Table Of Content

Manipulating the Probabilistic Serial Rule Haris Aziz, Serge Gaspers, Simon Mackenzie, Nicholas Mattei NICTA and UNSW, Kensington2033, Australia 5 Nina Narodytska 1 0 Carnegie Mellon University,Pittsburgh, PA 15213-3891, USA 2 Toby Walsh n a NICTA and UNSW, Kensington2033, Australia J 7 2 ] T Abstract G . Theprobabilisticserial(PS)ruleisoneofthemostprominentrandomizedrules s fortheassignmentproblem. Itiswell-knownforitssuperiorfairnessandwelfare c [ properties. However,PSisnotimmunetomanipulativebehaviourbytheagents. Weinitiatethestudyofthecomputationalcomplexityofanagentmanipulating 1 thePSrule. WeshowthatcomputinganexpectedutilitybetterresponseisNP- v 6 hard. On the other hand, we present a polynomial-time algorithm to compute 2 alexicographicbest response. Forthe caseoftwo agents,we showthatevenan 6 expected utility best response can be computed in polynomial time. Our result 6 for the case of two agents relies on an interesting connection with sequential 0 allocation of discrete objects. . 1 0 Keywords: Assignment problem, probabilistic serial mechanism, fair 5 allocation 1 JEL: C62, C63, and C78 : v i X 1. Introduction r a Theassignmentproblem isoneofthemostfundamentalandimportantprob- lemsineconomicsandcomputerscience[seee.g.,3,4,6,12,13]. Inthe setting, agents express preferences over objects and, based on these preferences, the objects are allocated to the agents. The model is applicable to many resource allocationorfairdivisionsettingswheretheobjectsmaybepublichouses,school seats, course enrollments, kidneys for transplant, car park spaces, chores, joint Email addresses: [email protected] (HarisAziz),[email protected] (SergeGaspers),[email protected] (SimonMackenzie), [email protected] (NicholasMattei), [email protected] (NinaNarodytska), [email protected] (TobyWalsh) 1 assets, or time slots in schedules. A randomized or fractional assignment rule takes the preferences of the agents into account in order to allocate each agent afractionoftheobject. Iftheobjectsareindivisiblebutallocatedinarandom- izedway,the fractioncanalsobe interpretedas the probabilityofreceiving the object. Randomizationiswidespreadinresourceallocationsinceitisoneofthe most natural ways to ensure procedural fairness [8]. Randomized assignments have been used to assign public land, radio spectra to broadcasting companies, and US permanent visas to applicants [Footnote 1, 8]. Amongthevariousrandomized/fractionalassignmentrules,theprobabilistic serial (PS) ruleisoneofthemostprominentrules[1,5,6,8,14,16,19,17]. PS worksas follows. Eachagentexpressesa linear order overthe set of houses (we use the term house throughout the paper though we stress any object could be allocated with these mechanisms). Each house is considered to have a divisible probability weight of one, and agents simultaneously and with the same speed eat the probability weight of their most preferred house. Once a house has been eaten by a subset of agents, these agents proceed to eat their next most preferred house that has not been completely eaten. The procedure terminates after all the houses have been eaten. The random allocation of an agent by PS is the amount of each object he has eaten. Although PS was originally defined for the setting where the number of houses is equal to the number of agents, it canbe usedwithout anymodificationfor fewer ormore housesthanagents[see e.g., 6, 16]. The probabilistic serial (PS) rule fares better than any other random assignment rule in terms of fairness and welfare [5, 6, 8, 16, 19]. In particular, it satisfiesstrongenvy-freenessandefficiencywithrespecttobothstochastic dominance (SD) and downward lexicographic (DL) relations [6, 18, 16]. SD is one of the most fundamental relations between fractional allocations because one allocation is SD-preferred over another if for every utility function consistent withthe ordinalpreferences,the former yields atleastasmuchexpected utility asthe latter. DL is a refinementofSD and basedonlexicographiccomparisons betweenfractionalallocations. Generalizationsofthe PSrule havebeenrecom- mended in many settings [see e.g., 8]. The PS rule also satisfies some desirable incentive properties. If the number of objects is at most the number of agents, thenPSis weakSD-strategyproof[6]. Anotherwell-establishedrulerandom serial dictator (RSD) is notenvy-free,notas efficientas PS[6]andthe fractional allocations under RSD are #P-complete to compute [2]. However,unlike RSD, PS is not strategyproof. In this paper, we examine the following natural question for the first time: what is the computational complexity of an agent computing a different preference to report so as to get a better PS outcome? This problem of computing the optimal manipulation has already been studied in great depth for voting rules [see e.g., 11]. Ekici and Kesten [10] showed that when agents are not truthful, the outcome of PS may not satisfy desirable properties related to ef- ficiency and envy-freeness. Hence, it is important to check that even if agents caninprinciple manipulate, howhardit is tocompute a beneficialmisreportof their preferences. The complexity of manipulationof the PSrule is also related 2 to the study of Nash dynamics and better responses. Efficient algorithms to compute best responses can be used to understand Nash dynamics under the mechanism. Inordertocomparerandomallocations,anagentneedstoconsiderrelations between them. We consider three well-knownrelations between randomalloca- tions [see e.g., 6, 18, 17, 9]: (i) expected utility (EU), (ii) stochastic dominance (SD),and(iii)downward lexicographic (DL).ForEU,anagentseeksadifferent allocation that yields more expected utility. For DL, an agent seeks an allocation that gives a higher probability to the most preferred alternative that has differentprobabilitiesinthe twoallocations. Throughoutthe paper, we assume that agents express strict preferences, i.e., they are not indifferent between any two houses. Contributions. We initiate the study of computing best responses for the PS mechanism — one of the most established randomized rules for the assignment problem. The study is additionally motivated by complementing experimental work where we observe that as the number of houses relative to the number of agents grows, the percentage of manipulable profiles (for which atleastoneagenthasincentivetomanipulate)increases,maximizingataround 99%. We present a polynomial-time algorithm to compute the DL best response for multiple agents and houses. For the case of two agents, we present a polynomial-time algorithm to compute an EU best response for any utilities consistent with the ordinal preferences. The two-agent case is also of special importance since various disputes arise between two parties. The result for the EU best response relies on an interesting connection between the PS rule and the sequential allocation rule for indivisible objects. In a sequential allocation, a picking sequence is specified for the agents and agent get his most preferred available object when his turns comes. For general n, we show that computing an EU best response is NP-hard. The result contrasts sharply with the recentresultof BouveretandLang [7] that a best response canbe computed in polynomial time for sequential allocation. 2. Preliminaries Anassignment problem (N,H,≻)consistsofasetofagentsN ={1,...,n}, a set of houses H = {h ,...,h } and a preference profile ≻= (≻ ,...,≻ ) 1 m 1 n in which ≻ denotes a complete, transitive and strict ordering on H repre- i senting the preferences of agent i over the houses in H. A fractional assignment is an (n×m) matrix [p(i)(h )] such that for all i ∈ N, and j 1≤i≤n,1≤j≤m h ∈ H, 0 ≤ p(i)(h ) ≤ 1; and for all j ∈ {1,...,n}, p(i)(h ) = 1. j j Pi∈N j The value p(i)(h ) is the fraction of house h that agent i gets. Each row j j p(i)=(p(i)(h ),...,p(i)(h )) represents the allocation of agent i. A fractional 1 m assignment can also be interpreted as a random assignment where p(i)(h ) is j the probability of agent i getting house h . j AstandardmethodtocomparerandomallocationsistousetheSD(stochastic dominance) relation. Given two random assignments p and q, p(i)≻SD q(i) i i.e., a player i SD prefers allocation p(i) to q(i) if p(i)(h ) ≥ Phj∈{hk:hk≻ih} j 3 q(i)(h ) for all h ∈ H and p(i)(h ) > Phj∈{hk:hk≻ih} j Phj∈{hk:hk≻ih} j q(i)(h ) for some h∈H. Phj∈{hk:hk≻ih} j Given two random assignments p and q, p(i) ≻DL q(i) i.e., a player i i DL (downward lexicographic) prefers allocation p(i) to q(i) if p(i) 6= q(i) and for the most preferred house h such that p(i)(h) 6= q(i)(h), we have that p(i)(h)>q(i)(h). When agents are considered to have cardinal utilities for the objects, we denote by u (h) the utility that agenti gets from househ. We will assume that i the total utility of an agent equals the sum of the utilities that he gets from each of the houses. Given two random assignments p and q, p(i)≻EU q(i) i.e., i a player i EU (expected utility) prefers allocation p(i) to q(i) if u (h)· Ph∈H i p(i)(h)> u (h)·q(i)(h). Ph∈H i Since for all i ∈ N, agent i compares assignment p with assignment q only with respect to his allocations p(i) and q(i), we will sometimes abuse the notation and use p ≻SD q for p(i) ≻SD q(i). A random assignment rule takes as i i inputanassignmentproblem(N,H,≻)andreturnsarandomassignmentwhich specifies what fraction or probability of each house is allocated to each agent. 3. The Probabilistic Serial Rule and its Manipulation TheProbabilisticSerial(PS)rule isarandomassignmentalgorithminwhich weconsidereachhouseasinfinitely divisible[6,16]. Ateachpointintime,each agent is eating (consuming the probability mass of) his most preferred house thathasnotbeen completelyeatenandeachagenteatsatthe sameunit speed. Hence all the houses are eaten at time m/n and each agent receives a total of m/n units of houses. The probability of house h being allocated to i is the j fraction of house h that i has eaten. The following example adapted from j [Section 7, 6] shows how PS works. Example1 (PSrule). Consider anassignmentproblem withthefollowing preference profile. ≻ : h ,h ,h ≻ : h ,h ,h ≻ : h ,h ,h 1 1 2 3 2 2 1 3 3 2 3 1 Agents 2 and 3 start eating h simultaneously whereas agent 1 eats h . When 2 2 1 and 3 finish h , agent 1 has only eaten half of h . The timing of the eating can 2 1 be seen below. h h h Agent 1 1 1 3 h h h Agent 2 2 1 3 h h h Agent 3 2 3 3 0 1 3 1 2 4 Time The final allocation computed by PS is 3/4 0 1/4   PS(≻ ,≻ ,≻ )= 1/4 1/2 1/4 . 1 2 3  0 1/2 1/2 4 Consider the assignment problem in Example 1. If agent 1 misreports his preferences as follows: ≻′: h ,h ,h , then 1 2 1 3 1/2 1/3 1/6 PS(≻′,≻ ,≻ )=1/2 1/3 1/6. 1 2 3  0 1/3 2/3 Then, if u (h ) = 7, u (h ) = 6, and u (h ) = 0, then agent 1 gets more 1 1 1 2 1 3 expected utility when he reports ≻′. In the example, although truth-telling is 1 a DL best response, it is not necessarily an EU best response for agent 1. Examples1and2ofKojima[16]showthatmanipulatingthePSmechanism canleadtoanSDimprovementwheneachagentcanbeallocatedmorethanone house. Inlightofthe factthatthe PSrulecanbe manipulated,weexaminethe complexityofasingleagentcomputingamanipulation,inotherwords,thebest responseforthePSrule.1 Forapreferenceprofile≻,wedenoteby(≻ ,≻′)the −i i preference profile obtained from ≻ by replacing agent i’s preference by ≻′. For i E ∈ {SD,EU,DL}, we define the problem E-BR: Given (N,H,≻), compute a preference ≻′ for agent 1 such that there exists no preference ≻′′ such that 1 1 PS(N,H,(≻′′,≻ ))≻E PS(N,H,(≻′,≻ )). 1 −1 1 1 −1 Foraconstantm,the problemE-BRcanbe solvedbybrute forceby trying out each of the m! preferences. Hence we will not assume that m is a constant. We establish some more notation and terminology for the rest of the paper. WewilloftenrefertothePSoutcomesforpartiallistsofhousesandpreferences. WewilldenotebyPS(≻L,≻ )(i),theallocationthatagentireceiveswhenhis i −i preferenceisaccordingtoorderedlistL. Notethatpreferencesandorderedlists are interchangeable,except that a list need not contain all houses in H. When an agent runs out of houses in his preference list, he stops eating. The length ofa list L is denoted |L|,and we refer to the kth house in L as L(k). In the PS rule,the eating start time ofa houseisthe time pointatwhichthe housestarts to be eaten by some agent. In Example 1, the eating start times of h ,h and 1 2 h are 0,0 and 0.5, respectively. 3 4. Lexicographic best response In this section, we present a polynomial-time algorithm for DL-BR. Lexi- cographic preferences are well-establishedin the assignment literature [see e.g., 17, 18, 9]. Let (N,H,≻) be an assignment problem where N = {1,...,n} and H ={h ,...,h }. We will show how to compute a DL best response for agent 1 m 1 ∈ N. It has been shown that when m ≤ n, then truth-telling is the DL best response but if m>n, then this need not be the case [17, 18, 16]. Recallthatapreference≻′ isaDLbestresponseforagent1ifthefractional 1 allocation agent 1 receives by reporting ≻′ is DL preferred to any fractional 1 1Notethatifanagentisrisk-averseanddoesnothaveinformationabouttheotheragent’s preferences,thenhismaximinstrategyistobetruthful. Thereasonisthatifallagents have thesamepreferences,thentheoptimalstrategyistobetruthful. 5 allocationagent1 receives by reporting anotherpreference. Thatis, there is no preference ≻′′ such that his share of a house h when reporting ≻′′ is strictly 1 1 larger than when reporting ≻′ while the share of all houses he prefers to h 1 (according to his true preference ≻ ) is the same whether reporting ≻′ or ≻′′. 1 1 1 Our algorithm will iteratively construct a partial preference list for the i most preferred houses of agent 1. Without loss of generality, denote ≻ : 1 h ,h ,...,h . 1 2 m For any i,1 ≤ i ≤ m, denote H = {h ,...,h }. A preference of agent 1 i 1 i restricted to H is a preference over a subset of H . For the preference of agent i i 1 restricted to H , the PS rule computes an allocation where the preference of i agent 1 is replaced with this preference and the preferences of all other agents remainunchanged. ThenotionsofDLbestresponseandDLpreferredfractional assignments with respect to a subset of houses H are defined accordingly for i restricted preferences of agent 1. For a house h ∈ H, let PS1(L,h) = (PS(≻L,≻ )(1))(h) denote the frac- 1 −1 tionofhousehthatthePSruleassignstoagent1whenhereportsthe(partial) preference L. We start with a simple lemma showing that a DL best response foragent1 forthe whole setH canbe no better andno worseonH thana DL i best response for H . i Lemma 1. Let i∈{1,...,m}. A DL best response for agent 1 on H gives the same fractional assignment to the houses in H as a DL best response for agent i 1 on H . i Our algorithm will compute a list L such that L ⊆ H .2 The list L will be i i i i a DL best response for agent 1 with respect to H . Suppose the algorithm has i computed L . Then, when considering H = H ∪{h }, it needs to make i−1 i i−1 i sure that the new fractionalallocationrestrictedto the houses in H remains i−1 the same (due to Lemma 1). For the preference to be optimal with respect to H , the algorithm needs to maximize the fractional allocation of h to agent 1 i i under the previous constraint. Our algorithm will compute a canonical DL best response that has several additional properties. A preference L for H is no-0 if L contains no house h i i i withPS1(L ,h)=0. AnyDLbestresponseforagent1forH canbeconverted i i intoano-0DLbestresponsebyremovingthehousesforwhichagent1obtainsa fractionof0. Forano-0preferenceL forH ,thestingyordering forapositionj i i isdeterminedbyrunningthePSrulewiththepreferenceL (1)⊕···⊕L (j−1)for i i agent 1, where ⊕ denotes concatenation. It orders the houses from |Li|L (k) byincreasingeatingstarttimes,andwhentwohousesh,h′ havethesSamk=ejeatiing start time, we order h before h′ iff h ≻ h′. Intuitively, houses occurring early 1 in this ordering are the most threatened by the other agents at the time point when agent 1 comes to position j. The following definition takes into account that the eating start times of later houses may change depending on agent 1’s ordering of earlier houses. 2Whenwetreatalistasasetwerefertothesetofallelements occurringinthelist. 6 Algorithm 1 DL best response for n agents Input: (N,H,≻) Output: DLBestresponseofagent1 L1←h1 //Bestresponseforagent1w.r.t. H1={h1} fori=2tondo //Computeabestresponsew.r.t. H2,...,Hn p←0 if ∃q∈{1,...,i−1}suchthat0<PS1(Li−1,Li−1(q))<1then p←max{q∈{1,...,i−1}:0<PS1(Li−1,Li−1(q))<1} endif forq←p+1to|Li|+1do //Newhousehi insertedafterpositionp LwqiheSi←slet←L←|Li{−qihE|1∈S(≤1TL)|(L⊕iN−i−·1,·H1\·|,⊕Ld(qiLoLqi:i,−e≻s1t(2(q,h.−).i.1s,)≻m⊕ni/nh)/i)imCuomm}pletethelistaccordingtothestingyordering hs←firsthouseamongS in≻1 Lqi ←Lqi⊕hs endwhile if PLSqi1←(LLqi,i−hi1)=0then endif endfor q←p //DeterminewhichLq isstingy worse[p−1]←true i finished←false whilefinished=falsedo if ∃whor∈se[Hq]i−←1tsruucehthatPS1(Lqi,h)6=PS1(Li−1,h)then q←q+1 else worse[q]←false if PifSw1o(rLseqi[,qh−1)1>]=0faanlsdetPhSen1(Lqi,h1)<1then q←q−1 endif finished←true elsieefsit∃f←hP∈SE1{S(LTLqiqi((N,qh+,1H)1,=)(,L.1.qi.t(h1,L)e⊕nqi(|·L·qi·|⊕)}Lsqiu(cqh−th1a)t,≻es2t,(.h.).,≤≻ens)t)(hi)then q←q+1 else finished=true endif endif endif endwhile Li←Lqi endfor return Ln A preference L for H is stingy if it is a no-0 DL best response for agent i i 1 on H , and for every j ∈ {1,...,i}, L (j) is the first house in the stingy i i ordering for this position such that there exists a DL best response starting with L (1)⊕···⊕L (j). We note that, due to Lemma 1, there is a unique i i stingy preference for each H . i Example 2. Consider the following assignment problem. ≻ :h ,h ,h ,h ,h ,h ≻ :h ,h ,h ,h ,h ,h 1 1 2 3 4 5 6 2 3 6 4 5 1 2 The preferences h ,h ,h ,h and h ,h ,h ,h are both no-0 DL best responses 3 1 4 2 3 2 4 1 for agent 1 with respect to H , allocating p(1)(h )=1,p(1)(h )=1,p(1)(h )= 4 1 2 3 7 1/2,p(1)(h ) = 1/2 to agent 1. When running the PS rule with h as the 4 3 preference list, h ’s eating start time comes first among {h ,h ,h }. However, 4 1 2 4 there is no DL best response for H starting with h ,h . The next house in the 4 3 4 stingy ordering is h . The preference h ,h ,h ,h is the stingy preference for 1 3 1 4 2 H . 4 The next lemma shows that when agent 1 receives a house partially (a fraction different from 0 and 1) in a DL best response, a stingy preference would not order a less preferred house before that house. Lemma 2. Let L be a stingy preference for H . Suppose there is a h ∈ H i i j i such that 0<PS1(L ,h )<1. Then, P ⊆H , where L =P ⊕h ⊕S. i j j i j The next lemma shows how the houses allocated completely to agent 1 are ordered in a stingy preference. Lemma 3. Let L be a stingy preference for H . If h ,h ∈H are two houses i i j k i such that PS1(L ,h )=PS1(L ,h )=1, with L =P ⊕h ⊕M⊕h ⊕S, then i j i k i j k either the eating start time of h is smaller than h ’s eating start time when j k agent 1 reports P, or it is the same and h ≻ h . j 1 k Proof. Suppose not. But then, L is not stingy since swapping h and h in L i j k i gives the same fractional allocation to agent 1. WenowshowthatwheniteratingfromasetofhousesH toH ,the previous i−1 i solution can be reused up to the last house that agent 1 receives partially. Lemma4. LetL and L bestingypreferences for H andH , respectively. i−1 i i−1 i Suppose there is a h∈H such that 0<PS1(L ,h)<1. Then the prefixes i−1 i−1 of L and L coincide up to h. i−1 i We are now ready to describe how to obtain L from L . See Algorithm 1 i i−1 for the pseudocode. The subroutine EST(N,H,≻) executes the PS rule for (N,H,≻) and for each item, records the first time point where some agent starts eating it. It returns the eating start times est(h) for each house h∈H. Let p be the last position in L such that the house L (p) is partially i−1 i−1 allocated to agent 1. In case agent 1 receives no house partially, set p := 0 and interpret L (p) as an imaginary house before the first house of L . By i−1 i−1 Lemma 4, we have that L (s) = L (s) for all s ≤ p. By Lemma 1, we have i−1 i thatthe fractionalassignmentresultingfromL mustwhollyallocateallhouses i L (p+1),...,L (|L |) to agent 1, and allocate a share of 0 to all houses i−1 i−1 i−1 in H \L . i−1 i−1 It remains to find the right ordering for {L (s) : p+1 ≤ s ≤ |L |}∪ i−1 i−1 {h }. By Lemmas 2 and 3, the prefixes of L and L coincide up to h. We i i−1 i will describe in the next paragraph how to determine the position q where h i should be inserted. Having determined this position one may then need to re-order the subsequent houses. This is because inserting h in the list may i change the eating start times of the subsequent houses. This leads us to the following insertion procedure. The list Lq obtained from L by inserting h i i−1 i 8 at position q, with p < q ≤ |L | + 1, is determined as follows. Start with i Lq := L (1)⊕···⊕L (q −1)⊕h . While |Lq| ≤ |L |, we append to i i−1 i−1 i i i−1 the end of Lq the first house among L \Lq in the stingy ordering for this i i−1 i position. After the while-loop terminates, run the PS rule for the resulting list Lq. In case we obtain that PS1(Lq,h ) = 0, we remove h again from this list i i i i (and actually obtain Lq =L ). i i−1 The position q where h is inserted is determined as follows. Start with i q := p. We have an array worse keeping track of whether the lists Lp,...,Li i i produceaworseoutcomeforagent1thanthelistL . Setworse[p−1]:=true. i−1 As long as the list L has not been determined, proceed as follows. Obtain i Lq from L by inserting h at position q, as described earlier. Consider the i i−1 i allocation of agent 1 when he reports Lq. If this allocation is not the same for i the houses in H as when reporting L , then set worse[q]:=true, otherwise i−1 i−1 setworse[q]:=false. Ifworse[q],thenincrementq. Thisisbecause,byLemma1, thispreferencewouldnotbe aDLbestresponsewithrespecttoH . Otherwise, i if 0 < PS1(Lq,h ) < 1, then we can determine h ’s position. If worse[q−1], i i i then set L := Lq, otherwise set L := Lq−1. This position for h is optimal i i i i i since moving h later in the list woulddecrease its shareto agent1. Otherwise, i we have that worse[q] = false and PS1(Lq,h ) ∈ {0,1}. This will be the share i i agent 1 receives of h . If PS1(Lq,h ) = 0, then set L := L . Otherwise i i i i i−1 (PS1(Lq,h ) = 1), it still remains to check whether the current position for i i h gives a stingy preference. For this, run the PS rule with the preference i Lq(1)⊕···⊕Lq(q−1) for agent 1. If h ’s eating start time is smaller than the i i i eating start time of each house Lq(r) with r > q, then set L := Lq, otherwise i i i increment q. Thus, given L , the preference L can be computed by executing the PS i−1 i rule O(m) times. The DL best response computed by the algorithm is L . m Sincethe PSrulecanbe implementedtoruninlineartime O(nm), therunning time of this DL best response algorithm is O(nm3). Theorem 1. DL-BR can be solved in O(nm3) time. Example 3. Consider the following instance. ≻ :h ,h ,h ,h ,h ,h ,h ,h ,h ,h 1 1 2 3 4 5 6 7 8 9 10 ≻ :h ,h ,h ,h ,h ,h ,h ,h ,h ,h 2 8 3 5 2 10 1 6 7 4 9 ≻ :h ,h ,h ,h ,h ,h ,h ,h ,h ,h 3 9 4 7 1 2 6 5 3 8 10 After having computed L =h ,h , the algorithm is now to consider H . Since 2 1 2 3 PS1(L ,h ) = PS1(L ,h ) = 1, the algorithm first considers L1 = h ,h ,h . 2 1 2 2 3 3 2 1 Note that h and h have been swapped with respect to L since agent 2 starts 1 2 2 eating h before agent 3 starts eating h when agent 1 reports the preference 2 1 list consisting of only h . It turns out that PS1(L1,h ) = PS1(L1,h ) = 3 3 1 3 2 PS1(L1,h ) = 1. Thus, worse[1] = false. Since h does not come first in the 3 3 3 stingy ordering, the algorithm needs to verify whether moving h later will still 3 give a DL best response with respect to H . It then considers L2 = h ,h ,h . 3 3 1 3 2 However,thisallocatesonlyhalfofh toagent1,implyingworse[2]=true. Since 3 9 worse[1]=false, the algorithm sets L =L1. The DL best response computed by 3 3 the algorithm is L =h ,h ,h ,h . 10 3 2 1 6 We note thata DL bestresponse is alsoanSD best response. One maywonder whetheranalgorithmtocomputetheDLbestresponsealsoprovidesuswithan algorithm to compute an EU best response. However, a DL best response may not be an EU best response for three or more agents. Consider the preference profile in Example 1. Since the number of houses is equal to the number of agents, reporting the truthful preference is a DL best response [18]. However, we have shown a different preference for agent 1 where he may obtain higher utility. 5. Expected utility best response In this section, we consider the problem of expected utility best response. 5.1. Case of two agents We first show that for the case of two agents, an EU best response can be computed in linear time. The result hinges on a close connection that we identify between PS and discrete allocation of objects to agents via sequential allocation. Inthesequentialallocationsetting(N,O,≻′,π),thereisanagentset N,anobjectsetO ={o1,...om′},apreferenceprofile≻′ thatspecifiesforeach agent i∈ N his preferences ≻′ over O, and a policy π : {1,...,m′} →N. The i sequentialallocationruleworksasfollows. Startingfromj =1tom′,agentπ(j) getshismostpreferredobjectthatisnotyetallocated. Ifnounallocatedobject isonthepreferencelistoftheagent,thentheagentdoesnotgetanyobjectwhen his turn comes. The assignment as a result of sequential allocation is denoted bySA(N,O,≻′,π). We willrestrictourselvestothe casewhereN ={1,2}and will only consider the alternating policy π∗ = 1212... in which agent 1 starts firstandthentheagentskeepalternating. Thesequentialallocationsettingwas introducedby KohlerandChandrasekaran[15]wheretheyshowedthatthebest response can be computed in linear time when |N| = 2 and the policy is the alternating sequence. Recently, Bouveret and Lang [7] generalized their result tothecaseofanynumberofagents,anypolicy,andwherethemanipulatormay be indifferent between objects. We highlight a close connection between sequential allocation and PS and therebybetweenallocationmechanismsforindivisible anddivisible houses. For the random assignment setting ({1,2},H,≻), the half-house reduction gives us thesequentialallocationsetting({1,2},O,≻′,π∗). Inthereduction,eachhouse h ∈H is cloned so that we have two half-houses h1 and h2 for each house h : j j j j O={h1,h2:j =1,...,m}. Both agents have preferences over half-houses that j j areconsistentwith their preferencesoverhouses andforeachhouse,eachagent prefersthefirsthalf-houseslightlymorethanthesecondhalf-house: ifh ≻ h , j i k then h1 ≻′ h2 ≻′ h1 ≻′ h2. We show that for n = 2, the assignment under j i j i k i k PS is ‘essentially’ the same as the assignment obtained by applying sequential allocation to the setting resulting from the half-house reduction: 10