Adjacent non-manipulability and strategy-proofness in voting domains: equivalence results

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1 Adjacent non-manpulablty and strategy-proofness n votng domans: equvalence results Souvk Roy, Arunava Sen, Sonal Yadav and Huaxa Zeng February 14, 2016 Abstract A plausble ncentve compatblty requrement based on behavoral consderatons, s that agents only consder manpulatons that are close to ther true preference. In a fnte votng envronment, ths s nterpreted as preference orderngs that are at a Kemeny dstance of one from the true preference orderng. A socal choce functon (SCF) s adjacent manpulaton (AM) proof, f t s mmune to such manpulatons. A doman satsfes equvalence f every AM-proof SCF defned on ths doman, s also strategy-proof. Sato (2013) provdes a necessary condton and a separate, stronger suffcent condton for equvalence. We extend Sato s results n several drectons. We dentfy a condton whch s necessary for equvalence; n conjuncton wth Sato s necessary condton, t s suffcent for equvalence for SCF s satsfyng unanmty. We also provde an addtonal, stronger suffcent condton for equvalence wthout unanmty. Fnally we show that every unanmous and AM-proof SCF s also group strategy-proof provded equvalence holds. Indan Statstcal Insttute, Kolkata, Inda. Indan Statstcal Insttute, New Delh, Inda. Unversty of Padua, Padua, Italy. Sngapore Management Unversty. 1

2 1 Introducton In any model where the agents have prvate nformaton, the prmary objectve of the mechansm desgner s to desgn rules that provde agents the ncentve to reveal ther prvate nformaton truthfully. Incentve compatblty guarantees that every agent truthfully reveals hs prvate nformaton, rrespectve of the announcements made by other agents. Incentve compatblty assumes that every feasble preference s a canddate for manpulaton. Thus the task of desgnng truthful rules may be too demandng n many settngs, as t requres the mechansm desgner to check all possble ncentve constrants. On the other hand f the rule s mmune to canddate manpulatons that are near or close to the true preference of an agent, then the rule satsfes local ncentve compatblty. Every ncentve compatble rule s locally ncentve compatble. However t s possble that a rule s locally ncentve compatble, but not ncentve compatble. The queston that we are nterested n s when a locally ncentve compatble rule s also ncentve comptable. In any settng where local ncentve compatblty mples ncentve compatblty, the desgner s task s now only nvolves checkng the local ncentve constrants. In a fnte votng envronment, ths s nterpreted as preference orderngs that are at a Kemeny dstance of one from the true preference orderng. 1 A socal choce functon s adjacent non-manpulable, f t s mmune to such local manpulatons. A doman satsfes equvalence f every adjacent non-manpulable socal choce functon defned on ths doman, s also strategy-proof or non manpulable on the doman. Several papers have examned ths ssue n dfferent settngs (we refer to some of ths lterature later n ths Secton.). Sato (2013) consders the standard votng model wthout transfers. He provdes a suffcent condton and a weaker necessary condton for the equvalence. For the smplcty of descrpton, we refer to the necessary condton n Sato (2013) as weak connectedness. We nduce an addtonal necessary condton called top-smlarty. The combnaton of weak connectedness and top-smlarty s referred to as consstent connectedness. We show that consstent connectedness s also suffcent for equvalence for socal choce functons satsfyng the addtonal property of unanmty. Unanmty s a natural axom wdely studed n the lterature. It requres the socal outcome to be the peak of preferences on whch all voters agree n a profle. Applyng a result establshed n Chatterj and Zeng (2015), unanmty and local strategy-proofness mply the tops-only property on consstently connected domans. Ths mples that an outcome at a profle s entrely determned by the peaks of voter preferences n a preference profle. Ths property reduces the possblty of manpulaton substantally and helps restore equvalence. 1 Kemeny (1959) frst ntroduced the noton of dstance between two dstnct preferences whch s the number of pars of alternatves wth opposte relatve rankng. Accordngly, the dstance between two adjacent preferences s one. A smlar noton s used n Gbbard (1977) referred to as a swtch. 2

3 We provde an example to show that consstent connectedness s not suffcent for equvalence for socal choce functons that do not satsfy unanmty. We provde a suffcent condton for equvalence for arbtrary socal choce functons. Ths condton s weaker than the suffcency condton n Sato (2013). We also provde an extenson of our result and an applcaton. We show that n a consstently connected doman where each alternatve s the peak of some preference, no coalton of voters can strctly beneft through a cooperatve msrepresentaton of preferences n a unanmous and locally strategy-proof socal choce functon. Consequently, we assert that every unanmous and locally strategy-proof socal choce functon s also mmune to group manpulaton once the equvalence of local strategy-proofness and strategy-proofness s acheved. In the lterature, sngle-peaked domans (Mouln (1980), Demange (1982)) are the most attractve restrcted domans whch admt a large class of well-behaved strategy-proof socal choce functons satsfyng some addtonal desrable propertes lke anonymty. Recently, some papers explored the followng converse queston: does sngle-peakedness restrcton naturally emerge as a consequence of the exstence of a well-behaved strategy-proof socal choce functon? 2 Chatterj et al. (2013) provde one answer to ths doman mplcaton problem by showng that n a sutably rch doman, sem-sngle-peakedness, whch s a weaker formulaton of sngle-peakedness, must be mpled by the admsson of a unanmous, anonymous, tops-only and strategy-proof socal choce functon wth even number of voters. As an applcaton, we study a smlar doman mplcaton problem n our model, and show that n the class of consstently connected domans, sngle-peakedness (on a tree) s unquely characterzed by unanmty, anonymty and local strategy-proofness. In related lterature, Carroll (2012) consders both votng model and allocaton models wth transfers. In votng models, he shows that a number of specfc domans such as the full doman, doman of all sngle peaked preferences etc are domans where local ncentve compatblty mples ncentve compatblty. However, he does not provde a general condton for votng domans. In models wth transfers, he shows that convexty of the type space s suffcent and almost necessary. Mshra et al. (2015) consder allocaton models wth transfers. They show that convexty of the type space s not necessary, f an addtonal assumpton on transfers s made. Archer and Klenberg (2014) examne an allocaton model wth money and a dfferent noton of local ncentve compatblty. Cho (2015) studes the equvalence of adjacent strategy-proofness and strategy proofness for random socal choce functons, when the preferences from sure alternatves to lotteres can be extended by stochastc domnance, downward lexcographc, or upward lexcographc extenson. In case of stochastc domanance, they show that the suffcent condton n Sato (2013) s also suffcent for the equvalence for random socal choce functons. 2 For more detals, please refer to Gul s conjecture n Barberà (2011). 3

4 The paper s organzed as follows. Secton 2 gves basc notaton and defntons. Secton 3 descrbes the exstng results. 2 Basc Notaton and Defnatons Let A = {a, b,... } be a fnte set of alternatves wth A = m 3, and I = {1,..., N} be a fnte set of voters wth I = N 2. Each voter has a (preference) order P over A whch s antsymmetrc, complete and transtve,.e., a lnear order. Gven a, b A, ap b s nterpreted as a s strctly preferred to b accordng to P. The set contanng all preferences s referred to as the complete doman P. Let D P be the set of admssble preferences over A, referred to as the doman of preferences. We assume that doman s dentcal for all voters. We wrte a preference profle P = (P 1,..., P N ) = (PÎ, P Î ), where Î I and Î. Gven P D, let r k (P ), k = 1,..., m, denote the kth ranked alternatve n P. For notatonal smplcty, let D a = {P D r 1 (P ) = a} denote the set of preferences wth peak a. We start from the noton of adjacency between two preferences n a doman. A par of preferences P, P D s adjacent, denoted P A P, f there exst x, y A such that the followng two condtons are satsfed () x = r k (P ) = r k+1 (P ) and y = r k+1 (P ) = r k (P ) for some 1 k m 1; () r t (P ) = r t (P ) for all t / {k, k + 1}. In other words, two preferences are adjacent f there are two alternatves locally swtched n the relatve rankng. Let A(P, P ) denote the ordered par of alternatves whch are swapped n P to obtan P. A Socal Choce Functon (SCF) s a map f : D N A, N 2. The axom of unanmty requres the socal planner to choose the peak as the desrable socal outcome whenever all voters agree on the peak n a preference profle,.e., a SCF f : D N A s unanmous f for all a A and P D N, [r 1 (P ) = a for all I] [f(p ) = a]. The standard noton of ncentve compatblty s strategy-proofness. Defnton 1 A SCF f : D N A s strategy-proof f for all I; P, P P D N 1, ether f(p, P ) = f(p, P ) or f(p, P )P f(p, P ). D and If a rule s strategy-proof, an agent cannot manpulate.e. cannot get a strctly preferred alternatve by msrepresentng her true preference. A local noton of ncentve compatblty ntroduced by Sato (2013) and Carroll (2012) s AM-proofness (Adjacency Manpulable proofness) or local strategy-proofness. 4

5 Defnton 2 A SCF f : D N A s locally strategy-proof or AM-proof f for all I; P, P D wth P A P and P D N 1, ether f(p, P ) = f(p, P ) or f(p, P )P f(p, P ). The AM-proofness of a rule can be characterzed by some elementary propertes. We descrbe these below usng the termnology of Gbbard (1977). Defnton 3 The rule f : D N A s local and non-perverse f for every P, P every P D N 1 wth P A P and A(P, P ) = (x, y) we have () [f(p, P ) = y] = [f(p, P ) = y] () [f(p, P ) = x] = [f(p, P ) {x, y}] () [f(p, P ) = z] = [f(p, P ) = z] D and P P z z.. x [y] [y] x.. P P z z.. [x] [y] y [x].. P P [z] [z].. x y y x.. Table 1: AM-proofness Proposton 1 A rule f s AM-proof ff t s local and non-perverse. The proof of ths standard and can be found n Sato (2013). It s clear that a strategy-proof rule s also AM-proof. The goal of ths paper s to analyze domans where the converse holds. Defnton 4 A doman D satsfes equvalence f every AM-proof rule s also strategy-proof. 3 Exstng results Carroll (2012) shows that several well known domans satsfy the equvalence for both determnstc and random rules. Sato (2013) nvestgates the queston more generally. He gves a suffcent condton on domans that ensures the equvalence. He also provdes a weaker necessary condton. We llustrate the ssues nvolved and the results n Sato (2013). 5

6 Defnton 5 Gven a par of dstnct preferences P, P D, the sequence of preferences {P k } t k=1 D, t 2, s referred to as a path connectng P and P f P 1 = P, P t = P and P k A P k+1, k = 1,..., t 1. Let (P, P ) denote the set of all paths between P and P P, P D are connected n D f there exsts a path between P and P connected f for all P, P D, there exsts a path between P and P n D. Two preferences n D. A doman D s n D. Proposton 2 (Sato(2013)) If Doman D satsfes equvalence then D s connected. Connectedness s necessary for the equvalence of AM-proofness and strategy-proofness on a doman. However connectedness s not suffcent for equvalence. We descrbe below a stronger noton of connectedness where for every par of preferences n the doman, there exsts a path whch satsfes the no-restoraton property. Defnton 6 Doman D s weakly connected f gven dstnct P, P D and dstnct a, b A, there exsts a path {P k } t k=1 D connectng P and P, and satsfyng the no-restoraton property wth respect to a and b: [ap k b and bp k+1 a for some 1 k t 1] [ap l b, 1 l k, and bp l a, k + 1 l t]. On a path, f the no-restoraton property s volated wth respect to some par of alternatves, such par s referred to as a restored par. Proposton 3 (Sato(2013)) If Doman D satsfes equvalence then D s weakly connected. However weak connectedness of a doman s not suffcent for the doman to satsfy equvalence. Sato (2013) provdes an example to llustrate ths fact. We now state the suffcenct condton n Sato (2013). Theorem 1 (Sato (2013)) Suppose that for each P, P D, there exsts a path (P 1,..., P L ) n D from P to P whch satsfes the followng property: f there exsts x, y A such that the path (P 1,..., P L ) s wth {x, y} restoraton and xp y, then for each h {1,..., L} such along whch x overtakes no that yp h x and xp h+1 y, there exsts a path from P to P h+1 alternatve. 6

7 4 Equvalence wth Unanmty In ths secton, we characterze domans that satsfy equvalence for socal choce functons that satsfy unanmty. We observe that n a weakly connected doman, there mght exst some par of dstnct preferences P, P D wth the same peak, say r 1 (P ) = r 1 (P ) = a, such that n every path connectng them, the peak of some preference n the path s not a. 3 Such a phenomenon never appears n a doman wth no-restoraton. Once ths phenomenon occurs, doman D can be parttoned nto two non-empty parts: D and D\ D satsfyng the followng two condtons: () D D a and [D\ D] D a, () for all P D and P D a [D\ D], P s not adjacent to P. Observe that gven P D and P D, f P A P, then ether P D, or P D\ D and r 1 (P ) = r 2 (P ). Then, we fx a voter and construct a SCF whch pcks her second best alternatve f she presents a preference n D, and pcks her best alternatve otherwse. In ths SCF, the fxed voter determnes the socal outcome and all others are dummes. Evdently, f the socal outcome s the peak of the fxed voter, the fxed voter has no ncentve to manpulate. Accordng to the observaton above, when the socal outcome s the second best alternatve of the fxed voter, an adjacent devaton wll ether brng a worse socal outcome f the adjacent preference s n D, or does not change the socal outcome f the adjacent preference belongs to D\ D. Therefore, ths SCF s locally strategy-proof. However, ths SCF s not strategy-proof, e.g., f the fxed voter s true preference falls nto D, she wll msreport a preference n D a [D\ D]. Therefore, we dentfy an addtonal necessary condton: top-smlarty, for the equvalence of local strategy-proofness and strategy-proofness whch ensures the absence of the phenomenon mentoned above. Defnton 7 Doman D s top-smlar f for all a A and all dstnct P, P D a, there exsts a path {P k } t k=1 Da connectng P and P. Proposton 4 If doman D satsfes equvalence, then t s top-smlar. Proof : Snce D satsfes equvalence, we know that D s weakly connected by Proposton 3. Suppose that D volates top-smlarty. Then, there must exst a A and P, P D a such that for every path {P k } t k=1 D connectng P and P, r 1 (P k ) a for some 1 < k < t. Lemma 1 There exsts D D a such that the followng two condtons are satsfed: () 0 < D < D a ; 3 Please refer to Example 3.2 n Sato (2013). 7

8 () for every P D a \ D, there exsts no P D such that P A P. Proof : We frst ntroduce an algorthm to dentfy a subset D of D a. The algorthm conssts of the followng steps: Step 1. Pck arbtrary P 1 D a. Let D 1 = {P 1 }. Step t 2. If there exsts P t D a \ D t 1 such that P t A P j for some P j D t 1, then let D t = D t 1 {P t }. Otherwse, the algorthm termnates and let D = D t 1. Snce D a s fnte, the algorthm must termnate n fnte steps. Assume that the algorthm stops at step t. Accordng to Step 1, snce D a, D. Accordng to the hypothess, by the algorthm, we know that there exsts P D a such that P / D. Therefore, 0 < D < D a. Ths completes the verfcaton of statement (). Statement () s evdently mpled by the constructon of the algorthm. Now, to acheve a contradcton, we construct a SCF and show that t s locally strategyproof but not strategy-proof. Fxng voter 1, let f : D N A be a SCF such that for all P D N, () [P 1 D] [f(p ) = r 2 (P 1 )] and () [P 1 / D] [f(p ) = r 1 (P 1 )]. Clam 1 : SCF f s locally strategy-proof. Suppose that f s not locally strategy-proof. Snce voter 1 determnes the socal outcome and all other voters are dummes, there exst P 1, P 1 D wth P 1 A P 1 and P 1 D N 1 such that f(p 1, P 1 ) x, f(p 1, P 1 ) y and yp 1 x. Evdently, x r 1 (P 1 ). Hence, the constructon of f mples that P 1 D and x = r 2 (P 1 ). Then, yp 1 x mples that y = r 1 (P 1 ) = a. Now, f(p 1, P 1 ) = y = a mples that P 1 / D. Consequently, a = f(p 1, P 1 ) = r 1 (P 1) and hence P 1 D a \ D whch contradcts statement () of Lemma 1. Therefore, f s locally strategy-proof. Ths completes the verfcaton of the clam. Clam 2 : SCF f s not strategy-proof. Accordng to the hypothess, we know that there exst P 1 D and P 1 D\ D. Gven arbtrary P 1 D N 1, snce f(p 1, P 1 ) a and f(p 1, P 1 ) = a, voter 1 wll manpulate at (P 1, P 1 ) va P 1. Ths completes the verfcaton of the clam and hence the proposton. Henceforth, a doman satsfyng weak connectedness and tops-smlarty s referred to as a consstently connected doman. We provde the followng example to llustrate consstent connectedness. Example 1. Let A = {x, y, a, b, c, d, α, β}. Doman D s specfed n the followng table. Accordngly, the structure of adjacency between preferences s specfed n the followng fgure: 8

9 P 1 P 2 P 3 P 4 P 5 P 6 P 7 P 8 P 9 P 10 P 11 P 12 P 13 P 14 x x x x x x y y y y y y y y y y y y y y x x x x x x x x [a] [b] [b] [b] [b] [d] [d] [d] b b b b [a] [a] b a a d d b b b [d] [c] [c] [a] b b c c d a c c c c c d a c c c d d c c a a a a a a d d d d α α α α α α α β β β β β β α β β β β β β β α α α α α α β Table 2: Doman D Frstly, doman D satsfes the top-smlarty,.e., () r 1 (P k ) = x, 1 k 6, and P k A P k+1, 1 k 5; () r 1 (P k ) = y, 7 k 14, and P k A P k+1, 7 k 13. Next, doman D s also weakly connected. Note that gven a par of dstnct preferences, there are exactly two paths connectng them. Moreover, on both paths, the ntersecton of both sets of restored pars s empty, e.g., between P 2 and P 10, on the path {P k } 10 k=2, [c, d] and [b, d] are restored pars, whle on the path {P 2, P 1, P 14, P 13, P 12, P 11, P 10 }, [a, b] s the only restored par. Evdently, the two paths connectng P 2 and P 10 ndcates that strong connectedness s volated on D. Unfortunately, consstent connectedness s not suffcent for equvalence. For nstance, fxng I, let f : D N A be a SCF choosng the alternatve n the square brackets n Table 2 accordng to voter s reported preference. 4 SCF f s locally strategy-proof, but not strategy-proof, e.g., voter wth true preference P 10 wll manpulate va P 2. Observe that the SCF proposed n Example 1 to show the nequvalence of local strategyproofness and strategy-proofness s unlateral (Defnton 1 n Gbbard (1977)) snce a fxed voter determnes the outcome on every preference profle whle all other voters play dummy. 5 Consequently, the axom of unanmty s volated by the SCF n Example 1. Under consstent connectedness, once unanmty s mposed on a locally strategy-proof SCF, we can obtan the tops-only property by Chatterj and Zeng (2015) whch mples that all socal outcomes are determnes entrely by all voters peaks of preference profles,.e., a SCF f : D N A satsfes the tops-only property f for all P, P D, [r 1 (P ) = r 1 (P ) for all I] [f(p ) = f(p )]. 6 Furthermore, snce the tops-only property decreases 4 For nstance, for all P D N 1, [P = P 10 ] [f(p, P ) = c] and [P = P 2 ] [f(p, P ) = b]. 5 A SCF f : D N A s unlateral f there exsts I such that for all P, P D N, [P = P ] [f(p ) = f(p )]. 6 Chatterj and Zeng (2015) study the tops-only property n random socal choce functons. Even though they adopt the noton of strategy-proofness, ther argument stll holds wth respect to local strategy-proofness 9

10 the degree of potental ndvdual devatons tremendously from D 1 to A 1, t helps enhance local strategy-proofness to strategy-proofness. Fnally, we assert that consstent connectedness s suffcent n the class of unanmous SCFs for the equvalence of local strategy-proofness and strategy-proofness. Gven a doman D, let F LSP (D) and F SP (D) denote the set of unanmous SCFs satsfyng local strategyproofness and strategy-proofness respectvely. Theorem 2 Consder doman D such that D s consstently connected. Then F LSP (D) = F SP (D). Proof : Frstly, by Chatterj and Zeng (2015), we know that every SCF n F LSP (D) satsfes the tops-only property. Gven a unanmous and locally strategy-proof SCF f : D N A, suppose that t s not strategy-proof. Thus, there exst I; P, P D and P D N 1 such that f(p, P ) a, f(p, P ) b and bp a. Snce f satsfes the tops-only property, we know r 1 (P ) r 1 (P ). Snce D s weakly connected, there must exst a path {P k } t k=1 D connectng P and P, and satsfyng the no-restoraton property wth respect to (a, b). We can assume that f(p k, P ) b for all 1 k t 1. 7 Next, we partton the path {P k } t k=1 nto q + 1 parts accordng to the peaks of preferences: { P 1,..., P s 1 ; P s 1+1,..., P s 2 Part 1 Part 2 ;... ; P s l 1+1,..., P s l Part l ;... ; P s q 1+1,..., P sq Part q ; P sq+1,..., P t Part q + 1 We assume that () r 1 ( P ) = x l and f( P, P ) = y l for all 1 l q + 1 and every P n Part l of the partton, and () x l x l+1 for all 1 l q. Note that there mght exst 1 l < l q + 1 such that l l 2 and x l = x l. Snce f(p k, P ) b for all 1 k t 1, t must be case that Part q + 1 s a sngleton set,.e., P s q+1 Frstly, snce f(p s q, P ) f(p t, P ) = f(p s q+1 = P t., P ), P s q A P s q+1 and x q x q+1, local strategy-proofness mples that y q = f(p s q, P ) = x q and y q+1 = f(p s q+1, P ) = x q+1. Thus, b = y q+1 = x q+1 = r 1 (P t ). Now, bp 1 a and bp t a. Then, on the path {P k } t k=1, bp k a for all 1 k t by the no-restoraton property wth respect to (a, b). Secondly, we analyze from Part 1 to Part q + 1. Snce r 1 (P 1 ) = x 1 and f(p 1, P ) = a = y 1, bp 1 a mples that y 1 x 1. Next, we provde an nducton hypothess: gven 2 l q + 1, for all 1 l < l, y l x l and a = y 1 = = y l. We wll show that y l x l and a = y 1 = = y l 1 = y l. Snce f(p s l 1, P ) = y l 1 and f(p s l 1+1, P ) = y l, local strategy-proofness mples that ether y l 1 P s l 1 y l and y l P s l 1+1 y l 1, or y l = y l 1. Suppose y l 1 P s l 1 y l and y l P s l 1+1 y l 1. Then, P s l 1 A P s l 1+1 and r 1 (P s l 1 ) = x l 1 x l = r 1 (P s l 1+1 ) mply that y l 1 = x l 1. under consstent connectedness whch s a specal case of ther suffcent condton. 7 If there exsts 1 < t < t such that f(p t P t, and the subsequence {P k }t k=1, P ) = b, then we have a smlar manpulaton at (P, P ) va stll holds the no-restoraton property wth respect to (a, b). 10 }.

11 Contradcton to the nducton hypothess! Therefore, y l = y l 1 and nducton hypothess mples that y l = y l 1 = = y 1 = a. Moreover, bp s l 1+1 a and x l = r 1 (P s l 1+1 ) mply that y l = a x l. Ths completes the verfcaton of the nducton hypothess. Consequently, we have b = f ( ) P t, P = yq+1 = = y 1 = a. Contradcton! Therefore, f s strategy-proof. Ths completes the verfcaton of the theorem. 5 Equvalence wthout Unanmty: A suffcent Condton We have observed that consstent connectedness does not guarantee equvalence wthout restrctng attenton to unanmous socal choce functons. In ths secton, we provde a condton that acheves ths objectve. We are however not able to show that t s necessary, although we beleve t to be true. Our condton s weaker than the suffcent condton provded by Sato; therefore our result s stronger. We defne the condton, whch we call admssble sequence (AS) condton below. We llustrate all concepts and defntons by means of Example 2. Fx a path σ(p, P ) = (P 1,..., P L ) (P, P ). The path σ(p, P ) s assocated wth a sequence of ordered pars of alternatves, S(σ(P, P )) = {A(P s, P s+1 ) = (u s, u s+1 ) : s {1,..., L 1}} Example 2 Let A = {x, y, v, z, w, u} and ˆD be the followng doman. Fx alternatves x, y A and preferences P 1, P 10 ˆD. There s only one path between P 1 and P 10. So (P 1, P 10 ) = {(P 1, P 2, P 3, P 4, P 5, P 6, P 7, P 8, P 9, P 10 )}. P 1 P 2 P 3 P 4 P 5 P 6 P 7 P 8 P 9 P 10 x x v v v v v v x x y v x x z z x x v y v y y z x x z y y v z z z y y y y z z z w w w w w u u u u u u u u u u w w w w w Table 3: Doman ˆD In Example 2, S(P 1, P 2,..., P 10 ) s descrbed below. S((P 1, P 2,..., P 10 )) = {(y, v), (x, v), (y, z), (x, z), (w, u), (z, x), (z, y), (v, x), (v, y)} 11

12 Observe that (y, v) s the contguous swap between P 1 and P 2, (x, v) the contguous swap between P 2 and P 3, (y, z) the contguous swap between P 3 and P 4 and so on. Fx a, b A. An essental sequence between the alternatves a, b on a path s a sequence of contguous swaps that begn wth a and end n b on ths path. Formally σ e (σ(p, P ); a, b) s an ordered selecton {(u s, u s +1 ) : {1,..., H}} from S(σ(P, P )) such that () u s 1 = a and u s H = b and () u s+1 = u s +1 for all s. The set of all such essental sequences s denoted by e (σ(p, P ); a, b). In Example 2, observe that {(x, v), (v, y)} s an essental sequence snce t s a selecton from S((P 1, P 2,..., P 10 )) that begns wth x and ends wth y. Smlarly {(x, z), (z, y)} s also an essental sequence. Therefore e ((P 1, P 2,..., P 10 ); x, y) = {{(x, v), (v, y)}, {(x, z), (z, y)}}. Defnton 8 (Condton AS) Let P 1, P h, P h+1 D and x, y A be such that xp 1 y, P h+1 A(P h ) and A(P h, P h+1 ) = (y, x). Let Z(x, y) = {z A\{x, y} : xp 1 z, zp 1 y and yp h+1 z}. For all w Z(x, y) {y}, there exsts a path σ (P, P h+1 ) such that e (σ; w, x) = and e (σ; x, w) =. The formal proof for the suffcency of Condton AS s provded below. Theorem 3 If D satsfes Condton AS, t satsfes equvalence. The proof uses the followng lemma. Lemma 2 Fx a, b A and preferences P, P D such that [ap b and ap b]. Suppose there exsts σ (P, P ) such that e (σ; b, a) =. Then [f(p ) = b] = [f(p ) a]. Proof : Suppose the clam s false,.e f(p ) = b and f(p ) = a. Let σ(p, P ) = (P 1, P 2,..., P l ). Snce f(p ) b and f s AM-proof, there exsts k 1 {1,..., l 1} such that A(P k 1, P k 1+1 ) = (b, u 1 ) for some u 1 A\{b}. Proposton 1 mples that f(p k 1+1 ) = u 1. If u 1 = a then (b, a) e (σ(p, P ); b, a), whch s n contradcton wth the assumpton that e (σ(p, P ); b, a) =. If u 1 a (note that u 1 A \ {b, a}) then AM-proofness of f and f(p ) = a mply that: there exsts k 2 {k 1 + 1,..., l 1} and u 2 A \ {b, u 1 } such that A(P k 2, P k 2+1 ) = (u 1, u 2 ). Ths follows from Proposton 1 and the fact that f(p k 1+1 ) f(p ). Proposton 1 also mples that f(p k 2+1 ) = u 2. If u 2 = a = f(p ) then (b, u 1 ), (u 1, a) e (σ(p, P ); b, a) whch results n a contradcton. However f u 2 a, then there exsts k 3 {k 2 + 1,..., l 1} and u 3 A \ {b, u 1, u 2 } such that A(P k 3, P k 3+1 ) = (u 2, u 3 ). 8 8 Note that u 1, u 2 and u 3 are dstnct from each other. 12

13 In ths manner, at each step we obtan k and the correspondng alternatve u such that A(P, P +1 ) = (u 1, u ). Snce S(σ) s a fnte set, thus there exsts K < l 1 such that u K = a. Ths mples that {(b, u 1 ), (u 1, u 2 ),..., (u K 1, a)} e (σ(p, P ); b, a), whch contradcts our assumpton that ths set s null. We now begn the proof of the theorem. Proof : Let D be a connected doman satsfyng Condton AS and f be an AM-proof rule on D. It s suffcent to show that f s strategy-proof. Let P, P D. The lower contour set for alternatve x at the preference P s denoted by L(P, x) = {y A : xp y} {x}. Our objectve s to prove that: f(p ) L(P, f(p )). Because D s connected, there exsts a path between P and P n D, (P 1,..., P L ) where P 1 = P and P L = P. Step 1: At each step from P h to P h+1 (h {1,..., L 1}) n the path (P 1,..., P l ), the outcome can change only when A(P h, P h+1 ) = (f(p h ), a) where a s the alternatve rght below f(p h ) n P h. In all other cases the outcome does not change (By Proposton 1). Step 2: For each k {1,..., l 1}, f [ f(p k ) L(P, f(p )) and f(p ) L(P k, f(p k )) ] then [ f(p k+1 ) L(P, f(p )) and f(p ) L(P k+1, f(p k+1 )) ]. By Step 1, t suffces to consder the case where a s the alternatve rght below f(p k ) n P k and A(P k, P k+1 ) = (f(p k ), a). By AM-proofness, f(p k+1 ) {f(p k ), a}. Case 1: f(p k+1 ) = a. Snce f(p ) L(P k, f(p k )) and A(P k, P k+1 ) = (f(p k ), x). L(P k+1, f(p k+1 )). We clam that f(p k+1 ) L(P, f(p )). Thus we have f(p ) Suppose the clam s false.e. ap f(p ). Snce f(p )P f(p k ), ths mples that ap f(p h ). We note that P, P k, P k+1 and the alternatves x, f(p h ) A are such that 1. P k+1 A(P k ). 2. A(P k, P k+1 ) = (f(p k, a). 3. f(p ) Z(a, f(p k )). Thus comparng wth Condton AS, we deduce that P = P 1, P k = P h and P k+1 = P h+1. Also a = x and f(p k ) = y. Case 1.1: f(p ) = f(p k ). Snce the doman satsfes Condton AS, there exsts σ (P, P k+1 ) such that e (σ; f(p k ), a) =. Ths along wth Lemma 2 leads to a contradcton. Case 1.2: f(p ) f(p k ). Snce the doman satsfes Condton AS, there exsts σ (P, P k+1 ) such that e (σ ; f(p ), a) =. Ths along wth Lemma 2 leads to a contradcton. 13

14 Case 2: f(p k+1 ) = f(p k ) (where f(p k ) a). In ths case, t trvally follows that f(p k+1 ) L(P, f(p )). Now, L(P k, f(p k )) = L(P k+1, f(p k+1 )) {a}. To complete the proof, we need to show that f(p ) a. Let us suppose by way of contradcton that f(p ) = a. We have assumed that the outcome at P k+1 s f(p k ). Comparng wth Condton AS, we know the followng: 1. x = f(p ) = a 2. y = f(p k 3. P 1 = P, P h = P k and P h+1 = P k ap f(p k ) and A(P k, P k+1 ) = (f(p k, a). Thus there exsts a path σ Σ(P, P k+1 ) such that Σ e (σ, a, f(p k )) =. Ths observaton along wth Lemma 5 leads to a contradcton. Example 3 below satsfes Condton AS and thereby admts equvalence. Example 3 Let X = {a, x, y, z, v, w, u} and D s be the followng doman. P 1 P 2 P 3 P 4 P 5 P 6 P 7 P 8 P 9 P 10 a a a a a x x x x a x y y y x a a a a x y x x x y y y z z z z z z z z z z y y y v v v u u v v v u u w w u v v w u u v v u u w w w u w w w w Table 4: Doman D s It can be verfed that the only problem path n ths example s (P 1, P 2, P 3, P 4, P 5 ). Consder preferences P 1, P 4 and P 5. In ths example xp 1 y, A(P 4, P 5 ) = (y, x) and Z(x, y) =. The path (P 1, P 6, P 7, P 8, P 9, P 10, P 5 ) does not contan any essental sequence for the alternatves y, x. Ths path s the alternatve antdote path whch elmnates the non strategyproof rules generated by the problem path (P 1, P 2, P 3, P 4, P 5 ). Therefore Condton α s satsfed. 14

15 We note that a doman satsfyng Condton AS also satsfes Sato (2013) suffcent condton. Proposton 5 Suppose that for each P, P D, there exsts a path (P 1,..., P L ) n D from P to P whch satsfes the followng property: f there exsts x, y A such that the path (P 1,..., P L ) s wth {x, y} restoraton and xp y, then for each h {1,..., L} such that yp h x and xp h+1 y, there exsts a path from P to P h+1 along whch x overtakes no alternatve. Then D satsfes Condton AS. Proof : Consder preferences P 1, P h and P h+1 on the path (P 1,..., P L ). We know xp 1 y, P h+1 A(P h ) and A(P h, P h+1 ) = (y, x). Thus there exsts an alternatve path σ from P 1 to P h+1 along whch x does not overtake any alternatve.e. ths path does not contan the contguous swap (q, x) for any q A \ {x}. We clam that e (σ; w, x) = for all w Z(x, y) {y} for the path σ. To see ths, consder any w Z(x, y) {y}. Suppose e (σ; w, x) and {(w, u 1 ), (u 1, u 2 ),..., (u K, x)} belongs to ths set. However (u K, x) contradcts the assumpton that σ s a path along whch x does not overtake any alternatve. Our result therefore mples Sato (2013) suffcency result. Our result s strctly stronger. The doman n Example 3 satsfes Condton AS and therefore satsfes equvalence. However we clam that t does not satsfy the suffcency condton s Sato. We conjecture that Condton AS s also necessary for equvalence. Suppose ths condton does not hold. Then there exsts P 1, P h+1 D and x, y A such that xp 1 y and xp h+1 y and all paths between P 1 and P h+1 contan an essental sequence between y and x. We conjecture that t s possble n ths case to construct a socal choce functon where equvalence breaks down. Note that t s possble to assgn alternatves to all preferences on any path between P 1 and P h+1 wth f(p 1 ) = y and f(p h+1 ) = x and satsfyng AM-proofness. Strategy-proofness s volated because there exsts a manpulaton at P 1 va P h+1. If all paths between P 1 and P h+1 have no preferences n common except P 1 and P h+1, then the argument s complete. However f the paths have a preference n common, a more refned argument s requred. We beleve that such a socal choce functon can be constructed. 6 Group strategy-proofness In ths secton, we push our study further to the coaltonal msrepresentaton ssue. We follow the same noton of group manpulaton studed n Le Breton and Zaporozhets (2009) and Barberà et al. (2010): once a devaton of a coalton causes a change of the socal outcome, every voter n the coalton strctly benefts accordng to her true preference. In a votng model wth strct preference, t s natural to adopt such an all-devators-strctly-benefttng noton of group manpulaton. Formally, a SCF f : D N A s group manpulable f there 15

16 exst Î I wth Î ; P Î, P Î D Î and P Î DN Î such that f(p Î, P Î)P f(pî, P Î ) for all Î. Accordngly, SCF f s group strategy-proof f t s not group manpulable. Gven a doman D, let F GSP (D) denote the set of unanmous SCFs satsfyng group strategyproofness. The proposton below shows that consstent connectedness s also suffcent for the equvalence of local strategy-proofness and group strategy-proofness, provded that the doman s mnmally rch, (.e., D a for all a A). Proposton 6 If doman D s mnmally rch and consstently connected, then F LSP (D) = F GSP (D). Proof : Fxng a unanmous and locally strategy-proof SCF f : D N A, we know that f satsfes the tops-only property and strategy-proofness by the proof of suffcency part of Theorem 2. For notatonal convenence, let ( (a, Î), P Î) denote a profle of preferences where Î, P Î DN Î, and every voter n Î presents a preference wth peak a. Furthermore, we ntroduce the noton of strong connectedness between two alternatves. Gven a, b A wth a b, f there exst P D a and P D b such that P A P, we say that alternatves a and b are strongly connected, denoted a b. 9 Lemma 3 For all dstnct x, y A, there exsts a sequence {x k } l k=1 A such that x 1 = x, x l = y and x k x k+1, k = 1,..., l 1. Proof : Gven x, y A, snce D s mnmally rch, we have P, P D wth r 1 (P ) = a and r 1 (P ) = b. Next, snce D s weakly connected, there exsts a path {P k } t k=1 D connectng P and P. We partton {P k } t k=1 accordng to the peaks of preferences: { P 1,..., P s 1 ; P s 1+1,..., P s 2 Part 1 Part 2 ;... ; P s l 1+1,..., P s l Part l ;... ; P s q 1+1,..., P sq Part q ; P sq+1,..., P t Part q + 1 We assume that () r 1 ( P ) = x l for all 1 l q +1 and every P n Part l of the partton, and () x l x l+1 for all 1 l q. Hence, x 1 = x, x q+1 = y and x k x k+1, k = 1,..., q. Note that there mght exst 1 l < l q + 1 such that l l 2 and x l = x l. Now, accordng to the sequence [x k ] q+1 k=1, we can extract out a sequence {x k s } l s=1 A such that x k1 = x, x kl = y and x ks x ks+1, s = 1,..., l 1. Frstly, strategy-proofness mples that for all Î I wth Î = 1; P Î, P D and Î P Î DN 1, [ f(pî, P Î ) f(p, P Î Î)] [ f(pî, P Î )P f(p, P Î Î) for some Î]. Next, we provde an nducton hypothess on the cardnalty of Î. Inducton hypothess (): Gven 2 t N, for all Î I; P Î, P Î D Î and P Î DN Î, 9 Strong connectedness s ntroduced by Chatterj et al. (2013). It s a partcular form of adjacency whch only concerns the flp of the top two ranked alternatves across two preferences. 16 }.

17 [ 1 Î < t and f(pî, P Î ) f(p Î, P Î)] [ f(pî, P Î )P f(p Î, P Î) for some Î]. Gven Î I wth Î = t; P Î, P Î D Î and P Î DN Î, assumng that f(pî, P Î ) = a, f(p Î, P Î) = a and a a, we wll show that ap a for some Î. Snce f(p Î, P Î) = a, strategy-proofness mples that f ( (a, Î), P Î) = a. 10 Suppose that a P a for all Î. Thus, t must be the case Î 2. Clam 1 : f ( (a, {}), P ) / {a, a } for all Î. Fxng Î, suppose that f( (a, {}), P ) = a. Snce f ( (a, Î), P Î) = a, nducton hypothess () mples that there must exst j Î\{} such that ap ja. Contradcton! Next, suppose that f ( (a, {}), P ) = a. Snce f(p, P ) = a, strategy-proofness mples that ap a. Contradcton! Ths completes the verfcaton of the clam. Fx Î and assume that r 1(P ) = x. Snce f(p, P ) = a, Clam 1 mples that x a. By Lemma 3, we have a sequence {x k } s k=1 A, s 2, such that x 1 = x, x s = a and x k x k+1, k = 1,..., s 1. Clam 2 : There exst 1 k s 2 and k + 1 k s 1 such that x k = a and f ( ) (x ν, {}), P = x k x ν x k f 1 ν k f k + 1 ν k f k + 1 ν s Snce f ( ) (x 1, {}), P = f(pî, P Î ) = a and f( ) ( ) (x s, {}), P = f (a, {}), P a by Clam 1, there exsts 1 k s 1 such that f ( ) (x ν, {}), P = a, 1 ν k and f ( ) (x k+1, {}), P a. Snce xk x k+1 and f ( ) ( ) (x k, {}), P f (xk+1, {}), P, strategy-proofness mples that a = f ( ) (x k, {}), P = xk and f ( ) (x k+1, {}), P = xk+1. Furthermore, snce f ( ) ( ) (x s, {}), P = f (a, {}), P a = x s by Clam 1, t must be the case that k + 1 s 1. Thus, 1 k s 2 and f ( ) (x ν, {}), P = xk, 1 ν k. Next, snce f ( ) (x k+1, {}), P = xk+1 and f ( ) ( ) (x s, {}), P = f (a, {}), P a = x s by Clam 1, there exsts k + 1 k s 1 such that f ( ) (x ν, {}), P = xν, k + 1 ν k, and f ( ) (x k+1, {}), P x k+1. Snce x k x k+1, strategy-proofness mples that f ( ) ( ) (x k+1, {}), P = f (x k, {}), P = x k. Furthermore, followng the sequence {x ν } s ν= k+1 and applyng the symmetrc argument step by step, we have f ( ) (x ν, {}), P = x k, k + 1 ν s. Ths completes the verfcaton of the clam. Clam 3 : x k P x ν, ν = k + 1,..., k. Ths clam follows from the strategy-proofness of f and Clam 2. Clam 4 : For every j Î\{}, f( (x ν, {j}), P j ) = xν, ν = k,..., k. 10 Mnmal rchness mples D a =. Smlar argument s made n the rest of the proof. 17

18 Fx j Î\{}. Snce f(p j, P j ) = f(pî, P Î ) = a = x k by Clam 2, strategy-proofness mples that f ( (x k, {j}), P j ) = xk. Then, we provde another nducton hypothess. Inducton hypothess (): Gven k + 1 l k, for every j Î\{}, we have [k l < l] [ f ( (x l, {j}), P j ) = xl ]. Snce f(pî, P Î ) = x k and f ( (x l 1, {ι}), P Î) = xl 1 for all ι Î\{} by nducton hypothess (), strategy-proofness mples that x k P ι x l 1 for all ι Î\{}, provded l > k + 1. Moreover, combnng wth Clam 3, we have x k P x l 1 for all ι Î, provded l > k + 1. Gven j Î\{}, we wll show that f( ) (x l, {j}), P j = xl. Snce x l x l 1 and f ( ) (x l 1, {j}), P j = xl 1 by nducton hypothess (), strategy-proofness mples that ether f ( ) (x l, {j}), P j = xl 1, or f ( ) (x l, {j}), P j = xl. Suppose that f ( ) (x l, {j}), P j = xl 1. Followng the sequence {x ν } s ν=l, applyng strategy-proofness step by step, we have that f ( ) (x ν, {j}), P j = xl 1 for all l ν s. Thus, f ( ) ( ) (a, {j}), P j = f (xs, {j}), P j = xl 1. Then, Clam 1 mples that x l 1 a. Snce f ( (a, Î), P Î) = a, nducton hypothess () mples that there exsts k Î\{j} such that x l 1P k a. Snce a = x k and a P ι a for all ι Î by hypothess, t must be the case that l > k +1. Moreover, we have x l 1 P k a P k x k and hence x l 1 P k x k. Contradcton! Therefore, f ( ) (x l, {j}), P j = xl. Ths completes the verfcaton of nducton hypothess () and hence the clam. Clam 5 : For all j Î\{}, x kp j x ν, ν = k + 1,..., k. Ths clam follows from the strategy-proofness of f and Clam 4. Lastly, snce f ( (x s, {}), P ) = x k by Clam 2 and f ( (x s, Î), P Î) = f ( (a, Î), P Î) = a, nducton hypothess () mples that there exsts j Î\{} such that x kp j a. Then, Clam 5 mples that ap j x kp j a and hence ap j a. Contradcton! Ths completes the verfcaton of nducton hypothess (). Therefore, SCF f s group strategy-proof. Ths completes the verfcaton of the theorem. 7 An applcaton: a doman mplcaton problem The semnal Gbbard-Satterthwate theorem (Gbbard (1973); Satterthwate (1975)) suggests that a unanmous, strategy-proof and non-dctatoral socal choce rule never exsts on the complete doman. As a polar opposton to dctatorshp, we add the axom of anonymty nto our study whch ensures that the socal outcome s ndependent of denttes of voters,.e., a SCF f : D N A s anonymous f for all P D N and every permutaton σ : I I, f(p 1,..., P N ) = f(p σ(1),..., P σ(n) ). It s well-known n the lterature that varous formulatons of sngle-peaked preference domans are admssble for the combnaton of unanm- 18

19 ty, anonymty and strategy-proofness (see Mouln (1980); Demange (1982); Barberà et al. (1993); Nehrng and Puppe (2007); Chatterj et al. (2013); Chatterj and Massó (2015)). In ths secton, we restrct our attenton on the class of consstently connected domans and address a doman mplcaton problem: what are the domans that admt a unanmous, anonymous and locally strategy-proof SCF? The answer turns out to be sngle-peakedness (on a tree). A graph G s a combnaton of the set of vertces and the set of edges. A graph-path n G s a sequence of vertces {x k } t k=1, t 2, such that (x k, x k+1 ), k = 1,..., t 1, s an edge n G. In partcular, a graph G s a tree f between every par of dstnct vertces a, b, there exsts a unque graph-path, denoted a, b. Defnton 9 Fx a tree G where the set of vertces s A. Doman D s sngle-peaked on G f for all P D and a, b A wth a b, [ a r 1 (P ), b ] [ ap b ]. Accordngly, a doman s sngle-peaked f there exsts a tree on whch every preference s sngle-peaked. By a smlar argument n Secton 4.2 n?, we know that the sngle-peaked doman on a tree whch contans all admssble preferences, s consstently connected. The followng proposton mples that n the class of consstently connected domans wth mnmal rchness, sngle-peaked domans (on a tree) are unque for the compatblty of unanmty, anonymty and local strategy-proofness. Proposton 7 Let a mnmally rch doman be consstently connected. If t admts a unanmous, anonymous and locally strategy-proof SCF, t s sngle-peaked (on a tree). Proof : Let f : D N A be a unanmous, anonymous and locally strategy-proof SCF. By Chatterj and Zeng (2015) and Proposton 6, we know that f satsfes the tops-only property and group strategy-proofness. Gven P D N, let r 1 (P ) = I {r 1 (P )} denote the set of peaks n P. We frst nduce a graph. Let G (D) denote a graph such that the set of vertces s A and for all a, b A wth a b, (a, b) s an edge n G (D) f a b. We wll show that G (D) s a tree. By Lemma 3, t suffces to show that there exsts no cycle n G (D). In partcular, f the number of voters s even, Lemma 3 n Chatterj et al. (2013) mples that G (D) s a tree. Therefore, n the rest of the proof, we consder the case that N s an odd number. Lemma 4 Gven x, y A wth x y; Î I wth Î and P Î DN Î, we have [ ( ) ( )] [ ( ) ( ) ] f (x, Î), P Î f (y, Î), P Î f (x, Î), P Î = x and f (y, Î), P Î = y. Proof : Snce x y, let P k D x and P k Dy be such that P k A P k. Evdently, r 2(P k ) = y and r 2 (P k ) = x. For all Î, let P = P k and P = P k. Thus, f(p Î, P Î) = f( (x, Î), P Î) and f(p, P Î Î) = f( (y, Î), P ( ( Î). Snce f (x, Î), P Î) f (y, Î), P Î), group strategy-proofness 19

20 mples that f(pî, P Î )P f(pî, P Î ) and f(p, P Î Î) P j f(pî, P Î ) for some, j Î. Then, P A P j, xp y and yp jx mply that f(pî, P Î ) = x and f(p, P Î Î) = y. For Lemma 5 below, we fx a sequence {x k } s k=1, s 2, such that x k x k+1, k = 1,..., s 1. Gven P D N wth r 1 (P ) {x k } s k=1, let max(r 1(P )), mn(r 1 (P )) {1,..., s} denote the maxmum and mnmum ndces,.e., [x k r 1 (P )] [mn(r 1 (P )) k max(r 1 (P ))]. Lemma 5 For all P D N, [ r 1 (P ) {x k } s k=1] [ f(p ) {xmn(r1 (P )),..., x max(r1 (P ))} ]. Proof : Gven P D N, f max(r 1 (P )) = mn(r 1 (P )), the lemma follows from unanmty. Next, we provde an nducton hypothess: gven 1 t s 1, for all P D N wth r 1 (P ) {x k } s k=1, [ 0 max(r 1 (P )) mn(r 1 (P )) < t ] [ f(p ) {x mn(r1 (P )),..., x max(r1 (P ))} ]. Gven P D N wth r 1 (P ) {x k } s k=1 and max(r 1(P )) mn(r 1 (P )) = t, we wll show that f(p ) {x mn(r1 (P )),..., x max(r1 (P ))}. To smplfy the notaton, let k = max(r 1 (P )) and k = mn(r 1 (P )). Thus, k k = t 1. Assume that P = ( (x k, Î), P ) Î where Î and r 1 (P Î ) {x k} k 1 k=k. Let P = ( (x k 1, Î), P ) Î. Thus, r1 (P ) {x k } s k=1 and max(r 1(P )) mn(r 1 (P )) = t 1. Then, nducton hypothess mples that f(p ) = x k for some k k k 1. If f(p ) = f(p ), t s evdent that f(p ) {x k,..., x k}. If f(p ) f(p ), then Lemma 4 mples that f(p ) = x k. Ths completes the verfcaton of nducton hypothess and hence the lemma. Lemma 6 Graph G (D) s a tree. Proof : Suppose that t s not true. Then, there must exst a cycle {x k } t k=1 A, t 3, such that x k x k+1, k = 1,..., t, where x t+1 = x 1. Clam 1 : Gven 1 k < k t and Î I wth Î 0 and I\Î = 0, f( (x k, Î), (x k, I\Î)) {x k, x k}. Snce {x k, x k+1,..., x k} {x k, x k 1,..., x 1, x t, x t 1,..., x k+1, x k} = {x k, x k}, Lemma 5 mples that f ( (x k, Î), (x k, I\Î)) {x k, x k}. Ths completes the verfcaton of the clam. N 1 Let Ī = {1,..., }. Clam 1 mples that f( (x 2 1, Ī), (x 2, I\Ī)) = {x 1, x 2 }. We consder two cases: () f ( (x 1, Ī), (x 2, I\Ī)) = x 1 and () f ( (x 1, Ī), (x 2, I\Ī)) = x 2. Assume that case () occurs. Followng the sequence {x k } t k=2 and applyng Lemma 4 step by step, we have f ( (x 1, Ī), (x t, I\Ī)) = x 1. Next, snce x 2 x 1, Lemma 4 and Clam 1 mply that f ( (x 2, Ī), (x t, I\Ī)) = x 2. Furthermore, by Lemma 4, x t x 1 mples that f ( (x 2, Ī), (x 1, I\Ī)) = x 2. Consequently, snce I\Ī > Ī, anonymty and group strategyproofness mply f ( (x 1, Ī), (x 2, I\Ī)) = x 2. Contradcton to the hypothess of case ()! Next, assume that case () occurs. Let I = { N+1} and 2 I = { N+3,..., N}. Thus, we 2 partton I nto three parts: Ī, I and I. We clam that f ( (x 1, Ī), (x 1, I ), (x 3, I ) ) = 20

21 x 1. Frstly, Clam 1 mples that f ( (x 1, Ī), (x 1, I ), (x 3, I ) ) {x 1, x 3 }. Suppose that f ( (x 1, Ī), (x 1, I ), (x 3, I ) ) = x 3. Snce x 2 x 1, f ( (x 2, Ī), (x 2, I ), (x 3, I ) ) = x 3 by Lemma 4 and Clam 1. Relabelng the cycle such that a 1 = x 3, a 2 = x 2, a 3 = x 1 and a k = x t+4 k, k = 4,..., t, we have f ( (a 2, Ī), (a 2, I ), (a 1, I ) ) = f ( (x 2, Ī), (x 2, I ), (x 3, I ) ) = x 3 = a 1. Snce I = Ī, anonymty mples that f( (a 1, Ī), (a 2, I\Ī)) = a 1 whch s dentcal to case () and hence causes a contradcton. Therefore, f ( (x 1, Ī), (x 1, I ), (x 3, I ) ) = x 1. Then, by Lemma 4 and Clam 1, x 2 x 1 mples that f ( (x 2, Ī), (x 2, I ), (x 3, I ) ) = x 2. Now, snce x 1 x 2, Lemma 4 mples that f ( (x 1, Ī), (x 2, I ), (x 3, I ) ) {x 1, x 2 }. Suppose that f ( (x 1, Ī), (x 2, I ), (x 3, I ) ) = x 1. Then, by Lemma 4, x 2 x 3 mples that f ( (x 1, Ī), (x 2, I\Ī)) = f ( (x 1, Ī), (x 2, I ), (x 2, I ) ) = x 1. Contradcton to case ()! Therefore, f ( (x 1, Ī), (x 2, I ), (x 3, I ) ) = x 2. Followng the sequence {x k } t+1 k=3 and applyng Lemma 4 step by step, we have f ( (x 1, Ī), (x 2, I ), (x 1, I ) ) = f ( (x 1, Ī), (x 2, I ), (x t+1, I ) ) = x 2. Snce Ī I, anonymty and group strategy-proofness mply that f ( (x 2, Ī), (x 1, I\Ī)) = x 2. Relabelng the cycle such that a 1 = x 2, a 2 = x 1 and a k = x t+3 k, k = 3,..., t, we have f ( (a 1, Ī), (a 2, I\Ī)) = f ( (x 2, Ī), (x 1, I\Ī)) = x 2 = a 1 whch s dentcal to case () and hence causes a contradcton. Therefore, there exsts no cycle and hence G (D) s a tree. Lemma 7 Doman D s sngle-peaked on G (D). Proof : Gven P D, assume that r 1 (P ) = x. Gven a, b A wth a b and a x, b, we wll show that ap b. If a = x, the result holds evdently. We assume that a x. Let x, b = {x k } t k=1 where x 1 = x, x t = b and x k x k+1, k = 1,..., t 1. Thus, a {x k } t k=1. Suppose that bp a. Snce D s mnmally rch, we have P D b. Then, consstent connectedness mples that there exsts a path {P k } s k=1 D connectng P and P satsfyng the no-restoraton property wth respect to (a, b). Thus, bp a and bp a mply that bp k a for all 1 k s. Consequently, a r 1 (P k ) for any 1 k s. Moreover, accordng to the proof of Lemma 3, we nduce a sequence {y k } l k=1 A such that () y 1 = x, y l = b, y k y k+1, k = 1,..., l 1; and () for each 1 k l, y k = r 1 (P k ) for some 1 k s. Hence, a / {y k } l k=1. Consequently, we have two dstnct graph-paths: {x k} t k=1 and {y k} l k=1, between x and b n G (D), whch contradcts Lemma 6. Therefore, ap b. Ths completes the verfcaton of the lemma and hence the proposton. 8 Concluson Ths paper refnes and extends the results n Sato (2013). We provde a complete characterzaton for domans satsfyng equvalence f attenton s restrcted to unanmous socal choce functons. In the case of arbtrary socal choce functons, we provde a suffcent condton, 21

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