Report

+ Random Tie Breaking Toby Walsh NICTA and UNSW + Random Tie Breaking Haris Aziz, Serge Gaspers, Nick Mattei, Nina Narodytska, Toby Walsh NICTA 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 Random vote E.g. UK general elections 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 tiebreaking 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 tiebreaking with a random vote is NP-complete, and vice versa Proof: Consider Borda voting, and a single manipulator, then tiebreaking 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 tiebreaking with a random vote is NP-complete, and vice versa Proof: Consider Borda voting, and a single manipulator, then tiebreaking with a random candidate is polynomial [Obraztsova, Elkind, and Hazon 2011]. But when tie-breaking with a random vote, manipulation is NPcomplete [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 NPhard 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- Copeland, approval, Borda, maximin Bucklin, ranked pairs, STV Random candidate All scoring rules (plurality, veto, kapproval, 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- Copeland, approval, Borda, maximin Bucklin, ranked pairs, STV Random candidate All scoring rules (plurality, veto, kapproval, 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- Copeland, approval, Borda, maximin Bucklin, ranked pairs, STV Random candidate All scoring rules (plurality, veto, kapproval, 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