complexity
3.2 Rules based on the weighted majority graph
We now consider tournament-based rules. LetP be a profile. NP(x, y) denotes the number voters in P preferring x to y. The majority graph MP is the directed graph whose set of vertices is X and containing an edge from x to y if and only if NP(x, y) > NP(y, x). The weighted majority graph MP is the same as MP, where each edge from x to y is weighted by N(x, y) (note that there is no edge in MP between x and y if and only if
NP(x, y) =NP(y, x).) A voting rule risbased on the majority graph(abridged into “MG- rule” ) if for any profile P, r(P) can be computed fromMP, and based on the weighted majority graph (abridged into “WMG-rule” ) if for any profile P, r(P) can be computed fromMP. Obviously, a MG-rule isa fortioria WMG-rule. A candidatexis theCondorcet winner for a profile P if it dominates every other candidate in MP. A voting rule r is Condorcet-consistentif it elects the Condorcet winner whenever there exists one.
Lemma 4 Let rbe a WMG-rule rule. IfMP =MP0 thenP ∼rP0.
Proof: For any Q, MP∪Q is fully determined from MP and MQ, because
NP∪Q(x, y) = NP(x, y) + NQ(x, y). If r is a WMG-rule then r(P ∪Q) is fully de- termined from MP∪Q, therefore fromMP andMQ, and a fortiori, from MP andQ.
Note that for rules based on the (non-weighted) majority graph, we still need theweighted majority graph ofP andP0 to coincide – having only the majority graph coinciding is not sufficient forP ∼rP0, sinceMP∪Q is generally not fully determined fromMP and MQ.
Lemma 4 gives an upper bound on the compilation complexity of a WMG-rule. Let
T(m, p) be the set of all weighted tournaments onX that can be obtained as the weighted majority graph of somem-voter profile.
Proposition 6 If ris a WMG-rule thenC(r)≤logT(m, p).
Getting a lower bound is not possible without a further assumption on r. After all, constant rules are based on the majority graph, yet they have a compilation complexity of 0. We say that a WMG-ruleris properifP ∼rP0 impliesMP =MP02. It is easy to find
a natural sufficient condition for a WMG-rule to be proper:
Lemma 5 If ris a Condorcet-consistent rule thenP ∼rP0 impliesMP =MP0.
Proof: Let r be a Condorcet-consistent rule. Assume MP 6= MP0, i.e., there exists
(x, y) ∈ X with NP(x, y) 6= NP0(x, y). W.l.o.g., NP(x, y) = NP0(x, y) + k (hence
NP(y, x) = NP0(y, x) − k), with k > 0. Let Q be a set of m + 1 voters where:
m+ 1−NP(x, y) voters prefer x to y and y to anyone else; NP(x, y) voters prefer y to x and x to anyone else. As we have NP∪Q(x, y) = NP(x, y) +NQ(x, y) = m+ 1; for any z 6= x, y, NP∪Q(x, z) = NP(x, z) + m + 1 ≥ m + 1, x is Condorcet winner in P ∪ Q (which contains 2m + 1 voters) and r(P ∪ Q) = x. But
NP0∪Q(y, x) = NP0(y, x) + NQ(y, x) = NP(y, x) + k + NP(x, y) = m +k, and for
anyz6=x, y,NP0∪Q(y, z) =NP0(y, z) +m+ 1≥m+ 1, soy is Condorcet winner inP0∪Q
andr(P0∪Q) =y. HenceP 6∼rP0.
This gives us the following lower bound.
2Examples of WMG-rules that are not proper: constant rules; dictatorial rules; strange rules such as r(P) = firstxi(wrt a fixed orderingx1> ... > xpon candidates) such that for allxj6=xithere is at least
one voter who prefersxitoxj, andxpif there is no suchxi; “restricted” rules such asr(P) being defined
Proposition 7 If ris a Condorcet-consistent rule thenC(r)≥logT(m, p). From Propositions 6 and 7 we get
Proposition 8 If ris a Condorcet-consistent WMG-rule, then C(r) = logT(m, p). Corollary 3 The compilation complexity of the following rules is exactly logT(m, p): Copeland,Simpson (maximin),Slater,Banks,uncovered set,Schwartz.
We now have to compute T(m, p). We easily get the following upper bound. Proposition 9 logT(m, p)≤p(p−1)2 log(m+ 2).
Proof: From Lemma 4 we know that it is enough to store MP. Let>be a fixed ordering on the candidates. Storing MP can be done by storing, for every pair (x, y) of distinct candidates such that x > y, (a) a single bit indicating whether NP(x, y) > NP(y, x) or
NP(x, y) ≤ NP(y, x) and (b) min(NP(x, y), NP(y, x)). Since the latter number can vary between 0 and m2 if m is even, and between 0 and m−12 is m is odd, storing this number requires at most log m
2 + 1
bits. This makes a total of 1 + log m 2 + 1
bits, that is, log(m+ 2) bits. We have p(p−1)2 pairs of distinct candidates, hence the result.
This bound is not necessarily reached: for any x, y, z ∈ X and any profile P we have
NP(x, z) ≥ NP(x, y) +NP(y, z)−m (e.g, if m = 3 and NP(x, y) = NP(y, z) = 2, then
NP(x, z) cannot be 0). Lemma 6 ConsiderVs
t the set of vectors ofsnon-negative integers whose sum is lower or equal to t. Then T(m, p)≥|V p(p−1) 2 m 2 |.
Proof: Assume m is even. Let {ci,j | 1 ≤ i < j ≤ p} be any set non-negative integers such that P
i<jci,j ≤ m2. We will show how to build a profile such that N(i, j) = 2ci,j, where N(i, j) indicates how many voters prefer i to j. Let us divide voters into p(p−1)2 groups gi,j with 1≤i < j ≤pand a final groupg0, such that each group gi,j is assumed to contain exactly 2ci,j voters andg0 contains the rest of the voters (i.e. m−Pi<j2ci,j). In each group gi,j, set the profile of half of the voters to i j x1 x2 . . . xp−2, and the other half to xp−2 xp−1 . . . x1 i j, where x1. . . xp−2 refer to the candidates other than i and j in an arbitrary order. In group g0 set half of the voters to
x1x2. . .xp and the other half toxp. . .x1. LetNg(x, y) denote the number of voters in groupg preferringxtoy. Clearly,Ngij(x, y) =N(x, y) ifx=i andy=j; and 0
otherwise; andNg0(x, y) = 0. Thus,N(x, y) =PNgi,j(x, y) = 2ci,j.
From the previous Lemma, and using a technique similar to the one used in Corollary 2 to enumerateVs
t, we obtain the compilation complexity of this family of rules: Corollary 4 Ifr is a Condorcet-consistent WMG-rule thenC(r) = Θ(p2logm).
3.3
Plurality with runoff
Plurality with runoff is the voting rule (denoted by r2) consisting of two rounds: the first round keeps only the two candidates with maximum plurality scores (with some tie-breaking mechanism), and the second round is simply the majority rule.
Proposition 10 Let r2 be the plurality-with-runoff rule. P ∼r2 Q holds if and only if for
every x,ntop(P, x) =ntop(Q, x)and for everyx, y,NP(x, y) =NQ(x, y).
Lemma 7 If for everyx,ntop(P, x) =ntop(Q, x)and for everyx, y,NP(x, y) =NQ(x, y), thenP ∼r2 Q.
Proof: For every x ∈ X, since ntop(P, x) = ntop(Q, x), we also have ntop(P ∪R, x) =
ntop(Q∪R, x): the two plurality winners are the same inP∪RandQ∪R. Letxandy be these two plurality winners. SinceNP(x, y) =NQ(x, y), we haveNP∪R(x, y) =NQ∪R(x, y), therefore,MP∪R(x, y) if and only ifMQ∪R(x, y) and hencer2(P∪R) =r2(Q∪R).
Lemma 8 If for some x ntop(P, x)6=ntop(Q, x), thenP 6∼r2 Q.
Proof: If p = 2, this is a corollary of Lemma 1. Assume p ≥ 3, and w.l.o.g., assume
ntop(P, x) > ntop(Q, x). Because P
c∈Xntop(P, c) = P
c∈Xntop(Q, c)(= m), there exists any 6=xsuch thatntop(P, y)< ntop(Q, y). Almost w.l.o.g., assume xhas priority overy
for tie-breaking3. Letz 6=x, y(which is possible becausep≥3). We now construct anR such that in P∪R, the two finalists are xand z, and the winner isx, and in Q∪R, the two finalists arey andz (therefore the winner cannot bex). LetRbe the following partial profile containing 14m+ntop(P, y)−ntop(P, x) new votes:
ntop(P, y)−ntop(P, x) + 4mvotes: x. . .
4mvotes: yxz. . .
6mvotes: z. . .
The plurality scores in P∪Rare:
• sP∪R(x) =ntop(P, x) +ntop(P, y)−ntop(P, x) + 4m=ntop(P, y) + 4m;
• sP∪R(y) =ntop(P, y) + 4m;
• sP∪R(z) =ntop(P, z) + 6m.
• for every other candidatec, sP∪R(c) =ntop(P, c).
Since ntop(P, c) ≤ m holds for every c 6= x, y, z, we have sP∪R(z) > sP∪R(x) =
sP∪R(y) > sP∪R(c) for every c 6= x, y, z. Because x has priority over y, the two candi- dates remaining for the second round are z and x. Now, the number of voters in P ∪R
preferringxtozisN(P∪R, x, z) =N(P, x, z) +ntop(P, y)−ntop(P, x) + 8m≥8m(because
N(P, x, z)≥ntop(P, x)); andN(P∪R, z, x) =N(P, z, x) + 6m≤7m. Hencer2(P∪R) =x. The plurality scores inQ∪Rare:
• sQ∪R(x) =ntop(Q, x) +ntop(P, y)−ntop(P, x) + 4m > ntop(P, y) + 4m;
• sQ∪R(y) =ntop(Q, y) + 4m;
• sQ∪R(z) =ntop(Q, z) + 6m.
• for every other candidatec, sQ∪R(c) =ntop(Q, c). 3The proof in the opposite case is very similar and we omit it.
sQ∪R(y) −sQ∪R(x) = ntop(Q, y) −ntop(P, y) +ntop(P, x) −ntop(Q, x). Now, by assumption we have ntop(Q, y) > ntop(P, y) and ntop(P, x) > ntop(Q, x), therefore
sQ∪R(y)> sQ∪R(x).
sQ∪R(z)−sQ∪R(x) = ntop(Q, z)−ntop(Q, x)−ntop(P, y) +ntop(P, x) + 2m. Now,
ntop(P, x)> ntop(Q, x), thereforesQ∪R(z)−sQ∪R(x)> ntop(Q, z)−ntop(P, y) + 2m >0, that is,sQ∪R(y)> sQ∪R(x).
Because the plurality scores of bothy andz inQ∪Rare larger than the plurality score ofx,xdoes not pass the first round, thereforer2(P∪R)6=x.
Lemma 9 If for some x, y∈X,N(P, x, y)=6 N(Q, x, y) thenP 6∼r2 Q.
Proof: Assume w.l.o.g. that N(P, x, y)> N(Q, x, y). We are going to complete P and Q
such that in bothP∪R andQ∪R, the finalists arexandy, withxwinning inP∪Rand
y inQ∪R. LetRbe composed of the following 2N(P, y, x) + 3m+ 1 votes: 2N(P, y, x) +m+ 1 votes: xy. . .
2mvotes: yx. . .
Obviously, the plurality scores inP∪RverifysP∪R(x)> m,sP∪R(y)> m, and for any
c6=x, y, sP∪R(c)≤m, therefore, the finalists arexandy. Things are the same forQ∪R. Now, N(P ∪R, x, y) = N(P, x, y) + 2N(P, y, x) +m+ 1 = m+N(P, y, x) +m+ 1 = N(P, y, x)+2m+1; andN(P∪R, x, y) = 2N(P, y, x)+4m+1−N(P∪R, x, y) =N(P, y, x)+ 2m. Therefore,r2(P∪R) =x. Lastly,N(Q∪R, x, y) =N(Q, x, y)+2N(P, y, x)+m+1, andN(Q∪R, x, y) =N(Q, y, x)+ 2m. We have nowN(Q∪R, y, x)−N(Q∪R, x, y) =N(Q, y, x)+m−N(Q, x, y)−2N(P, y, x)− m−1 =N(Q, y, x) +m−N(Q, x, y)−2N(P, x, y)−1 = 2(N(Q, y, x)−N(P, y, x))−1. Now,
N(Q, y, x)> N(P, y, x), therefore,N(Q∪R, y, x)> N(Q∪R, x, y), that is,r2(Q∪R) =y.
Proposition 10 is now a corollary from Lemmas 7, 8 and 9, and it follows that:
Proposition 11 The compilation complexity of plurality with runoff is logL(m, p) + logT(m, p).
4
Conclusion
This paper has introduced a notion that we believe to be of primary importance in many practical situations: the compilation of incomplete profiles. In particular, the amount of information that a given polling station needs to transmit to the central authority is a good indicator of the difficulty of the verification process. We have established a general technique which allows us to derive the compilation complexity of a voting rule, and have related it to other issues in communication complexity. We have derived a number of results for specific classes of voting rules. A question that we have only sketched in this paper and that we plan to consider more carefully concerns the situations where the number k of remaining voters is fixed. In this case, a different approach can be taken: instead of compiling the partial profiles as provided by themvoters, it may be more efficient to compile the possible completion of this partial profile together with the associated outcome, or, in other words, to compile the function that takes the remaining k profiles as input (there are (p!)k such inputs) and return the outcome. As there are (p!)k possible profiles, the number of such functions is p(p!)k. This tells us that a general upper bound for C(r, k) is ≤ (p!)k.logp.
Hence, overall we have C(r, k) ≤ min(m.log(p!),(p!)k.logp) (note that the second term becomes interesting only whenmis big enough, andpandksmall enough).
The general problem of dealing with incomplete profiles opens a host of related questions, for instance the probability that a voting process could be stopped after onlymvoters have expressed their opinions, or (a closely related question), the probability that the central authority would make a mistake were it forced to commit on a winner in situations where no candidate is yet guaranteed to prevail. Another interesting issue for further research would consist in designing new ways of computing NP-hard voting rules using an off-line compilation step so that their on-line computation time becomes polynomial in the size of the initial profile. This, of course, implies that the size ofσ(P) may be much larger (possibly exponentially) than the size of the input, which means that the computation ofρ may be logarithmic in the size of the compilationσ(P).
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Y. Chevaleyre, N. Maudet, G. Ravilly-Abadie LAMSADE, Univ. Paris-Dauphine
75775 Paris, France Email:{chevaleyre,maudet,ravilly-abadie}@lamsade.dauphine.fr J. Lang IRIT, CNRS 31062 Toulouse, France Email:[email protected]