45 lines
2.0 KiB
TeX
45 lines
2.0 KiB
TeX
\documentclass[12pt]{article}
|
|
% \usepackage{chao}
|
|
\usepackage[sans]{xenotes}
|
|
% \usepackage{natbib}
|
|
|
|
\title{Exercises in Sariel Har-Peled's \\ \textit{Geometric Approximation Algorithms}}
|
|
\author{}
|
|
\date{}
|
|
|
|
\begin{document}
|
|
\maketitle
|
|
% \tableofcontents
|
|
% \newpage
|
|
|
|
For errata and more stuff, see \url{https://sarielhp.org/book/}
|
|
|
|
\section{Grid}
|
|
\begin{exercise}\label{ex1.1}
|
|
Let $P$ be a max cardinality point set contained in the $d$-dimensional unit hypercube such that the smallest distance of point pairs in $P$ is 1. Prove that
|
|
\[\left( \floor{\sqrt{d}}+1 \right)^d \leq |P|\leq \left( \ceil{\sqrt{d}}+1 \right)^d. \]
|
|
\end{exercise}
|
|
hmm... the first exercise in this book is wrong. See \url{https://sarielhp.org/book/errata.pdf}.
|
|
The stated lowerbound is actually an upperbound.
|
|
\begin{proof}
|
|
We evenly partition the $[0,1]$ interval into $m=\left( \floor{\sqrt{d}}+1 \right)$ small segments for each of the $d$ axes. The unit hypercube is partitioned into $m^d$ cells. The length of each cell's diagonal is $\sqrt{\frac{d}{m^2} }< 1$. Thus there is at most one point of $P$ in each cell and there are $\left( \floor{\sqrt{d}}+1 \right)^d$ cells.
|
|
|
|
For lowerbound, one can construct a solution of size $2^d$ by selecting vertices of the hypercube. For sufficient large $d$ one can find a solution of size $(\sqrt{d}/5)^d$.\footnote{Exercise 1.1 (C) in \url{https://sarielhp.org/book/chapters/min_disk.pdf}}
|
|
Let point set $P$ be the optimal solution and let $n=|P|$. We place a $d$-dimensional unit sphere around each point of $P$. These $n$ spheres must cover the unit hypercube since otherwise we can add more points into $P$. Thus one has $n\vol(1b^d)\geq 1$.
|
|
\begin{equation*}
|
|
\begin{aligned}
|
|
n &\geq 1/\vol(1b^d)\\
|
|
&= \frac{\Gamma(d/2+1)}{\pi^{d/2}}\\
|
|
&\geq \sqrt{2\pi/(d/2+1)} (\frac{\sqrt{d}}{\sqrt{2e\pi}})^{d}
|
|
\end{aligned}
|
|
\end{equation*}
|
|
The last line is greater than $(\sqrt{d}/5)^d$ for large enough $d$.
|
|
\end{proof}
|
|
|
|
\begin{exercise}
|
|
Compute clustering radius
|
|
\end{exercise}
|
|
|
|
|
|
\end{document}
|