4-Connected Triangulations on Few Lines

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.


Introduction
Given a planar graph G we denote by π(G) the minimum number such that G has a plane straight-line drawing in which the vertices can be covered by a collection of lines. Clearly π(G) = 1 if and only if G is a forest of paths. The set of graphs with π(G) = 2, however, is already surprisingly rich, it contains trees, outerplanar graphs and subgraphs of grids, see [1,8].
The parameter π(G) has received some attention in recent years, here is a list of known results: It is NP-complete to decide whether π(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 π(G) are d G + 1 and d G + 2 respectively, see Biedl et al. [2] and for the lower bound also Eppstein [7,Thm. 16.13].
The main result of this paper is the following theorem. Theorem 1 If G is a 4-connected plane triangulation on n vertices, then π(G) p 2n.
The result is not far from optimal since, using a small number of additional vertices and many additional edges, the graph G mentioned above can be transformed into a 4-connected plane triangulation, i.e., in the class we have graphs with π(G) P Ω(n 1/3 ). Figure 1 shows an section of such an extension of G . The proof of the Theorem 1 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 Section 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 Section 2.2. Dimension of posets and the connection with planarity are covered in Section 2.3. In Section 2.4 we survey orthogonal partitions of posets. The theory implies that every poset on n elements can be covered by at most p 2n − 1 subsets such that each of the subsets is a chain or an antichain.
In Section 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 p 2n − 1 lines. All the lines used for the construction are either horizontal or vertical.
Finally in Section 4 we prove the main result: transversal structures can be drawn on at most p 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.

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 transversal structure 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 2 illustrates the properties and shows an example. Transversal structures have been studied in [17], [12], 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 x W, 0 y H, and H + W 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 bipolar orientation source s and a unique sink t, cf. [5]. Bipolar orientations of plane graphs are also required to have s and t incident to the outer face. 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 dene 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 0 ) is an edge, then there is no directed path v, u 1 , . . . , u k , v 0 with k ! 1.
(F) For every blue edge e P E B 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.

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 cover relation write x 0 y to denote a cover relation of the two elements. A diagram (a.k.a. Hasse diagram) 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 0 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 gures. A poset is said to be planar if it has a planar diagram. planar 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)¢(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. [6] 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.

Dimension of planar posets
Let P = (X, <) be a poset. A realizer of P is a collection L 1 , L 2 , . . . , L t of linear extensions of realizer P such that P = L 1 L 2 ¡ ¡ ¡ L t . The dimension of P = (X, <), denoted dim(P), is the least dimension 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 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 nite poset which is also a lattice, i.e., which has well dened meet and join operations, always has both a zero and a one.
The following result may be considered part of the folklore of the subject.  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 L 2 . For each x P 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 P X at the point in the plane with integer coordinates (x 1 , x 2 ) and joining points x and y with a straight line 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 0 y and x 0 0 y 0 are two covers whose edges cross, then x y 0 and x 0 y whence x and x 0 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.
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 drawing on an (n − 1) ¢ (n − 1) grid.
To the best of our knowledge the area problem for diagrams of general planar posets is open. 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 xed ordered realizer [L 1 , L 2 ]. In the case of the lattice, xing the realizer can be interpreted as xing a plane drawing of the diagram. By xing the realizer of P we also have a well-dened primary conjugate, this is the primary conjugate poset Q on X with realizer [L 1 , L 2 ], where L 2 is the reverse of L 2 . Dene the left of relation 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.

Orthogonal partitions of posets
Let P be a nite 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. Dene a kantichain to be a family of k pairwise disjoint antichains. where the maximum is taken over all k-antichains A and the minimum over all chain partitions C of P.
Greene [15] stated the dual of this theorem. Let a -chain be a family of pairwise disjoint -chain chains.
Theorem 4 For any partially ordered set P and any positive integer where the maximum is taken over all -chains C and the minimum over all antichain partitions 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].
To a partially ordered set P with n elements there is an associated partition λ of n, such that for the Ferrer's diagram G(P) corresponding to λ we get: Theorem 5 The number of squares in the longest columns of G(P) equals the maximal number of elements covered by an -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 | = 14. Several proofs of Greene's results are known, e.g. [9], [11], and [18]. For a not so recent, but at its time comprehensive survey we recommend [22].
The approach taken by Andr as Frank [11] is particularly elegant. Following Frank we call a chain family C and an antichain family A an orthogonal pair i orthogonal pair 1.
If C is orthogonal to a k-antichain A and C + is obtained from C by adding the rest of P as singletons, then Thus C + is a k optimal chain partition in the sense of Theorem 3. Similarly an optimal antichain partition in the sense of Theorem 4 can be obtained from an orthogonal pair A, C where C is an -chain. Using the minimum cost ow algorithm of Ford and Fulkerson [10], 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, ) from the boundary of G(P) we get an orthogonal pair A, C such that A is a k-antichain and C an -chain, see Figure 5. Since G(P) is the Ferrer's diagram of a partition of n we can nd a point (k, ) on the boundary of G(P) with k + p 2n − 1 (This is because every Ferrer's shape of a partition of m which contains no point (x, y) with x + y s on the boundary contains the shape of the partition (1, 2, . . . , s + 1). From m ! s+2 2 ¡ we get s + 1 < p 2m).
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 A, C where A is a k-antichain and C an -chain and k + p 2n − 1. Figure 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 C are blue, antichains of A are red, green, and yellow.
For our application we will need some additional structure on the antichains and chains of an orthogonal pair A, C.
The canonical antichain partition of a poset P = (X, <) is constructed by recursively canonical antichain partition removing all minimal elements from P and make them one of the antichains of the partition. More explicitely A 1 = Min(X) and A j = Min X \ Note that by denition for each element y P A j with j > 1 there is some x P 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 A, C be an orthogonal pair of P = (X, <) and let P A be the order induced by P on the set X A = {A : A P A}. If A 0 is the canonical antichain partition of P A , then A 0 , C is again an orthogonal pair of P Proof. Let A be the family A 1 , . . . , 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 C forms an orthogonal pair. At the end of the process the family of antichains will be the canonical antichain partition of P A .
The rst 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 P A i : there is an x P A j with x < y} and B j = {x P A j : there is a y P A i with x < y}.
It is easy to see that A 0 i and A 0 j are antichains and that the family obtained by replacing A i , A j by A 0 i , A 0 j is orthogonal to C. This results in a family of k antichains such that if i < j and x P A i and y P A j are comparable, then x < y.
The second phase is the push-down phase. We iteratively It is again easy to see that A 0 i and A 0 i+1 are antichains and that the family obtained by replacing A i , A i+1 by A 0 i , A 0 i+1 is orthogonal to C. This results in a family of k antichains such that if y P A i+1 , then there is an x P A i with x < y. This implies that A j = Min(X A \ whence the family is the canonical antichain partition. Let P = (X, <) be a 2-dimensional poset with realizer [L 1 , L 2 ] and recall that the primary conjugate has realizer [L 1 , 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 canonical chain partition C 1 , C 2 , . . . , C w of P can be characterized by the property that for each 1 i < j w and each element y P C j there is some x P 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 A, C be an orthogonal pair of the 2-dimensional P = (X, <). Applying the proof of Lemma 1 to the orthogonal pair C, A of Q we obtain: Lemma 2 Let A, C be an orthogonal pair of P = (X, <) and let P C be the order induced by P on the set X C = {C : C P C}. If C 0 is the canonical chain partition of P C , then C 0 , 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 , . . . , 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 canonical chain cover is obtained by successively adding to C i all compatible elements of C i−1 , C i−2 , . . . in this order. Alternatively C + i can be described by looking at the conjugate of P with realizer [L 1 , L 2 ] (this is the dual of the primary conjugate Q), and dening C + i as the rst chain in the canonical chain partition of the order induced by {C j : 1 j i}). The maximality of C + i follows from the characterization of the canonical chain partition given above.

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 p 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 xed plane drawing of its diagram. Proposition 4 For any planar lattice L = (X, <) with an extension h : X → IR of L there is a plane straight-line drawing Γ of the diagram D L of L such that each element x P 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.
Proof. Let C 1 , C 2 , . . . , C w be the canonical chain partition and C + 1 , C + 2 , . . . , C + w be the corresponding canonical chain cover. Dene S i as the suborder of L induced by {C j : 1 j i} and note that S i is a sublattice of L with left boundary chain C 1 = C + 1 and right boundary chain C + i . Embed the elements of C 1 on a vertical line g 1 (e.g. the line y = 0) with points as prescribed by h. This is a drawing Γ 1 of S 1 . Suppose that a drawing Γ i of the diagram S i is constructed. The right boundary path γ i of Γ i is a polygonal y-monotone path. Embed the elements of C i+1 on a vertical line g i+1 with points as prescribed by h. We need a position for g i+1 to the right of γ i such that all the diagram edges connecting C i+1 to C + i can be inserted to obtain a crossing free drawing Γ i+1 of the diagram of S i+1 .
Let E i be the set of diagram edges connecting C i+1 to C + i . For each e P E i there are points p P γ i and q P g i+1 representing the endpoints. Let K p be an open cone with apex p which intersects γ i only at p and contains a horizontal ray to the right. Let b e be the minimal horizontal distance of γ i and g i+1 such that q P K p . Let β = max(b e : e P E i ). If we place γ i and g i+1 at horizontal distance β, then the edges of E i can be drawn such that they do not interfere (introduce crossings) with γ i . We claim that there is no crossing of edges of E i . Let (p, q) and (p 0 , q 0 ) be two drawn edges from E i . Since they are edges of a planar diagram and have endpoints on two chains, we know, that h(p) h(p 0 ) implies h(q) h(q 0 ). Edge (p, q) is drawn in the cone K p . If (p 0 , q 0 ) intersects the edge and p 0 is above p on γ i , then q 0 has to be below q on g i+1 , in contradiction to h(q) h(q 0 ). Hence we have a planar drawing Γ i+1 of the diagram of S i+1 . With induction we obtain the drawing Γ = Γ w of D L . Proof. Let A, C be an orthogonal pair of L such that A is a k-antichain, C an -chain, and k + p 2n − 1. It follows from Corollary 1 that such a pair exists. Since L has a xed ordered realizer [L 1 , L 2 ], we can apply Lemma 1 to A and Lemma 2 to C to get an orthogonal pair (A 1 , . . . , A k ), (C 1 , . . . , C ) where the antichain family and the chain family are both canonical. Fix an extension h : X → IR of L with the property that h(x) = i for all x P A i .
In the following we will construct a drawing Γ of D L such that each element x P 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 g i for 1 i . By Property 1 of orthogonal pairs, for each x P X there is an i such that x P A i or a j such that x P C j or both. Therfore, Γ will be a drawing such that the k horizontal lines y = i with i = 1, . . . , k together with the vertical lines g j with j = 1, . . . , cover all the elements of X. Property 2 of orthogonal pairs implies the second extra property mentioned in the theorem.
If the number of chains is zero, then we get a drawing Γ with all the necessary properties from Proposition 4. Now let > 0.
The chain family C 1 , . . . , C is the canonical chain partition of the order induced on X C = {C i : i = 1 . . . }. Let C + 1 , . . . , C + be the corresponding canonical chain covering of X C . Let X i for 1 i be the set of all elements which are left of some element of C + i in L, and let X +1 = X. Dene 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: 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 Γ 1 of S 1 with the right boundary chain being aligned vertically is obtained by applying Proposition 4 to the vertical reection of the diagram D L [X 1 ] and reecting the resulting drawing vertically.
We construct the drawing Γ of D L in phases. In phase i we aim for a drawing Γ i+1 of S i+1 extending the given drawing Γ i of S i , i.e., we need to construct a drawing Λ i of T i such that (1) The left boundary chain of Λ i matches the right boundary chain of Γ i . (2) In Λ i all elements of C i+1 are on a common vertical line g i+1 . The construction of Λ 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 rst two stages can be combined to yield the drawing Λ i .
To avoid extensive use of indices let Y = Y i , T = T i , C + = C + i , and let γ be a copy of the y-monotone polygonal right boundary of Γ i , i.e., γ is a drawing of C. We initialize Λ 0 = γ.
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 left ear 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 γ. We represent the right boundary z 0 < z 1 < . . . < z l excluding z 0 and z l of F on a vertical line g by points q 1 , . . . , 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 γ. Then we place g at some distance β to the right of γ. The value of β 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 γ. Let Λ 0 be the drawing augmented by the polygonal path q 0 , q 1 , . . . , q l−1 , q l and let C + again refer to the right boundary chain γ of Λ 0 . 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 Λ 0 which can be glued to the right side of Γ i . Let γ 0 be the right boundary chain of Λ 0 . Now let C = C i+1 . Initialize a new drawing Λ 00 by placing the elements of C on a vertical line g at the heights prescribed by h and connect consecutive ones by an edge whenever the order relation is indeed a cover relation of L. The initial drawing may thus be disconnected and if so this will remain the case throughout this stage. We now consider right ears from T . A right ear of T corresponding to a face F is feasible if the left boundary chain of F contains no element of γ 0 . The left boundary chain of a feasible ear can be drawn as a y-monotone polygonal chain left of the left boundary γ 00 of Λ 00 . Update γ 00 to be the new left boundary of the augmented Λ 00 and remove the elements of the ear from Y and T . The shelling of right ears from T is iterated until there remains no feasible right ear. Note that γ 00 is y-monotone but it may consist of several components.
In the nal stage we have to combine the drawings Λ 0 , Λ 00 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 γ 0 and γ 00 . This will be possible because we can shift γ 0 and γ 00 as far apart horizontally as necessary.
First we draw all the edges connecting the two chains. Let E be the set of edges connecting the left and right boundary chains of T . For each e P E there are points p P γ 0 and q P γ 00 representing the endpoints. Let K p be an open cone with apex p which intersects γ 0 only at p and contains a horizontal ray to the right and let K q be an open cone with apex q which intersects γ 00 only at q and contains a horizontal ray to the left. Let b e be the minimal horizontal distance of γ 0 and γ 00 such that p P K q and q P K p . Let β = max(b e : e P E). If we place γ 0 and γ 00 at horizontal distance β, then the edges of E can be drawn such that they do not interfere (introduce crossings) with γ 0 and γ 00 . We claim that there is no crossing of edges of E. Let (p, q) and (p 0 , q 0 ) be two drawn edges from E. Since they are diagram edges with endpoints on two chains we know that h(p) h(p 0 ) implies h(q) h(q 0 ). Edge (p, q) is drawn in K p K q . If (p 0 , q 0 ) intersects the edge and p 0 is above p on γ 0 , then q 0 has to be below q on γ 00 , in contradiction to h(q) h(q 0 ). Placing Λ 0 and Λ 00 such that β is the distance between their outer chains and drawing the edges of E yields a drawing Λ of some lattice. An important feature of the drawing is that if we move the two subdrawings Λ 0 and Λ 00 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, and the drawing is crossing-free. Now assume that T contains elements which are not represented in Λ. 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 Λ. Let δ 0 and δ 00 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 Λ 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 P δ 0 and the minimum of C has an incoming edge from an element represented by q P δ 00 . We represent the elements of C as points on the prescribed heights on a line segment ζ 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 γ 0 and γ 00 with a uniform stretch factor. There may be additional edges between elements of C and elements on δ 0 and δ 00 . They can also be drawn without crossing when the distance of δ 0 and δ 00 exceeds some value b.
Stretching the whole drawing between γ 0 and γ 00 allows us to draw the segment ζ and additional edges inside of F B because of the following invariant.
For each face F of the drawing Λ 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 γ 0 or γ 00 . When including a chain C in the drawing Λ, we place the elements of C at the prescribed heights on a common line segment ζ. 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 γ 0 or γ 00 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 γ 0 and γ 00 do not admit ear extensions we know that not both of p and q belong to one of γ 0 and γ 00 . If the segment from p to q is obstructed, then the invariant ensures that with sucient horizontal stretch the segment ζ connecting p and q will be inside F. Hence, chain C can be drawn and Λ 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 ζ B .
In this situation B is kind of a big ear over ζ B . We next describe how to draw B, but note, that doing this we will not maintain or need the invariant.
By construction all elements of ζ B belong to a common chain C B . Consider the union B + C B and note that this is a planar lattice L B , moreover, C B is either the left or the right boundary chain of L B . Assume that C B is the left boundary chain of L B . Now use Proposition 4 to get a drawing Λ B of L B with C B aligned vertically. Using an ane transformation we can map Λ B into Λ such that the line containing C B in Λ B is mapped to the line supporting the segment ζ B . Since elements of C B are at their prescribed heights their representing points in Λ B are mapped to the representing points of Λ. The ane map also has to compress Λ B horizontally so that it is placed in a narrow strip on the right side of ζ B . This strip can be chosen narrow enough to make sure that all of B is mapped to the face of Λ where it belongs.
Glueing the drawings Λ 0 with Λ at the polygonal path γ 0 and Λ with Λ 00 at γ 00 (a ymonotone collection of paths) yields a drawing Λ i of T i . The drawing Λ i can be glued to Γ i to form a drawing Γ i+1 of S i+1 . Eventually the drawing Γ will be constructed. From there the drawing Γ = Γ +1 is obtained by adding some left ears. Proof. Fix a transversal structure of G and consider the red graph G R = (V, E + R ). From Proposition 1 and (R) we know that G R is bipolar and transitively reduced. This implies that there is a planar lattice L = (V, <) such that a diagram of L is an upward drawing of G R . The relation < is dened as v < v 0 if and only if there is a directed path from v to v 0 in G R .
We would like to use Theorem 6 to draw G R on p 2n − 1 lines and then include the blue edges of the transversal structure in the drawing. This, however, may yield crossings. Instead we go through the proof of Theorem 6 and include blue edges while constructing the drawing of the red graph.
When adding a left feasible ear, i.e., when adding the right boundary of a face F, we draw all the blue edges corresponding to the face F. If e has to connect p P γ and q P g dene b e as the minimal horizontal distance of γ and g such that q P K p . When placing g at a distance β from γ which exceeds all the values b e , the blue edges can be drawn crossing free. When adding a right feasible ear the situation is symmetric. Now let us consider the stage where a left and right drawing Λ 0 and Λ 00 with boundary chains γ 0 and γ 00 have to be combined. When drawing edges connecting the two chains we include the set of all blue edges with one end on γ 0 and one on γ 00 . Then we complete the combination on the basis of the red edges. Only in the`bad' case we have to be careful. First, when drawing L B using Proposition 4 we also include the blue edges in the drawing. This only requires to choose the distances β as maxima over larger sets of values b e . Second, when placing the drawing Λ B in a narrow strip on the side of ζ B we have to be carefull that this does not obstruct a visibility from the left side of the face to the right side. Finally, all the remaining blue edges have to be drawn in the faces between γ 0 and γ 00 . Due to the invariant this is possible if we stretch the drawing between the two chains suciently. It remains to see how Theorem 1 follows from Theorem 7. Let G be a 4-connected triangulation and let G 0 be obtained from G by deleting one of the outer edges. Now G 0 is an internally 4-connected inner triangulation of a 4-gon. Label the outer vertices of G 0 such that the deleted edge is the edge s, t. Slightly stretching Theorem 7 we prescribe h(s) = −∞ and h(t) = ∞, this yields a planar straight-line drawing Γ of G 0 such that the vertices except s and t are represented by points on a set of at most p 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 g be a vertical line which is disjoint from Γ . On g we nd 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 nite 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 π(G) P O( p 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( p n) lines?