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7 votes
0 answers
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Approximating any convex shape in $\mathbb{R}^d$ with a polytope having $\mathrm{poly}(d)$ facets

We denote by $V(A)$ the $d$-volume of any convex set $A$. Furthermore, given any two convex sets $A,B\in\mathbb{R}^d$, we denote by $V_{A,B}$ the $d$-volume of the symmetric difference $V\left(A \...
Penelope Benenati's user avatar
7 votes
0 answers
122 views

Discrepancy of the finite approximation of the Lebesgue measure

Let $\mu$ be a probabilistic measure on the unit square $Q$ which is the average of $N$ delta-measures in some points in this square; let $\lambda$ denote the Lebesgue measure on $Q$. What is the rate ...
Fedor Petrov's user avatar
4 votes
0 answers
144 views

Approximation of a convex shape in the $d$-dimensional Euclidean space for $d\gg 1$

We are given a convex shape $C$ lying inside the hypercube $[0,1]^d$ in the $d$-dimensional Euclidean space. Let the volume of $C$ be $\tfrac12$ (I guess nothing changes for any other fixed constant ...
Penelope Benenati's user avatar
4 votes
0 answers
94 views

Finding closest set of K disjoint hyperspheres to a point in $\mathbb{R}^n$ with uniform radius

I am interested in the following problem: in $\mathbb{R}^n$, we have $N$ overlapping hyperspheres all with the same radius. Given a point $p$ in $\mathbb{R}^n$, the objective is to find the $K$ non ...
eagle34's user avatar
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3 votes
0 answers
187 views

Approximating any $d$-dimensional convex shape that occupies a constant fraction of its bounding box with a polytope having $\mathrm{poly}(d)$ facets

Given any convex set $A\in\mathbb{R}^d$, we denote by $V(A)$ its $d$-volume. Furthermore, given any two convex sets $A_1,A_2\in\mathbb{R}^d$, we denote by $V_{A_1,A_2}$ the $d$-volume of the symmetric ...
Penelope Benenati's user avatar
2 votes
0 answers
59 views

Totally distance non-preserving transformations

JL lemma (https://en.wikipedia.org/wiki/Johnson%E2%80%93Lindenstrauss_lemma) guarantees if you have a set of $K$ points in $\Bbb R^N$ a random transformation guarantees that the set can be projected ...
Turbo's user avatar
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