Random Tie Breaking
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Transcript of Random Tie Breaking
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Random Tie BreakingToby WalshNICTA and UNSW
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Random Tie BreakingHaris Aziz, Serge Gaspers, Nick Mattei, Nina Narodytska, Toby WalshNICTA and UNSW
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+Ties matter
Manipulators can only change result if election is close!
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+Ties matter
Manipulators can only change result if election is close! How we deal with ties often matters critically Typical assumption is ties broken in favour of the
manipulators
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+Ties matter
Manipulators can only change result if election is close! How we deal with ties often matters critically Typical assumption is ties broken in favour of the
manipulators In real elections, ties broken randomly, by the chair, by age
….
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+Ties matter
Manipulators can only change result if election is close! How we deal with ties often matters critically Typical assumption is ties broken in favour of the
manipulators In real elections, ties broken randomly, by the chair, by age
…. Tie breaking can itself be a source of computational
complexity 2nd order Copeland, Copeland with weighted votes: polynomial to manipulate
if ties are scored 1, but NP-hard if ties are scored 0 [Faliszewski, Hemaspaandrea,
Schnoor 08]
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+Unique and co-winner problems
Unique winner manipulation problem Equivalent to tie-breaking against manipulator Can we construct a strategic vote so given candidate is the
unique winner of the election?
Co-winner manipulation problem Equivalent to tie-breaking in favour of the manipulator Can we construct a strategic vote so given candidate is one
of the co-winners of the election?
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+Tie-breaking in practice
Random candidate E.g. UK general elections
Random vote E.g. Schulze voting breaks ties according to order of
candidates in a random vote
By the chair
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+Tie-breaking with a random candidate See [Obraztsova, Elkind, Hazon AAMAS 2011],
[Obratzsova, Elkind IJCAI 2011] Agents assign utilities to candidates Look to maximize expected utility of result
Can get a large way though with a simple model of just asking if a given candidate can win with probability > p?
Equivalent to u(a)=1, u(b)=0 for all other candidates Avoids the difficult problem of having to assign utilities!
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+Tie-breaking with a random candidate Several common rules have been shown to be
(in)tractable THM: When tie-breaking with a random candidate, all
scoring rules (including Borda) are polynomial to manipulate, as are plurality with runoff and Bucklin
THM: When tie-breaking with a random candidate, Copeland and Maximin are NP-hard to manipulate
[Obraztsova & Elkind 2011]
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+Tie-breaking with a random vote In case of a tie, pick a vote uniformly at random
Order candidates according to this vote
In some forthcoming work, we’ve shown that this has different computational properties to tie-breaking with a random candidate In practice, it seems harder Indeed, it is often proposed as a barrier to manipulation Suppose you vote strategic to get a preferred candidate to
win, but then your strategic vote may actually make them loose!
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+Tie-breaking with a random vote Candidates can have quite different probabilities of
winning than tie-breaking with a random candidate Suppose we use Borda scoring
Half voters vote a>b>c Half voters vote c>b>a
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+Tie-breaking with a random vote Candidates can have quite different probabilities of
winning than tie-breaking with a random candidate Suppose we use Borda scoring
Half voters vote a>b>c Half voters vote c>b>a
Tie-breaking with a random vote a or c win with probability 1/2
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+Tie-breaking with a random vote Candidates can have quite different probabilities of
winning than tie-breaking with a random candidate Suppose we use Borda scoring
Half voters vote a>b>c Half voters vote c>b>a
Tie-breaking with a random vote a or c win with probability 1/2
Tie-breaking with a random candidate a, b, or c win with probability 1/3
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+Tie-breaking with a random vote Formally incomparable to tie-breaking with a random
candidate
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+Tie-breaking with a random vote Formally incomparable to tie-breaking with a random
candidate THM: exists voting rule, such that the manipulation problem
when tie-breaking with a random candidate is polynomial but tie-breaking with a random vote is NP-complete, and vice versa
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+Tie-breaking with a random vote Formally incomparable to tie-breaking with a random
candidate THM: exists voting rule, such that the manipulation problem
when tie-breaking with a random candidate is polynomial, but tie-breaking with a random vote is NP-complete, and vice versa Proof: Consider Borda voting, and a single manipulator,
then tie-breaking with a random candidate is polynomial [Obraztsova, Elkind, and Hazon 2011]
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+Tie-breaking with a random vote Formally incomparable to tie-breaking with a random
candidate THM: exists voting rule, such that the manipulation problem
when tie-breaking with a random candidate is polynomial, but tie-breaking with a random vote is NP-complete, and vice versa Proof: Consider Borda voting, and a single manipulator,
then tie-breaking with a random candidate is polynomial [Obraztsova, Elkind, and Hazon 2011].But when tie-breaking with a random vote, manipulation
is NP-complete [forthcoming 2013]
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+Tie-breaking with a random vote Tie-breaking with a random vote is incomparable to the
unique and co-winner manipulation problems
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+Tie-breaking with a random vote Tie-breaking with a random vote is incomparable to the
unique and co-winner manipulation problems THM: exists voting rule, such that the co-winner and unique
winner manipulation problems are polynomial, but tie-breaking with a random vote is NP-complete, and vice versa
Contrast this with tie-breaking with a random candidate If unique winner or co-winner manipulation problems are
NP-hard then tie-breaking with a random candidate is also
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+Random vote versus Random candidate
Polynomial NP-complete Open
Random vote
Plurality, veto, (fixed) k-approval, plurality with runoff
(unbounded) k-approval, Borda, Bucklin, ranked pairs, STV
Copeland, maximin
Random candidate
All scoring rules (plurality, veto, k-approval, Borda), plurality with runoff, Bucklin
Copeland, maximin, ranked pairs, STV
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+Random vote versus Random candidate
Polynomial NP-complete Open
Random vote
Plurality, veto, (fixed) k-approval, plurality with runoff
(unbounded) k-approval, Borda, Bucklin, ranked pairs, STV
Copeland, maximin
Random candidate
All scoring rules (plurality, veto, k-approval, Borda), plurality with runoff, Bucklin
Copeland, maximin, ranked pairs, STV
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+Random vote versus Random candidate
Polynomial NP-complete Open
Random vote
Plurality, veto, (fixed) k-approval, plurality with runoff
(unbounded) k-approval, Borda, Bucklin, ranked pairs, STV
Copeland, maximin
Random candidate
All scoring rules (plurality, veto, k-approval, Borda), plurality with runoff, Bucklin
Copeland, maximin, ranked pairs, STV
How you break ties impacts on the computational complexity!
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+Control by breaking ties
Somewhat related problem If I am chair, how do I control the result by breaking
ties? Tie-breaking only once (between co-winners), this is trivial Pick the person you want to win Tie-breaking even just twice, control can be NP-hard!
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+Control by breaking ties
Control by tie-breaking with two stage rules THM: Exists a two stage rule combining veto and plurality
for which control by tie-breaking is NP-hard Proof: Consider rule that eliminates half the candidates
using veto, then elects the plurality winner
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+Control by breaking ties
Control by tie-breaking with two stage rules THM: Exists a two stage rule combining veto and plurality
for which control by tie-breaking is NP-hard Proof: Consider rule that eliminates half the candidates
using veto, then elects the plurality winner
Control by tie-breaking with multi-stage rules THM: Control by tie-breaking with STV, Baldwin and Coombs
is NP-hard
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+Control by breaking ties
Control by tie-breaking with two stage rules THM: Exists a two stage rule combining veto and plurality
for which control by tie-breaking is NP-hard Proof: Consider rule that eliminates half the candidates
using veto, then elects the plurality winner
Control by tie-breaking with multi-stage rules THM: Control by tie-breaking with STV, Baldwin and Coombs
is NP-hard THM: Control by tie-breaking with Nanson is polynomial
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+Control by breaking ties
Polynomial NP-hardScoring rules, Cup, STV, Baldwin,Nanson, Copeland, maximin, Ranked pairs,Bucklin, fallback Coombs
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+Control by breaking ties
Incomparable to the manipulation problem when breaking ties with a random candidate, or in a fixed order THM: Exists voting rule such that control by tie-breaking is
polynomial but manipulation problem breaking ties at random/in a fixed order is NP-complete, and vice versa
E.g. control by breaking ties for Copeland is polynomial, but manipulation when breaking ties at random is NP-hard
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+Conclusions
Ties do matter Breaking ties with a random vote somewhat more
computationally challenging than with a random candidate
For two and multi-stage rules, it can be NP-hard for the chair to control result by breaking ties
Of course, these are all worst case observations and we need to consider the difficulty of breaking ties in practice/on average/…
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+Questions? PS I’m recruiting PhD students and a PostDoc