Proceedings of the Twenty-Sixth AAAI Conference on Artificial Intelligence
The Price of Neutrality for the Ranked Pairs Method
Markus Brill Felix Fischer Institut¨ fur¨ Informatik Statistical Laboratory Technische Universitat¨ Munchen¨ University of Cambridge 85748 Garching, Germany Cambridge CB3 0WB, UK
Abstract To remedy this situation, a large number of SPFs have been suggested, together with a variety of criteria that a The complexity of the winner determination problem has been studied for almost all common voting rules. A notable reasonable SPF should satisfy (see Arrow, Sen, and Suzu- exception, possibly caused by some confusion regarding its mura 2002 for an overview). Neutrality and anonymity, for exact definition, is the method of ranked pairs. The original instance, are basic fairness criteria which require, loosely version of the method, due to Tideman, yields a social pref- speaking, that all alternatives and all voters are treated erence function that is irresolute and neutral. A variant intro- equally. Another criterion we will be interested in is the duced subsequently uses an exogenously given tie-breaking computational effort required to evaluate an SPF. Compu- rule and therefore fails neutrality. The latter variant is the tational tractability of the winner determination problem is one most commonly studied in the area of computational so- obviously a significant property of any SPF: the inability cial choice, and it is easy to see that its winner determina- to efficiently compute social preferences would render the tion problem is computationally tractable. We show that by method virtually useless, at least for large problem instances contrast, computing the set of winners selected by Tideman’s original ranked pairs method is NP-complete, thus revealing that do not exhibit additional structure. As a consequence, a trade-off between tractability and neutrality. In addition, computational aspects of preference aggregation have re- several known results concerning the hardness of manipula- ceived tremendous interest in recent years (see, e.g., Fal- tion and the complexity of computing possible and necessary iszewski et al. 2009; Conitzer 2010). winners are shown to follow as corollaries from our findings. In this paper, we study the computational complexity of the ranked pairs method (Tideman 1987). To the best of our 1 Introduction knowledge, this question has not been considered before, which is particularly surprising given the extensive litera- The fundamental problem of social choice theory can be ture that is concerned with computational aspects of ranked concisely described as follows: given a number of individ- pairs.2 A possible reason for this gap might be the confu- uals, or voters, each having a preference ordering over a set sion of two variants of the method, only one of which sat- of alternatives, how can we aggregate these preferences into isfies neutrality. In Section 2, we address this confusion a collective, or social, preference ordering that is in some and describe both variants as well as a closely related SPF sense faithful to the individual preferences? By a preference known as Kemeny’s rule. After introducing the necessary ordering we here understand a (transitive) ranking of all al- notation in Section 3, we show in Section 4 that deciding ternatives, and a function aggregating individual preferences whether a given alternative is a ranked pairs winner for the orderings into social preference orderings is called a social 1 neutral variant is NP-complete. Section 5 shows that a num- preference function (SPF). ber of known results follow as corollaries from our results, A natural idea to construct an SPF is by letting an al- and Section 6 discusses variants of the ranked pairs method ternative a be socially preferred to another alternative b if that are not anonymous. Finally, Section 7 concludes with a and only if a majority of voters prefers a to b. However, complexity-theoretic comparison of the ranked pairs method it was observed as early as the 18th century that this ap- and Kemeny’s rule. proach might lead to paradoxical situations: the collective preference relation may be cyclic even when all individual preferences are transitive (de Condorcet 1785). 2 Kemeny’s Rule and Two Variants of the Ranked Pairs Method Copyright c 2012, Association for the Advancement of Artificial Intelligence (www.aaai.org). All rights reserved. In this section we address the difference between two vari- 1In contrast to a social welfare function as studied by Ar- ants of the ranked pairs method that are commonly studied row (1951), an SPF can output multiple social preference order- ings with the interpretation that all those rankings are tied for win- 2Typical problems include the hardness of manipulation (Bet- ner. The rationale behind this is to allow for a symmetric outcome zler, Hemmann, and Niedermeier 2009; Xia and Conitzer 2011) when individual preferences are symmetric, like in the case of two and the complexity of computing possible and necessary winners individuals with diametrically opposed preferences. (Xia et al. 2009; Obraztsova and Elkind 2011).
1299 in the literature. Both variants are anonymous, i.e., treat all We will henceforth denote this variant by RP. voters equally. Non-anonymous variants of the ranked pairs In a subsequent paper, Zavist and Tideman (1989) showed method have been suggested by Tideman (1987) and Zavist that a tie-breaking rule is in fact necessary in order to achieve and Tideman (1989), and will be discussed in Section 6. the property of independence of clones, which was the main It will be instructive to first consider Kemeny’s rule (Ke- motivation for introducing the ranked pairs method. While meny 1959; Young 1995). The latter, which is also known Zavist and Tideman (1989) proposed a way to define a tie- as the Kemeny-Young method, chooses the ranking or rank- breaking rule based on the preferences of a distinguished ings with maximal Kemeny score, where the Kemeny score voter (see Section 6 for details), the variant that is most of a ranking measures the extent to which it agrees with indi- commonly studied in the literature considers the tie-breaking vidual preferences: for every pair (a, b) of alternatives, each rule to be exogenously given and fixed for all profiles. This voter i contributes one point to the Kemeny score of a rank- variant of ranked pairs will be denoted by RPT. Whereas ing if and only if i ranks a and b in the same order as the RP may output a set of rankings, with the interpretation that ranking does. As the number of possible rankings grows all the rankings in the set are tied for winner, RPT always rapidly with the number of alternatives (for n alternatives outputs a single ranking. In social choice terminology, RP there are n! rankings), computing all Kemeny scores is com- is an irresolute SPF, and RPT is a resolute one. It is straight- putationally demanding. Indeed, it was shown by Bartholdi, forward to see that RP is neutral, i.e., treats all alternatives III, Tovey, and Trick (1989) that finding the so-called Ke- equally, and that RPT is not. An easy example for the latter meny rankings, i.e., the rankings with a maximal Kemeny statement is the case of two alternatives and two voters who score, is NP-hard.3 This is commonly seen as strong evi- each prefer a different alternative. dence that no efficient algorithm exists for this problem. Rather than completely ranking all alternatives, it is often The method of ranked pairs, introduced by Tide- sufficient to identify the socially “best” alternatives. This man (1987) and extended by Zavist and Tideman (1989), can is the purpose of a social choice function (SCF). An SCF be viewed as a heuristic to find a ranking with high—if not has the same input as an SPF, but returns alternatives instead maximal—Kemeny score. This is not to say that the main of rankings. Each SPF gives rise to a corresponding SCF motivation for ranked pairs was the approximation of Ke- that returns the top elements of the rankings instead of the meny’s rule. Actually, the ranked pairs method is interesting rankings themselves, and we will frequently switch between in its own right and satisfies a number of desirable proper- these two settings. Interestingly, deciding whether a given ties, some of which are not satisfied by Kemeny’s rule (see ranking is chosen by an SPF can be considerably easier than Tideman 1987; Lamboray 2009; Parkes and Xia 2012). deciding whether a given alternative is chosen by the corre- The easiest way to describe the ranked pairs method is to sponding SCF (see Table 1). formulate it as a procedure. The procedure first defines a pri- From a computational perspective, RPT is easy: con- ority ordering over the set of all (ordered) pairs (a, b) of al- structing the ranking for a given tie-breaking rule takes ternatives by giving priority to pairs (a, b) with a larger num- time polynomial in the size of the input (see Proposition 1). ber of voters preferring a to b. Then, it constructs a ranking For RP, however, the picture is different: as the number of of the alternatives by starting with the empty ranking and it- tie-breaking rules is exponential, executing the iterative pro- eratively considering pairs in order of priority. When a pair cedure for every single tie-breaking rule is infeasible. Of (a, b) is considered, the ranking is extended by fixing that course, this does not preclude the existence a clever algo- a precedes b—unless fixing this pairwise comparison would rithm that efficiently computes the set of all alternatives that create a cycle together with the previously fixed pairs, in are the top element of some chosen ranking.6 Our main which case the pair (a, b) is discarded. This procedure is result states that such an algorithm does not exist unless P guaranteed to terminate with a complete ranking of all alter- equals NP. natives. What is missing from the above description is a tie- breaking rule for cases where two or more pairwise com- 3 Preliminaries parisons have the same support from the voters. This turns For a finite set X, let L(X) denote the set of all rankings out to be a rather intricate issue. In principle, it is possi- of X, where a ranking of X is a complete, transitive, and ble to employ an arbitrary tie-breaking rule. However, each asymmetric relation on X. The top element of a ranking fixed tie-breaking rule biases the method in favor of some L L(X), denoted by top(L), is the unique element x X alternative and thereby destroys neutrality.4 In order to re- such2 that xLyfor all y X x . 2 pair this flaw, Tideman (1987) originally defined the ranked 2 \{ } pairs method to return the set of all those rankings that re- breaking (PUT), can also be used to “neutralize” other voting rules sult from the above procedure for some tie-breaking rule.5 that involve tie-breaking (Conitzer, Rognlie, and Xia 2009). PUT can be interpreted as a possible winner notion: if the ranked pairs 3Later, Hemaspaandra, Spakowski, and Vogel (2005) strenght- method is used with an unknown tie-breaking rule, the PUT ver- ened this result by showing that identifying Kemeny rankings is sion of ranked pairs selects exactly those alternatives that have a p NP complete for ⇥2 = P , the class of problems solvable via parallel chance to be chosen in the actual election. || access to NP. 6As the number of chosen rankings might be exponential (for 4 Neutrality can be maintained if the tie-breaking rule varies completely tied instances such as R⇤ in the proof of Proposition 4, with the individual preferences (see Section 6). all rankings are chosen), it immediately follows that computing all 5This definition, sometimes called parallel universes tie- of them requires exponential time in the worst case.
1300 Let N = 1,...,n be a set of voters with prefer- L of A, and two alternatives a and b, we say that a at- ences over a finite{ set A} of alternatives. The preferences tains b through L if there exists a sequence of distinct al- of voter i N are represented by a ranking R on A. The ternatives a ,a ,...,a , where ` 2, such that a = a, 2 i 1 2 ` 1 interpretation of (a, b) Ri, usually denoted by aRi b, is a` = b, ai Lai+1, and nR(ai,ai+1) nR(b, a) for all i that voter i strictly prefers2 a to b.Apreference profile is an with 1 i<`. In this case, we will say that a attains b ordered list containing a ranking for each voter. via (a1,a2,...,a`). A ranking L is called a stack if for any A social choice function (SCF) f associates with every pair of alternatives a and b it holds that aLbimplies that a preference profile R a non-empty set f(R) A of alterna- attains b through L. f ✓ tives. A social preference function (SPF) associates with Lemma 1 (Zavist and Tideman 1989). A ranking of A is a every preference profile R a non-empty set f(R) L(A) ranked pairs ranking if and only if it is a stack. of rankings of A. ✓ An SCF or SPF is neutral if permuting the alternatives It follows that an alternative is a ranked pairs winner if in the individual rankings also permutes the set of chosen and only if it is the top element of a stack. alternatives, or the set of chosen rankings, in the exact same way. Formally, f is neutral if f(⇡(R)) = ⇡(f(R)) for all 4 Results preference profiles R and all permutations ⇡ of A. An SCF We are now ready to study the computational complexity or SPF is anonymous if the set of chosen alternatives, or the of RP. We first consider the SPF setting and observe that set of chosen rankings, does not change when the voters are finding and checking ranked pairs rankings is easy. This permuted. also provides an efficient way to find some ranked pairs win- For a given preference profile R =(R1,...,Rn) and two ner, i.e., some alternative that is chosen in the SCF setting. distinct alternatives a, b A, we denote by nR(a, b) the The problem of deciding whether a particular alternative is 2 difference between the number of voters who prefer a to b a ranked pairs winner, on the other hand, turns out to be and the number of voters who prefer b to a, i.e., NP-complete. We then demonstrate that for the ranked pairs method neutrality and tractability are incompatible in a more n (a, b)= i N : aR b i N : bR a . R |{ 2 i }| |{ 2 i }| general sense. Finally, we extend the hardness result to a The resolute variant of the ranked pairs method takes as variant of the winner determination problem that asks for input a preference profile R and a tie-breaking rule ⌧ unique winners. Some easy proofs are omitted due to space L(A A). A ranking R of A A is constructed by or-2 constraints and can be found in the full version of the paper. ⇥ ⌧ ⇥ dering all pairs in accordance with nR( , ), using ⌧ to break ties: (a, b) R (c, d) if and only if n ·(a,· b) >n (c, d) or 4.1 Ranked Pairs Rankings ⌧ R R (nR(a, b)=nR(c, d) and (a, b) ⌧ (c, d)). It can easily be seen that an arbitrary ranked pairs ranking R The relation L⌧ on A is constructed by iteratively con- can be found efficiently. sidering the pair ranked highest by R among all pairs that ⌧ Finding a ranked pairs ranking is in P. have not been considered so far. The pair is then added to the Proposition 1. R R relation L⌧ unless this addition would create an L⌧ -cycle Deciding whether a given ranking is a ranked pairs rank- with the pairs that have been added before.7 After all pairs ing is also feasible in polynomial time, by checking whether in A A have been considered, LR is a ranking of A. The the given ranking is a stack. ⇥ ⌧ resolute variant of ranked pairs, interpreted as an SCF, re- Deciding whether a given ranking is a R Proposition 2. turns the top element of L⌧ . ranked pairs ranking is in P. Definition 1. RPT(R,⌧ )= top(LR) . { ⌧ } It is worth noting that Proposition 2 can also be shown The outcome of RPT depends on the choice of ⌧, and directly, without referring to stacks. For a given ranking L, RPT is not neutral. Tideman (1987) therefore defined an ir- define a tie-breaking rule ⌧L such that (a, b) ⌧L (c, d) for all resolute and neutral variant that chooses all alternatives that (a, b) L and (c, d) / L. It can be shown that L is a ranked R 2 2 R are at the top of L⌧ for some tie-breaking rule ⌧. pairs ranking if and only if L = L⌧L . The advantage of this alternative proof is that for each “yes” instance it constructs Definition 2. RP(R)= a A : there exists ⌧ L(A A) such that a = top(LR{) . 2 2 ⇥ a witnessing tie-breaking rule. ⌧ } The alternatives in RP(R) are called ranked pairs winners 4.2 Ranked Pairs Winners R LR : for . In the SPF setting, RP returns the rankings ⌧ We now consider the SCF setting. As every ranked pairs ⌧ L(A A) { , which are henceforth called ranked pairs ranking yields a ranked pairs winner, Proposition 1 imme- 2 ⇥ R } rankings for . diately implies that an arbitrary element of RP(R) can be We will work with an alternative characterization of found efficiently. ranked pairs rankings that was introduced by Zavist and Tideman (1989). Given a preference profile R, a ranking Proposition 3. Finding a ranked pairs winner is in P. Deciding whether a given alternative is a ranked pairs 7A P -cycle of a relation P A A is a set of pairs ✓ ⇥ winner, on the other hand, turns out to be NP-complete. (a1,a2), (a2,a3),...,(a` 1,a`) with a` = a1 and ai Pai+1 {for all i with 1 i<`. Pairs of} the form (a, a) are considered Theorem 1. Deciding whether a given alternative is a R cycles of length 1, and are therefore never added to L⌧ . ranked pairs winner is NP-complete.
1301 d
v10 v20
v1 v¯1 v¯2 v2
v¯10 v¯20
y1 y2 y3
Figure 1: Graphical representation of n ( , ) for the Boolean formula ' = v , v¯ v ,v v¯ ,v . The relation 2 is R' · · { 1 2}^{ 1 2}^{ 1 2} represented by arrows, and 4 is represented by double-shafted arrows. For all pairs (a, b) that are not connected by an arrow, we have n(a, b)=n(b, a)=0 .
4 2 2 Membership in NP follows from Proposition 2. For hard- v¯ v0 v . For each clause C , 1 j k, let v y i i i j i j ness, we give a reduction from the NP-complete Boolean if variable vi V appears in clause Cj as a positive literal, 2 2 satisfiability problem (SAT, see, e.g., Papadimitriou 1994). and v¯i yj if variable vi appears in clause cj as a negative An instance of SAT consists of a Boolean formula ' = 2 2 literal. Finally let yj d for 1 j k and d vi0 and C C 2 1 k in conjunctive normal form over a finite d v¯0 for 1 i m. For all pairs (a, b) for which n(a, b) set ^V ···= ^v ,...,v of variables. Denote by X = i { 1 m} has not been specified so far, let n(a, b)=n(b, a)=0. An v1, v1,...,vm, vm the set of all literals, where a literal example is shown in Figure 1. { } clause C is either a variable or its negation. Each j is a set As n(a, b) 4, 2, 0, 2, 4 for all a, b A', De- of literals. An assignment ↵ X is a subset of the literals bord’s (1987) theorem2{ guarantees} the existence2 of a prefer- with the interpretation that all✓ literals in ↵ are set to “true.” ence profile R' with nR' (a, b)=n(a, b) for all a, b A', Assignment ↵ is valid if ` ↵ implies `/↵ for all ` X, 2 2 2 2 and such a profile can in fact be constructed efficiently. and ↵ satisfies clause Cj if Cj ↵ = . A valid assign- d \ 6 ; The following two lemmata show that alternative is a ment that satisfies all clauses of ' will be called a satisfying ranked pairs winner for R' if and only if the formula ' is assignment for ', and a formula that has a satisfying assign- satisfiable. ment will be called satisfiable. Lemma 2. If d RP(R'), then ' is satisfiable. For a particular Boolean formula ' = C1 Ck over 2 a set V = v ,...,v of variables, we will^··· construct^ a Proof. Assume that d is a ranked pairs winner for R and let { 1 m} ' preference profile R' over a set A' of alternatives such that L be a stack with top(L)=d. Consider an arbitrary j with a particular alternative d A is a ranked pairs winner for 1 j k. As L is a stack and dLy, d attains y through 2 ' j j R' if and only if ' is satisfiable. L, i.e., there exists a sequence Pj =(a1,a2,...,a`) with Let us first define the set A of alternatives. For each a1 = d and a` = yj such that ai Lai+1 and n(ai,ai+1) 2 ' variable v V , 1 i m, there are four alternatives v , for all i with 1 i<`. If d attains yj via several sequences, i 2 i v¯i, vi0, and v¯i0. For each clause Cj, 1 j k, there is one fix one of them arbitrarily. alternative y . Finally, there is one alternative d for which The definition of n( , ) implies that j · · we want to decide membership in RP(R'). Pj =(d, `0, `,` 0,`,yj) or Instead of constructing R' explicitly, we will specify a number n(a, b) for each pair (a, b) A A . De- Pj =(d,` 0,`,yj), 2 ' ⇥ ' bord (1987) has shown that there exists a preference pro- where ` is some literal. The former is in fact not possible file R such that n (a, b)=n(a, b) for all a, b, as long R `0 ` L P as n(a, b)= n(b, a) for all a, b and all the numbers because does not attain through . Therefore, each j P =(d,` ,`,y ) ` X n(a, b) have the same parity.8 In order to conveniently de- is of the form j 0 j for some . ↵ 2 fine n( , ), the following notation will be useful: for a nat- Now define assignment as the set of all literals that are P , 1 j k ↵ = ural number· · w, a w b denotes setting n(a, b)=w and contained in one of the sequences j , i.e., k n(b, a)= w. X ( j=1 Pj). We claim that ↵ is a satisfying assignment 4 2 \ For each variable vi V , 1 i m, let vi v¯i0 for '. 2 In orderS to show that ↵ is valid, suppose there exists a 8See also Le Breton (2005). literal ` X such that both ` and ` are contained in ↵. 2
1302 This implies that there exist i and j such that d attains y set T L(A A) of tie-breaking rules such that the corre- i ✓ ⇥ via Pi =(d,` 0,`,yi) and d attains yj via Pi =(d, `0, `, yj). sponding variant is both neutral and tractable. Formally, for a preference profile R and a set T of tie- In particular, ` L`and `0 L `. As L is transitive and 0 breaking rules, define asymmetric, it follows that either `0 L ` or `0 L`. However, R 0 RPT(R)= a A : ⌧ T such that a = top(L ) . neither does `0 attain ` through L, nor does ` attain ` through { 2 9 2 ⌧ } L, a contradiction. Thus, in particular, RP( )=RPL(A A)( ) and RPT( ,⌧)= In order to see that ↵ satisfies ', consider an arbitrary · ⇥ · · RP ⌧ ( ). While RPT is anonymous for all T L(A A), clause Cj. As d attains yj via Pj =(d,` 0,`,yj) and { } · ✓ ⇥ other properties of RPT obviously depend on T. We say n(yj,d)=2, we have that n(`, yj) 2. By definition that T is neutral if RPT is neutral. Furthermore, we call of n( , ), this implies that ` Cj. · · 2 T intractable if deciding membership in RPT(R) is NP- Lemma 3. If ' is satisfiable, then d RP(R'). complete, and tractable if the problem is in P. Our previous 2 results imply that L(A A) is neutral but intractable, and Proof. Assume that ' is satisfiable and let ↵ be a satisfying that ⌧ is tractable but⇥ not neutral for any ⌧ L(A A). { } 2 ⇥ assignment. Let V = v , v¯ ,v0, v¯0 , 1 i m, and It turns out that RP is the only variant in this framework i { i i i i} Y = y1,y2,...,yk . We define a ranking L of A' as that is neutral. follows,{ using BLC}as shorthand for bLcfor all b B Proposition 4. If T is neutral, then RPT = RP. and c C. 2 2 Proof. Let T be neutral. We have to show that RPT(R)= For all 1 i m, let dLV and V LY. • i i RP(R) for all preference profiles R. The inclusion from left For all 1 i
1303 all a A d . Within A , R0 coincides with R .We problem is equivalent to the (unique) winner determination 2 ' \{ } ' ' ' show that RP(R'0 )= d⇤ if and only if ' is unsatisfiable. problem in the special case when the preference profile is For the direction from{ left} to right, assume for contradic- completely specified. tion that RP(R )= d and ' is satisfiable. Consider a '0 ⇤ The unweighted coalitional manipulation (UCM) prob- satisfying assignment{↵ and} let L be the ranking of A de- ' lem asks whether it is possible for a group of voters to cast fined in the proof of Lemma 3. Define the ranking L of A 0 '0 their votes in a way that achieves a given outcome. by Corollary 4 (Xia et al. 2009). The UCM problem under L0 = L (d, d⇤) (d⇤,a):a A' d . ranked pairs is NP-complete for any number of manipula- [{ }[{ 2 \{ }} tors.10 That is, L0 extends L by inserting the new alternative d⇤ in the second position. As in the proof of Lemma 3, it Proof. Membership in NP is again straightforward. Hard- can be shown that L0 is a stack. It follows that top(L0)= ness follows from the fact that the UCM problem with zero d RP(R0 ), contradicting the assumption that RP(R0 )= manipulators is equivalent to the unique winner determina- 2 ' ' d⇤ . tion problem. { For} the direction from right to left, assume for contradic- tion that ' is unsatisfiable and RP(R0 ) = d⇤ . Then there 6 Non-Anonymous Variants ' 6 { } R'0 As mentioned in Section 2, Tideman (1987) and Zavist and exists a tie-breaking rule ⌧ such that top(L⌧ )=a = d⇤. 6 4 Tideman (1989) suggested ways to use the preferences of a From the definition of R0 it follows that a = d, as d⇤ b ' distinguished voter, say, a chairperson, to render the ranked for all b A d and there are no 4-cycles. By the same 2 ' \{ } pairs method resolute. There are essentially two ways to argument as in the proof of Lemma 2, it can be shown that ' achieve this, which differ in the point in time when ties are is satisfiable, contradicting our assumption. broken. For the sake of simplicity, we only consider the SCF setting in this section. 5 Alternative Proofs for Known Results The a priori variant uses the preferences of the chairper- In this section we briefly review results from the literature son in order to construct a tie-breaking rule ⌧ L(A A), that follow from our findings. All results concern the neutral which is then used to compute RPT( ,⌧). The2a posteriori⇥ variant RP. We refer to the respective papers for formal variant first computes RP( ) and then· chooses the alterna- definitions of the computational problems. tive from this set that is most· preferred by the chairperson. An alternative a is a possible winner for a partially speci- Both variants are neutral: if the alternatives are permuted in fied preference profile R if there exists a completion R0 of R each ranking, including the ranking of the chairperson, the such that a is a winner for R0. It is a necessary winner if it is tie-breaking rule and thus the chosen alternative will change a winner for every completion of R. Both the possible and accordingly. the necessary winner problem have a variant that requires an Whereas the a priori variant is a special case of RPT and alternative to be the unique winner for the completion. therefore efficiently computable, the a posteriori variant is intractable by the results in Section 4. It follows that neu- Corollary 2 (Xia and Conitzer 2011). Computing possible (unique) ranked pairs winners for partially specified prefer- trality and tractability can be reconciled at the expense of ence profiles is NP-complete. anonymity. By moving to non-deterministic SCFs, one can even regain anonymity: choosing the chairperson for the a Proof. Membership in NP is straightforward. Hardness fol- priori variant uniformly at random results in a procedure that lows from the fact that the possible (unique) winner problem is neutral, anonymous, and tractable, for appropriate gener- is equivalent to the (unique) winner determination problem alizations of anonymity and neutrality to the case of non- in the special case when the preference profile is completely deterministic SCFs. The winner determination problem for specified. the a posteriori variant remains intractable when the chair- person is chosen randomly. Corollary 3 (Xia and Conitzer 2011). The following hold: Computing necessary unique ranked pairs winners for 7 Conclusion • partially specified preference profiles is coNP-complete. We have studied the complexity of the ranked pairs method. Computing necessary ranked pairs winners for partially While some ranked pairs winner is easy to find, deciding • specified preference profiles is coNP-hard.9 whether a given alternative is a winner turns out to be NP- complete. If one is interested in ranked pairs rankings, Proof. The “unique” variant is in coNP because for every both problems are computationally easy. Table 1 summa- “no” instance there is a completion and a tie-breaking rule rizes these results and contrasts them with the correspond- that produces a different winner. Hardness of both vari- ing results for Kemeny’s rule. Interestingly, all four prob- ants follows from the fact that the necessary (unique) winner lems are computationally harder for the latter under plausi- ble complexity-theoretic assumptions. 9Xia and Conitzer (2011) state that the “non-unique” variant is coNP-complete as well. However, their argument for membership 10The proof of Theorem 4.1 by Xia et al. (2009) actually works in coNP is incorrect since it assumes that winner determination is for both RP and RPT (Xia, personal communication, March 29, in P. 2012).
1304 Ranked pairs Kemeny’s rule mation. In Proceedings of the 21st International Joint Con- ference on Artificial Intelligence (IJCAI), 109–115. find ranking in P NP-harda find winner in P NP-harda Conitzer, V. 2010. Making decisions based on the pref- is ranking in P coNP-completeb erences of multiple agents. Communications of the ACM p 53(3):84–94. is winner NP-complete ⇥ -completeb 2 de Condorcet, M. 1785. Essai sur l’application de l’analyse a Bartholdi, III, Tovey, and Trick (1989) a` la probabilite´ des decisions´ rendues a` la pluralite´ des voix. b Hemaspaandra, Spakowski, and Vogel (2005) Imprimerie Royale. Facsimile published in 1972 by Chelsea Table 1: Computational aspects of the ranked pairs method Publishing Company, New York. and Kemeny’s rule Debord, B. 1987. Caracterisation´ des matrices des pref´ erences´ nettes et methodes´ d’agregation´ associees.´ Mathematiques´ et sciences humaines 97:5–17. From a practical point of view, the ranked pairs method Faliszewski, P.; Hemaspaandra, E.; Hemaspaandra, L.; and is easier than Kemeny’s rule as well. The reason is that the Rothe, J. 2009. A richer understanding of the complex- expected number of ties among two or more pairs is rather ity of election systems. In Ravi, S., and Shukla, S., eds., small. This is particularly true when the number of voters is Fundamental Problems in Computing: Essays in Honor of large compared to the number of alternatives, which is the Professor Daniel J. Rosenkrantz. Springer-Verlag. case in many realistic settings. It is therefore to be expected Hemaspaandra, E.; Spakowski, H.; and Vogel, J. 2005. The that ranked pairs winners are easy to compute on average complexity of Kemeny elections. Theoretical Computer Sci- for most reasonable distributions of individual preferences. ence 349(3):382–391. Our results reveal a trade-off between neutrality and tractability in the context of the ranked pairs method: while Kemeny, J. G. 1959. Mathematics without numbers. the efficiently computable variant RPT fails neutrality, the Daedalus 88:577–591. neutral variant RP is intractable. In fact, this tension cannot Lamboray, C. 2009. A prudent characterization of the be resolved even when moving to an arbitrary set of fixed ranked pairs rule. Social Choice and Welfare 32(1):129– tie-breaking rules. A very similar trade-off can be observed 155. for the single transferable vote rule (Conitzer, Rognlie, and Le Breton, M. 2005. On the uniqueness of equilibrium in Xia 2009; Wichmann 2004). symmetric two-player zero-sum games with integer payoffs. We have finally discussed variants of the ranked pairs Economie´ publique 17(2):187–195. method that achieve neutrality at the expense of anonymity, Obraztsova, S., and Elkind, E. 2011. On the complexity by using individual preferences to break ties. The tractabil- of voting manipulation under randomized tie-breaking. In ity of those variants depends on the point in time ties are Proceedings of the 22nd International Joint Conference on broken. Artificial Intelligence (IJCAI), 319–324. AAAI Press. Papadimitriou, C. H. 1994. Computational Complexity. Acknowledgements Addison-Wesley. This material is based on work supported by the Deutsche Parkes, D., and Xia, L. 2012. A complexity-of-strategic- Forschungsgemeinschaft under grants BR 2312/10-1 and behavior comparison between Schulze’s rule and ranked FI 1664/1-1. We thank three anonymous reviewers for valu- pairs. In Proceedings of the 26th AAAI Conference on Arti- able comments and suggestions. ficial Intelligence (AAAI). AAAI Press. Forthcoming. References Tideman, T. N. 1987. Independence of clones as a criterion for voting rules. Social Choice and Welfare 4(3):185–206. Arrow, K. J.; Sen, A. K.; and Suzumura, K., eds. 2002. Handbook of Social Choice and Welfare, volume 1. 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