Ioannis Caragiannis, Jason A. Covey, Michal Feldman, Christopher M. Homan, Christos Kaklamanis, Nikos Karanikolask, Ariel D. Procaccia, Je ff rey S. Rosenschein.

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Ioannis Caragiannis, Jason A. Covey, Michal Feldman, Christopher M. Homan, Christos Kaklamanis, Nikos Karanikolask, Ariel D. Procaccia, Je ff rey S. Rosenschein. Presented by Noa Avigdor-Elgrabli

Agenda Basic definitions Dodgson score upper bounds Randomized approximation algorithm Definition Analysis Deterministic approximation algorithm What’s better? Lower bounds

Basic Definitions Input: Example: m alternatives {a, b, c} n agents {1,2,3} Ranking of each agent Condorcet winner: An alternative that beats every other alternative in pairwise election Dodgson score of a*: The least number of exchanges between adjacent alternatives needed to make a* a Condorcet winner Young score of a*: The size of the largest subset of agents for whom a* is a Condorcet winner Agent 1Agent 2Agent 3 aba bac ccb Condorcet winner: a abc 013 Dodgson score: abc 310 Young score:

Remarks Algorithm that Usually computes the Dodgson score (under the assumption that the number of agents is significantly large than the number of)

Randomized Algorithm for the Dodgson score of a* Construct an integer linear program for the Dodgson score (not solvable in polynomial time  ). Relax it to a linear program (solvable but the solution is fractional  ). Round the fractional solution to an integer solution using some randomness. Prove that the solution is “good” (O (log m) - approximation) with high probability.

Notations def(a) = The number of additional agents that must rank a* above a in order for a* to beat a in a pair wise election. e i ja = 1 iff pushing a* by j positions in the ranking of agent i makes a* gain additional vote against a. (e i ja = 0 otherwise)

Linear Program [BTT89] iff a* is “pushed” by j positions in the ranking of agent i The cost of the solution We choose for each i exactly one j (from 0 to m-1) we improve the position of a* against each a in at least def(i) agents

Randomized Rounding Algorithm 1. Solve the relaxed LP – We get a solution x 2. For each agent i assign X i  j with probability 3. Repeat (2) t times (t would be defined latter) 4. X i max  the maximum of all t received X i ’s 5. Return ∑ i X i max Theorem For t=8 log m the randomized rounding algorithm returns t-approximation of the Dodgson score with probability at least 1/2

Analysis Remainder: Lemma (Jogdeo and Samuels) Let Y 1,..., Y n be independent heterogeneous Bernoulli trials. Then

Greedy Algorithm Live = {a | def(a) > 0} While Live is not empty: Do the most “cost- effective” push: Theorem: Greedy is O(log m)- approximation

Example Live={a1,a2,a3,a4} Greedy(a*)=7 (=Opt) R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 a4 a3a2a1 a3 b4b9b13b16 a2 b5b10b14b17 a1 b6b11b15a*.b1b7b12a*..b2b8a*...b3a* /1 4/1 7/4

Desirable Properties Algorithmic properties: Running Time, randomness, approximation factor. Social choice properties: Truthfulness – the agents cannot benefit by lying. Condorcet Property – a* is a Condorcet winner  a* wins. Monotonicity Property (weak) – pushing an alternative in the preference of an agent cannot worsen its score. Truthfulness  Dictator or Existence of alternative that can’t win [Gibbard-Satterthwaite]

Greedy - Non Monotonic ds R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 a4 a3a2a1 a3 b4b9b13b16 a2 b5b10b14b17 a1 b6b11b15a*.b1b7b12a*..b2b8a*...b3a* /4 R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 a4 a3a2a1 a3 b4b9b13a* a2 b5b10a*b16 a1 b6a*b14b17.b1a*b11b15..b2b7b12...b3b8....a* /4 Cost(a*)=7 2 7/3 3 7/27 Cost(a*)= =9 4

Rounding or Greedy? Approximation factor Running time RandomnessCondorcet Property Monotonicity Roundinglog (m)BadRandomizedYes Greedylog (m)GoodDeterministicYesNo

Lower Bounds Dodgson Score: For any ε > 0, there is no polynomial-time Ω ( (1/2 – ε) ln m)-approximation for the Dodgson score of a given alternative*  Greedy is optimal up to factor 2 Young Score: It is NP-hard to approximate the Young score by any factor *under the assumption that an NP problems can’t be solved in quasi-poly time

Thank You!