Presentation is loading. Please wait.

Presentation is loading. Please wait.

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:

Similar presentations


Presentation on theme: "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:"— Presentation transcript:

1 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)

2 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

3 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:

4 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

5 Outline Intro Algorithm Analysis Application (Kemeny Rank Aggregation)

6 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:

7 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

8 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…)

9 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…)

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

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

12 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…)

13 Outline Intro Algorithm Analysis Application (Kemeny Rank Aggregation)

14 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

15 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…)

16 Example 2 kABCDE EACBD ABCDE EABCD

17 Example 3 kABCDEF BECADF ABECDF

18 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

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

20 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

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

22 Bounding (1/2) Defn:ABCDE BCEAD

23 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)

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

25 Outline Intro Algorithm Analysis Application (Kemeny Rank Aggregation)

26 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

27 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


Download ppt "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:"

Similar presentations


Ads by Google