Commit 17ff0baa authored by Zdenek Dvorak's avatar Zdenek Dvorak
Browse files

Added the treewidth argument.

parent 0b81ca9b
......@@ -138,12 +138,82 @@ In fact, we will prove the following stronger fact (TODO: Was this published som
\begin{lemma}\label{lemma-tw}
Let $(T,\beta)$ be a tree decomposition of a graph $G$ of width $t$.
Then $G$ has a touching representation $h$ by hypercubes in $R^{t+1}$ such that
for $u,v\in V(G)$, if $p(u)\preceq p(v)$, then $h(u)\sqsubseteq h(v)$.
Then $G$ has a touching representation $h$ by axis-aligned hypercubes in $R^{t+1}$ such that
for $u,v\in V(G)$, if $p(u)\prec p(v)$, then $h(u)\sqsubset h(v)$.
Moreover, the representation can be chosen so that no two hypercubes have the same size.
\end{lemma}
\begin{proof}
...
Without loss of generality, we can assume that the root has a bag of size one
and that for each $x\in V(T)$ with parent $y$, we have $|\beta(x)\setminus\beta(y)|=1$
(if $\beta(x)\subseteq \beta(y)$, we can contract the edge $xy$; if $|\beta(x)\setminus\beta(y)|>1$,
we can subdivide the edge $xy$, introducing $|\beta(x)\setminus\beta(y)|-1$ new vertices,
and set their bags appropriately). It is now natural to relabel the vertices of $G$
so that $V(G)=V(T)$, by giving the unique vertex in $\beta(x)\setminus\beta(y)$
the label $x$. In particular, $p(x)=x$. Furthermore, we can assume that $y\in\beta(x)$.
Otherwise, if $y$ is not the root of $T$, we can replace the edge $xy$ by the edge from $x$
to the parent of $y$. If $y$ is the root of $T$, then the subtree rooted in $x$ induces a
union of connected components in $G$, and we can process this subtree separately from the
rest of the graph (being careful to only use hypercubes smaller than the one representing $y$
and of different sizes from those used on the rest of the graph).
Let us now greedily color $G$ by giving $x$ a color different from the colors of
all other vertices in $\beta(x)$; such a coloring $\varphi$ uses only colors
$\{1,\ldots,t+1\}$.
Let $D=4\Delta(T)+1$. Let $V(G)=V(T)=\{x_1,x_2,\ldots, x_n\}$, where
for every $i<j$, $x_i$ and $x_j$ are either incomparable in $\prec$ or
$x_j\prec x_i)$; in particular, $x_1$ is the root of $T$. Let $\varepsilon=D^{-n-1}$.
Let $s_i=D^{-i}$; we will represent $x_i$ by a hypercube $h(x_i)$ with edges of length $s_i$.
Additionally, we will need to consider larger hypercubes around $h(x_i)$; let $h'(x_i)$
be the hypercube with sides of length $2s_i$ and with $\min(h'(x_i)[j])=\min(h(x_i)[j])$
for $j\in\{1,\ldots, t+1\}$, and $h''(x_i)$ the hypercube with sides of length $2s_i+\epsilon$
and with $\min(h''(x_i)[j])=\min(h(x_i)[j])-\varepsilon$. We will construct the representation $h$ so that the following
invariant is satisfied:
\begin{itemize}
\item[(a)] For each $x,z\in V(T)$ such that $x\prec z$, we have $h'(x)\subset h''(z)$.
\item[(b)] For each $y\in V(T)$ and distinct children $x$ and $z$ of $y$, we have $h''(x)\cap h''(z)=\emptyset$.
\end{itemize}
Note that this ensures that if $x$ and $z$ are vertices of $T$ and $h(x)\cap h(z)\neq\emptyset$, then $x\prec z$ or $z\prec x$.
We now construct the representation $h$. For the root $x_1$ of $T$, $h(r)$ is an arbitrary hypercube with sides
of length $s_1$. Assuming now we have already selected $h(y)$ for a vertex $y\in V(T)$, the hypercube $h(x_i)$ with sides of length $s_i$
for a child $x_i$ of $y$ is chosen as follows. For $j\in\{1,\ldots, t+1\}$,
\begin{itemize}
\item[(i)] if $j=\varphi(w)$ for $w\in\beta(x_i)\setminus\{x_i\}$, we choose $h(x_i)[j]$ so that
$\min(h(x_i)[j])=\max(h(w)[j])$ if $xw\in E(G)$ and so that $\min(h(x_i)[j])=\max(h(w)[j]) + \varepsilon$ otherwise.
\item[(ii)] if $j$ is different from the colors of all vertices in $\beta(x_i)\setminus\{x_i\}$,
then we choose $h(x_i)[j]$ so that $h''(x_i)[j]$ is a subset of the interior of $h(y)[j]$. The interval $h''(x_i)[[j]$
is furthermore chosen to be disjoint from $h''(x_m)[j]$ for any other child $x_m$ of $y$;
this is always possible by the choice of $D$, $s_i$, and $s_m$.
\end{itemize}
Note that (ii) always applies for $j=\varphi(x_i)$ and this ensures that the invariant (b) holds.
For the invariant (a), note that in the case (ii), we ensure $h''(x_i)[[j]\subseteq h(y)[j]$ and
we have $h(y)[j]\subseteq h''(z)[j]$ by the invariant (a) for $y$ and $z$. In the case (i),
if $z\prec w$, then we have $w\in\beta(z)\setminus\{z\}$ and
$\min(h(x_i)[j]),\min(h(z)[j])\in\{\max(h(w)[j]),\max(h(w)[j])+\varepsilon\}$.
If $w\preceq z$, then note we choose $h'(x_i)[j]\subseteq h'(w)[j]$ by (i) and that
we have $h'(w)[j]\subset h''(z)[j]$ by (a). This verifies that the invariant (a)
also holds at $x_i$.
Consider now two adjacent vertices of $G$, say $x_i$ and $w$. Note that any two
adjacent vertices are comparable in $\prec$, and thus we can assume $x_i\prec w$
and $w\in\beta(x_i)$.
By (i), for $j=\varphi(w)$, the intervals $h(x_i)[j]$ and $h(w)[j]$
intersect in a single point. If $j\neq \varphi(w)$, then let $w_1$ be the child of $w$ on the path in $T$ from
$w$ to $x_i$. If no vertex in $\beta(w_1)\setminus\{w_1\}$ has color $j$, then by (ii), we have $h''(w_1)[j]\subset h(w)[j]$,
Otherwise, $z\in \beta(w_1)\setminus\{w_1\}$ such that $\varphi(z)=j$; clearly, $w\prec z$ and $z\in\beta(w)\setminus\{w\}$.
We have $\min(h(w_1)[j]),\min(h(w)[j])\in\{\max(h(z)[j]),\max(h(z)[j])+\varepsilon\}$ by (i).
Hence, $\min(h(w)[j])-2\epsilon\le \min(h''(w_1)[j])\le \max(h''(w_1)[j])\le \max(h(w)[j])$.
Since $h(x_i)[j]\subseteq h''(w_1)[j]$ by (a) and the length of the interval $h(x_i)[j]$ is greater than $2\varepsilon$,
we conclude that $h(x_i)[j]\cap h(w)[j]\neq\emptyset$. Therefore, the boxes $h(w)$ and $h(x_i)$ touch.
Consider now two non-adjacent vertices of $G$, say $x_i$ and $w$. As we noted before, if $x_i$ and $w$ are
incomparable in $\prec$, then (a) and (b) implies that the boxes $h(x_i)$ and $h(w)$ are disjoint.
Suppose now that say $x_i\prec w$, and let $j=\varphi(w)$. If $w\in\beta(x_i)$, then $h(x_i)[j]$ and $h(w)[j]$ are disjoint by (i).
Otherwise, let $y$ be the last vertex on the path from $w$ to $x_i$ in $T$ such that $w\in\beta(y)$ and let $z$ be the child of $y$
on this path. As we argued in the first paragraph, $y\neq w$, and by (i), the interior of $h(y)[j]$ is disjoint from $h(w)[j]$.
By (ii), $h''(z)[j]$ is contained in the interior of $h(y)[j]$. By (a), we conclude that $h(x_i)[j]\subseteq h''(z)[j]$,
implying that the boxes $h(x_i)$ and $h(w)$ are disjoint. Therefore, $h$ is a touching representation of $G$.
\end{proof}
Next, let us deal with clique-sums. A \emph{clique-sum} of two graphs $G_1$ and $G_2$ is obtained from their disjoint union
......@@ -191,14 +261,14 @@ and thus $p(x)=x$.
We are now ready to deal with the clique-sums.
\begin{theorem}
\begin{theorem}\label{thm-cs}
If $G$ is obtained from graphs in a class $\GG$ by clique-sums, then there exists a graph $G'$ such that
$G\subseteq G'$ and $\cbdim(G')\le (\cbdim(\GG)+1)(\omega(\GG)+1)$.
\end{theorem}
\begin{proof}
Let $(T,\beta)$ be a tree decomposition of $G$ over $\GG$; the adhesion $a$ of $(T,\beta)$ is at most $\omega(\GG)$.
By Lemma~\ref{lemma-legraf}, $(T,\pi)$ is a tree decomposition of $T_\beta$ of width at most $a$. By Lemma~\ref{lemma-tw},
$T_\beta$ has a touching representation $h$ by hypercubes in $\mathbb{R}^{a+1}$ such that
$T_\beta$ has a touching representation $h$ by axis-aligned hypercubes in $\mathbb{R}^{a+1}$ such that
$h(x)\sqsubseteq h(y)$ whenever $x\preceq y$.
Since $T_\beta$ has treewidth at most $a$, it has a proper coloring $\varphi$ by colors $\{0,\ldots,a\}$.
For every $x\in V(T)$, let $f_x$ be a touching representation of the torso $G_x$ of $x$ by comparable boxes in $\mathbb{R}^d$,
......@@ -233,6 +303,8 @@ Moreover, $x\prec y$, implying that $E_i(u)\cap E_i(v)\neq\emptyset$ for $i=0,\l
$xy\in E(T_\beta)$. Hence, again we have $f(u)\cap f(v)\neq\emptyset$.
\end{proof}
Note that in Theorem~\ref{thm-cs}, we only get a representation of a supergraph
of $G$.
%We will need the fact that the chromatic number is at most exponential in the comparable box dimension;
%this follows from~\cite{subconvex} and we include the argument to make the dependence clear.
......@@ -249,7 +321,7 @@ $xy\in E(T_\beta)$. Hence, again we have $f(u)\cap f(v)\neq\emptyset$.
%\end{proof}
\section{Exploiting the product structure}
\section{The product structure and minor-closed classes}
\subsection*{Acknowledgments}
This research was carried out at the workshop on Geometric Graphs and Hypergraphs organized by Yelena Yuditsky and Torsten Ueckerdt
......
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