A counterexample to the DemyanovRyabova conjecture 25 August 2018 - - PowerPoint PPT Presentation

A counterexample to the Demyanov–Ryabova conjecture

25 August 2018 AVOCADO, Newcastle

Vera Roshchina UNSW Sydney v.roshchina@unsw.edu.au

Demyanov Converter

For a finite set of polytopes Ω, C(Ω) =

• conv

P∈Ω Arg max x∈P

x, g, g = 0

• .

Arg max

x∈P1

x, g Arg max

x∈P1

x, g P1 P2 g Application: conversion between upper and lower envelopes, min

P∈Ω max u∈P x, u =

max

P∈C(Ω) min v∈Qx, v.

The Demyanov–Ryabova conjecture

The sequence Ω0, Ω1, Ω2, . . . , where Ωi+1 = C(Ωi), eventually cycles, so ΩN+L = ΩN for some N, L ≥ 0. Conjecture: The length of the minimal cycle is at most 2. True when

• all vertices are affinely independent (T. Sang);
• all vertices are extreme points of conv Ω0 and Ω0 contains all poly-

topes of minimal cardinality (A. Daniilidis and C. Petitjean).

Demyanov, Vladimir F .; Ryabova, Julia A. Exhausters, coexhausters and converters in nonsmooth analysis. Discrete Contin. Dyn. Syst. (2011). Sang, Tian On the conjecture by Demyanov-Ryabova in converting finite exhausters.

• J. Optim. Theory Appl. (2017).

Daniilidis, Aris; Petitjean, Colin A partial answer to the Demyanov-Ryabova con-

• jecture. Set-Valued Var. Anal. (2018).

Counterexample

P1 P2 P3 P4 y x

For Ω0 = {P1, P2, P3, P4} we have Ω1 = Ω5, but Ω1 = Ω3.

Roshchina, V ., The Demyanov–Ryabova conjecture is false, Optim. Lett. (2018).

Ω1 = Ω5

2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3

Ω3

2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3 2 2 1 1 2 3