From 17c5d79825eed833df2401ce7d7d2424294e4b6f Mon Sep 17 00:00:00 2001
From: "jane.tan" <jane.tan@maths.ox.ac.uk>
Date: Mon, 14 Mar 2022 18:41:20 +0000
Subject: [PATCH] tiny extra edits in tw section
---
arxiv_cbd.tex | 14 +++++++-------
1 file changed, 7 insertions(+), 7 deletions(-)
diff --git a/arxiv_cbd.tex b/arxiv_cbd.tex
index 83e5749..927607e 100644
--- a/arxiv_cbd.tex
+++ b/arxiv_cbd.tex
@@ -858,9 +858,9 @@ Before going further, let us recall some notions about treewidth.
$(T,\beta)$, where $T$ is a rooted tree and $\beta:V(T)\to 2^{V(G)}$
assigns a \emph{bag} to each of its nodes, such that
\begin{itemize}
-\item for each $uv\in E(G)$, there exists $x\in V(T)$ such that
+\item for each edge $uv\in E(G)$, there exists $x\in V(T)$ such that
$u,v\in\beta(x)$, and
-\item for each $v\in V(G)$, the set $\{x\in V(T):v\in\beta(x)\}$ is
+\item for each vertex $v\in V(G)$, the set $\{x\in V(T):v\in\beta(x)\}$ is
non-empty and induces a connected subtree of $T$.
\end{itemize}
For nodes $x,y\in V(T)$, we write $x\preceq y$ if $x=y$ or $x$ is a descendant of $y$ in $T$.
@@ -877,13 +877,13 @@ how to prove that disk graphs of thickness $t$ are fractionally treewidth-fragil
We first consider unit disk graphs.
By partitioning the plane with a random grid $\HH$ with square cells of side-length $2k$, any unit disk has probability $1/2k$
of intersecting a vertical (resp. horizontal) line of the grid. Using a union bound, any disk has probability at most $1/k$ of intersecting
-the grid. Using this probability distribution, we show that removing the disks intersected by the grid leads to a
+the grid. Using this probability distribution, we show that removing the disks that intersect the grid leads to a
unit disk graph of bounded treewidth. Indeed, in such a graph any connected component corresponds to unit disks contained in the
same cell of the grid. Each cell has area bounded by $4k^2$, so there are at most $16tk^2/\pi$ disks contained in a cell.
-This bounds the size of any connected component, and so the treewidth is also bounded.
+This bounds the size of any connected component, and the treewidth is also bounded.
-Note that the above distribution also works if we are given
-disks whose diameter lie in a certain range. That is, for any diameter $\delta$ with $1/c \le \delta \le 1$, applying the same process
+The preceding distribution also works if we are given
+disks whose diameter lie in a certain range. For any diameter $\delta$ with $1/c \le \delta \le 1$, applying the same process
with a random grid of $2k\times 2k$ cells ensures that any disk is deleted with probability at most $1/k$, and the
connected components have size at most $4tc^2k^2/\pi$. Dealing with arbitrary disk graphs (with any diameter $\delta$ in the range
$0< \delta \le 1$) necessitates deleting more disks. This can be handled by partitioning each $(2k\times 2k)$-cell in a quadtree-like manner.
@@ -891,7 +891,7 @@ Now a disk with diameter between $\ell /2$ and $\ell$ (with $\ell =1/2^i$ for so
in a $(2k\ell \times 2k\ell)$-cell of a quadtree. It is straightforward to see that each disk is deleted with probability at most $1/k$.
To prove that the remaining graph has bounded treewidth, one should consider the following tree decomposition $(T,\beta)$. Here, the tree $T$ is obtained by linking the roots of the quadtrees used (as trees) to a new common root.
Then for a $(2k\ell \times 2k\ell)$-cell $C$, $\beta(C)$ contains all disks of diameter at least $\ell/2$ intersecting $C$.
-To see that such bag is bounded, consider the $((2k+1)\ell \times (2k+1)\ell)$ square $C'$ centered on $C$, and note that any
+To see that such bag is bounded, consider the $(2k+1)\ell \times (2k+1)\ell$ square $C'$ centered on $C$, and note that any
disk in $\beta(C)$ intersects $C'$ on an area at least $\pi\ell^2/16$. This implies that $|\beta(C)| \le 16t(2k+1)^2 / \pi$.
Let us now give a detailed proof in a more general setting.
--
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