Commit 5ada1937 authored by Zdenek Dvorak's avatar Zdenek Dvorak
Browse files

Added details to the proof of Lemma 10.

parent 0d75f91f
...@@ -210,58 +210,65 @@ Then $h$ is a touching representation of $G$ by comparable boxes in $\mathbb{R}^ ...@@ -210,58 +210,65 @@ Then $h$ is a touching representation of $G$ by comparable boxes in $\mathbb{R}^
Let $G\boxtimes H$ denote the \emph{strong product} of the graphs $G$ Let $G\boxtimes H$ denote the \emph{strong product} of the graphs $G$
and $H$, i.e., the graph with vertex set $V(G)\times V(H)$ and with and $H$, i.e., the graph with vertex set $V(G)\times V(H)$ and with
distinct vertices $(u_1,v_1)$ and $(u_2,v_2)$ adjacent if and only if, distinct vertices $(u_1,v_1)$ and $(u_2,v_2)$ adjacent if and only if
either $u_1=u_2$ or $u_1u_2\in E(G)$, and either $v_1=v_2$ or $u_1$ is equal to or adjacent to $u_2$ in $G$
$v_1v_2\in E(G)$. To obtain a touching representation of $G\boxtimes and $v_1$ is equal to or adjacent to $v_2$ in $H$.
H$ it suffice to take a product of representations of $G$ and $H$, but To obtain a touching representation of $G\boxtimes
the obtained representation may contain uncomparable boxes. Thus, H$ it suffices to take a product of representations of $G$ and $H$, but
bounding $\cbdim(G\boxtimes H)$ in terms of $\cbdim(G)$ and the resulting representation may contain incomparable boxes.
$\cbdim(H)$ seems to be a complicated task. In the following lemma we Indeed, $\cbdim(G\boxtimes H)$ in general is not bounded by a function
overcome this issue, by constraining one of the representations. of $\cbdim(G)$ and $\cbdim(H)$; for example, every star has comparable box dimension
at most two, but the strong product of the star $K_{1,n}$ with itself contains
$K_{n,n}$ as an induced subgraph, and thus its comparable box dimension is at least $\Omega(\log n)$.
However, as shown in the following lemma, this issue does not arise if the representation of $H$ consists of translates
of a single box; by scaling, we can without loss of generality assume this box is a unit hypercube.
\begin{lemma}\label{lemma-sp} \begin{lemma}\label{lemma-sp}
Consider a graph $H$ having a touching representation $h$ in Consider a graph $H$ having a touching representation $h$ in
$\mathbb{R}^{d_H}$ with hypercubes of unit size. Then for any graph $\mathbb{R}^{d_H}$ by axis-aligned hypercubes of unit size. Then for any graph
$G$, the strong product of these graphs is such that $G$, the strong product $G\boxtimes H$ of these graphs has comparable box dimension at most
$\cbdim(G\boxtimes H) \le \cbdim(G) + d_H$. $\cbdim(G) + d_H$.
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
The proof simply consists in taking a product of the two The proof simply consists in taking a product of the two
representations. Indeed, consider a touching respresentation with representations. Indeed, consider a touching respresentation $g$ of $G$ by
comparable boxes $g$ of $G$ in $\mathbb{R}^{d_G}$, with comparable boxes in $\mathbb{R}^{d_G}$, with
$d_G=\cbdim(G)$, and the representation $h$ of $H$. Let us define a $d_G=\cbdim(G)$, and the representation $h$ of $H$. Let us define a
representation $f$ of $G\boxtimes H$ in $\mathbb{R}^{d_G+d_H}$ as representation $f$ of $G\boxtimes H$ in $\mathbb{R}^{d_G+d_H}$ as
follows. follows.
$$f((u,v))[i]=\begin{cases} $$f((u,v))[i]=\begin{cases}
g(u)[i]&\text{ if $i\le d_G$}\\ g(u)[i]&\text{ if $i\le d_G$}\\
h(u)[i-d_G]&\text{ if $i > d_G$} h(v)[i-d_G]&\text{ if $i > d_G$}
\end{cases}$$ \end{cases}$$
Notice first that the boxes of $f$ are comparable as $f((u,v)) Consider distinct vertices $(u,v)$ and $(u',v')$ of $G\boxtimes H$.
\sqsubset f((u',v'))$ if and only if $g(u) \sqsubset g(u')$. The boxes $g(u)$ and $g(u')$ are comparable, say $g(u)\sqsubseteq g(u')$. Since $h(v')$
is a translation of $h(v)$, this implies that $f((u,v))\sqsubseteq f((u',v'))$. Hence, the boxes
Now let us observe that for any two vertices $u, u'$ of $G$, there of the representation $f$ are pairwise comparable.
is an hyperplane separating the interiors of $g(u)$ and $g(u')$, and
similarly for $h$ and $H$. This implies that the boxes in $f$ are The boxes of the representations $g$ and $h$ have pairwise disjoint interiors.
interiorly disjoint. Indeed, the same hyperplane that separates $g(u)$ Hence, if $u\neq u'$, then there exists $i\le d_G$ such that the interiors
and $g(u')$, when extended to $\mathbb{R}^{d_G+d_H}$, now separates of the intervals $f((u,v))[i]=g(u)[i]$ and $f((u',v'))[i]=g(u')[i]$ are disjoint;
any two boxes $f((u,v))$ and $f((u',v'))$. This implies that $f$ is and if $v\neq v'$, then there exists $i\le d_H$ such that the interiors
a touching representation of a subgraph of $G\boxtimes H$. of the intervals $f((u,v))[i+d_G]=h(v)[i]$ and $f((u',v'))[i+d_G]=h(v')[i]$ are disjoint.
Consequently, the interiors of boxes $f((u,v))$ and $f((u',v'))$ are pairwise disjoint.
Similarly, one can also observe that there is a point $p$ in the Moreover, if $u\neq u'$ and $uu'\not\in E(G)$, or if $v\neq v'$ and $vv'\not\in E(G)$,
intersection of $f((u,v))$ and $f((u',v'))$, if and only if there is then the intervals discussed above (not just their interiors) are disjoint for some $i$;
a point $p_G$ in the intersection of $g(u)$ and $g(u')$, and a point hence, if $(u,v)$ and $(u',v')$ are not adjacent in $G\boxtimes H$, then $f((u,v))\cap f((u',v'))=\emptyset$.
$p_H$ in the intersection of $h(v)$ and $h(v')$. Indeed, one can Therefore, $f$ is a touching representation of a subgraph of $G\boxtimes H$.
obtain $p_G$ and $p_H$ by projecting $p$ in $\mathbb{R}^{d_G}$ or in
$\mathbb{R}^{d_H}$ respectively, and conversely $p$ can be obtained Finally, suppose that $(u,v)$ and $(u',v')$ are adjacent in $G\boxtimes H$.
by taking the product of $p_G$ and $p_H$. Thus $f$ is indeed a Then there exists a point $p_G$ in the intersection of $g(u)$ and $g(u')$,
touching representation of $G\boxtimes H$. since $u=u'$ or $uu'\in E(G)$ and $g$ is a touching representation of $G$;
and similarly, there exists a point $p_H$ in the intersection of $h(v)$ and $h(v')$.
Then $p_G\times p_H$ is a point in the intersection of $f((u,v))$ and $f((u',v'))$.
Hence, $f$ is indeed a touching representation of $G\boxtimes H$.
\end{proof} \end{proof}
\subsection{Subgraph} \subsection{Taking a subgraph}
Examples show that the comparable box dimension of a graph $G$ may be The comparable box dimension of a subgraph of a graph $G$ may be larger than $\cbdim(G)$, see the end of this
larger than the one a subgraph $H$ of $G$. However we show that the section for an example. However, we show that the
comparable box dimension of a subgraph is at most exponential in the comparable box dimension of a subgraph is at most exponential in the
comparable box dimension of the whole graph. This is essentially comparable box dimension of the whole graph. This is essentially
Corollary~25 in~\cite{subconvex}, but since the setting is somewhat Corollary~25 in~\cite{subconvex}, but since the setting is somewhat
...@@ -273,24 +280,31 @@ If $G$ is a subgraph of a graph $G'$, then $\cbdim(G)\le \cbdim(G')+\chi^2_s(G') ...@@ -273,24 +280,31 @@ If $G$ is a subgraph of a graph $G'$, then $\cbdim(G)\le \cbdim(G')+\chi^2_s(G')
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
As we can remove the boxes that represent the vertices, we can assume $V(G')=V(G)$. As we can remove the boxes that represent the vertices, we can assume $V(G')=V(G)$.
Let $f$ be a touching representation by comparable boxes in $\mathbb{R}^d$, where $d=\cbdim(G')$. Let $\varphi$ Let $f$ be a touching representation of $G'$ by comparable boxes in $\mathbb{R}^d$, where $d=\cbdim(G')$. Let $\varphi$
be a star coloring of $G'$ using colors $\{1,\ldots,c\}$, where $c=\chi_s(G')$. be a star coloring of $G'$ using colors $\{1,\ldots,c\}$, where $c=\chi_s(G')$.
For any distinct colors $i,j\in\{1,\ldots,c\}$, let $A_{i,j}\subseteq V(G)$ consist of vertices $u$ of color $i$ For any distinct colors $i,j\in\{1,\ldots,c\}$, let $A_{i,j}\subseteq V(G)$ be the set of vertices $u$ of color $i$
such that there exists a vertex $v$ of color $j$ such that $uv\in E(G')$ and $uv\not\in E(G)$. such that there exists a vertex $v$ of color $j$ such that $uv\in E(G')\setminus E(G)$. For each $u\in A_{i,j}$,
let $a_j(u)$ denote such a vertex $v$ chosen arbitrarily.
Let us define a representation $h$ by boxes in $\mathbb{R}^{d+\binom{c}{2}}$ by starting from the representation $f$ and, Let us define a representation $h$ by boxes in $\mathbb{R}^{d+\binom{c}{2}}$ by starting from the representation $f$ and,
for each pair $i<j$ of colors, adding a dimension $d_{i,j}$ and setting for each pair $i<j$ of colors, adding a dimension $d_{i,j}$ and setting
$h(v)[d_{i,j}]=[1/3,4/3]$ for $v\in A_{i,j}$, $h(v)[d_{i,j}]=[-4/3,-1/3]$ for $v\in A_{j,i}$, $$h(v)[d_{i,j}]=\begin{cases}
and $h(v)[d_{i,j}](v)=[-1/2,1/2]$ otherwise. Note that the boxes in this extended representation are comparable, [1/3,4/3]&\text{if $v\in A_{i,j}$}\\
[-4/3,-1/3]&\text{if $v\in A_{j,i}$}\\
[-1/2,1/2]&\text{otherwise.}
\end{cases}$$
Note that the boxes in this extended representation are comparable,
as in the added dimensions, all the boxes have size $1$. as in the added dimensions, all the boxes have size $1$.
Suppose $uv\in E(G)$, where $\varphi(u)=i$ and $\varphi(v)=j$ and say $i<j$. The boxes $f(u)$ and $f(v)$ touch. Suppose $uv\in E(G)$, where $\varphi(u)=i$ and $\varphi(v)=j$ and say $i<j$.
We cannot have $u\in A_{i,j}$ and $v\in A_{j,u}$, as then $G'$ would contain a 4-vertex path in colors $i$ and $j$. We cannot have $u\in A_{i,j}$ and $v\in A_{j,i}$, as then $a_j(u)uva_i(v)$ would be a 4-vertex path in $G'$ in colors $i$ and $j$.
Hence, in any added dimension $d'$, at least one of $h(u)$ and $h(v)$ is represented by the interval $[-1/2,1/2]$, Hence, in any added dimension $d'$, we have $h(u)[d']=[-1/2,1/2]$ or $h(v)[d']=[-1/2,1/2]$,
and thus $h(u)[d']\cap h(v)[d']\neq\emptyset$. Therefore, the boxes $h(u)$ and $h(v)$ touch. and thus $h(u)[d']\cap h(v)[d']\neq\emptyset$.
Since the boxes $f(u)$ and $f(v)$ touch, it follows that the boxes $h(u)$ and $h(v)$ touch as well.
Suppose now that $uv\not\in E(G)$. If $uv\not\in E(G')$, then $f(u)$ is disjoint from $f(v)$, and thus $h(u)$ is disjoint from Suppose now that $uv\not\in E(G)$. If $uv\not\in E(G')$, then $f(u)$ is disjoint from $f(v)$, and thus $h(u)$ is disjoint from
$h(v)$. Hence, we can assume $uv\in E(G')$, $\varphi(u)=i$, $\varphi(v)=j$ and $i<j$. Then $u\in A_{i,j}$, $v\in A_{j,i}$, $h(v)$. Hence, we can assume $uv\in E(G')\setminus E(G)$, $\varphi(u)=i$, $\varphi(v)=j$ and $i<j$. Then $u\in A_{i,j}$, $v\in A_{j,i}$,
$h(u)[d_{i,j}]=[1/3,4/3]$, $h(v)[d_{j,i}]=[-4/3,-1/3]$, and $h(u)\cap h(v)=\emptyset$. $h(u)[d_{i,j}]=[1/3,4/3]$, $h(v)[d_{j,i}]=[-4/3,-1/3]$, and $h(u)\cap h(v)=\emptyset$.
Consequently, $h$ is a touching representation of $G$ by comparable boxes in dimension $d+\binom{c}{2}\le d+c^2$. Consequently, $h$ is a touching representation of $G$ by comparable boxes in dimension $d+\binom{c}{2}\le d+c^2$.
...@@ -302,7 +316,7 @@ Let us now combine Lemmas~\ref{lemma-chrom} and \ref{lemma-subg}. ...@@ -302,7 +316,7 @@ Let us now combine Lemmas~\ref{lemma-chrom} and \ref{lemma-subg}.
If $G$ is a subgraph of a graph $G'$, then $\cbdim(G)\le \cbdim(G')+4\cdot 81^{\cbdim(G')}\le 5\cdot 81^{\cbdim(G')}$. If $G$ is a subgraph of a graph $G'$, then $\cbdim(G)\le \cbdim(G')+4\cdot 81^{\cbdim(G')}\le 5\cdot 81^{\cbdim(G')}$.
\end{corollary} \end{corollary}
Let us remark that an exponential increase in the dimension is unavoidable: We have $\cbdim{K_{2^d}}=d$, Let us remark that an exponential increase in the dimension is unavoidable: We have $\cbdim(K_{2^d})=d$,
but the graph obtained from $K_{2^d}$ by deleting a perfect matching has comparable box dimension $2^{d-1}$. but the graph obtained from $K_{2^d}$ by deleting a perfect matching has comparable box dimension $2^{d-1}$.
...@@ -324,162 +338,206 @@ an arbitrary number of clique-sums. We thus introduce the notion of ...@@ -324,162 +338,206 @@ an arbitrary number of clique-sums. We thus introduce the notion of
\emph{clique-sum extendable} representations. \emph{clique-sum extendable} representations.
\begin{definition} \begin{definition}
Consider a graph $G$ with a distinguished clique $C^*$, called the Consider a graph $G$ with a distinguished clique $C^\star$, called the
\emph{root clique} of $G$. A touching representation (with comparable \emph{root clique} of $G$. A touching representation $h$ of $G$
boxes or not) $h$ of $G$ in $\mathbb{R}^d$ is called by (not necessarily comparable) boxes in $\mathbb{R}^d$ is called
\emph{$C^*$-clique-sum extendable} if the following conditions hold. \emph{$C^\star$-clique-sum extendable} if the following conditions hold for every sufficiently small $\varepsilon>0$.
\begin{itemize} \begin{itemize}
\item[{\bf(vertices)}] There are $|V(C^*)|$ dimensions, denoted $d_u$ for each \item[{\bf(vertices)}] For each $u\in V(C^\star)$, there exists a dimension $d_u$,
vertex $u\in V(C^*)$, such that: such that:
\begin{itemize} \begin{itemize}
\item[(v1)] for each vertex $u\in V(C^*)$, $h(u)[d_u] = [-1 ,0]$ and \item[(v0)] $d_u\neq d_{u'}$ for distinct $u,u'\in V(C^\star)$,
$h(u)[i] = [0,1]$, for any dimension $i\neq d_u$, and \item[(v1)] each vertex $u\in V(C^\star)$ satisfies $h(u)[d_u] = [-1,0]$ and
\item[(v2)] for any vertex $v\notin V(C^*)$, $h(v) \subset [0,1)^d$. $h(u)[i] = [0,1]$ for any dimension $i\neq d_u$, and
\item[(v2)] each vertex $v\notin V(C^\star)$ satisfies $h(v) \subset [0,1)^d$.
\end{itemize} \end{itemize}
\item[{\bf(cliques)}] For every clique $C$ of $G$ we define a point \item[{\bf(cliques)}] For every clique $C$ of $G$, there exists
$p(C)\in I_C \cap [0,1)^d$, where $I_C =\cap_{v\in V(C)} h(v)$, and a point $$p(C)\in [0,1)^d\cap \bigcap_{v\in V(C)} h(v)$$
we define the box $h^\epsilon(C)$, for any $\epsilon > 0$, by such that, defining the \emph{clique box} $h^\varepsilon(C)$
$h^\epsilon(C)[i] = [p(C)[i],p(C)[i]+\epsilon]$, for every dimension by setting
$i$. Furthermore, for a sufficiently small $\epsilon > 0$ these $$h^\varepsilon(C)[i] = [p(C)[i],p(C)[i]+\varepsilon]$$ for every dimension
\emph{clique boxes} verify the following conditions. $i$, the following conditions are satisfied:
\begin{itemize} \begin{itemize}
\item[(c1)] For any two cliques $C_1\neq C_2$, $h^\epsilon(C_1) \cap \item[(c1)] For any two cliques $C_1\neq C_2$, $h^\varepsilon(C_1) \cap
h^\epsilon(C_2) = \emptyset$ (i.e. $p(C_1) \neq p(C_2)$). h^\varepsilon(C_2) = \emptyset$ (equivalently, $p(C_1) \neq p(C_2)$).
\item[(c2)] A box $h(v)$ intersects $h^\epsilon(C)$ if and only if \item[(c2)] A box $h(v)$ intersects $h^\varepsilon(C)$ if and only if
$v\in V(C)$, and in that case their intersection is a facet of $v\in V(C)$, and in that case their intersection is a facet of
$h^\epsilon(C)$ incident to $p(C)$ (i.e. if we denote this $h^\varepsilon(C)$ incident to $p(C)$. That is, there exists a dimension $i_{C,v}$
intersection $I$, then $I[i] = \{p(C)[i] \}$ for some dimension such that for each dimension $j$,
$i$, and $I[j] = [p(C)[j],p(C)[j]+\epsilon]$ for the other $$h(v)[j]\cap h^\varepsilon(C)[j] = \begin{cases}
dimensions $j\neq i$). \{p(C)[i_{C,v}] \}&\text{if $j=i_{C,v}$}\\
[p(C)[j],p(C)[j]+\varepsilon]&\text{otherwise.}
\end{cases}$$
\end{itemize} \end{itemize}
\end{itemize} \end{itemize}
\end{definition} \end{definition}
Note that we may consider that the root clique is empty, that is the Note that the root clique can be empty, that is the
empty subgraph with no vertices. In that case the clique is denoted empty subgraph with no vertices. In that case the clique is denoted
$\emptyset$. Let $\ecbdim(G)$ be the minimum dimension such that $G$ $\emptyset$. Let $\ecbdim(G)$ be the minimum dimension such that $G$
has a $\emptyset$-clique-sum extendable touching representation by has an $\emptyset$-clique-sum extendable touching representation by
comparable boxes. The following lemma ensures that clique-sum comparable boxes.
Let us remark that a clique-sum extendable representation in dimension $d$ implies
such a representation in higher dimensions as well.
\begin{lemma}\label{lemma-add}
Let $G$ be a graph with a root clique $C^\star$ and let $h$ be
a $C^\star$-clique-sum extendable touching representation of $G$ by comparable boxes in $\mathbb{R}^d$.
Then $G$ has such a representation in $\mathbb{R}^{d'}$ for every $d'\ge d$.
\end{lemma}
\begin{proof}
It clearly suffices to consider the case that $d'=d+1$.
Note that the \textbf{(vertices)} conditions imply that $h(v')\sqsubseteq h(v)$ for every $v'\in V(G)\setminus V(C^\star)$
and $v\in V(C^\star)$. We extend the representation $h$
by setting $h(v)[d+1] = [0,1]$ for $v\in V(C^\star)$ and $h(v)[d+1] = [0,\frac12]$ for $v\in V(G)\setminus V(C^\star)$.
The clique point $p(C)$ of each clique $C$ is extended by setting $p(C)[d+1] = \frac14$.
It is easy to verify that the resulting representation is $C^\star$-clique-sum extendable.
\end{proof}
The following lemma ensures that clique-sum
extendable representations behave well with respect to full extendable representations behave well with respect to full
clique-sums. clique-sums.
\begin{lemma}\label{lem-cs} \begin{lemma}\label{lem-cs}
Consider two graphs $G_1$ and $G_2$, given with a $C^*_1$- and a Consider two graphs $G_1$ and $G_2$, given with a $C^\star_1$- and a
$C^*_2$-clique-sum extendable representations with comparable boxes $C^\star_2$-clique-sum extendable representations $h_1$ and $h_2$ by comparable boxes
$h_1$ and $h_2$, in $\mathbb{R}^{d_1}$ and $\mathbb{R}^{d_2}$ in $\mathbb{R}^{d_1}$ and $\mathbb{R}^{d_2}$,
respectively. Let $G$ be the graph obtained after performing a full respectively. Let $G$ be the graph obtained by performing a full
clique-sum of these two graphs on any clique $C_1$ of $G_1$, and on clique-sum of these two graphs on any clique $C_1$ of $G_1$, and on
the root clique $C^*_2$ of $G_2$. Then $G$ admits a $C^*_1$-clique the root clique $C^\star_2$ of $G_2$. Then $G$ admits a $C^\star_1$-clique
sum extendable representation by comparable boxes $h$ in sum extendable representation $h$ by comparable boxes in
$\mathbb{R}^{\max(d_1,d_2)}$. $\mathbb{R}^{\max(d_1,d_2)}$.
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
By Lemma~\ref{lemma-add}, we can assume that $d_1=d_2$; let $d=d_1$.
The idea is to translate (allowing also exchanges of dimensions) and The idea is to translate (allowing also exchanges of dimensions) and
scale $h_2$ to fit in $h_1^\epsilon(C_1)$. Consider an $\epsilon >0$ scale $h_2$ to fit in $h_1^\varepsilon(C_1)$. Consider an $\varepsilon >0$
sufficiently small so that, $h_1^\epsilon(C_1)$ verifies all the sufficiently small so that $h_1^\varepsilon(C_1)$ satisfies all the
(cliques) conditions, and such that $h_1^\epsilon(C_1) \sqsubseteq \textbf{(cliques)} conditions, and such that $h_1^\varepsilon(C_1) \sqsubseteq
h_1(v)$ for any vertex $v\in V(G_1)$. Without loss of generality, h_1(v)$ for any vertex $v\in V(G_1)$. Let $V(C_1)=\{v_1,\ldots,v_k\}$;
let us assume that $V(C_1)=\{v_1,\ldots,v_k\}$, and we also assume without loss of generality, we can assume $i_{C_1,v_i}=i$ for $i\in\{1,\ldots,k\}$,
that $h_1(v_i)$ and $h_1^\epsilon(C_1)$ touch in dimension $i$ (i.e. and thus
$h_1(v_i)[i] \cap h_1^\epsilon(C_1)[i] = \{p(C_1)[i]\}$, and $$h_1(v_i)[j] \cap h_1^\varepsilon(C_1)[j] = \begin{cases}
$h_1(v_i)[j] \cap h_1^\epsilon(C_1)[j] = \{p_1(C_1)[i]\}&\text{ if $j=i$}\\
[p(C_1)[j],p(C_1)[j]+\epsilon] $ for $j\neq i$. [p_1(C_1)[j],p_1(C_1)[j]+\varepsilon]&\text{ otherwise.}
\end{cases}$$
Now let us consider $G_2$ and its representation $h_2$. Here the Now let us consider $G_2$ and its representation $h_2$. Here the
vertices of $C^*_2$ are also denoted $v_1,\ldots,v_k$, and let us vertices of $C^\star_2$ are also denoted $v_1,\ldots,v_k$, and
denote $d_{v_i}$ the dimension in $h_2$ that fulfills condition (v1) without loss of generality, the \textbf{(vertices)} conditions are
with respect to $v_i$. satisfied by setting $d_{v_i}=i$ for $i\in\{1,\ldots,k\}$
Let $d=\max(d_1,d_2)$. We are now ready for defining $h$. For the We are now ready to define $h$. For $v\in V(G_1)$, we set $h(v)=h_1(v)$.
vertices of $G_1$ it is almost the same representation as $h_1$, as We now scale and translate $h_2$ to fit inside $h_1^\varepsilon(C_1)$.
we set $h(v)[i] = h_1(v)[i]$ for any dimension $i\le d_1$. If $d=d_2 That is, we fix $\varepsilon>0$ small enough so that
> d_1$, then for any dimension $i>d_1$ we set $h(v)[i] = [0,1]$ if \begin{itemize}
$v\in V(C^*_1)$, and $h(v)[i] = [0,\frac12]$ if $v\in \item the conditions \textbf{(cliques)} hold for $h_1$,
V(G_1)\setminus V(C^*_1)$. Similarly the clique points $p_1(C)$ \item $h_1^\varepsilon(C_1)\subset [0,1)^d$, and
become $p(C)$ by setting $p(C)[i] = p_1(C)[i]$ for $i\le d_1$, and \item $h_1^\varepsilon(C_1)\sqsubseteq h_1(u)$ for every $u\in V(G_1)$,
$p(C)[i] = \frac14$ for $i> d_1$. \end{itemize}
and for each $v\in V(G_2) \setminus V(C^\star_2)$,
For $h_2$ and the vertices in $V(G_2) \setminus \{v_1,\ldots,v_k\}$ we set $h(v)[i]=p_1(C_1)[i] + \varepsilon h_2(v)[i]$ for $i\in\{1,\ldots,d\}$.
we have to consider a mapping $\sigma$ from $\{1,\ldots,d_2\}$ to Note that the condition (v2) for $h_2$ implies $h(v)\subset h_1^\varepsilon(C_1)$.
$\{1,\ldots,d\}$ such that $\sigma(d_{v_i}) = i$. This mapping Each clique $C$ of $H$ is a clique of $G_1$ or $G_2$.
describes the changes of dimension we have to perform. We also have If $C$ is a clique of $G_2$, we set $p(C)=p_1(C_1)+\varepsilon p_2(C)$,
to perform a scaling in order to make $h_2$ fit inside otherwise we set $p(C)=p_1(C)$. In particular, for subcliques of $C_1=C^\star_2$,
$h_1^\epsilon(C_1)$. This is ensured by multiplying the coordinates we use the former choice.
by $\epsilon$. More formally, for any vertex $v\in V(G_2) \setminus
\{v_1,\ldots,v_k\}$, we set $h(v)[\sigma(i)] = p(C_1)[\sigma(i)] + Let us now check that $h$ is a $C^\star_1$-clique sum extendable
\epsilon h_2(v)[i]$ for $i\in \{1 ,\ldots,d_2\}$, and $h(v)[j] =
[p(C_1)[j], p(C_1)[j] +\epsilon/2]$, otherwise (for any $j$ not in the
image of $\sigma$). Note that if we apply the same mapping from
$h_2$ to $h$, to the boxes $h_2(v_i)$ for $i\in \{1,\ldots,k\}$, then the
image of $h_2(v_i)$ fits inside the (previously defined) box
$h(v_i)$. Similarly the clique points $p_2(C)$ become $p(C)$ by
setting $p(C)[\sigma(i)] = p(C_1)[\sigma(i)] + \epsilon p_2(C)[i]$
for $i\in \{1 ,\ldots,d_2\}$, and $p(C)[j] = p(C_1)[j] + \epsilon/4$, otherwise.
Note that we have defined (differently) both $h^\epsilon(C_1)$
(resp. $p(C_1)$) and $h^\epsilon(C^*_2)$ (resp. $p(C^*_2)$), despite
the fact that those cliques were merged. In the following we use
$h^\epsilon(C_1)$ and $p(C_1)$ only for the purpose of the
proof. The point and the box corresponding to this clique in $h$ is
$p(C^*_2)$ and $h^\epsilon(C^*_2)$.
Let us now check that $h$ is a $C^*_1$-clique sum extendable
representation by comparable boxes. The fact that the boxes are representation by comparable boxes. The fact that the boxes are
comparable follows from the fact that those of $V(G_1)$ comparable follows from the fact that those of $h_1$ and $h_2$
(resp. $V(G_2)$) are comparable in $h_1$ (resp. $h_2$) with the are comparable and from the scaling of $h_2$: By construction both
boxes of $V(C^*_1)$ (resp. $V(C^*_2)$) being hypercubes of side one, $h_1(v) \sqsubseteq h_1(u)$ and $h_2(v) \sqsubseteq h_2(u)$ imply
and the other boxes being smaller. Clearly, by construction both $h(v) \sqsubseteq h(u)$, and for any vertex $u\in V(G_1)$ and any
$h_1(u) \sqsubseteq h_1(v)$ or $h_2(u) \sqsubseteq h_2(v)$, imply vertex $v\in V(G_2) \setminus V(C^\star_2)$, we have $h(v) \subset h_1^\varepsilon(C_1) \sqsubseteq h(u)$.
$h(u) \sqsubseteq h(v)$, and for any vertex $u\in V(G_1)$ and any
vertex $v\in V(G_2) \setminus \{v_1,\ldots,v_k\}$, we have $h(v)
\sqsubseteq h^\epsilon(C_1) \sqsubseteq h(u)$.
We now check that $h$ is a contact representation of $G$. For $u,v We now check that $h$ is a contact representation of $G$. For $u,v
\in V(G_1)$ (resp. $u,v \in V(G_2) \setminus \{v_1,\ldots,v_k\}$) it \in V(G_1)$ (resp. $u,v \in V(G_2) \setminus V(C^\star_2)$) it
is clear that $h(u)$ and $h(v)$ have disjoint interiors, and that they is clear that $h(u)$ and $h(v)$ have disjoint interiors, and that they
intersect if and only if $h_1(u)$ and $h_1(v)$ intersect (resp. if intersect if and only if $h_1(u)$ and $h_1(v)$ intersect (resp. if
$h_2(u)$ and $h_2(v)$ intersect). Consider now a vertex $u \in $h_2(u)$ and $h_2(v)$ intersect). Consider now a vertex $u \in
V(G_1)$ and a vertex $v \in V(G_2) \setminus \{v_1,\ldots,v_k\}$. As V(G_1)$ and a vertex $v \in V(G_2) \setminus V(C^\star_2)$. As
$h(v)$ fits inside $h^\epsilon(C_1)$, we have that $h(u)$ and $h(v)$ $h(v)\subset h^\varepsilon(C_1)$, the condition (v2) for $h_1$ implies
have disjoint interiors. Furthermore, if they intersect then $u\in that $h(u)$ and $h(v)$ have disjoint interiors.
V(C_1)$, say $u=v_1$, and $h(v)[1] = [p(C_1)[1], p(C_1)[1]+\alpha]$
for some $\alpha>0$. By construction, this implies that $h_2(v_1)$ Furthermore, if $uv\in E(G)$, then $u\in V(C_1)=V(C^\star_2)$, say $u=v_1$.
and $h_2(v)$ intersect. Since $uv\in E(G_2)$, the intervals $h_2(u)[1]$ and $h_2(v)[1]$ intersect,
and by (v1) and (v2) for $h_2$, we conclude that $h_2(v)[1]=[0,\alpha]$ for some positive $\alpha<1$.
Finally for the $C^*_1$-clique-sum extendability, one can easily Therefore, $p_1(C_1)[1]\in h(v)[1]$. Since $p_1(C_1)\in \bigcap_{x\in V(C_1)} h_1(x)$,
check that the (vertices) conditions hold. For the (cliques) we have $p_1(C_1)\in h(u)$, and thus $p_1(C_1)[1]\in h(u)[1]\cap h(v)[1]$.
conditions, as the mapping from $p_2(C)$ to $p(C)$ (extended to a For $i\in \{2,\ldots,d\}$, note that $i\neq 1=i_{C_1,u}$, and thus
mapping from $\mathbb{R}^{d_2}$ to $\mathbb{R}^{d}$) is injective, by (c2) for $h_1$, we have $h_1^\varepsilon(C_1)[i]\subseteq h_1(u)[i]=h(u)[i]$.
we have that (c1) clearly holds. For (c2) one has to notice that if Since $h(v)[i]\subseteq h_1^\varepsilon(C_1)[i]$, it follows that $h(u)$ intersects $h(v)$.
$d=d_2$, then the mapping from $h_2$ to $h$ extended to the clique boxes
would lead to the same clique boxes $h^{\epsilon'}(C)$, with the same point $p(C)$ in their lower corner. Finally, let us consider the $C^\star_1$-clique-sum extendability. The \textbf{(vertices)}
If there are extra dimensions, that is if $d> d_2$, then for any such conditions hold, since (v0) and (v1) are inherited from $h_1$, and
dimension $j$ that is not in the image of $\sigma$, we have that (v2) is inherited from $h_1$ for $v\in V(G_1)\setminus V(C^\star_1)$
$h^{\epsilon'}(C)[j] = [p(C_1)[j] + \epsilon/4, p(C_1)[j] + \epsilon/4 + \epsilon'] and follows from the fact that $h(v)\subseteq h_1^\varepsilon(C_1)\subset [0,1)^d$
\subset [p(C_1)[j], p(C_1)[j] + \epsilon/2] = h(v)[j]$. for $v\in V(G_2)\setminus V(C^\star_2)$. For the \textbf{(cliques)} condition (c1),
the mapping $p$ inherits injectivity when restricted to cliques of $G_2$,
or to cliques of $G_1$ not contained in $C_1$. For any clique $C$ of $G_2$,
the point $p(C)$ is contained in $h_1^\varepsilon(C_1)$, since $p_2(C)\in [0,1)^d$.
On the other hand, if $C'$ is a clique of $G_1$ not contained in $C_1$, then there
exists $v\in V(C')\setminus V(C_1)$, we have $p(C')=p_1(C')\in h_1(v)$, and
$h_1(v)\cap h_1^\varepsilon(C_1)=\emptyset$ by (c2) for $h_1$.
Therefore, the mapping $p$ is injective, and thus for sufficiently small $\varepsilon'>0$,
we have $h^{\varepsilon'}(C)\cap h^{\varepsilon'}(C')=\emptyset$ for any distinct
cliques $C$ and $C'$ of $G$.
The condition (c2) of $h$ is (for sufficiently small $\varepsilon'>0$)
inherited from the property (c2) of $h_1$ and $h_2$
when $C$ is a clique of $G_2$ and $v\in V(G_2)\setminus V(C^\star_2)$, or
when $C$ is a clique of $G_1$ not contained in $C_1$ and $v\in V(G_1)$.
If $C$ is a clique of $G_1$ not contained in $C_1$ and $v\in V(G_2)\setminus V(C^\star_2)$,
then by (c1) for $h_1$ we have $h_1^\varepsilon(C)\cap h_1^\varepsilon(C_1)=\emptyset$,
and since $h^{\varepsilon'}(C)\subseteq h_1^\varepsilon(C)$ and $h(v)\subseteq h_1^\varepsilon(C_1)$,
we conclude that $h(v)\cap h^{\varepsilon'}(C)=\emptyset$.
It remains to consider the case that $C$ is a clique of $G_2$ and $v\in V(G_1)$.
Note that $h^{\varepsilon'}(C)\subseteq h_1^\varepsilon(C_1)$.
\begin{itemize}
\item If $v\not\in V(C_1)$, then by the property (c2) of $h_1$, the box $h(v)=h_1(v)$ is disjoint from $h_1^\varepsilon(C_1)$,
and thus $h(v)\cap h^{\varepsilon'}(C)=\emptyset$.
\item Otherwise $v\in V(C_1)=V(C^\star_2)$, say $v=v_1$.
Note that by (v1), we have $h_2(v)=[-1,0]\times [0,1]^{d-1}$.
\begin{itemize}
\item If $v\not\in V(C)$, then by the property (c2) of $h_2$, the box $h_2(v)$ is disjoint from $h_2^\varepsilon(C)$.
Since $h_2^\varepsilon(C)[i]\subseteq[0,1]=h_2(v)[i]$ for $i\in\{2,\ldots,d\}$,
it follows that $h_2^\varepsilon(C)[1]\subseteq (0,1)$, and thus $h^{\varepsilon'}(C)[1]\subseteq h_1^\varepsilon(C_1)[1]\setminus\{p(C_1)[1]\}$.
By (c2) for $h_1$, we have $h(v)[1]\cap h_1^\varepsilon(C_1)[1]=h_1(v)[1]\cap h_1^\varepsilon(C_1)[1]=p(C_1)[1]$,
and thus $h(v)\cap h^{\varepsilon'}(C)=\emptyset$.
\item If $v\in V(C)$, then by the property (c2) of $h_2$, the intersection of
$h_2(v)[1]=[-1,0]$ and $h_2^\varepsilon(C)[1]\subseteq [0,1)$ is the single point $p_2(C)[1]=0$,
and thus $p(C)[1]=p_1(C_1)[1]$ and $h^{\varepsilon'}(C)[1]=[p_1(C_1)[1],p_1(C_1)[1]+\varepsilon']$.
Recall that the property (c2) of $h_1$ implies $h(v)[1]\cap h_1^\varepsilon(C_1)[1]=\{p(C_1)[1]\}$,
and thus $h(v)[1]\cap h^{\varepsilon'}(C)[1]=\{p(C)[1]\}$. For $i\in\{2,\ldots, d\}$,
the property (c2) of $h_1$ implies $h_1^\varepsilon(C_1)[i]\subseteq h_1(v)[i]=h(v)[i]$, and
since $h^{\varepsilon'}(C)[i]\subseteq h_1^\varepsilon(C_1)[i]$, it follows that
$h^{\varepsilon'}(C)[i]\subseteq h(v)[i]$.
\end{itemize}
\end{itemize}
\end{proof} \end{proof}
The following lemma shows that any graphs has a $C^*$-clique-sum The following lemma shows that any graphs has a $C^\star$-clique-sum
extendable representation in $\mathbb{R}^d$, for $d= \omega(G) + extendable representation in $\mathbb{R}^d$, for $d= \omega(G) +
\ecbdim(G)$ and for any clique $C^*$. \ecbdim(G)$ and for any clique $C^\star$.
\begin{lemma}\label{lem-apex-cs} \begin{lemma}\label{lem-apex-cs}
For any graph $G$ and any clique $C^*$, we have that $G$ admits a For any graph $G$ and any clique $C^\star$, we have that $G$ admits a
$C^*$-clique-sum extendable touching representation by comparabe $C^\star$-clique-sum extendable touching representation by comparabe
boxes in $\mathbb{R}^d$, for $d = |V(C^*)| + \ecbdim(G\setminus boxes in $\mathbb{R}^d$, for $d = |V(C^\star)| + \ecbdim(G\setminus
V(C^*))$. V(C^\star))$.
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
The proof is essentially the same as the one of The proof is essentially the same as the one of
Lemma~\ref{lemma-apex}. Consider a $\emptyset$-clique-sum Lemma~\ref{lemma-apex}. Consider a $\emptyset$-clique-sum
extendable touching representation $h'$ of $G\setminus V(C^*)$ by extendable touching representation $h'$ of $G\setminus V(C^\star)$ by
comparable boxes in $\mathbb{R}^{d'}$, with $d' = \cbdim(G\setminus comparable boxes in $\mathbb{R}^{d'}$, with $d' = \cbdim(G\setminus
V(C^*))$, and let $V(C^*) = \{v_1,\ldots,v_k\}$. We now construct V(C^\star))$, and let $V(C^\star) = \{v_1,\ldots,v_k\}$. We now construct
the desired representation $h$ of $G$ as follows. For each vertex the desired representation $h$ of $G$ as follows. For each vertex
$v_i\in V(C^*)$ let $h(v_i)$ be the box fulfilling (v1) with $v_i\in V(C^\star)$ let $h(v_i)$ be the box fulfilling (v1) with
$d_{v_i} = i$. For each vertex $u\in V(G)\setminus V(C^*)$, if $i\le $d_{v_i} = i$. For each vertex $u\in V(G)\setminus V(C^\star)$, if $i\le
k$ then let $h(u)[i] = [0,1/2]$ if $uv_i \in E(G)$, and $h(u)[i] = k$ then let $h(u)[i] = [0,1/2]$ if $uv_i \in E(G)$, and $h(u)[i] =
[1/4,3/4]$ if $uv_i \notin E(G)$. For $i>k$ we have $h(u)[i] = [1/4,3/4]$ if $uv_i \notin E(G)$. For $i>k$ we have $h(u)[i] =
\alpha_i h'(u)[i-k]$, for some $\alpha_i>0$. The values $\alpha_i>0$ \alpha_i h'(u)[i-k]$, for some $\alpha_i>0$. The values $\alpha_i>0$
are chosen suffciently small so that $h(u)[i] \subset [0,1)$, whenever $u\notin V(C^*)$. are chosen suffciently small so that $h(u)[i] \subset [0,1)$, whenever $u\notin V(C^\star)$.
We proceed similarly for the clique points. For any We proceed similarly for the clique points. For any
clique $C$ of $G$, if $i\le k$ then let $p(C)[i] = 0$ if $v_i \in clique $C$ of $G$, if $i\le k$ then let $p(C)[i] = 0$ if $v_i \in
V(C)