Ties Matter: - PowerPoint Presentation

Slide1

Ties Matter:Complexity of Voting Manipulation Revisited

based on joint work with Svetlana Obraztsova (NTU/PDMI)and Noam Hazon (CMU)

Edith Elkind

(Nanyang

Technological University, Singapore

)Slide2

SynopsisWe will talk about voting and, in particular, voting

manipulationWe will focus on a frequently neglected aspect of voting: tie-breaking rulesWe will show thatties matterSlide3

SetupAn election is given bya set of candidates C,

|C| = ma list of voters V = {1, ..., n}for each voter i in V, a preference order Rieach Ri is a total order over Ca voting rule F

:

for each list of voters’ preference orders,

F

outputs a candidate in

CSlide4

Examples of Voting Rules (1/2)Scoring rules:

any vector s = (s1, ..., sm) defines a scoring rule Fs : each candidate receives s

i

points from

each

voter

who

ranks him in

position

ia candidate’s score is his total # of pointsthe candidate with the highest score winsExamples:Plurality: (1, 0, ..., 0)Borda: (m-1, m-2, ..., 2, 1, 0)

a b c d e

7 5

2

1

0Slide5

Examples of Voting Rules (2/2)Copeland:

for a, b  C, we say that a beats b in a pairwise election if more than half of the voters rank a above b the score of a candidate

c

is

# of pairwise elections

c

wins

- # of pairwise elections c losesMaximin:for a, b  C

, let S(a, b) = # of voters who prefer a to

b

the

score

of a candidate

c

is

min

a



C

\{c}

S(c, a)

the number of votes

c

gets against his

toughest opponentSlide6

ApplicationsPolitical electionsHiring new facultyPrizes

Decision-making in multi-agent systemsvoting over joint plansSlide7

ManipulationA voting rule is manipulable if there exists

a preference profile s.t. some voter has an incentive to lie about their preferencesi prefers F(R1, ..., R’i, ..., Rn) to F(R

1

, ...,

R

i

, ...,

R

n

)Gibbard’73, Satterthwaite’75: for |C|>2, any non-dictatorial voting rule is manipulableBut maybe manipulations are hard to compute?Bartholdi, Tovey

, Trick’89: given a profile, one can find a beneficial manipulation in poly-time for most voting rules

Plurality, Borda, Copeland,

maximinSlide8

A ComplicationThe common voting “rules” are voting correspondences:

several candidates may have the top scoreTies need to be brokenBTT’89 assumes that ties are broken in favor of manipulatorThe algorithm extends to any lexicographic tie-breaking rulei.e., one that uses a priority order over C What if the tie-breaking rule is not lexicographic?Slide9

This WorkWe consider two types of tie-breaking rules:randomized tie-breaking:

the winner is selected from the tied candidatesuniformly at randomthe manipulator assigns utilities to candidates, maximizes his E[utility]arbitrary poly-time tie-breakingtie-breaking rule is given by an oracleQuestion: do easiness results of BTT’89 still hold under these tie-breaking rules? Slide10

Results: Randomized Tie-Breaking, Scoring RulesTheorem: for any scoring rule

the manipulation problem is poly-time solvable under randomized tie-breaking

What is the best outcome

where winners have

t

points,

for each

feasible

t

?

manipulator’s vote

non-manipulator’ votes

tSlide11

Results: Randomized Tie-Breaking, Maximin, “Special” UtilitiesTheorem

: for Maximin, if the manipulator’s utility is given by u(p)=1, u(c)=0 for c ≠ p, then the manipulation problem is poly-time solvable under randomized tie-breakingProof sketch:let s(c) denote

c

’s

score before we

vote

our

vote changes each score by at most

1:

c’s score goes up iff c appears before each of its toughest opponentsif s(c) > s(p)+1 for some c ≠ p, we

lose; suppose this is not the caserank p firstSlide12

Maximin Proof, Continuedc is

good if s(c) < s(p) c is bad if s(c) = s(p)c is ugly if s(c) = s(p)+1G = directed graph with vertex set C s.t. there is an edge

from

a

to

b

iff

a

is

b’s toughest opponentGoal: sort G so that each ugly vertex and as many bad vertices as possible have an incoming edge can be done in poly-timeSlide13

Results: Randomized Tie-Breaking, Maximin, General UtilitiesTheorem

: for Maximin the manipulation problem is NP-hard under randomized tie-breakingeven if u(d)=0, u(c)=1 for c ≠ dProof idea:set up the instance so that

d

necessarily wins

need to maximize the number of winners

i.e., sort

G

so that

as few vertices as possible

have an incoming edgereduction from Feedback Vertex SetSlide14

Results: Randomized Tie-Breaking, Copeland, General Utilities

Theorem: for Copeland the manipulation problem is NP-hard under randomized tie-breakingeven if u(c)  {0, 1} for all c  CReduction from Independent Set extends to a hardness of approximation result :assuming

P ≠ NP

, no

poly-time

algorithm can find

a manipulative vote such that the manipulator’s utility is within a

constant

factor from optimalSlide15

Arbitrary Poly-Time Tie-Breaking Theorem: there exists a poly-time

computable tie-breaking rule T s.t. its combinations with Borda, Copeland and maximin are NP-hard to manipulateT depends on the set of tied candidates onlyif we allow tie-breaking rules that depend on the rest of the profile, even Plurality is NP-hard to manipulate

Proof idea

:

the winning set encodes a

B

oolean formula

f

and a truth assignment a for fthe manipulator can affect a, but not f the tie-breaking rule checks if

a satisfies fSlide16

Conclusions and Future WorkApproximation algorithms/inapproximability results for Maximin?Complexity

of control/bribery/coalitional manipulation under randomized tie-breaking?TIES MATTER!