160 likes | 348 Views
Approximability and Inapproximability of Dodgson and Young Elections. Ariel D. Procaccia, Michal Feldman and Jeffrey S. Rosenschein. Voting: reminder?. Set of voters V={1,...,n}. Set of Candidates C={a,b,c...}; |C|=m. Voters (strictly) rank the candidates.
E N D
Approximability and Inapproximability of Dodgson and Young Elections Ariel D. Procaccia, Michal Feldman and Jeffrey S. Rosenschein
Voting: reminder? • Set of voters V={1,...,n}. • Set of Candidates C={a,b,c...}; |C|=m. • Voters (strictly) rank the candidates. • Preference profile: a vector of rankings. a b a b a c c c b
Condorcet winner • a beats b in a pairwise election if the majority of voters prefers a to b. • a is a Condorcet winner if a beats any other candidate in a pairwise election.
The Condorcet Paradox c a b c a b b a c
Condorcet voting rules • Copeland: a’s score is num of other canidates a beats in a pairwise election. • If a is a Condorcet winner, score = m-1, and for any b≠a, score < m-1. • P(a,b) = |{iN: a >i b}| • Maximin:a’s score is minbP(a,b) • If a is a condorcet winner, minbP(a,b) > n/2, any for any b≠a, P(b,a) < n/2. • Voting trees.
Dodgson’s voting rule • (Dodgson) • Find candidate closest to a Condorcet winner. • distance/score of c = number of exchanges between pairwise candidates needed for c to become a Condorcet winner. • Alternatively: number of places each voter has to push c. • Elect candidate with minimal distance/score.
Example b d e c e
Hardness and Approximation • Bartholdi, Tovey and Trick 89: NP-hard to compute Dodgson score. • Hemaspaandra et al. 97: Even harder to compute Dodgson winner. (Why not in NP?) • Poly time if either n or m is constant. • We want to approximate the Dodgson score. • Discussion: essentially gives us a new voting rule (can satisfy desiderata). • Existing lower bound: log(m). Also works for random algs, unless NP = RP.
Trivial alg • Given: profile, c*. • Alg: • Let C’ be the candidates not beaten by c* in a pairwise election. • While C’ is not empty: • Choose some a in C’. • Perform minimal number of exchanges needed to make c* beat a. • Recalculate C’. • Step 2 in while: d(a) is deficit w.r.t. a; sufficient to choose d(a) voters which require smallest number of exchanges.
Trivial claim about trivial alg • Claim: alg gives m-approx. • Proof: • Let a be the candidate which requires the max number t of exchanges to get c* to beat a. • Score of c* >= t. • Each iteration of the while loop performs <= t flips. There are at most m iterations. • Trivial alg which gives n-approx: at every stage, each voter pushes c* one place up.
LP for Dodgson • Notations: • Variables xij: binary, 1 iffi pushed c* j positions. • d(a) – deficit of c* w.r.t. a. • constants eija: binary, 1 iff pushing c* j positions by i gives c* additional vote against a. • (Example) • ILP is NP-hard.
Randomized Rounding Alg • Solve relaxed LP to obtain solution x. • For k=1,..., log(m): for all i, randomly and independently choose Xik = j w. prob. xij. • For all i, Ximax = maxkXik. • i pushes c* by Ximax.
Young’s rule • Also chooses candidate “closest” to Condorcet winner. • Score of c*: maximum subset of voters for which c* is a Condorcet winner. • 0 is no nonempty subset. • Alternatively: minimum number of voters one has to remove.
About Young • Same hardness results. • Can also formulate as LP. • Young is nonmonotonic: if it is possible to make c* a winner on k voters, it doesn’t mean that it’s possible on 0< r < k voters. • Theorem: NP-hard to approximate the Young score to any factor. • Specifically: It is NP-hard to determine whether there is a nonempty subset of voters on which c* is a Condorcet winner. • Discussion.
Related work • Not... • Ask me if you’re interested.