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How to Rank with Fewer Errors A PTAS for Minimum Feedback Arc Set in Tournaments Warren Schudy Claire Mathieu, Warren Schudy Brown University Thanks to: Nir Ailon, Marek Karpinski, and Eli Upfal (for useful discussions) Cora Borradaile, Aparna Das, Micha Elsner, Dave McClosky, Fabio Vandin, and Matt Wronka (for useful comments on practice talks)

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Feedback arc set in tournaments FAS problem: minimize number of upsets NP-hard [Ailon Charikar Newman 05, Alon 06, Charbit Thomassè Yeo 07] Applications – –Ranking by pairwise comparisons – –Kemeny rank aggregation A B C DABCD 1 upset (Animating…) Best…Worst

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Algorithms for FAS-T Slater and Alway 1961 – –Heuristic algorithm Fernandez de la Vega 96, Arora Frieze Kaplan 96 and many others – –small additive error Ailon Charikar Newman 05 – –Quicksort 3 approx – –LP + quicksort 2.5 approx Coppersmith Fleischer Rudra 06 – –Sort by Wins 5-approx Our Main Result: a PTAS for FAS-T:

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Aside: FAS in general graphs Log n log log n approx [Seymour 95, Even, Noar, Rao, Schieber 95] At least as hard as vertex cover, which cannot be approximated better than 1.36 [Karp 72, Dinur 02] AB C D

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Outline Intro Algorithm Analysis Application (Kemeny Rank Aggregation)

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Towards our algorithm (1/6) An easy instance: when there is an ordering with zero upsets BDAC 1 win Algorithm 1: Sort by Wins 5-approx [CFR] A B C D 3 wins 0 wins 2 wins This means David beat Charlie 2310 Wins:

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Towards our algorithm (2/6) A difficult instance for sort by wins Cost Θ(n²), so additive approx Θ(εn²) works! Algorithm 2: return better of additive approximation and sort by wins Sort by Wins ABCD 2222 E 2 EDCB 2222 A 2 OPT

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Towards our algorithm (3/6) A hard instance for Algorithm 2: Algorithm 3: Sort by wins. Recursively divide in two until Cost=Ω(n²), then run additive approximation EFGHABCDIJ AEDCB F G JIH Sorted by wins: Additive Approx (Animating…)

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Towards our algorithm (4/6) Algorithm 4: like Algorithm 3 but divide into two parts of random sizes EFGHABCDIJ Uh-oh, we just committed ourselves to paying 3 but OPT=2 A hard instance for algorithm 3: Sorted by wins: (Animating…)

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Towards our algorithm (5/6) EFGHABCDIJ KLMNO Sorted by wins Uh-oh, with constant probability we commit to paying Θ(n) (Animating…)

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EFGABCDIJ KLMNOHH Final ingredient: after sorting by wins do single vertex move local optimization Towards our algorithm (6/6) (Animating…)

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Our algorithm Algorithm: Sort by Wins Single vertex moves Result ordering: ImproveRec(all) ImproveRec(vertex set S): If (Cost(S, )Θ(ε²|S|²)) Choose k at random between 1 and |S| Recurse on the first k vertices of S Recurse on the other vertices else Run additive approx. Θ(ε³|S|²) on S EFGHABCDIJ AEDCB Etc… (Animating…)

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Outline Intro Algorithm Analysis Application (Kemeny Rank Aggregation)

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Proof plan After first divide step commit to putting some vertices before others. For talk, will show there is a good ordering consistent with the first divide step Therefore Defined later Lemmas Non-Stopping conditionConstant-factor approx

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Example 1 BCFDE ABCkDEF BCFDAEABCDAEF For this section of talk, align permutations. Ignore this detail. It is cheaper to put A to the left of B,C,D than to the right (Animating…)

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Example 2 kABCDE EACBD ABCDE EABCD

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Example 3 kABCDEF BECADF ABECDF

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Second order pairs Ex. 1 BCFDAE ABCDEF ABCDE EACBD ABCDEF BECADF Ex. 2 Ex. 3 AFAF alternating, A split -> {A, F} 2 nd order pair EBBE not alternating -> {B, E} not 2 nd order pair AEAE alternating, A split -> {A, E} 2 nd order pairABCDEF BECADF Ex. 3 AEAE alternating, neither split -> {A, E} not 2 nd order pair Definition: {u,v} are second order pair if u and v are alternating and at least one of them is split are the root of all evil error

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Bound this Bounding (1/2) ABCDE EACBD ABCDE ACBDE EABCD (Example 2 again)

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BCEAD BAECD ABCDE ACBDE BCADE If {u,v} second-order pair, bound by 3 Otherwise, show zero by cases: – –e.g. {E,D} (Neither split) Proof: by defns and – –e.g. {C,D} (u__
__

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Outline Intro Algorithm Analysis – – Application (Kemeny Rank Aggregation)

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Bounding (1/2) Defn:ABCDE BCEAD

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Claim: Proof: – –[Diaconis and Graham 77]: – –If the pair are in different orders, at least one of the orders has to pay for the pair (This is where the restriction to tournament graphs helps)ABCDE BCEAD Bounding (2/2)

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Reminder: overall proof Lemmas Non-Stopping conditionConstant-factor approx Questions on proof?

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Outline Intro Algorithm Analysis Application (Kemeny Rank Aggregation)

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An application to voting Kemeny Rank Aggregation: (1959) – –Voters submit rankings of candidates – –Translate rankings into tournaments – –Add those tournaments together – –Find feedback arc set of resulting weighted tourney – –Nice properties, e.g. ranks Condorcet winner first [Young & Levenglick 78, Young 95] Our PTAS generalizes to this! A>B>C A B C C>A>B A B C A>C>B A B C A B C A B C

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Open Question Real rankings often have ties, e.g. restaurant guides with ratings 1-5 Exists 1.5-approx [Ailon 07]; is there a PTAS? A B C A: 5 C: 4 B: 5 D: 3 D

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