How To Calculate An Approxmaton Factor Of 1 1/E

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1 Approxmaton algorthms for allocaton problems: Improvng the factor of 1 1/e Urel Fege Mcrosoft Research Redmond, WA [email protected] Jan Vondrák Prnceton Unversty Prnceton, NJ [email protected] Abstract Combnatoral allocaton problems requre allocatng tems to players n a way that maxmzes the total utlty. Two such problems receved attenton recently, and were addressed usng the same lnear programmng LP) relaxaton. In the Maxmum Submodular Welfare SMW) problem, utlty functons of players are submodular, and for ths case Dobznsk and Schapra [SODA 2006] showed an approxmaton rato of 1 1/e. In the Generalzed Assgnment Problem GAP) utlty functons are lnear but players also have capacty constrants. GAP admts a 1 1/e)- approxmaton as well, as shown by Flescher, Goemans, Mrrokn and Svrdenko [SODA 2006]. In both cases, the approxmaton rato was n fact shown for a more general verson of the problem, for whch mprovng 1 1/e s NPhard. In ths paper, we show how to mprove the 1 1/e approxmaton rato, both for SMW and for GAP. A common theme n both mprovements s the use of a new and optmal Far Contenton Resoluton technque. However, each of the mprovements nvolves a dfferent roundng procedure for the above mentoned LP. In addton, we prove APX-hardness results for SMW such results were known for GAP). An mportant feature of our hardness results s that they apply even n very restrcted settngs, e.g. when every player has nonzero utlty only for a constant number of tems. 1 Introducton Allocaton problems. Combnatoral allocaton problems arse n stuatons such as combnatoral auctons, where tems or goods are to be allocated to competng players or bdders ) by a central authorty, wth the goal of maxmzng the total utlty provded to the players. In the work done whle at Mcrosoft Research most general settng, there s a fnte set I of m tems and n players who have possbly dfferent utlty functons w :2 I R, representng ther utlty derved from subsets of I. We always assume monotoncty: S T ; 0 = w ) w S) w T ). The goal s to fnd dsjont sets S 1,S 2,...,S n I to be allocated to the n players such that ther total utlty n =1 w S ) s maxmzed. In ths generalty, the problem s NP-hard to approxmate wthn any reasonable factor, such as m 1/2 ɛ even for sngle-mnded bdders [10, 11]. Postve results can be acheved only under stronger assumptons on the utlty functons. The followng propertes are commonly consdered, n the decreasng order of generalty: 1. Subaddtvty. w S T ) w S)+w T ). 2. Fractonal subaddtvty. w S) k α kw T k ), whenever 0 α k 1 and k:j T k α k 1 for each j S. As proved n [6], ths s equvalent to the XOS property : w S) = max t T f t S), where each f t s lnear. 3. Submodularty. w S T )+w S T ) w S)+ w T ). 4. Lnearty. w S) = j S w j. Another ssue s how the utlty functons are represented and accessble to an algorthm. An explct table of values would requre exponental sze, whle we would lke to acheve runnng tme polynomal n n and m. Compact representatons are possble n the case of lnearty and n other specal cases. In general, we assume an oracle model to access the utlty functon. In the weakest model, only a value oracle s avalable, whch returns the value of a gven set for a gven player. Followng [3, 4, 6], we use the noton of a demand oracle: Gven prces p j to ndvdual tems, the oracle returns a set maxmzng w S) j S p j. In ths paper, we consder three specal types of allocaton problems:

2 Maxmum Submodular Welfare SMW). The allocaton problem where utlty functons are monotone and submodular. We assume the avalablty of a demand oracle to query the utlty functons. Generalzed Assgnment Problem GAP). The allocaton problem wth lnear utlty functons, gven explctly by tem values w j. In addton, each player has a capacty constrant defnng feasble sets: j S s j 1. Here, s j s the sze of tem j, possbly dfferent for each player. Maxmum Fractonally Subaddtve Welfare FSAW). The allocaton problem wth fractonally subaddtve functons, accessble usng a demand oracle. Note. In the context of GAP, players are usually referred to as bns. Throughout ths paper, players and bns refer to the same enttes. We use bns only when talkng specfcally about GAP. Capacty constrants can be smulated usng a fractonally subaddtve functon: For each set S t feasble for player, defne f t S) = j S S t w j and w S) = max t f t S). Under ths utlty functon, t does not brng any advantage to allocate an nfeasble set S to player because we mght as well replace t by S t for some feasble S t S). Therefore, the FSAW problem wth utlty functons w S) = max t f t S) s equvalent to GAP. It s also known that submodular functons are fractonally subaddtve. I.e., both SMW and GAP are specal cases of FSAW. Prevous work. Allocaton problems have receved a lot of attenton recently. The Maxmum Submodular Welfare problem was consdered frst by Lehmann, Lehmann and Nsan [9] who presented a 1/2-approxmaton algorthm usng a value oracle only. In ths model, Khot et al. proved NP-hardness of approxmaton for any factor better than 1 1/e [8]. The factor of 1 1/e has been recently acheved by Dobznsk and Schapra [4], but n the stronger demand oracle model. For the more general class of subaddtve utlty functons, Fege [6] developed a 1/2-approxmaton algorthm, whch s optmal unless P = NP. NoAPX- hardness result was known for submodular utltes n the demand oracle model. Regardng the Generalzed Assgnment Problem, a 1/2- approxmaton algorthm was mplct n the work of Shmoys and Tardos [12], as observed by Chekur and Khanna [2]. The approxmaton factor has been recently mproved to 1 1/e by Flescher, Goemans, Mrrokn and Svrdenko [5]. Although the ttle of [5] mght suggest otherwse, t remaned an open queston whether 1 1/e s ndeed the optmal approxmaton factor for GAP. The authors n [5] proved hardness of 1 1/e + ɛ)-approxmaton for SAP, an assgnment problem wth more general feasblty constrants n partcular, for ts specal case, the Dstrbuted Cachng Problem). For GAP, only hardness of 1 ɛ)- approxmaton for some small ɛ>0 was known. Both 1 1/e)-approxmaton algorthms for GAP and SMW) use a smlar LP-based approach. The factor of 1 1/e arses because of the same reason: both algorthms use randomzed roundng wth ndependent choces for dfferent players. The 1 1/e)-approxmaton algorthm n [4] also apples to the more general FSAW problem, but only n an even stronger model usng an XOS oracle. In [6], t s shown that the XOS oracle s actually not needed and FSAW has a 1 1/e)-approxmaton usng only demand queres. Ths s optmal unless P = NP [3]. Stll, t was not known whether 1 1/e s optmal for GAP and SMW. Our results. Our man result s that the factor of 1 1/e s suboptmal for both GAP and SMW. In both cases, we develop randomzed approxmaton algorthms whch acheve an approxmaton factor of 1 1/e + ɛ for some absolute constant ɛ>0. Both algorthms are based on the same LP as used n [4, 6]. The roundng technques are, however, qute dfferent n the two cases. The ntutve reason why an mprovement s possble s that the 1 1/e)-approxmaton algorthm, usng ndependent choces for dfferent players, leaves an expected 1/e-fracton of all tems unclamed. These tems can be possbly allocated to some players, ncreasng ther utltes. However, two ssues arse: n the case of submodular utlty functons, t s not clear whether addtonal tems always brng addtonal proft and ndeed, sometmes they do not). In the case of GAP, t s not clear whether addtonal tems can be added to any player wthout volatng the capacty constrant. These two obstacles need to be treated dfferently; we explan the case of SMW n Secton 3 and the case of GAP n Secton 4. We prove that n both cases, the LP ntegralty gap s bounded away from 1 1/e. On the other hand, we present smple examples of ntegralty gap 5/6 for SMW and 4/5 for GAP see Secton 2.1). Far Contenton Resoluton. The frst step n both of our algorthms s a new technque to resolve conflcts between the random choces of dfferent players. We beleve that ths technque may be of ndependent nterest: Suppose that n players request an tem ndependently wth probabltes p 1,p 2,...,p n. Ths mght result n several players requestng the tem smultaneously. We show how to resolve contenton among the competng players, so that condtoned on any fxed player competng, she obtans the tem wth the same probablty 1 1 p ))/ p. Ths s optmal snce 1 1 p ) s the probablty that at least one player competes for the tem. Ths technque mples an approxmaton factor of 1 1 1/n) n for the FSAW problem wth n players. For detals, see Secton

3 Hardness of approxmaton. We prove that there s ɛ>0 such that t s NP-hard to approxmate the SMW problem wthn a factor of 1 ɛ. Our reducton has the property that each player s only nterested n a constant number of tems. We also show that 1 ɛ)-approxmaton s NP-hard n the case of n players wth the same utlty functon of polynomal sze, and n the case of two players whose utlty functons are separable, meanng that ws) = ws C ) where C are dsjont classes of constant sze. In all these cases, demand queres can be answered effcently, n contrast to prevously known reductons. More detals can be found n Secton 5. 2 The Confguraton LP The followng lnear program has been used to develop approxmaton algorthms for several allocaton problems [3,4,5,6]. LP = max j; ; S I x,s w S); S I x :j S,S 1, S I x,s 1, x,s 0. Here, I denotes the collecton of feasble sets for player. In the case of SMW, ths s the collecton of all sets of tems, whle n the case of GAP, I contans the sets respectng the capacty constrant for player. The varable x,s can be nterpreted as the extent to whch set S s allocated to player. For an nteger soluton x,s {0, 1}, the constrants requre each player to choose at most one set, and each tem to be allocated to at most one player. Ths LP has exponentally many varables but there s an optmal soluton wth polynomally many nonzero varables. We do not address the ssue of solvng the LP here. For allocaton problems wth a demand oracle, the LP can be solved snce the demand oracle provdes a separaton oracle for the dual [3]. For the GAP problem, the LP can be solved to an arbtrary precson see [5] for more detals). In ths work, we suppose a fractonal soluton s gven to us and we develop randomzed roundng technques to convert ths nto an ntegral soluton. 2.1 Integralty gaps We present two smple examples for ths LP, showng ntegralty gap 5/6 for SMW and 4/5 for GAP. The examples use only 2 players and 4 or 3 tems, respectvely. We can also prove that our gaps are the worst possble for 2 players wth a half-ntegral fractonal soluton; we do not present the proof here. Example for SMW, ntegralty gap 5/6. Consder 4 tems {a, b, c, d} and 2 players. Each sngleton has value 3 and every set of at least 3 elements has value 6. For pars of tems, defne the utlty functon of player 1 as follows: w 1 a, d) =w 1 b, c) =5 w 1 a, c) =4 w 1 b, d) =4 w 1 a, b) =6 a b w 1 c, d) =6 c d Symmetrcally, the utlty functon of player 2 s the same except that w 2 a, b) =w 2 c, d) =4and w 2 a, c) = w 2 b, d) = 6. Ths can be verfed to be a submodular functon. The optmal fractonal soluton s x 1,{a,b} = x 1,{c,d} = x 2,{a,c} = x 2,{b,d} =1/2 whch has value 12, whle any ntegral soluton has value at most 10. Example for GAP, ntegralty gap 4/5. Consder 3 tems {a, b, c} and 2 bns. The followng table shows the values and szes for the two bns: Item sze 1 value 1 sze 2 value 2 a b c The optmal fractonal soluton s x 1,{a,b} = x 1,{c} = 1/2 and x 2,{a} = x 2,{b,c} =1/2 whch has value 5. Any ntegral soluton has value at most The Far Roundng Algorthm We start by gvng an alternatve algorthm to acheve the factor of 1 1/e for the FSAW problem. Ths roundng technque combnes all the favorable propertes of prevous approaches: 1) It s effcent and oblvous,.e. t does not depend on the utlty functons. 2) It acheves an approxmaton factor of 1 1 1/n) n whch s strctly better than 1 1/e for any fxed number of players. 3) It beats 1 1/e even more sgnfcantly, f the fractonal soluton s unbalanced ; ths wll be useful later. 4) It treats all players equally, wth the same approxmaton guarantee per player. The frst step n a randomzed roundng algorthm s to nterpret the varables x,s as probabltes. Snce S x,s 1, ths s a vald probablty dstrbuton for each player. We say that a player samples a random set from her dstrbuton, f S s chosen wth probablty x,s. Observe that E[w S)] = S x,sw S) s exactly the player s share n the LP. The Far Roundng Algorthm. 1. Tentatve Choces. Let each player sample ndependently a random tentatve set S from her dstrbuton. Each player competes for the tems n her tentatve set. 3

4 2. Far Contenton Resoluton. Consder an tem j. Denote by A j the random set of players competng for tem j. Letp j =Pr[ A j ]= S I :j S x,s and s j = A j p j / n =1 p j. If A j =, do not allocate the tem. If A j = {k}, allocate the tem to player k. If A j > 1, choose randomly one of two contenton resoluton schemes: a) wth probablty 1 s j, b) wth probablty s j. a) Allocate tem j to a unformly random player k A j. b) Allocate tem j to a random player k A j, wth probabltes proportonal to A p j\{k} j. Remark. The unform contenton resoluton scheme a) has been prevously consdered [6] and t s known that t acheves factor 1/2 for fractonally subaddtve functons. In order to mprove ths, t s necessary to penalze players who request the tem wth hgh probablty, whch s the purpose of scheme b). Lemma 1. Let players compete for an tem ndependently wth probabltes p 1,p 2,...,p n. Condtoned on player k competng, the Far Contenton Resoluton technque allocates t to her wth probablty ρ = 1 n =1 1 p ) n =1 p. Proof. Condton on A beng the set of players competng for the tem and assume A > 1. In scheme a), each player n A obtans the tem wth probablty 1/ A. In b), we get by normalzng that the probablty for player k s 1/ A 1) A\{k} p / A p. By averagng these two schemes wth weghts 1 s and s, where s = A p / n =1 p, we obtan that player k gets the tem wth probablty r A,k = 1 n =1 p A\{k} p p A 1 + 1) A / A condtoned on the set of competng players A. Suppose for now that we allocate the tem to each player k A wth probablty r A,k, even when A = {k}. For the sake of the proof, we nterpret the sum over A \{k} for A = {k} as zero, even though the summand s undefned.) Then the total probablty that player k receves the tem would be q k = p k E A [r A,k k A]. 2) However, when A = {k}, our technque actually allocates the tem to player k wth probablty 1, rather than k r {k},k = p n =1 p =1 p k n =1 p. p So player k gans an addtonal probablty k n Pr[A = =1 p {k}] whch makes the total probablty that player k obtans the tem equal to q k = q k + p k n =1 p Pr[A = {k}]. 3) We would lke to show that q k = p k p Pr[A ]. Ths means that q k should be equal to p k p Pr[A \{k} ]. Let s defne B =[n] \{k} and let A = A \{k} be the set of players competng wth k. The probablty of a partcular set A occurrng s pa ) = A p B\A 1 p ). Let s expand 2) as a weghted sum over all possble subsets A B: q k = p k A B pa ) r A {k},k = p k n A =1 p B pa p ) A A + p B\A A +1 Ideally, we would lke to get q k = p k p A B pa ) but we have to perform a certan redstrbuton of terms to acheve ths. Observe that p pa ) A = 1 p pa {}) +1 p A +1 = pa 1 p {}) A {}. Usng ths dentty to replace the terms for B \ A,we get q k p k = n =1 p pa ) p A A B A + pa 1 p ) {}) A {} A B B\A p k = n =1 p pa ) p A A B A + pa ) 1 p ) A A B A and jonng the terms where A = A, we get q k = p k n A =1 p B pa p ) A A + ) 1 p A A = p k n A =1 p B pa )= p k n Pr[A \{k} ]. =1 p So ndeed, we obtan from 3) that player k receves the tem wth probablty q k = p k n Pr[A ] = p k 1 n n =1 p =1 p =1 1 p )). p ). 4

5 In the case of two players who compete for an tem wth probabltes p 1 + p 2 =1, ths algorthm resolves a possble contenton by allocatng the tem wth reversed probabltes: p 2 for player 1 and p 1 for player 2. Ths acheves an approxmaton factor of 3/4 for two players, even for fractonally subaddtve utltes, as observed already n [6]. For n players, ths technque guarantees an approxmaton factor of 1 1 1/n) n > 1 1/e. Ths follows just lke n [6] from the fact that each player receves every requested tem wth condtonal probablty ρ 1 1 1/n) n. Ths matches exactly the known ntegralty gap for the FSAW problem - for fractonally subaddtve utltes, ths technque cannot be mproved. However, our man goal s to obtan an absolute constant larger than 1 1/e, ndependent of n, forthesmw and GAP problems. 2.3 Unbalanced fractonal solutons Observe that the Far Contenton Resoluton technque actually gves a value of ρ better than 1 1 1/n) n,f some players take the tem wth a large probablty p j = S I x :j S,S. Let s fx an ɛ 1 > 0 and call an tem jɛ 1 - unbalanced f p j >ɛ 1 for some, orf p j < 1 ɛ 1. An elementary estmate gves that n both cases, each player receves tem j wth condtonal probablty ρ j = 1 n =1 1 p j) n =1 p 1 j ɛ2 1 ) 1 e. If a substantal fracton of the LP value comes from unbalanced tems, then we gan compared to 1 1/e. However, we have to defne carefully what we mean by the contrbuton of an tem n the case of submodular utlty functons. Defnton 2. Fx an orderng of the tems [m] = {1, 2,...,m} and denote [j] ={1, 2,...,j}. Defne the expected contrbuton of tem j to player as σ j = E[w S [j]) w S [j 1])] where S s a random set sampled from player s dstrbuton. Note that j σ j = E[w S )] and,j σ j = LP.Ths way of parttonng the LP value s useful because of the followng lemma. Lemma 3. Let S be a random set sampled from player s dstrbuton, and let X be a random set such that condtoned on any S and for any tem j, Pr[j X S ] ρ j. Assume w monotone and submodular. Then takng the expectaton over both S and X, E[w S X)] m ρ j σ j. j=1 Proof. Usng the property of decreasng margnal values, we obtan m w S X) = w S X [j]) w S X [j 1])) j=1 m w S [j 1] X {j}))) w S [j 1])). j=1 Condtoned on S, j appears n X wth probablty at least ρ j, so takng expectaton over X yelds E X [w S X) S ] m ρ j w S [j]) w S [j 1])) j=1 and therefore E S,X[w S X)] m j=1 ρ jσ j. Note. Ths apples to GAP even more easly, snce tems have ndvdual values w j and we have σ j = p j w j. Now we can argue that we gan f ɛ 1 -unbalanced tems contrbute a sgnfcant fracton of the LP value. As a consequence, we can assume that there are no ɛ 1 -unbalanced tems; we call such a fractonal soluton ɛ 1 -balanced. Lemma 4. Suppose that for any ɛ 1 -balanced fractonal soluton, there s a roundng technque for ether SMW or GAP) whch acheves an approxmaton factor of 1 1/e+ɛ, 0 < ɛ < 1/2e). Then there s a roundng technque for any fractonal soluton whch acheves factor at least 1 1/e + ɛɛ 2 1/2e). Proof. Let U denote the set of ɛ 1 -unbalanced tems. Frst suppose that j U σ j ɛ LP and apply the Far Roundng Algorthm. For each unbalanced tem j U, each player competng for t receves t wth condtonal probablty at least 1 1/e + ɛ 2 1/2e). By Lemma 3, the expected total proft from ths roundng s at least ρ j σ j 1 1 ) σ j + ɛ2 1 σ j e 2e j 1 1 e + ɛɛ2 1 2e j ) LP. j U Otherwse, we have j U σ j <ɛ LP. Then let s remove the unbalanced tems. In the worst case ths ncurs a factor of 1 ɛ) on the value of the fractonal soluton. By assumpton, there s a technque to be shown later) whch we can apply to the balanced soluton, that acheves approxmaton factor at least 1 1/e+ɛ)1 ɛ) = 1 1/e + ɛ/e ɛ 2 whch s at least 1 1/e + ɛ/2e) for ɛ<1/2e). The arguments so far were equally vald for SMW and GAP. We can assume now that the fractonal soluton s ɛ 1 - balanced. Applyng the Far Roundng Algorthm or n fact 5

6 any roundng procedure where every player samples ndependently a random set and contenton s resolved n some way), we see that tem j remans unclamed wth probablty n =1 1 p j) 1/e. On the average, 1/e of all tems are avalable to be allocated n a second stage, whch gves some hope that an mprovement over 1 1/e should be possble. Somewhat surprsngly, there exst nstances where the remanng tems do not brng any addtonal proft for SMW) or they do not ft for GAP), regardless of how we assgn them or how conflcts were resolved n the frst stage. Therefore we must redesgn even the frst stage of the algorthm n order to acheve some mprovement. Ths s the pont where the two solutons dverge. 3 Maxmum Submodular Welfare Our general dea s that each player should sample multple sets ndependently from ther dstrbuton. However, contenton must be resolved carefully among these sets, so that we can provably ncrease the total welfare. Our fnal roundng technque s qute complcated; t s nstructve to descrbe t frst on the example of two players. 3.1 Two players wth a balanced fractonal soluton Assume that for any tem j, S:j S x 1,S = T :j T x 2,T = 1/2. Ths s the worst case for the Far Roundng Algorthm, n whch case the approxmaton factor s exactly 3/4. We show that n fact we can mprove upon 3/4 n ths specal case. Algorthm for two players wth submodular utltes and a balanced fractonal soluton. Player 1 samples ndependently random sets S, S. Player 2 samples ndependently random sets T,T. Let Y =S T ) S T ),Z =T S ) T S ). We assgn tems randomly as follows: Probablty 1/3 1/3 1/6 1/6 Player 1 S T Y Y \ Z Player 2 S T Z \ Y Z Theorem 5. The algorthm for two balanced players gves expected proft at least 37/48 LP. Proof. We only sketch the mportant arguments. Note that each of the random sets S, T, S,T,Y,Z contans every tem wth probablty 1/2. For player 1, let α = E[w 1 S)] be hs share n the LP, whle E[w 1 T )] s what s left after player 2 chooses her set T frst. Snce every tem remans n T wth prob. 1/2, by monotoncty and Lemma 3 wth X = T and ρ j =1/2 we get E[w 1 T )] E[w 1 S T )] α/2. If the nequaltes were tght, player 1 would get only 1/2 of hs share, followng player 2 s choce. Randomzng the orderng of the two players, ths would yeld a factor of 3/4. However, at ths moment the sets Y and Z come nto play. For Y, the condton of submodularty mples that E[w 1 Y )] + E[w 1 T )] E[w 1 Y T )] + E[w 1 Y T )] E[w 1 S)] + E[w 1 S T )] 3 2 α usng lnearty of expectaton and Lemma 3. I.e., f E[w 1 T )] s close to α/2, then E[w 1 Y )] α. Thesame holds for Z by the same reasonng for player 2. Thus n ths case, the sets Y,Z retan the complete value of the LP. Moreover, they are not ndependent lke S and T.The events j Y,j Z are negatvely correlated: Whle Pr[j Y ]=Pr[j Z ]=1/2, Pr[j Y Z ]=Pr[j S T ) S T )] = 3 16 rather than 1/4 whch s the probablty of appearance n S T. Thus we lose only 3/16 of the expected value by resolvng contenton between Y and Z. A detaled computaton yelds that the total expected proft from ths roundng technque s at least 37/48 LP. Smlarly to Lemma 4, ths mples an mprovement over 3/4 for two players n the general case. Let s remark that by a more careful analyss, we are able to combne the balanced and unbalanced case to acheve an approxmaton factor of 13/17 for two players. 3.2 n players wth a balanced fractonal soluton We adapt the deas from the two-player case to acheve a constant mprovement over 1 1/e for any number of players. Our approach s to dvde the players randomly nto two groups A, B and treat them as two superplayers. Let s assume agan that the fractonal soluton s ɛ 1 -balanced. Due to a concentraton result on sums of ndependent varables, we have for most tems j p j p j 1 2 ± O ɛ 1 ). A B For a collecton of sets {S : A}sampled by players n one group, we use Lemma 1 to resolve contenton. To dstrbute tems between the two groups, we use deas nspred by the two-player case. 6

7 Algorthm for n balanced submodular players. Let each player n group A sample two ndependent random sets S,S. Let each player n group B sample two ndependent random sets T,T. Let U = A S, U = A S, V = B T, V = B T. Let the players n A resolve contenton among S to obtan dsjont sets S. Smlarly, resolve contenton among S to obtan dsjont sets S. Let the players n B resolve contenton among T to obtan dsjont sets T and among T to obtan dsjont sets T. Let Y =S V ) S V ), Z =T U) T U), Y =S V ) S V ) and Z = T U ) T U ). We assgn the tems randomly accordng the followng table, wth p =1/1 + 2e 1/2 ): Prob. Player A Player B e 1/2 p S T T \ V )) \ U e 1/2 p S S \ U)) \ V T p/2 Y Z \ j A Y j p/2 Y \ j B Z j Z Theorem 6. For n players wth an ɛ 1 -balanced fractonal soluton, the algorthm above yelds expected proft at least 1 1/e +1/100 O ɛ 1 )) LP. We omt the proof from ths extended abstract. Due to Lemma 4, ths mples a 1 1/e + ɛ)-approxmaton for any fractonal soluton. 4 Generalzed Assgnment Problem Now we turn to the Generalzed Assgnment Problem GAP). Recall that n GAP, each tem j has an explct value w j for bn. The added complcaton s that only sets satsfyng the capacty constrant j S s j 1 can be allocated to bn. As a result, our approach for the SMW problem does not work here, snce we cannot pack two sets S,S nto the same bn. Instead, we proceed as follows. Recall that we have to deal only wth the case of ɛ 1 -balanced fractonal solutons, for some very small constant ɛ 1 > 0. Also, we wll show that we may further assume that tem values are unform, and there are no precous sets these terms wll be defned shortly). Thereafter, we wll be able to prove that t s possble to mprove the Far Roundng Algorthm by reallocatng certan tems and packng some addtonal tems. Non-unform tem values. Frst, we treat the case where tem values vary sgnfcantly between dfferent bns. We know that the Far Roundng Algorthm yelds a factor of at least 1 1/e. However, snce now the utlty functons are lnear, there s a smpler way to resolve conflcts - just allocate the tem to the bn that gves t the maxmum value w j. Ths s the technque used n [5]; let s call t the Greedy Roundng Algorthm. We show that f the tem values are sgnfcantly non-unform then Greedy Roundng gans sgnfcantly compared to 1 1/e. The followng lemma follows easly by comparng the two roundng methods. Lemma 7. Fx ɛ 2 > 0. For each tem j, let W j = p jw j and B j = { : w j < 1 ɛ 2 )W j }. Call the value of tem j non-unform f B j p j >ɛ 2. For any non-unform tem, the Greedy Roundng Algorthm retreves expected value at least 1 1/e +ɛ 2 /e) 2 )W j. We nterpret the set B j as bad bns for tem j. The meanng of ths lemma s that f many bns are bad for tem j, then we gan by placng t n the optmal bn rather than a bad bn. We choose ɛ 2 > 0 such that ɛ 1 = e 2 ɛ 3 2.Ifat least an ɛ 2 -fracton of the LP value comes from non-unform tems, we gan e 2 ɛ 3 2LP = ɛ 1 LP. If the contrbuton of non-unform tems s smaller than ɛ 2, we remove each tem from all ts bad bns, and we remove all non-unform value tems completely. We redefne the value of tem j to 1 ɛ 2 )W j. Ths decreases the value of the LP by at most 3ɛ 2 LP and after ths procedure, each tem has the same value for each bn where t s used and the new value s not hgher than the orgnal value). Precous sets. Now we can assume that the value of each tem s ndependent of the bn where t s placed, w j = w j. Next, we consder the dstrbuton of values among dfferent sets for a gven bn. The average value assgned to bn s V = S x,sws) = j p jw j. We call a set S precous for bn f ws) > 10V. We prove that f many sets are precous, we can gan by takng these sets wth hgher probablty. We set ɛ 3 = 400ɛ 2. The followng lemma can be proved by ncreasng the samplng probablty of each precous set by a factor of 2 and decreasng the samplng probabltes of non-precous sets. Lemma 8. Assume that precous sets contrbute value at least ɛ 3 LP. Then there s an algorthm whch acheves expected value at least 1 1/e + ɛ 3 /100) LP. Agan, we ether gan due to ths lemma, or we remove all precous sets from the LP. If most of a bn s value s composed of precous sets, we remove the bn from the soluton completely. For other bns, the expected value V mght go down by a factor of 2, but overall we lose at most 2ɛ 3 LP and any set possbly allocated to bn has value at most 20V. 7

8 Mgrants. Now the tem values are unform and there are no precous sets. Our fnal goal s to pack some addtonal tems nto the gaps created by contenton among bns. We process the bns n the order of V 1 V 2 V 3... and resolve contenton n favor of the frst bn that clams an tem n ths order. We call ths the Sequental Allocaton Algorthm. Ths would stll acheve an approxmaton factor of 1 1/e. However, we prove that ths algorthm can be mproved, snce some tems are gone due to contenton wth precedng bns and some addtonal tems can be packed n the avalable space. We refer to tems useful for ths purpose as mgrants. Defnton 9. Fx an ɛ 4 > 0. Amgrant for bn k s an tem j such that k 1 =1 p j 1 ɛ 4. We choose ɛ 4 such that ɛ 3 << ɛ 3 4 but ɛ 4 s stll very small. Therefore, at the moment t s consdered for bn k,a mgrant s avalable wth probablty roughly 1/e. Thsm- poses constrants on the other ɛ s and consequently our fnal mprovement ɛ 3 1). We can set for example ɛ 4 =0.001 whch mples ɛ and then our fnal mprovement s on the order of Note that beng mgrant s only defned wth respect to a certan bn. Every tem s mgrant for some bns, snce n =1 p j > 1 ɛ 1 balanced fractonal soluton). Therefore, we can assume that an Ωɛ 4 )-fracton of the LP value comes from mgrants. Consequently, at least an Ωɛ 4 )-fracton of bns by LP contrbutons) have an Ωɛ 4 )- fracton of ther value n mgrants. Let s call these bns flexble. We wll prove that each of them can gan a constant fracton of ts LP share n addton to what t would get under the Sequental Allocaton Algorthm. For each set S whch could be assgned to bn k, let s defne M k S) to be the mgrants for bn k n S. We wll drop the ndex when t s clear whch bn we are referrng to. We can assume that the value of these tems s ether wm k S)) = Ωɛ 4 )V k or M k S) =. For sets wth mgrants of value less than Ωɛ 4 )V k,wesetm k S) = ; ths decreases the contrbuton of mgrants only by a constant factor.) Let s call the sets wth nonempty M k S) flexble for bn k. Wehave x k,s wm k S)) = Ωɛ 4 )V k S F k where F k s the collecton of flexble sets for bn k. Suppose that the probablty of samplng a flexble set for bn k s S F k x k,s = r k. Note that snce mgrants should contrbute at least an Ωɛ 4 )- fracton of the bn s value V k, ths probablty must be r k =Ωɛ 4 ). Here, we use the fact that there are no sets of value sgnfcantly exceedng V k. The probablty of samplng each ndvdual set s at most ɛ 1 << ɛ 4 whch allows us to assume that we can splt the collecton of flexble sets nto three roughly) equal parts n terms of probablty. Let s select a porton of the flexble sets S F k wth the largest szes of MS) and denote them by A k. For each set S A k, order the remanng flexble sets S F k \A k by the sze of MS )\MS), and partton them nto two parts B k S), C k S), so that for any S B k S),S C k S),we have sze k MS )\MS)) sze k MS )\MS)). We make these parts roughly equal so that x k,s x k,s x k,s r k 4. S A k S B k S) S C k S) The purpose of ths parttonng s that we wll be able to swtch mgrants between sets: any MS ) for S B k S) fts nto the space occuped by MS),S A k, and any MS ) \ MS) for S C k S) fts nto the space of MS ) \ MS) for S B k S). Ths allows a scheme of swtchng mgrants that acheves an approxmaton factor strctly better than 1 1/e. The Improved Sequental Allocaton Algorthm. Order the bns by V 1 V 2 V 3...For bn k, let U k denote the tems allocated to prevous bns. Let bn k sample a random set S k from ts dstrbuton. If k s a flexble bn, sample two more sets S k,s k. If S k A k and S k B ks k ), pack S k MS k ))\U k f possble nto bn k. If S k A k, S k B k S k ), S k C k S k ) and S k MS k)) \ U k fts n bn k as n the prevous case but wth the roles of S k and S k exchanged), then pack S k \MS k )\MS k )) MS k )\MS k )))\U k nto bn k. Ths set s guaranteed to ft because sze k MS k ) \ MS k )) sze kms k ) \ MS k )) due to the propertes of B k S k ) and C ks k ). In all other cases, allocate S k \ U k to bn k and go to the next bn. Theorem 10. For any balanced fractonal soluton where tem values are unform and there are no precous sets, the Improved Sequental Allocaton Algorthm acheves expected value at least 1 1/e +Ωɛ 5 4)) LP. We only sketch the proof here. Consder a flexble bn k. For each par of sets S k A k, S k B ks), we calculate the expected sze of MS k ) MS k )) \ U k by averagng over the prevously allocated tems U k. Snce the sze of MS k ) s at most the sze of MS k), and each mgrant appears n U k wth probablty at least 1 e 1 ɛ4),wehave E Uk [sze k MS k ) MS k)) \ U k )] 8

9 e 1 ɛ4) sze k MS k ) MS k)) 4 5 sze kms k )). By Markov s nequalty, we get Pr[sze k S k MS k)) \ U k ) 1] 1 U k 5. Thus for each fxed par S k,s k ) lke ths, the algorthm succeeds n packng S k MS k )) \ U k wth constant probablty. We call U k favorable for the par S k,s k ) f ths occurs. Now consder what happens when the same par of sets s selected by the bn n the opposte order,.e. the roles of S k and S k are swtched. Now, S k A k and S k B k S k ). Suppose U k s favorable for S k,s k) and n the prevous case, we managed to pack MS k ) \ MS k )) \ U k n addton to what the standard algorthm would have packed. Now when the algorthm samples S k,s k ), we can afford to gve up ths set and take only S k \ MS k ) \ MS k ))) \ U k. By averagng the two cases, ths would brng us back to a random allocaton equvalent to the standard algorthm. In addton, however, our mproved algorthm allocates MS k ) \ MS k )) \ U k. Ths happens whenever S k A k, S k B k S k ), S k C ks k ) and U k s favorable for the par S k,s k). Snce the set S k C ks k ) s sampled ndependently of everythng else, we wll gan a defnte mprovement f there s at least a constant fracton of C k S k ) stll avalable for bn k. For ths purpose, we use the followng concentraton result. Lemma 11. Let A be a fnte set wth a weght functon f : A R +, fa) =1; α, λ > 0 and 0 <ɛ<1/2. Let S 1,S 2,...,S q be ndependent random subsets of A where Pr[j S ]=p j and ; j A; p j ɛ. j A; q =1 p j 1. We always have fs ) α. Then [ q ) ] Pr f S > 1 e 1+ɛ) + λ =1 < α + Oɛ) λ 2. Ths lemma can be proved usng the second moment and Chebyshev s nequalty. We apply t to a functon f whch represents the contrbuton of mgrants n C k S k ) to bn k, normalzed to sum up to 1. The sets S 1,S 2,... are the sets of tems allocated to precedng bns. The crucal pont here s that any set S allocated to bn < k has value at most 20V 20V k ; snce V k = j p kjw kj and p kj ɛ 1, the loss of LP value for bn k due to tems n S can be at most 20ɛ 1 V k. Ths s very small even compared to the value of mgrants n C k S k ) whch s Ωɛ 4V k ). Therefore, Lemma 11 mples that a constant fracton of the value n mgrants s stll avalable for bn k, say wth probablty 9/10 wth respect to U k ). Even f ths event s negatvely correlated wth the event that U k s favorable for S k,s k), whch happens wth probablty at least 1/5, there s probablty at least 1/10 that both events occur. Thus each flexble bn k gans a constant fracton of ts value V k. 5 Hardness of approxmaton for SMW In ths secton we show that there s some constant ɛ>0 such that t s NP-hard to approxmate the Maxmum Submodular Welfare problem wthn a rato better than 1 ɛ, even usng a demand oracle. Prevously t was shown n [8] that the maxmum submodular welfare problem s hard to approxmate wthn a rato better than 1 1/e; however, there the source of the hardness result s the complexty of ndvdual utlty functons: gven k, t s already NP-hard to approxmate wthn a rato better than 1 1/e the maxmum utlty that a sngle player can derve by choosng at most k tems. In partcular, t s NP-hard for players to answer demand queres. We remark that a powerful oracle model can bypass prevous hardness results; e.g., there s a polynomal tme ncentve compatble mechansm based on far dvson queres that extracts the maxmum welfare, provded only that all players have the same utlty functon submodular or not). Our new hardness results are the followng. Theorem 12. There s some constant ɛ>0 such that t s NP-hard to approxmate the Maxmum Submodular Welfare problem wthn a rato better than 1 ɛ n the followng cases. 1. When all players have constant sze utlty functons. In partcular, each player gets nonzero utlty only from 15 specfc tems. 2. When all players have the same utlty functon. Moreover, the utlty functon s constant for sets of sze more than 7. The value of ɛ here s 1/12 23). 3. When there are only two players. Moreover, ther utlty functons are separable;.e., the tems can be parttoned nto dsjont classes C j of constant sze such that w S) = j w S C j ). In all these cases, demand queres can be answered effcently. The proofs are based on reductons from Max 3- colorng-5 and Max k-cover. We present only the proof of the frst hardness result. We beleve that the use of constant sze utlty functons s the most natural way to rule out a possble approxmaton result even n a very strong oracle model. It s hard to magne any reasonable query that would be dffcult to answer regardng such a utlty functon, and hence queres wll not transfer the computatonal burden to the players. 9

10 Proof. We use the followng NP-hardness result [7]: There s an ɛ > 0 such that gven a 5-regular graph, t s NPhard to dstngush between the case where ts vertces can be legally 3-colored, and the case where every 3-colorng makes an ɛ-fracton of edges llegally colored. Gven a 5-regular graph wth n vertces and m edges hence 2m =5n) we reduce t to the followng SMW nstance. Wth every edge e we assocate three tems, e 1, e 2 and e 3, correspondng to the three colors {1, 2, 3}. Hence there are 3m tems. There wll be m edge players, one for every edge, and n vertex players, one for every vertex. The utlty functon of the player p e who s assocated wth edge e gves the player utlty 1 f she receves at least one of the three tems assocated wth the edge, and utlty 0 otherwse. The utlty functon of the player p v who s assocated wth vertex v s nonzero on 15 tems, the tems assocated wth the edges ncdent wth v. There are three specal sets of sze 5: the monochromatc subsets of these 15 tems. Each specal set has value 5, whle any other set of sze 5 has value 4.5. Anysetofszeb<5 has value b, and any set of sze b>5 has value 5. These utlty functons are submodular. On postve nstances, we can legally 3-color the graph. Then each vertex player gets the fve tems assocated wth her chosen color. Each edge player can get the unallocated tem assocated wth her edge. Altogether, the total welfare s 3m all tems are allocated and gve utlty 1 per tem), and all players are maxmally happy. On negatve nstances, we use the followng analyss. Wthout loss of generalty, we may assume that every edge player gets one tem because then the tem contrbutes margnal utlty 1, and there s no way t can contrbute more). Lkewse, we may assume that every vertex player gets exactly 5 tems, one from every ncdent edge otherwse a shftng argument would yeld an allocaton wth at least as much welfare). We defne a vertex colorng derved from the colors of these tems. For vertex players whose tems are all of the same color, assgn the same color to the vertex. Such players are maxmally happy. Assume that k vertex players are unhappy n the sense that ther 5 tems are not monochromatc. For such vertces, choose a majorty color among the 5 tems and assgn t to the vertex. For each unhappy vertex player, there can be up to 3 edges wth an llegal colorng. For all other edges, the colorng of the two endponts s derved from two tems of dfferent colors, therefore ther colorng s legal. The number of llegally colored edges s at least ɛm and at most 3k,.e. 3k ɛm. Each unhappy player loses value 1/2, hence the maxmum welfare s at most 3m ɛm/6, showng that maxmum submodular welfare cannot be approxmated wthn a factor better than 1 ɛ/18. References [1] Steven J. Brams, Alan D. Taylor. Far Dvson: From cake-cuttng to dspute resoluton. Cambrdge Unversty Press, [2] Chandra Chekur and Sanjeev Khanna. A PTAS for the multple knapsack problem. Proceedngs of SODA 2000: [3] Shahar Dobznsk, Noam Nsan and Mchael Schapra. Approxmaton algorthms for combnatoral auctons wth complement-free bdders. Proceedngs of STOC 2005: [4] Shahar Dobznsk and Mchael Schapra. An mproved approxmaton algorthm for combnatoral auctons wth submodular bdders. Proceedngs of SODA 2006: [5] Lsa Flescher, Mchel X. Goemans, Vahab Mrrokn and Maxm Svrdenko. Tght approxmaton algorthms for maxmum general assgnment problems. Proceedngs of SODA 2006: [6] Urel Fege. On maxmzng welfare when utlty functons are subaddtve. Proceedngs of STOC 2006: [7] Urel Fege, Magnus M. Halldorsson, Guy Kortsarz and Aravnd Srnvasan. Approxmatng the domatc number. SIAM J. Comput. 321): ). [8] Subhash Khot, Rchard Lpton, Evangelos Markaks and Aranyak Mehta. Inapproxmablty results for combnatoral auctons wth submodular utlty functons. Proceedngs of WINE [9] Benny Lehmann, Danel J. Lehmann and Noam Nsan. Combnatoral auctons wth decreasng margnal utltes. In ACM Conference on Electronc Commerce 2001: [10] Danel J. Lehmann, Ladan O Callaghan and Yoav Shoham. Truth revelaton n approxmately effcent combnatoral auctons. In ACM Conference on Electronc Commerce [11] Tuomas Sandholm. An algorthm for optmal wnner determnaton n combnatoral auctons. Proceedngs of IJCAI [12] Davd Shmoys and Eva Tardos. An approxmaton algorthm for the generalzed assgnment problem. Math. Programmng 623), ). 10