On absolutely and simply popular rankings
OOn absolutely and simply popular rankings
Sonja Kraiczy · ´Agnes Cseh · DavidManlove
Received: date / Accepted: date
Abstract
Van Zuylen et al. [29] introduced the notion of a popular rankingin a voting context, where each voter submits a strictly-ordered list of allcandidates. A popular ranking π of the candidates is at least as good as anyother ranking σ in the following sense: if we compare π to σ , at least half ofall voters will always weakly prefer π . Whether a voter prefers one ranking toanother is calculated based on the Kendall distance.A more traditional definition of popularity—as applied to popular matchings,a well-established topic in computational social choice—is stricter, becauseit requires at least half of the voters who are not indifferent between π and σ to prefer π . In this paper, we derive structural and algorithmic results inboth settings, also improving upon the results in [29]. We also point out strongconnections to the famous open problem of finding a Kemeny consensus with 3voters. Keywords majority rule · Kemeny consensus · complexity · preferenceaggregation · popular matching A 2-page extended abstract of this paper will appear at AAMAS 2021.Sonja KraiczyMerton College, University of Oxford, UKE-mail: [email protected]´Agnes CsehHasso-Plattner-Institute, University of Potsdam, GermanyE-mail: [email protected] ManloveSchool of Computing Science, University of Glasgow, UKE-mail: [email protected] a r X i v : . [ c s . G T ] F e b Sonja Kraiczy et al.
A fundamental question in preference aggregation is the following: given anumber of voters who rank candidates, can we construct a ranking that expressesthe preferences of the entire set of voters as a whole? A common way ofevaluating how close the constructed ranking is to a voter’s preferences is theKendall distance, which measures the pairwise disagreements between tworankings. Among others, a well-known rank aggregation method is the Kemenyranking method [22], in which the winning ranking minimises the sum of itsKendall distances to the voters’ rankings.For the preference aggregation problem, van Zuylen et al. [29] introducea new rank aggregation method called popular ranking, which is also basedon the Kendall distance. Each voter can compare two given rankings π and σ , and prefers the one that is closer to her submitted ranking in terms of theKendall distance. Van Zuylen et al. define π to be a winning ranking in aninstance if for any ranking σ , at least half of the voters prefer π to σ or areindifferent between them. This implies that there is no ranking σ such thatswitching to σ from π would benefit a majority of all voters.According to the definition of popularity in [29], even in a situation whereexactly half of the voters are indifferent between rankings π and σ , whilstthe other half of the voters prefer σ to π , σ is not more popular than π .This example demonstrates how hard it is for the dissatisfied voters to find aranking that overrules π —the definition requires them to find a profiting set ofvoters who build an absolute majority , that is, a majority of all voters for thisendeavour.A straightforward option would be to require only a simple majority , thisis, a majority of the non-abstaining voters, to profit from switching to σ from π . Excluding the abstaining voters in a pairwise majority voting rule iscommon practice [13]. It is also analogous to the classical popularity notionin the matching literature [1, 9, 26]. In this paper, we propose an alternativedefinition of a popular ranking. We define π to be a simply popular ranking iffor every ranking σ , at least half of the non-abstaining voters prefer π to σ .This means that switching from π to σ would harm at least as many voters asit would benefit. Related literature.
Aggregating voters’ preferences given as ordered lists hasbeen challenging researchers for decades. The most common approach to thisproblem is to search for a ranking that minimises the sum of the distances to thevoters’ rankings. If the Kendall distance [23] is used as the metric on rankings,then this optimality concept corresponds to the Kemeny consensus [22, 24].Deciding whether a given ranking is a Kemeny consensus is NP -complete,and calculating a Kemeny consensus is NP -hard [6] even if there are only 4voters [7, 14], or at least 6 of them [2]. Interestingly, the complexity of theproblem is open for 3 and 5 voters [2, 7].Majority voting rules offer another natural way of aggregating voters’preferences. The earliest reference for this might be from Condorcet [8], who uses n absolutely and simply popular rankings 3 pairwise comparisons to calculate the winning candidate, establishing his famousparadox on the smallest voting instance not admitting any majority winner.The absolute and simple majority voting rules have both been extensivelydiscussed in the setting where the goal is to choose the winning candidate [4, 5].Vermeule [30] focuses on strategic minorities and demonstrates the effect of thesimple majority rule compared to the absolute majority rule based on data fromdecisions made by the United States Congress. By undertaking a probabilisticanalysis, Dougherty and Edward [13] discuss the differences between the tworules. Felsenthal and Machover [25] generalise simple voting games to ternaryvoting games by adding the possibility to abstain.The concept of majority voting readily translates to other scenarios, wherevoters submit preference lists. One such field is the area of matchings underpreferences, where popular matchings [18, 1, 26, Chapter 7, 9] serve as a voting-based alternative concept to the well-known notion of stable matchings [3, 17] intwo-sided markets. In short, a popular matching M is a simple majority winneramong all matchings, because it guarantees that no matter what alternativematching is offered on the market, a weak majority of the non-abstaining agentswill opt for M .Besides two-sided matchings, majority voting has also been defined for thehouse allocation problem [1, 28], the roommates problem [15, 19], spanningtrees [10], permutations [29], the ordinal group activity selection problem [11],and very recently, to branchings [21]. The notion of popularity is aligned withsimple majority in all these papers, with one exception, namely [29], whichdefines popularity based on the absolute majority rule.A part of this work revisits the paper from van Zuylen et al. [29]. They showthat a popular ranking—according to their definition of popularity—need notnecessarily exist. More precisely, they show that the acyclicity of a structureknown as the majority graph is a necessary, but not sufficient condition forthe existence of a popular ranking. They also prove that if the majority graphis acyclic, then one can efficiently compute a ranking, which may or may notbe popular, but for which the voters have to solve an NP -hard problem tocompute a ranking that a majority of them prefer. Our contribution.
We study both the weaker notion of popularity from [29]and the stronger notion of popularity analogous to the one in the matchingliterature, which excludes abstaining voters. Our most important results are asfollows.1. For at most 5 voters, the two notions are equivalent, but with 6 voters thisdoes not hold anymore.2. We give a sufficient condition for the two notions to be equivalent for agiven ranking π .3. In the case of 2 or 3 voters, one can find a popular ranking of either kind andverify the absolute or simple popularity of a given ranking in polynomialtime. Sonja Kraiczy et al.
4. The problem of verifying the absolute or simple popularity of a given rankingfor 4 voters is polynomial-time solvable if and only if it is polynomial-timesolvable for 5 voters.5. If finding a ranking that is more popular in either sense than a given rankingin an instance with 4 (or 5) voters were polynomial-time solvable, thenthe famously open Kemeny consensus problem for 3 voters would also bepolynomial-time solvable.
Structure of the paper.
In Section 2 we introduce the necessary definitions andnotations used in the following sections. Section 3 deals with the relationshipbetween the two different popularity notions. In Section 4 we study the com-plexity of the problems of deciding whether a given ranking is absolutely orsimply popular with a small number of voters. We also demonstrate the strongconnection to the Kemeny consensus problem there. We lay out some problemsthat still remain to be answered in Section 5.
We start this section with the formal definitions of various standard notions invoting theory in Section 2.1. Then, in Section 2.2, we introduce absolutely andsimply popular rankings and the decision problems we will later analyse.2.1 RankingsWe are given a set C of m candidates and a set V of n voters. A ranking π is a permutation of the m candidates. When exhibiting a specific ranking, wewill often enclose parts of it in square brackets, e.g. we will write [1 , , [3 , , , ,
4. These brackets can be ignored and are simply used forbetter readability in instances with specific structural properties. Each voter v has her own ranking denoted by π v . An example is depicted in Figure 1. The rank of candidate a in ranking π is the position (counting from 1) it appearsat in π , and it is denoted by rank π [ a ]. We say that voter v prefers candidate a to candidate b if rank π v [ a ] < rank π v [ b ]. In Figure 1, voter v prefers candidate1 to candidate 2, and rank π v [5] = 6. A voting instance I comprises C , V , andthe rankings π v for each v ∈ V . For convenience, we use the set of rankingsand I interchangeably, since the former already contains all information on C and V . For candidates a and b , { a, b } denotes the unordered pair of them,while ( a, b ) is their ordered pair.We say that voter v (or ranking π v ) agrees with voter v (or with ranking π v ) in the order of two distinct candidates a and b if v and v either equivocallyprefer a to b or they equivocally prefer b to a . Otherwise they disagree in theorder of a and b . The similarity between two rankings can be measured byvarious metrics defined on permutations. Possibly the most common metric,the Kendall distance [23], is defined below. n absolutely and simply popular rankings 5 π v = [1 , , , [6 , , , [8 , , π v = [2 , , , [4 , , , [9 , , π v = [3 , , , [5 , , , [7 , , π v = [1 , , , [4 , , , [7 , , π v = [1 , , , [5 , , , [9 , , π v = [1 , , , [5 , , , [7 , , Fig. 1
A voting instance I with 6 voters v , v , . . . v and 9 candidates 1 , , . . . , Definition 1
The
Kendall distance K ( π, σ ) between two rankings π and σ isdefined as the number of pairwise disagreements between π and σ , or, formallyas K ( π, σ ) = |{ ( a, b ) : rank π [ a ] > rank π [ b ] and rank σ [ a ] < rank σ [ b ] }| + |{ ( a, b ) : rank π [ a ] < rank π [ b ] and rank σ [ a ] > rank σ [ b ] }| . The Kendall distance is also called bubble-sort distance, because it isequivalent to the number of swaps that the bubble sort algorithm [16] executeswhen converting ranking π to ranking σ . To be more precise, let us first definea total order on 1 , . . . , n such that under this order, ranking σ is sorted inincreasing order. We define the bubble swap path from a ranking π to σ as thesequence π := π, π , . . . , π k := σ of intermediate permutations obtained whensorting π using the bubble sort algorithm. We call the change π i → π i +1 a swap . Alternatively we denote the swap by the consecutive candidates a and b it interchanges: ( b, a ) → ( a, b ). We say that a swap ( b, a ) → ( a, b ) is good forvoter v if v prefers a to b , otherwise this swap is bad for v . Note that if the swap π i → π i +1 is good for v , then K ( π i +1 , π v ) = K ( π i , π v ) − v , then K ( π i +1 , π v ) = K ( π i , π v ) + 1.Let V ( a, b ) ⊆ V be the set of voters who prefer candidate a to b , i.e. V ( a, b ) = { v ∈ V : rank π v ( a ) < rank π v ( b ) } . The majority graph belonging toa voting instance is defined as the directed graph which has as vertices thecandidates and an arc from candidate a to candidate b if a majority of thevoters prefer a to b , i.e. | V ( a, b ) | > | V ( b, a ) | . As mentioned in the introduction,Condorcet observed that the majority graph may contain a directed cycle. Thishas come to be known as Condorcet paradox [8]. A tournament is a majoritygraph that is complete, or, in other words, for every a and b in its votinginstance either | V ( a, b ) | > | V ( b, a ) | or | V ( a, b ) | < | V ( b, a ) | holds. The majoritygraph of our instance in Figure 1 is depicted in Figure 2. As the edges form nocycle, it is an acyclic majority graph, but since it is not a complete graph, it isnot a tournament.Ranking π is a topologically sorted ranking in I if rank π [ a ] < rank π [ b ] holdsfor each pair of candidates a and b with | V ( a, b ) | > | V ( b, a ) | . Topologicallysorted rankings correspond to the graph-theoretical topological sort of thevertices in the majority graph, and thus only exist if the majority graphis acyclic. Acyclic tournaments trivially have a unique topologically sorted Sonja Kraiczy et al.7 8 94 5 61 2 3
Fig. 2
The majority graph of the instance from Figure 1. Each gray set of 3 candidatesdenotes a specific subgraph called component [27]. Arcs between these components symbolisethe complete set of 9 arcs, between any two vertices from different components. ranking. A topologically sorted ranking for the instance in Figure 1 with 9candidates is σ = [1 , , , [4 , , , [7 , , Kemeny rank of a ranking π for a given voting instance with voters v , . . . , v n is defined as (cid:80) ni =1 K ( π, π v i ). If ranking σ has minimum Kemenyrank over all rankings, then σ is a Kemeny consensus [22]. The followingwell-known observation [12] will be useful in our proofs.
Observation 1
Each topologically sorted ranking is a Kemeny consensusranking. For acyclic majority graphs, the set of topologically sorted rankingscoincides with the set of Kemeny consensus rankings. v prefers ranking σ to ranking π if K ( σ, π v ) < K ( π, π v ).Analogously, voter v abstains between π and σ if K ( σ, π v ) = K ( π, π v ). We willsimply call v an abstaining voter if π and σ are clear from the context.For example, let σ = [1 , , , [5 , , , [9 , ,
8] and σ = [1 , , , [4 , , , [7 , , v prefers σ to σ , since π v = σ and π v (cid:54) = σ , that is, K ( π v , σ ) = 0 < K ( π v , σ ). Voter v in the same instanceis an abstaining voter since K ( π v , σ ) = 4 = K ( π v , σ ).We now define the two different notions of popularity. The first notion ofan absolutely popular ranking corresponds to the popular ranking as definedin [29]. Definition 2
Ranking π (cid:48) is more popular than ranking π in the absolute sense if K ( π (cid:48) , π v ) < K ( π, π v ) for an absolute majority of all voters. Ranking π is absolutely popular in I if no ranking π (cid:48) is more popular than π in the absolutesense, in other words, if there is no ranking π (cid:48) such that K ( π (cid:48) , π v ) < K ( π, π v )for an absolute majority of all voters in I .If we consider σ = [2 , , , [4 , , , [7 , , σ = [1 , , , [4 , , , [7 , ,
9] is more popular than σ in the absolute sense.Notice that σ and σ only differ in their ordering of the pair of candidates n absolutely and simply popular rankings 7 { , } . So since five out of six voters prefer candidate 1 to candidate 2, theyform an absolute majority of all voters who prefer σ to σ .This definition requires more than half of the n voters to prefer π (cid:48) to π in order to declare π (cid:48) to be more popular than π . Abstaining voters make ithard to beat π in such a pairwise comparison. However, if π (cid:48) only needs toreceive more votes than π among the voters not abstaining between these tworankings, then it can beat π . This leads to the notion of simple popularity.Let V abs ( π, π (cid:48) ) be the set of voters who abstain in the vote between π and π (cid:48) , that is, v ∈ V abs ( π, π (cid:48) ) if and only if K ( π v , π ) = K ( π v , π (cid:48) ). In Figure 1, V abs ( σ , σ ) = { v , v , v } with σ and σ as before. Definition 3
Ranking π (cid:48) is more popular than ranking π in the simple sense if K ( π (cid:48) , π v ) < K ( π, π v ) for a majority of the non-abstaining voters V \ V abs ( π, π (cid:48) ).Ranking π is simply popular in I if no ranking π (cid:48) is more popular than π in the simple sense, in other words, if there is no ranking π (cid:48) such that K ( π (cid:48) , π v ) < K ( π, π v ) for a majority of the non-abstaining voters V \ V abs ( π, π (cid:48) ).It follows directly from the two definitions above that simply popular rankingsare absolutely popular as well, but absolutely popular rankings are not neces-sarily simply popular. In the instance in Figure 1, σ = [1 , , , [5 , , , [9 , , σ = [1 , , , [4 , , , [7 , ,
9] in the simple sense, since v and v prefer σ to σ , while v , v , and v abstain. Notice that σ is not morepopular than σ in the absolute sense, because two voters do not constitutean absolute majority of all 6 voters, only a majority of the non-abstaining 3voters.We now define a natural verification problem arising from the notionsof absolutely and simply popular rankings, starting with absolutely popularrankings. We are given an input ranking π and ask if there exists a rankingthat is preferred to π by a majority of voters. In other words, we ask whether π is not absolutely popular. n - ABSOLUTELY - UNPOPULAR - RANKING - VERIFY ( n - AURV )Input : A voting instance I and a ranking π . Question : Does there exist a ranking σ preferred by a majority of allvoters to π ?The variant where we ask whether there exists a ranking σ that is preferredto π by a majority of the non-abstaining voters V \ V abs ( π, σ ) is called ( n - SURV ). In this section, we study the relationship between absolutely and simply popularrankings. We first place absolutely and simply popular rankings in the context The term absolutely popular is a shortened version of absolute-majority popular, while simply popular abbreviates simple-majority popular . These abbreviations should be kept inmind when interpreting the observation, given later, that an absolutely popular ranking neednot be simply popular in general. Sonja Kraiczy et al. of Kemeny consensus rankings in Section 3.1. Then in Section 3.2 we show thatfor as few as six voters, the two notions of popularity may not be equivalent andthat at the heart of this lies Condorcet’s paradox. Building upon our 6-voterexample, in Section 3.3 we derive a sufficient, but not necessary condition foran absolutely popular ranking to be simply popular as well. This conditionopens a way to prove in Section 3.4 that for up to five voters the two notionsare equivalent.3.1 A subset of Kemeny consensus rankingsWe first revisit two results from [29], established for absolute popularity, andtranslate them for the notion of simple popularity.
Lemma 1
If a voting instance I has a majority graph with a directed cycle,then there does not exist a simply popular ranking. If I has an acyclic majoritygraph, then a topologically sorted ranking is not necessarily simply popular.Proof These two statements hold for absolute popularity [29, Lemmas 2 and 3]and hence also for simple popularity, because simply popular rankings are alsoabsolutely popular by definition. (cid:117)(cid:116)
Example 1
A voting instance I with 4 voters and 6 candidates. Ranking π v is a Kemeny consensus, but σ = [2 , , , [6 ,
5] is more popular than π v inthe absolute sense. π v = [2 , , [4 , , [5 , π v = [1 , , [4 , , [6 , π v = [2 , , [3 , , [6 , π v = [1 , , [3 , , [5 , Observation 2
Simply popular rankings form a subset of absolutely popu-lar rankings, absolutely popular rankings form a subset of topologically sortedrankings, and finally, topologically sorted rankings form a subset of Kemenyconsensus rankings. In voting instances with an acyclic majority graph, topo-logically sorted rankings coincide with Kemeny consensus rankings.
Figure 3 illustrates these relations. In voting instances with a cyclic majoritygraph, Kemeny consensus rankings offer a preference aggregation method byrelaxing the definition of topologically sorted rankings. Absolutely and simplypopular rankings do exactly the opposite: they restrict the set of topologicallysorted rankings in instances with an acyclic majority graph, in order to servethe welfare of the majority to an even larger degree than topologically sorted n absolutely and simply popular rankings 9 rankings do. Absolute and simple popularity are desirable robustness propertiesof a ranking. However, the set of absolutely popular rankings may be emptyeven if a topologically sorted ranking exists [29]. Analogous to Condorcetconsistency, in which a Condorcet winner is chosen whenever it exists, onecould require to always choose a simply popular ranking if one exists, or, failingthat, an absolutely popular ranking if that exists.
Fig. 3
The hierarchy of various optimality notions for rankings as a Venn diagram. AKemeny consensus ranking always exists, but its subset of topologically sorted rankingsmight be empty. n = 6, we present a useful technical lemma.Let { C , . . . , C k } be a partition of C into k sets. We say that a ranking π preserves { C , . . . , C k } if each voter prefers a to b whenever a ∈ C i and b ∈ C j for some i < j . Lemma 2
Let { C , . . . , C k } be a partition of C and π be a ranking that pre-serves { C , . . . , C k } . If a subset of voters V (cid:48) ⊆ V prefers a ranking σ to π ,then there exists a ranking ζ such that ζ preserves { C , . . . , C k } and all votersin V (cid:48) prefer ζ to π . Let τ i be a permutation of the candidates in C i . We denote by τ τ . . . τ n thepermutation of the candidates ∪ mi =1 C i = C , in which each candidate in C i ispreferred to each candidate in C j whenever i < j , and candidates within a set C i are ranked according to τ i . Proof
Let ζ i be the permutation on C i that orders candidates in C i accordingto their rank in σ , that is, rank ζ i [ a ] < rank ζ i [ b ] if rank σ [ a ] < rank σ [ b ]. Let ζ := ζ ζ . . . ζ k . So ζ preserves { C , . . . , C k } . Suppose v ∈ V prefers a to b , and ζ orders b before a . Then a, b ∈ C i for some 1 ≤ i ≤ m . Since the relative orderof candidates in C i is the same in ζ and σ by construction, σ also orders b before a . In particular, every pair of candidates that contributes 1 to K ( ζ, π v ) also contributes 1 to K ( σ, π v ). We conclude K ( ζ, π v ) ≤ K ( σ, π v ), as desired. (cid:117)(cid:116) Theorem 3
There exists a voting instance with six voters that has an abso-lutely popular ranking which is not simply popular.Proof
We prove this statement for the voting instance I from Figure 1. Claim 1 σ = [1 , , , [4 , , , [7 , , is absolutely popular.Proof Let C = { , , } , C = { , , } , C = { , , } . Note that all voterspreserve { C , C , C } . By Lemma 2, in order to check if σ is absolutelypopular it suffices to generate the 6 rankings that preserve { C , C , C } ,and compare each of them to σ . We checked all of these rankings usinga program code, which is available under https://github.com/SonjaKrai/PopularRankingsExampleCheck . (cid:117)(cid:116) Claim 2 σ = [1 , , , [4 , , , [7 , , is not simply popular.Proof Ranking σ = [1 , , , [5 , , , [9 , ,
8] is more popular than σ in thesimple sense, because voters v , v , v abstain, voter v prefers σ to σ , andfinally, voters v and v prefer σ to σ . Appendix 6.1 contains the detailedcalculations behind this. (cid:117)(cid:116) This finishes the proof of our theorem. (cid:117)(cid:116)
For the sake of completeness, we remark that the voting instance I fromFigure 1 has 4 further absolutely popular rankings, each of which is simplypopular as well. For an instance that admits an absolutely popular rankingbut does not admit a simply popular ranking, please consult Appendix 6.2.3.3 A conditional equivalenceWe now present a sufficient condition under which simple popularity followsfrom absolute popularity. We start by showing that for a ranking π that is notsimply popular, there is a condition under which we can compute a rankingthat is also more popular than π in the absolute sense. Theorem 4 If σ is more popular than π in the simple sense and the majoritygraph of the voters in V abs ( σ , π ) is acyclic, then there is a ranking σ thatis more popular than π in the absolute sense. Such a σ can be computed inpolynomial time.Proof The theorem immediately follows from Observation 2 if π is not atopologically sorted ranking. In case there is a topologically sorted ranking,then it will serve as σ , otherwise σ can be constructed by the algorithmdeveloped for this purpose in [29, Lemma 2]. Furthermore, if V abs ( σ , π ) = ∅ ,then σ is preferred to π by a majority of all voters, and thus the theorem isproved by taking σ := σ . n absolutely and simply popular rankings 11 From here on we therefore assume that π is topologically sorted and that V abs ( σ , π ) (cid:54) = ∅ . Let D ( σ , π ) (cid:54) = ∅ be the set of pairs of candidates in the orderof which σ and π disagree. Claim 3 If v ∈ V abs ( σ , π ) , then v agrees with π on the order of exactly half ofthe pairs in D ( σ , π ) , and disagrees on the other half. In particular, | D ( σ , π ) | is even.Proof Any pair of candidates { a, b } such that σ and π agree on the order( a, b ), adds 1 to both K ( π v , σ ) and K ( π v , π ) if π v has them in the relativeorder ( b, a ), and adds 0 to both otherwise. So to be impartial, i.e. to have K ( π v , σ ) = K ( π v , π ), voter v ∈ V abs ( σ , π ) must agree with σ on the orderof exactly half of the pairs in D ( σ , π ), and disagree on the other half. Since V abs ( σ , π ) (cid:54) = ∅ can be assumed, | D ( σ , π ) | must be even. (cid:117)(cid:116) Claim 4
There exist consecutive candidates ( a ∗ , b ∗ ) in σ such that at leasthalf of the voters in V abs ( σ , π ) prefer b ∗ to a ∗ .Proof Assume the contrary, i.e. that for any two consecutive candidates in σ , a majority of voters in V abs ( σ , π ) agree with σ . For an arbitrary pair ofcandidates { a, b } , at least half of the voters in V abs ( σ , π ) must then also agreewith their order in σ , as otherwise this would imply a directed cycle in theirmajority graph, which is acyclic by assumption.Since σ (cid:54) = π , there is some ordered pair ( a ∗ , b ∗ ) that is consecutive in σ and b ∗ is ordered somewhere before a ∗ in π , that is rank π [ b ∗ ] < rank π [ a ∗ ]. Notethat ( a ∗ , b ∗ ) ∈ D ( σ , π ). In particular, by our starting assumption, a majorityof voters in V abs ( σ , π ) agrees with σ on the order ( a ∗ , b ∗ ) and hence disagreeswith π .We now introduce the indicator variable I v,π, { a,b } , which is set to 1 if π v and π disagree on the order of candidates a and b , and it is set to 0 otherwise.We sum up the disagreements of voters in V abs ( σ , π ) with π over pairs in D ( σ , π ) and obtain a contradiction. | V abs ( σ , π ) | | D ( σ , π ) | (cid:88) { a,b }∈ D ( σ ,π ) (cid:88) v ∈ V abs ( σ ,π ) I v,π, { a,b } (1) > (cid:88) { a,b }∈ D ( σ ,π ) | V abs ( σ , π ) | | V abs ( σ , π ) | | D ( σ , π ) | π is exactly half of | D ( σ , π ) | , expressed on the left-hand sideof Line 1. The inequality in Line 2 follows, because for all pairs in D ( σ , π ),at least half of the abstaining voters disagree with π , and, additionally, thereexists a pair { a ∗ , b ∗ } such that a majority of voters in V abs ( σ , π ) disagreewith π , as we proved above. Finally, reordering the terms leads back to thesame formula as on the left-hand side in Line 1, creating a contradiction. (cid:117)(cid:116) The pair of candidates ( a ∗ , b ∗ ) in Claim 4 leads us to a suitable ranking σ .Let σ be the permutation we get from σ by the swap ( a ∗ , b ∗ ) → ( b ∗ , a ∗ ). InClaims 5 and 6 we will show that two groups of voters prefer this σ to π , andthat these two groups constitute a majority of all voters. Claim 5
All voters v ∈ V abs ( σ , π ) who prefer b ∗ to a ∗ also prefer σ to π .Proof If v ∈ V abs ( σ , π ) prefers b ∗ to a ∗ , then the following holds. K ( π v , σ ) = K ( π v , σ ) − < K ( π v , σ ) = K ( π v , π ) (4)Line 3 is true, because σ is obtained from σ by performing a swap that isgood for voter v . The equality in Line 4 holds since v abstains between σ and π . The inequality in Line 4 proves the claim. (cid:117)(cid:116) The second group of voters we investigate consists of voters who prefer σ to π . This group by assumption makes up a majority of the non-abstainingvoters V \ V abs ( σ , π ). Let voter v belong to this group. Claim 6 If K ( π v , σ ) < K ( π v , π ) for voter v ∈ V \ V abs ( σ , π ) , then K ( π v , σ )+2 ≤ K ( π v , π ) .Proof Only the pairs in D ( σ , π ) contribute differently to K ( π v , σ ) and to K ( π v , π ). In particular, a pair in D ( σ , π ) adds 1 to either K ( π v , σ ) or K ( π v , π ),and 0 to the other. If k is the number of pairs on whose order σ and π agree,but π v disagrees, then by the previous argument, K ( π v , σ ) + K ( π v , π ) = | D ( σ , π ) | + 2 k , which is even by Claim 3. Since K ( π v , σ ) < K ( π v , π ) byassumption and their sum | D ( σ , π ) | + 2 k is even, K ( π v , σ ) + 2 ≤ K ( π v , π )must hold. (cid:117)(cid:116) Since a swap of consecutive candidates in a ranking can increase the distanceto any other ranking by at most one, Claim 6 implies that for any voter v whoprefers σ to π , the following holds. K ( π v , σ ) ≤ K ( π v , σ ) + 1 < K ( π v , π )We conclude that voters who prefer σ to π also prefer σ to π .Claims 4 and 5 show that at least half of the voters in V abs ( σ , π ) prefer σ to π , and Claim 6 proves that more than half of the voters outside of V abs ( σ , π )prefer σ to π . The two sets of voters thus constitute a majority of all voterswho prefer σ to π . (cid:117)(cid:116) By rephrasing Theorem 4, we get the following.
Corollary 1
If ranking π is absolutely popular, and for any ranking σ , V abs ( σ, π ) has an acyclic majority graph, then π is also simply popular. Furthermore, if V abs ( σ, π ) has an acyclic majority graph for each absolutely popular ranking π and any ranking σ in an instance I , then absolutely and simply popularrankings in I coincide. n absolutely and simply popular rankings 13 The following example demonstrates that if σ is more popular than σ inthe simple sense, but the majority graph of the voters in V abs ( σ , σ ) is cyclic(contradicting our assumption in Theorem 4), then σ might be absolutelypopular. The calculations supporting the example can be found in Appendix 6.1. Example 2
Consider the instance described in Figure 1. The voters abstainingbetween σ = [1 , , , [5 , , , [9 , ,
8] and σ = [1 , , , [4 , , , [7 , ,
9] have acyclic majority graph. In this instance, σ is the unique absolutely popularranking but is not simply popular, since σ is more popular than σ in thesimple sense. Notice that for both σ and σ , for every pair of consecutivecandidates in each, a majority of the abstaining voters agree with the order ofthese candidates.3.4 At most 5 votersFrom Theorem 4 we can deduce that for a small number of voters, the twonotions of popularity are equivalent. This is due to the fact that we need atleast 3 abstaining voters in order for V abs ( σ, π ) to have a cyclic majority graph. Theorem 5
A ranking σ is absolutely popular in a voting instance I with atmost five voters if and only if it is simply popular in I .Proof From Observation 2 we know that simply popular rankings are alsoabsolutely popular, which allows us to concentrate only on one direction of thestatement, namely that if a ranking is not simply popular, then it also cannotbe absolutely popular. Let us assume that ranking π is not simply popular.By definition there exists a ranking σ that is preferred to π by a majority ofnon-abstaining voters V \ V abs ( σ, π ) (cid:54) = ∅ .If at least one voter prefers π to σ , then at least two voters must prefer σ to π , and the remaining at most two abstaining voters can only form anacyclic majority graph. From this follows by Theorem 4 that π is not absolutelypopular.On the other hand, if no voter prefers π to σ , then K ( π v , σ ) ≤ K ( π v , π )holds for all voters. For V \ V abs ( σ, π ) (cid:54) = ∅ , by assumption there is a voter v ∗ who prefers σ to π , that is, for whom K ( π v ∗ , σ ) < K ( π v ∗ , π ). We thushave (cid:80) ni =1 K ( π v i , σ ) < (cid:80) ni =1 K ( π v i , π ). This means that π is not a Kemenyconsensus and by Observation 2, π is not absolutely popular. (cid:117)(cid:116) n - aurv and n - surv In this section, we analyse the complexity of the verification problems for thetwo popularity notions. We prepare for this by giving a sufficient conditionfor absolute popularity in Section 4.1. This condition will then be used inSection 4.2, where we prove the polynomial solvability of both problems inthe case of n ≤
3. For 4 ≤ n ≤
5, we reach the same conclusion in Section 4.3, however, only for special instances. Then, in Section 4.4, NP -hardness is provedfor n = 6. We finish by drawing attention to the connection with the complexityof the famous Kemeny consensus problem in Section 4.5. This explains whywe only succeeded to prove polynomial-time solvability for special cases of4 ≤ n ≤ P for n = 4 or n = 5 in thegeneral case, then one can find a Kemeny consensus for 3 voters in polynomialtime. We reach these results via both Karp and Turing reductions.4.1 A sufficient condition for n - aurv We call a ranking π c -sorted for 0 < c ≤ { a, b } with rank π [ a ] < rank π [ b ], at least a c -fraction of the voters prefers a to b . Atopologically sorted ranking π is by definition -sorted. In [29], van Zuylen etal. show that even a topologically sorted ranking is not necessarily absolutelypopular. Here we ask for which constant < c ≤ n - aurv . Theorem 6 c = is the smallest constant c for which the following holds: Ifa voting instance I admits a c -sorted ranking π , then π is absolutely popular.Proof We first show that any -sorted ranking is absolutely popular. Claim 7
Let I be a voting instance and π be a ranking. If for every pair ofcandidates { a, b } with rank π [ a ] < rank π [ b ] , at least a -fraction of the votersagree with the order ( a, b ) , then π is absolutely popular.Proof Let σ be any ranking different from π . We will show that there is novoter set of cardinality (cid:98) n (cid:99) +1, in which every voter prefers σ to π . Let V (cid:48) be anarbitrary set of (cid:98) n (cid:99) +1 voters. On the bubble swap path π := π, π , . . . π k := σ from π to σ , each swap π i → π i +1 is bad for at least34 | V | = 3 n ≥ (cid:98) n (cid:99) + 12 = 12 | V (cid:48) | voters. Hence for a swap π i → π i +1 , we have that (cid:80) v ∈ V (cid:48) K ( π i +1 , π v ) − K ( π i , π v ) ≥
0. Summing over all swaps we get a telescoping sum that re-duces to (cid:88) v ∈ V (cid:48) K ( σ, π v ) − K ( π, π v ) ≥ . In particular, not every v ∈ V (cid:48) can prefer σ to π . Since V (cid:48) was an arbitraryset of size (cid:98) n (cid:99) + 1, no voter set of at least this size can prefer σ to π , and thus, π must be absolutely popular. (cid:117)(cid:116) We now show that in fact c = is tight, meaning that for any c < , we canconstruct an instance I and a c -sorted ranking π such that π is not absolutelypopular in I . n absolutely and simply popular rankings 15 Claim 8
For arbitrary c < there exists a voting instance I and a topologicallysorted, but not absolutely popular ranking π in it, such that a c -fraction ofthe voters agree with ( a, b ) for every pair of candidates { a, b } with rank π [ a ] < rank π [ b ] .Proof We create voting instance I with 4 j voters and 4 j + 2 candidates. Let c = − ε for some ε >
0. Next, choose j large enough such that ( + ε ) | V | ≥ j +1,which is equivalent to j ≥ ε . We create a set V of 2 j − π = [1 , , [3 , , . . . , [4 j + 1 , j + 2]. Note that we group thecandidates into 2 j + 1 blocks of 2 candidates each, in the form [ a, b ]. The set ofthe remaining 2 j + 1 voters is called V , and their set of rankings is depictedin Table 1. For informal intuition, the first ranking orders the first j + 1 blocks[ a, b ] in decreasing order and the remaining j blocks in increasing order. Witheach new row, the j + 1 decreasingly ordered blocks are shifted by 1 to theright. When the end of the row is reached, the j + 1 decreasingly ordered blocksare wrapped around to continue at the beginning of the row. This constructionis similar to the one in [29, Lemma 3]. [2 ,
1] [4 , . . . [2 j, j −
1] [2 j + 2 , j + 1] [2 j + 3 , j + 4] [2 j + 5 , j + 6] . . . [4 j + 1 , j + 2][1 ,
2] [4 , . . . [2 j, j −
1] [2 j + 2 , j + 1] [2 j + 4 , j + 3] [2 j + 5 , j + 6] . . . [4 j + 1 , j + 2]... ...[1 ,
2] [3 , . . . [2 j − , j ] [2 j + 2 , j + 1] [2 j + 4 , j + 3] [2 j + 6 , j + 5] . . . [4 j + 2 , j + 1]][2 ,
1] [3 , . . . [2 j − , j ] [2 j + 1 , j + 2] [2 j + 4 , j + 3] [2 j + 6 , j + 5] . . . [4 j + 2 , j + 1]... ...[2 ,
1] [4 , . . . [2 j, j −
1] [2 j + 1 , j + 2] [2 j + 3 , j + 4] [2 j + 5 , j + 6] . . . [4 j + 2 , j + 1] Table 1
The blocks in decreasing order are highlighted.
For each (2 i − , i ) for 1 ≤ i ≤ j + 1, every v ∈ V agrees with this pairand also exactly j of the voters in V agree with it. In total, 3 j − j − j = 34 − j ≥ − ε It is trivial to see for all other pairs of voters { a, b } that if a < b then all votersprefer a to b . Hence in particular π is (cid:0) − ε (cid:1) -sorted.We now show that ranking σ = [2 , , [4 , , . . . [4 j + 2 , j + 1] is preferred to π by a majority of all voters, namely by all 2 j + 1 voters in V . Each voter in V has exactly j + 1 blocks with decreasing order and j blocks with increasingorder. Thus, each of them would rather have all pairs in decreasing order thanall pairs in increasing order. (cid:117)(cid:116) This finishes the proof of our theorem. (cid:117)(cid:116)
As an aside, following on from Theorem 6, it is natural to ask about theexistence of ε -popular rankings: rankings that are preferred to all other rankingsby some ε -fraction of voters. The following theorem (whose proof can be found in Appendix 6.3) shows that for any ε >
0, there is a voting instance that doesnot admit any ε -popular ranking. Theorem 7
There is a voting instance with n voters and n candidates suchthat for any ranking π , there exists another ranking σ such that n − votersprefer σ to π and only one voter prefers π to σ . n ≤ n ≤
5, we will refer to them as popularity if n ≤ I .We first show that for n ≤
3, the problems n - aurv and n - surv arepolynomial-time solvable. We establish this by proving that for at most threevoters, the set of topologically sorted rankings coincides with the set of popu-lar rankings. Since verifying whether a given ranking is topologically sortedcan be carried out in polynomial time, n - aurv and n - surv turn out to bepolynomial-time solvable for n = 2 , Lemma 3
In a voting instance I with n = 2 , a ranking is popular if and onlyif it is topologically sorted.Proof From Observation 2 we know that all popular rankings must be topologi-cally sorted. Let D ( π v , π v ) be the set of pairs of candidates { a, b } that π v and π v order differently. Consider any ranking π . Each pair { a, b } ∈ D ( π v , π v )adds 1 to either K ( π v , π ) or K ( π v , π ), and each pair { a, b } / ∈ D can at bestadd 0 to both and at worst 1 to both K ( π v , π ) and K ( π v , π ). From thisfollows that | D ( π v , π v ) | ≤ K ( π v , π ) + K ( π v , π ) . If σ is a topologically sorted ranking, then by definition there is no pair ofcandidates { a, b } that adds 1 to both K ( π v , σ ) and K ( π v , σ ). Thus, K ( π v , σ )+ K ( π v , σ ) = | D ( π v , π v ) | . If π is preferred to σ by a majority, then both votersprefer π to σ , which leads to | D ( π v , π v ) | ≤ K ( π v , π ) + K ( π v , π ) In a voting instance I with n = 3 , a ranking is popular if and onlyif it is topologically sorted.Proof Just as for the n = 2 case, Observation 2 implies here as well that allpopular rankings must be topologically sorted. Let π be a topologically sortedranking in I . Note that whenever rank π [ a ] < rank π [ b ] holds for candidates n absolutely and simply popular rankings 17 a and b , at least half of the voters, that is, at least two voters prefer a to b .So for any two voters, at least one of them prefers a to b , implying that π isalso topologically sorted for any two of the three voters. In particular, π isabsolutely popular for any two voters by Lemma 3, showing that there is noranking that they both prefer to π . Hence π must be absolutely popular in I and by Theorem 5, also simply popular. (cid:117)(cid:116) Using the fact that verifying whether a given ranking is topologically sortedcan be done in polynomial time, we derive the following result. Theorem 8 For n ≤ , n - aurv and n - surv are polynomial-time solvable. n = 4 and n = 5If we have 4 or 5 voters, it turns out a topologically sorted ranking may not beabsolutely or simply popular anymore. In the case of an acyclic tournament asthe majority graph, finding and verifying a popular ranking are both polynomial-time solvable. We further show that 4- aurv (4- surv ) in general is polynomial-time solvable if and only if 5- aurv (5- surv ) is. Lemma 5 If a voting instance I with n = 4 has an acyclic tournament asits majority graph, then the unique topologically sorted ranking is the uniquepopular ranking.Proof Since the majority graph of I is a tournament, the unique topologicallysorted ranking π is -sorted. The lemma then follows from Theorem 6 andObservation 2. (cid:117)(cid:116) Lemma 6 In a voting instance I with n = 4 and an acyclic majority graph,if π is popular in at least one of the instances formed by three of the voters,then π is popular in I .Proof Let π be absolutely popular in the instance formed by v , v , and v .Then by definition there exists no ranking σ that is preferred by two of v , v , v ,as this would contradict the fact that π is absolutely popular in the instanceformed by voters v , v , v . In particular, there does not exist a ranking σ preferred by a majority of v , v , v , v , as this would require at least 3 votersand hence at least 2 of v , v , v . We conclude that π is absolutely popular in I .By Theorem 5 it follows that π is simply popular as well. (cid:117)(cid:116) We now present an example in which there is ranking π that is not topolog-ically sorted such that π is more popular than a topologically sorted ranking. Observation 9 In a voting instance I with n = 4 and an acyclic majoritygraph, a ranking that is not topologically sorted can be absolutely (and simply)more popular than a topologically sorted ranking. Proof Consider the instance I below, with n = 4 and m = 10. π v = [1 , , [3 , , [5 , , [7 , , [9 , π v = [1 , , [4 , , [6 , , [7 , , [10 , π v = [1 , , [4 , , [6 , , [8 , , [9 , π v = [2 , , [3 , , [5 , , [8 , , [10 , π v is a topological sorted ranking of I . However, σ = [2 , , [4 , , [6 , , [8 , , [10 , 9] is preferred by v , v , and v to π , since K ( π v i , π v ) = 3 and K ( π v i , σ ) = 2 for 2 ≤ i ≤ 4. Since a majority ofvoters prefer candidate 1 to candidate 2, σ is not topologically sorted. Forthe sake of completeness, we remark that the topologically sorted ranking[1 , , [4 , , [6 , , [8 , , [10 , 9] is popular. (cid:117)(cid:116) We now discuss a strongly related decision problem called n - all-closer-ranking , which will come useful when establishing results for the cases n = 4and n = 5. For a voting instance with n voters and a given ranking π , we askwhether there is a ranking that all the voters prefer to π . n - ALL - CLOSER - RANKING Input : A voting instance I and a ranking π . Question : Does there exist a ranking σ preferred by each of the n voters to π ?The next theorem reveals some features of this problem. Theorem 10 In a voting instance I with n = 3 and an acyclic majority graph,we can decide in polynomial time if there exists a ranking preferred by all votersto a given ranking π and if it does, output it.Proof Our first two technical observations apply for the case of one and twovoters, respectively. Observation 11 If K ( σ, π v ) > for a voter v and a ranking σ , then thereexists a swap in σ that is good for v .Proof Suppose there is no swap that is good for v . So σ is a topologically sortedranking for v , and for one voter this means π v = σ , i.e. K ( σ, π v ) = 0. (cid:117)(cid:116) Observation 12 If there is no swap in ranking π that is good for both v and v , then there is no ranking σ preferred to π by both v and v .Proof Since there is no swap that is good for both v and v , π is a topologicallysorted ranking for v and v . By Lemma 3 this means that π is absolutelypopular, so there exists no ranking preferred by a majority, that is, preferredby both voters. (cid:117)(cid:116) We are now ready to proceed to the case of 3 voters. First we computethe unique topologically sorted ranking σ in I in polynomial time [20]. Wedistinguish four cases, based on how many of the three voters prefer σ to π ,which can be checked in polynomial time. n absolutely and simply popular rankings 19 1. If σ is preferred to π by all 3 voters, then we are done.2. Suppose that two of the voters, v and v , prefer σ to π . Suppose that v and v are distances d = K ( π, π v ) − K ( σ, π v ) and d = K ( π, π v ) − K ( σ, π v )closer to σ than to π , respectively, and that v is further away from it bydistance d = K ( σ, π v ) − K ( π, π v ). Notice that we have d i ≥ ≤ i ≤ 3. Without loss of generality we can further assume that d ≤ d .Let π := π, . . . , π k := σ be the bubble sort swap path from π to σ . Claim 9 For the above defined distances, d − d ≥ and similarly, d − d ≥ hold.Proof Firstly, no swap is bad for both v and v , since every swap π i → π i +1 that is bad for v must be good for both v and v , because σ is thetopologically sorted ranking of I . Secondly, there is at least one swap thatis good for both v and v . If there was no swap that is good for both v and v , then there cannot exist a ranking preferred to π by all voters, sincethere cannot exist a ranking preferred by both v and v by Observation 12.Good swaps for voters v , v , and v add 1 to d and d , and subtract 1from d , respectively. It follows that d − d ≥ d ≥ d , thesame argument also implies d − d ≥ (cid:117)(cid:116) We are now ready to transform σ to a ranking that is preferred by all 3voters to π . Procedure Let σ = σ . In the i th round we search for a swap in σ i − that is good for v and if found, perform the swap to obtain σ i for i ≥ 1. Otherwise weoutput an error message. We stop the procedure in round i = d + 1 andoutput σ d +1 . Claim 10 If the process terminates outputting σ d +1 , then v , v , and v prefer σ d +1 to π . Otherwise, there does not exist a ranking preferred by allvoters to π .Proof The process terminates before reaching σ d +1 if and only if K ( σ j , π v ) =0 for some integer j < d + 1, otherwise by Observation 11, we can find aswap that is good for v . Again by Observation 11, K ( π v , σ ) ≤ d . Since K ( π v , σ ) − K ( π v , π ) = d , K ( π v , π ) = 0 i.e. π v = π . So clearly therecannot exist a ranking preferred by all voters, including v , to π .If we successfully obtain σ d +1 , it will be at most d + 1 swaps away from σ .So σ d +1 is closer to π v than π by d (cid:48) := K ( π, π v ) − K ( σ d +1 , π v ) ≥ K ( π, π v ) − K ( σ, π v ) − ( d + 1) = d − d − > . Similarly d (cid:48) > 0, that is, π v is closer to σ d +1 than to π .Also, by construction d (cid:48) := K ( π, π v ) − K ( σ d +1 , π v ) = K ( π, π v ) − K ( σ, π v )+ K ( σ, π v ) − K ( σ d +1 , π v ) = − d + d + 1 = 1 . So π v is closer to σ d +1 than to π . That is, all of v , v and v prefer to σ d +1 to π . (cid:117)(cid:116) 3. Suppose that only one voter, v prefers σ to π . We show that this casecannot occur. There exists a bubble sort swap on the path from π to σ thatis good for both v and v , else there cannot be a ranking preferred by allvoters by Observation 12. Since every bubble sort swap is good for at leastone of v and v , without loss of generality, let v be the voter for whomat least half of the bubble sort swaps are good. This means that v hasmore good swaps on the path than bad swaps, i.e. v also prefers σ to π , acontradiction to v being the only voter who prefers σ to π .4. Finally, suppose that no voter prefers σ to π , i.e. K ( π v i , σ ) ≥ K ( π v i , π ) forall 1 ≤ i ≤ 3. Since σ is topologically sorted and hence a Kemeny consensus(see Observation 1), (cid:80) i =1 K ( π v i , σ ) ≤ (cid:80) i =1 K ( π v i , π ) holds. From thesetwo inequalities follows that K ( π v i , σ ) = K ( π v i , π ) for all 1 ≤ i ≤ 3. Thatis, π is also a Kemeny consensus, hence there does not exist a rankingpreferred to π by all voters.Having discussed all 4 cases, we now can output a ranking preferred by allthe voters to a given ranking π —or a proof for its non-existence—in polynomialtime. (cid:117)(cid:116) Lemma 7 For n ≥ , if at least one of (2 n − - surv , (2 n − - aurv , and (2 n − - aurv is polynomial-time solvable, then n - all-closer-ranking ispolynomial-time solvable.Proof Assume first that (2 n − aurv ∈ P holds. Consider an instance of n - all-closer-ranking with input ranking π and voters v , . . . , v n . To thisinstance of n - all-closer-ranking we construct the following instance of(2 n − aurv . We copy π as the given input ranking and create 2 n − n − π and the other n voters corresponding to v , . . . , v n .Since voters with ranking π clearly prefer π to any other ranking, if there existsa ranking preferred by a majority (at least n ) of the 2 n − n voters must be v , . . . , v n . If a ranking is preferred by v , . . . , v n to π , then itis a solution to n - all-closer-ranking . Hence there is a ranking σ preferredby v , . . . , v n to π if and only if σ is a solution to n - all-closer-ranking . For(2 n − aurv and (2 n − surv we simply add another voter with ranking π ,and otherwise keep the proof intact. (cid:117)(cid:116) Lemma 8 Let n ≥ be a constant. If n - all-closer-ranking is polynomial-time solvable, then (2 n − - aurv and (2 n − - aurv are both polynomial-timesolvable.Proof If n - all-closer-ranking has a polynomial-time algorithm A , then wecan solve (2 n − aurv by applying A to each of the (cid:0) n − n (cid:1) voter groups ofsize n , which itself is a polynomial-time procedure if n is a constant. If one ofthe calls to A returns yes, return yes, else return no. It is easy to see that thisprocedure returns yes if and only if there is some group of n voters that prefers n absolutely and simply popular rankings 21 another ranking i.e. if and only if there is a majority that prefers anotherranking. A similar argument can be applied for (2 n − aurv . (cid:117)(cid:116) An immediate consequence of Lemmas 7 and 8 is the following result. Theorem 13 - aurv (resp. - surv ) is polynomial-time solvable if and onlyif - aurv (resp. - surv ) is polynomial-time solvable.Proof With n = 3 in Lemma 8, 5- aurv is ∈ P implies 3- all-closer-ranking ∈ P . Due to Lemma 7, 3- all-closer-ranking ∈ P implies 4- aurv ∈ P . By asimilar argument 4- aurv ∈ P implies 5- aurv ∈ P . By Theorem 5, an analogousresult holds for 4- surv and 5- surv . (cid:117)(cid:116) NP -hardness for n = 6We now improve upon the NP -hardness result of [29, Theorem 4] on the searchversion of 7- aurv from 7 voters to 6 voters, and also extend it to simply popularrankings with 6 or 7 voters. Theorem 14 The search versions of - aurv , - surv , and - surv are all NP -hard.Proof We modify the proof from [29, Theorem 4], which shows that the searchversion of 7- aurv is NP -hard. In that proof, 4 of the 7 voters have rankings π , π , π , π , respectively, and the remaining 3 voters have ranking L ( σ ). Theauthors (implicitly) prove that it is NP -hard to construct a ranking ζ thatall 4 voters with rankings π , π , π , π prefer to L ( σ ). We will first show that6- aurv and 7- surv both lead to this NP -hard problem.We start with 7- surv . For any ranking ζ (cid:54) = L ( σ ), the 3 voters with lists L ( σ ) must prefer L ( σ ) to ζ . In order for ζ to be more popular than L ( σ ) in thesimple sense, ranking ζ must be preferred to L ( σ ) by more than 3 voters. Thishappens if and only if all 4 voters with rankings π , π , π , π prefer ζ to L ( σ ).For 6- aurv , we have two voters with lists L ( σ ) instead of three. The samereduction holds as for 7- surv , since a majority of all 6 voters, that is, the 4voters with rankings π , π , π , π must prefer ζ to L ( σ ).We keep the same instance for 6- surv . Now only two voters prefer L ( σ ) to ζ , and thus ζ is more popular than L ( σ ) in the simple sense if and only if atleast 3 of the remaining 4 voters prefer ζ to L ( σ ), and none of these four votersprefer L ( σ ) to ζ . Even though it is not directly observed by van Zuylen et al.,their NP -hardness proof carries over to this case without modification. (cid:117)(cid:116) aurv /5- aurv (4- surv /5- surv ) is polynomial-time solvable, then so is thewell-known Kemeny consensus problem for 3 voters, whose complexity is currently unknown. Consider the following problem: we are given a ranking π as well as three voters’ rankings π v , π v , π v . Our task is to output a ranking σ that has smaller Kemeny rank than π , or reports that none exists. We callthis search problem n - smaller-Kemeny-rank . n - SMALLER -K EMENY - RANK Input : A voting instance I and a ranking π . Output : A ranking σ with smaller Kemeny rank than π , that is, (cid:80) ni =1 K ( σ, π v i ) < (cid:80) ni =1 K ( π, π v i ) or a statement that no such rankingexists. Theorem 15 A Kemeny consensus for n voters can be computed in polynomialtime if and only if n - smaller-Kemeny-rank is in P .Proof Assume that n - smaller-Kemeny-rank has a polynomial-time algo-rithm A . We simply choose an arbitrary ranking π for the Kemeny consensusproblem and apply A to find π with smaller Kemeny rank than π , andcontinue this way until we have found a Kemeny consensus. The number ofcalls to A can be naively bounded by n m ( m − , which is the maximum Kemenyrank of a ranking. Similarly, if we can find a Kemeny consensus for n voters inpolynomial time, then we can check if it has smaller Kemeny rank than π inthe input of the n - smaller-Kemeny-rank problem. (cid:117)(cid:116) By an argument similar to the one in [29, Theorem 5], we prove the followingresult on the complexity of 3- smaller-Kemeny-rank . Theorem 16 If 3- smaller-Kemeny-rank is NP -hard, then the search ver-sion of 3- all-closer-ranking is also NP -hard.Proof We create an instance I (cid:48) of 3- all-closer-ranking as follows. Thethree voters in I (cid:48) will be v (cid:48) , v (cid:48) , and v (cid:48) . We create a set of 3 m candidates as C (cid:48) = C ∪ C ∪ C , where C j = { c jr : 1 ≤ r ≤ n } for each 1 ≤ j ≤ 3. Intuitively, C (cid:48) consists of three distinguishable copies of C . Given any ranking σ of the m candidates in I , let σ j be the ranking obtained from σ by replacing eachcandidate c r by c jr , preserving the original order in σ . Let π i be a permutationof the candidates in C i . We denote by π π . . . π m the permutation of thecandidates, in which each candidate in C i is preferred to each candidate in C j whenever i < j , and candidates within a set C i are ranked according to π i .Now we form the preference lists of the voters in I (cid:48) as follows. π v (cid:48) = π v π v π v π v (cid:48) = π v π v π v π v (cid:48) = π v π v π v Finally we create ranking π (cid:48) = π π π for the input of 3- all-closer-ranking ,besides I (cid:48) . n absolutely and simply popular rankings 23 Claim 11 Ranking σ in I has a smaller Kemeny rank than π if and only ifits counterpart in I (cid:48) , ranking σ (cid:48) is preferred by all of v (cid:48) , v (cid:48) , and v (cid:48) to π (cid:48) .Proof Suppose first that σ has a smaller Kemeny rank than π in I , thatis, (cid:80) p =1 K ( π v p , σ ) < (cid:80) p =1 K ( π v p , π ). Let σ (cid:48) = σ σ σ . Note that for each1 ≤ i ≤ K ( π v (cid:48) i , σ (cid:48) ) = (cid:88) p =1 K ( π v p , σ ) < (cid:88) p =1 K ( π v p , π ) = K ( π v (cid:48) i , π (cid:48) ) . So indeed each v (cid:48) i for 1 ≤ i ≤ σ (cid:48) to π (cid:48) .For the converse direction, we first informally summarise the argument.We will argue that if there is a ranking σ (cid:48) in I (cid:48) preferred to π (cid:48) by all voters,then we can extract a ranking σ in I with smaller Kemeny rank than π intwo steps. First of all, by a previous lemma, we can break up σ (cid:48) into threedifferent rankings, each defined on a different candidate set. Secondly, oneof these rankings translated back to our instance I will be a ranking with asmaller Kemeny rank than π , as we will argue using the averaging principle.This argument relies on every π v i , 1 ≤ i ≤ 3, appearing once in each “column”of I (cid:48) , hence justifying the cyclic shift used in I (cid:48) .Suppose that σ (cid:48) is preferred to π (cid:48) by all three voters v (cid:48) , v (cid:48) , v (cid:48) . By Lemma 2we can assume that σ (cid:48) preserves { C , C , C } . So we can also assume that σ (cid:48) = ζ ζ ζ , where ζ jj is a ranking of the candidates in C j for 1 ≤ j ≤ ζ nj be the ranking that isobtained from ζ jj by replacing candidate c jr with candidate c nr for 1 ≤ j, n ≤ ≤ r ≤ n , preserving the original order in ζ jj . Let τ j = ζ j ζ j ζ j , so thatintuitively, we copy the same ranking three times, on different candidate sets.We will show that for some 1 ≤ j ≤ τ j is also preferred to π (cid:48) by all thevoters v (cid:48) , v (cid:48) , and v (cid:48) .Notice that (cid:80) i =1 K ( σ (cid:48) , π v (cid:48) i ) = (cid:80) j =1 K ( τ j , π v (cid:48) ). Since K ( σ (cid:48) , π v (cid:48) i ) < K ( π (cid:48) , π v (cid:48) i )for all 1 ≤ i ≤ K ( π (cid:48) , π v (cid:48) ) = K ( π (cid:48) , π v (cid:48) ) = K ( π (cid:48) , π v (cid:48) ), it follows that (cid:88) j =1 K ( τ j , π v (cid:48) ) = (cid:88) i =1 K ( σ (cid:48) , π v (cid:48) i ) < (cid:88) i =1 K ( π (cid:48) , π v (cid:48) i ) = 3 K ( π (cid:48) , π v (cid:48) ) . (6)So there must exist an index j , 1 ≤ j ≤ 3, such that K ( τ j , π v (cid:48) ) < K ( π (cid:48) , π v (cid:48) ).But then (cid:88) i =1 K ( ζ j , π v i ) = (cid:88) i =1 K ( ζ ij , π iv i ) = K ( τ j , π v (cid:48) ) We observe that a slight tweak to the above proofs lets us show NP -hardnessfor 4 problems related to 3- all-closer-ranking . Observation 17 If finding a ranking with a smaller Kemeny rank than agiven ranking π for voters is NP -hard, then finding a ranking ζ that exactlyone / at least one / exactly two / at least two of the voters prefer π , whileno voter prefers π to ζ is also NP -hard.Proof We only need to argue why the converse direction still holds with theweaker assumption in Theorem 16. Note that in the proof of Theorem 16,Inequality 6 still holds if only one / at least one / exactly two / at least two ofthe three voters is non-abstaining and prefers σ (cid:48) to π (cid:48) , while the other votersabstain. (cid:117)(cid:116) Corollary 2 If finding a ranking that is more popular than a given ranking π in an instance of 4- aurv , 4- surv , 5- aurv , or 5- surv were polynomial-timesolvable, then we could find a Kemeny consensus for 3 voters in polynomialtime.Proof This proof is illustrated in Figure 4. By Lemma 7, if the search version of4- aurv or 5- aurv is in P , then the search version of 3- all-closer-ranking isalso in P . Now, if the latter is true, then by Theorem 16, 3- smaller-Kemeny-rank is also in P . This would finally imply that finding a Kemeny consensusfor 3 voters is in P , by Theorem 15. An analogous result holds for 4- surv and5- surv by Theorem 5. (cid:117)(cid:116) aurv , 5- aurv , 4- surv , 5- surv search version of 3- all-closer-ranking smaller-Kemeny-rank finding a Kemeny consensus for 3 votersLemma 7Theorem 16Theorem 15 Fig. 4 If any of 4- aurv , 5- aurv , 4- surv , and 5- surv is solvable in polynomial time, thenvia the depicted implications, finding a Kemeny consensus for 3 voters is ∈ P . We studied absolutely popular rankings, defined in [29], and also introducedthe notion of simply popular rankings analogous to the concept of popularity n absolutely and simply popular rankings 25 found in the matching literature which ignores abstaining voters. Then weshowed that a ranking π is absolutely popular if and only if it is simply popularassuming that the majority graph of the abstaining voters between π and anyother ranking σ is acyclic. Using this result we also proved that the two notionsof popularity are equivalent for voting instances with at most five voters. Forinstances with six voters, however, we showed that this equivalence does nothold anymore.We found the smallest constant c , for which any c -sorted ranking for aninstance is absolutely popular. For two or three voters, a topologically sortedranking turned out always to be popular with respect two both of the popularitynotions. For four voters this also holds as long as the majority graph of thevoters is a tournament, but it does not hold in general. We explained thatthe problem of deciding whether there exists a ranking σ that is preferredto a given ranking π by simple or absolute a majority of voters for instanceswith four of five voters boils down to the problem of deciding for three voterswhether there is a ranking σ they all prefer to π . This problem, we showed, ispolynomial-time solvable if the majority graph of the three voters is acyclic, butopen in general. Importantly, if it were polynomial-time solvable, this wouldimply the polynomial-time solvability of the well-known Kemeny consensusproblem for three voters, whose complexity is currently open.The study of popular rankings can be extended into various directions. Wenow list some open problems that our work poses, starting with a questionalready raised by van Zuylen et al. [29], which we made some progress on.1. Determine the complexity of deciding whether an absolutely or simplypopular ranking exists for an instance with arbitrary n . Our Lemmas 3,4, and 5 show that for at most 3 voters and for 4 voters with an acyclictournament as the majority graph, the existence of absolutely/simplypopular rankings can be checked efficiently. Besides this, Theorem 6 gives asufficient condition for the existence of an absolutely popular ranking forinstances with arbitrary n .2. Determine the complexity of 3- all-closer-ranking .3. Construct an example showing that for any n > 5, the two notions ofpopularity are not equivalent. Theorem 6 might prove to be helpful here.Finally, popular rankings can be defined and studied in instances whereties in the rankings are allowed, or the rankings are not necessarily complete.Besides the Kendall distance, other metrics on rankings can also be applied,such as the Spearman distance. Acknowledgement. We thank Markus Brill for fruitful discussions, and thereviewers of the conference version of this paper for their suggestions. SonjaKraiczy was supported by Undergraduate Research Bursary 19-20-66 fromthe London Mathematical Society, and by the School of Computing Science,University of Glasgow. ´Agnes Cseh was supported by the Federal Ministry ofEducation and Research of Germany in the framework of KI-LAB-ITSE (project number 01IS19066). David Manlove was supported by grant EP/P028306/1from the Engineering and Physical Sciences Research Council. References 1. D. J. Abraham, R. W. Irving, T. Kavitha, and K. Mehlhorn. Popularmatchings. SIAM Journal on Computing , 37:1030–1045, 2007.2. G. Bachmeier, F. Brandt, C. Geist, P. Harrenstein, K. Kardel, D. Peters,and H. G. Seedig. k -majority digraphs and the hardness of voting witha constant number of voters. Journal of Computer and System Sciences ,105:130–157, 2019.3. M. Ba¨ıou and M. Balinski. The stable allocation (or ordinal transportation)problem. Mathematics of Operations Research , 27(3):485–503, 2002.4. M. Balinski and R. Laraki. Judge: Don’t vote! Operations Research ,62(3):483–511, 2014.5. M. Balinski and R. Laraki. What should ‘majority decision’ mean? J.Elster et S. Novak (eds.), Majority Decisions , pages 103–131, 2014.6. J. Bartholdi, C. A. Tovey, and M. A. Trick. Voting schemes for which itcan be difficult to tell who won the election. Social Choice and Welfare ,6(2):157–165, 1989.7. T. Biedl, F. J. Brandenburg, and X. Deng. On the complexity of crossingsin permutations. Discrete Mathematics , 309(7):1813–1823, 2009.8. J. A. N. d. C. d. Condorcet. Essai sur l’application de l’analyse `a laprobabilit´e des d´ecisions rendues `a la pluralit´e des voix . L’ImprimerieRoyale, 1785.9. ´A. Cseh. Popular matchings. Trends in Computational Social Choice ,105(3), 2017.10. A. Darmann. Popular spanning trees. International Journal of Foundationsof Computer Science , 24(05):655–677, 2013.11. A. Darmann. A social choice approach to ordinal group activity selection. Mathematical Social Sciences , 93:57–66, 2018.12. A. Davenport and J. Kalagnanam. A computational study of the Kemenyrule for preference aggregation. In AAAI , volume 4, pages 697–702, 2004.13. K. L. Dougherty and J. Edward. The properties of simple vs. absolutemajority rule: cases where absences and abstentions are important. Journalof Theoretical Politics , 22(1):85–122, 2010.14. C. Dwork, R. Kumar, M. Naor, and D. Sivakumar. Rank aggregationmethods for the web. In Proceedings of the 10th International Conferenceon World Wide Web , pages 613–622, 2001.15. Y. Faenza, T. Kavitha, V. Powers, and X. Zhang. Popular matchings andlimits to tractability. In Proceedings of the Thirtieth Annual ACM-SIAMSymposium on Discrete Algorithms , pages 2790–2809. SIAM, 2019.16. E. H. Friend. Sorting on electronic computer systems. Journal of the ACM ,3(3):134–168, 1956. n absolutely and simply popular rankings 27 17. D. Gale and L. S. Shapley. College admissions and the stability of marriage. American Mathematical Monthly , 69:9–15, 1962.18. P. G¨ardenfors. Match making: assignments based on bilateral preferences. Behavioural Science , 20:166–173, 1975.19. S. Gupta, P. Misra, S. Saurabh, and M. Zehavi. Popular matching inroommates setting is NP-hard. In Proceedings of the Thirtieth AnnualACM-SIAM Symposium on Discrete Algorithms , pages 2810–2822. SIAM,2019.20. A. B. Kahn. Topological sorting of large networks. Communications of theACM , 5(11):558–562, Nov. 1962.21. T. Kavitha, T. Kir´aly, J. Matuschke, I. Schlotter, and U. Schmidt-Kraepelin.Popular branchings and their dual certificates. In International Conferenceon Integer Programming and Combinatorial Optimization , pages 223–237.Springer, 2020.22. J. Kemeny. Mathematics without numbers. Daedalus , 88:571–591, 1959.23. M. G. Kendall. A new measure of rank correlation. Biometrika , 30(1/2):81–93, 1938.24. A. Levenglick. Fair and reasonable election systems. Behavioral Science ,20(1):34–46, 1975.25. M. Machover, D. S. Felsenthal, et al. Ternary voting games. InternationalJournal of Game Theory , 26(3):335–351, 1997.26. D. F. Manlove. Algorithmics of Matching Under Preferences . WorldScientific, 2013.27. H. G. Seedig. Majority Relations and Tournament Solutions . PhD thesis,Technische Universit¨at M¨unchen, 2015.28. C. T. Sng and D. F. Manlove. Popular matchings in the weighted capaci-tated house allocation problem. Journal of Discrete Algorithms , 8:102–116,2010.29. A. van Zuylen, F. Schalekamp, and D. P. Williamson. Popular ranking. Discrete Applied Mathematics , 165:312–316, 2014.30. A. Vermeule. Absolute majority rules. British Journal of Political Science ,pages 643–658, 2007. π v = [1 , , , [6 , , , [8 , , π v = [2 , , , [4 , , , [9 , , π v = [3 , , , [5 , , , [7 , , π v = [1 , , , [4 , , , [7 , , 9] = σ π v = [1 , , , [5 , , , [9 , , π v = [1 , , , [5 , , , [7 , , σ = [1 , , , [5 , , , [9 , , 8] and σ = [1 , , , [4 , , , [7 , , v prefers σ to σ . Voters v , v , v abstain in the vote between σ and σ . We first discuss thethree impartial voters and justify why they indeed are impartial between σ and σ . Note that each of v , v , v agrees on one triple with σ , and onone triple with σ . For the remaining one or two triples, each of these threevoters agrees with neither the corresponding triple in σ , nor the one in σ ,but instead have distance 2 to both of these. For example, voter v agreeson the first triple with both σ and σ , but agrees on the other two tripleswith neither of them, and K ([8 , , , [9 , , K ([8 , , , [7 , , K ([6 , , , [4 , , K ([6 , , , [5 , , v agrees with σ on thesecond triple, with σ on the third triple, while the distances to those triplesshe disagrees with are K ([4 , , , [5 , , K ([9 , , , [7 , , σ and σ agree on the first triple, these both have the same distance to the firstriple in π v , K ([2 , , , [1 , , v .Notice that there are three directed cycles in the majority graph of thesethree voters, one for candidates 1 , , 3, one for candidates 4 , , , , 9. So the abstaining voters have a cyclic majority graph,justifying the condition in Theorem 4. Voters preferring σ to σ . Finally, we argue that v and v prefer σ to σ .Both of them agree with the first triple with both of σ and σ , as well withone other triple in σ , being 2 swaps away from this triple in σ . The remainingtriple is distance 2 from both the corresponding triples in σ and σ away. Absolutely but not simply popular ranking. By Theorem 3, σ is absolutelypopular, and by the above argument it is not simply popular, since σ is morepopular than σ in the simple sense. This example also justifies the acyclicitycondition for the majority graph of the abstaining voters in Theorem 4. n absolutely and simply popular rankings 29 π v = [1 , , , [6 , , , [8 , , π v = [2 , , , [4 , , , [9 , , π v = [3 , , , [5 , , , [7 , , π v = [1 , , , [4 , , , [8 , , π v = [1 , , , [5 , , , [7 , , π v = [2 , , , [4 , , , [7 , , π v = [2 , , , [5 , , , [8 , , π v = [2 , , , [5 , , , [8 , , 7] = π v Our computer simulations (available under https://github.com/SonjaKrai/PopularRankingsExampleCheck ) confirmed that the unique absolutely popularranking is π v = π v = [2 , , , [5 , , , [8 , , , , , [4 , , , [7 , , π v in the simple sense.6.3 Proof of Theorem 7 Proof Consider the extended Condorcet paradox, i.e. n voters with rankings of n candidates as follows: π v = 1 , , . . . , n − , nπ v = 2 , , . . . , n, π v n = n, , . . . , n − , n − π be an arbitrary ranking of the n candidates. We will show that thereis a ranking σ preferred by n − n − A := { (1 , , (2 , , . . . ( n − , n ) , ( n, } . We also know that there must be at least one ordered pair in A with which π disagrees. Let ( a, b ) be such a pair.Let σ be the ranking obtained from π by swapping a and b . Let c be acandidate ranked between a and b in π . Each voter v who prefers b to a mustalso prefer c to a or b to c . So together the pairs ( a, c ) and ( b, c ) add at leastone to K ( π v , π ) and at most one to K ( π v , σ ). Pair ( a, b ) adds 1 to K ( π v , π )and 0 to K ( π v , σ ). Since σ ranks b higher than a , it must also prefer c to a or b to c . This implies that at most one of the pairs ( a, c ) and ( b, c ) adds one to K ( π v , σ ). From definition of σ follows that K ( π v , σ ) < K ( π v , π ).).