z-logo
Premium
The longest segment in the complement of a packing
Author(s) -
Böröczky K.,
Tardos G.
Publication year - 2002
Publication title -
mathematika
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.955
H-Index - 29
eISSN - 2041-7942
pISSN - 0025-5793
DOI - 10.1112/s002557930001603x
Subject(s) - complement (music) , citation , mathematics , combinatorics , library science , computer science , chemistry , biochemistry , complementation , gene , phenotype
Let K be a compact convex body in R not contained in a hyperplane, and denote the norm whose unit ball is 1 2 (K − K) by ‖ · ‖K . Given a translative packing of K, we are interested in how long segments (with respect to ‖ · ‖K) lie in the complement of the interiors of the translates. The main result of this note is showing the existence of a translative packing such that the length of the longest segments avoiding it is only exponential in the dimension n (see below). We start here with a lower bound showing that this bound is close to optimal for balls. We show that any packing of the unit Euclidean ball B avoids a segment of length exponential in n. It is a rather interesting question to find how long segments necessarily exist that avoid any packing of any convex, open body in R . Our lower bound proof does not work for bodies allowing dense packings. Let | · | denote the n–dimensional Lebesgue measure. Let us consider any packing of B, and denote the area and the packing density of the unit ball by κn and δ(B ), respectively. Choose a unit segment s, and denote the projection of B into some hyperplane orthogonal to s by B, and set

This content is not available in your region!

Continue researching here.

Having issues? You can contact us here