Minimal manipulability: anonymity and unanimity

Soc Choice Welfare (2007) 29:247 269 DOI 10.1007/s00355-006-0202-3 ORIGINAL PAPER Minimal manipulability: anonymity and unanimity Stefan Maus Hans Peters Ton Storcken Received: 6 July 2004 / Accepted: 14 September 2006 / Published online: 11 October 2006 Springer-Verlag 2006 Abstract This paper is concerned with the minimal number of profiles at which a unanimous and anonymous social choice function for three alternatives is manipulable. The lower bound is derived and examples of social choice functions attaining the lower bound are given. It is conjectured that these social choice functions are in fact all minimally manipulable social choice functions. Since some of these social choice functions are Pareto optimal, it follows that the lower bound also holds for Pareto optimal and anonymous social choice functions. Some of the minimally manipulable Pareto optimal and anonymous social choice functions can be interpreted as status quo voting. 1 Introduction The well-known result of Gibbard (1973) and Satterthwaite (1975) states that any strategy-proof surjective social choice function for more than two alternatives is dictatorial. This implies that the combination of anonymity and unanimity is not compatible with strategy-proofness. In this paper we investigate, for the case of three alternatives, how incompatible these conditions are, i.e., how much manipulability anonymous and unanimous social choice functions minimally admit. S. Maus H. Peters (B) T. Storcken University of Maastricht, Department of Quantitative Economics, P.O. Box 616, 6200 MD Maastricht, The Netherlands e-mail: h.peters@ke.unimaas.nl S. Maus e-mail: s.maus@ke.unimaas.nl T. Storcken e-mail: t.storcken@ke.unimaas.nl

248 S. Maus et al. There is a small literature on the degree of vulnerability to strategic behavior of classical social choice functions, such as Borda, Plurality, etc. The paper by Aleskerov and Kurbanov (1999) contains simulation and enumeration results on 26 different social choice functions for different indices of manipulability. Slinko (2002) counts the number of instable profiles of classical social choice functions, which is an upper bound for the number of manipulable profiles of these social choice functions. On the other hand, little is known about the extent of manipulability of social choice functions satisfying certain properties, in particular on lower bounds. A first investigation was carried out by Kelly (1988), who found the minimal number of manipulable profiles for nondictatorial surjective social choice functions with three alternatives and two agents and also formulated several conjectures. This line of research was continued by Fristrup and Keiding (1998), who determined the minimal number of manipulable profiles for two agents and any number of alternatives. Maus et al. (2004b) consider among other things the three alternative case and show that for any number of agents larger than two there are six minimally manipulable nondictatorial surjective social choice functions, and these are even anonymous. However, they also consider minimally manipulable unanimous and nondictatorial social choice functions. These turn out to be nondictatorial only in a minimal sense, namely at only one profile. Here we exclude such social choice functions by demanding, besides unanimity, that agents are treated anonymously. (Maus et al. 2004a contains results for this case with surjectivity instead of unanimity.) We look for the minimal number of manipulations. It is well known that in case of two alternatives any monotonic social choice function is strategy-proof. As, in that case, monotonicity and anonymity are compatible (see e.g. May 1952), the minimal number of manipulations is zero. We consider the three alternatives case and show that for n agents the minimal number is 2 n 2. Furthermore, we give examples of social choice functions reaching the lower bound and point out that some of these examples are Pareto optimal. Thus, 2 n 2 is also the lower bound for Pareto optimal and anonymous social choice functions. We argue that status quo voting is an example of a Pareto optimal and anonymous social choice function which is minimally manipulable for three alternatives. It should be noted that the way in which we measure manipulability, namely by counting manipulable profiles, is not the only possible one. There are many interesting variations, for example by taking the number of agents that can manipulate or the severity of the manipulations into account. The already mentioned paper by Aleskerov and Kurbanov (1999) contains an overview of such variations. The paper is organized as follows. Section 2 contains the basic definitions. Section 3 contains special results about the manipulability of two and three agent social choice functions that are used in the proof of the main result. The main result: any unanimous and anonymous social choice function for three alternatives and n agents has at least 2 n 2 manipulable profiles, is proved in Sect. 4. Examples of social choice functions attaining the lower bound are presented in Sect. 5. Section 6, finally, considers Pareto optimality instead of unanimity.

Minimal manipulability, anonymity, unanimity 249 2 Preliminaries We denote the cardinality of a finite set S by S. LetN be the set of natural numbers, and let N 0 := N {0}. Let denote the upper entier: for n N 0, n2 := min{l N0 l n 2 }. Let A be a finite set of alternatives, m := A 3, and N ={1,..., n} be a finite set of agents, n 2. Let t A A be a binary relation. We call t complete if (x, y) t or (y, x) t for all x, y A. Note that completeness of t implies (x, x) t for all x A. We call t transitive if (x, y) t and (y, z) t implies (x, z) A for all x, y, z A. We call t antisymmetric if (x, y) t and (y, x) t implies x = y for all x, y A. A preference t A A is a linear order on A, i.e., a complete, transitive, antisymmetric binary relation. Let P denote the set of all preferences. Let A ={x 1, x 2,..., x m }. By completeness, transitivity and antisymmetry we can write conveniently t = x 1 x 2...x m for the preference t such that (x i, x j ) t if and only if i j, i, j {1, 2,..., m}, t =...x...y... if we want to express only that x is strictly preferred to y, and t = x... if we want to express only that x is preferred to all other alternatives. A profile is a map p : N P. Thus, a profile assigns to every agent i a preference p(i) over the alternatives. For a nonempty subset S of N we denote by p S the restriction of the map p to the set S. The set of all profiles is denoted by P N. For a permutation σ of N and a profile p P N let p σ be the profile defined by (p σ)(i) := p(σ (i)) for all i N. Two profiles p, q P N are called anonymously equivalent if there is a permutation σ of N such that q = p σ. Then p 1 (t) = q 1 (t) for all t P. We denote anonymous equivalence between p and q in P N by p q and write [p] := {q P N q p}. For all p P N, [p] := n! t P p 1 (t)!. A social choice function is a function f : P N A. Hence, a social choice function selects a unique alternative f (p) at every profile p. A social choice function is surjective if any alternative in A is chosen at least once, i.e. if f (P N ) = A. In the literature this is also known as citizen-sovereignty. A social choice function is unanimous if f (p) = x for all profiles p P N such

250 S. Maus et al. that p(i) = x... for all i N. Clearly, unanimity is stronger than surjectivity. Throughout this paper we assume that any social choice function is (at least) surjective. A social choice function is anonymous if f ([p]) ={f (p)} for all p P N. Thus, anonymous social choice functions are symmetric in the arguments. In a sense they treat agents equally. In contrast to anonymity, the following social choice functions respect only the preference of a single agent. Let d N and for any profile p let dict d (p) := x where p(d) = x...,sox is the most preferred alternative of agent d in p. The social choice functions f = dict d, d N, aredictatorial social choice functions, and agent d is a dictator. A social choice function f is nondictatorial if there is no agent d such that f = dict d. We are interested in strategic behavior of individuals when facing cooperative decision-making as captured by social choice functions. This is formalized by the following definitions. Let f : P N A be a social choice function. Let p P N be a profile. Then each profile q such that q N {i} = p N {i} for some i N is an i-devation from p. Letting t := q(i) we use the notation q = (p i, t). If it is not important which agent deviates from p to q we call q a deviation from p. A profile p is manipulable (under f ) if there is an agent that is better off by being dishonest about his preference, i.e. if there is an i N and an i-deviation q such that (f (q), f (p)) p(i). In this case we say that p is manipulable towards q (under f ). Let M f :={p p N p is manipulable under f }. Note that p M f implies [p] M f if f is anonymous. A social choice function is strategy-proof if M f =, otherwise it is said to be manipulable. Let ˆF be a nonempty set of social choice functions. We call f minimally manipulable (in ˆF) if f arg min f ˆF M f. The prominence of the dictatorial rules arises from the following impossibility result due to Gibbard (1973) and Satterthwaite (1975). Theorem 2.1 Let f : P N A be a nondictatorial surjective social choice function. Then M f 1. This theorem makes no statement about the number of manipulable profiles such social choice functions admit. This question has been solved by Kelly (1988) for two agents and three alternatives, and by Fristrup and Keiding (1998) for two agents, and any number of alternatives larger than 3. Maus et al. (2004b) consider the case of unanimous and nondictatorial social choice functions for

Minimal manipulability, anonymity, unanimity 251 any numbers of agents and alternatives larger than or equal to three, and the case of surjective and nondictatorial social choice functions for three alternatives. Maus et al. (2004a) consider the case of anonymous and surjective social choice functions, but leave some cases open. The case of anonymous voting where the social choice function takes only the top elements of the agents preferences into account is considered in Maus et al. (2005). In the present paper, we address the case of unanimous and anonymous social choice functions for the case of three alternatives. We let F be the set of all such social choice functions with a set of agents N such that N =n 2. In Sect. 4 we prove that the minimal number of manipulable profiles for social choice functions in F is 2 n 2, and in Sect. 5 that this lower bound is tight. In the next section we first derive some auxiliary technical results for the cases of two and three agents. 3 Special results for two and three agents Since the connection of the results of this section with the main result (in Sect. 4) is far from obvious we sketch the proof of the main result beforehand. Let f F. In Sect. 4 we prove that there are natural numbers k 1, k 2 such that n 2 k 1 k 2 n, and profiles q k i M f for all n 2 k k i 1, k N, i {1, 2}, with the following properties. Each such q k i shows that there are at least ( n k) manipulable profiles in Mf, which are distinct from the ones induced by another q k i unless k = n 2 for both profiles. Furthermore, we show that if k i < n, i {1, 2}, then there is a p i M f that yields at least another n 1 ) different j=k i ( n j manipulable profiles in M f. Altogether we are going to have in any case n 1 ( ) n M f = 2 n 2, j j=1 which is our main result. We find the q k i by applying Theorem 3.1 to two agent social choice functions f k, n 2 k < k 2, derived from f. To show that the manipulable profiles arising from the same f k, k = n 2, are different we need Lemma 3.2. To apply Theorem 3.1 and Lemma 3.2 we need that the f k, k < k 2, are nondictatorial. This is shown by applying Lemma 3.3 to a three person social choice function g defined from f. This will also show that the p i exist if k i < n. Throughout the rest of this section let N ={1, 2} if there are two agents, and N ={1, 2, 3} if there are three agents. Two agent social choice functions can be depicted in 6 6 tables since there are only six preferences by letting agent 1 s preference be constant in the rows and agent 2 s preference be constant in the columns. We use this device at a few places in the proofs. The following result is taken from Kelly (1988). It can also be proved by considering 6 6 tables.

252 S. Maus et al. Theorem 3.1 Let f : P N A be a surjective two agent social choice function. Then f is either dictatorial or M f 2. Let D f : M f 2 N P be the correspondence assigning to each manipulable profile the set of agents that can manipulate and the preferences they can use to manipulate, i.e., for all p M f, D f (p) := {(i, t) N P (f (p i, t), f (p)) p(i)}. The first result shows some properties of D f for unanimous and nondictatorial two agent social choice functions. Lemma 3.2 Let f : P N A be a unanimous and nondictatorial two agent social choice function. Then: (i) For every p M f there is an (i, t p ) D f (p) such that t p / p(n). (ii) At least one of the following three statements is true: (a) There are p, q M f and (i, t p ) D f (p), (j, t q ) D f (q), satisfying (i), (b) such that i = j. There are p, q M f and (i, t p ) D f (p), (j, t q ) D f (q), satisfying (i), such that p(n) {t p } = q(n) {t q }. (c) For all p, q M f, there is a labelling of the alternatives in A ={a, b, c} such that p = (bca, cab), (i, cba) D f (p) q = (cba, bca), (i, cab) D f (q) f (p) = a. Proof (i) Let p M f and suppose without loss of generality (1, t p ) D f (p). Suppose t p p(n), hence p = (p(1), t p ).Let t p = t p be the preference such that t p = x... and t p = x... Then f ( t p, t p ) = f (t p, t p ) by unanimity of f.so (1, t p ) D f (p) as (1, t p ) D f (p), and t p / p(n) ={p(1), t p }. This proves (i). (ii) Suppose that (a) and (b) are not true. Note that, since (a) does not hold and by the proof of (i), all profiles in M f are manipulable by the same agent i. Without loss of generality let i = 1. Fix p, q M f. Assume (1, t p ) D f (p) and (1, t q ) D f (q) areasin(i).we show (c). Let O 2 (t) := {f (p) p(1) = t} be the set of alternatives that agent 2 can induce if the preference of agent 1 is fixed at t. LetA ={x, y, z}. Note that, for any preference t with top element u A, we have u O 2 (t) by unanimity. We now fix p(1) = xyz and in several steps analyze the set O 2 ( ). Step 0: x O 2 (xyz). As already observed, this is a consequence of unanimity. Step 1: O 2 (xyz) = {x}. If O 2 (xyz) ={x} then f (p) = x hence agent 1 cannot manipulate at p since p(1) = xyz, a contradiction.

Minimal manipulability, anonymity, unanimity 253 Step 2: O 2 (xzy) = {x}. Suppose to the contrary that O 2 (xzy) = {x}. ByStep1thereisau O 2 (xyz) {x}. Lett 1 = t 2 be the two preferences such that t i = u..., i = 1, 2. Then f (xyz, t i ) = u for i {1, 2} since agent 2 cannot manipulate, and so (xyz, t i ) is manipulable by agent 1 towards (xzy, t i ) for i {1, 2}. Since {xyz, t 1, xzy} = {xyz, t 2, xzy}, (b) holds, a contradiction. Step 3: z O 2 (xzy). Suppose to the contrary using Step 2 that O 2 (xzy) = {x, y}. Then f (xzy, zyx) = y since agent 2 cannot manipulate, and f (zxy, zyx) = z by unanimity. So r = (xzy, zyx) is manipulable by agent 1 towards (zxy, zyx) = (t r, r(2)). Since p(1) = xyz / r(n) {t r } this contradicts our assumption that (b) does not hold. Step 4: z O 2 (xyz). Suppose to the contrary using Step 1 thato 2 (xyz) ={x, y}. Then, by Step 2 and since agent 2 cannot manipulate, f (xzy, zxy) = z and, again since agent 2 cannot manipulate, f (xyz, zxy) = x. Hence, r = (xzy, zxy) is manipulable by agent 1 towards (xyz, zxy) = (t r, r(2)). Since (b) does not hold we have p(n) {t p }={xyz, p(2), t p }=r(n) {t r }={xyz, xzy, zxy}. Ifp(2) {xyz, xzy} then f (p) = x by unanimity. If p(2) = zxy, then f (p) = x since O 2 (xyz) ={x, y} and agent 2 cannot manipulate. But f (p) = x contradicts the manipulability of p by agent 1. Step 5: z O 2 (yxz) and z O 2 (yzx). Suppose to the contrary that there is a t {yxz, yzx} such that z / O 2 (t), so O 2 (t) {x, y}. Letr 1 = (xyz, zxy), r 2 = (xyz, zyx). By Step 4 we have z O 2 (xyz), and so f (r 1 ) = f (r 2 ) = z. Since z / O 2 (t) we have f (r 1 1, t) {x, y} and f (r 1 2, t) {x, y}, sor1 and r 2 are manipulable towards (r 1 1, t) and (r 1 2, t) respectively. Clearly, this contradicts the assumption that (b) is not satisfied. Step 6: O 2 (xzy) ={x, z}. Suppose to the contrary using Steps 0 and 3 that O 2 (xzy) = A. By Steps 0, 2, and 4 either O 2 (xyz) ={x, z} or O 2 (xyz) = A. Consider first the case in which O 2 (xyz) ={x, z}. Letr 1 = (xzy, yxz), r 2 = (xzy, yzx). Since O 2 (xzy) = A we have f (r 1 ) = f (r 2 ) = y, whereas f (r 1 1, xyz) = x and f (r2 1, xyz) = z since O 2 (xyz) ={x, z}. Hence, r 1 and r 2 are manipulable by agent 1 towards (r 1 1, xyz) and (r 1 2, xyz) respectively. But then (b) is satisfied, a contradiction. Thus, suppose that O 2 (xyz) = A. By Step 5, unanimity and since agent 2 cannot manipulate, we have f (P {zxy, zyx}) = {z}, hence no manipulation by agent 1 can occur if p(2) {zxy, zyx}. By unanimity, agent 1 cannot manipulate if p(2) {xyz, xzy}. Hence, p(2) {yxz, yzx}. Then f (p) = y since O 2 (xyz) = A and agent 2 cannot manipulate, and f (t p, p(2)) = x since agent 1 manipulates p towards (t p, p(2)). Since O 2 (xzy) = A implies that f (xzy, p(2)) = y (as agent 2 cannot manipulate), and unanimity implies that f (yxz, p(2)) = f (yzx, p(2)) = y, we obtain t p {zxy, zyx}. Since f (t p, p(2)) = x the profile r = (xzy, p(2)) is

254 S. Maus et al. also manipulable by agent 1, since f (r) = y (because O 2 (xyz) = A and agent 2 cannot manipulate). Then (b) holds for p and r, a contradiction. Step 7: y / O 2 (zyx). Suppose to the contrary that y O 2 (zyx). Then f (r) = y for r = (zyx, yzx) since agent 2 cannot manipulate, so r is manipulable towards (xzy, yzx) = (t r, r(2)), asf (xzy, yzx) = z by Step 6 and since agent 2 cannot manipulate. Since p(1) = xyz / r(n) {t r } this contradicts the assumption that (b) does not hold. Step 8: O 2 (zyx) ={z}. By Step 7, y / O 2 (zyx), and by unanimity, z O 2 (zyx). Suppose to the contrary O 2 (zyx) ={x, z}. Then r = (zyx, yxz), withf (r) = x as agent 2 cannot manipulate, is manipulable towards (yzx, yxz) = (t r, r(2)), since f (yzx, yxz) = y by unanimity. Since p(1) = xyz / r(n) {t r } this contradicts the assumption that (b) does not hold. Step 9: O 2 (zxy) ={z}. Suppose to the contrary that there is a t P such that f (zxy, t) = z. By unanimity, t / {zxy, zyx}. Then r = (zxy, t) is manipulable towards (zyx, t) = (t r, r(2)) by Step 8. The assumption that (b) does not hold implies p(1) = t = xyz and {p(2), t p }={zxy, zyx}. In that case f (p 1, t p ) = f (t p, p(2)) = z by unanimity, but agent 1 in p = (xyz, p(2)) cannot manipulate to a profile (p 1, t p ) where outcome z obtains. Thus, this contradicts the assumption that agent 1 manipulates at p towards (p 1, t p ). Step 10: O 2 (yzx) ={y, z}. Suppose to the contrary using Step 5 and unanimity that O 2 (yzx) = A. Then r = (yzx, xzy), withf (r) = x since agent 2 cannot manipulate, is manipulable towards (zxy, xzy) = (t r, r(2)) by Step 9. As p(1) = xyz / r(n) {t r } this contradicts the assumption that (b) does not hold. We summarize the consequences of Steps 0 10 in Table 1. The rows correspond to player 1 s preferences and the columns to player 2 s preferences. We finish the proof by considering the two cases O 2 (xyz) = {x, z} and O 2 (xyz) = A. If O 2 (xyz) ={x, z} then f (xyz, yxz) = x since agent 2 cannot manipulate. Then r = (xyz, yzx), f (r) = z, is manipulable towards (yxz, yzx) = (t r, r(2)), and r is the only manipulable profile such that r(1) = xyz. Hence, p = r. Assume Table 1 0 10 Summary of Steps 1 2 xyz xzy yxz yzx zxy zyx xyz x x z z xzy x x x z z z yxz y y z z yzx y z y y z z zxy z z z z z z zyx z z z z z z

Minimal manipulability, anonymity, unanimity 255 q = p (this is possible by Theorem 3.1). Since (b) is not satisfied we must have q(n) {t q }={xyz, yxz, yzx}. Since q(1) = xyz we have q(1) {yxz, yzx}. Since q is manipulable, by considering Table 1 we obtain to q = (yxz, xyz), t q = yzx. But then (c) holds with i = 1, x = b, y = c and z = a. On the other hand, if O 2 (xyz) = A, then r = (xyz, yxz), f (r) = y, is manipulable towards (xzy, yxz) = (t r, r(2)), and r is the only manipulable profile such that r(1) = xyz. Hence, p = r. If (b) is not satisfied we must have q(n) {t q }={xyz, xzy, yxz}. Since q(1) = xyz and clearly also q(1) = xzy (otherwise q {(xzy, xyz), (xzy, yxz)}, sof (q) = x) we have q(1) = yxz. So, q {(yxz, xyz), (yxz, xzy)}. Now,ifq = (yxz, xyz) then t q = xzy, so agent 1 manipulates q towards (xzy, xyz), where he obtains the outcome x = f (xzy, xyz). This implies that x is preferred to f (q) under the preference q(1) = yxz, hence f (q) = z. This is a contradiction, since then agent 2 can manipulate q as well. So, only the case q = (yxz, xzy), t q = xyz, remains. For the same reasons as in the other case the manipulability of q towards (xyz, xzy) by agent 1 implies that f (q) = z. But then (c) holds with i = 1, x = c, y = b, z = a, and exchanging the roles of p and q. This completes the proof of the lemma. The next result is about manipulability of unanimous three agent social choice functions where coalitions are dictators. The nonempty coalition S N is a dictator if for all t = xyz P and q S P N S we have In this case, f is S-dictatorial. f (t S, q S ) = x. Lemma 3.3 Let f : P N A be a unanimous three agent social choice function. Suppose that f is {1, 2}-dictatorial and {1, 3}-dictatorial. Then, one of the following is true. (a) (b) Agent 1 is a dictator. There are anonymously inequivalent p, q M f, manipulable towards (p i, t p ), (q j, t q ), i, j {1, 2, 3}, respectively, such that p(n) = q(n) =3 and p(n) {t p } = q(n) {t q }. Proof Let ft 2 : P {1,2} A, t P, be the two agent social choice function defined by ft 2(p) := f (p(1), p(2), t) for all p P{1,2}.Letft 3 : P {1,3} A, t P,bethetwo agent social choice function defined by ft 3(p) := f (p(1), t, p(3)) for all p P{1,3}. Suppose that (b) is not true. We show that all ft 2 are dictatorial with dictator 1, so that (a) holds. By unanimity and {1, 3}-dictatoriality ({1, 2}-dictatoriality) of f we have some information about ft 2 and ft 3.LetA ={x, y, z}. We summarize this information in Table 2 for t = xyz, where agent 1 s preference is described by the rows, as before. The other agent can be either agent 2 or agent 3, and the remaining of the three agents is fixed to the preference xyz. The proof proceeds in a number of steps, in which Table 2 is gradually completed. In Steps 1 4 we write f t ( ) instead of f 2 ( ).

256 S. Maus et al. Table 2 f 2 t and f 3 t for t = xyz 1 2(3) xyz xzy yxz yzx zxy zyx xyz x x x x x x xzy x x yxz y yzx y zxy z zyx z Step 1: f xyz (xzy, t) = x for all t P. (i.e., the cells in the second row of the table can contain only x s.) Suppose to the contrary that u := f xyz (xzy, t) {y, z} for some t P. Let A ={u, v, w}, and p = (xzy, uvw, xyz), q = (xzy, uwv, xyz). Iff (p) = u then p is manipulable towards (p 2, t) =: (p 2, t p ), and if f (p) = u then p is manipulable towards (p 1, xyz) =: (p 1, t p ). The same holds for q. Clearly, p and q are anonymously inequivalent and satisfy p(n) = q(n) =3. Furthermore, p(n) {t p }=q(n) {t q } implies that t p = uwv and t q = uvw. Asu = x this implies that agent 2 is manipulating at p and q and so t p = t = t q, contradicting t p = uwv, t q = uvw. Step 2: f xyz is unanimous. It remains to be shown that f xyz (yxz, yzx) = f xyz (yzx, yxz) = y and f xyz (zxy, zyx) = f xyz (zyx, zxy) = z. But, for example, by applying a similar argument as in Step 1 to fyxz 3 we obtain f xyz(yzx, yxz) = f (yzx, yxz, xyz) = fyxz 3 (yzx, xyz) = y. The other parts of unanimity of f xyz follow in the same way. Step 3: f xyz (r) = z if r(1) {yxz, yzx}. (I.e., in the third and fourth rows z cannot be chosen.) Suppose to the contrary that r(1) {yxz, yzx} and f xyz (r) = z. Letp = (r(1), zxy, xyz) and q = (r(1), zyx, xyz). Iff (p) = z then p is manipulable towards (p 2, r(2)) =: (p 2, t p ), and if f (p) = z then p is manipulable towards (p 3, r(1)) =: (p 3, t p ), since by {1, 3}-dictatoriality of f we have f (r(1), zxy, r(1)) = y. The same holds for q. Clearly, p and q are anonymously inequivalent and satisfy p(n) = q(n) =3. Furthermore, p(n) {t p }=q(n) {t q } implies that t p = zyx and t q = zxy. Asr(1) {yxz, yzx} this implies that agent 2 is manipulating at p and q and t p = r(2) = t q, contradicting t p = zyx, t q = zxy. Hence, (b) is satisfied, a contradiction. Step 4: f xyz (yzx, zxy) = y. Let p = (yzx, zxy, xyz) and q = (yzx, zxy, xzy). By Step 3 we have f (p) {x, y}. Suppose contrary to the claim of Step 4 that f (p) = x. Then p is manipulable towards (p 1, zxy) =: (p 1, t p ) by {1, 2}-dictatoriality of f.iff (q) = x then q is manipulable towards p = (q 3, xyz) =: (q 3, t q ), and if f (q) = x then q is manipulable towards (q 1, zxy) by {1, 2}-dictatoriality of f. Clearly,

Minimal manipulability, anonymity, unanimity 257 p and q are anonymously inequivalent and satisfy p(n) = q(n) =3. Since xzy q(n) but xzy / p(n) {t p } it follows that (b) is satisfied, a contradiction. Step 5: f xyz (yzx, t) = y for all t P. (I.e., in the fifth row y has to be chosen.) We show Step 5 for fxyz 2. The proof for f xyz 3 is analogous. By Step 3 we have that fxyz 2 (yzx, t) {x, y} for all t P. Furthermore, f xyz 2 (yzx, zyx) = f zyx 3 (yzx, xyz) {x, y} {y, z} = {y} by Step 3 applied to fxyz 2 and (the analogue of Step 3) to fzyx 3. Hence, in view of Step 4, the claim is true if t {yxz, yzx, zxy, zyx}. Suppose, contrary to the claim, that fxyz 2 (yzx, t) = x for some t {xyz, xzy}.let p = (yzx, zxy, xyz). Then p is manipulable towards (p 2, t) =: (p 2, t p ),asf(p) = fxyz 2 (yzx, yxz) = y and f (p 2, t) = fxyz 2 (yzx, t) = x. Letq1 = (yxz, zxy, xyz), q 2 = (yzx, xzy, xyz) and q 3 = (yxz, xzy, xyz). ByStep3,f (q i ) {x, y} for all i {1, 2, 3}. We show that for all possible combinations of f (q i ) {x, y}, i {1, 2, 3}, (b) is satisfied, contradicting our assumptions. If f (q 1 ) = x then q 1 is manipulable towards (q 1 1, yzx) =: (q1 1, t q1) = (yzx, zxy, xyz), asf (yzx, zxy, xyz) = y. Clearly, p and q 1 are anonymously inequivalent and satisfy p(n) = q 1 (N) = 3. As q 1 (1) = yxz / p(n) {t p } {yzx, zxy, xyz, xzy}, (b) is satisfied. Hence, we must have f (q 1 ) = y. If f (q 2 ) = y then q 2 is manipulable towards (q 2 2, t) =: (q2 2, t q2) = (yzx, t, xyz), asf (yzx, t, xyz) = x. Clearly, p and q 2 are anonymously inequivalent and satisfy p(n) = q 2 (N) =3. As p(2) = zxy / q 2 (N) {t q 2} {yzx, xzy, xyz}, (b) is satisfied, a contradiction. Hence, we must have f (q 2 ) = x. If f (q 3 ) = y then q 2 is manipulable towards (q 2 1, yxz) =: (q2 1, t q 2) = q3, as f (q 2 ) = x. Then p and q 2 are anonymously inequivalent, satisfy p(n) = q 2 (N) =3, and p(2) = zxy / q 2 (N) {t q 2} {yzx, xzy, xyz, yxz}, so(b)is satisfied. Hence, f (q 3 ) = x. But then q 1 is manipulable towards (q 1 2, xzy) =: (q 1 2, t q 1) = q3,asf(q 1 ) = y. Then p and q 1 are anonymously inequivalent, satisfy p(n) = q 2 (N) =3, and q 1 (1) = yxz / p(n) {t p } {yzx, zxy, xyz, xzy}, so (b) is satisfied. This completes the proof of Step 5. Step 6: f xyz (yxz, t) = y for all t P. (I.e., in the fourth row y has to be chosen.) We show Step 6 for fxyz 2. The proof for f xyz 3 is analogous. By Step 3, f xyz 2 (yxz, t) {x, y} for all t P. Furthermore, fxyz 2 (yxz, zyx) = f zyx 3 (yxz, xyz) {x, y} {y, z} ={y} by Step 3 applied to fxyz 2 and (the analogue of Step 3) to f zyx 3. Hence, the claim is true if t {yxz, yzx, zyx}. We extend this to {xzy, yxz, yzx, zxy, zyx}. So suppose, to the contrary, that t {xzy, zxy} and fxyz 2 (yxz, t) = x. Then, by Step 5, p = (yxz, t, xyz) is manipulable by agent 1 towards (yzx, t, xyz) =: (t p, t, xyz). Ast {xzy, zxy}, p(n) = 3. Consider t {xzy, zxy} {t} and let q = (yxz, t, xyz). Iff (q) = x then q is manipulable by agent 1 towards (yzx, t, xyz) =: (t q, t, xyz). Iff (q) = y then q is manipulable by agent 2 towards (yxz, t, xyz) =: (yxz, t q, xyz). In both cases q(n) =3 since t {xzy, zxy}. Furthermore, p(n) {t p } = q(n) {t q },as t q(n) but t / p(n) {t p }= {yxz, yzx, t, xyz}. Hence, (b) is satisfied, a contradiction. So, f xyz (yxz, t) = y

258 S. Maus et al. for all t P {xyz}. It remains to be shown that f xyz (yxz, xyz) = y. Suppose to the contrary that f xyz (yxz, xyz) = x. Then p = (yxz, xzy, xyz) and q = (yxz, zxy, xyz), where by Step 5 we have f (p) = f (q) = y, are both manipulable by agent 2 towards (yxz, xyz, xyz). So (b) is satisfied, a contradiction. This proves Step 6. Step 7: fxyz 2 can be described by Table 3. Rows 1 4 and the four cells in the lower right corner follow by Steps 1, 5 and 6 applied to fxyz 2. By Steps 5 and 6 applied to f yzx 3 we see that fxyz 2 (t, yzx) = f yzx 3 (t, xyz) = z for all t {zxy, zyx}. Likewise, f xyz 2 (t, xzy) = z for all t {zxy, zyx} by applying Steps 5 and 6 to fxzy 3. We prove now that f is dictatorial with dictator 1. This is equivalent to dictatoriality of ft 2 for all t P. To the contrary, suppose that ft 2 is nondictatorial for some t P. Without loss of generality t {xyz, yxz}. Note that Table 3 is similar for both these t. In view of Step 7 this implies that there is an r V := {zxy, zyx} {xyz, yxz} such that u = ft 2 (r) {x, y}. Without loss of generality r(1) = zxy, otherwise exchange the roles of x and y, which is allowed since we only fixed t {xyz, yxz}. LetA ={u, v, z}, p = (r(1), uzv, t). Then p is manipulable towards (p 2, r(2)) =: (p 2, t p ) = (r, t), since ft 2 (r) = u and f (p) = z. Ifthereisan r {zyx} {xyz, yxz} such that ũ = ft 2 ( r) {x, y}, A ={ũ, z, ṽ} then, likewise, one obtains that q = ( r(1), ũzṽ, t) is manipulable towards (q 2, r(2)) =: (q 2, t q ) = ( r, t). Clearly, such p, q, t p and t q satisfy (b), a contradiction. Hence, we must have f 2 t (zyx, t) = z for all t P. Let q = (zxy, yxz, t). Iff (q) = z then q is manipulable towards (q 2, r(2)) =: (q 2, t q ) = (r, t),asf (r, t) {x, y}. Clearly, p and q are anonymously inequivalent and satisfy p(n) = q(n) =3. Since p(2) = uzv / q(n) {t q } {zxy, yxz, xyz} it follows that (b) is satisfied, a contradiction. Hence, f (q) {x, y}. But then q is manipulable towards (q 1, zyx) =: (q 1, t q ), since f (zyx, q) = ft 2 (zyx, q(2)) = z. Again, p and q are anonymously inequivalent, satisfy p(n) = q(n) =3, and p(2) = uzv / q(n) {t q } {zxy, yxz, xyz, zyx}. Hence, (b) is satisfied, and we have obtained contradictions in all cases. So, there is no nondictatorial f t and this completes the proof. Table 3 Step 7 1 2 xyz xzy yxz yzx zxy zyx xyz x x x x x x xzy x x x x x x yxz y y y y y y yzx y y y y y y zxy z z z z zyx z z z z

Minimal manipulability, anonymity, unanimity 259 4 The lower bound In this section we prove the lower bound min f F M f 2 n 2, using the results from Sect. 3. Define k : P N { n 2,..., n} by { k(p) := min p 1 (T) T P, p 1 (T) n } 2 and l : P N {0,..., n 2 } by l(p) := n k(p). We first state a simple combinatorial observation. Lemma 4.1 Let k, l N. Then Proof k!l! (k + l 1)! (k + l 1)! =k!(k + 1) (k + l 1) k! 2 3 (k + l 1 (k 1)) = k! l!. The following combinatorial inequality holds for profiles that contain at least three different preferences. Lemma 4.2 Let p be a profile such that p(n) 3. Then [p] l(p) 1 j=0 ( ) n. k(p) + j Proof Let T be such that p 1 (T) =k(p). Then n! [p] = t P p 1 (t)! n! = t T p 1 (t)! t P T p 1 (t)! n! (k(p) T +1)!(l(p) P + T +1)! ( ) k(p)! l(p)! n = (k(p) T +1)! (l(p) P + T +1)! k(p) ( ) l(p) 1 n ( ) n l(p). k(p) k(p) + j j=0

260 S. Maus et al. Here, the first inequality holds since p 1 (t)! (k(p) T +1)! t T for all T P by applying Lemma 4.1 repeatedly. Furthermore, by p(n) 3 either P T 2 and or T 2 and l(p)! (l(p) P + T +1)! l(p)! (l(p) 1)! = l(p) k(p)! (k(p) T +1)! k(p)! = k(p) l(p), (k(p) 1)! which justifies the second inequality. Finally the third inequality holds as ( ) ( ) n n k(p) k(p) + j for all j {0,..., l(p) 1}, since k(p) n 2. Let Mf 3 :={p M f p(n) 3}, and if Mf 3 = let p 1 arg min k(p), k 1 := k(p 1 ), l 1 := l(p 1 ), p Mf 3 and if M 3 f = let k 1 = k 2 = n. Furthermore, if k 1 < n and M 3 f [p 1] = let p 2 arg min p M 3 f [p 1] k(p), k 2 := k(p 2 ), l 2 := l(p 2 ), and if Mf 3 [p 1]= let k 2 = n. The next lemma shows, using the results from Sect. 3, that for all k { n 2,..., k1 1} we have two profiles q k 1, qk 2 M f such that q k 1 (N) = qk 2 (N) = 2 and k(q k 1 ) = k(qk 2 ) = k. Ifk = n 2 we show that these two profiles are anonymously inequivalent. For all k {k 1,..., k 2 1} we show that there is one profile q k 1 M f such that q k 1 (N) =2 and k(qk 1 ) = k. Now note that profiles p, q P N, such that p(n) = q(n) =2 and k(p) = k(q) are anonymously

Minimal manipulability, anonymity, unanimity 261 inequivalent. Altogether, this shows that, if n is odd, M f = k 2 1 k= n 2 k 2 1 k= n 2 k 2 1 k= n 2 k 1 1 [q k 1 ] ( ) n k ( ) n k + + k= n 2 k 1 1 k= n 2 k 1 1 k= n 2 n 1 ( ) n = = 2 n 2, k k=1 [q k 2 ] [p 1] [p 2 ] ( ) n + [p 1 ] + [p 2 ] k ( ) n + k l 1 1 j=0 ( ) n + k 1 + j l 2 1 j=0 ( ) n k 2 + j where we use Lemma 4.2 for the second inequality. Likewise one shows that M f 2 n 2ifnis even, the only difference is that the two profiles q n 2 1, q n 2 2 do not have to be anonymously inequivalent. Lemma 4.3 Let f : P N A be a unanimous and anonymous social choice function. Let k 1, k 2, p 1, p 2 be as defined above. Then for all k { n 2,..., k1 1} there are two profiles q k 1, qk 2 M f such that q k 1 (N) = qk 2 (N) =2 and k(qk 1 ) = k(q k 2 ) = k. If k = n 2 these two profiles are anonymously inequivalent. For all k {k 1,..., k 2 1} there is a profile q k 1 M f such that q k 1 (N) = 2 and k(q k 1 ) = k. Proof We will apply Lemma 3.2 to the following two agent social choice functions to show the existence of the q k i. Let k K := { n 2,..., n}, and l := n k. Define fk : P {1,2} A by f k (p) := f (p(1) k, p(2) l ) for all p P {1,2}, where (p(1) k, p(2) l ) denotes a profile q in P N such that q 1 (p(1)) =k and q 1 (p(2)) =l. Since f is anonymous, f k is well defined and since f is unanimous, f k is unanimous. To apply Lemma 3.2 to an f k, k k 2, we have to show that f k is nondictatorial. Let d be the smallest k K such that f k is dictatorial, d := min{k K f k is dictatorial}. Note that f n is dictatorial with dictator 1, so d n. We show that k 2 d, which implies that all f k, k k 2, are nondictatorial.

262 S. Maus et al. Claim: k 2 d. If d = n this is trivial. Suppose that d < n. Letl := n d. Since f is anonymous, d > n 2 l. Define the three agent social choice function g : P{1,2,3} A by g(p) := f (p(1) d l, p(2) l, p(3) l ), where (p(1) d l, p(2) l, p(3) l ) must be read similarly to (p(1) k, p(2) l ) above. Since f is anonymous, g is well defined, and since f is unanimous, g is unanimous. Furthermore, since f d is dictatorial, g is {1, 2}- and {1, 3}-dictatorial. So we can apply Lemma 3.3 to g. If part (a) of Lemma 3.3 holds, i.e., if agent 1 is a dictator, then f d, where d := max{2l, d l} K, is dictatorial. If d < d then, as d n 2, this contradicts the minimality of d. So d d, which implies that 2l d. If2l = d then g is anonymous, contradicting dictatoriality. So, l > d l. Consider the profiles r i := (abc d l+i, bca l i, cab l ), i {0,...,2l d}. Then, f (r 0 ) = a, f (r 2l d ) = b,asf is anonymous and g is dictatorial with dictator 1. But any r i+1 is a j-deviation of r i for an agent j such that (b, a) r i (j). Hence, one of the r i is manipulable. Clearly, r i Mf 3 and k(ri ) d for this r i.forthe same reasons an r j, r j := (acb d l+j, bca l j, cab l ), j {0,...,2l d}, is manipulable, in Mf 3, and k(rj ) d for this r j.asthisr j is anonymously inequivalent to r i we have k 2 max{k(r i ), k(r j )} d. So, suppose that part (b) of Lemma 3.3 is true. Then we have anonymously inequivalent p, q M g, manipulable towards (p i, t p ), (q j, t q ), i, j {1, 2, 3}, respectively, such that p(n) = q(n) =3 and p(n) {t p } = q(n) {t q }.If i {2, 3} then, by anonymity of f, without loss of generality i = 2. The same holds for j.let r s := { (p(1) d l s, tp s, p(2)l, p(3) l ) if i = 1, s {0,..., d l 1}, (p(1) d l, p(2) l s, tp s, p(3)l ) if i = 2, s {0,..., l 1}, and r u := { (q(1) d l s, tq s, q(2)l, q(3) l ) if j = 1, u {0,..., d l 1}, (q(1) d l, q(2) l s, tq s, q(3)l ) if j = 2, u {0,..., l 1}. By manipulability of p and q under g there are such r s and r u that are manipulable under f.as p(n) = q(n) =3, r s, r u Mf 3.Asp and q are anonymously inequivalent and p(n) {t p } = q(n) {t q } we have [r s ] [r u ].So, k 2 max{k(r s ), k(r u )} d. This completes the proof of the Claim. Let k { n 2,..., k2 1}. By the Claim, f k is nondictatorial. So, by Theorem 3.1, there are p, q M fk. By the first part of Lemma 3.2 there are (i, t p )

Minimal manipulability, anonymity, unanimity 263 D fk (p), (j, t q ) D fk (q) such that t p / p(n), t q / q(n). Let { r s (p(1) k s, p(2) l, tp s ) if i = 1, s {0,..., k 1}, := (p(1) k, p(2) l s, tp s ) if i = 2, s {0,..., l 1}, and { r u (q(1) k u, q(2) l, tq u ) if i = 1, u {0,..., k 1}, := (q(1) k, q(2) l u, tq u ) if i = 2, u {0,..., l 1}. By manipulability of p and q under f k there are s and u such that r s, r u M f are manipulable towards r s+1 or r u+1, respectively. Let s be the smallest index s such that an r s is manipulable, i.e., the manipulation does not have to occur towards r s+1.letu be the smallest index u such that r u is manipulable towards r u+1. The reason for these choices will become apparent in case (c). Suppose that k { n },..., k 1 1. 2 As k(r s ) k < k 1, k(r u ) k < k 1, r s (N) 3ifs 1 and r u (N) 3if u 1, we must have s = 0 and u = 0. So q k 1 := (p(1)k, p(2) l ) = r 0 M f and q k 2 := (q(1)k, q(2) l ) = r 0 M f satisfy q k 1, qk 2 M f, q k 1 (N) = qk 2 (N) =2 and k(q k 1 ) = k(qk 2 ) = k. Ifk = n 2 we are done. If k = n 2 then k > l and choosing p = q implies that q k 1 and qk 2 are anonymously inequivalent. So, for k { n 2,..., k1 1} we are done. Suppose that k {k 1,..., k 2 1}. We are done if s = 0oru = 0. So, suppose to the contrary that s > 0 and u > 0. Using cases (a) to (c) of Lemma 3.2 we show that p and q can be chosen in such a way that r s r u.ask(r s ) k, k(r u ) k, r s (N) 3and r u (N) 3ifu 1, this contradicts k < k 2. Case (a): We can choose p and q in M fk that are manipulable by different agents, say without loss of generality p by agent 1 and q by agent 2. (Note that here we have to allow for p = q.) Then so r s r u. max t P (rs ) 1 (t) =max{k s, l, s} k 1 < k = (r u ) 1 (q(1)), Case (b): We can choose p and q in M fk such that p(n) {t p } = q(n) {t q }. Then r s r u is obvious.

264 S. Maus et al. Case (c): Either i = j = 1 and or i = j = 2 and p = (bca, cab), (1, cba) D fk (p) q = (cba, bca), (1, cab) D fk (q) f k (p) = a p = (cab, bca), (2, cba) D fk (p) q = (bca, cba), (2, cab) D fk (q) f k (p) = a. Suppose that i = j = 1 and r s r u. Then r s = (bca k s, cba s, cab l ) (cba k u, cab u, bca l ) = r u. This implies that k s = l, s = k u and l = u. Note that f (r 0 ) = f k (p) = a is the least preferred alternative of the manipulating agent, who has the preference bca. By the minimality of s and the anonymity of f this implies that f (r 0 ) = f (r 1 ) =...f (r s 1 ) = f (r s ) = f (r u ) = a. Furthermore, as (cba k u, cab u, bca l ) is manipulable by an agent with preference cba, we have f (cba k u 1, cab u+1, bca l ) {b, c}. Hence, f (cba k u 1, cab u+1, bca l ) = f (cba s 1, cab l+1, bca k s ) {b, c}. But then r s 1 is manipulable, as f (r s 1 ) = f (bca k s+1, cba s 1, cab l ) = a, and by deviating to cab any agent with preference bca can manipulate. This contradicts the minimality of s. Suppose that i=j=2. Then r s =(bca k, cab l s, cba s ), r u = (cba k, bca l u, cab u ). Since k > l and hence k > l u, we have r s r u. Hence, in any of the cases (a) to (c), p and q can be chosen in such a way that r s r u. This contradicts k < k 2 if s > 0 and u > 0. So, without loss of generality u = 0, and letting q k 1 = r0 M f we have q k 1 (N) =2 and k(qk 1 ) = k. This completes the proof of the lemma. By the remarks before Lemma 4.3 we can now state the following theorem. Theorem 4.4 Let f : P N A be a unanimous and anonymous social choice function. Then M f 2 n 2. 5 Social choice functions attaining the lower bound Let F :={f F M f =2 n 2} be the set of all social choice functions attaining the lower bound. In this section we formulate a conjecture about F, saying that f F, n 4, if and only if f belongs to one of three classes of

Minimal manipulability, anonymity, unanimity 265 social choice functions which we describe completely. The situation is somewhat different for n {2, 3}. First, one class is separately defined for n 3, since its definition changes slightly, compared to n 4. Furthermore, for n = 3 a fourth extra class of social choice functions is needed. Altogether we define four classes C i, i {1, 2, 3, 4}, of social choice functions such that C i F.For any x A and p P N let Pa(x, p) := {y A {x} (y, x) p(i) for all i N} be the set of all alternatives that Pareto dominate x in p. In 1 5 below it is to be understood that indefiniteness is resolved in such a way that anonymity and unanimity hold. Let A ={x, y, z}. 1. C 1 F is the set of all social choice functions f such that z f (p) = x y if Pa(z, p) = if Pa(z, p) ={x} if Pa(z, p) ={y} and f (p) = f (q) {x, y, z} for all q P N such that Pa(z, p) = Pa(z, q) ={x, y}, [q] =[p]. 2. C 2 F is the set of all social choice functions f such that z x f (p) = y z x if Pa(z, p) = if Pa(z, p) ={x} if Pa(z, p) ={y} and p / [(xyz n 1, yzx)] if p [(xyz n 1, yzx)] if p [(xyz n 1, yxz)] and f (p) = f (q) {x, y, z} for all p, q P N such that Pa(z, p) = Pa(z, q) = {x, y}, [p] =[q] = [(xyz n 1, yxz)]. 3. For n 4, C 3 F is the set of all social choice functions f such that z if Pa(z, p) = x if Pa(z, p) ={x} and p / [(yxz n 1, xzy)] y if Pa(z, p) ={y} and p / [(xyz n 1, yzx)] f (p) = z if p [(yxz n 1, xzy)] z if p [(xyz n 1, yzx)] x if p [(xyz n 1, yxz)] [(xyz n 2, yxz 2 )] y if p [(yxz n 1, xyz)] [(yxz n 2, xyz 2 )] and f ([p]) {x, y, z} if Pa(z, p) ={x, y} and p / [(xyz n 1, yxz)] [(xyz n 2, yxz 2 )] [(yxz n 1, xyz)] [(yxz n 2, xyz 2 )].

266 S. Maus et al. 4. For n = 3, C 3 F is the set of all social choice functions f such that z x y f (p) = z z x y if Pa(z, p) = if Pa(z, p) ={x} and p / [(yxz 2, xzy)] if Pa(z, p) ={y} and p / [(xyz 2, yzx)] if p [(yxz 2, xzy)] if p [(xyz 2, yzx)] if p [(xyz 2, yxz)] if p [(yxz 2, xyz)] and f (p) {x, y, z} if Pa(z, p) ={x, y} and p / [(xyz 2, yxz)] [(yxz 2, xyz)]. 5. For n = 3, C 4 F is the set of all social choice functions f for which there are x, y A, x = y, such that f (p) is determined by majority voting between x and y, i.e. { x if {i N (x, y) p(i)} > {i N (y, x) p(i)}, f (p) = y if {i N (y, x) p(i)} > {i N (x, y) p(i)}, for all profiles p where f (p) is not determined already by unanimity. The following lemma can be proved by counting manipulable profiles and evaluating the freedom left in the definitions of the C i. The counting argument can be simplified by noting that no agent wants to manipulate to a Pareto dominated alternative. Since the rest is elementary counting we do not present the proof here. Lemma 5.1 C 1 C 2 F if n = 2, C 1 C 2 C 3 C 4 F if n = 3 and C 1 C 2 C 3 F if n 4. Furthermore, the C i that are defined in a particular case are disjoint. We have C 1 =3 n, C 2 =2 3 n 1, and C 3 =3 n 2, C 4 =3 if they are defined, and 15 if n = 2, F 51 if n = 3, 3 n + 2 3 n 1 + 3 n 2 if n 4. We conjecture that equalities hold in the lemma, and that the proof of the main result in Section 4 can be used to show this, as follows. First of all we note that in the statement of Lemma 4.2 [p i ] l(p i ) 1 j=0 ( ) n, p i Mf 3 k(p i ) + j, equality cannot hold. Going through the proof of that statement one can see that equality would imply that l(p i ) = 1, k(p i ) = n 1, and p i (N) =2, a contradiction. So, M 3 f = for any f F. This also implies that d = n. Hence,

Minimal manipulability, anonymity, unanimity 267 all manipulable profiles arise from the f k. In particular for each f k we must have that it yields only two anonymously inequivalent manipulable profiles. Now, classifying all such f k, we know the possible M f s and what f can look like on {p P N p(n) 2}. Connecting this information by strategy-proofness, in particular on B 3 :={p P N p(n) 3}, it might be possible to characterize F. The reason why we do not pursue this here is the classification of all such f k. There are 125 unanimous and nondictatorial two agent social choice functions with two manipulable profiles which have to be checked first, and there might be even more with more manipulable profiles, but only two that give anonymously inequivalent profiles for f. We mention also that we have confirmed the following conjecture up to 8 agents by computer, which is also how we found the 125 unanimous and nondictatorial two agent social choice functions over three alternatives. Conjecture 5.2 C 1 C 2 if n = 2, F = C 1 C 2 C 3 C 4 if n = 3, C 1 C 2 C 3 if n 4, Furthermore, as the C i are disjoint, 9 + 6 = 15 if n = 2, F = 27 + 18 + 3 + 3 = 51 if n = 3, 3 n + 2 3 n 1 + 3 n 2 if n 4. 6 Pareto optimality and anonymity In this section we show how our results and the conjecture apply when unanimity is replaced by the stronger requirement of Pareto optimality. A social choice function is called Pareto optimal if it does not choose Pareto dominated alternatives, i.e. if Pa(x, p) =, x A, implies that f (p) = x. It is easy to see that Pareto optimality is a (strictly) stronger requirement than unanimity. Let G be the set of Pareto optimal and anonymous social choice functions. Then G F. Let C 1 G be the set of all social choice functions g G for which there is a labeling of the alternatives A ={x, y, z} such that z g(p) = x y if Pa(z, p) = if Pa(z, p) ={x} if Pa(z, p) ={y} and g(p) = g(q) {x, y} for all q P N such that Pa(z, p) = Pa(z, q) ={x, y}, [q] =[p].

268 S. Maus et al. Note that C 1 G F and C 1 C 1 F So the following lemma is implied by Lemma 5.1 and by counting the number of elements of C 1. Lemma 6.1 Let G be the set of minimally manipulable social choice functions in G. Then C 1 G, C 1 =3 2 n and G 3 2 n. Furthermore, G F implies the following corollary to Theorem 4.4. Corollary 6.2 Let g : P N A be a Pareto optimal and anonymous social choice function. Then M g 2 n 2. It is easily checked that there are no Pareto optimal social choice functions in C i, i {2, 3, 4}, whenever these sets are defined. So we can state the following conjecture based on Conjecture 5.2. Conjecture 6.3 Let G be the set of all minimally manipulable Pareto optimal and anonymous social choice functions for a fixed size of the agent set n. Then and G = C 1, G =3 2 n. Note that for any g C 1 there is an alternative z that is chosen whenever it is not Pareto dominated by one of the other alternatives. We call this alternative the status quo. Ifz is Pareto dominated, then the choice is taken among one of the dominating alternatives in an arbitrary way. In practice this arbitrariness might be replaced by a voting rule between x and y that seems reasonable in the given situation. For example it might be that y is declared as a second status quo, so that x is only chosen if p = x... n. These explanations show that our result has a nice interpretation if a status quo z among the alternatives exists, e.g., a current jurisdiction z about which there is consensus that it should only be altered to one of two new jurisdictions x and y if every voter prefers the new jurisdiction. Then, the least manipulable social choice functions that guarantee anonymity and Pareto optimality of the outcome are as follows. Choose z in the situation where there is consensus that no change should be made. In the other situations choose an alternative Pareto dominating z; if necessary, i.e., if there are two such alternatives, choose according to any unanimous social choice function over two alternatives. References Aleskerov F, Kurbanov E (1999) Degree of manipulability of social choice procedures. In: Proceedings of the third international meeting of the society for the advancement of economic theory. Springer, Berlin Heidelberg New York

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