Communication Complexity of Common Voting Rules∗

Vincent Conitzer Tuomas Sandholm Carnegie Mellon University Carnegie Mellon University 5000 Forbes Avenue 5000 Forbes Avenue Pittsburgh, PA 15213, USA Pittsburgh, PA 15213, USA [email protected] [email protected]

for all the agents to report all of their preference information. Clever protocols that elicit the agents’ preferences partially ABSTRACT and sequentially have the potential to dramatically reduce We determine the communication complexity of the com- the required communication. This has at least the following mon voting rules. The rules (sorted by their communication advantages: complexity from low to high) are plurality, plurality with It can make preference aggregation feasible in settings runoff, single transferable vote (STV), Condorcet, approval, • Bucklin, cup, maximin, Borda, Copeland, and ranked pairs. where the total amount of preference information is For each rule, we first give a deterministic communication too large to communicate. protocol and an upper bound on the number of bits com- Even when communicating all the preference informa- municated in it; then, we give a lower bound on (even the • tion is feasible, reducing the communication require- nondeterministic) communication requirements of the vot- ments lessens the burden placed on the agents. This is ing rule. The bounds match for all voting rules except STV especially true when the agents, rather than knowing and maximin. all their preferences in advance, need to invest effort (such as computation or information gathering) to de- Categories and Subject Descriptors termine their preferences [16]. F.2 [Theory of Computation]: Analysis of Algorithms It preserves (some of) the agents’ privacy. and Problem Complexity; J.4 [Computer Applications]: • Social and Behavioral Sciences—Economics Most of the work on reducing the communication burden in preference aggregation has focused on resource allocation General Terms settings such as combinatorial auctions, in which an auc- tioneer auctions off a number of (possibly distinct) items Algorithms, Economics, Theory in a single event. Because in a combinatorial auction, bid- ders can have separate valuations for each of an exponen- Keywords tial number of possible bundles of items, this is a setting in which reducing the communication burden is especially Communication Complexity, Voting crucial. This can be accomplished by supplementing the auctioneer with an elicitor that incrementally elicits parts 1. INTRODUCTION of the bidders’ preferences on an as-needed basis, based on One key factor in the practicality of any preference ag- what the bidders have revealed about their preferences so gregation rule is its communication burden. To successfully far, as suggested by Conen and Sandholm [5]. For example, aggregate the agents’ preferences, it is usually not necessary the elicitor can ask for a bidder’s value for a specific bundle (value queries), which of two bundles the bidder prefers (or- ∗This material is based upon work supported by the Na- der queries), which bundle he ranks kth or what the rank of tional Science Foundation under ITR grants IIS-0121678 and IIS-0427858, and a Sloan Fellowship. The authors thank a given bundle is (rank queries), which bundle he would pur- Ilya Segal for helpful comments. chase given a particular vector of prices (demand queries), etc.—until (at least) the final allocation can be determined. Experimentally, this yields drastic savings in preference rev- elation [11]. Furthermore, if the agents’ valuation functions Permission to make digital or hard copies of all or part of this work for are drawn from certain natural subclasses, the elicitation personal or classroom use is granted without fee provided that copies are problem can be solved using only polynomially many queries not made or distributed for profit or commercial advantage and that copies even in the worst case [23, 4, 13, 18, 14]. For a review bear this notice and the full citation on the first page. To copy otherwise, to of preference elicitation in combinatorial auctions, see [17]. republish, to post on servers or to redistribute to lists, requires prior specific Ascending combinatorial auctions are a well-known special permission and/or a fee. EC’05, June 5–8, 2005, Vancouver, British Columbia, Canada. form of preference elicitation, where the elicitor asks demand Copyright 2005 ACM 1-59593-049-3/05/0006 ...$5.00. queries with increasing prices [15, 21, 1, 9]. Finally, resource allocation problems have also been studied from a communi- bound. After obtaining the upper bounds, we show that cation complexity viewpoint, thereby deriving lower bounds they are tight by giving matching lower bounds on (even the on the required communication. For example, Nisan and Se- nondeterministic) communication complexity of each voting gal show that exponential communication is required even rule. There are two exceptions: STV, for which our upper to obtain a surplus greater than that obtained by auction- and lower bounds are apart by a factor log m; and maximin, ing off all objects as a single bundle [14]. Segal also studies for which our best deterministic upper bound is also a factor social choice rules in general, and shows that for a large log m above the (nondeterministic) lower bound, although class of social choice rules, supporting budget sets must be we give a nondeterministic upper bound that matches the revealed such that if every agent prefers the same outcome lower bound. in her budget set, this proves the optimality of that out- come. Segal then uses this characterization to prove bounds on the communication required in resource allocation as well 2. REVIEW OF VOTING RULES as matching settings [20]. In this section, we review the common voting rules that In this paper, we will focus on the communication re- we study in this paper. A voting rule2 is a function map- quirements of a generally applicable subclass of social choice ping a vector of the n voters’ votes (i.e. preferences over rules, commonly known as voting rules. In a voting setting, candidates) to one of the m candidates (the winner) in the there is a set of candidate outcomes over which the voters candidate set C. In some cases (such as the Condorcet rule), express their preferences by submitting a vote (typically, the rule may also declare that no winner exists. We do not a ranking of the candidates), and the winner (that is, the concern ourselves with what happens in case of a tie be- chosen outcome) is determined based on these votes. The tween candidates (our lower bounds hold regardless of how communication required by voting rules can be large either ties are broken, and the communication protocols used for because the number of voters is large (such as, for example, our upper bounds do not attempt to break the ties). All of in national elections), or because the number of candidates the rules that we study are rank-based rules, which means is large (for example, the agents can vote over allocations of that a vote is defined as an ordering of the candidates (with a number of resources), or both. Prior work [8] has studied the exception of the plurality rule, for which a vote is a sin- elicitation in voting, studying how computationally hard it gle candidate, and the approval rule, for which a vote is a is to decide whether a winner can be determined with the in- subset of the candidates). formation elicited so far, as well as how hard it is to find the We will consider the following voting rules. (For rules that optimal sequence of queries given perfect suspicions about define a score, the candidate with the highest score wins.) the voters’ preferences. In addition, that paper discusses scoring rules. Let α~ = α1,... ,αm be a vector of strategic (game-theoretic) issues introduced by elicitation. • h i integers such that α1 α2 ... αm. For each voter, a In contrast, in this paper, we are concerned with the ≥ ≥ candidate receives α1 points if it is ranked first by the voter, worst-case number of bits that must be communicated to ex- α2 if it is ranked second etc. The score sα~ of a candidate ecute a given voting rule, when nothing is known in advance is the total number of points the candidate receives. The about the voters’ preferences. We determine the communi- Borda rule is the scoring rule with α~ = m 1,m 2,... ,0 . cation complexity of the common voting rules. For each rule, The plurality rule is the scoring rule withh α~−= 1,−0,... ,0 .i we first give an upper bound on the (deterministic) commu- h i nication complexity by providing a communication protocol single transferable vote (STV). The rule proceeds through a series• of m 1 rounds. In each round, the candidate with for it and analyzing how many bits need to be transmitted − in this protocol. (Segal’s results [20] do not apply to most the lowest plurality score (that is, the least number of voters voting rules because most voting rules are not intersection- ranking it first among the remaining candidates) is elimi- monotonic (or even monotonic).1) For many of the voting nated (and each of the votes for that candidate “transfer” rules under study, it turns out that one cannot do better to the next remaining candidate in the order given in that than simply letting each voter immediately communicate all vote). The winner is the last remaining candidate. her (potentially relevant) information. However, for some plurality with run-off. In this rule, a first round elimi- rules (such as plurality with runoff, STV and cup) there is nates• all candidates except the two with the highest plurality a straightforward multistage communication protocol that, scores. Votes are transferred to these as in the STV rule, with some analysis, can be shown to significantly outperform and a second round determines the winner from these two. the immediate communication of all (potentially relevant) approval. Each voter labels each candidate as either information. Finally, for some rules (such as the Condorcet approved• or disapproved. The candidate approved by the and Bucklin rules), we need to introduce a more complex greatest number of voters wins. communication protocol to achieve the best possible upper Condorcet. For any two candidates i and j, let N(i, j)be • 1 the number of voters who prefer i to j. If there is a candidate For two of the rules that we study that are intersection- i that is preferred to any other candidate by a majority of monotonic, namely the approval and Condorcet rules, Se- the voters (that is, N(i, j) >N(j, i) for all j = i—that is, i gal’s results can in fact be used to give alternative proofs 6 of our lower bounds. We only give direct proofs for these wins every pairwise election), then candidate i wins. rules here because 1) these direct proofs are among the eas- ier ones in this paper, 2) the alternative proofs are nontrivial 2The term voting protocol is often used to describe the same even given Segal’s results, and 3) a space constraint applies. concept, but we seek to draw a sharp distinction between However, we hope to also include the alternative proofs in a the rule mapping preferences to outcomes, and the commu- later version. nication/elicitation protocol used to implement this rule. maximin (aka. Simpson). The maximin score of i is will know which two candidates have the highest plurality • s(i) = minj=i N(i, j)—that is, i’s worst performance in a scores). In general, when the voters receive such informa- pairwise election.6 The candidate with the highest maximin tion, it may give them incentives to vote differently than score wins. they would have in a single-stage communication protocol in which all voters declare their entire votes simultaneously. Copeland. For any two distinct candidates i and j, let Of course, even the single-stage communication protocol is C(•i, j)=1ifN(i, j) >N(j, i), C(i, j)=1/2ifN(i, j)= not strategy-proof4 for any reasonable voting rule, by the N(j, i) and C(i, j)=0ifN(i, j) approval voting rule introduces strategic voting, Bucklin. For any candidate i and integer l, let B(i, l) and give principles for designing elicitation protocols that be• the number of voters that rank candidate i among the do not introduce strategic problems.) Now that we have reviewed voting rules, we move on to a top l candidates. The winner is arg mini(min l : B(i, l) > n/2 ). That is, if we say that a voter “approves”{ her top brief review of communication complexity theory. l candidates,} then we repeatedly increase l by 1 until some candidate is approved by more than half the voters, and this 3. REVIEW OF SOME COMMUNICATION candidate is the winner. COMPLEXITY THEORY ranked pairs. This rule determines an order on all In this section, we review the basic model of a commu- the• candidates, and the winner is the candidate at the top nication problem and the lower-bounding technique of con- of this order. Sort all ordered pairs of candidates (i, j)by structing a fooling set. (The basic model of a communication N(i, j), the number of voters who prefer i to j. Starting problem is due to Yao [22]; for an overview see Kushilevitz with the pair (i, j) with the highest N(i, j), we “lock in” and Nisan [12].) the result of their pairwise election (i j). Then, we move Each player 1 i n knows (only) input xi. Together, ≤ ≤ to the next pair, and we lock the result of their pairwise they seek to compute f(x1,x2,... ,xn). In a deterministic election. We continue to lock every pairwise result that does protocol for computing f, in each stage, one of the players not contradict the ordering established so far. announces (to all other players) a bit of information based on her own input and the bits announced so far. Even- We emphasize that these definitions of voting rules do not tually, the communication terminates and all players know concern themselves with how the votes are elicited from the f(x ,x ,... ,x ). The goal is to minimize the worst-case voters; all the voting rules, including those that are sug- 1 2 n (over all input vectors) number of bits sent. The determin- gestively defined in terms of “rounds”, are in actuality just istic communication complexity of a problem is the worst- functions mapping the vector of all the voters’ votes to a case number of bits sent in the best (correct) deterministic winner. Nevertheless, there are always many different ways protocol for it. In a nondeterministic protocol, the next of eliciting the votes (or the relevant parts thereof) from the bit to be sent can be chosen nondeterministically. For the voters. For example, in the plurality with runoff rule, one purposes of this paper, we will consider a nondeterministic way of eliciting the votes is to ask every voter to declare her protocol correct if for every input vector, there is some se- entire ordering of the candidates up front. Alternatively, we quence of nondeterministic choices the players can make so can first ask every voter to declare only her most preferred that the players know the value of f when the protocol ter- candidate; then, we will know the two candidates in the minates. The nondeterministic communication complexity runoff, and we can ask every voter which of these two candi- of a problem is the worst-case number of bits sent in the dates she prefers. Thus, we distinguish between the voting best (correct) nondeterministic protocol for it. rule (the mapping from vectors of votes to outcomes) and the We are now ready to give the definition of a fooling set. communication protocol (which determines how the relevant parts of the votes are actually elicited from the voters). The Definition 1. A fooling set is a set of input vectors goal of this paper is to give efficient communication proto- 1 1 1 2 2 2 k k k (x1,x2,... ,xn),(x1,x2,... ,xn),... ,(x1,x2,... ,xn) such cols for the voting rules just defined, and to prove that there { i i i that for any i, f(x1,x2,... ,xn)=f0 for some constant f0, do not exist any more efficient communication protocols. but for any i = j, there exists some vector (r1,r2,... ,rn) It is interesting to note that the choice of the communica- n 6 r1 r2 rn ∈ i, j such that f(x1 ,x2 ,... ,xn )=f0. (That is, we can tion protocol may affect the strategic behavior of the voters. mix{ } the inputs from the two input vectors6 to obtain a vector Multistage communication protocols may reveal to the vot- with a different function value.) ers some information about how the other voters are voting (for example, in the two-stage communication protocol just It is known that if a fooling set of size k exists, then log k given for plurality with runoff, in the second stage voters is a lower bound on the communication complexity (even the nondeterministic communication complexity) [12]. 3Balanced means that the difference in depth between two 4A strategy-proof protocol is one in which it is in the players’ leaves can be at most one. best interest to report their preferences truthfully. For the purposes of this paper, f is the voting rule that Sometimes for the purposes of a proof the internal ranking maps the votes to the winning candidate, and xi is voter of a subset of the candidates does not matter, and in this i’s vote (the information that the voting rule would require case we will not specify it. For example, if S = c2,c3 , { } from the voter if there were no possibility of multistage then c1 S c4 indicates that either the ranking c1 communication—i.e. the most preferred candidate (plural- c2 c3 c4 or the ranking c1 c3 c2 c4 can be used ity), the approved candidates (approval), or the ranking of for the proof. all the candidates (all other protocols)). However, when we We first give a universal upper bound. derive our lower bounds, f will only signify whether a dis- tinguished candidate a wins. (That is, f is 1 if a wins, and Theorem 1. The deterministic communication complex- 0 otherwise.) This will strengthen our lower bound results ity of any rank-based voting rule is O(nm log m). (because it implies that even finding out whether one given Proof. This bound is achieved by simply having every- candidate wins is “hard”).5 Thus, a fooling set in our con- one communicate their entire ordering of the candidates (in- text is a set of vectors of votes so that a wins (does not win) dicating the rank of an individual candidate requires only with each of them; but for any two different vote vectors in O(log m) bits, so each of the n voters can simply indicate the set, there is a way of taking some voters’ votes from the the rank of each of the m candidates). first vector and the others’ votes from the second vector, so that a does not win (wins). The next lemma will be useful in a few of our proofs. To simplify the proofs of our lower bounds, we make as- sumptions such as “the number of voters n is odd” in many Lemma 1. If m divides n, then log(n!) m log((n/m)!) of these proofs. Therefore, technically, we do not prove the n(log m 1)/2. − ≥ lower bound for (number of candidates, number of voters) − pairs (m, n) that do not satisfy these assumptions (for exam- Proof. If n/m = 1 (that is, n = m), then this expres- ple, if we make the above assumption, then we technically n sion simplifies to log(n!). We have log(n!) = log i do not prove the lower bound for any pair (m, n)inwhich i=1 ≥ nis even). Nevertheless, we always prove the lower bound n P for a representative set of (m, n) pairs. For example, for ev- log(i)dx, which, using integration by parts, is equal to x=1 ery one of our lower bounds it is the case that for infinitely nRlog n (n 1) >n(log n 1) = n(log m 1) >n(log m many values of m, there are infinitely many values of n such 1)/2. So,− we− can assume that− n/m 2.− We observe that− that the lower bound is proved for the pair (m, n). n n/m 1 m ≥ n/m 1 m log(n!) = log i = − log(im+j) − log(im) i=1 i=0 j=1 ≥ i=1 j=1 4. RESULTS n/m 1P P P P n/mP = m − log(im), and that m log((n/m)!) = m log(i). We are now ready to present our results. For each voting i=1 i=1 rule, we first give a deterministic communication protocol n/m 1 P − P for determining the winner to establish an upper bound. Therefore, log(n!) m log((n/m)!) m log(im) − ≥ − Then, we give a lower bound on the nondeterministic com- i=1 n/m n/m 1 P munication complexity (even on the complexity of deciding m log(i)=m(( − log(im/i)) log(n/m)) = m((n/m whether a given candidate wins, which is an easier ques- i=1 i=1 − − tion). The lower bounds match the upper bounds in all but 1) logP m log n+log mP)=nlog m m log n. Now, using the − − two cases: the STV rule (upper bound O(n(log m)2); lower fact that n/m 2, we have m log n = n(m/n) log m(n/m)= ≥ bound Ω(n log m)) and the maximin rule (upper bound n(m/n)(log m + log(n/m)) n(1/2)(log m + log 2). Thus, ≤ O(nm log m), although we do give a nondeterministic pro- log(n!) m log((n/m)!) n log m m log n n log m − ≥ − ≥ − tocol that is O(nm); lower bound Ω(nm)). n(1/2)(log m + log 2) = n(log m 1)/2. − When we discuss a voting rule in which the voters rank the Theorem candidates, we will represent a ranking in which candidate c1 2. The deterministic communication complex- is ranked first, c2 is ranked second, etc. as c1 c2 ...cm. ity of the plurality rule is O(n log m). Proof. Indicating one of the candidates requires only 5One possible concern is that in the case where ties are pos- sible, it may require much communication to verify whether O(log m) bits, so each voter can simply indicate her most a specific candidate a is among the winners, but little com- preferred candidate. munication to produce one of the winners. However, all the fooling sets we use in the proofs have the property that if Theorem 3. The nondeterministic communication com- a wins, then a is the unique winner. Therefore, in these plexity of the plurality rule is Ω(n log m) (even to decide fooling sets, if one knows any one of the winners, then one whether a given candidate a wins). knows whether a is a winner. Thus, computing one of the winners requires at least as much communication as veri- Proof. We will exhibit a fooling set of size n0! where n0 m fying whether a is among the winners. In general, when a (( m )!) communication problem allows multiple correct answers for n0 =(n 1)/2. Taking the logarithm of this gives log(n0!) a given vector of inputs, this is known as computing a rela- − − m log((n0/m)!), so the result follows from Lemma 1. The tion rather than a function [12]. However, as per the above, we can restrict our attention to a subset of the domain where fooling set will consist of all vectors of votes satisfying the the voting rule truly is a (single-valued) function, and hence following constraints: lower bounding techniques for functions rather than rela- tions will suffice. For any 1 i n 0 , voters 2i 1 and 2i vote the same. • ≤ ≤ − Every candidate receives equally many votes from the Every candidate is ranked at the top of equally many • • first 2n0 = n 1 voters. of the first 4n0 = n 2 votes. − − The last voter (voter n) votes for a. Voter 4n0 +1 = n 1 ranks the candidates a C a . • • − −{ } Candidate a wins with each one of these vote vectors be- Voter 4n0 +2=nranks the candidates b C b . cause of the extra vote for a from the last voter. Given that • −{ } m divides n0, let us see how many vote vectors there are in Candidate a wins with each one of these vote vectors: the fooling set. We need to distribute n0 voter pairs evenly because of the last two votes, candidates a and b are one vote over m candidates, for a total of n0/m voter pairs per can- ahead of all the other candidates and continue to the runoff, didate; and there are precisely n0! ways of doing this.6 and at this point all the votes that had another candidate (( n0 )!)m m ranked at the top transfer to a, so that a wins the runoff. All that remains to show is that for any two distinct vectors Given that m0 divides n0, let us see how many vote vectors of votes in the fooling set, we can let each of the voters vote there are in the fooling set. We need to distribute n0 groups according to one of these two vectors in such a way that a of four voters evenly over the m0 pairs of candidates, and loses. Let i be a number such that the two vote vectors dis- n0! (as in the proof of Theorem 3) there are n ways of agree on the candidate for which voters 2i 1 and 2i vote. (( 0 )!)m0 m0 Without loss of generality, suppose that in− the first vote doing this. All that remains to show is that for any two vector, these voters do not vote for a (but for some other distinct vectors of votes in the fooling set, we can let each candidate, b, instead). Now, construct a new vote vector by of the voters vote according to one of these two vectors in taking votes 2i 1 and 2i from the first vote vector, and such a way that a loses. Let i be a number such that ck(i) is − 1 the remaining votes from the second vote vector. Then, b not the same in both of these two vote vectors, that is, ck(i) 2 receives 2n0/m + 2 votes in this newly constructed vote vec- (ck(i) in the first vote vector) is not equal to ck(i) (ck(i) in tor, whereas a receives at most 2n0/m+1 votes. So, a is not the second vote vector). Without loss of generality, suppose 1 the winner in the newly constructed vote vector, and hence ck(i) = a. Now, construct a new vote vector by taking votes we have a correct fooling set. 4i 63, 4i 2, 4i 1, 4i from the first vote vector, and the remaining− − votes from− the second vote vector. In this newly Theorem 4. The deterministic communication complex- 1 1 constructed vote vector, c and d each receive 4n0/m+2 ity of the plurality with runoff rule is O(n log m). k(i) k(i) votes in the first round, whereas a receives at most 4n0/m+1 Proof. First, let every voter indicate her most preferred votes. So, a does not continue to the runoff in the newly candidate using log m bits. After this, the two candidates constructed vote vector, and hence we have a correct fooling in the runoff are known, and each voter can indicate which set. one she prefers using a single additional bit. Theorem 6. The nondeterministic communication com- Theorem 5. The nondeterministic communication com- plexity of the Borda rule is Ω(nm log m) (even to decide plexity of the plurality with runoff rule is Ω(n log m) (even whether a given candidate a wins). to decide whether a given candidate a wins). n Proof. We will exhibit a fooling set of size (m0!) 0 where Proof. n0! We will exhibit a fooling set of size n (( 0 )!)m0 m m0 = m 2 and n0 =(n 2)/4. This will prove the theorem 0 − − n where m0 = m/2 and n0 =(n 2)/4. Taking the loga- because m0!isΩ(mlog m), so that log((m0!) 0 )=n0log(m0!) − rithm of this gives log(n0!) m0 log((n0/m0)!), so the result is Ω(nm log m). For every vector (π1,π2,... ,π ) consisting − n0 follows from Lemma 1. Divide the candidates into m0 pairs: of n0 orderings of all candidates other than a and another (c ,d ),(c ,d ),... ,(c ,d ) where c = a and d = b. 1 1 2 2 m0 m0 1 1 fixed candidate b (technically, the orderings take the form of The fooling set will consist of all vectors of votes satisfying a one-to-one function πi : 1, 2,... ,m0 C a, b with { }→ −{ } the following constraints: πi(j)=cindicating that candidate c is the jth in the order represented by πi), let the following vector of votes be an For any 1 i n0 , voters 4i 3 and 4i 2 rank element of the fooling set: • the candidates≤ c≤ a C − a, c ,− for some k(i) −{ k(i)} candidate ck(i). (If ck(i) = a then the vote is simply For 1 i n0 , let voters 4i 3 and 4i 2 rank the • ≤ ≤ − − a C a .) candidates a b πi(1) πi(2) ... πi(m0). −{ } For any 1 i n0 , voters 4i 1 and 4i rank the can- For 1 i n0 , let voters 4i 1 and 4i rank the • didates d ≤ ≤a C a, d− (that is, their most • ≤ ≤ − k(i) k(i) candidates πi(m0) πi(m0 1) ... πi(1) b a. preferred candidate is the−{ candidate} that is paired with − the candidate that the previous two voters vote for). Let voter 4n0 +1 = n 1 rank the candidates a • − b π (1) π (2) ... π (m0) (where π is an 6An intuitive proof of this is the following. We can count 0 0 0 0 arbitrary order of the candidates other than a and b the number of permutations of n0 elements as follows. First, divide the elements into m buckets of size n0/m, so that if which is the same for every element of the fooling set). x is placed in a lower-indexed bucket than y, then x will Let voter 4n0 +2=n rank the candidates π0(m0) be indexed lower in the eventual permutation. Then, decide • on the permutation within each bucket (for which there are π0(m0 1) ... π0(1) a b. − (n0/m)! choices per bucket). It follows that n0! equals the n0 number of ways to divide n0 elements into m buckets of size We observe that this fooling set has size (m0!) , and that m n0/m, times ((n0/m)!) . candidate a wins in each vector of votes in the fooling set (to 2 2 2 1 2 see why, we observe that for any 1 i n0, votes 4i 3 and (π1 ,π2,... ,πn ). For some i, we must have πi = πi , so that ≤ ≤ − 0 1 1 6 2 1 4i 2 rank the candidates in the exact opposite way from for some j 1,2,... ,m0 , we have (πi )− (j) < (πi )− (j). votes− 4i 1 and 4i, which under the Borda rule means they Now, construct∈{ a new vote} vector by taking vote 2i 1from cancel out;− and the last two votes give one more point to a the first vote vector, and the remaining votes from− the sec- than to any other candidate—besides b who gets two fewer ond vote vector. a’s Copeland score remains unchanged. Let points than a). All that remains to show is that for any two us consider the score of lj . We first observe that the rank of distinct vectors of votes in the fooling set, we can let each of lj in vote 2i 1 in the newly constructed vote vector is at the voters vote according to one of these two vectors in such least 2 higher− than it was in the second vote vector, because 1 1 2 1 1 a way that a loses. Let the first vote vector correspond to (πi )− (j) < (πi )− (j). Let D (lj ) be the set of candidates 1 1 1 the vector (π1 ,π2,... ,πn ), and let the second vote vector in L R that voter 2i 1 ranked lower than lj in the first 0 2 2 2 ∪ 1 − 1 correspond to the vector (π1 ,π2,... ,πn ). For some i,we vote vector (D (lj )= c L R:lj 2i 1 c), and let 1 2 0 2 { ∈ ∪ − } must have πi = πi , so that for some candidate c/ a, b , D (lj ) be the set of candidates in L R that voter 2i 1 1 1 6 2 1 ∈{1 } ∪ 2 − (πi )− (c) < (πi )− (c) (that is, c is ranked higher in πi than ranked lower than lj in the second vote vector (D (lj )= 2 2 in πi ). Now, construct a new vote vector by taking votes c L R:lj 2i 1 c). Then, it follows that in the { ∈ ∪ − } 4i 3 and 4i 2 from the first vote vector, and the remaining newly constructed vote vector, lj defeats all the candidates − − 1 2 votes from the second vote vector. a’s Borda score remains in D (lj ) D (lj ) in their pairwise elections (because lj re- 1 − unchanged. However, because c is ranked higher in πi than ceives an “extra” vote in each one of these pairwise elections 2 in πi , c receives at least 2 more points from votes 4i 3 and relative to the second vote vector), and loses to all the candi- − 2 1 4i 2 in the newly constructed vote vector than it did in dates in D (lj ) D (lj ) (because lj loses a vote in each one the− second vote vector. It follows that c has a higher Borda of these pairwise− elections relative to the second vote vec- 1 2 score than a in the newly constructed vote vector. So, a is tor), and ties with everyone else. But D (lj ) D (lj) 2, 1 2 2 | 1 |−| |≥ not the winner in the newly constructed vote vector, and and hence D (lj ) D (lj ) D (lj) D (lj) 2. Hence, | − |−| − |≥ hence we have a correct fooling set. in the newly constructed vote vector, lj has at least two more pairwise wins than pairwise losses, and therefore has Theorem 7. The nondeterministic communication com- at least 1 more point than if lj had tied all its pairwise elec- plexity of the Copeland rule is Ω(nm log m) (even to decide tions. Thus, lj has a higher Copeland score than a in the whether a given candidate a wins). newly constructed vote vector. So, a is not the winner in the n newly constructed vote vector, and hence we have a correct Proof. We will exhibit a fooling set of size (m0!) 0 where fooling set. m0 =(m 2)/2 and n0 =(n 2)/2. This will prove the − − n theorem because m0!isΩ(mlog m), so that log((m0!) 0 )= Theorem 8. The nondeterministic communication com- n0log(m0!) is Ω(nm log m). We write the set of candidates plexity of the maximin rule is O(nm). as the following disjoint union: C = a, b L R where { }∪ ∪ Proof. The nondeterministic protocol will guess which L = l1,l2,... ,lm and R = r1,r2,... ,rm . For every { 0} { 0} candidate w is the winner, and, for each other candidate vector (π1,π2,... ,πn ) consisting of n0 permutations of the 0 c, which candidate o(c) is the candidate against whom c integers 1 through m0 (πi : 1, 2,... ,m0 1,2,... ,m0 ), let the following vector of votes{ be an element}→{ of the fooling} receives its lowest score in a pairwise election. Then, let set: every voter communicate the following: for each candidate c = w, whether she prefers c to w; For 1 i n0 , let voter 2i 1 rank the candidates • 6 • a b≤ l≤ r −l r ... πi(1) πi(1) πi(2) πi(2) for each candidate c = w, whether she prefers c to o(c). l r . • 6 πi(m0) πi(m0) We observe that this requires the communication of 2n(m For 1 i n0 , let voter 2i rank the candidates − • ≤ ≤ 1) bits. If the guesses were correct, then, letting N(d, e)be rπ (m ) lπ (m ) rπ (m 1) lπ (m 1) ... i 0 i 0 i 0− i 0− the number of voters preferring candidate d to candidate e, rπ (1) lπ (1) b a. i i we should have N(c, o(c)) n/2 votes). All that remains to show is that for any two distinct vectors Theorem 13. The nondeterministic communication com- of votes in the fooling set, we can let each of the voters vote plexity of the approval rule is Ω(nm) (even to decide whether according to one of these two vectors in such a way that a given candidate a wins). a loses. Let the first vote vector correspond to the vector 1 1 1 Proof. n0m0 (S1 ,S2,... ,Sn ), and let the second vote vector correspond We will exhibit a fooling set of size 2 where 0 2 2 2 to the vector (S1 ,S2,... ,Sn ). For some i, we must have m0 = m 1 and n0 =(n 1)/4. For every vector 1 2 1 0 2 2 1 − − (S ,S ,... ,S ) consisting of n0 subsets S C a , let Si = Si , so that either Si * Si or Si * Si . Without loss 1 2 n0 i 6 1 2 the following vector of votes be an element⊆ of the−{ fooling} of generality, suppose Si * Si , and let c be some candidate 1 2 set: in Si Si . Now, construct a new vote vector by taking votes 4−i 3 and 4i 2 from the first vote vector, and the − − For 1 i n0 , let voters 4i 3 and 4i 2 approve remaining votes from the second vote vector. In this newly • ≤ ≤ − − Si a . constructed vote vector, a is ranked higher than c by only ∪{ } 2n0 1 voters, for the following reason. Whereas voters 4i 3 For 1 i n 0 , let voters 4i 1 and 4i approve C − − • ≤ ≤ − − and 4i 2 do not rank c higher than a in the second vote (Si a ). − 2 ∪{ } vector (because c/Si), voters 4i 3 and 4i 2dorank chigher than a in∈ the first vote vector− (because− c S1). Let voter 4n0 +1=napprove a . i • { } Moreover, in every one of b’s pairwise elections, b is ranked∈ m0 n0 n0m0 higher than its opponent by at least 2n0 voters. So, a has We observe that this fooling set has size (2 ) =2 , a lower maximin score than b, therefore a is not the winner and that candidate a wins in each vector of votes in the in the newly constructed vote vector, and hence we have a fooling set (a is approved by 2n0 + 1 voters, whereas each correct fooling set. other candidate is approved by only 2n0 voters). All that remains to show is that for any two distinct vectors of votes Theorem 10. The deterministic communication complex- in the fooling set, we can let each of the voters vote ac- ity of the STV rule is O(n(log m)2). cording to one of these two vectors in such a way that a Proof. loses. Let the first vote vector correspond to the vector Consider the following communication protocol. 1 1 1 (S1 ,S2,... ,Sn ), and let the second vote vector correspond Let each voter first announce her most preferred candidate 0 2 2 2 to the vector (S1 ,S2,... ,Sn ). For some i, we must have (O(n log m) communication). In the remaining rounds, we 1 2 1 0 2 2 1 Si = Si , so that either Si * Si or Si * Si . Without loss will keep track of each voter’s most preferred candidate 6 1 2 among the remaining candidates, which will be enough to of generality, suppose Si * Si , and let b be some candidate 1 2 implement the rule. When candidate c is eliminated, let in Si Si . Now, construct a new vote vector by taking − each of the voters whose most preferred candidate among the votes 4i 3 and 4i 2 from the first vote vector, and the − − remaining candidates was c announce their most preferred remaining votes from the second vote vector. In this newly candidate among the candidates remaining after c’s elimina- constructed vote vector, a is still approved by 2n0 + 1 votes. tion. If candidate c was the ith candidate to be eliminated However, b is approved by 2n0 + 2 votes, for the following (that is, there were m i + 1 candidates remaining before reason. Whereas voters 4i 3 and 4i 2 do not approve b in − − − 2 c’s elimination), it follows that at most n/(m i + 1) voters the second vote vector (because b/Si), voters 4i 3 and − ∈ − 1 had candidate c as their most preferred candidate among 4i 2 do approve b in the first vote vector (because b Si ). − ∈ the remaining candidates, and thus the number of bits to be It follows that b’s score in the newly constructed vote vector communicated after the elimination of the ith candidate is is b’s score in the second vote vector (2n0), plus two. So, a O((n/(m i+1)) log m).7 Thus, the total communication in is not the winner in the newly constructed vote vector, and − m 1 hence we have a correct fooling set. this communication protocol is O(n log m + − (n/(m i + i=1 − Interestingly, an Ω(m) lower bound can be obtained even m 1 mP 1)) log m). Of course, − 1/(m i +1)= 1/i, which is for the problem of finding a candidate that is approved by i=1 − i=2 more than one voter [20]. O(log m). SubstitutingP into the previous expression,P we find that the communication complexity is O(n(log m)2). Theorem 14. The deterministic communication complex- ity of the Condorcet rule is O(nm). Theorem 11. The nondeterministic communication com- plexity of the STV rule is Ω(n log m) (even to decide whether Proof. We maintain a set of active candidates S which a given candidate a wins). is initialized to C. At each stage, we choose two of the active candidates (say, the two candidates with the lowest indices), Proof. We omit this proof because of space constraint. and we let each voter communicate which of the two candi- 7Actually, O((n/(m i + 1)) log(m i + 1)) is also correct, dates she prefers. (Such a stage requires the communication but it will not improve− the bound. − of n bits, one per voter.) The candidate preferred by fewer voters (the loser of the pairwise election) is removed from Theorem 17. The nondeterministic communication com- S. (If the pairwise election is tied, both candidates are re- plexity of the cup rule is Ω(nm) (even to decide whether a moved.) After at most m 1 iterations, only one candidate given candidate a wins). is left (or zero candidates− are left, in which case there is no Proof. n0m0 Condorcet winner). Let a be the remaining candidate. To We will exhibit a fooling set of size 2 where m0 =(m 1)/2 and n0 =(n 7)/2. Given that m + 1 is a find out whether candidate a is the Condorcet winner, let − − each voter communicate, for every candidate c = a, whether power of 2, so that one candidate gets a bye (that is, does not she prefers a to c. (This requires the communication6 of at face an opponent) in the first round, let a be the candidate most n(m 1) bits.) This is enough to establish whether with the bye. Of the m0 first-round matchups, let lj denote a won each− of its pairwise elections (and thus, whether a is the one (“left”) candidate in the jth matchup, and let rj be the other (“right”) candidate. Let L = lj :1 j m0 the Condorcet winner). { ≤ ≤ } and R = rj :1 j m0 , so that C = L R a . Theorem 15. The nondeterministic communication com- { ≤ ≤ } ∪ ∪{ } plexity of the Condorcet rule is Ω(nm) (even to decide whether a given candidate a wins). Proof. We will exhibit a fooling set of size 2n0m0 where . . m0 = m 1 and n0 =(n 1)/2. For every vector . . − − (S ,S ,... ,S ) consisting of n0 subsets S C a , let . . 1 2 n0 i the following vector of votes be an element⊆ of the−{ fooling} set: a For 1 i n0 , let voter 2i 1 rank the candidates • ≤ ≤ − ... Si a C Si. − lrlr1122 lm' rm' For 1 i n 0 , let voter 2i rank the candidates C • ≤ ≤ − Si a Si. Figure 1: The schedule for the cup rule used in the proof of Theorem 17. Let voter 2n0 +1 = n rank the candidates a C a . • −{ } m0 n0 n0m0 For every vector (S1,S2,... ,S ) consisting of n0 subsets We observe that this fooling set has size (2 ) =2 , n0 Si R, let the following vector of votes be an element of and that candidate a wins in each vector of votes in the ⊆ fooling set (a wins each of its pairwise elections by a single the fooling set: vote). All that remains to show is that for any two distinct For 1 i n0 , let voter 2i 1 rank the candidates vectors of votes in the fooling set, we can let each of the • ≤ ≤ − Si L a R Si. voters vote according to one of these two vectors in such a − way that a loses. Let the first vote vector correspond to For 1 i n 0 , let voter 2i rank the candidates R the vector (S1,S1,... ,S1 ), and let the second vote vector • ≤ ≤ − 1 2 n0 Si L a Si. correspond to the vector (S2,S2,... ,S2 ). For some i,we 1 2 n0 1 2 1 2 2 1 Let voters 2n0 +1 = n 6, 2n0+2 = n 5, 2n0+3 = n 4 must have Si = Si , so that either Si * Si or Si * Si . • − − − 6 1 2 rank the candidates L a R. Without loss of generality, suppose Si * Si , and let b be 1 2 some candidate in Si Si . Now, construct a new vote Let voters 2n0 +4=n 3,2n0 +5=n 2 rank the − • − − vector by taking vote 2i 1 from the first vote vector, and candidates a r1 l1 r2 l2 ... rm lm . the remaining votes from− the second vote vector. In this 0 0 newly constructed vote vector, b wins its pairwise election Let voters 2n0 +6=n 1,2n0 +7=n rank the candi- • − against a by one vote (vote 2i 1 ranks b above a in the dates rm lm rm 1 lm 1 ... r1 l1 a. 0 0 0− 0− newly constructed vote vector because− b S1, whereas in i m n n m the second vote vector vote 2i 1 ranked a∈above b because We observe that this fooling set has size (2 0 ) 0 =2 0 0. 2 − Also, candidate a wins in each vector of votes in the fooling b/Si). So, a is not the Condorcet winner in the newly constructed∈ vote vector, and hence we have a correct fooling set, for the following reasons. Each candidate rj defeats its set. opponent lj in the first round. (For any 1 i n0, the net effect of votes 2i 1 and 2i on the pairwise≤ ≤ election Theorem 16. The deterministic communication complex- − between rj and lj is zero; votes n 6,n 5,n 4 prefer ity of the cup rule is O(nm). − − − lj to rj , but votes n 3,n 2,n 1,n all prefer rj to lj .) − − − Proof. Consider the following simple communication pro- Moreover, a defeats every rj in their pairwise election. (For tocol. First, let all the voters communicate, for every one of any 1 i n0, the net effect of votes 2i 1 and 2i on the ≤ ≤ − the matchups in the first round, which of its two candidates pairwise election between a and rj is zero; votes n 1,n − they prefer. After this, the matchups for the second round prefer rj to a, but votes n 6,n 5,n 4,n 3,n 2 all − − − − − are known, so let all the voters communicate which candi- prefer a to rj .) It follows that a will defeat all the candidates date they prefer in each matchup in the second round–etc. that it faces. Because communicating which of two candidates is preferred All that remains to show is that for any two distinct vec- requires only one bit per voter, and because there are only tors of votes in the fooling set, we can let each of the voters m 1 matchups in total, this communication protocol re- vote according to one of these two vectors in such a way that quires− O(nm) communication. a loses. Let the first vote vector correspond to the vector 1 1 1 n0m0 (S1 ,S2,... ,Sn ), and let the second vote vector correspond Proof. We will exhibit a fooling set of size 2 where 0 2 2 2 to the vector (S1 ,S2,... ,Sn ). For some i, we must have m0 =(m 1)/2 and n0 = n/2. We write the set of candidates 1 2 1 0 2 2 1 − Si = Si , so that either Si * Si or Si * Si . Without as the following disjoint union: C = a L R where 6 loss of generality, suppose S1 S2, and let r be some can- L = l ,l ,... ,l and R = r ,r ,...{ }∪ ,r ∪. For any i * i j 1 2 m0 1 2 m0 1 2 { } { } didate in S S . Now, construct a new vote vector by subset S 1,2,... ,m0 , let L(S)= li :i S and let i − i ⊆{ } { ∈ } taking vote 2i from the first vote vector, and the remaining R(S)= ri :i S. For every vector (S1,S2,... ,Sn ) { ∈ } 0 votes from the second vote vector. We note that, whereas consisting of n0 sets Si 1,2,... ,m0 , let the following ⊆{ } in the second vote vector vote 2i preferred rj to lj (because vector of votes be an element of the fooling set: 2 rj R Si ), in the newly constructed vote vector this is no ∈ − 1 longer the case (because rj S ). It follows that, whereas For 1 i n0 , let voter 2i 1 rank the candidates ∈ i • ≤ ≤ − in the second vote vector, r defeated l in the first round L(Si) R R(Si) a L L(Si) R(Si). j j − − by one vote, in the newly constructed vote vector, lj de- For 1 i n0 , let voter 2i rank the candidates L feats rj in the first round. Thus, at least one lj advances • ≤ ≤ − L(Si) R(Si) a L(Si) R R(Si). to the second round after defeating its opponent rj .Now, − we observe that in the newly constructed vote vector, any We observe that this fooling set has size (2m0 )n0 =2n0m0, lk wins its pairwise election against any rq with q = k. This 6 and that candidate a wins in each vector of votes in the fool- is because among the first 2n0 votes, at least n0 1 prefer lk − ing set, for the following reason. Each candidate in C a to rq; votes n 6,n 5,n 4 prefer lk to rq; and, because −{ } − − − is ranked among the top m0 candidates by exactly half the q = k, either votes n 3,n 2 prefer lk to rq (if kq). Thus, at least − at the voters’ top m0 +1 candidates, and a is ranked m0 +1th n0 +4=(n+1)/2>n/2 votes prefer lk to rq. Moreover, by all voters. All that remains to show is that for any two any lk wins its pairwise election against a. This is because distinct vectors of votes in the fooling set, we can let each of only votes n 3 and n 2 prefer a to lk. It follows that, − − the voters vote according to one of these two vectors in such after the first round, any surviving candidate lk can only lose a matchup against another surviving l , so that one of a way that a loses. Let the first vote vector correspond to k0 1 1 1 the l must win the election. So, a is not the winner in the the vector (S1 ,S2,... ,Sn ), and let the second vote vector k 0 2 2 2 newly constructed vote vector, and hence we have a correct correspond to the vector (S1 ,S2,... ,Sn ). For some i,we 1 2 1 0 2 2 1 must have Si = Si , so that either Si * Si or Si * Si . fooling set. 6 Without loss of generality, suppose S1 S2, and let j be Theorem 18. The deterministic communication complex- i * i some integer in S1 S2. Now, construct a new vote vector ity of the Bucklin rule is O(nm). i i by taking vote 2i − 1 from the first vote vector, and the Proof. Let l be the minimum integer for which there is remaining votes from− the second vote vector. In this newly a candidate who is ranked among the top l candidates by constructed vote vector, a is still ranked m0 + 1th by all more than half the votes. We will do a binary search for l. votes. However, lj is ranked among the top m0 candidates At each point, we will have a lower bound l which is smaller L by n0 +1 = n/2 + 1 votes. This is because whereas vote than l (initialized to 0), and an upper bound lH which is at 2i 1 does not rank lj among the top m0 candidates in the least l (initialized to m). While lH lL > 1, we continue − 2 2 − second vote vector (because j/Si, we have lj / L(Si )), by finding out whether (lH l)/2 is smaller than l, after ∈ ∈ b − c vote 2i 1doesranklj among the top m0 candidates in the which we can update the bounds. − 1 1 first vote vector (because j Si , we have lj L(Si )). So, a To find out whether a number k is smaller than l,we is not the winner in the newly∈ constructed vote∈ vector, and determine every voter’s k most preferred candidates. Ev- hence we have a correct fooling set. ery voter can communicate which candidates are among her k most preferred candidates using m bits (for each candi- Theorem 20. The nondeterministic communication com- date, indicate whether the candidate is among the top k or plexity of the ranked pairs rule is Ω(nm log m) (even to de- not), but because the binary search requires log m iterations, cide whether a given candidate a wins). this gives us an upper bound of O((log m)nm), which is not Proof. strong enough. However, if lL