generality that for each of the uncertainagents 1 , . . . , k , each of the possible preferences for these agents occurs withprobability ℓd , where the numerator ℓ can vary between different agents anddifferent possible preferences, but where the denominator d is common among allagents and all possible preferences. In other words, all probabilities mentionedin the instance are rational numbers that share a common denominator d . Ifthis were not the case, we could straightforwardly transform the instance inpolynomial time to an equivalent instance that does satisfy this property.Also, assume without loss of generality that there exists no trading cycle thatinvolves only the agents o k +1 , . . . , o n . If this were the case, the assignment isPareto optimal with probability zero, and we can filter out such trivial instancesusing a polynomial-time preprocessing procedure.We now how to compute the probability that the given assignment is Paretooptimal in fixed-parameter tractable time. Our computation will proceed inthree stages:(1) We construct a directed graph G with O ( ku k ) vertices, where the edgesare weighted. Here u denotes the maximum number of possible preferencesfor any uncertain agent. 142) We count the number of homomorphisms f of a directed path P k +2 oflength 2 k + 2 to this graph G , where each homomorphism is countedmultiple times according to (the product of) the weights on the edgesin f ( P k +2 ). This counting can be done in polynomial time using anextension of a known algorithm [13, 14].(3) We divide the weighted total number of homomorphisms of P k +2 to G by the number d k to obtain the probability that the given assignment isPareto optimal.We begin with phase (1), and we construct the weighted, directed graph G .Let Π = { o , . . . , o k } be the set of all possible pairs ( o i , o j ) of objectsamong o , . . . , o k . We define the set V of vertices of G as follows. First, wedefine an auxiliary set V ′ : V ′ = { , . . . , k + 1 } ∪ { ( i, ≻ i,j ) | i ∈ [ k + 1] , j ∈ [ u i ] } . Then, we define the set V of vertices as follows: V = { s, t } ∪ { ( v ′ , Π ′ ) | v ′ ∈ V ′ , Π ′ ⊆ Π } . That is, the graph G has vertices s and t , and 2 k copies of each element in V ′ (one for each Π ′ ⊆ Π). Intuitively, the vertices s and t will act as source andtarget for each homomorphism of P k +2 to G .The sets Π ′ ⊆ Π will intuitively be used to memorize the ‘trading paths’(i.e., paths in the trading cycle graph) that result from particular choices of thepreference orders ≻ i,j chosen for the agents 1 , . . . , k . That is, each ( o i , o j ) ∈ Π ′ corresponds to a path from o i to o j in the directed graph with vertices o , . . . , o n where there is an edge from o i ′ to o i ′′ if and only if agent i ′ prefers object o i ′′ to object o i ′ .We construct the set E of (weighted and directed) edges as follows. • We add an edge with weight 1 from s to (1 , ∅ ). • For each i ∈ [ k ], each j ∈ [ u i ], and each Π ′ ⊆ Π, we add an edge from ( i, Π ′ )to ( i, ≻ i,j , Π ′ ). This edge has weight ℓ , where the possible preferenceorder ≻ i,j for agent i occurs with probability ℓd . • For each i ∈ [ k ], each j ∈ [ u i ], and each Π ′ ⊆ Π, we add an edge withweight 1 from ( i, ≻ i,j , Π ′ ) to the vertex ( i + 1 , Π ′′ ), for some Π ′ ⊆ Π ′′ ⊆ Π. The choice of Π ′′ is determined as follows. Consider the follow-ing graph G Π ′ , ≻ i,j . The vertices of this graph are o , . . . , o n . For eachpair ( o i ′ , o i ′′ ) of vertices among o k +1 , . . . , o n , there is an edge from o i ′ to o i ′′ if and only if agent j prefers object o i ′′ to object o i ′ . Moreover,for each ( o i ′ , o i ′′ ) ∈ Π ′ , we add an edge from o i ′ to o i ′′ . Finally, for eachagent o i ′ among o k +1 , . . . , o n , we add an edge from o i to o i ′ if and onlyif o i ′ ≻ i,j o i . We then let Π ′′ ⊆ Π be the set of all pairs ( o i ′ , o i ′′ ) such thatthere is a path from o i ′ to o i ′′ in G Π ′ , ≻ i,j . Clearly, Π ′ ⊆ Π ′′ .15 For each Π ′ ⊆ Π such that ( o i , o i ) Π ′ for all i among 1 , . . . , k , we addan edge with weight 1 from ( k + 1 , Π ′ ) to t .Clearly, any homomorphism f from the directed path P k +2 of length 2 k + 2to G must map the first vertex of the path to s and the last vertex of the pathto t . Each such homomorphism must map the (2 i )-th vertex of the path to somevertex ( i, Π ′ ) and the (2 i + 1)-th vertex of the path to some vertex ( i, ≻ i,j , Π ′ ).Also, the (2 k +2)-th vertex of the path must be mapped to some vertex ( k +1 , Π ′ )where Π ′ contains no pair ( o i , o i ). These observations follows directly from theconstruction of G .Moreover, each homomorphism f ′ from the directed path P k +1 oflength 2 k + 1 to G that maps the first vertex of the path to s is uniquely deter-mined by some series of choices ≻ ,j , . . . , ≻ k,j k for the possible preferences ofthe uncertain agents 1 , . . . , k . We argue that such a homomorphism f ′ can beextended to a homomorphism f from P k +2 to G if and only if the correspondingpreferences ≻ ,j , . . . , ≻ k,j k lead to a trading cycle. The homomorphism f ′ mapsthe (2 k + 2)-th vertex of the path to some pair ( k + 1 , Π ′ ). Here Π ′ is the set ofpairs ( o i , o j ) ∈ { o , . . . , o k } such that the preferences ≻ ,j , . . . , ≻ k,j k lead to atrading path from o i to o j . By our assumption that there exists no trading cyclethat involves only the agents o k +1 , . . . , o n , we know that the set Π ′ contains somepair ( o i , o i ) if and only if there exists a trading cycle. Therefore, by constructionof the edges between ( k + 1 , Π ′ ) and t , we know that the choices ≻ ,j , . . . , ≻ k,j k of preferences for the agents 1 , . . . , k that make the assignment Pareto optimalare in one-to-one correspondence with the homomorphisms f from P k +2 to G .We count each such homomorphism f in a weighted fashion as follows—this is phase (2). Take a homomorphism f from P k +2 to G . Its weight inthe grand total is the product of the weights for each edge in f ( P k +2 ). Theonly edges in f ( P k +2 ) that have weigth > i, Π ′ ) to ( i, ≻ i,j , Π ′ ). Such an edge has weight ℓ , where the probability that ≻ i,j occurs is ℓd .From this, it is straightforward to verify that the total weighted sum of allhomomorphisms is equal to p · d k , where p is the probability that the givenassignment is Pareto optimal. Therefore, in order to compute p , we only needto take the weighted sum of all homomorphisms, and divide it by d k —this isphase (3) of the algorithm.All that remains is to show how we can compute the weighted sum of allhomomorphisms f from P k +2 to G in polynomial time. We can do this byextending a known polynomial-time algorithm to count the number of homo-morphisms of a graph whose treewidth is bounded by a fixed constant intoanother graph [14, Theorem 14.7]. Since paths have treewidth 1, counting thenumber of homomorphisms from a path to another graph can be done in poly-nomial time using this algorithm. This algorithm uses a dynamic programmingapproach to count the number of homomorphisms. This dynamic programmingtechnique can straightforwardly be extended to take into account the weightsof the homomorphisms. (We omit a detailed description of the extended algo-rithm.)This concludes our proof that k - PO-Probability can be solved in fixed-16arameter tractable time for the lottery model.
6. Conclusions
Computing Pareto optimal outcomes is an active line of research in eco-nomics and computer science. In this paper, we examined the problem for anassignment setting where the preferences of the agents are uncertain. Our cen-tral technical results are computational hardness results. We see that as wemove from deterministic preferences to uncertain preferences, the complexityof computing Pareto optimal outcomes jumps significantly. The computationalhardness results carry over to more complex models in which there may be moreitems than agents, agents may have capacities, and items may have copies. Forfuture work, we are also starting to consider other uncertainty models [5]. Ifwe consider the compact indifference model [5] which is an independent uncer-tainty model, then the results in Section 4 apply to it. If we allow for intransitivepreferences, even a possibly Pareto optimal assignment may not exist and theproblem of checking whether a possible Pareto optimal assignment exists be-comes interesting. An orthogonal but equally interesting direction will be toconsider other fairness, stability, or efficiency desiderata [4].