+

Random Tie Breaking

Toby WalshNICTA and UNSW

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Random Tie Breaking

Haris Aziz, Serge Gaspers, Nick Mattei, Nina Narodytska, Toby WalshNICTA and UNSW

+Ties matter

Manipulators can only change result if election is close!

+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

+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

….

+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]

+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?

+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

+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!

+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]

+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!

+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

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 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

+Tie-breaking with a random vote

Formally incomparable to tie-breaking with a random candidate

+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

+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]

+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]

+Tie-breaking with a random vote

Tie-breaking with a random vote is incomparable to the unique and co-winner manipulation problems

+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

+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

+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

+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!

+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!

+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 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

+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

+Control by breaking ties

Polynomial NP-hard

Scoring rules, Cup, STV, Baldwin,

Nanson, Copeland, maximin, Ranked pairs,

Bucklin, fallback Coombs

+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

+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/…

+Questions?

PS I’m recruiting PhD students and a PostDoc