Abstract

Related DatabasesWeb of Science You must be logged in with an active subscription to view this.Article DataHistorySubmitted: 9 January 2019Accepted: 16 February 2021Published online: 27 May 2021Keywordsonline algorithms, convex optimization, finite metric spaceAMS Subject Headings68W27Publication DataISSN (print): 0097-5397ISSN (online): 1095-7111Publisher: Society for Industrial and Applied MathematicsCODEN: smjcat

Highlights

  • Let (X, d) be a finite metric space with | X| = n > 1

  • An online algorithm is a sequence of mappings \bfitrho = \langle \rho 1, \rho 2, . . . , \rangle where, for every t \geqslan 1, \rho t : (\BbbR X+ )t \rightar X maps a sequence of cost functions \langle c1, . . . , ct\rangle to a state

  • Since one can assume that D \leqslan O(log n) for an n-point hierarchically separated trees (HSTs) metric, the mirror descent framework yields an arguably simpler O((log n)2)-competitive algorithm for arbitrary HSTs that, satisfies the refined guarantees of Theorem 1.2

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Summary

Introduction

The cost of the offline optimum, denoted \sansc \sanso \sans \sanst \ast (\bfitc ), is the infimum of \sum t\geqslan1[ct(\rho t) + d(\rho t - 1, \rho t)] over any sequence \langle \rho t : t \geqslan 1\rangle of states. A randomized online algorithm \bfitrho is said to be \alpha -competitive if for every \rho 0 \in X there is a constant \beta > 0 such that for all cost sequences \bfitc. There is an O(D log n)-competitive randomized algorithm for MTS on any n-point tree metric with combinatorial depth D. Since one can assume that D \leqslan O(log n) for an n-point HST metric (see [BBMN15]), the mirror descent framework yields an arguably simpler O((log n)2)-competitive algorithm for arbitrary HSTs that, satisfies the refined guarantees of Theorem 1.2. Competitive algorithms for unfair task systems are useful in constructing algorithms for HSTs, where rx is a proxy for the competitive ratio of an algorithm on MTS instances defined in a subtree rooted at x

METRICAL TASK SYSTEMS ON TREES
Lagrange multiplier corresponding to the constraint
We prove that for almost all t
Iu has measure zero for each
We conclude
We also note that
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