Skip to content
GitLab
Projects
Groups
Snippets
/
Help
Help
Support
Community forum
Keyboard shortcuts
?
Submit feedback
Contribute to GitLab
Sign in
Toggle navigation
Menu
Open sidebar
Zdenek Dvorak
Comparable box dimension
Commits
17ff0baa
Commit
17ff0baa
authored
Sep 08, 2021
by
Zdenek Dvorak
Browse files
Added the treewidth argument.
parent
0b81ca9b
Changes
1
Hide whitespace changes
Inline
Sidebyside
comparableboxdimension.tex
View file @
17ff0baa
...
...
@@ 138,12 +138,82 @@ In fact, we will prove the following stronger fact (TODO: Was this published som
\begin{lemma}
\label
{
lemmatw
}
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
)
\prec
eq
p
(
v
)
$
, then
$
h
(
u
)
\sqsubset
eq
h
(
v
)
$
.
Then
$
G
$
has a touching representation
$
h
$
by
axisaligned
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
$
2
s
_
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
$
2
s
_
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 nonadjacent 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 cliquesums. A
\emph
{
cliquesum
}
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 cliquesums.
\begin{theorem}
\begin{theorem}
\label
{
thmcs
}
If
$
G
$
is obtained from graphs in a class
$
\GG
$
by cliquesums, 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
{
lemmalegraf
}
,
$
(
T,
\pi
)
$
is a tree decomposition of
$
T
_
\beta
$
of width at most
$
a
$
. By Lemma~
\ref
{
lemmatw
}
,
$
T
_
\beta
$
has a touching representation
$
h
$
by hypercubes in
$
\mathbb
{
R
}^{
a
+
1
}$
such that
$
T
_
\beta
$
has a touching representation
$
h
$
by
axisaligned
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
{
thmcs
}
, 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 t
he product structure
}
\section
{
T
he product structure
and minorclosed classes
}
\subsection*
{
Acknowledgments
}
This research was carried out at the workshop on Geometric Graphs and Hypergraphs organized by Yelena Yuditsky and Torsten Ueckerdt
...
...
Write
Preview
Supports
Markdown
0%
Try again
or
attach a new file
.
Cancel
You are about to add
0
people
to the discussion. Proceed with caution.
Finish editing this message first!
Cancel
Please
register
or
sign in
to comment