How Hard Is It To Manipulate Voting? Edith Elkind, U. of Warwick Helger Lipmaa, Tartu U.

Short Bio High school diploma, School № 15, Tallinn, 1993 M.Sc., Moscow State University, Department of Mathematics, 1998 Ph.D, Princeton University, 2005 Now: Postdoctoral researcher, U. of Warwick, UK Research interests: algorithmic game theory, voting, algorithms, complexity…

Bib Info Small Coalitions Cannot Manipulate Voting, Financial Cryptography 05 Hybrid Voting Protocols And Hardness of Manipulation, ISAAC 05, to appear

What Is Manipulation? In a small country far, far away there is an election coming up…

Manipulation: Example 99 voters, 3 candidates (Red, Blue, Green). –49 voters: R > B > G. –48 voters: B > R > G. –2 voters (Edith and Helger): G > B > R. Aggregation rule: Plurality –each voter casts a vote for one candidate. –the candidate with the largest number of votes wins. –draws are resolved by a coin toss.

What Will Edith and Helger Do? If I vote for G, R will get elected, so I’d rather vote for B R: 49 votes B: 48 votes G > B > R If Edith and Helger vote B > G > R, they can guarantee that B is elected

Why Manipulation Is Bad Aggregation rules are designed with certain social welfare criteria in mind. Misrepresentation of preferences results in a suboptimal choice w.r.t. these criteria. Also, election results do not reflect true distribution of preferences in the society: –maybe, in fact, in % of the U.S. population prefered Nader to Gore to Bush?

Single Transferable Vote: This time, Edith and Helger are better off voting honestly, but this will not always be the case… Other popular voting schemes (Borda, Copeland) suffer from the same problem R > B > G B > R > G G > B > R What If We Change Aggregation Rule? 1 st round 49 votes 48 votes 2 votes R > B B > R 2 nd round 49 votes 50 votes B wins

Formal Setup n voters m candidates c 1, …, c m Preference of a voter i: a permutation  i of c 1, …, c m (best to worst). Aggregation rule S:  1, …,  n → c j.

Voting Schemes: Examples Borda: a candidate gets –m points for each voter who ranks him 1 st, –m-1 point for each voter who ranks him 2 nd, etc. Copeland: –candidate that wins the largest # of pairwise elections Maximin: –c’s score against d: # of voters that prefer c to d; –c’s # of points: min score in any pairwise election. many, many others…

Voting Schemes: Properties Pareto-optimality: if everyone prefers a to b, b does not win Condorcet-consistency: if there is a candidate that wins every pairwise election, this candidate wins Majority: if there is a candidate that is ranked first by a majority of voters, this candidate wins Monotonicity: it is impossible to cause a winning candidate to lose by moving it up in one’s vote Arrow’s theorem: there is no perfect scheme

Manipulation: Definition A voter i can manipulate a voting scheme S if there is –a preference vector   = (  1,…,  i, …,  n ) –a permutation  i ’ s.t. S(  1,…,  i ’, …,  n ) > i S(  ). Theorem (Gibbard-Satterthwaite, 1971): every non-dictatorial aggregation rule with ≥3 candidates is manipulable.

How Do We Get Around The Impossibility Result? We cannot make manipulation impossible… But we can try to make it hard! How do you manipulate Plurality? – vote for your favorite candidate among those tied for the top position. How do you manipulate Borda? –rank your favorite feasible candidate highest, move his competitors to the bottom of your vote. How do you manipulate STV? – try all m! possible ballots…

What Is Known? 2 nd order Copeland is NP-hard to manipulate (Bartholdi, Tovey, Trick 1989) STV is NP-hard to manipulate (Bartholdi, Orlin 1991) these rules may not reflect the welfare goals (why is there so many voting rules out there?) Want a universal method to turn any voting protocol into a hard-to-manipulate one.

Adding a Preround (Conitzer- Sandholm’03) AliceBob CarlDianaErnestFrank Retains some of the flavor of the original protocol. Is NP-hard to manipulate for many base protocols. Still, the outcome may be very different from the original protocol… Original protocol AliceDiana Ernest PreroundPreround Do most voters prefer Alice to Bob?

Binary Cup R1 A A B C C D F E F G G H Do most voters prefer A to B? C F C R2 R3 Binary Cup itself is easy to manipulate.

Our Work: Hybrid Protocols Protocols with a preround can be viewed as hybrids of BC and other protocols –how about other hybrids? Hyb(X k, Y): execute k steps of X, then apply Y to the remaining candidates. –Hyb(Plurality k, Borda): eliminate k candidates with the lowest Plurality scores, then compute Borda scores w.r.t. survivors. Observation: Hyb(Plurality 1, …, Plurality m ) = STV. abcdeabcde

New Results New protocols that are hard to manipulate: –Hyb(X k, STV), Hyb(STV k, Y) (for any reasonable X, Y) –Hyb(Borda k, Plurality), Hyb(Maximin k, Plurality) Generally, Hyb(X k, X) ≠ X (and may be much harder to manipulate) –manipulating Hyb(Borda k, Borda) is NP-hard Extensions to utlity-based voting (voters rate candidates rather that rank them) –manipulating Hyb(HighScore k, HighScore) is NP-hard

Proof Idea Set Cover: G = {g 1, …, g n }, S = {s 1, …, s m }, each s i is a subset of G. Is there a cover of size K, i.e, C = {s i1, …, s iK } s.t. each g i is contained in some s j ? Set Cover is NP-hard. Can we reduce Set Cover to protocol manipulation? G s1s1 s3s3 s4s4 s2s2 Suppose someone can manipulate Hyb (X k, Y). We want to show that he can also find a set cover.

Proof Idea (continued) Candidates: g 1, …, g n, s 1, …, s m, p, d 1, …, d t Voters: –( p, … ): A ballots –for each g i (g i, …, ): A - 1 ballot (s j1, …, s jr, g i, … ) s.t. g i is in s j1, …, s jr : B ballots –some ballots of the form (d j, …, ) k = m – K s 1, …, s m are tied for the last place under X k –manipulator’s vote decides which K candidates survive the preround manipulator’s prefered candidate

Proof Idea (continued) p wins iff none of g i gains votes after X k g i gains votes iff all s j s.t. g i is in s j are eliminated no g i gains votes iff surviving s j cover all g i manipulation is successful iff manipulator can find a set cover of size K s1s5s8g1…s1s5s8g1… s1s5s8g1…s1s5s8g1… s3s4s5g2…s3s4s5g2… s3s4s5g2…s3s4s5g2… g1…………g1………… g1…………g1………… p …... … p…………p………… p…………p………… d1…………d1…………

Is 3-Coloring hard on a random graph? –likely to contain just say “no” –poly-time, works for almost all graphs We embed a problem that is sometimes hard into some class of preference profiles Want a reduction that 1.works for (almost) all preference profiles 2.reduces from a problem that is (almost) always hard We show how to do (2) using one-way functions; (1) is an open problem. Worst-Case vs. Average-Case Hardness

One-Way Functions f is a one-way function (OWF) if it is easy to compute: for any x, f(x) is polynomial time computable. hard to invert: –x is drawn at random from {0, 1} n –we are given y = f(x) –have to guess x, or any z s.t. y = f(z) –average-case hardness: any poly-time algorithm has negligible chance of success (over the choice of x). y x OWF

OWF: Properties It is not known if OWF exist (implies P ≠ NP). –Multiplication of large primes is conjectured to be a OWF (factoring is believed to be hard) OWF are widely used in cryptography –adversary’s task must be hard almost always, not just sometimes. Can we make manipulation as hard as inverting OWF? –sort of: we embed inverting OWF into some preference profiles.

Recall: Preround (CS’03) AliceBob CarlDianaErnestFrank Retains some of the flavor of the original protocol. Is NP-hard to manipulate for many base protocols. Still, the outcome may be very different from the original protocol… Original protocol AliceDiana Ernest PreroundPreround Do most voters prefer Alice to Bob?

Our Scheme: Preround With a Twist Idea: use votes themselves to select preround schedule … XOR 01101… 11100… 10111… Alice Bob …  1 Bob Carl …  2 Carl Alice …  n Alice ↕ Frank Bob ↕ Ernest Carl ↕ Diana Preround schedule OWF 00100…

Key Properties No trusted source of randomness is needed. Manipulating this protocol is as hard as inverting the underlying one-way function –even for a large coalition of conspiring would-be manipulators Proof proceeds by constructing a vector of honest voters’ preferences s.t. there is a unique preround schedule that makes manipulators’ protégé a winner –still a worst-case construction…

Average-Case Hardness? Can we make manipulation hard with high probability over honest voters’ preferences? Do we want to make assumptions about the distribution of preference profiles? In real life, not all preference profiles are equally likely…

Hyb(BC, Plurality): Easy for Any Preround Schedule Candidates a, b, c 1, …, c 2t+1, p k+1 voters: p > a > b > C, k ≈ n/3 k voters: a > b > p > C n-2k-2 voters: C > p > a > b Manipulator: C > a > b > p Under any preround schedule –p survives –a c j survives –either a or b survives In the final round, p competes against a or b –manipulator is better off voting a > b > p > C

Open Problems Average-case hardness seems hard to achieve using a preround. Other approaches? What is the maximum fraction of manipulators we can tolerate? –for most base protocols, we can prove security against 1/6 of all voters. Is it optimal? Our results are for specific protocols. More general proofs?