Linear Size Universal Point Sets for Classes of Planar Graphs

TL;DR

This paper introduces a universal point set of size 2n-2 for bipartite and degree-3 planar graphs, enabling efficient straight-line embeddings and improving previous bend-drawing bounds.

Contribution

It presents a universal point set of size 2n-2 for bipartite and degree-3 planar graphs, and introduces the exploding double chain for broader graph embeddings.

Findings

Universal point set of size 2n-2 for bipartite planar graphs

Universal point set of size 2n-2 for degree-3 planar graphs

Improved bend-drawing bounds to 4n-6

Abstract

A finite set $P$ of points in the plane is $n$-universal with respect to a class $\mathcal{C}$ of planar graphs if every $n$-vertex graph in $\mathcal{C}$ admits a crossing-free straight-line drawing with vertices at points of $P$. For the class of all planar graphs the best known upper bound on the size of a universal point set is quadratic and the best known lower bound is linear in $n$. Some classes of planar graphs are known to admit universal point sets of near linear size, however, there are no truly linear bounds for interesting classes beyond outerplanar graphs. In this paper, we show that there is a universal point set of size $2n-2$ for the class of bipartite planar graphs with $n$ vertices. The same point set is also universal for the class of $n$-vertex planar graphs of maximum degree $3$. The point set used for the results is what we call an exploding double chain, and we prove that this point set allows planar straight-line embeddings of many more planar graphs, namely of all subgraphs of planar graphs admitting a one-sided Hamiltonian cycle. The result for bipartite graphs also implies that every $n$-vertex plane graph has a $1$-bend drawing all whose bends and vertices are contained in a specific point set of size $4n-6$, this improves a bound of $6n-10$ for the same problem by L\"offler and T\'oth.

Figures

Equations

$${\mathcal{C}}^{\prime }:=\{\mathcal{G}:\mathcal{G}\text{ is POSH}\}.$$

$$H(p_i,q_j)=\{\begin{array}{ll}(p_k{)}_{k\le j}\cup (p_k{)}_{k>i}\cup (q_{\ell }{)}_{\ell <j}& \text{if}i>j\\ (p_k{)}_{k<i}\cup (q_{\ell }{)}_{\ell <i}& \text{if}i=j\\ (p_k{)}_{k<i}\cup (q_{\ell }{)}_{\ell \le i}\cup (q_{\ell }{)}_{\ell >j}& \text{if}i<j\end{array}$$

Full text

Linear Size Universal Point Sets for Classes of Planar Graphs

Stefan Felsner

  

Hendrik Schrezenmaier

  

Felix Schröder

  

Raphael Steiner

For the class of all planar graphs the best known upper bound on the size of a universal point set is quadratic and the best known lower bound is linear in $n$.

Some classes of planar graphs are known to admit universal point sets of near linear size, however, there are no truly linear bounds for interesting classes beyond outerplanar graphs.

In this paper, we show that there is a universal point set of size $2n-2$ for the class of bipartite planar graphs with $n$ vertices. The same point set is also universal for the class of $n$-vertex planar graphs of maximum degree $3$. The point set used for the results is what we call an exploding double chain, and we prove that this point set allows planar straight-line embeddings of many more planar graphs, namely of all subgraphs of planar graphs admitting a one-sided Hamiltonian cycle.

The result for bipartite graphs also implies that every $n$-vertex plane graph has a 1-bend drawing all whose bends and vertices are contained in a specific point set of size $4n-6$, this improves a bound of $6n-10$ for the same problem by Löffler and Tóth.

1 Introduction

Given a family $\mathcal{C}$ of planar graphs and a positive integer $n$, a point set $P\subseteq\mathbb{R}^{2}$ is called an $n$-universal point set for the class $\mathcal{C}$ or simply $n$-universal for $\mathcal{C}$ if for every graph $G\in\mathcal{C}$ on $n$ vertices there exists a straight-line crossing-free drawing of $G$ such that every vertex of $G$ is placed at a point of $P$.

To determine the minimum size of universal sets for classes of planar graphs is a fundamental problem in geometric graph theory, see e.g. Problem [17] in the Open Problem Garden. More specifically, the quest is for good bounds on the minimum size $f_{\mathcal{C}}(n)$ of an $n$-universal point set for a class $\mathcal{C}$.

Schnyder [21] showed that for $n\geq 3$ the $[n-1]\times[n-1]$-grid forms an $n$-universal point set for planar graphs, even if the combinatorial embedding of the planar graph is prescribed. This shows that $f(n):=f_{\mathcal{P}}(n)\leq n^{2}\in O(n^{2})$, where $\mathcal{P}$ is the class of all planar graphs. Asymptotically, the quadratic upper bound on $f(n)$ remains the state of the art. Only the multiplicative constant in this bound has seen some improvement, the current upper bound is $f(n)\leq\frac{1}{4}n^{2}+O(n)$ by Bannister et al. [5]. For several subclasses $\mathcal{C}$ of planar graphs, better upper bounds are known: A classical result by Gritzmann et al. [13] is that every outerplanar $n$-vertex graph embeds straight-line on any set of $n$ points in general position, and hence $f_{\textrm{out-pl}}(n)=n$. Near-linear upper bounds of $f_{\mathcal{C}}(n)=O(n\;\text{polylog}(n))$ are known for $2$-outerplanar graphs, simply nested graphs, and for the classes of bounded pathwidth [4, 5]. Finally, for the class $\mathcal{C}$ of planar $3$-trees (also known as Apollonian networks or stacked triangulations), $f_{\mathcal{C}}(n)=O(n^{3/2}\log n)$ has been proved by Fulek and Tóth [12].

As for lower bounds, the trivial bounds $n\leq f_{\mathcal{C}}(n)\leq f(n)$ hold for all $n\in\mathbb{N}$ and all planar graph classes $\mathcal{C}$. The current lower bound $f(n)\geq 1.293n-o(n)$ from [20] has been shown using planar $3$-trees, we refer to [6, 8, 9, 15] for earlier work on lower bounds.

Choi, Chrobak and Costello [7] recently proved that point sets chosen uniformly at random from the unit square must have size $\Omega(n^{2})$ to be universal for $n$-vertex planar graphs with high probability. This suggests that universal point sets of size $o(n^{2})$ -if they exist- will not look nice, e.g., they will have a large ratio between shortest and largest distances.

In this paper we study a specific ordered point set $H$ (the exploding double chain) and denote the initial piece of size $2n-2$ in $H$ as $H_{n}$. Let $\cal{C}$ be the class of all planar graphs $G$ which have a plane straight-line drawing on the point set $H_{n}$ where $n=|V(G)|$. That is, $H_{n}$ forms an $n$-universal point set for $\cal{C}$.

A graph is POSH (partial one-sided Hamiltonian) if it is a spanning subgraph of a graph admitting a plane embedding with a one-sided Hamiltonian cycle (for definitions see Section 2). Triangulations with a one-sided Hamiltonian cycle have been studied before by Alam et al. [2] in the context of cartograms. They conjectured that every plane 4-connected triangulation has a one-sided Hamiltonian cycle. Later Alam and Kobourov [3] found a plane 4-connected triangulation on 113 vertices which has no one-sided Hamiltonian cycle.

Our main result (Theorem 3.1) is that every POSH graph is in $\cal{C}$. We let

[TABLE]

Theorem 3.1 motivates further study of $\cal{C}^{\prime}$. On the positive side we show that every bipartite plane graph is POSH (proof in Section 4). We proceed to use the construction for bipartite graphs to show that subcubic planar graphs have a POSH embedding in Section 5. On the negative side, we also show that not all $2$-trees are POSH. We conclude with some conjectures and open problems in Section 7.

An exploding double chain was previously used by Löffler and Tóth [16]. They show that every planar graph with $n$ vertices has a 1-bend drawing on a subset $S_{n}$ of $H$ with $|S_{n}|=6n-10$. Our result about bipartite graphs implies a better bound:

Corollary 1.1**.**

There is a point set $P=H_{2n-2}$ of size $4n-6$ such that every $n$-vertex planar graph admits a 1-bend drawing with bends and vertices on $P$.

Proof 1.2**.**

The dual of a plane triangulation is a bridgeless 3-regular graph of $2n-4$ vertices; it has a perfect matching by Petersen’s Theorem [19]. Hence, subdividing at most $n-2$ edges can make any planar graph on $n$ vertices bipartite. Thus $H_{n+n-2}$ of size $2(n+n-2)-2=4n-6$ is sufficient to accomodate 1-bend drawings of all $n$-vertex planar graphs.

Universality for 1-bend and 2-bend drawings with no restriction on the placement of bends has been studied by Kaufmann and Wiese [14], they show that every $n$-element point set is universal for 2-bend drawings of planar graphs.

2 The point set and the class of POSH graphs

In this section we define the exploding double chain $H$ and the class $\cal{C}^{\prime}$ of POSH graphs and show that for every $n\geq 2$ the initial part $H_{n}$ of size $2n-2$ of $H$ is $n$-universal for $\cal{C}^{\prime}$.

A sequence $(y_{i})_{i\in\mathbb{N}{}}$ of real numbers satisfying $y_{1}=0$, $y_{2}=0$ is exploding and the corresponding point set $H=\{p_{i},q_{i}|{i\in\mathbb{N}}\}$, where $p_{i}=(i,y_{i}),q_{i}=(i,-y_{i})$, is an exploding double chain, if for all $n\in\mathbb{N}$, $y_{n+1}$ is large enough that all intersections of lines going through two points of $H_{n}=\{p_{i},q_{i}|{i\in[n]}\}$ with the line $x=n+1$ lie strictly between $y_{n+1}$ and $-y_{n+1}$. It is $p_{1}=q_{1}$ and $p_{2}=q_{2}$, thus $|H_{n}|=2n-2$. Figure 1 shows $H_{6}$. This fully describes the order type of the exploding double chain. Note that the coordinates given here can be made integers, but the largest coordinate of $H_{n}$ is exponential in $n$, which is unavoidable for the order type. However, the ratio of largest to smallest distance does not have to be: We can alter the construction setting $y_{i}=i$, but letting the $x$-coordinates grow slowly enough as to achieve the same order type, but with a linear ratio.

An explicit construction of a point set $H$ in this order type is given now.

A sequence $Y=(y_{i})_{i\geq 1}$ of real numbers satisfying $y_{1}=0$, $y_{2}=0$, and $y_{i+1}>2y_{i}+y_{i-1}$ for all $i\geq 2$ is exploding. Note that if $\alpha>1+\sqrt{2}$, then $y_{1}=y_{2}=0$ and $y_{i}=\alpha^{i-3}$ for $i\geq 3$ is an exploding sequence, e.g. $\alpha=3$. Given an exploding sequence $Y$ let $P(Y)=(p_{i})_{i\geq 1}$ be the set of points with $p_{i}=(i,y_{i})$ and let $\bar{P}(Y)=(q_{i})_{i\geq 1}$ be the set of points with $q_{i}=(i,-y_{i})$, i.e., the point set reflected at the $x$-axis.

Let $H=H(Y)$ for some exploding sequence $Y$. For two points $p$ and $q$ let $H(p,q)$ be the set of points of $H$ in the open right half-plane of the directed line $\overrightarrow{pq}$. Note that111In cases where $i$ or $j$ are in $\{1,2\}$ the following may list one of the two points defining the halfspace with its second name as member of the halfspace. For correctness such listings have to be ignored.

[TABLE]

Moreover, if $i<j$ then $H(q_{i},q_{j})=H(p_{i},q_{j})\setminus\{q_{i}\}$ and if $i>j$ then $H(p_{i},p_{j})=H(p_{i},q_{j})\setminus\{p_{j}\}$. These sidedness conditions characterize the order type of the exploding double chain.

A plane graph $G$ has a one-sided Hamiltonian cycle with special edge $vu$ if it has a Hamiltonian cycle $(v=v_{1},v_{2},\ldots,v_{n}=u)$ such that $vu$ is incident to the outer face and for every $j=2,\ldots,n$, the two edges incident to $v_{j}$ in the Hamiltonian cycle, i.e., edges $v_{j-1}v_{j}$ and $v_{j+1}v_{j}$, are consecutive in the rotation of $v_{j}$ in the subgraph induced by $v_{1},\ldots,v_{j},v_{j+1}$ in $G$. In particular, the one-sided condition depends on the Hamiltonian cycle, its direction and its special edge. A more visual reformulation of the second condition is obtained using the closed bounded region $D$ whose boundary is the Hamiltonian cycle. It is that in the embedding of $G$ for every $j$ either all the back-edges $v_{i}v_{j}$ with $i<j$ are drawn inside $D$ or in the open exterior of $D$. We let $V_{I}$ be the set of vertices $v_{j}$ which have a back-edge $v_{i}v_{j}$ with $i<j-1$ drawn inside $D$ and $V_{O}=V\setminus V_{I}$. The set $V_{I}$ is the set of vertices having back-edges only inside $D$ while vertices in $V_{O}$ have back-edges only outside $D$.

Recall that $\cal{C}^{\prime}$ is the class of planar graphs which are spanning subgraphs of plane graphs admitting a one-sided Hamiltonian cycle. It is worth noting all subgraphs are POSH.

Proposition 2.1**.**

Any subgraph of a POSH graph is POSH.

Proof 2.2**.**

As edge deletions preserve the POSH property by definition, it suffices to show that deleting a vertex preserves it as well. Let $G$ be a POSH graph and let $G^{\prime}$ be its supergraph with a one-sided Hamiltonian cycle. Now after deleting $v$ from $G^{\prime}$, adding an edge between its neighbours on the Hamiltonian cycle (if it does not exist) can be done along the two edges of $v$ along the cycle. This is a supergraph of $G\setminus v$ with a one-sided Hamiltonian cycle.

3 The embedding strategy

Our interest in POSH graphs is motivated by the following theorem.

Theorem 3.1**.**

Let $G^{\prime}$ be POSH and let $v_{1},\ldots,v_{n}$ be a one-sided Hamiltonian cycle of a plane supergraph $G$ of $G^{\prime}$ on the same vertex set. Then there is a crossing-free embedding of $G^{\prime}$ on $H_{n}$ with the property that $v_{i}$ is placed on either $p_{i}$ or $q_{i}$.

Proof 3.2**.**

It is sufficient to describe the embedding of the supergraph $G$ on $H_{n}$. For the proof we assume that in the plane drawing of $G$ the sequence $v_{1},\ldots,v_{n}$ traverses the boundary of $D$ in counter-clockwise direction. For each $i$ vertex $v_{i}$ is embedded at $\bar{v}_{i}=p_{i}$ if $v_{i}\in V_{I}$ and at $\bar{v}_{i}=q_{i}$ if $v_{i}\in V_{O}$.

Let $G_{i}=G[v_{1},\ldots,v_{i}]$ be the subgraph of $G$ induced by $\{v_{1},\ldots,v_{i}\}$. The path $\Lambda_{i}=v_{1},\ldots,v_{i}$ separates $G_{i}$. The left part $GL_{i}$ consists of the intersection of $G_{i}$ with $D$, the right part $GR_{i}$ is $G_{i}$ minus all edges which are interior to $D$. The intersection of $GL_{i}$ and $GR_{i}$ is $\Lambda_{i}$ and their union is $G_{i}$. The counter-clockwise boundary walk of $G_{i}$ consists of a path $\partial R_{i}$ from $v_{1}$ to $v_{i}$ which is contained in $GR_{i}$ and a path from $v_{i}$ to $v_{1}$ which is contained in $GL_{i}$, let $\partial L_{i}$ be the reverse of this path.

Let $\bar{G}_{i}$ be the straight-line drawing of the plane graph $G_{i}$ obtained by placing each vertex $v_{j}$ at the corresponding $\bar{v}_{j}$. A vertex $\bar{v}$ of $\bar{G}_{i}$ is said to see a point $p$ if there is no crossing between the segment $\bar{v}p$ and an edge of $\bar{G}_{i}$. By induction on $i$ we show:

The drawing $\bar{G}_{i}$ is plane, i.e., non-crossing. 2. 2.

$\bar{G}_{i}$* and $G_{i}$ have the same outer boundary walks.* 3. 3.

Every vertex of $\partial L_{i}$ in $\bar{G}_{i}$ sees all the points $p_{j}$ with $j>i$ and every vertex of $\partial R_{i}$ in $\bar{G}_{i}$ sees all the points $q_{j}$ with $j>i$.

For $i=2$ the graph $G_{i}$ is just an edge and the three claims are immediate, for Property 3 just recall that the line spanned by $p_{1}$ and $p_{2}$ separates the $p$-side and the $q$-side of $H_{n}$.

Now assume that $i\in\{3,\ldots,n\}$, the properties are true for $\bar{G}_{i-1}$ and suppose that $v_{i}\in V_{I}$ (the argument in the case $v_{i}\in V_{O}$ works symmetrically). This implies that all the back-edges of $v_{i}$ are in the interior of $D$ whence all the neighbors of $v_{i}$ belong to $\partial L_{i-1}$. Since $v_{i}\in V_{I}$ we have $\bar{v}_{i}=p_{i}$ and Property 3 of $\bar{G}_{i-1}$ implies that the edges connecting to $\bar{v}_{i}$ can be added to $\bar{G}_{i-1}$ without introducing a crossing. This is Property 1 of $\bar{G}_{i}$.

Since $G_{i-1}$ and $\bar{G}_{i-1}$ have the same boundary walks and $v_{i}$ (respectively $\bar{v}_{i}$) belong to the outer faces of $G_{i}$ (respectively $\bar{G}_{i}$) and since $v_{i}$ has the same incident edges in $G_{i}$ as $\bar{v}_{i}$ in $\bar{G}_{i}$, the outer walks of $G_{i}$ and $\bar{G}_{i}$ again equal each other, i.e., Property 2.

Let $j$ be minimal such that $v_{j}v_{i}$ is an edge and note that $\partial L_{i}$ is obtained by taking the prefix of $\partial L_{i-1}$ whose last vertex is $v_{j}$ and append $v_{i}$. The line spanned by $\bar{v}_{j}$ and $\bar{v}_{i}=p_{i}$ separates all the edges incident to $\bar{v}_{i}$ in $\bar{G}_{i}$ from all the segments $\bar{v}_{\ell}p_{k}$ with $\ell<j$ and $\bar{v}_{\ell}\in\partial L_{i}$ and $k>i$. This shows that every vertex of $\partial L_{i}$ in $\bar{G}_{i}$ sees all the points $p_{k}$ with $k>i$. For the proof of the second part of Property 3 assume some edge $\bar{v}_{i}\bar{v}_{j}$ crosses the line of sight from $\bar{v}_{l}$ to $q_{k},k>i$, we refer to Figure 3. First note that this is only possible if $l\leq j$, since otherwise $\bar{v}_{j}\bar{v}_{l}$ separates $\bar{v}_{i}=p_{i}$ and $q_{k}$, because $p_{i}$ is on the left as can be seen at $x=i$ and $q_{k}$ is on the right as can be seen at $x=k$ by definition. Since $j=l$ is impossible by construction, we are left with the case $l<j$. Then one of $\bar{v}_{i}$ and $\bar{v}_{l}$, say $\bar{v}$, lies to the right of the oriented line $\bar{v}_{j}q_{k}$. However that implies that $\bar{v}_{j}\bar{v}$ has $q_{k}$ on its left, which is a contradiction to the definition of $q_{k}$ at $x=k$. This completes the proof of Property 3 and thus the inductive step.

Finally, Property 1 for $\bar{G}_{n}$ implies the theorem.

4 Plane bipartite graphs

In this section we consider bipartite plane graphs and show that they are POSH.

Theorem 4.1**.**

Every bipartite plane graph $G=(V,E)$ is a subgraph of a plane graph $G^{\prime}$ on the same vertex set $V$ which has a one-sided Hamiltonian cycle, i.e., $G$ is POSH.

Proof 4.2**.**

Quadrangulations are the plane graphs with all faces of degree four. Equivalently they are the maximal plane bipartite graphs, i.e., any bipartite plane graph except stars is a subgraph of a quadrangulation. Thus since POSH graphs are closed under taking subgraphs, it suffices to prove the theorem for quadrangulations.

Let $Q$ be a quadrangulation and let $V_{B}$ and $V_{W}$ be the black and white vertices of a 2-coloring. Label the two black vertices of the outer face as $s$ and $t$. Henceforth, when talking about a quadrangulation we think of an embedded quadrangulation endowed with $s$ and $t$. A separating decomposition is a pair $D=(Q,Y)$ where $Q$ is a quadrangulation and $Y$ is an orientation and coloring of the edges of $Q$ with colors red and blue such that:

The edges incident to $s$ and $t$ are incoming in color red and blue, respectively. 2. 2.

Every vertex $v\not\in\{s,t\}$ is incident to a non-empty interval of red edges and a non-empty interval of blue edges. If $v$ is white, then, in clockwise order, the first edge in the interval of a color is outgoing and all the other edges of the interval are incoming. If $v$ is black, the outgoing edge is the clockwise last in its color (see Figure 4).

Separating decompositions of a quadrangulation $Q$ have been defined by de Fraysseix and Ossona de Mendez [18]. They show a bijection between separating decompositions and $2$-orientations (orientations of the edges of $Q$ such that every vertex $v\not\in\{s,t\}$ has out-degree $2$) and show the existence of a $2$-orientation of $Q$ with an argument related to flows and matchings. An inductive proof for the existence of separating decompositions was given by Felsner et al. [11], this proof is based on identifying pairs of opposite vertices on faces.

In a separating decomposition the red edges form a tree directed towards $s$, and the blue edges form a tree directed towards $t$. Each of the trees connects all the vertices $v\not\in\{s,t\}$ to the respective root. Felsner et al. ([10, 11]) show that the edges of the two trees can be separated by a curve which starts in $s$, ends in $t$, and traverses every vertex and every inner face of $Q$. This curve is called the equatorial line.

If $Q$ is redrawn such that the equatorial line is mapped to the $x$-axis with $s$ being the left end and $t$ being the right end of the line, then the red tree and the blue tree become alternating trees ([11], defined below) drawn in the upper respectively lower half-plane defined by the $x$-axis. Note that such a drawing of $Q$ is a 2-page book embedding, we call it an alternating 2-page book embedding to emphasize that the graphs drawn on the two pages of the book are alternating trees.

An alternating tree is a plane tree $T$ with a plane drawing such that the vertices of $T$ are placed at different points of the $x$-axis and all edges are embedded in the half-plane above the $x$-axis (or all below). Moreover, for every vertex $v$ it holds that all its neighbors are on one side, either they are all left of $v$ or all right of $v$. In these cases we call the vertex $v$ respectively a right or a left vertex of the alternating layout. Note that every vertex is a left vertex in one of the two trees and a right vertex in the other.

Let $Q$ be a plane quadrangulation on $n$ vertices and let $S$ be a separating decomposition of $Q$. Let $s=v_{1},v_{2},\ldots,v_{n}=t$ be the spine of the alternating 2-page book embedding of $Q$ based on $S$. Let $Q^{+}$ be obtained from $Q$ by adding $v_{n}v_{1}$ and all the edges $v_{i}v_{i+1}$ which do not yet belong to the edge set of $Q$. By construction $v_{1},v_{2},\ldots,v_{n}$ is a Hamiltonian cycle of $Q^{+}$ and since the trees are alternating, black vertices have only blue edges to the left and white vertices have only red edges to the left. Thus this Hamiltonian cycle is one-sided with reverse edge $v_{n}v_{1}=ts$. Hence $Q$ is POSH.

It is worth noting that the Hamiltonian cycle read in the reverse direction, i.e., as $v_{n},v_{n-1},\ldots,v_{1}$, is again one-sided, now the reverse edge is $v_{1}v_{n}=st$.

5 Planar subcubic graphs

In this section we identify another large subclass of the $\cal{C}^{\prime}$. Recall that 3-regular graphs are also known as cubic graphs and in subcubic graphs all vertices have degree at most 3.

Theorem 5.1**.**

Every planar subcubic graph $G$ is a spanning subgraph of a planar graph $G^{\prime}$ which has an embedding with a one-sided Hamiltonian cycle, i.e., $G$ has a POSH embedding.

Remark 5.2**.**

Note that we do not claim the theorem for all plane subcubic graphs. However, we are not aware of any connected subcubic plane graph, which is not POSH.

To prove this, we use Theorem 4.1 and the following lemmas:

Lemma 5.3**.**

Let $G$ be a subcubic graph. Then $G$ admits a matching $M$ such that contracting all the edges of $M$ results in a bipartite multi-graph.

Proof 5.4**.**

Let $(X,Y)$ be a partition the vertex-set of $G$ such that the size of the cut, i.e., the number of edges in $G$ with one endpoint in $X$ and one endpoint in $Y$, is maximized. We claim that the induced subgraphs $G[X]$ and $G[Y]$ of $G$ are matchings. Suppose that a vertex $v\in X$ has at least two neighbors in $G[X]$. Then $v$ has at most one neighbor in $Y$, and hence moving $v$ from $X$ to $Y$ increases the size of the cut by at least one, a contradiction. The same argument works for $G[Y]$.

Let $M$ be the matching in $G$ consisting of all the edges in $G[X]$ and $G[Y]$. Contracting the edges in $M$ transforms $G[X]$ and $G[Y]$ into independent sets, and hence results in a bipartite multi-graph $G/M$.

A separating $k$-cycle of a plane graph $D$ is a simple cycle of length $k$, i.e., $k$ edges, such that there are vertices of $D$ inside the cycle.

Lemma 5.5**.**

Let $G$ be a subcubic planar graph. Then $G$ admits a plane embedding $D_{G}$ and a matching $M$ such that contracting all the edges of $M$ in $D_{G}$ results in a bipartite multi-graph without separating 2-cycles.

Proof 5.6**.**

Let $G$ be a subcubic planar graph. Without loss of generality $G$ is connected, otherwise we just deal with the components first, then embed $G$ in a way that all components are incident to the outer face.

Note that a 2-cycle can only arise by contracting one matching edge of a triangle or two matching edges of a quadrilateral. Consider an embedding $D$ of $G$ which minimizes the number of separating 3-cycles and among those minimizes the number of separating 4-cycles.

Claim 1**.**

$D$* has no separating 3-cycle.*

Proof 5.7**.**

For illustration, see Figure 6. We will first show $D$ has no separating diamond, that is, two triangles sharing an edge $e=uv$, at least one of which is a separating 3-cycle. Otherwise place $u$ very closely to $v$. Now $e$ is short and we reroute the other two edges of $u$ such that they stay close to the corresponding edge of $v$. Since one of the triangles containing $e$ was assumed to be separating the new drawing has fewer separating 3-cycles, a contradiction.

We are ready to show $D$ has no separating 3-cycle. If $T$ is a separating 3-cycle some edge has to go from a vertex $v$ of $T$ into its interior. Since $v$ has degree at most 3 it has no edge to the outside of $T$. We can then redraw the edge $e$ of $T$ not incident to $v$ outside of $T$ closely to its two other edges. Again the new drawing has fewer separating 3-cycles: indeed, if the redrawn edge would be part of another 3-cycle, $T$ is part of a separating diamond.

Now choose an edge set $M$ of minimum cardinality, such that contracting it yields a bipartite multi-graph. The proof of Lemma 5.3 implies that $M$ is a matching. Among those matchings, we choose $M$ such that the number of separating 4-cycles which have 2 edges in $M$ is minimized. Such separating 4-cycles are said to be covered by $M$.

Claim 2**.**

$M$* covers no separating 4-cycle.*

Proof 5.8**.**

Suppose $Q=v_{1}v_{2}v_{3}v_{4}$ is a separating 4-cycle such that $v_{1}v_{2}$ and $v_{3}v_{4}\in M$ and $v_{1}$ has an edge $e_{I}$ to the inside, thus no edge to the outside.

If $v_{4}$ has no edge to the outside either, we change $D$ to a drawing $D^{\prime}$ by redrawing the part $\Gamma$ of $D$ inside $Q$ outside of it reflected across $v_{1}v_{4}$, see Figure 7. In $D^{\prime}$ the original separating 4-cycle is no longer separating. We claim that no new separating 3-cycle or 4-cycle that is covered by $M$ was created. The claim contradicts the choice of $D$ or $M$.

To prove the claim note that $S=\{v_{2},v_{3}\}$ is a 2-separator, unless $Q$ is the outer face of $D$, so let’s assume first that it is not. Thus a separating 3- or 4-cycle has to live on one side of $S$, since the shortest path between them in $Q\cup\Gamma$ except their edge is of length 3 except if both $v_{2}$ and $v_{3}$ are adjacent to the same vertex of $\Gamma$, in which case $Q$ is the outer face, a contradiction. Let $X$ be the component of $G\setminus S$ containing $\Gamma$. Then the number of vertices inside 3- or 4-cycles that are not part of $X$ is unchanged in $D^{\prime}$, since the face $X$ is located in is still the same. The only 3- or 4-cycles in $X\cup S$ that were not reflected in their entirety are the ones containing the edge $v_{2}v_{3}$. Since $Q$ is assumed not to be the outer face, at least one of $v_{2}$ and $v_{3}$ is not connected to $\Gamma$. Thus such a cycle $C$ is a 4-cycle consisting of $v_{2},v_{3}$, one of $v_{1}$ or $v_{4}$ as well as a common neighbour of $v_{2}$ and $v_{4}$ or $v_{1}$ and $v_{3}$ in $\Gamma$. However $v_{1}v_{2}$ or $v_{3}v_{4}$ respectively would be the only edge in $M\cap C$. This is a contradiction to the fact that contracting $M$ yields a bipartite graph.

Now if $Q$ is the outer face of $D$, it is still true that the only cycles not reflected in their entirety contain $v_{2}v_{3}$. However $v_{2}$ and $v_{3}$ could both be adjacent to a vertex in $\Gamma$, either a common neighbour for a 3-cycle or two adjacent neighbours for a 4-cycle. Since $v_{2}$ and $v_{3}$ are already covered by $M$, this 3-cycle would contain no edge in $M$, whereas the 4-cycle would contain at most one. Therefore both of these contradict the definition of $M$.

Therefore, we know that $v_{4}$ has an edge $e_{O}$ to the outside. This edge does not go to any vertex of the quadrilateral, because the only candidate left would be $v_{2}$, but this would yield that one of the triangles $v_{2}v_{3}v_{4}$ and $v_{1}v_{2}v_{4}$ is separating.

Change the matching $M$ to an edge set $M^{\prime}$ by removing $v_{1}v_{2}$ and $v_{3}v_{4}$ from it and adding $e_{O}$ and $e_{I}$. Contracting $M^{\prime}$ still results in a bipartite graph, because the same four facial cycles that contained our previous edges contain exactly one new edge each as well, so their size after contraction does not change. Thus $M^{\prime}$ is a matching, because it has the same cardinality as $M$ and is therefore minimal as well. We conclude $M^{\prime}$ does not cover $v_{2}$ or $v_{3}$, because $M$ did not contain any other edge than $v_{1}v_{2}$ and $v_{3}v_{4}$ at them either. Since $M^{\prime}$ does not contain two edges from quadrilateral $v_{1},\ldots,v_{4}$ but $M$ is minimal, there has to be a separating quadrilateral, of which $M^{\prime}$ contains two edges, but $M$ doesn’t. If such a separating quadrilateral $Q$ contains $e_{I}$, then it has to contain another edge incident to $v_{1}$. It cannot contain $v_{1}v_{2}$, because we know $v_{2}$ is not covered by $M^{\prime}$. Therefore it contains $v_{1}v_{4}$ and consequently $e_{O}$. The same argumentation works to show that if it contains $e_{O}$, then it also contains $e_{I}$. This is a contradiction to the existence of $M^{\prime}$ because the endpoints of $e_{O}$ and $e_{I}$ are on the outside and the inside of the quadrilateral respectively and therefore non-adjacent.

So we proved that our choice of $M$ makes sure that no separating 2-cycles will be present in the contracted plane bipartite multi-graph.

Remark 5.9**.**

The embedding $D$ and the matching $M$ can be constructed starting from an arbitrary embedding and matching by iterative application of the operations used in the proof.

Proof 5.10** (Proof of Theorem 5.1).**

Now let $B$ be the plane bipartite multi-graph obtained from $G$ by contracting the edges in $M$ without changing the embedding any further. Let $B^{\prime}$ be the underlying simple graph of $B$ and let $Q$ be a quadrangulation or a star which has $B^{\prime}$ as a spanning subgraph. The proof of Theorem 4.1 shows that there is a left to right placement $v_{1},\ldots,v_{s}$ of the vertices of $Q$ on the $x$-axis such that for each $i\in[s]$ all the edges $v_{j}v_{i}$ with $j<i-1$ are in one half-plane and all edges $v_{i}v_{j}$ with $j>i+1$ are in the other half-plane. Delete all the edges from $Q$ which do not belong to $B^{\prime}$, and duplicate the multi-edges of $B$ in the drawing. This yields a 2-page book embedding $\Gamma$ of $B$.

Let $v$ be a contracted vertex of $B$. Vertex $v$ was obtained by contracting an edge $uw\in M$. If $u$ and/or $w$ did not have degree 3, we add edges at the appropriate places into the embedding that end in leaves, see Figure 8. To add an edge to $u$ for instance, choose a face $f$ incident to $u$ that is not contracted into a 2-cycle. Let $e$ and $e^{\prime}$ be the two edges incident to both $v$ and $f$. If the angle between $e$ and $e^{\prime}$ contains part of the spine (the $x$-axis), we put the leaf on the spine close to $v$ connected to $v$ with a short edge below or above the spine, in a way to accomodate the local vertex condition of $v$. If it doesn’t, assume without loss of generality it is in the upper half-plane and that edge $e$ is the edge closer to the spine. This edge is unique because both edges at $v$ delimiting $f$ go upwards and therefore both to the same side, say right of $v$. Route the new edge closely along $e$ then put the leaf just next to the other endpoint $x$ of $e$. Edges that would cross this new edge cannot cross $e$, thus the only possibility are edges incident to $x$ that emanate into the upper halfspace. However those edges have to go to the left of $x$ by its local vertex condition. These edges do not exist, as any such edge would have to cross $e^{\prime}$, see the dashed line in Figure Figure 8. Thus the new edge is uncrossed. This procedure will be done to every vertex first. Note that the resulting graph stays bipartite and the local vertex conditions are still fulfilled, but now every contracted vertex has degree 4. This makes the case distinction of splitting the vertices easier.

We now show how to undo the contractions, i.e., split vertices, in the drawing $\Gamma$ in such a way that at the end we arrive at a one-sided 2-page book drawing $\Gamma^{\star}$ of $G$, that is, a $2$-book embedding of $G$ with vertex-sequence $v_{1},\ldots,v_{n}$ such that for every $j\in\{1,\ldots,n\}$ the incident back-edges $v_{i}v_{j}$ with $1\leq i<j$ are all drawn either on the spine or on the same page of the book embedding (all above or all below the spine). Once we have obtained such a book embedding, we can delete the artificial added leaves, then add the spine edges (including the back edge from the rightmost to the leftmost vertex) to $G$ to obtain a supergraph $G^{+}$ of $G$ which has a one-sided Hamiltonian cycle, showing that $G$ is POSH.

Before we advance to show how we split a single vertex $v$ of degree four into an edge $uw\in M$, we first want to give an overview of the order in which the different splits, the far splits and local splits are applied. We will then describe what these different splits actually mean. To split all the degree four vertices we proceed as follows:

First we split all vertices which are subject to a far split, from the outside inwards. More precisely, define a partially ordered set on the edges incident222 There will be a clarification later as to what this means exactly. to vertices subject to a far split in the following way: Every edge $e$ defines a region $R_{e}$ which is enclosed by $e$ and the spine. Now order the edges by the containment order of regions $R_{e}$. From this poset, choose a maximum edge and then a vertex that needs a far split incident to that edge. When no further far split is possible we do all the local splits. These splits are purely local, so they cannot conflict with each other. Therefore their order can be chosen arbitrarily.

We label the edges of $v$ in clockwise order as $e_{1},e_{2},e_{3},e_{4}$ such that in $G$ the edges $e_{1},e_{2}$ are incident to $u$ and $e_{3},e_{4}$ are incident to $w$. If the two angles $\angle e_{2}e_{3}$ and $\angle e_{4}e_{1}$ together take part of both half-planes defined by the spine, then it is possible to select two points left and right of the point representing $v$ in $\Gamma$ and to slightly detour the edges $e_{i}$ such that no crossings are introduced and one of the two points is incident to $e_{1},e_{2}$ and the other to $e_{3},e_{4}$. The addition of an edge connecting the two points completes the split of $v$ into the edge $uw\in M$. Figure 9 shows a few instances of this local split.

The above condition about the two angles is not fulfilled if and only if all four edges of $v$ emanate into the same halfspace, say the upper one, and the clockwise numbering starting at the $x$-axis is either $e_{4},e_{1},e_{2},e_{3}$ or $e_{2},e_{3},e_{4},e_{1}$. The two cases are the same up to exchanging the names of $u$ and $w$, therefore we can assume the first one. A more important distinction is whether most $e_{i}$ end to the left or right of $v$. Note that in the ordering given by $\Gamma$, all $e_{i}$ go to the same side, since they are all in the same halfplane. However, if $v$ is not the first vertex we are splitting, it may happen, that a single edge on the spine to the other side exists, see Figure 10. For all $i\in[4]$ let $v_{i}$ be the other endpoint of $e_{i}$ than $v$. While it can happen that some of the $v_{i}$ coincide due to multi-edges, we will first discuss the case that they don’t. In the left case we put $u$ slightly left of $v_{1}$ while in the right case $u$ is put slightly right of $v_{2}$, connecting $u$ to this close vertex by a spine edge. In both cases we leave $w$ at the former position of $v$. Figure 10 shows the right case and Figure 11 the left.

To see that in the left case edges $uv_{2}$ and $uw$ are completely free of crossings, observe that we can route them close to the path $v_{2}vv_{1}$ and the edge $v_{1}v$ respectively in the original drawing (dashed in Figure 11). It is important to note here, that due to the order in which we chose to do the splits, $v_{1}$ and $v_{2}$ are still original vertices of $B$, that is, they have not been split in the upper half-plane and thus still don’t have two edges emanating to the upper half-plane to both sides. Therefore, similarly to the argumentation for adding leaves, no edge incident to $v_{1}$ crosses $uw$ or $uv_{2}$. The right case is analogous, just exchange the roles of $v_{1}$ and $v_{2}$.

This kind of split is a far split. For the purposes of incidence in the poset structure mentioned above, vertices are not only considered incident to any edge they are an endpoint of, but the spine neighbour of $u$ ($v_{1}$ or $v_{2}$) is also considered to be incident to the edge $uw$. For illustration, consider the outermost black edge in Figure 10 (left), it is considered incident to $v$.

In the following we describe how the different kinds of splits are affected by the presence of multi-edges. The first thing to note is that local splits can be done in the same way, since we did not mention the end vertices at all.

Concerning the far splits, firstly we talk about the case that exactly two edges go from one vertex to another: As depicted in Figures  10 and 11 the case $v_{2}=v_{3}$ and/or $v_{4}=v_{1}$ is unproblematic, in this case we keep the dashed line(s) in the drawing. Double-edges are consecutive because non-consecutive double-edges are separating 2-cycles, which we avoided in the construction. Thus the last case of a double-edge to consider is $v_{1}=v_{2}$. In this case, we follow the same strategy of placement of $u$ and $w$, but this results in a double-edge on the spine between $u$ and $v_{1}=v_{2}$, see Figure 12. As in later local splits, we might be interested what half-space the angle between the two spine edges is part of, we interpret one of these edges as a spine edge and the other as an edge which is above or below the spine depending on the right vertex of the two. This might be $u$ or $v_{1}$, depending on whether we are in the left or right case. It is important for the one-sidedness condition to choose this direction so that all left neighbours of the right vertex of the two are reached by edges emanating into the same halfspace and/or spine edges.

Secondly, if there are three edges between a left vertex $v_{\ell}$ and a right vertex $v_{r}$, say in the upper half-plane, we will split both simultaneously, for illustration, see Figure 13. Since three edges go between these two vertices, there is just one more edge $e$ left for $v_{\ell}$. Therefore we can find a place on the spine just to the right or to the left of $v_{\ell}$ which is free, because the edge $e$ is on the other side. Now we split $v_{\ell}$ into $u_{\ell}$ and $w_{\ell}$ and $v_{r}$ into $u_{r}$ and $w_{r}$ simultaneously where $w_{\ell}$ and $w_{r}$ are the vertices with the edge that goes somewhere else on both sides. From left to right we put $u_{r}$ then $u_{\ell}$ just left of the position of $v_{l}$, which is the new position of $w_{\ell}$. The three of them are connected by spine edges, just $u_{r}$ and $w_{\ell}$ have an edge in the lower half-plane. These edges are not crossed, because the vertices are close enough together. Finally we put $w_{r}$ at the position of $v_{r}$ and add edges to $w_{r}$ and $w_{\ell}$ in the upper half-plane. These edges are not crossed, because any edge crossing them would have crossed the triple edge in the original drawing.

This kind of split is a double split. These splits are purely local, so they can be performed together with the local splits in the end.

The last case is that all four edges of a given vertex go to the same vertex, this is a full connected component of the bipartite graph, because it has maximum degree 4. This component goes back to a $K_{4}$ component in the cubic graph that had two independent edges contracted. A one-sided Hamiltonian cycle of $K_{4}$ is illustrated in Figure 2. We apply another local double split which consists of replacing the 4 parallel edges by this drawing, embedded close to the place of one of the original vertices.

This completes the proof of Theorem 5.1.

6 2-Trees

From the positive results in Sections 4 and 5 one might expect that “sufficiently sparse” planar graphs are POSH. This section shows that 2-trees are not.

A $2$-tree is a graph which can be obtained, starting from a $K_{3}$, by repeatedly selecting an edge of the current graph and adding a new vertex which is made adjacent to the endpoints of that edge. We refer to this operation as stacking a vertex over an edge.

From the recursive construction it follows that a $2$-tree on $n$ vertices is a planar graph with $2n-3$ edges. We also mention that $2$-trees are series-parallel planar graphs. Another well studied class which contains $2$-trees as a subclass is the class of (planar) Laman graphs.

Fulek and Tóth have shown that planar 3-trees admit $n$-universal point sets of size $O(n^{3/2}\log n)$. Since every $2$-tree is an induced subgraph of a planar 3-tree the bound carries over to this class.

Theorem 6.1**.**

There is a 2-tree $G$ on 499 vertices that is not POSH.

Proof 6.2**.**

Throughout the proof we assume that a $2$-tree $G$ is given together with a left to right placement $v_{1},\ldots,v_{n}$ of the vertices on the $x$-axis such that adding the spine edges and the reverse edge $v_{n}v_{1}$ to $G$ yields a plane graph with a one-sided Hamiltonian cycle.

For an edge $e$ of $G$ we let $X(e)$ be the set of vertices which are stacked over $e$ and $S(e)$ the set of edges which have been created by stacking over $e$, i.e., each edge in $S(e)$ has one vertex of $e$ and one vertex in $X(e)$. We partition the set $X(e)$ of an edge $e=v_{i}v_{j}$ with $i<j$ into a left part $X\kern-2.0ptL(e)=\{v_{k}\in X(e):k<i\}$, a middle part $X\kern-2.0ptM(e)=\{v_{k}\in X(e):i<k<j\}$, and a right part: $X\kern-2.0ptR(e)=\{v_{k}\in X(e):j<k\}$.

Claim 3**.**

For every edge $|X\kern-2.0ptR(e)|\leq 2$.

Suppose that $|X\kern-2.0ptR(e)|\geq 3$. Each vertex in this set has all its back-edges on the same side. Two of them use the same side for the back edges to the vertices of $e$. This implies a crossing pair of edges, a contradiction.

Claim 4**.**

If for all $e^{\prime}\in S(e)$ we have $|X(e^{\prime})|\geq 3$, then $|X\kern-2.0ptM(e)|\leq 3$.

Suppose that $e=v_{i}v_{j}$ with $i<j$ is in the upper half-plane and there are four vertices $x_{1},x_{2},x_{3},x_{4}$ in $X\kern-2.0ptM(e)$. One-sidedness implies that the four edges $x_{k}v_{j}$ are in the upper half-plane. Thus if $x_{1},x_{2},x_{3},x_{4}$ is the left to right order, then the edges $v_{i}x_{2}$, $v_{i}x_{3}$, and $v_{i}x_{4}$ have to be in the lower half-plane. Now let $e^{\prime}=v_{i}x_{3}$ and consider the three vertices in $X(e^{\prime})$. Two of them, say $y_{1},y_{2}$, are on the same side of $x_{3}$. First suppose $y_{1},y_{2}\in X(e^{\prime})$ are left of $x_{3}$. The edges of $v_{i}x_{2}$ and $x_{2}v_{j}$ enforce that $y_{1},y_{2}$ are between $x_{2}$ and $x_{3}$. Due to edge $x_{2}v_{j}$ the edges $v_{i}y_{1},v_{i}y_{2}$ are in the lower half-plane. One-sidedness at $x_{3}$ requires that $y_{1}x_{3}$ and $y_{2}x_{3}$ are also in the lower half-plane. This makes a crossing unavoidable.

Now suppose that $y_{1},y_{2}\in X(e^{\prime})$ are right of $x_{3}$. The edges $v_{i}x_{4}$ and $x_{4}v_{j}$ enforce that $y_{1},y_{2}$ are between $x_{3}$ and $x_{4}$. Due to the edge $x_{3}v_{j}$ the edges $v_{i}y_{1}$ and $v_{i}y_{2}$ are in the lower half-plane. Now let $y_{1}$ be left of $y_{2}$. One-sidedness at $y_{2}$ requires that $x_{3}y_{2}$ is also in the lower half-plane, whence, there is a crossing between $v_{i}y_{1}$ and $x_{3}y_{2}$. This completes the proof of the claim.

Claim 5**.**

If $X\kern-2.0ptL(e)\geq 2$ and $x$ is the rightmost element of $X\kern-2.0ptL(e)$, then $X\kern-2.0ptL(e^{\prime})\leq 1$ for some $e^{\prime}\in S(e)$ incident with $x$ and $X\kern-2.0ptR(e^{\prime})=\emptyset$ for both.

Suppose that $e=v_{i}v_{j}$ with $i<j$ is in the upper half-plane and there are two vertices $x_{1},x_{2}$ in $X\kern-2.0ptL(e)$. We assume that $x_{2}$ is the rightmost element of $X\kern-2.0ptL(e)$. From one-sidedness at $v_{j}$ we know that $x_{1}v_{j}$ and $x_{2}v_{j}$ are in the upper half-plane. Now $x_{1}v_{i}$ and hence also $x_{2}v_{i}$ are in the lower half-plane. All the vertices of $X(x_{2}v_{i})$ and $X(x_{2}v_{j})$ are in the region bounded by $x_{1}v_{j},v_{j}v_{i},v_{i}x_{1}$, in particular $X\kern-2.0ptR(e^{\prime})=\emptyset$ for both. Suppose for contradiction that we have $y_{1},y_{2}\in X\kern-2.0ptL(x_{2}v_{i})$ and $z_{1},z_{2}\in X\kern-2.0ptL(x_{2}v_{j})$. By one-sidedness the edges from $x_{2}$ to the four vertices $y_{1},y_{2},z_{1},z_{2}$ are in the same half-plane. If they are in the lower half-plane and $y_{1}$ is left of $y_{2}$ there is a crossing between $y_{1}x_{2}$ and $y_{2}v_{i}$. If they are in the upper half-plane and $z_{1}$ is left of $z_{2}$ there is a crossing between $z_{1}x_{2}$ and $z_{2}v_{j}$. The contradiction shows that $X\kern-2.0ptL(x_{2}v_{i})\leq 1$ or $X\kern-2.0ptL(x_{2}v_{j})\leq 1$, since $x=x_{2}$ this completes the proof of the claim.

We are ready to define the graph $G$ and then use the claims to prove that $G$ is not POSH. The graph $G$ contains a base edge $e$ and seven vertices stacked on $e$, i.e., $|X(e)|=7$. For each edge $e^{\prime}\in S(e)$ there are five vertices stacked on $e^{\prime}$. Finally, for each edge $e^{\prime\prime}$ introduced like that three vertices are stacked on $e^{\prime\prime}$. Note that there are $7\cdot 2=14$ edges $e^{\prime}$, $14\cdot 5\cdot 2=140$ edges $e^{\prime\prime}$ and $140\cdot 3\cdot 2=840$ edges introduced by stacking on an edge $e^{\prime\prime}$. In total the number of edges is $995=2n-3$, hence, the graph has 499 vertices.

Now suppose that $G$ is POSH and let $v_{1},\ldots,v_{n}$ be the order of vertices on the spine of a certifying 2-page book embedding. Let $e=v_{i}v_{j}$ with $i<j$ be the base edge. Assume by symmetry that $e$ is in the upper half-plane. From Claim 3 we get $|X\kern-2.0ptR(e)|\leq 2$ and from Claim 4 we get $|X\kern-2.0ptM(e)|\leq 3$, it follows that $|X\kern-2.0ptL(e)|\geq 2$. Let $x_{1}$ and $x_{2}$ be elements of $X\kern-2.0ptL(e)$ such that $x_{2}$ is the rightmost element of $X\kern-2.0ptL(e)$. Let $e^{\prime}=x_{2}v_{i}$ and $e^{\prime\prime}=x_{2}v_{j}$ then $X\kern-2.0ptR(e^{\prime})=\emptyset=X\kern-2.0ptR(e^{\prime\prime})$ by Claim 5. From Claim 4 applied to $e^{\prime}$ and $e^{\prime\prime}$ we deduce that $|X\kern-2.0ptM(e^{\prime})|\leq 3$ and $|X\kern-2.0ptM(e^{\prime\prime})|\leq 3$. Hence $|X\kern-2.0ptL(e^{\prime})|\geq 2$ and $|X\kern-2.0ptL(e^{\prime\prime})|\geq 2$. This is in contradiction with Claim 3. Thus there is no spine ordering for $G$ which leads to a one-sided crossing-free 2-page book embedding.

7 Concluding remarks

We have examined the exploding double chain as a special point set (order type) and shown that the initial part $H_{n}$ of size $2n-2$ is $n$-universal for graphs on $n$ vertices that are POSH. We believe that the class of POSH graphs is quite rich. On the sparse side, the result on bipartite graphs might be generalized, while for triangulations, the sheer number of Hamiltonian cycles in 5-connected graphs [1] makes it likely one of them is one-sided.

Conjecture 7.1**.**

Every triangle-free planar graph is POSH.

Conjecture 7.2**.**

Every 5-connected planar triangulation is POSH.

We have shown that $2$-trees and their superclasses series-parallel and planar Laman graphs are not contained in the class $\cal{C}^{\prime}$ of POSH graphs. The question whether these classes admit universal point sets of linear size remains intriguing.