Complexity of Voting Manipulation Revisited b ased on joint work with Svetlana Obraztsova NTUPDMI and Noam Hazon CMU Edith Elkind Nanyang Technological University Singapore ID: 337641
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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!