... ... @@ -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