Halving by a Thousand Cuts or Punctures

$\newcommand{\Arr}{\mathcal{A}} \newcommand{\numS}{k} \newcommand{\ArrX}[1]{\Arr(#1)} \newcommand{\eps}{\varepsilon} \newcommand{\opt}{\mathsf{o}}$ For point sets $P_1, \ldots, P_\numS$, a set of lines $L$ is halving if any face of the arrangement $\ArrX{L}$ contains at most $|P_i|/2$ points of $P_i$, for all $i$. We study the problem of computing a halving set of lines of minimal size. Surprisingly, we show a polynomial time algorithm that outputs a halving set of size $O(\opt^{3/2})$, where $\opt$ is the size of the optimal solution. Our solution relies on solving a new variant of the weak $\eps$-net problem for corridors, which we believe to be of independent interest. We also study other variants of this problem, including an alternative setting, where one needs to introduce a set of guards (i.e., points), such that no convex set avoiding the guards contains more than half the points of each point set.

No reviews yet.

No comments yet.