Internally-Convex Drawings of Outerplanar Graphs in Small Area

TL;DR

This paper presents improved algorithms for drawing outerplanar graphs with convex internal faces in significantly smaller area, advancing visualization techniques for such graphs.

Contribution

It introduces an algorithm for embedding-preserving convex drawings of outerplanar graphs in O(n^{1.5}) area and explores strictly-convex drawings with optimized area for graphs with a path dual.

Findings

Convex drawings in O(n^{1.5}) area for outerplanar graphs

Strictly-convex drawings with area Θ(nk^2) for graphs with a path dual

Improved area bounds over previous O(n^2) results

Abstract

A well-known result by Kant [Algorithmica, 1996] implies that n-vertex outerplane graphs admit embedding-preserving planar straight-line grid drawings where the internal faces are convex polygons in $O(n^2)$ area. In this paper, we present an algorithm to compute such drawings in $O(n^{1.5})$ area. We also consider outerplanar drawings in which the internal faces are required to be strictly-convex polygons. In this setting, we consider outerplanar graphs whose weak dual is a path and give a drawing algorithm that achieves $\Theta(nk^2)$ area, where $k$ is the maximum size of an internal facial cycle.

Figures

Equations

$$f(d)=\underset{d_a,d_b\in {\mathbb{N}}_0:d_a^p+d_b^p\le (1-\delta )d^p}{\max }\{f(d_a)+f(d_b)\}+2\sqrt{d}.$$

Full text

1 Introduction

Drawing outerplanar graphs in small area is a core topic in Graph Drawing which has been the subject of intense work. The first one we are aware of dates back to 1980, when Leiserson provided an algorithm for the construction of linear-area VLSI layouts of bounded-degree outerplanar graphs [18]. In the early 90’s, de Fraysseix, Pach, and Pollack [9] and Schnyder [19] proved that $n$-vertex planar graphs, and thus $n$-vertex outerplanar graphs, admit $O(n^{2})$-area planar straight-line grid drawings. This bound is the best possible for planar graphs, however whether a subquadratic area upper bound can be achieved for outerplanar graphs has been an intriguing question asked repeatedly (see, e.g., [4, 6, 13, 16]). A positive answer was first provided by Di Battista and Frati [11]. They presented an algorithm which, given an $n$-vertex maximal outerplanar graph $G$ with weak dual tree $T$, uses a structural decomposition of binary trees by Chan [7] in order to construct an $O(n^{1.48})$ area “star-shaped” drawing of $T$; this is a planar straight-line grid drawing which can be augmented to an outerplanar straight-line drawing of $G$ within the same area bound. The same strategy has later led to an almost-linear area bound, namely $O\left(n2^{\sqrt{2\log n}}\sqrt{\log n}\right)$ [15], and also to an $O(dn\log n)$ bound for outerplanar graphs of maximum degree $d$ [14].

In this paper we study the area requirements of convex drawings of outerplanar graphs, in which the boundary of every face (including the outer face) is a convex polygon. It was surprising to us that this question has not been studied before, considering the prominent place that convex drawings occupy in the graph drawing literature. We only mention here the result on convex drawings that is most relevant to our paper. Namely, Chrobak and Kant [8, 17] and, independently, Felsner [12] have presented extensions of the algorithms by de Fraysseix, Pach, and Pollack [9] and Schnyder [19], proving that every $3$-connected planar graph admits a convex drawing in quadratic area, which is asymptotically optimal. Specifically, Felsner’s upper bound is $(f-1)\times(f-1)$, where $f$ denotes the number of faces.

Strictly-convex drawings, in which no three vertices on the boundary of a face are allowed to be collinear, require larger area. Indeed, an $n$-vertex cycle cannot be drawn as a strictly-convex polygon on a grid of size $o(n^{3})$ [1]. This bound is not yet matched by a corresponding upper bound. Bárány and Rote [2] proved that $n$-vertex $3$-connected planar graphs admit strictly-convex grid drawings in $O(n^{4})$ area. Bekos, Gronemann, Montecchiani, and Symvonis [3] provided an alternative algorithm with the same asymptotic area bound but with significantly smaller hidden constants.

The construction of small-area convex grid drawings of outerplanar graphs encounters two natural barriers. First, a planar graph admits a convex drawing only if it is biconnected, hence one is required to consider biconnected outerplanar graphs only. Second, we observe that there exist $n$-vertex biconnected outerplanar graphs that require $\Omega(n^{3})$ area in any convex grid drawing (Section˜2); note that an $O(n^{3})$ upper bound for (even strictly-)convex drawings of $n$-vertex biconnected outerplanar graphs can be achieved by drawing the cycle delimiting the outer face as a strictly-convex polygon [1] and drawing the internal edges as chords. We thus relax the convexity requirement and require only the internal faces to be convex, that is, we consider internally-convex drawings. For outerplanar graphs, this is a natural variation of the convex drawing style, taking into account that the outer face of an outerplanar graph is special: All the vertices are incident to it and this property defines the graph class. Chrobak and Kant’s algorithms [8, 17] allow us to obtain internally-convex drawings of $n$-vertex outerplanar graphs on a grid of size $O(n^{2})$ (via a suitable augmentation of the input graph to a 3-connected planar graph, by adding vertices and edges in its outer face). This restores in the internally-convex setting the pursuit for a sub-quadratic area upper bound for grid drawings of outerplanar graphs. We also consider the, in our opinion equally interesting, question of determining the area requirements of internally-strictly-convex drawings of outerplanar graphs.

Our contributions. Our main result is an algorithm that constructs an internally-convex grid drawing with $O(n^{1.5})$ area for $n$-vertex outerplanar graphs (Section˜3). We employ some ideas and tools from a recent paper [5]. The main geometric component of our algorithm consists of constructing a “leveled” drawing, where consecutive levels are drawn on consecutive horizontal grid lines, up to a level where there are sufficiently few vertices so that the drawing can “make a turn” and extend horizontally rather than vertically. We remark that ours is the first algorithm that constructs outerplanar straight-line grid drawings in sub-quadratic area by drawing directly the outerplanar graph, without passing through a star-shaped drawing of its weak dual tree. This aspect might be of independent interest.

For internally-strictly-convex grid drawings, the same arguments as in the strictly-convex setting prove that $\Theta(n^{3})$ is a tight bound for the area requirements. However, the natural question to consider here is how the area relates not only to the number $n$ of vertices but also to the maximum size $k$ of an internal face. We show that there exist outerplanar graphs requiring $\Omega(nk^{2})$ area in any internally-strictly-convex grid drawing and that the lower bound can be matched by a corresponding upper bound for outerpaths, that is, biconnected outerplanar graphs whose weak dual is a path (Section˜4).

We conclude in Section˜5 with several open problems.

2 Preliminaries

We assume familiarity with Graph Drawing, see, e.g. [10].

A drawing of a graph maps each vertex to a distinct point in the plane and each edge to a Jordan arc between its endpoints. In a grid drawing vertices have integer coordinates. Let $\Gamma$ be a grid drawing of a graph. Let $\max_{q}(\Gamma)$ and $\min_{q}(\Gamma)$ be the maximum and minimum $q$-coordinate of a vertex in $\Gamma$, with $q\in\{x,y\}$, respectively. The height $h(\Gamma)$ and width $w(\Gamma)$ of $\Gamma$ are the positive integers $h(\Gamma)=\max_{y}(\Gamma)-\min_{y}(\Gamma)+1$ and $w(\Gamma)=\max_{x}(\Gamma)-\min_{x}(\Gamma)+1$, respectively. In other words, $h(\Gamma)$ and $w(\Gamma)$ represent the number of horizontal and vertical grid lines intersecting the smallest axis-parallel rectangle enclosing $\Gamma$, respectively. The area of a drawing is the product of its width and height.

A drawing is planar if no two edges intersect, except at a common end-point. In a planar drawing, a vertex or an edge is external if it is incident to the outer face, and internal otherwise. A graph is planar if it admits a planar drawing. A graph is outerplanar if it admits a planar drawing in which each vertex is external. A planar drawing partitions the plane into topologically connected regions called faces. The unbounded face is the outer face, whereas the bounded faces are the internal faces. Two planar drawings of a connected graph are equivalent if they induce (i) the same counter-clockwise order of the edges incident to each vertex and (ii) the same counter-clockwise order of the edges along the boundaries of their outer faces. The equivalence classes of equivalent outerplanar drawings are called outerplanar embeddings, or in this paper just embeddings. An outerplane graph is a planar graph equipped with an outerplanar embedding. A biconnected outerplanar graph has a unique outerplanar embedding (up to an inversion of all the orders defining the embedding), hence we often talk about faces of a biconnected outerplane graph, referring to faces of its unique outerplanar embedding. Also, a drawing of a graph is embedding-preserving if it respects an embedding associated to the graph.

In a straight-line drawing, edges are drawn as straight-line segments. In a planar straight-line drawing, a face is convex (strictly-convex) if it is delimited by a polygon whose internal angles are not larger than $180^{\circ}$ (resp. are smaller than $180^{\circ}$). A convex drawing (a strictly-convex drawing) is a straight-line planar drawing in which every face, including the outer face, is convex (resp., strictly-convex). An internally-convex drawing (an internally-strictly-convex drawing) is a straight-line planar drawing in which every internal face, but not necessarily the outer face, is convex (resp., strictly-convex). When dealing with outerplanar graphs, we do not require convex drawings to be outerplanar per se. Our results however hold in the strongest possible sense: The lower bounds on the area hold true even for drawings that are not outerplanar, while the upper bounds are proved by constructing outerplanar drawings.

The following theorem motivates the study of internally-convex drawings.

Theorem 2.1**.**

There exists an $n$-vertex biconnected outerplanar graph that requires $\Omega(n^{3})$ area in every convex grid drawing.

Proof 2.2**.**

Let $G$ be the $n$-vertex biconnected outerplanar graph obtained from a cycle $\mathcal{C}$ with $n/2$ vertices by attaching a degree-$2$ vertex incident to the end-vertices of each edge of $\mathcal{C}$. First, every convex drawing of $G$ respects the unique outerplanar embedding of $G$, as moving a degree-$2$ vertex inside $\mathcal{C}$ would create an angle larger than $180^{\circ}$ in an internal face at that vertex. Second, in every outerplanar convex drawing of $G$, the drawing of $\mathcal{C}$ is strictly-convex, as the angle at each vertex $v$ of $\mathcal{C}$ in the interior of the cycle delimiting the outer face of $G$ is at most $180^{\circ}$ and part of this angle is used by the two triangular faces incident to $v$. Since every strictly-convex drawing of a cycle requires cubic area [1], the lower bound follows.

Let $G$ be a biconnected outerplane graph with $n\geq 3$ vertices; refer to Fig.˜1. Note that $G$ has at least one internal face. The weak dual of $G$, denoted by $T^{*}$, is the tree having a vertex for each internal face of $G$ and an edge between any two vertices corresponding to internal faces sharing an edge. For simplicity, we use the same label to denote a vertex of $T^{*}$ and the corresponding face of $G$. If $T^{*}$ is a path, then $G$ is an outerpath. Let $(u,v)$ be an external edge of $G$ such that $v$ immediately precedes $u$ in counter-clockwise order along the outer face of $G$. The outerplane graph $G$ with the designated edge $(u,v)$, denoted by $G[u,v]$, is said to be rooted at $(u,v)$. Let $(x,y)$ be an internal edge of $G[u,v]$ such that $u,x,y,v$ appear in this counter-clockwise order along the outer face of $G$ (possibly $u=x$ or $y=v$ may hold). Consider the outerplane subgraph $G^{\prime}$ of $G$ induced by the vertices $x$, $y$, and the vertices that are encountered when traversing the outer face of $G$ in counter-clockwise direction from $x$ to $y$. Observe that the edge $(x,y)$ is an external edge of $G^{\prime}$ such that $y$ immediately precedes $x$ in counter-clockwise order along the outer face of $G^{\prime}$. Then we denote by $G[x,y]$ the outerplanar graph $G^{\prime}$ rooted at the edge $(x,y)$.

Let $G$ be rooted at $(u,v)$ and let $f_{1}$ be the internal face of $G$ incident to $(u,v)$. The extended weak dual tree $T$ of $G[u,v]$ is the tree with root $f_{1}$ obtained from the weak-dual tree $T^{*}$ of $G$ by inserting a new leaf $\ell_{e}$ for each external edge $e$ of $G$ different from $(u,v)$ and an edge $(f,\ell_{e})$ if $f$ is the internal face of $G$ incident to $e$. Let $d$ be the number of vertices of $T$. Note that $d\geq n$, as $T$ contains the root $f_{1}$ plus one leaf for each of the $n-1$ external edges of $G$ different from $(u,v)$. Also, $d\leq 2n-3$, since $G$ has $n-1$ external edges different from $(u,v)$ and at most $n-2$ internal faces. Let $\pi=(f_{1},f_{2},\dots,f_{p})$ be a root-to-leaf path in $T$. We denote by $G[\pi]$ the outerpath induced by the vertices incident to the faces $f_{1},f_{2},\dots,f_{p-1}$. We call $G[\pi]$ the outerpath dual to $\pi$. Denote by $e_{p}$ the edge of $G$ dual to $(f_{p-1},f_{p})$. Let $u=z_{1},z_{2},\dots,z_{a},z_{a+1},\dots,z_{m}=v$ be the counter-clockwise order of the vertices along the outer face of $G[\pi]$, where $e_{p}=(z_{a},z_{a+1})$. We call the rooted outerplanar graphs $G[z_{1},z_{2}],\dots,G[z_{a-1},z_{a}]$ left subgraphs of $G$ at $\pi$ and the rooted outerplanar graphs $G[z_{a+1},z_{a+2}],\dots,G[z_{m-1},z_{m}]$ right subgraphs of $G$ at $\pi$.

3 Internally-Convex Drawings

In this section we prove the following theorem, which is our main result.

Theorem 3.1**.**

Every $n$-vertex outerplane graph admits an embedding-preserving internally-convex grid drawing in $O(n^{1.5})$ area.

Our proof is inspired by the proof that every outer-1-plane graph admits an embedding-preserving visibility representation in $O(n^{1.5})$ area [5]. In particular, we are going to use the following theorem, which generalizes a well-known result by Chan [7].

Theorem 3.2**.**

([5, Theorem 1]) Let $p=0.48$. Given any rooted tree with $n$ vertices, there exists a root-to-leaf path $\pi$ such that for any left subtree $\alpha$ and for any right subtree $\beta$ of $\pi$, $|\alpha|^{p}+|\beta|^{p}\leq(1-\delta)n^{p}$, for some constant $0<\delta<1$.

The following lemma is similar to the combination of Definition 1 and Observation 3 from [5], although with different constants. For the sake of completeness, we provide a proof, which follows the one in [5] almost verbatim. Recall that $p=0.48$ and $0<\delta<1$ are the constants from Theorem˜3.2. We define the constant $c$ as $c:=2/\delta$. Observe that $c>2$.

Lemma 3.3**.**

Let $f(d)$ be the recursive function defined on $\mathbb{N}_{0}$ as follows. First, $f(0)=0$ and $f(1)=1$. Second, for $d\in\mathbb{N}_{0}$ with $d\geq 2$, we have that

[TABLE]

Then $f(d)\leq c\sqrt{d}$.

Proof 3.4**.**

We prove the statement by induction on $d$. The statement is obviously true if $d\leq 1$, given that $c>2$. Suppose next that $d\geq 2$. Let $d_{a}$ and $d_{b}$ be non-negative integers at which the maximum in the definition of $f(d)$ is achieved, i.e., $d_{a}^{p}+d_{b}^{p}\leq(1-\delta)d^{p}$ and $f(d)=f(d_{a})+f(d_{b})+2\sqrt{d}$. Since $\delta>0$, we have $d_{a}<d$ and $d_{b}<d$. Dividing $d_{a}^{p}+d_{b}^{p}\leq(1-\delta)d^{p}$ by $d^{p}$ we get $\left(\frac{d_{a}}{d}\right)^{p}+\left(\frac{d_{b}}{d}\right)^{p}\leq(1-\delta)$. By $d_{a}<d$, $d_{b}<d$, and $p<1/2$, we get $\left(\frac{d_{a}}{d}\right)^{1/2}+\left(\frac{d_{b}}{d}\right)^{1/2}<\left(\frac{d_{a}}{d}\right)^{p}+\left(\frac{d_{b}}{d}\right)^{p}$, which implies that $\left(\frac{d_{a}}{d}\right)^{1/2}+\left(\frac{d_{b}}{d}\right)^{1/2}<(1-\delta)$, that is, $\sqrt{d}_{a}+\sqrt{d}_{b}<(1-\delta)\sqrt{d}$. Therefore, from $f(d)=f(d_{a})+f(d_{b})+2\sqrt{d}$ by induction we get $f(d)\leq c\sqrt{d}_{a}+c\sqrt{d}_{b}+2\sqrt{d}$, which by the previous inequality is smaller than $c(1-\delta)\sqrt{d}+2\sqrt{d}$, which is equal to $c\sqrt{d}$ for $c=2/\delta$.

Let $G$ be an $n$-vertex outerplane graph; refer to Fig.˜2. In what follows, we assume $G$ to be biconnected. Indeed, if it is not, it can be augmented to a biconnected outerplane graph $G^{\prime}$ by introducing edges in its outer face. Then the restriction of an embedding-preserving internally-convex drawing $\Gamma^{\prime}$ of $G^{\prime}$ yields an embedding-preserving internally-convex drawing of $G$ that inherits the area bounds of $\Gamma^{\prime}$. Let $u$ and $v$ be any two vertices such that $v$ immediately precedes $u$ in counter-clockwise order along the outer face of $G$. We root $G$ at the edge $(u,v)$. Recall that $G[u,v]$ denotes the outerplanar graph $G$ rooted at the edge $(u,v)$.

We show that $G[u,v]$ admits a (small-area) $uv$-separated drawing. This is an embedding-preserving internally-convex grid drawing $\Gamma$ of $G$ satisfying the following properties:

1. P.1

$u$ and $v$ lie on a horizontal line $h_{0}$; 2. P.2

all neighbors of $u$ and $v$ lie on the horizontal line $h_{1}$ one unit below $h_{0}$; 3. P.3

all the other vertices of $G$ lie not above $h_{1}$; and 4. P.4

all vertices different from $u$ (from $v$) are to the right of $u$ (resp. to the left of $v$).

The following property is a direct consequence of the definition of $uv$-separated drawing.

Property 1**.**

Let $\Gamma$ be a $uv$-separated drawing of $G[u,v]$. Vertices $u$ and $v$ can be shifted arbitrarily by integer amounts to the left and right, respectively, while maintaining $\Gamma$ an outerplanar internally-convex grid drawing.

If $n\geq 3$, then let $T$ be the extended weak dual tree of $G[u,v]$ and let $d$ be the number of vertices of $T$. If $n=2$, then $T$ is not defined and we set $d=1$. This choice is justified by the size of the subtrees of $T$ we generate. Indeed, if $n\geq 3$, we will select a root-to-leaf path $\pi$ in $T$ and consider left and right subgraphs of $G$ at $\pi$. If any of such subgraphs is just an edge, it is associated with a subtree of $T$ which consists of a single vertex, from which the correspondence between $n=2$ and $d=1$ stems.

We show an algorithm that, by induction on $n$, constructs a $uv$-separated drawing $\Gamma$ of $G[u,v]$ satisfying the following two properties:

• •

Property W: every vertical grid line intersecting $\Gamma$ contains at least one vertex of $G$ (and thus the width of $\Gamma$ is at most $n$); and

• •

Property H: every horizontal grid line intersecting $\Gamma$ contains at least one vertex of $G$ and the height of $\Gamma$ is at most $f(d)$.

The statement implies Theorem˜3.1, given that $f(d)\in O(\sqrt{d})$ by Lemma˜3.3 and since $d\leq 2n-3$.

Denote by $h(d)$ the maximum value of $h(\Gamma)$, over all drawings $\Gamma$ constructed by our algorithm for rooted outerplane graphs whose extended weak dual has $d$ vertices.

In the base case, we have $n=2$ (and thus $d=1$). Hence, $G$ is the edge $(u,v)$ and a drawing $\Gamma$ satisfying Properties W and H is constructed by placing $u$ on a grid point one unit to the left of $v$. In particular, note that $h(\Gamma)=1$ and that $f(1)=1$.

In the inductive case, we have $n>2$, hence $G$ has internal faces, since it is biconnected. Let $f_{1}$ be the root of $T$, which is the internal face of $G$ incident to $(u,v)$. Let $\pi=(f_{1},f_{2},\dots,f_{p})$ be the root-to-leaf path in $T$ from Theorem˜3.2, where $e_{p}$ is the edge of $G$ dual to $(f_{p-1},f_{p})$.

We introduce some notation and definitions. Call $e^{*}_{0}=(u,v)$ a transition edge; let $V_{0}:=\{u,v\}$, let the path $P_{0}=(v^{0}_{0},v^{0}_{1})$ coincide with $(u,v)$, and let the rooted outerplane graph $G^{*}_{0}$ coincide with $G[u,v]$. Suppose that, for some $i\geq 1$, a transition edge $e^{*}_{i-1}$ has been defined that is part of a path $P_{i-1}$. Consider the set of vertices that do not belong to $V_{0}\cup\dots\cup V_{i-1}$ and that belong to the faces of $G$ incident to the end-vertices of $e^{*}_{i-1}$; these vertices induce a path $P_{i}=(v^{i}_{1},\dots,v^{i}_{\ell_{i}})$, where the labels of the vertices are such that $u,v^{i}_{1},v^{i}_{\ell_{i}},v$ appear in this counter-clockwise order along the outer face of $G$. See again Fig.˜2, where, for example, $P_{2}$ is the path induced by the vertices that do not belong to $V_{0}\cup V_{1}$ (roughly speaking, by the vertices that are on the “correct side” of $e^{*}_{1}$) and that belong to the faces of $G$ incident to the end-vertices of $e^{*}_{1}$ (only one of such faces is incident to both the end-vertices of $e^{*}_{1}$, the other ones are incident to a single end-vertex of $e^{*}_{1}$). Only one of the edges of $P_{i}$ might be dual to an edge of $\pi$; this edge of $P_{i}$ is the transition edge $e^{*}_{i}$ with end-vertices $v^{i}_{s_{i}}$ and $v^{i}_{s_{i}+1}$, for some $1\leq s_{i}\leq\ell_{i}-1$. The set $V_{i}$ comprises the vertices $v^{i}_{1},\dots,v^{i}_{\ell_{i}}$ of $P_{i}$, as well as the vertices of the subgraphs $G^{i}_{j}:=G[v^{i}_{j},v^{i}_{j+1}]$ of $G$, for $j=1,\dots,\ell_{i}-1$ with $j\neq s_{i}$. Observe that, for $j=1,\dots,s_{i}-1$, the graph $G^{i}_{j}$ is a left subgraph of $G$ at $\pi$ or a proper subgraph of a left subgraph of $G$ at $\pi$. Similarly, for $j=s_{i}+1,\dots,\ell_{i}-1$, the graph $G^{i}_{j}$ is a right subgraph of $G$ at $\pi$ or a proper subgraph of a right subgraph of $G$ at $\pi$. We denote by $G^{*}_{i}$ the graph $G[v^{i}_{s_{i}},v^{i}_{s_{i}+1}]$. For the last-defined path $P_{k}$, we have that either the transition edge $e^{*}_{k}$ does not exist or, if it does, it is the edge $e_{p}$, which is incident to the outer face of $G$ and thus $G^{*}_{k}$ coincides with the edge $e^{*}_{k}$. An example of a situation in which $e^{*}_{k}$ does not exist is the one in which $P_{k}$ is a single vertex. This happens if $f_{p-1}$, the second-to-last vertex of $\pi$, is a triangular face delimited by the end-vertices of $e^{*}_{k-1}$ and by one additional vertex, which coincides with $P_{k}$. In this case $e_{p}$, the edge of $G$ dual to the last edge $(f_{p-1},f_{p})$ of $\pi$, connects a vertex of $e^{*}_{k-1}$ with the unique vertex of $P_{k}$ and $e^{*}_{k}$ does not exist.

Ideally, we would like the $uv$-separated drawing we construct to have the following features. The vertices $v^{i}_{1},\dots,v^{i}_{\ell_{i}}$ of each path $P_{i}$ should appear on a common horizontal line $h_{i}$ in this left-to-right order, and $h_{i}$ should lie one unit above $h_{i+1}$. The subgraphs $G^{i}_{j}$ with $j\neq s_{i}$ should be recursively drawn and placed below $h_{i}$, except for the edge $(v^{i}_{j},v^{i}_{j+1})$ which lies on $h_{i}$, in such a way that any two of these subgraphs are horizontally disjoint, that is, there exists a vertical line that keeps the drawings of the two subgraphs on different sides, except, possibly, for a single vertex shared by the two subgraphs which lies on the line. The subgraph $G^{i}_{s_{i}}$ is not drawn recursively, as it coincides with the subgraph $G^{*}_{i}$ on which the level-by-level construction we are describing is further applied. Recall that $k$ is the index of the path $P_{k}$ defined last in the procedure discussed above. If $k\leq\sqrt{n}$ (Case 1), this plan can be accomplished. Otherwise (Case 2), by a simple vertex counting argument, there exists an index $2\leq t\leq 1+\sqrt{n}$ such that the set $V_{t}$ has at most $\sqrt{n}$ vertices. In this case, we draw the vertices in $V_{t}$ differently than before, so that the transition edge $e^{*}_{t}$ is vertical. The small size of $V_{t}$ allows us to “turn” from horizontal to vertical without increasing the height dramatically. From $e^{*}_{t}$ towards the end of $\pi$, the construction proceeds by drawing the vertices of the outerpath dual to $\pi$ on two horizontal lines, and by placing the recursively-constructed drawings of the right and left subgraphs of $G$ at $\pi$ above and below it, respectively.

We now formalize this argument. As sketched above, we distinguish two cases. In Case 1, we have $k\leq\sqrt{n}$; refer to Fig.˜3. For $i=1,\dots,k$ and for $j=1,\dots,\ell_{i}-1$ with $j\neq s_{i}$, we recursively construct a $v^{i}_{j}v^{i}_{j+1}$-separated drawing $\Gamma^{i}_{j}$ of $G^{i}_{j}[v^{i}_{j},v^{i}_{j+1}]$. We construct, for $i=k-1,k-2,\dots,0$, a $v^{i}_{s_{i}}v^{i}_{s_{i}+1}$-separated drawing $\Gamma^{*}_{i}$ of $G^{*}_{i}[v^{i}_{s_{i}},v^{i}_{s_{i}+1}]$ as described in the following. Notice that the order of the indices ensures that, if $G^{*}_{i+1}$ exists, then its drawing $\Gamma^{*}_{i+1}$ has been already constructed once $\Gamma^{*}_{i}$ is about to be constructed. In order to start the iteration, if $e^{*}_{k}$ does not exist, then $G^{*}_{k}$ does not exist either, and hence $\Gamma^{*}_{k}$ does not need to be defined. If $e^{*}_{k}$ exists, then $G^{*}_{k}$ coincides with such an edge and $\Gamma^{*}_{k}$ is constructed by placing $v^{k}_{s_{k}}$ on a grid point one unit to the left of $v^{k}_{s_{k}+1}$. In order to construct $\Gamma^{*}_{i}$, we place the recursively constructed drawings $\Gamma^{i+1}_{j}$ with $j=1,\dots,\ell_{i+1}-1$ with $j\neq s_{i+1}$, as well as the drawing $\Gamma^{*}_{i+1}$ (if $G^{*}_{i+1}$ exists), side by side, so that the placement of a vertex $v^{i+1}_{j}$ coincides in two drawings it belongs to; these are $\Gamma^{i+1}_{j-1}$ and $\Gamma^{i+1}_{j}$ if $v^{i+1}_{j}$ is neither $v^{i+1}_{s_{i+1}}$ nor $v^{i+1}_{s_{i+1}+1}$, or are $\Gamma^{i+1}_{s_{i+1}-1}$ and $\Gamma^{*}_{i+1}$ if $v^{i+1}_{j}$ is $v^{i+1}_{s_{i+1}}$, or are $\Gamma^{*}_{i+1}$ and $\Gamma^{i+1}_{s_{i+1}+1}$ if $v^{i+1}_{j}$ is $v^{i+1}_{s_{i+1}+1}$. This ensures that the vertices of $P_{i+1}$ all lie on the same horizontal line $h_{i+1}$. Further, we place $v^{i}_{s_{i}}$ one unit above and one unit to the left of the leftmost vertex of $P_{i+1}$, and $v^{i}_{s_{i}+1}$ one unit above and one unit to the right of the rightmost vertex of $P_{i+1}$. This results in the desired $v^{i}_{s_{i}}v^{i}_{s_{i}+1}$-separated drawing $\Gamma^{*}_{i}$ of $G[v^{i}_{s_{i}},v^{i}_{s_{i}+1}]$. When $i=0$, the resulting drawing $\Gamma:=\Gamma^{*}_{0}$ is a $uv$-separated drawing of $G[u,v]$, as established by the following lemma.

Lemma 3.5**.**

In Case 1, the constructed drawing $\Gamma$ is a $uv$-separated drawing of $G[u,v]$ satisfying Properties W and H.

Proof 3.6**.**

First, it trivially comes from the construction that $\Gamma$ is a straight-line grid drawing and that it preserves the given embedding, provided its planarity.

We now prove that $\Gamma$ is planar. The paths $P_{0},P_{1},\dots,P_{k}$ do not cross each other in $\Gamma$, as they lie on distinct horizontal lines $h_{0},h_{1},\dots,h_{k}$. The other edges that are directly drawn by the algorithm (we call them binding edges) connect vertices on a path $P_{i}$ and a path $P_{i+1}$, hence they do no cross any of the paths $P_{0},P_{1},\dots,P_{k}$. Also, no two binding edges cross each other in $\Gamma$, given that the left-to-right order of the vertices of $P_{i}$ and $P_{i+1}$ on $h_{i}$ and $h_{i+1}$ is the same as the order in which such vertices are encountered when traversing in counter-clockwise direction the outer face of $G$ from $u$ to $v$. Any two recursively drawn graphs $G^{i}_{j}$ do not cross each other in $\Gamma$, as they are horizontally disjoint, by construction. Likewise, for any $l>i$, any recursively drawn graph $G^{i}_{j}$ does not cross any edge of $P_{l}$ or any binding edge incident to a vertex of $P_{l}$, as $G^{i}_{j}$ and such an edge are horizontally-disjoint in $\Gamma$. Also, for any $l<i$, any recursively drawn graph $G^{i}_{j}$ does not cross any edge of $P_{l}$ or any binding edge incident to a vertex of $P_{l}$, as the horizontal line $h_{i}$ keeps $G^{i}_{j}$ and such an edge on different sides in $\Gamma$; the fact that $G^{i}_{j}$ lies below $h_{i}$ is a consequence of Properties˜P.1, P.2 and P.3 of a $v^{i}_{j}v^{i}_{j+1}$-separated drawing. Finally, $P_{i}$ and $G^{i}_{j}$ share the edge $(s^{i}_{j},s^{i}_{j+1})$ and nothing else.

The convexity of each internal face $f$ of $G$ in $\Gamma$ follows by induction if $f$ is internal to some recursively drawn graph $G^{i}_{j}$. Otherwise, $f$ is convex as it is drawn as a triangle or a trapezoid in $\Gamma$, since its incident vertices lie on two horizontal lines $h_{i}$ and $h_{i+1}$, for some $0\leq i\leq k-1$. Properties˜P.1, P.2, P.3 and P.4 of a $uv$-separated drawing are easily implied by our construction. In particular, $u$ and $v$ are the vertices last placed by our construction, and their placement satisfies Property˜P.1 and Property˜P.4. The neighbors of $u$ and $v$ all lie on $h_{1}$, which is one unit below the horizontal line $h_{0}$ through $u$ and $v$, thus satisfying Property˜P.2. Finally, each vertex of $G$ different from $u$ and $v$ lies on $h_{1},\dots,h_{k}$ or belongs to a recursively constructed drawing whose top side is on one of such lines. This ensures Property˜P.3.

In order to complete the proof, it remains to prove Property W and Property H.

Property W* easily follows by induction, given that every vertical grid line intersecting $\Gamma$ intersects a recursively constructed drawing or a vertex directly placed by the algorithm.*

The first part of Property H also follows by induction, given that every horizontal grid line intersecting $\Gamma$ intersects a recursively constructed drawing or a vertex directly placed by the algorithm. Concerning the second part, we have that $h(\Gamma)$ is at most $k$ plus the maximum height of a recursively constructed drawing $\Gamma^{i}_{j}$. Indeed, the paths $P_{0},P_{1},\dots,P_{k}$ occupy $k+1$ consecutive horizontal lines $h_{0},h_{1},\dots,h_{k}$, respectively. No vertex of $\Gamma$ lies above $h_{0}$. Furthermore, every recursively constructed drawing $\Gamma^{i}_{j}$ is placed in $\Gamma$ so that it intersects at least one of the lines $h_{0},h_{1},\dots,h_{k}$. Note that $k\leq\sqrt{n}$ by assumption and hence $k\leq\sqrt{d}$, given that $d\geq n$. Assume that $i$ and $j$ are the indices such that $h(\Gamma^{i}_{j})$ is maximum. Let $d^{i}_{j}$ be the number of vertices in the extended weak dual tree of $G^{i}_{j}$. Hence, we have $h(\Gamma)\leq\sqrt{d}+h(\Gamma^{i}_{j})$ and thus, by induction, $h(\Gamma)\leq\sqrt{d}+f(d^{i}_{j})$. Since $G^{i}_{j}$ is a subgraph of a left subgraph of $G$ at $\pi$ or of a right subgraph of $G$ at $\pi$, by Theorem˜3.2, it holds true that $(d^{i}_{j})^{p}\leq(1-\delta)d^{p}$. By the definition of $f(d)$ in Lemma˜3.3 (with $d_{a}=d^{i}_{j}$ and $d_{b}=0$), we have that $\sqrt{d}+f(d^{i}_{j})\leq f(d)$, hence $h(\Gamma)\leq f(d)$, which proves Property H.

We now describe the construction of Case 2. Recall that, in this case, we have $k\geq 1+\sqrt{n}$ We start the description with the following simple combinatorial lemma.

Lemma 3.7**.**

There exists an index $t$ with $2\leq t\leq 1+\sqrt{n}$ such that $|V_{t}|\leq\sqrt{n}$.

Proof 3.8**.**

Since the sets $V_{2},V_{3},\dots,V_{1+\sqrt{n}}$ are disjoint and since there are $\sqrt{n}$ of them, the smallest among them contains at most $\sqrt{n}$ vertices.

The construction for Case 2 proceeds in three steps. In Step 1, we construct a drawing $\Gamma^{*}_{t}$ of $G^{*}_{t}$; in Step 2, we augment $\Gamma^{*}_{t}$ to a drawing $\Gamma^{*}_{t-1}$ of $G^{*}_{t-1}$; finally, in Step 3, we augment $\Gamma^{*}_{t-1}$ to the desired drawing $\Gamma$ of $G$.

Step 1:

We construct a drawing $\Gamma^{*}_{t}$ of $G^{*}_{t}$; refer to Fig.˜4. Recall that the transition edge $e^{*}_{t}$ has end-vertices $v^{t}_{s_{t}}$ and $v^{t}_{s_{t}+1}$ and that $G^{*}_{t}$ is the graph $G[v^{t}_{s_{t}},v^{t}_{s_{t}+1}]$. Let $(f_{j},f_{j+1})$ be the edge of $\pi$ dual to $e^{*}_{t}$, let $\pi^{\prime}$ be the subpath $(f_{j+1},f_{j+2},\dots,f_{p})$ of $\pi$, and let $(u_{0}=v^{t}_{s_{t}},u_{1},\dots,u_{x},w_{y},w_{y-1},\dots,w_{1},w_{0}=v^{t}_{s_{t}+1})$ be the counter-clockwise order of the vertices along the outer face of $G[\pi^{\prime}]$, where $(u_{x},w_{y})$ is the edge $e_{p}$. We initialize $\Gamma^{*}_{t}$ by drawing $e^{*}_{t}$ as a vertical segment of height $1$. For $j=1,\dots,x$, we recursively construct a drawing $\Lambda^{u}_{j}$ of $G[u_{j-1},u_{j}]$. We place the drawings $\Lambda^{u}_{1},\dots,\Lambda^{u}_{x}$ side by side, so that the placement of a vertex $u_{j}$ coincides in the drawings $\Lambda^{u}_{j}$ and $\Lambda^{u}_{j+1}$ where it appears, for $j=1,\dots,x-1$, and so that the placement of $u_{0}$ coincides in $\Lambda^{u}_{1}$ and in the drawing of $e^{*}_{t}$. Analogously, for $j=1,\dots,y$, we recursively construct a drawing $\Lambda^{w}_{j}$ of $G[w_{j},w_{j-1}]$. Unlike for $\Lambda^{u}_{1},\dots,\Lambda^{u}_{x}$, we rotate the drawings $\Lambda^{w}_{1},\dots,\Lambda^{w}_{y}$ by $180^{\circ}$, and place them side by side, so that the placement of a vertex $w_{j}$ coincides in $\Lambda^{w}_{j}$ and $\Lambda^{w}_{j+1}$, for $j=1,\dots,y-1$, and so that the placement of $w_{0}$ coincides in $\Lambda^{w}_{1}$ and in the drawing of $e^{*}_{t}$. This completes the construction of $\Gamma^{*}_{t}$.

Step 2:

We now augment $\Gamma^{*}_{t}$ to a drawing $\Gamma^{*}_{t-1}$ of $G^{*}_{t-1}$; refer to Fig.˜5. Recall that the path $P_{t}$ has vertices $(v^{t}_{1},v^{t}_{2},\dots,v^{t}_{s_{t}},v^{t}_{s_{t}+1},\dots,v^{t}_{\ell_{t}})$. For $j=1,\dots,\ell_{t}-1$ with $j\neq s_{t}$, we recursively construct a $v^{t}_{j}v^{t}_{j+1}$-separated drawing $\Gamma^{t}_{j}$ of $G[v^{t}_{j},v^{t}_{j+1}]$. We place the recursively constructed drawings $\Gamma^{t}_{1},\dots,\Gamma^{t}_{s_{t}-1}$ side by side, so that the placement of a vertex $v^{t}_{j}$ coincides in $\Gamma^{t}_{j-1}$ and $\Gamma^{t}_{j}$, for $j=2,\dots,s_{t}-1$, and so that the placement of $v^{t}_{s_{t}}$ coincides in $\Gamma^{t}_{s_{t}-1}$ and in $\Gamma^{*}_{t}$. Next, we rotate the drawings $\Gamma^{t}_{s_{t}+1},\dots,\Gamma^{t}_{\ell_{t}-1}$ in counter-clockwise direction by $90^{\circ}$, and we stack them one on top of the other, so that the placement of a vertex $v^{t}_{j}$ coincides in $\Gamma^{t}_{j-1}$ and $\Gamma^{t}_{j}$, for $j=s_{t}+2,\dots,\ell_{t}-1$, and so that the placement of $v^{t}_{s_{t}+1}$ in $\Gamma^{t}_{s_{t}+1}$ is on the same vertical line as in $\Gamma^{*}_{t}$ and on the same horizontal line as the highest vertex in $\Gamma^{*}_{t}$ (refer to the left side of Fig.˜6).

This might provide a double placement for $v^{t}_{s_{t}+1}$ (unless $v^{t}_{s_{t}+1}$ is the highest vertex in $\Gamma^{*}_{t}$). The issue is resolved by keeping for $v^{t}_{s_{t}+1}$ the position it has in $\Gamma^{*}_{t}$; note that, in $\Gamma^{t}_{s_{t}+1}$, this amounts to shifting $v^{t}_{s_{t}+1}$ downwards (refer to the right side of Fig.˜6). The drawing of $\Gamma^{*}_{t-1}$ is completed by placing $v^{t-1}_{s_{t-1}}$ and $v^{t-1}_{s_{t-1}+1}$ on the same horizontal line, one unit higher than the highest vertex in $\Gamma^{*}_{t-1}$ so far (this is either $v^{t}_{\ell_{t}}$ if $s_{t}\leq\ell_{t}-2$ or the highest vertex in $\Gamma^{*}_{t}$ otherwise), so that $v^{t-1}_{s_{t-1}+1}$ is one unit to the left of the vertical line through $e^{*}_{t}$ and $v^{t-1}_{s_{t-1}}$ is one unit to the left of the leftmost vertex in $\Gamma^{*}_{t-1}$ so far (this is either $v^{t}_{1}$ if $s_{t}\geq 2$ or $v^{t-1}_{s_{t-1}+1}$ otherwise).

Step 3:

Finally, we show how to augment $\Gamma^{*}_{t-1}$ to a $uv$-separated drawing of $G[u,v]$. This is done similarly to Case 1. Namely, we construct, for $i=t-2,t-3,\dots,0$, a $v^{i}_{s_{i}}v^{i}_{s_{i}+1}$-separated drawing $\Gamma^{*}_{i}$ of $G[v^{i}_{s_{i}},v^{i}_{s_{i}+1}]$ from an already constructed $v^{i+1}_{s_{i+1}}v^{i+1}_{s_{i+1}+1}$-separated drawing $\Gamma^{*}_{i+1}$ of $G[v^{i+1}_{s_{i+1}},v^{i+1}_{s_{i+1}+1}]$. Recall that, in Step 2, we constructed $\Gamma^{*}_{t-1}$. As in Case 1, we place the recursively constructed drawings $\Gamma^{i+1}_{j}$ with $j=1,\dots,\ell_{i+1}-1$ with $j\neq s_{i+1}$, as well as the drawing $\Gamma^{*}_{i+1}$, side by side, so that the placement of a vertex $v^{i+1}_{j}$ coincides in two drawings it belongs to, with one exception in the case $i=t-2$; refer to Fig.˜7. Namely, the drawing $\Gamma^{t-1}_{s_{t-1}+1}$ of $G^{t-1}_{s_{t-1}+1}$ is embedded so that $v^{t-1}_{s_{t-1}+1}$ is on the same horizontal line as in $\Gamma^{*}_{t-1}$ and on the same vertical line as the rightmost vertex in $\Gamma^{*}_{t-1}$. This provides a double placement for $v^{t-1}_{s_{t-1}+1}$. The issue is resolved by keeping for $v^{t-1}_{s_{t-1}+1}$ the position it has in $\Gamma^{*}_{t-1}$; note that, in $\Gamma^{t-1}_{s_{t-1}+1}$, this amounts to shifting $v^{t-1}_{s_{t-1}+1}$ leftwards. The desired $v^{i}_{s_{i}}v^{i}_{s_{i}+1}$-separated drawing $\Gamma^{*}_{i}$ of $G[v^{i}_{s_{i}},v^{i}_{s_{i}+1}]$ is completed by placing $v^{i}_{s_{i}}$ one unit above and one unit to the left of the leftmost vertex of $P_{i+1}$, and $v^{i}_{s_{i}+1}$ one unit above and one unit to the right of the rightmost vertex of $P_{i+1}$. When $i=0$, the obtained drawing $\Gamma:=\Gamma^{*}_{0}$ is a $uv$-separated drawing of $G[u,v]$, as proved in the following.

Lemma 3.9**.**

In Case 2, the constructed drawing $\Gamma$ is a $uv$-separated drawing of $G[u,v]$ satisfying Property W and H.

Proof 3.10**.**

First, it trivially comes from the construction that $\Gamma$ is a straight-line grid drawing and that it preserves the given embedding, provided its planarity.

We now prove that $\Gamma$ is planar. In order to do that, we prove that the drawing produced at each step of the algorithm’s construction is planar. As in the proof of Lemma˜3.5, our arguments use (without explicitly mentioning it) the fact that every recursively constructed $z_{1}z_{2}$-separated drawing lies, apart from the edge $(z_{1},z_{2})$, below the horizontal line through $z_{1}$ and $z_{2}$. This is a consequence of Properties˜P.1, P.2 and P.3 of a $z_{1}z_{2}$-separated drawing.

• •

Step 1.* The drawing $\Gamma^{*}_{t}$ is easily proved to be planar. Indeed, the paths $(v^{t}_{s_{t}}=u_{0},u_{1},\dots,u_{x})$ and $(v^{t}_{s_{t}+1}=w_{0},w_{1},\dots,w_{y})$ on the boundary of $G[\pi^{\prime}]$ lie on two horizontal lines in $\Gamma^{*}_{t}$. This implies that they do not cross each other, that they do not cross any edge $(u_{i},w_{j})$, and that no two edges $(u_{i},w_{j})$ and $(u_{i^{\prime}},w_{j^{\prime}})$ cross each other. Each graph $G[u_{i},u_{i+1}]$ does not cross any graph $G[u_{j},u_{j+1}]$, as such graphs are horizontally disjoint in $\Gamma^{*}_{t}$, does not cross any subgraph $G[w_{j},w_{j+1}]$ as it is vertically disjoint from it, and does not cross $G[\pi^{\prime}]$ as it is vertically disjoint from it, except for the edge $(u_{i},u_{i+1})$ which is shared by the two graphs. Similarly, each graph $G[w_{j},w_{j+1}]$ is free of crossings in $\Gamma^{*}_{t}$.*

• •

Step 2.* We now deal with $\Gamma^{*}_{t-1}$. Path $P_{t}$ is represented in $\Gamma^{*}_{t-1}$ as a polygonal line composed of a vertical segment on a line $\ell$ and of a horizontal segment to the left of $\ell$. Since $G^{*}_{t}$ lies entirely to the right of $\ell$ in $\Gamma^{*}_{t-1}$, except for the edge $(v^{t}_{s_{t}},v^{t}_{s_{t}+1})$ which is shared with $P_{t}$, it follows that $P_{t}$ does not cross $G^{*}_{t}$. The recursively drawn graphs $G^{t}_{j}$ with $j<s_{t}$ do not cross each other as they are horizontally disjoint, and do not cross $G^{*}_{t}$ and the recursively drawn graphs $G^{t}_{j}$ with $j>s_{t}$ as they are horizontally disjoint from them in $\Gamma^{*}_{t-1}$. Also, the recursively drawn graphs $G^{t}_{j}$ with $j>s_{t}$ do not cross each other as they are vertically disjoint, and the recursively drawn graphs $G^{t}_{j}$ with $j>s_{t}+1$ do not cross $G^{*}_{t}$ as they are vertically disjoint from it. The proof that $G^{t}_{s_{t}+1}$ and $G^{*}_{t}$ do not cross each other in $\Gamma^{*}_{t-1}$ uses the properties of a $v^{t}_{s_{t}+1}v^{t}_{s_{t}+2}$-separated drawing (and in fact, the avoidance of such crossings is the driving force behind the definition of our drawing invariant). Indeed, by Property˜P.4 and by the $90^{\circ}$ counter-clockwise rotation of $\Gamma^{t}_{s_{t}+1}$, we have that $v^{t}_{s_{t}+1}$ is the lowest vertex in $\Gamma^{t}_{s_{t}+1}$. Since $\Gamma^{t}_{s_{t}+1}$ is placed so that $v^{t}_{s_{t}+1}$ is on the same horizontal line as the highest vertex in $\Gamma^{*}_{t}$, it follows that the only edges of $G^{t}_{s_{t}+1}$ that might cross $G^{*}_{t}$ are those incident to $v^{t}_{s_{t}+1}$. However, since all the vertices of $G^{t}_{s_{t}+1}$ different from $v^{t}_{s_{t}+1}$ lie above $\Gamma^{*}_{t}$ and since all the neighbors of $v^{t}_{s_{t}+1}$ in $G^{t}_{s_{t}+1}$ lie on the vertical line one unit to the right of $\ell$, by Property˜P.2 and by the $90^{\circ}$ counter-clockwise rotation of $\Gamma^{t}_{s_{t}+1}$, it follows that the edges of $G^{t}_{s_{t}+1}$ incident to $v^{t}_{s_{t}+1}$ lie above every edge of $G^{*}_{t}$ with which they have a horizontal overlap. We conclude the proof of planarity of $\Gamma^{*}_{t-1}$ by remarking that, by Property˜1, placing $v^{t}_{s_{t}+1}$ at the point where it is placed in $\Gamma^{*}_{t}$ maintains the planarity of $\Gamma^{t}_{s_{t}+1}$.*

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Step 3.* That the paths $P_{0},P_{1},\dots,P_{t-1}$, the binding edges connecting vertices on such paths, and the recursively drawn graphs $G^{i}_{j}$ (as well as the graph $G^{*}_{t-1}$) sharing an edge with these paths do not cross each other in $\Gamma$ is proved by the same arguments as in Case 1, with one exception. Namely, $G^{*}_{t-1}$ is not horizontally disjoint in $\Gamma$ from $G^{t-1}_{s_{t-1}+1}$. To prove that these two graphs do not cross each other in $\Gamma$, we use the same argument as the one employed for Step 2. By Property˜P.4 and since $\Gamma^{t-1}_{s_{t-1}+1}$ is placed so that $v^{t-1}_{s_{t-1}+1}$ is on the same vertical line as the rightmost vertex in $\Gamma^{*}_{t-1}$, the only edges of $G^{t-1}_{s_{t-1}+1}$ that might cross $G^{*}_{t-1}$ are those incident to $v^{t-1}_{s_{t-1}+1}$. However, since all the vertices of $G^{t-1}_{s_{t-1}+1}$ different from $v^{t-1}_{s_{t-1}+1}$ lie to the right of $\Gamma^{*}_{t-1}$ and since all the neighbors of $v^{t-1}_{s_{t-1}+1}$ in $G^{t-1}_{s_{t-1}+1}$ lie on the horizontal line one unit below $h_{t-1}$, by Property˜P.2, it follows that the edges of $G^{t-1}_{s_{t-1}+1}$ incident to $v^{t-1}_{s_{t-1}+1}$ lie to the right of every edge of $G^{*}_{t-1}$ with which they have a vertical overlap. Finally, note that, by Property˜1, placing $v^{t-1}_{s_{t-1}+1}$ at the point where it is placed in $\Gamma^{*}_{t-1}$ maintains the planarity of $\Gamma^{t-1}_{s_{t-1}+1}$.*

The convexity of each internal face $f$ of $G$ in $\Gamma$ follows by induction if $f$ is internal to some recursively drawn subgraph $G^{i}_{j}$ of $G$. If $f$ is an internal face of $G[\pi^{\prime}]$ or if $f$ has vertices on two distinct paths among $P_{0},P_{1},\dots,P_{t-1}$, then $f$ is drawn as a triangle or a trapezoid in $\Gamma$, hence it is convex. The only faces that do not fit in any of these two categories are those incident to $v^{t-1}_{s_{t-1}}$ and/or $v^{t-1}_{s_{t-1}+1}$ and incident to at least one vertex in $P_{t}$. The convexity of these faces follows from two facts.

• •

First, the polygon $Q$ delimited by the edge $(v^{t-1}_{s_{t-1}},v^{t-1}_{s_{t-1}+1})$ and by $P_{t}$ is a convex pentagon $(v^{t-1}_{s_{t-1}},v^{t-1}_{s_{t-1}+1},v^{t}_{\ell_{t}},v^{t}_{s_{t}},v^{t}_{1})$ (if $s_{t}>1$) or a convex quadrilateral $(v^{t-1}_{s_{t-1}},v^{t-1}_{s_{t-1}+1},v^{t}_{\ell_{t}},v^{t}_{1})$ (if $s_{t}=1$). This is a direct consequence of the construction, which ensures that $v^{t-1}_{s_{t-1}}$ and $v^{t-1}_{s_{t-1}+1}$ lie on a horizontal line higher than $v^{t}_{\ell_{t}}$, $v^{t}_{s_{t}}$, and $v^{t}_{1}$, with $v^{t-1}_{s_{t-1}+1}$ to the right of $v^{t-1}_{s_{t-1}}$, that $v^{t}_{\ell_{t}}$ and $v^{t}_{s_{t}}$ lie on a vertical line to the right of $v^{t-1}_{s_{t-1}}$ and $v^{t-1}_{s_{t-1}+1}$ (and $v^{t}_{1}$ if $s_{t}>1$), with $v^{t}_{\ell_{t}}$ higher than $v^{t}_{s_{t}}$, and that, if $s_{t}>1$, $v^{t}_{1}$ and $v^{t}_{s_{t}}$ lie on a horizontal line lower than $v^{t}_{\ell_{t}}$, $v^{t-1}_{s_{t-1}+1}$, and $v^{t-1}_{s_{t-1}}$, with $v^{t}_{s_{t}}$ to the right of $v^{t}_{1}$.

• •

Second, the faces incident to $v^{t-1}_{s_{t-1}}$ and/or $v^{t-1}_{s_{t-1}+1}$ and incident to at least one vertex in $P_{t}$ form a convex subdivision of $Q$.

The fact that $\Gamma$ satisfies Properties˜P.1, P.2, P.3 and P.4 of a $uv$-separated drawing can be proved as in Lemma˜3.5, given that the “topmost” part of $\Gamma$ is constructed as in Case 1, with the only addendum that $\Gamma^{*}_{t-1}$ lies entirely below or on $h_{t-1}$, by construction; this is needed to ensure Property˜P.3 for $\Gamma$.

Finally, we prove Property W and Property H. Actually, Property W and the first part of Property H are proved as in the proof of Lemma˜3.5. Concerning the second part of Property H, we start by proving that $h(\Gamma)$ is at most $t$, plus $\sqrt{n}$, plus the maximum height of a recursively constructed drawing $\Gamma^{i}_{j}$ of a subgraph $G^{i}_{j}$ of a left subgraph of $G$ at $\pi$, plus the maximum height of a recursively constructed drawing $\Gamma^{i^{\prime}}_{j^{\prime}}$ of a subgraph $G^{i^{\prime}}_{j^{\prime}}$ of a right subgraph of $G$ at $\pi$. First, no vertex of $\Gamma$ lies above $h_{0}$; this fact, which is trivial in the proof of Lemma˜3.5, uses here the observation that $\Gamma^{*}_{t-1}$ lies entirely below or on $h_{t-1}$, by construction. The horizontal grid lines $h_{0},h_{1},\dots,h_{t-1}$ on which $P_{0},P_{1},\dots,P_{t-1}$ are placed give rise to the $t$ term in the upper bound for $h(\Gamma)$. The $\sqrt{n}$ term accounts for the horizontal grid lines between the highest line intersecting $\Gamma^{t}_{\ell_{t}-1}$ (which is one unit below $h_{t-1}$) and the lowest line intersecting $\Gamma^{t}_{s_{t}+1}$ before $v^{t}_{s_{t}+1}$ is moved to the position it has in $\Gamma^{*}_{t}$. Indeed, since the drawings $\Gamma^{t}_{s_{t}+1},\Gamma^{t}_{s_{t}+2},\dots,\Gamma^{t}_{\ell_{t}-1}$ are constructed recursively, and hence satisfy Property W, and are rotated by $90^{\circ}$, and since the graphs $G^{t}_{s_{t}+1},G^{t}_{s_{t}+2},\dots,G^{t}_{\ell_{t}-1}$ have a total of at most $\sqrt{n}$ vertices, by Lemma˜3.7, it follows that the drawings $\Gamma^{t}_{s_{t}+1},\Gamma^{t}_{s_{t}+2},\dots,\Gamma^{t}_{\ell_{t}-1}$ intersect, before moving $v^{t}_{s_{t}+1}$, at most $\sqrt{n}$ grid lines. Next, we observe that $h(\Gamma^{*}_{t})$ is given by the sum of the maximum height of a graph $G[u_{j},u_{j+1}]$ and the maximum height of a graph $G[w_{j},w_{j+1}]$. Since the former are left subgraphs of $G$ at $\pi$ and the latter are right subgraphs of $G$ at $\pi$, this gives rise to the term maximum height of a recursively constructed drawing $\Gamma^{i}_{j}$ of a subgraph $G^{i}_{j}$ of a left subgraph of $G$ at $\pi$ and the term maximum height of a recursively constructed drawing $\Gamma^{i^{\prime}}_{j^{\prime}}$ of a subgraph $G^{i^{\prime}}_{j^{\prime}}$ of a right subgraph of $G$ at $\pi$ in the upper bound for $h(\Gamma)$. The proof of the claimed upper bound on $h(\Gamma)$ is concluded by considering the recursively drawn graphs that were not dealt with yet. Every recursively constructed drawing $\Gamma^{i}_{j}$ with $i\leq t-1$ is placed in $\Gamma$ so that it intersects at least one of the lines $h_{0},h_{1},\dots,h_{t-1}$, hence the number of horizontal grid lines between $h_{0}$ and the lowest line intersecting one of such drawings $\Gamma^{i}_{j}$ is at most $t-1+h(\Gamma^{i}_{j})$, which is smaller than the claimed upper bound. Finally, the recursively drawn graphs $G^{t}_{j}$ with $j<s_{t}$ have their top side, defined by the edge $(v^{t}_{j},v^{t}_{j+1})$, on the horizontal line through vertices $u_{0},u_{1},\dots,u_{x}$. Since these graphs are left subgraphs of $G$ at $\pi$, same as the graphs $G[u_{j},u_{j+1}]$, it follows that the number of horizontal grid lines between $h_{0}$ and the lowest line intersecting $G^{t}_{j}$ with $j<s_{t}$ is at most the claimed upper bound on $h(\Gamma)$, given that the same is true for $G[u_{j},u_{j+1}]$.

Note that $t\leq\sqrt{n}$ by construction and hence $t\leq\sqrt{d}$, given that $d\geq n$. Assume that $i$ and $j$ are indices such that $G^{i}_{j}$ is a recursively drawn subgraph of a left subgraph of $G$ at $\pi$ and $h(\Gamma^{i}_{j})$ is maximum. Also, assume that $i^{\prime}$ and $j^{\prime}$ are indices such that $G^{i^{\prime}}_{j^{\prime}}$ is a recursively drawn subgraph of a right subgraph of $G$ at $\pi$ and $h(\Gamma^{i^{\prime}}_{j^{\prime}})$ is maximum. Let $d^{i}_{j}$ and $d^{i^{\prime}}_{j^{\prime}}$ be the number of vertices in the extended weak dual trees of $G^{i}_{j}$ and $G^{i^{\prime}}_{j^{\prime}}$, respectively. By the upper bound proved above, we have $h(\Gamma)\leq 2\sqrt{d}+h(\Gamma^{i}_{j})+h(\Gamma^{i^{\prime}}_{j^{\prime}})$ and thus, by induction, $h(\Gamma)\leq 2\sqrt{d}+f(d^{i}_{j})+f(d^{i^{\prime}}_{j^{\prime}})$. By Theorem˜3.2, it holds true that $(d^{i}_{j})^{p}+(d^{i^{\prime}}_{j^{\prime}})^{p}\leq(1-\delta)d^{p}$. By the definition of $f(d)$ in Lemma˜3.3 (with $d_{a}=d^{i}_{j}$ and $d_{b}=d^{i^{\prime}}_{j^{\prime}}$), we have that $2\sqrt{d}+f(d^{i}_{j})+f(d^{i^{\prime}}_{j^{\prime}})\leq f(d)$, hence $h(\Gamma)\leq f(d)$, which proves Property H.

4 Internally-Strictly-Convex Drawings of Outerpaths

In this section, we prove a tight bound on the area required by internally-strictly-convex grid drawings of outerpaths. More precisely, we prove that every $n$-vertex outerpath whose internal faces have size at most $k$ admits an internally-strictly-convex grid drawing in area $O(nk^{2})$, which we also show to match a general lower bound for the internally-strictly-convex grid drawings of outerplanar graphs. We start by proving the lower bound.

Theorem 4.1**.**

Every $n$-vertex outerplane graph with $\Omega(\frac{n}{k})$ internal faces of size $k$ requires $\Omega(nk^{2})$ area in any internally-strictly-convex grid drawing.

Proof 4.2**.**

Let $G$ be an $n$-vertex outerplane graph with $\Omega(\frac{n}{k})$ internal faces of size $k$. First, note that any internally-strictly-convex grid drawing of $G$ must be outerplanar. Indeed, any embedding of an outerplanar graph which is not an outerplanar embedding contains a degree-$2$ vertex as an internal vertex. However, such a vertex would create an angle larger than or equal to $180^{\circ}$ in an internal face. The area lower bound is obtained by a simple packing argument. Since any internally-strictly-convex grid drawing of $G$ is outerplanar, it contains $\Omega(\frac{n}{k})$ internal faces of size $k$. Each of these faces occupies $\Omega(k^{3})$ area [1], hence the total area of the drawing is $\Omega(\frac{n}{k}\cdot k^{3})$, which gives the bound claimed by the theorem.

In the rest of the section, we prove a matching upper bound for the case of outerpaths. Note that there exist $n$-vertex outerpaths with $\Omega(\frac{n}{k})$ internal faces of size $k$, hence the lower bound of Theorem˜4.1 applies to outerpaths as well. We introduce some notation and definitions. Let $G$ be an outerpath and let $\pi=(f_{1},f_{2},\dots,f_{p-1})$ be the weak-dual of $G$, which is a path; refer to Fig.˜8. Let $\hat{e}$ be the edge of $G$ dual to the edge $(f_{1},f_{2})$ of $\pi$ and let $\tilde{e}$ be the edge of $G$ dual to the edge $(f_{p-2},f_{p-1})$ of $\pi$. Moreover, let $\hat{g}=(u_{0},v_{0})$ be any of the two external edges incident to $\hat{e}$ that belong to the face $f_{1}$ of $G$, where $u_{0}$ immediately follows $v_{0}$ when traversing the outer face of $G$ in counter-clockwise order. Also, let $\tilde{g}$ be any of the two external edges incident to $\tilde{e}$ that belong to the face $f_{p-1}$ of $G$. We augment $\pi$ with two new vertices $f_{0}$ and $f_{p}$, and edges $(f_{0},f_{1})$ and $(f_{p-1},f_{p})$. We interpret $(f_{0},f_{1})$ and $(f_{p-1},f_{p})$ as edges dual to $\hat{g}$ and $\tilde{g}$, respectively.

We now describe a decomposition of $G$ into a sequence of smaller outerpaths such that any two consecutive outerpaths in the sequence share exactly one internal edge of $G$. We let $e^{*}_{0}$ be the edge $\hat{g}$. Let $G_{0}$ be the plane graph induced by the vertices delimiting the internal faces of $G$ incident to the end-vertex common to $e^{*}_{0}$ and $\hat{e}$ (see the vertex $u_{0}$ in Fig.˜8). Let $e^{*}_{1}$ be the external edge of $G_{0}$, other than $e^{*}_{0}$, dual to an edge of $\pi$. We call $G_{0}$ a fan graph with gate edges $e^{*}_{0}$ and $e^{*}_{1}$. Suppose that, for some $i\geq 1$, a fan graph $G_{i-1}$ with gate edges $e^{*}_{i-1}$ and $e^{*}_{i}$ has been defined. Let $u_{i}$ and $v_{i}$ be the end-vertices of $e^{*}_{i}$, where $u_{i}$ is encountered before $v_{i}$ when traversing the outer face of $G$ in counter-clockwise direction from $u_{0}$ to $v_{0}$. We define the fan graph $G_{i}$ as follows. Let $V_{i-1}$ be defined as $\bigcup^{i-1}_{j=0}V(G_{j})\setminus\{u_{i},v_{i}\}$. Then $G_{i}$ is the outerplane graph induced by the vertices that do not belong to $V_{i-1}$ and that are incident to faces that have $u_{i}$ or $v_{i}$ on their boundary. Let $e^{*}_{i+1}$ be the external edge of $G_{i}$, other than $e^{*}_{i}$, dual to an edge of $\pi$. The edges $e^{*}_{i}$ and $e^{*}_{i+1}$ are the gate edges of $G_{i}$. Eventually, the decomposition defines a fan graph $G_{r}$ with gate edges $e^{*}_{r}$ and $e^{*}_{r+1}$ such that $e^{*}_{r+1}=\tilde{g}$.

We will show that each fan graph $G_{i}$ with gate edges $e^{*}_{i}$ and $e^{*}_{i+1}$ admits a block-drawing. This is an outerplanar internally-strictly-convex grid drawing $\Gamma_{i}$ of $G_{i}$ satisfying the following properties:

1. B.1

The two gate edges $e^{*}_{i}$ and $e^{*}_{i+1}$ are drawn as vertical segments of unit length such that $y(u_{i})=y(u_{i+1})=0$ and $y(v_{i})=y(v_{i+1})=1$; 2. B.2

the vertices $u_{i}$ and $v_{i}$ are the leftmost vertices of $\Gamma_{i}$ and no other vertex of $G_{i}$ has the same $x$-coordinate as these vertices; and 3. B.3

the vertices $u_{i+1}$ and $v_{i+1}$ are the rightmost vertices of $\Gamma_{i}$ and no other vertex of $G_{i}$ has the same $x$-coordinate as these vertices.

We proceed as follows. Lemmas˜4.3 and 4.5 are two technical lemmas that yield special internally-strictly-convex grid drawings of cycles. We use these two lemmas to compute a block-drawing $\Gamma_{i}$ of each fan graph $G_{i}$ with appropriate area bounds. We do so by drawing each of its faces using either Lemma˜4.3 or Lemma˜4.5, and we appropriately “merge” them together (Lemma˜4.7). The final drawing $\Gamma$ of $G$ is obtained by “gluing” the block-drawings $\Gamma_{0},\dots,\Gamma_{r}$ (Theorem˜4.9) at their shared gate edges.

In the following, by half-plane of a line we mean an open half-plane bounded by the line.

Lemma 4.3**.**

Let $C=(v_{1},v_{2},\dots,v_{h})$ with $h\geq 3$ be a cycle. Let $s_{1}$ be a segment representing $e_{1}=(v_{1},v_{2})$ such that $v_{1}$ is at the origin $(0,0)$, $x(v_{2})$ is a non-negative integer, and $y(v_{2})=1$. Then $C$ admits an internally-strictly-convex grid drawing $\Gamma_{C}$ such that $C$:

1. I.1

$s_{1}$* represents $e_{1}$ in $\Gamma_{C}$;* 2. I.2

the vertex $v_{h}$ is placed at $(x(v_{2})+h-2,1)$; 3. I.3

let $\ell$ be the line that passes through $s_{1}$ oriented in the direction from $v_{1}$ to $v_{2}$. Then, all the vertices $v_{3},\dots,v_{h}$ lie in the right half-plane of $\ell$; 4. I.4

the width of $\Gamma_{C}$ is $x(v_{2})\cdot(h-2)+\frac{(h-3)(h-2)}{2}+1$; and 5. I.5

the height of $\Gamma_{C}$ is $h-1$.

Proof 4.4**.**

We show how to construct the drawing $\Gamma_{C}$; refer to Fig.˜9. Vertices $v_{1}$ and $v_{2}$ are at $(0,0)$ and $(x(v_{2}),1)$ in $\Gamma_{C}$. For $j=3,\dots,h-1$, we place $v_{j}$ at $(x(v_{2})\cdot(j-1)+\frac{(j-2)(j-1)}{2},j-1)$. Finally, we place vertex $v_{h}$ at $(x(v_{2})+h-2,1)$. This completes the construction of $\Gamma_{C}$.

Note that $\Gamma_{C}$ satisfies Properties˜I.1 and I.2 by construction. Furthermore, observe that the $x$-coordinate of $v_{h-1}$ is $x(v_{2})\cdot(h-2)+\frac{(h-3)(h-2)}{2}$, while its $y$-coordinate is $h-2$. The leftmost grid line intersecting $\Gamma_{C}$ passes through $v_{0}$ and thus this is the line $x=0$, whereas the rightmost grid line intersecting $\Gamma_{C}$ passes through $v_{h-1}$ and thus this is the line $x=x(v_{2})\cdot(h-2)+\frac{(h-3)(h-2)}{2}$. Therefore, $\Gamma_{C}$ has width equal to $x(v_{h-1})+1$, and Property˜I.4 holds. The lowest horizontal grid line intersecting $\Gamma_{C}$ passes through $v_{0}$ and thus this is the line $y=0$, whereas the highest horizontal grid line intersecting $\Gamma_{C}$ passes through $v_{h-1}$ and thus this is the line $y=h-2$. Therefore, $\Gamma_{C}$ has height $y(v_{h-1})+1$, and Property˜I.5 holds.

Next, we show that $\Gamma_{C}$ is strictly-convex. This is trivial if $h=3$. If $h>3$, the strict convexity follows by these observations:

• •

For $i=1,\ldots,h-2$, the slope of the edge $(v_{i},v_{i+1})$ is $\frac{1}{x(v_{2})+i-1}$, that is, the slopes decrease with $i$, hence the interior angles at vertices $v_{2},\ldots,v_{h-2}$ are smaller than $180^{\circ}$.

• •

By construction, the slopes of the edges $(v_{1},v_{h})$ and $(v_{h-2},v_{h-1})$ are both equal to $\frac{1}{x(v_{2})+h-2}$. Since these edges are also of the same length, vertices $v_{1},v_{h-2},v_{h-1},v_{h}$ form a parallelogram $\cal P$. This implies that the interior angles of $\cal P$ (and thus of $C$) at $v_{h-1}$ and at $v_{h}$ are smaller than $180^{\circ}$. Since the edge $(v_{1},v_{2})$ has positive slope, by hypothesis, and the slope of the edge $(v_{1},v_{h})$ is positive and smaller than the slope of $(v_{1},v_{2})$ by construction, the angle of $C$ at $v_{1}$ is also smaller than $180^{\circ}$.

Finally, the strict convexity of $\Gamma_{C}$ and the fact that the slope of the edge $(v_{1},v_{h})$ is smaller than the slope of the edge $(v_{1},v_{2})$ imply Property˜I.3.

The following property is a consequence of the constraints of the drawings of Lemma˜4.3.

Property 2**.**

Let $\Gamma_{C}$ be the drawing of a cycle $C=(v_{1},v_{2},\dots,v_{h})$ produced by Lemma˜4.3. Then the drawing obtained by shifting vertex $v_{h}$ in $\Gamma_{C}$ by any integer amounts to the right and/or to the bottom is an internally-strictly-convex grid drawing of $C$.

Lemma 4.5**.**

Let $C=(v_{1},v_{2},\dots,v_{h})$ with $h>3$ be a cycle. Let $e_{1}=(v_{1},v_{2})$ and $e_{2}=(v_{i},v_{i+1})$ be two non-adjacent edges of $C$. Let $s_{1}$ be a segment that represents $(v_{1},v_{2})$ such that $v_{1}$ is at the origin $(0,0)$, $x(v_{2})$ is a non-negative integer, and $y(v_{2})=1$. Then $C$ admits an internally-strictly-convex grid drawing $\Gamma_{C}$ such that:

1. J.1

$s_{1}$* represents $e_{1}$ in $\Gamma_{C}$;* 2. J.2

the edge $e_{2}$ is represented by a vertical segment such that $x(v_{i})>x(v_{2})$, $y(v_{i})=1$, and $y(v_{i+1})=0$; 3. J.3

the width of $\Gamma_{C}$ is $1+\max\{x(v_{2})\cdot(i-2)+\frac{(i-3)(i-2)}{2},\frac{(h-i-1)(h-i)}{2}\}$; and 4. J.4

the height of $\Gamma_{C}$ is $h-2$.

Proof 4.6**.**

We show how to construct the drawing $\Gamma_{C}$; refer to Fig.˜10. First, we define two cycles $C_{1}=(v_{1},v_{2},\dots,v_{i})$ and $C_{2}=(v_{1},v_{h},v_{h-1},\dots,v_{i+1})$. Observe that the length of $C_{1}$ is $i$ and that the length of $C_{2}$ is $h-i+1$. We apply Lemma˜4.3 to obtain a drawing $\Gamma_{C_{1}}$ of $C_{1}$ in which the edge $(v_{1},v_{2})$ is represented by $s_{1}$ and to obtain a drawing $\Gamma_{C_{2}}$ of $C_{2}$ in which the edge $(v_{1},v_{h})$ is represented by the segment connecting points $(0,0)$ and $(1,1)$. Let $M$ be the maximum $x$-coordinate of a vertex in $\Gamma_{C_{1}}$ and $\Gamma_{C_{2}}$. We leverage Property˜2 to obtain a drawing $\Gamma^{\prime}_{C_{1}}$ of $C_{1}$ by shifting the vertex $v_{i}$ horizontally and rightward in $\Gamma_{C_{1}}$ so that its $x$-coordinate is $1+M$. Analogously, we leverage Property˜2 to obtain a drawing $\Gamma^{\prime}_{C_{2}}$ of $C_{2}$ by shifting the vertex $v_{i+1}$ downward and rightward in $\Gamma_{C_{2}}$ so that its $x$-coordinate is $1+M$ and its $y$-coordinate is [math]. To obtain the drawing $\Gamma_{C}$ we then proceed as follows. We first initialize $\Gamma_{C}$ to $\Gamma^{\prime}_{C_{1}}$. Then we construct a vertically-mirrored copy $\Gamma^{\prime\prime}_{C_{2}}$ of $\Gamma^{\prime}_{C_{2}}$, that is, we flip the sign of the $y$-coordinate of each vertex, and we insert $\Gamma^{\prime\prime}_{C_{2}}$ in $\Gamma_{C}$ so that the placement of $v_{1}$ is the same in both drawings. Finally, we remove from $\Gamma_{C}$ the drawing of the edges $(v_{1},v_{i})$ and $(v_{1},v_{i+1})$ and draw the edge $(v_{i},v_{i+1})$ as a vertical segment of unit length. This concludes the construction of $\Gamma_{C}$.

By construction, $\Gamma_{C}$ satisfies Property˜J.1 and Property˜J.2. Since the length of $C_{1}$ is $i$, it follows by Lemma˜4.3 that $w(\Gamma_{C_{1}})=x(v_{2})\cdot(i-2)+\frac{(i-3)(i-2)}{2}$, and that $h(\Gamma_{C_{1}})=i-1$. Also, since the length of $C_{2}$ is $h-i+1$, it follows by Lemma˜4.3 that $w(\Gamma_{C_{2}})=(h-i-1)+\frac{(h-i-2)(h-i-1)}{2}=\frac{(h-i-1)(h-i)}{2}$, and that $h(\Gamma_{C_{2}})=h-i$. The width of $\Gamma_{C}$ is given by $1$ plus the maximum between $w(\Gamma_{C_{1}})$ and $w(\Gamma_{C_{2}})$, which implies Property˜J.3. Also, the height of $\Gamma_{C}$ is given by $h(\Gamma_{C_{1}})+h(\Gamma_{C_{2}})-1$, which implies Property˜J.4.

To complete the proof, it remains to prove that $\Gamma_{C}$ is an internally-strictly-convex drawing of $C$. By Lemma˜4.3, the interior angles of $C$ at $v_{2},\ldots,v_{i-1},v_{i+2},\ldots,v_{h}$ are strictly smaller than $180^{\circ}$. Since the slope of $(v_{1},v_{2})$ ranges in $(0,+\infty]$ while the slope of $(v_{1},v_{h})$ is either $-1$ (if $h\neq i+1$) or [math] (otherwise), the interior angle of $C$ at $v_{1}$ is strictly smaller than $180^{\circ}$. Since $v_{i}$ and $v_{i+1}$ are the rightmost vertices of $\Gamma_{C}$, it follows that also each of the interior angles of $C$ at $v_{i}$ and $v_{i+1}$ is also strictly smaller than $180^{\circ}$, completing the proof.

Property 3**.**

Let $\Gamma_{C}$ be the drawing of a cycle $C=(v_{1},v_{2},\dots,v_{i},v_{i+1},\dots,v_{h})$ produced by Lemma˜4.5. Then the drawing obtained by shifting vertices $v_{i}$ and $v_{i+1}$ in $\Gamma_{C}$ by the same integer amount to the right is an internally-strictly-convex grid drawing of $C$.

We next show how to construct a block-drawing of a fan graph with appropriate area bounds.

Lemma 4.7**.**

Every $n_{i}$-vertex fan graph $G_{i}$ with internal faces of size at most $k_{i}$ admits a block-drawing whose width is $O(n_{i}k_{i})$ and whose height is $O(k_{i})$.

Proof 4.8**.**

Let $G_{i}$ be an $n_{i}$-vertex fan graph with gate edges $e^{*}_{i}=(u_{i},v_{i})$ and $e^{*}_{i+1}=(u_{i+1},v_{i+1})$. To prove the statement, we distinguish two cases based on whether $G_{i}$ contains internal edges or not; refer to Fig.˜11.

In the latter case (see, e.g., $G_{1}$ in Fig.˜11), $G_{i}$ is a cycle $C=(u_{i},v_{i},\dots,v_{i+1},u_{i+1},\dots)$ with $n_{i}\geq 4$. Then a block drawing $\Gamma_{i}$ of $G_{i}$ can be obtained by applying Lemma˜4.5 to cycle $C$ with $e_{1}$ being $(u_{i},v_{i})$, $e_{2}$ being $(v_{i+1},u_{i+1})$, and with $s_{1}$ being the vertical segment connecting $(0,0)$ and $(0,1)$. In particular, Properties˜J.1 and J.2 of Lemma˜4.5 immediately imply Property˜B.1 of $\Gamma_{i}$. Also, Properties˜B.2 and B.3 follow from the strict-convexity of $\Gamma_{i}$. By Property˜J.3 and Property˜J.4 of Lemma˜4.5, the width of $\Gamma_{i}$ is in $O(k_{i}^{2})$, which is in $O(n_{i}k_{i})$, and the height of $\Gamma_{i}$ is in $O(k_{i})$.

If $G_{i}$ contains internal edges, we further distinguish two sub-cases based on which, between $u_{i}$ and $v_{i}$, is incident to an internal edge of $G_{i}$. First, we discuss the case in which $u_{i}$ is incident to internal edges (see, e.g., $G_{0}$ in Fig.˜11). Let $f^{\prime}$ be the internal face of $G_{i}$ incident to $e^{*}_{i+1}$ and let $f^{\prime\prime}$ be the internal face of $G_{i}$ that shares an edge with $f^{\prime}$. We denote the edge shared by $f^{\prime}$ and $f^{\prime\prime}$ as $e^{\prime}$, and note that $e^{\prime}$ is incident to $u_{i}$; also, we denote as $e^{\prime\prime}$ the edge of $f^{\prime\prime}$ distinct from $e^{\prime}$ and incident to $u_{i}$. We temporarily remove the edge $e^{\prime}$ from $G_{i}$. We treat the resulting graph $G^{-}_{i}$ as a fan graph with gate edges $e^{*}_{i}$ and $e^{*}_{i+1}$.

If $G^{-}_{i}$ is a simple cycle, which occurs if $e^{\prime}$ is the only internal edge of $G_{i}$ and thus $e^{\prime\prime}=e^{*}_{i}$, we compute a block drawing of $G_{i}$ as follows. First, we compute a block drawing $\Gamma^{-}_{i}$ of $G^{-}_{i}$ by means of Lemma˜4.5, as in the case in which $G_{i}$ is a simple cycle described above. Then, since $\Gamma^{-}_{i}$ is a strictly-convex polygon, we obtain a block drawing of $G_{i}$ by simply drawing the edge $e^{\prime}$ as a straight-line chord of this polygon. Since in this case we have $n_{i}\leq 2k_{i}-2$, by Property˜J.3 and Property˜J.4 of Lemma˜4.5 the width of $\Gamma_{i}$ is in $O(k_{i}^{2})$, which is in $O(n_{i}k_{i})$, and the height of $\Gamma_{i}$ is in $O(k_{i})$.

Suppose now that $G^{-}_{i}$ contains internal edges (incident to $u_{i}$). Let $g_{1},g_{2},\dots,g_{t+1}$ be the internal faces of $G^{-}_{i}$ in clockwise order around $u_{i}$, where $g_{1}$ is incident to $e^{*}_{i}$ and $g_{t+1}$ is incident to $e^{\prime\prime}$ and to $u_{i+1}$. Note that $g_{t+1}$ is the face resulting from “merging” $f^{\prime}$ and $f^{\prime\prime}$ after the removal of $e^{\prime}$. First, we compute a block drawing $\Gamma^{-}_{i}$ of $G^{-}_{i}$ as follows. We initialize $\Gamma^{-}_{i}$ to a drawing obtained by applying Lemma˜4.3 to the cycle bounding $g_{1}$ with $e_{1}$ being $e^{*}_{i}$ and $s_{1}$ being the vertical segment of unit length between $(0,0)$ and $(0,1)$. Then, for $j=2,\dots,t$, we assume to have already extended $\Gamma^{-}_{i}$ to an internally-strictly convex grid drawing of the graph formed by the boundaries of the faces $g_{1},g_{2},\dots,g_{j-1}$ and let $(u_{i},z_{j})$ be the unique edge shared by $g_{j-1}$ and $g_{j}$. Moreover, we assume that $\Gamma^{-}_{i}$ lies in the left half-plane of the line $\ell_{j}$ passing through $u_{i}$ and $z_{j}$ when going from $u_{i}$ to $z_{j}$; also, we assume that the segment representing $(u_{i},z_{j})$ is such that $y(z_{j})=1$ and $x(z_{j})\geq 0$. We extend $\Gamma^{-}_{i}$ further by adding a drawing of the cycle bounding $g_{j}$ obtained by applying Lemma˜4.3 to such a cycle with $e_{1}$ being $(u_{i},z_{j})$ and $s_{1}$ being the segment representing $(u_{i},z_{j})$ in the drawing $\Gamma^{-}_{i}$ computed so far. Observe that, by Lemma˜4.3, the drawing of the cycle bounding $g_{j}$ is strictly-convex and that the edges of $g_{j}$ do not cross any of the edges of $\Gamma^{-}_{i}$ by Property˜I.3 of Lemma˜4.3. Therefore, $\Gamma^{-}_{i}$ is an internally-strictly convex grid drawing. Moreover, by Property˜I.2 of Lemma˜4.3, $\Gamma^{-}_{i}$ lies in the left half-plane of the line $\ell_{j+1}$ passing through $u_{i}$ and $z_{j+1}$ when going from $u_{i}$ to $z_{j+1}$; also, the segment representing $(u_{i},z_{j+1})$ is such that $y(z_{j+1})=1$ and $x(z_{j+1})\geq 0$. Once $\Gamma^{-}_{i}$ has been extended to an internally-strictly convex grid drawing of the graph formed by the boundaries of the faces $g_{1},g_{2},\dots,g_{t}$, we have that the edge $e^{\prime\prime}$ belongs to $\Gamma^{-}_{i}$, as it belongs to $g_{t}$. Then we augment $\Gamma^{-}_{i}$ with a drawing of the cycle $C_{t+1}$ bounding the face $g_{t+1}$ obtained by applying Lemma˜4.5 with $e_{1}$ being $e^{\prime\prime}$, $e_{2}$ being $e^{*}_{i+1}$, and $s_{1}$ being the segment representing $e^{\prime\prime}$ in the drawing $\Gamma^{-}_{i}$ computed so far. Furthermore, by Property˜3, we shift (by the same amount) to the right vertices $u_{i+1}$ and $v_{i+1}$ so that they lie one unit to the right of the rightmost of the remaining vertices of $G_{i}$. Note that this shift might be needed when the boundaries of some of the faces $g_{1},\ldots,g_{t}$ drawn by Lemma˜4.3 are “long” (e.g., of length $k$), while the boundary of the last face $g_{t+1}$ drawn by Lemma˜4.5 is “short” (e.g., of length $4$); in this situation, $u_{i+1}$ and $v_{i+1}$ might not be the rightmost vertices of $\Gamma^{-}_{i}$ and this requires shifting $u_{i+1}$ and $v_{i+1}$. Furthermore, observe that the edge $e^{\prime\prime}$ lies along $\ell_{t+1}$ and that $C_{t+1}$ is drawn as a strictly-convex polygon $P_{t+1}$ with one side along $\ell_{t+1}$ (Property˜J.1) that lies to the right of this line (Property˜J.2). Therefore, $\Gamma^{-}_{i}$ is planar, and thus it is an internally-strictly convex grid drawing of $G^{-}_{i}$. Finally, to obtain the desired block drawing $\Gamma_{i}$ of $G_{i}$, we simply draw the edge $e^{\prime}$ as a straight-line chord of $P_{t+1}$ in $\Gamma^{-}_{i}$.

Since all the applications of Lemma˜4.3 and Lemma˜4.5 use segments incident to $u_{i}$, it follows that the height of the computed drawing is in $O(k_{i})$.

We now prove an upper bound on the width of $\Gamma_{i}$. Since $u_{i+1}$ and $v_{i+1}$ are the rightmost vertices of $\Gamma_{i}$, it suffices to compute the width of the drawing of the face $g_{t+1}$ drawn using Lemma˜4.5, which is the merge of two faces of length at most $k_{i}$. To do so, we first upper bound the $x$-coordinate of vertex $z_{t+1}$. By Property˜I.2 of Lemma˜4.3, we have that the $x$-coordinate of a vertex $z_{p}$ is at most $\sum^{p-1}_{j=1}|g_{p}|$, with $2\leq p\leq t+1$. Therefore, $x(z_{t+1})\leq n_{i}$. It follows by Property˜J.3 of Lemma˜4.5, that the width of the drawing of $g_{t+1}$ is at most $1+\max\{x(z_{t+1})\cdot(a-2)+\frac{(a-3)(a-2)}{2},\frac{(h-a-1)(h-a)}{2}\}$, with $a\leq k_{i}$ and $h\leq 2k_{i}$. Since $x(z_{t+1})\leq n_{i}$, we get that the width of $\Gamma_{i}$ is in $O(n_{i}k_{i}+k_{i}^{2})$, and hence in $O(n_{i}k_{i})$.

Finally, the case in which $v_{i}$ is incident to an internal edge (see, e.g., $G_{2}$ in Fig.˜11) can be handled by first flipping the outerplanar embedding of $G_{i}$, afterward exchanging the roles of $u_{i}$ and $v_{i}$, then computing a block drawing $\Gamma_{i}$ of $G_{i}$ as described above, and finally vertically flipping $\Gamma_{i}$ around the axis $y:=0.5$.

We are now ready to prove the main theorem of this section.

Theorem 4.9**.**

Every $n$-vertex outerpath whose internal faces have size at most $k$ admits an internally-strictly-convex grid drawing in $O(nk^{2})$ area.

Proof 4.10**.**

We compute block drawings $\Gamma_{0},\dots,\Gamma_{r}$ of the fan graphs $G_{0},\dots,G_{r}$ of the outerpath $G$ so that each block drawing has height $O(k_{i})$ and width $O(n_{i}k_{i})$ by Lemma˜4.7, where $n_{i}$ is the number of vertices of $G_{i}$ and $k_{i}$ is the maximum size of any face of $G_{i}$ ($0\leq i\leq r)$. We iteratively construct $\Gamma$ by “gluing” together the block drawings $\Gamma_{0},\dots,\Gamma_{r}$ at their common gate edges. More precisely, we proceed as follows. We initialize $\Gamma$ to $\Gamma_{0}$. Then, for $i=1,\dots,r$, we augment $\Gamma$ with a copy of $\Gamma_{i}$ so that the drawing of the gate edge $e^{*}_{i}$ in $\Gamma_{i}$ coincides with the drawing of $e^{*}_{i}$ in the drawing constructed so far. It follows that the height of $\Gamma$ is $O(\underset{0\leq i\leq r}{\max}\{k_{i}\})$, which is in $O(k)$, while its width is $O(\sum_{i=0}^{r}n_{i}k_{i})$, which is in $O(nk)$.

5 Conclusions

In this paper we have proved the first sub-quadratic area upper bound for internally-convex grid drawings of outerplanar graphs. We also presented algorithms to construct internally-strictly-convex grid drawings of outerpaths whose area is asymptotically optimal, with respect to the number of vertices of the graph and the maximum size of an internal face.

Several intriguing questions are left open by our research:

• •

A first, natural, research direction is to narrow the gap between the $O(n^{1.5})$ upper bound and the $\Omega(n)$ lower bound for internally-convex grid drawings of $n$-vertex outerplanar graphs. In particular, any super-linear lower bound would be an important step towards proving an analogous result for (not necessarily convex) planar straight-line grid drawings.

• •

Concerning internally-strictly-convex grid drawings of (general) $n$-vertex outerplane graphs whose faces have size at most $k$, for which we proved an $\Omega(nk^{2})$ lower bound, we are not aware of any upper bound better than $O(n^{3})$, which comes from the literature on strictly-convex polygons [1]. Tightening this gap is a nice goal.

• •

The case $k=4$, for which there is an $O(n^{2})$ upper bound (a consequence of results in [2]), is especially interesting. Indeed, any upper bound would translate to an upper bound for the area requirements of straight-line drawings of outer-1-planar graphs.

• •

Finally, we have observed that, for convex grid drawings of $n$-vertex outerplane graphs (in which the outer face is also required to be convex), $\Theta(n^{3})$ is a tight bound for the area requirements. However, if the maximum size of an internal face is bounded, then our lower bound does not hold anymore. We believe that an $\Omega(n\log n)$ lower bound can be proved, given that the outer face would require a dimension to have $\Omega(n)$ length in order to be convex and given that there exist $n$-vertex trees that require $\Omega(\log n)$ length in both dimensions in any planar straight-line drawing (see, e.g., [13]). However, this lower bound is far from the $O(n^{3})$ upper bound and it would be interesting to close this gap.

We conclude by mentioning that we believe that the techniques we developed in order to construct internally-strictly-convex grid drawings of outerpaths could lead to non-trivial area bounds for outerplanar graphs whose weak dual tree has bounded diameter. We plan to investigate the truth of this intuition in the near future.