Commit 0b81ca9b authored by Zdenek Dvorak's avatar Zdenek Dvorak
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Improved readability of the clique-sum argument.

parent e743a8cd
......@@ -68,7 +68,8 @@ For example, all complete bipartite graphs $K_{n,m}$ are touching graphs of axis
one part are represented by $m\times 1\times 1$ boxes and the vertices of the other part are represented by $1\times n\times 1$
boxes (a \emph{box} is the cartesian product of intervals of non-zero length).
Dvo\v{r}\'ak, McCarty and Norin~\cite{subconvex} noticed that this issue disappears if we forbid such a combination of
long and wide boxes: Two boxes are \emph{comparable} if a translation of one of them is a subset of the other one.
long and wide boxes: For two boxes $B_1$ and $B_2$, we write $B_1 \sqsubseteq B_2$ if a translation of $B_1$ is a subset of $B_2$.
We say that $B_1$ and $B_2$ are \emph{comparable} if $B_1\sqsubseteq B_2$ or $B_2\sqsubseteq B_1$.
A \emph{touching representation by comparable boxes} of a graph $G$ is a touching representation $f$ by boxes
such that for every $u,v\in V(G)$, the boxes $f(u)$ and $f(v)$ are comparable. For a graph $G$, let the \emph{comparable box dimension} $\cbdim(G)$
of $G$ be the smallest integer $d$ such that $G$ has a touching representation by comparable boxes in $\mathbb{R}^d$.
......@@ -112,20 +113,6 @@ where $M$ is chosen large enough so that $f(u)\subseteq [-M,M] \times \cdots \ti
Then $h$ is a touching representation of $G$ by comparable boxes in $\mathbb{R}^{d+1}$.
\end{proof}
%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.
%\begin{lemma}\label{lemma-chrom}
%If $G$ has a comparable box representation $f$ in $\mathbb{R}^d$, then $G$ is $3^d$-colorable.
%\end{lemma}
%\begin{proof}
%We actually show that $G$ is $(3^d-1)$-degenerate. Since every induced subgraph of $G$ also
%has a comparable box representation in $\mathbb{R}^d$, it suffices to show that the minimum degree of $G$
%is less than $3^d$. Let $v$ be a vertex of $G$ such that $f(v)$ has the smallest volume. For every neighbor $u$ of $v$,
%there exists a translation $B_u$ of $f(v)$ such that $B_u\subseteq f(u)$ and $B_u$ touches $f(v)$.
%Note that $f(v)\cup \bigcup_{u\in N(v)} B_u$ is a union of internally disjoint translations of $f(v)$ contained in
%a box obtained from $f(v)$ by scaling it by a factor of three, and thus $1+|N(v)|\le 3^d$.
%\end{proof}
We need a bound on the clique number in terms of the comparable box dimension.
For a box $B=I_1\times \cdots I_d$ and $i\in\{1,\ldots,d\}$, let $B[i]=I_i$.
\begin{lemma}\label{lemma-cliq}
......@@ -142,6 +129,7 @@ such that
\item for each $uv\in E(G)$, 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 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 $T$.
For each vertex $v\in V(G)$, let $p(v)$ be the node $x\in V(T)$ such that $v\in \beta(x)$ nearest to the root of $T$.
The \emph{adhesion} of the tree decomposition is the maximum of $|\beta(x)\cap\beta(y)|$ over distinct $x,y\in V(T)$,
and its \emph{width} is the maximum of the sizes of the bags minus $1$. The \emph{treewidth} of a graph is the minimum
......@@ -151,8 +139,8 @@ 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 $p(u)\neq p(v)$ and $p(u)$ is an ancestor of $p(v)$ in $T$,
then $\vol(h(u))>\vol(h(v))$.
for $u,v\in V(G)$, if $p(u)\preceq p(v)$, then $h(u)\sqsubseteq h(v)$.
Moreover, the representation can be chosen so that no two hypercubes have the same size.
\end{lemma}
\begin{proof}
...
......@@ -178,6 +166,7 @@ For each note $x\in V(T)$, let $\pi(x)=\{x\}\cup \{p(v):v\in\beta(x)\}$. Let $T
$xy\in E(T_\beta)$ if and only if $x\in\pi(y)$ or $y\in\pi(x)$.
\begin{lemma}\label{lemma-legraf}
If $(T,\beta)$ is a tree decompositon of $G$ of adhesion $a$, then $(T,\pi)$ is a tree decomposition of $T_\beta$ of width at most $a$.
Moreover, $\pi(x)$ is a clique in $T_\beta$ and $p(x)=x$ for each $x\in V(T)$.
\end{lemma}
\begin{proof}
For each edge $xy\in E(T_\beta)$, we have $x,y\in \pi(x)$ or $x,y\in \pi(y)$ by definition.
......@@ -191,6 +180,13 @@ Consider a node $x\in V(T)$. Note that for each $v\in \beta(x)$, the vertex $p(
In particular, if $x$ is the root of $T$, then $\pi(x)=\{x\}$. Otherwise, if $y$ is the parent of $x$ in $T$, then
$p(v)=x$ for every $v\in \beta(x)\setminus\beta(y)$, and thus $|\pi(x)|\le |\beta(x)\cap \beta(y)|+1\le a+1$.
Hence, the width of $(T,\pi)$ is at most $a$.
Suppose now that $y$ and $z$ are distinct vertices in $\pi(x)$. Then both $y$ and $z$ are ancestors of $x$ in $T$,
and thus without loss of generality, we can assume that $y\preceq z$. If $y=x$, then $yz\in E(T_\beta)$ by definition.
Otherwise, there exist vertices $u,v\in \beta(x)$ such that $p(u)=y$ and $p(v)=z$. Since $v\in \beta(x)\cap\beta(z)$
and $y$ is on the path in $T$ from $x$ to $z$, we also have $v\in\beta(y)$. This implies $z\in\pi(y)$ and $yz\in E(T_\beta)$.
Hence, $\pi(x)$ is a clique in $T_\beta$. Moreover, note that $x\not\in \pi(w)$ for any ancestor $w\neq x$ of $x$ in $T$,
and thus $p(x)=x$.
\end{proof}
We are now ready to deal with the clique-sums.
......@@ -202,37 +198,56 @@ $G\subseteq G'$ and $\cbdim(G')\le (\cbdim(\GG)+1)(\omega(\GG)+1)$.
\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 $\mathbb{R}^{a+1}$. Moreover,
letting $\prec$ be a linear ordering on $V(T)$ in a non-decreasing order according to the volume of the hypercubes assigned by $h$,
we have that $x\prec y$ whenever $x$ is a descendant of $y$ in $T$.
$T_\beta$ has a touching representation $h$ by 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 of $x$ by comparable boxes in $\mathbb{R}^d$
for $d=\cbdim(\GG)$. We scale and translate the representations so that for every $x\in V(T)$ and $i\in\{0,\ldots,a\}$,
For every $x\in V(T)$, let $f_x$ be a touching representation of the torso $G_x$ of $x$ by comparable boxes in $\mathbb{R}^d$,
where $d=\cbdim(\GG)$. We scale and translate the representations so that for every $x\in V(T)$ and $i\in\{0,\ldots,a\}$,
there exists a box $E_i(x)$ such that
\begin{itemize}
\item whenever $x\prec y$, we have $E_i(x)\subseteq E_i(y)$ and a translation of $E_i(x)$ is a subset of every box of
the representation $f_y$ whenever $x\prec y$,
\item if $i=\varphi(x)$, then all boxes of $f_x$ are subsets of $E_i(x)$, and
\item if $i\neq p(v)$ and $K=\{v\in\beta(x):\varphi(p(v))=i\}$, then letting $y=p(v)$ for $v\in K$ (and noting that this $y$
is unique, since $\varphi$ is a proper coloring of $T_\beta$ and that $K$ is a clique in $G_y$), the box $E_i(x)$
contains a point belonging to $\bigcap_{v\in K} f_y(v)$.
\item[(a)] for distinct $x,y\in V(T)$, if $h(x)\sqsubset h(y)$, then $E_i(x) \sqsubset E_i(y)$ and $E_i(x) \sqsubset f_y(v)$ for every $v\in \beta(y)$,
\item[(b)] if $x\prec y$, then $E_i(x)\subseteq E_i(y)$,
\item[(c)] if $i=\varphi(x)$, then $f_x(v)\subseteq E_i(x)$ for every $v\in\beta(x)$, and
\item[(d)] if $i\neq \varphi(x)$ and $K=\{v\in\beta(x):\varphi(p(v))=i\}$ is non-empty, then letting $y=p(v)$ for $v\in K$,
the box $E_i(x)$ contains a point belonging to $\bigcap_{v\in K} f_y(v)$.
\end{itemize}
Some explanation is in order for the last point: Firstly, since $\pi(x)$ is a clique in $T_\beta$, there
exists only one vertex $y\in \pi(x)$ of color $i$, and thus $y=p(v)$ for all $v\in K$.
Moreover, $K$ is a clique in $G_y$, and thus $\bigcap_{v\in K} f_y(v)$ is non-empty. Lastly,
note that if $x\prec z\prec y$, then $K=\beta(x)\cap\beta(y)\subseteq \beta(z)\cap \beta(y)$,
and thus $E_i(z)$ was also chosen to contain a point of $\bigcap_{v\in K} f_y(v)$;
hence, a choice of $E_i(x)$ satisfying $E_i(x)\subseteq E_i(z)$ as required by (b) is possible.
Let us now define $f(v)=h(p(v))\times E_0(v)\times \cdots\times E_a(v)$ for each $v\in V(G)$,
where $E_i(v)=f_{p(v)}(v)$ if $i=\varphi(p(v))$ and $E_i(v)=E_i(p(v))$ otherwise. We claim this gives a touching representation
of a supergraph of $G$ by comparable boxes in $\mathbb{R}^{(d+1)(a+1)}$. First, note that the boxes are indeed comparable;
if $p(u)=p(v)$, then this is the case since $f_{p(v)}$ is a representation by comparable boxes, and if say
$p(u)\prec p(v)$, then this is due to the scaling of $f_{p(u)}$. Next, let us argue $f(u)$ and $f(v)$ have disjoint
$p(u)\prec p(v)$, then this is due to (a) and (c). Next, let us argue $f(u)$ and $f(v)$ have disjoint
interiors. If $p(u)=p(v)$, this is the case since $f_{p(v)}$ is a touching representation, and if $p(u)\neq p(v)$,
then this is the case because $h$ is a touching representation. Finally, suppose that $uv\in E(G)$. Let $x$
be the node of $T$ nearest to the root such that $u,v\in \beta(x)$. Without loss of generality, $p(u)=x$.
Let $y=p(v)$. If $x=y$, then $f(u)\cap f(v)\neq\emptyset$, since $f_x$ is a touching representation of $G_x$.
Let $y=p(v)$. If $x=y$, then $f(u)\cap f(v)\neq\emptyset$, since $f_x$ is a touching representation of $G_x$ and $uv\in E(G_x)$.
If $x\neq y$, then $y\in\pi(x)$ and $xy\in E(T_\beta)$, implying that $h(x)\cap h(y)\neq\emptyset$.
Moreover, $x\prec y$, implying that $E_i(u)\cap E_i(v)\neq\emptyset$ for $i=0,\ldots,a$.
Hence, again we have $f(u)\cap f(v)\neq\emptyset$.
Moreover, $x\prec y$, implying that $E_i(u)\cap E_i(v)\neq\emptyset$ for $i=0,\ldots,a$ by (b) if $\varphi(x)\neq i\neq \varphi(y)$,
(b) and (c) if $\varphi(x)=i\neq\varphi(y)$, and (d) if $\varphi(x)\neq i=\varphi(y)$ (we cannot have $\varphi(x)=i=\varphi(y)$, since
$xy\in E(T_\beta)$. Hence, again we have $f(u)\cap f(v)\neq\emptyset$.
\end{proof}
%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.
%\begin{lemma}\label{lemma-chrom}
%If $G$ has a comparable box representation $f$ in $\mathbb{R}^d$, then $G$ is $3^d$-colorable.
%\end{lemma}
%\begin{proof}
%We actually show that $G$ is $(3^d-1)$-degenerate. Since every induced subgraph of $G$ also
%has a comparable box representation in $\mathbb{R}^d$, it suffices to show that the minimum degree of $G$
%is less than $3^d$. Let $v$ be a vertex of $G$ such that $f(v)$ has the smallest volume. For every neighbor $u$ of $v$,
%there exists a translation $B_u$ of $f(v)$ such that $B_u\subseteq f(u)$ and $B_u$ touches $f(v)$.
%Note that $f(v)\cup \bigcup_{u\in N(v)} B_u$ is a union of internally disjoint translations of $f(v)$ contained in
%a box obtained from $f(v)$ by scaling it by a factor of three, and thus $1+|N(v)|\le 3^d$.
%\end{proof}
\section{Exploiting the product structure}
......
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