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. Voter 1Voter 2Voter 3 a b c a c b b a c 2

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. 3

The Condorcet Paradox Voter 1Voter 2 Voter 3 c b a a c b b a c 4 abc

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) = |{i  N: a > i b}| Maximin: a’s score is min b P(a,b) – If a is a condorcet winner, min b P(a,b) > n/2, any for any b≠a, P(b,a) < n/2. Voting trees. 5

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 bcdea dbedc eacbe aeaab cdbcd bcdea dbedc eacbe aeaab cdbcd bcdea dbedc eacae aeabb cdbcd bcdea dbeac eacde aeabb cdbcd bcdea dbeac aacde eeabb cdbcd bbdea dcedc eacbe aeaab cdbcd bbdea dcedc eacbb aeaae cdbcd bde ce

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 x ij : binary, 1 iff i pushed c* j positions. – d(a) – deficit of c* w.r.t. a. – constants e ij a : 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 X i k = j w. prob. x ij. For all i, X i max = max k X i k. i pushes c* by X i max.

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.

Example bcdea aaedc dbcbe ceaab edbcd 125 bca aac dbe ceb edd

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.

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