4-Connected Triangulations on Few Lines

Abstract

We show that 4-connected plane triangulations can be redrawn such that edges are represented by straight segments and the vertices are covered by a set of at most $\sqrt{2n}$ lines each of them horizontal or vertical. The same holds for all subgraphs of such triangulations.

The proof is based on a corresponding result for diagrams of planar lattices which makes use of orthogonal chain and antichain families.

The full version of the paper with complete proofs is available at http://page.math.tu-berlin.de/~felsner/Paper/grid-lines.pdf and on arXiv:1908.04524 as version 2.

Partially supported by DFG grant FE-340/11-1.

Figures in this paper use colors which can be seen in the online versions.

1 Introduction

Given a planar graph G we denote by $\pi (G)$ the minimum number $\ell$ such that G has a plane straight-line drawing in which the vertices can be covered by a collection of $\ell$ lines. Clearly $\pi (G) =1$ if and only if G is a forest of paths. The set of graphs with $\pi (G)=2$, however, is already surprisingly rich, it contains trees, outerplanar graphs and subgraphs of grids, see [1, 7].

The parameter $\pi (G)$ has received some attention in recent years, here is a list of known results:

• It is NP-complete to decide whether $\pi (G)=2$ (Biedl et al. [2]).

• For a stacked triangulation G, a.k.a. planar 3-tree or Apollonian network, let $d_G$ be the stacking depth (e.g. $K_4$ has stacking depth 1). On this class lower and upper bounds on $\pi (G)$ are $d_G+1$ and $d_G+2$ respectively, see Biedl et al. [2] and for the lower bound also Eppstein [6, Thm. 16.13].

• Eppstein [7] constructed a planar, cubic, 3-connected, bipartite graph $G_\ell$ on $O(\ell ^3)$ vertices with $\pi (G_\ell ) \ge \ell$.

Related parameters have been studied by Chaplick et al. [3, 4].

The main result of this paper is the following theorem.

Theorem 1

If G is a 4-connected plane triangulation on n vertices, then $\pi (G) \le \sqrt{2n}$.

The result is not far from optimal since, using a constant number of additional vertices and many additional edges, the graph $G_\ell$ mentioned above can be transformed into a 4-connected plane triangulation, i.e., in the class we have graphs with $\pi (G) \in \varOmega (n^{1/3})$.

The proof of the theorem makes use of transversal structures, these are special colorings of the edges of a 4-connected inner triangulation of a 4-gon with colors red and blue.

In Sect. 2.1 we survey transversal structures. The red subgraph of a transversal structure can be interpreted as the diagram of a planar lattice. Background on posets and lattices is given in Sect. 2.2. Dimension of posets and the connection with planarity are covered in Sect. 2.3. In Sect. 2.4 we survey orthogonal partitions of posets. The theory implies that every poset on n elements can be covered by at most $\sqrt{2n}-1$ subsets such that each of the subsets is a chain or an antichain.

In Sect. 3 we prove that the diagram of a planar lattice on n elements has a straight-line drawing with vertices placed on a set of $\sqrt{2n}-1$ lines. All the lines used for the construction are either horizontal or vertical.

Finally in Sect. 4 we prove the main result: transversal structures can be drawn on at most $\sqrt{2n}-1$ lines. In fact, the red subgraph of the transversal structure has such a drawing by the result of the previous section. It is rather easy to add the blue edges to this drawing. Theorem 1 is obtained as a corollary.

Fig. 1.

The two local properties.

2 Preliminaries

2.1 Transversal Structures

Let G be an internally 4-connected inner triangulation of a 4-gon, in other words G is a plane graph with quadrangular outer face, triangular inner faces, and no separating triangle. Let s, a, t, b be the outer vertices of G in clockwise order. A transversal structure for G is an orientation and 2-coloring of the inner edges of G such that

• (1) All edges incident to s, a, t and b are red outgoing, blue outgoing, red incoming, and blue incoming, respectively.

• (2) The edges incident to an inner vertex v come in clockwise order in four non-empty blocks consisting solely of red outgoing, blue outgoing, red incoming, blue incoming edges, respectively.

Figure 1 illustrates the properties and Fig. 2 shows an example. Transversal structures have been studied in [12, 17], and [13]. In particular it has been shown that every internally 4-connected inner triangulation of a 4-gon admits a transversal structure. Fusy [13] used transversal structures to prove the existence of straight-line drawings with vertices being placed on integer points (x, y) with $0\le x \le W$, $0\le y \le H$, and $H+W \le n-1$.

An orientation of a graph G is said to be acyclic if it has no directed cycle. Given an acyclic orientation of G, a vertex having no incoming edge is called a source, and a vertex having no outgoing edge is called a sink. A bipolar orientation is an acyclic orientation with a unique source s and a unique sink t, cf. [11]. Bipolar orientations of plane graphs are also required to have s and t incident to the outer face.

Fig. 2.

An example of a transversal structure.

A bipolar orientation of a plane graph has the property that at each vertex v the outgoing edges form a contiguous block and the incoming edges form a contiguous block. Moreover, each face f of G has two special vertices $s_f$ and $t_f$ such that the boundary of f consists of two non-empty oriented paths from $s_f$ to $t_f$.

Let $G=(V,E)$ be an internally 4-connected inner triangulation of a 4-gon with outer vertices s, a, t, b in clockwise order, and let $E_R$ and $E_B$ respectively be the red and blue oriented edges of a transversal structure on G. We define $E_R^+ = E_R \cup \{(s,a),(s,b),(a,t),(b,t)\}$ and $E_B^+ = E_B\cup \{(a,s),(a,t),(s,b),(t,b)\}$, i.e., we think of the outer edges as having both, a red direction and a blue direction. The following has been shown in [17] and in [12].

Proposition 1

The red graph $G_R=(V,E_R^+)$ and the blue graph $G_B=(V,E_B^+)$ are both bipolar orientations. $G_R$ has source s and sink t, and $G_B$ has source a and sink b.

The following two properties are easy consequences of the previous discussion.

• (R) The red and the blue graph are both transitively reduced, i.e., if $(v,v')$ is an edge, then there is no directed path $v,u_1,\ldots ,u_k,v'$ with $k\ge 1$.

• (F) For every blue edge e there is a face f in the red graph such that e has one endpoint on each of the two oriented $s_f$ to $t_f$ paths on the boundary of f.

2.2 Posets

We assume basic familiarity with concepts and terminology for posets, referring the reader to the monograph [20] and survey article [21] for additional background material. In this paper we consider a poset $P=(X,<)$ as being equipped with a strict partial order.

A cover relation of P is a pair (x, y) with $x < y$ such that there is no z with $x< z < y$, we write $x\prec y$ to denote a cover relation of the two elements. A diagram (a.k.a. Hasse diagram) of a poset is an upward drawing of its transitive reduction. That is, X is represented by a set of points in the plane and a cover relation $x\prec y$ is represented by a y-monotone curve going upwards from x to y. In general these curves (edges) may cross each other but must not touch any vertices other than their endpoints. A diagram uniquely describes a poset, therefore, we usually show diagrams in our figures. A poset is said to be planar if it has a planar diagram.

It is well known that in discussions of graph planarity, we can restrict our attention to straight-line drawings. In fact, using for example a result of Schnyder [19], if a planar graph has n vertices, then it admits a planar straight-line drawing with vertices on an $(n-2)\times (n-2)$ grid. Discussions of planarity for posets can also be restricted to straight-line drawings; however, this may come at some cost in visual clarity. Di Battista et al. [5] have shown that an exponentially large grid may be required for upward planar drawings of directed acyclic planar graphs with straight lines. In the next subsection we will see that for certain planar posets the situation is more favorable.

2.3 Dimension of Planar Posets

Let $P=(X,<)$ be a poset. A realizer of P is a collection $L_1,L_2,\dots ,L_t$ of linear extensions of P such that $P=L_1\cap L_2\cap \dots \cap L_t$. The dimension of $P=(X,<)$, denoted $\dim (P)$, is the least positive integer t such that P has a realizer of size t. Obviously, a poset P has dimension 1 if and only if it is a chain (total order). Also, there is an elementary characterization of posets of dimension at most 2 that we shall use.

Proposition 2

A poset $\mathbf {P}=(X,P)$ has dimension as most 2 if and only if its incomparability graph is also a comparability graph.

There are a number of results concerning the dimension of posets with planar order diagrams. Recall that an element is called a zero of a poset P when it is the unique minimal element. Dually, a one is a unique maximal element. A finite lattice, i.e., a poset which has well defined meet and join operations, always has a zero and a one.

The following result may be considered part of the folklore of the subject.

Theorem 2

Let P be a finite lattice. Then P is planar if and only if it has dimension at most 2.

Fig. 3.

Diagrams of a planar poset of dimension 3 (left), a non-planar lattice (middle), and a planar lattice (right).

For the reverse direction in the theorem, let P be a lattice of dimension at most 2. Let $L_1$ and $L_2$ be linear orders on X so that $P=L_1\cap L_2$. For each $x\in X$, and each $i=1,2$, let $x_i$ denote the height of x in $L_i$. Then a planar diagram of P is obtained by locating each $x\in X$ at the point in the plane with integer coordinates $(x_1,x_2)$ and joining points x and y with a straight segment when one of x and y covers the other in P. A pair of crossing edges in this drawing would violate the lattice property, indeed if $x\prec y$ and $x'\prec y'$ are two covers whose edges cross, then $x\le y'$ and $x'\le y$ whence x and $x'$ have no unique least upper bound.

A planar digraph D with a unique sink and source, both of them on the outer face, and no transitive edges is the digraph of a planar lattice. Hence, the above discussion directly implies the following classical result.

Fig. 4.

The planar lattice from Fig. 3 with a realizer $L_1,L_2$.

Proposition 3

A planar digraph D on n vertices with a unique sink and source on the outer face and no transitive edges has an upward planar drawing on an $(n-1)\times (n-1)$ grid.

In this paper we will, henceforth, use the terms 2-dimensional poset and planar lattice respectively to refer to a poset $P=(X,<)$ together with a fixed ordered realizer $[L_1,L_2]$. In the case of the lattice, fixing the realizer can be interpreted as fixing a plane drawing of the diagram. By fixing the realizer of P we also have a well-defined primary conjugate, this is the poset Q on X with realizer $[L_1,\overline{L_2}]$, where $\overline{L_2}$ is the reverse of $L_2$. Define the left of relation on X such that x is left of y if and only of $x=y$ or x and y are incomparable in P and $x<y$ in Q.

2.4 Orthogonal Partitions of Posets

Let P be a finite poset, Dilworth’s theorem states that the maximum size of an antichain equals the minimum number of chains partitioning the elements of P.

Greene and Kleitman [16] found a nice generalization of Dilworth’s result. Define a k -antichain to be a family of k pairwise disjoint antichains.

Theorem 3

For any partially ordered set P and any positive integer k

$$\max \sum _{A\in \mathcal {A}}|A| = \min \sum _{C \in \mathcal {C}} \min (|C|,k) $$

where the maximum is taken over all k-antichains $\mathcal {A}$ and the minimum over all chain partitions $\mathcal {C}$ of P.

Greene [15] stated the dual of this theorem. Let an $\ell$-chain be a family of $\ell$ pairwise disjoint chains.

Theorem 4

For any partially ordered set P and any positive integer $\ell$

$$\max \sum _{C\in \mathcal {C}}|C| = \min \sum _{A \in \mathcal {A}} \min (|A|,\ell )$$

where the maximum is taken over all $\ell$-chains $\mathcal {C}$ and the minimum over all antichain partitions $\mathcal {A}$ of P.

A further theorem of Greene [15] can be interpreted as a generalization of the Robinson-Schensted correspondence and its interpretation given by Greene [14]. With a partially ordered set P with n elements there is an associated partition $\lambda$ of n, such that for the Ferrer’s diagram G(P) corresponding to $\lambda$ we get:

Theorem 5

The number of squares in the $\ell$ longest columns of G(P) equals the maximal number of elements covered by an $\ell$-chain of P and the number of squares in the k longest rows of G(P) equals the maximal number of elements covered by a k-antichain.

Figure 5 shows an example, in this case the Ferrer’s diagram G(P) corresponds to the partition $6+3+3+1+1\,{\models }\,14$. Several proofs of Greene’s results are known, e.g. [8, 10], and [18]. For a not so recent, but at its time comprehensive survey we recommend [22].

The approach taken by Frank [10] is particularly elegant. Following Frank we call a chain family $\mathcal {C}$ and an antichain family $\mathcal {A}$ an orthogonal pair iff

• 1. $\qquad \displaystyle P = \Bigl (\bigcup _{A\in \mathcal {A}} A\Bigr ) \ \cup \ \Bigl (\bigcup _{C\in \mathcal {C}} C\Bigr )$, and

• 2.       $|A\cap C| = 1$   for all $A \in \mathcal {A},\ C\in \mathcal {C}$.

If $\mathcal {C}$ is orthogonal to a k-antichain $\mathcal {A}$ and $\mathcal {C}^{+}$ is obtained from $\mathcal {C}$ by adding the rest of P as singletons, then

$$ \sum _{A\in \mathcal {A}} |A| = \sum _{C\in \mathcal {C}^{+}} \sum _{A\in \mathcal {A}} |A \cap C| = \sum _{C\in \mathcal {C}^{+}} \min (|C|,k). $$

Thus $\mathcal {C}^{+}$ is a k optimal chain partition in the sense of Theorem 3. Similarly an $\ell$ optimal antichain partition in the sense of Theorem 4 can be obtained from an orthogonal pair $\mathcal {A},\mathcal {C}$ where $\mathcal {C}$ is an $\ell$-chain.

Using the minimum cost flow algorithm of Ford and Fulkerson [9], Frank proved the existence of a sequence of orthogonal chain and antichain families. This sequence is rich enough to allow the derivation of the whole theory. The sequence consists of an orthogonal pair for every point from the boundary of G(P). With the point $(k,\ell )$ from the boundary of G(P) we get an orthogonal pair $\mathcal {A},\mathcal {C}$ such that $\mathcal {A}$ is a k-antichain and $\mathcal {C}$ an $\ell$-chain, see Fig. 5. Since G(P) is the Ferrer’s diagram of a partition of n we can find a point $(k,\ell )$ on the boundary of G(P) with $k+\ell \le \sqrt{2n}-1$ (This is because every Ferrer’s shape of a partition of m which contains no point (x, y) with $x+y \le s$ on the boundary contains the shape of the partition $(1,2,\ldots ,s+1)$. From $m \ge {s+2 \atopwithdelims ()2}$ we get $s+1 < \sqrt{2m})$.

Fig. 5.

The Ferrer’s shape of the lattice L from Fig. 4 together with two orthogonal pairs of L corresponding to the boundary points (1, 3) and (3, 1) of G(L); chains of $\mathcal {C}$ are blue, antichains of $\mathcal {A}$ are red, green, and yellow.

We will use the following corollary of the theory:

Corollary 1

Let $P=(X,<)$ be a partial order on n elements, then there is an orthogonal pair $\mathcal {A},\mathcal {C}$ where $\mathcal {A}$ is a k-antichain and $\mathcal {C}$ an $\ell$-chain and $k+\ell \le \sqrt{2n}-1$.

For our application we will need some additional structure on the antichains and chains of an orthogonal pair $\mathcal {A},\mathcal {C}$.

The canonical antichain partition of a poset $P=(X,<)$ is constructed by recursively removing all minimal elements from P and make them one of the antichains of the partition. More explicitely $A_1=\mathrm {Min}(X)$ and $A_j = \mathrm {Min}\big (X\setminus \bigcup \{A_i : 1\le i < j\}\big )$ for $j>1$. Note that by definition for each element $y\in A_j$ with $j>1$ there is some $x\in A_{j-1}$ with $x<y$. Due to this property there is a chain of h elements in P if the canonical antichain partition consists of h non-empty antichains. This in essence is the dual of Dilworth’s theorem, i.e., the statement: the maximal size of a chain equals the minimal number of antichains partitioning the elements of P.

Lemma 1

Let $\mathcal {A},\mathcal {C}$ be an orthogonal pair of $P=(X,<)$ and let $P_\mathcal {A}$ be the order induced by P on the set $X_\mathcal {A}= \bigcup \{A: A \in \mathcal {A}\}$. If $\mathcal {A}'$ is the canonical antichain partition of $P_\mathcal {A}$, then $\mathcal {A}',\mathcal {C}$ is again an orthogonal pair of P

Proof

Let $\mathcal {A}$ be the family $A_1,\ldots ,A_k$. Starting with this family we will change the antichains in the family while maintaining the invariant that the family of antichains together with $\mathcal {C}$ forms an orthogonal pair. At the end of the process the family of antichains will be the canonical antichain partition of $P_\mathcal {A}$.

The first phase of changes is the uncrossing phase. We iteratively choose two antichains $A_i,A_j$ with $i<j$ from the present family and let $B_i = \{y\in A_i : \text { there is an } x\in A_j \text { with } x<y\}$ and $B_j = \{x\in A_j : \text { there is a } y\in A_i \text { with } x<y\}$. Define $A'_i = A_i-B_i+B_j$ and $A'_j = A_j-B_j+B_i$. It is easy to see that $A'_i$ and $A'_j$ are antichains and that the family obtained by replacing $A_i,A_j$ by $A'_i,A'_j$ is orthogonal to $\mathcal {C}$. This results in a family of k antichains such that if $i<j$ and $x\in A_i$ and $y\in A_j$ are comparable, then $x<y$.

The second phase is the push-down phase. We iteratively choose $i\in [k-1]$ and let $B = \{y\in A_{i+1} : \text { there is no } x\in A_i \text { with } x<y\}$ and define $A'_{i+1} = A_{i+1}-B$ and $A'_i = A_i+B$. It is again easy to see that $A'_i$ and $A'_{i+1}$ are antichains and that the family obtained by replacing $A_i,A_{i+1}$ by $A'_i,A'_{i+1}$ is orthogonal to $\mathcal {C}$. This results in a family of k antichains such that if $y\in A_{i+1}$, then there is an $x\in A_i$ with $x<y$. This implies that $A_j = \mathrm {Min}(X_\mathcal {A}\setminus \bigcup \{A_i : 1\le i < j\})$, whence the family is the canonical antichain partition.    $\square$

Let $P=(X,<)$ be a 2-dimensional poset with realizer $[L_1,{L_2}]$ and recall that the primary conjugate has realizer $[L_1,\overline{L_2}]$. The order Q corresponds to a transitive relation on the complement of the comparability graph of P, in particular chains of P and antichains of Q are in bijection.

The canonical antichain partition of Q yields the canonical chain partition of P. The canonical chain partition $C_1,C_2,\ldots ,C_w$ of P can be characterized by the property that for each $1\le i < j\le w$ and each element $y\in C_j$ there is some $x\in C_{i}$ with $x\; ||\; y$ and in $L_1$ element x comes before y. In particular $C_1$ is a maximal chain of P.

Let $\mathcal {A},\mathcal {C}$ be an orthogonal pair of the 2-dimensional $P=(X,<)$. Applying the proof of Lemma 1 to the orthogonal pair $\mathcal {C},\mathcal {A}$ of Q we obtain:

Lemma 2

Let $\mathcal {A},\mathcal {C}$ be an orthogonal pair of $P=(X,<)$ and let $P_\mathcal {C}$ be the order induced by P on the set $X_\mathcal {C}= \bigcup \{C : C\in \mathcal {C}\}$. If $\mathcal {C}'$ is the canonical chain partition of $P_\mathcal {C}$, then $\mathcal {C}',\mathcal {A}$ is again an orthogonal pair of P

In a context where edges of the diagram are of interest, it is convenient to work with maximal chains. The canonical chain partition $C_1,C_2,\ldots ,C_w$ of a 2-dimensional P induces a canonical chain cover of P which consists of maximal chains. With chain $C_i$ associate a chain $C^+_i$ which is obtained by successively adding to $C_i$ all compatible elements of $C_{i-1},C_{i-2},\ldots$ in this order. Alternatively $C^+_i$ can be described by looking at the conjugate of P with realizer $[\overline{L_1},{L_2}]$ (this is the dual of the primary conjugate Q), and defining $C^+_i$ as the first chain in the canonical chain partition of the order induced by $\bigcup \{C_j : 1\le j \le i\}$. The maximality of $C^+_i$ follows from the characterization of the canonical chain partition given above.

3 Drawing Planar Lattices on Few Lines

In this section we prove that planar lattices with n elements have a straight-line diagram with all vertices on a set of $\sqrt{2n}-1$ horizontal and vertical lines. The following proposition covers the case where the lattice has an antichain partition of small size. We assume that a planar lattice is given with a realizer $[L_1,L_2]$ and, hence, with a fixed plane drawing of its diagram.

Proposition 4

For any planar lattice $L=(X,<)$ with an extension of L there is a plane straight-line drawing $\varGamma$ of the diagram $D_L$ of L such that each element $x\in X$ is represented by a point with y-coordinate h(x). Additionally all elements of the left boundary chain of $D_L$ are aligned vertically in the drawing.

The proof of this proposition can be found in the full version. The idea is to extend the drawing of a chain on a vertical line by ears. Such ear-extensions are also used in the proof of the next theorem.

Theorem 6

For every planar lattice $L=(X,<)$ with $|X|=n$, there is a plane straight-line drawing of the diagram such that the elements are represented by points on a set of at most $\sqrt{2n}-1$ lines. Additionally

• each of the lines is either horizontal or vertical,

• each crossing of a horizontal and a vertical line hosts an element of X.

Proof

Let $\mathcal {A},\mathcal {C}$ be an orthogonal pair of L such that $\mathcal {A}$ is a k-antichain, $\mathcal {C}$ an $\ell$-chain, and $k+\ell \le \sqrt{2n}-1$. It follows from Corollary 1 that such a pair exists.

Since L has a fixed ordered realizer $[L_1,L_2]$, we can apply Lemma 1 to $\mathcal {A}$ and Lemma 2 to $\mathcal {C}$ to get an orthogonal pair $(A_1,\ldots ,A_k),(C_1,\ldots ,C_\ell )$ where the antichain family and the chain family are both canonical. Fix an extension of L with the property that $h(x) = i$ for all $x\in A_i$.

In the following we will construct a drawing $\varGamma$ of $D_L$ such that each element $x\in X$ is represented by a point with y-coordinate h(x), and in addition all elements of chain $C_i$ lie on a common vertical line $\mathbf {g}_i$ for $1\le i\le \ell$. By Property 1 of orthogonal pairs, for each $x\in X$ there is an i such that $x\in A_i$ or a j such that $x\in C_j$ or both. Therefore, $\varGamma$ will be a drawing such that the k horizontal lines $y=i$ with $i=1,\ldots ,k$ together with the $\ell$ vertical lines $\mathbf {g}_j$ with $j=1,\ldots ,\ell$ cover all the elements of X. Property 2 of orthogonal pairs implies the second extra property mentioned in the theorem.

If the number $\ell$ of chains is zero, then we get a drawing $\varGamma$ with all the necessary properties from Proposition 4. Now let $\ell > 0$.

The chain family $C_1,\ldots ,C_\ell$ is the canonical chain partition of the order induced on $X_\mathcal {C}= \bigcup \{C_i : i=1\ldots \ell \}$. Let $C^+_1,\ldots ,C^+_\ell$ be the corresponding canonical chain covering of $X_\mathcal {C}$.

Let $X_i$ for $1\le i\le \ell$ be the set of all elements which are left of some element of $C^+_i$ in L, and let $X_{\ell +1}=X$. Define $S_i$ as the suborder of L induced by $X_i$. Also let $Y_i = X_{i+1} - X_{i} + C^+_i$ and let $T_i$ be the suborder of L induced by $Y_i$. Note the following properties of these sets and orders:

• $X_i\cap C_j = \emptyset$ for $1\le i<j\le \ell$.

• Each $S_i$ is a planar sublattice of L, its right boundary chain is $C^+_i$.

• $T_i$ is a planar sublattice of L.

A drawing $\varGamma _1$ of $S_1$ with the right boundary chain being aligned vertically is obtained by applying Proposition 4 to the vertical reflection of the diagram $D_L[X_1]$ and reflecting the resulting drawing vertically.

We construct the drawing $\varGamma$ of $D_L$ in phases. In phase i we aim for a drawing $\varGamma _{i+1}$ of $S_{i+1}$ extending the given drawing $\varGamma _{i}$ of $S_i$, i.e., we need to construct a drawing $\varLambda _i$ of $T_i$ such that

• (1) The left boundary chain of $\varLambda _i$ matches the right boundary chain of $\varGamma _{i}$.

• (2) In $\varLambda _i$ all elements of $C_{i+1}$ are on a common vertical line $\mathbf {g}_{i+1}$.

The construction of $\varLambda _i$ is done in three stages. First we extend $C^+_i$ to the right by adding ‘ears’. Then we extend $C^+_{i+1}$ to the left by adding ‘ears’. Finally we show that the left and the right part obtained from the first two stages can be combined to yield the drawing $\varLambda _i$.

To avoid extensive use of indices let $Y=Y_i$, $T=T_i$, $C^+=C^+_i$, and let $\gamma$ be a copy of the y-monotone polygonal right boundary of $\varGamma _i$, i.e., $\gamma$ is a drawing of C. We initialize $\varLambda '=\gamma$.

A left ear of T is a face F in the diagram $D_L[Y]$ of T such that the left boundary of F is a subchain of the left boundary chain $C^+$ of $D_L[Y]$. The ear is feasible if the right boundary chain contains no element of $C_{i+1}$. Given a feasible ear we use the method from the proof of Proposition 4 to add F to $\gamma$. We represent the right boundary $z_0< z_1< \ldots < z_{l}$ excluding $z_0$ and $z_l$ of F on a vertical line $\mathbf {g}$ by points $q_1,\ldots ,q_{l-1}$ with y-coordinates as prescribed by h. The points $q_0$ and $q_{l}$ representing $z_0$ and $z_l$ respectively are already represented on $\gamma$. Then we place $\mathbf {g}$ at some distance $\beta$ to the right of $\gamma$. The value of $\beta$ has to be chosen large enough to ensure that edges $q_0,q_1$ and $q_{l-1},q_l$ are drawn such that they do not interfere with $\gamma$. Let $\varLambda '$ be the drawing augmented by the polygonal path $q_0,q_1,\ldots ,q_{l-1},q_l$ and let $C^+$ again refer to the right boundary chain $\gamma$ of $\varLambda '$. Delete all elements of the left boundary of F except $z_0$ and $z_l$ from Y and T. This shelling of a left ear from T is iterated until there remains no left feasible ear. Upon stopping we have a drawing $\varLambda '$ which can be glued to the right side of $\varGamma _i$. Let $\gamma '$ be the right boundary chain of $\varLambda '$.

In the second stage the procedure is quite similar; starting from $C=C_{i+1}$ on a line $\mathbf {g}$ right ears taken from T are added to the drawing until there remains no feasible right ear. This yields a drawing $\varLambda ''$ with left boundary $\gamma ''$.

In the final stage we have to combine the drawings $\varLambda '$, $\varLambda ''$ into a single drawing. This is done by drawing the edges and chains which remain in T between the two boundary chains as straight segments between $\gamma '$ and $\gamma ''$. This will be possible because we can shift $\gamma '$ and $\gamma ''$ as far apart horizontally as necessary.

First we draw all the edges connecting the two chains. If we place $\gamma '$ and $\gamma ''$ at sufficient horizontal distance, then the edges of E can be drawn such that they do not interfere (introduce crossings) with $\gamma '$ and $\gamma ''$. Let $\varLambda$ be the drawing consisting of $\gamma '$, $\gamma ''$, and the deges between them. An important feature of $\varLambda$ is that if we move the two chains $\gamma '$ and $\gamma ''$ even further apart the drawing keeps the needed properties, i.e., the height of elements remains unaltered, vertices of a chain which should be vertically aligned remain vertically aligned because they are vertically aligned in either $\gamma '$ or $\gamma ''$ and the drawing is crossing-free.

Now assume that T contains elements which are not represented in $\varLambda$. Let B be a connected component of such elements where connectivity is with respect to $D_L$. All the elements of B have to be placed in a face $F_B$ of $\varLambda$. Let $\delta '$ and $\delta ''$ be the left and right boundary of $F_B$.

In the following we will repeat the choice of a component B and a chain C from B which is to be drawn in the corresponding face $F_B$ of $\varLambda$ such that the minimum and the maximum of C have connecting edges to the two sides of the boundary of $F_B$. Let us consider the case that in $D_L$ the maximum of C has an outgoing edge to an element which is represented by a point $p\in \delta '$ and the minimum of C has an incoming edge from an element represented by $q\in \delta ''$. We represent the elements of C as points on the prescribed heights on a line segment $\zeta$ with endpoints p and q. It may become necessary to stretch the face horizontally to be able to place C. In this case we stretch the whole drawing between $\gamma '$ and $\gamma ''$ with a uniform stretch factor. There may be additional edges between elements of C and elements on $\delta '$ and $\delta ''$. They can also be drawn without crossing when the distance of $\delta '$ and $\delta ''$ exceeds some value b.

Stretching the whole drawing between $\gamma '$ and $\gamma ''$ allows us to draw the segment $\zeta$ and additional edges inside of $F_B$ because of the following invariant. For each face F of the drawing $\varLambda$ and two points x and y from the boundary of F it holds that: if the segment x, y is not in the interior of F, then the parts of the boundary obstructing the segment x, y belong to $\gamma '$ or $\gamma ''$.

When including a chain C in the drawing $\varLambda$, we place the elements of C at the prescribed heights on a common line segment $\zeta$. This ensures that each new element contributes convex corners in all incident faces. Hence, new elements can not obstruct a visibility within a face. Therefore, obstructing corners correspond to elements of $\gamma '$ or $\gamma ''$ and the invariant holds.

Now consider the case where maximum and minimum of the chain C connect to two elements p and q on the same side of F. Since $\gamma '$ and $\gamma ''$ do not admit ear extensions we know that not both of p and q belong to one of $\gamma '$ and $\gamma ''$. If the segment from p to q is obstructed, then the invariant ensures that with sufficient horizontal stretch the segment $\zeta$ connecting p and q will be inside F. Hence, chain C can be drawn and $\varLambda$ can be extended.

When there remains no component B containing a chain C which can be included in the drawing using the above strategy, then either all elements of Y are drawn or we have the following: every component B only connects to elements of a line segment $\zeta _B$.

In this situation B is kind of a big ear over $\zeta _B$. The details of how to draw B are given in the full version. But note, that in this situation we will not maintain or need the invariant.

Glueing the drawings $\varLambda '$ with $\varLambda$ at the polygonal path $\gamma '$ and $\varLambda$ with $\varLambda ''$ at $\gamma ''$ (a y-monotone collection of paths) yields a drawing $\varLambda _i$ of $T_i$. The drawing $\varLambda _i$ can be glued to $\varGamma _i$ to form a drawing $\varGamma _{i+1}$ of $S_{i+1}$. Eventually the drawing $\varGamma _\ell$ will be constructed. From there the drawing $\varGamma =\varGamma _{\ell +1}$ is obtained by adding some left ears.    $\square$

4 Transversal Structures on Few Lines

Theorem 7

For every internally 4-connected inner triangulation of a 4-gon $G=(V,E)$ with n vertices there is a planar straight-line drawing such that the vertices are represented by points on a set of at most $\sqrt{2n}-1$ lines. Additionally

• each of the lines is either horizontal or vertical,

• each crossing point of a horizontal and a vertical line hosts a vertex.

Due to Theorem 6 it is possible to draw the red subgraph $G_R$ of a transversal structure on $\sqrt{2n}-1$ lines. It would be nice if we could include the blue edges of the transversal structure in the drawing. This, however, may yield crossings. Therefore we have to go through the proof of Theorem 6 and take care of blue edges while constructing the drawing of the red graph. The proof can be found in the full version. An example for an intermediate stage of the construction is shown in Fig. 6.

Fig. 6.

A partially drawn transversal structure. The figure shows a drawing of $\varGamma _4$, these are the vertices left of some element in $C^+_4$ together with the induced edges.

It remains to see how Theorem 1 follows from Theorem 7. Let G be a 4-connected triangulation and let $G'$ be obtained from G by deleting one of the outer edges. Now $G'$ is an internally 4-connected inner triangulation of a 4-gon. Label the outer vertices of $G'$ such that the deleted edge is the edge s, t. Slightly stretching Theorem 7 we prescribe $h(s)={-}\infty$ and $h(t)=\infty$, this yields a planar straight-line drawing $\varGamma$ of $G'$ such that the vertices except s and t are represented by points on a set of at most $\sqrt{2n}-1$ lines and the edges connecting to s and t are vertical rays. Moreover with every edge v, s or v, t there is an open cone K containing the vertical ray, such that the point representing v is the apex of K and this is the only vertex contained in K. Now let $\mathbf {g}$ be a vertical line which is disjoint from $\varGamma$. On $\mathbf {g}$ we find a point $p_s$ which is contained in all the upward cones and a point $p_t$ contained in all the downward cones. Taking $p_s$ and $p_t$ as representatives for s and t we can tilt the rays and make them finite edges ending in $p_s$ and $p_t$ respectively, and in addition draw the edge $p_s,p_t$.

We conclude with a remark and two open problems.

• Our results are constructive and can be complemented with algorithms running in polynomial time.

• Is $\pi (G) \in O(\sqrt{n})$ for every planar graph G on n vertices?

• What size of a grid is needed for drawings of 4-connected plane graphs on $O(\sqrt{n})$ lines?