1 / 13

New Perspectives on the Implementation of Functions by Voting Trees in Tournaments

This paper explores the implementation of functions by voting trees in tournaments, discussing Copeland scores, approximations, randomized models, upper and lower bounds, caterpillars, and open problems.

ehren
Télécharger la présentation

New Perspectives on the Implementation of Functions by Voting Trees in Tournaments

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Felix Fischer, Ariel D. Procaccia and Alex Samorodnitsky A New Perspective on Implementation by Voting Trees

  2. Tournaments • A = {1,...,m}: set of alternatives • A tournament is a complete and asymmetric relation T on A. T(A) set of tournaments • The Copeland score of i in T is its outdegree • Copeland Winner: max Copeland score in T 1 2 3 4 5 6

  3. Voting trees 1 2 ? 3 ? ? 3 1 3 ? 2 1

  4. Implementation by Voting trees • An alternative can appear multiple times in leaves of tree, or not appear (not surjective!) • Voting tree  implementsf:T(A)A if f(T)=(T) for all T • Which functionsf:T(A)A can be implemented by voting trees? • [Moulin 86] Copeland cannot be implemented when m  8 • [Srivastava and Trick 96] ... can be when m  7 • Can Copeland be approximated by trees?

  5. The two models • Si(T) = Copeland score of i in T • Deterministic model: a voting tree  has an -approx ratio if T, (S(T)(T) / maxiSi(T))   • Randomized model: • Randomizations over voting trees • Dist.  over trees has an -approx ratio if T, (E[S(T)(T)] / maxiSi(T))   • Randomization is admissible if its support contains only surjective trees

  6. Upper bound: Components • C A is a component of T if  i,jC, kC, iTkjTk • Lemma [Moulin 86]: T and T’ differ only inside a component C,  a voting tree, then (T)A\C(T)=(T’) 1 2 3 4 5 6

  7. ¾ deterministic upper bound • m = 3k • T is 3 cycle of regular components of size k • i, Si(T) k + k/2 • Let , choose (T) • One component in T’ is transitive • is.t. Si(T’)=k + (k-1), winner doesn’t change • The ratio tends to ¾ ’ T k = 5

  8. Randomized Upper Bound • Can we do very well in the randomized model? • Theorem. No randomization over trees can achieve approx ratio better than 5/6 + O(1/m) • Proof by using similar ideas plus Yao’s minimax principle

  9. Randomized lower bound • Main theorem.  admissible randomization over voting trees of polynomial size with an approximation ratio of ½-O(1/m) • Important to keep the trees small from CS point of view

  10. Spot the fake caterpillar • 1-Caterpillar is a singleton tree • k-Caterpillar is a binary tree where left child of root is (k-1)-caterpillar, and right child is a leaf • Voting k-caterpillar is a k-caterpillar whose leaves are labeled by A ? ? ? 1 ? 5 ? 4 3 4 2

  11. Randomized voting caterpillars • k-RSC: uniform distribution over surjective voting k-caterpillars • Main theorem reformulated. k-RSC with k=poly(m) has approx ratio of ½-O(1/m) • Sketchiest proof ever: • k-RSC close to k-RC • k-RC identical to k steps of Markov chain • k = poly(m) steps of chain close to stationary dist. of chain (rapid mixing, via spectral gap + conductance) • Stationary distribution of chain gives ½-approx of Copeland

  12. אם בארזים נפלה שלהבת • Permutation trees give (log(m)/m)-approx • Huge randomized balanced trees intuitively do very well • Theorem. Arbitrarily large random balanced voting trees give an approx ratio of at most O(1/m) 5 7 2 1 3 4 8 6 9

  13. Open problems • Paper contains many additional results • Randomized model: gap between LB of ½ (admissible, small) and UB of 5/6 (even inadmissible and large) • Deterministic: enigmatic gap between LB of (logm/m) and UB of ¾

More Related