# Planar CPG graphs

We show that for any k ≥ 0, there exists a planar graph which is B_k+1-CPG but not B_k-CPG. As a consequence, we obtain that B_k-CPG is a strict subclass of B_k+1-CPG.

## Authors

• 1 publication
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• ### On Morphing 1-Planar Drawings

Computing a morph between two drawings of a graph is a classical problem...
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• ### Quantum isomorphism is equivalent to equality of homomorphism counts from planar graphs

Over 50 years ago, Lovász proved that two graphs are isomorphic if and o...
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• ### Power series expansions for the planar monomer-dimer problem

We compute the free energy of the planar monomer-dimer model. Unlike the...
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• ### Multiscale Planar Graph Generation

The study of network representations of physical, biological, and social...
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• ### An SPQR-Tree-Like Embedding Representation for Upward Planarity

The SPQR-tree is a data structure that compactly represents all planar e...
08/01/2019 ∙ by Guido Brückner, et al. ∙ 0

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## 1 Introduction

A graph is a contact graph of paths on a grid (or CPG for short) if there exists a collection of interiorly disjoint paths on a grid in one-to-one correspondence with such that two vertices are adjacent in if and only if the corresponding paths touch; if furthermore every path has at most bends (i.e. 90-degree turn at a grid-point) for , then the graph is -CPG. The pair is a CPG representation of , and more specifically a -bend CPG representation of if every path in has at most bends. It was shown in [cpg] that not all planar graphs are CPG and that there exists no value of for which -CPG is a subclass of the class of planar graphs. In this note, we show that there exists no value of such that -CPG contains the class of planar CPG graphs. More specifically, we prove the following theorem.

###### Theorem 1.1

For any , there exists a planar graph in -CPG -CPG.

It immediately follows from the definition that -CPG -CPG but it was not known whether this inclusion is strict; Theorem 1.1 settles this question.

###### Corollary 1

For any , -CPG is strictly contain in -CPG, even within the class of planar graphs.

Note finally that Theorem 1.1 implies that the class of planar CPG graphs has an unbounded bend number (the bend number of a graph class is the smallest such that -CPG).

## 2 Preliminaries

Let be a CPG graph and be a CPG representation of . The path in representing some vertex is denoted by . An interior point of a path is a point belonging to and different from its endpoints; the interior of is the set of all its interior points. A grid-point is of type II.a if it is an endpoint of two paths and an interior point, different from a bendpoint, of a third path (see Fig. 0(a)); a grid-point is of type II.b if it is an endpoint of two paths and a bendpoint of a third path (see Fig. 0(b)).

## 3 Proof of Theorem 1.1

We show that the planar graph , with , depicted in Fig. 2 is in -CPG -CPG. We refer to the vertices , for , as the secondary vertices, and to the vertices , for and a given , as the -sewing vertices. In Fig. 3 is given a -bend CPG representation of (where the blue paths correspond to sewing vertices and the red paths correspond to secondary vertices). We next prove that in any CPG representation of , there exists a path with at least bends.

Let be a CPG representation of . A path in corresponding to a secondary vertex (resp. an -sewing vertex) is called a secondary path (resp. an -sewing path). A secondary path is said to be pure if no endpoint of or belongs to . We then have the following easy observation.

###### Observation 1

If a path is pure, then one endpoint of belongs to and the other endpoint belongs to .

###### Claim 1

Let and be two pure paths and let and be two -sewing vertices such that . If a grid-point belongs to , then corresponds to an endpoint of both and , and a bendpoint of either or .

It follows from Observation 1 and the fact that is non-adjacent to both and , that no endpoint of or belongs to . Consequently, one endpoint of belongs to and the other endpoint belongs to . We conclude similarly for . By definition, corresponds to an endpoint of at least one of and , which implies that belongs to or . But then, must be an endpoint of both and ; in particular, is a grid-point of type either II.a or II.b. Without loss of generality, we may assume that . We denote by (resp. ) the endpoint of (resp. ) belonging to . Now, suppose to the contrary that is of type II.a. The union of , and the portion of between and defines a closed curve , which divides the plane into two regions. Since and touch neither , nor (recall that is pure), and lie entirely in one of those regions; and, as and are adjacent, and in fact belong to the same region. On the other hand, since one endpoint of (resp. ) belongs to while the other endpoint belongs to , and both endpoints of are in , it follows that is the only contact point between (resp. ) and ; but and being non-adjacent, this implies that crosses only once and has therefore one endpoint in each region. But both endpoints of belong to which contradicts the fact that and lie in the same region. Hence, is of type II.b which concludes the proof.

###### Claim 2

If two paths and are pure, then one of them contains at least bends and the other one contains at least bends. Moreover, all of those bendpoints belong to (i,i+1)-sewing paths.

For all , consider a point . It follows from Claim 1 that is a bendpoint of either or . Since and are the endpoints of , one belongs to while the other belongs to . Therefore, is a subset of one of the considered secondary paths and is a subset of the other secondary path.

Finally, we claim that there exists an index such that , , and are all pure. Indeed, if it weren’t the case, then there would be at least secondary paths that are not pure; but at most secondary paths can contain endpoints of or , a contradiction. It now follows from Claim 2 that has at least bends (which belong to -sewing paths), and that has at least bends (which belong to -sewing paths). Furthermore, either or has at least bends which are endpoints of -sewing paths. Either way, there is a path with at least bends, which concludes the proof of Theorem 1.1.

## 4 Conclusion

In this note, we prove that the class of planar CPG graphs is not included in any -CPG, for . More specifically, we show that for any , there exists a planar graph which is -CPG but not -CPG. As a consequence, we also obtain that -CPG -CPG for any .