Computational social choice Lirong Xia Todays schedule - PowerPoint PPT Presentation

Computational social choice Lirong Xia

Today’s schedule Ø Computational social choice: the easy-to- compute axiom • voting rules that can be computed in P • satisfies the axiom • Kemeny: a full proof of NP-hardness • IP formulation of Kemeny 2

Kemeny’s rule Ø Kendall tau distance • K ( R , W )= # {different pairwise comparisons} K( b ≻ c ≻ a , a ≻ b ≻ c ) = 2 Ø Kemeny( P )=argmin W K( P , W ) = argmin W Σ V ∈ P K( V , W ) Ø For single winner, choose the top-ranked alternative in Kemeny( P ) 3

Example Profile P a ≻ b ≻ c ≻ d b ≻ a ≻ c ≻ d d ≻ a ≻ b ≻ c c ≻ d ≻ b ≻ a 1 1 2 2 a WMG b 2 2 2 2 2 d c K ( P , a ≻ b ≻ c ≻ d )=0+1+2*3+2*5=17 4

Computing the Kemeny winner Ø For each linear order W ( m ! iter) • for each vote R in D ( n iter) • compute K ( R , W ) Ø Find W* with the smallest total distance • W* = argmin W K ( D , W ) = argmin W Σ R ∈ D K ( R , W ) • top-ranked alternative at W* is the winner Ø Takes exponential O ( m ! n ) time! 5

Kemeny Ø Ranking W → direct acyclic complete graph G ( W ) a b a ≻ b ≻ c ≻ d c d Ø Given the WMG G ( P ) of the input profile P Ø K ( P , W ) = Σ a → b ∈ G ( W ) #{ V ∈ P : b ≻ a in V } = Σ a → b ∈ G ( W ) ( n + w ( b → a )) /2 = nm ( m- 1)/4 + Σ a → b ∈ G ( W ) w ( b → a ) /2 Ø argmin W K ( P , W ) = argmin W Σ a → b ∈ G ( W ) w ( b → a ) = argmin W Total weight on inconsistent edges in WMG 6

Example a b W= a ≻ b ≻ c ≻ d c d a b 20 Profile P : 12 16 6 14 8 c d Ø Total weight on inconsistent edges between W and P is: 20 7

Kemeny is NP-hard to compute Ø Reduction from feedback arc set (FAC) • Given a directed graph G and a number k • does there exist a way to eliminate no more than k edges to obtain an acyclic graph? a b d c 8 J. Bartholdi III, C. Tovey, M. Trick, Voting schemes for which it can be difficult to tell who won the election, Social Choice Welfare 6 (1989) 157–165.

Proof Ø The KendallDistance problem: • Given a profile P and a number k , • Does there exist a ranking W whose total Kendall distance is at most k ? P-time Instance of FAC Instance of KendallDistance a b a b 2 2 G 2 WMG( P ): 2 d c 2 d c k k’= 2 k+nm(m-1)/4-5 Yes Yes 9 No No

Constructing the profile Ø For any edge a → b ∈ G , define Ø P a → b = { a ≻ b ≻ others, Reverse(others) ≻ a ≻ b } 2 a b WMG( P a → b )= d c Ø P = ∪ a → b ∈ G P a → b 10

The easy-to-compute axiom Ø A voting rule satisfies the easy-to- compute axiom if computing the winner can be done in polynomial time • P: easy to compute • NP-hard: hard to compute • assuming P≠NP 11

The winner determination problem Ø Given: a voting rule r Ø Input: a preference profile P and an alternative c • input size: nm log m Ø Output: is c the winner of r under P ? 12

Computing positional scoring rules Ø If following the description of r the winner can be computed in p-time, then r satisfies the easy-to-compute axiom Ø Positional scoring rule • For each alternative ( m iter) • for each vote in D ( n iter) – find the position of m , find the score of this position • Find the alternative with the largest score ( m iter) • Total time O ( mn + m )= O ( mn ) 13

Computing the weighted majority graph Ø For each pair of alternatives c , d ( m ( m -1) iter) • let k = 0 • for each vote V ∈ P • if c > d add 1 to the counter k • if d > c subtract 1 from k • the weight on the edge c → d is k 14

Satisfiability of easy-to-compute Rule Complexity Positional scoring Plurality w/ runoff STV P Copeland Maximin Ranked pairs Kemeny Slater NP-hard Dodgson 15

Solving Kemeny in practice Ø For each pair of alternatives a , b there is a binary variable x ab Ø x ab = 1 if a > b in W Ø x ab = 0 if b > a in W Ø max Σ a , b w ( a → b ) x ab s.t. for all a , b, x ab +x ba =1 No edges in both directions for all a , b , c, x ab +x bc +x ca ≤ 2 No cycle of 3 vertices Ø Do we need to worry about cycles of >3 vertices? Homework 16

Manpulation 17

Manipulation under plurality rule (lexicographic tie-breaking) > > Alice > > Plurality rule Bob > > Carol > >

Strategic behavior (of the agents) Ø Manipulation: an agent (manipulator) casts a vote that does not represent her true preferences, to make herself better off Ø A voting rule is strategy-proof if there is never a (beneficial) manipulation under this rule

Using STV? > > × 4 > > × 2 × 2 > > Ø Voting time! Ø N>M>O à O>M>N 20

Any strategy-proof voting rule? Ø No reasonable voting rule is strategyproof • Gibbard-Satterthwaite Theorem [Gibbard Econometrica-73, Satterthwaite JET-75] : When there are at least three alternatives, no voting rules except dictatorships satisfy • non-imposition: every alternative wins for some profile • unrestricted domain: voters can use any linear order as their votes • strategy-proofness Ø Randomized voting rules [Gibbard Econometrica-77] Ø Multiple winners [Duggan &Schwartz SCW 00]

Proof idea Ø Using Arrow’s impossibility theorem • Arrow’s theorem: no ranking rule satisfies non-dictatorship, universal domain, unanimity, and IIA Ø Suppose G-S does not hold for r (strategy-proof + non-dictatorial), construct a contraduction f to Arrow’s theorem • 1. for any profile P and any pair of alternatives ( a,b ), raise them to the top-2 to obtain P ab . Let a > b in f ( P ) iff r ( P ab )= a • 2. prove that f ( P ) is a linear order • 3. prove that f satisfies non-dictatorship, universal domain, unanimity, and IIA è contradiction 22

A few ways out Ø Relax non-dictatorship: use a dictatorship Ø Restrict the number of alternatives to 2 Ø Relax unrestricted domain: mainly pursued by economists • Single-peaked preferences Participation 70% 10% 40% 23

Single-peaked preferences Ø There exists a social axis S • linear order over the alternatives Ø Each voter’s preferences V are compatible with the social axis S • there exists a “peak” a such that • [b ≺ c ≺ a in S] implies [c ≻ b in V] • [a ≻ c ≻ b in S] implies [c ≻ b in V] • alternatives closer to the peak are more preferred • different voters may have different peaks, but must share the same social axis 24

Examples rank Axis 25

Strategy-proof rules for single- peaked preferences Ø The median rule • given a profile of “peaks” • choose the median in the social axis Ø Theorem. The Median rule is strategy-proof. Ø The median rule with phantom voters [Moulin PC 80] • parameterized by a fixed set of “peaks” of phantom voters • chooses the median of the peaks of the regular voters and the phantom voters Ø Theorem. Any strategy-proof rule for single-peaked preferences are median rules with phantom voters 26