Journal of Graph Algorithms and Applications Drawing Planar Graphs with Reduced Height 434 Durocher and Mondal Drawing Planar Graphs with Reduced Height

A polyline (resp., straight-line) drawing Γ of a planar graph G on a set L k of k parallel lines is a planar drawing that maps each vertex of G to a distinct point on L k and each edge of G to a polygonal chain (resp., straight line segment) between its corresponding endpoints, where the bends lie on L k. The height of Γ is k, i.e., the number of lines used in the drawing. In this paper we establish new upper bounds on the height of polyline drawings of planar graphs using planar separators. Specifically , we show that every n-vertex planar graph with maximum degree ∆, having an edge separator of size λ, admits a polyline drawing with height 4n/9 + O(λ), where the previously best known bound was 2n/3. Since λ ∈ O(√ n∆), this implies the existence of a drawing of height at most 4n/9 + o(n) for any planar triangulation with ∆ ∈ o(n). For n-vertex pla-nar 3-trees, we compute straight-line drawings, with height 4n/9 + O(1), which improves the previously best known upper bound of n/2. All these results can be viewed as an initial step towards compact drawings of pla-nar triangulations via choosing a suitable embedding of the graph.


Introduction
A polyline drawing of a planar graph G is a planar drawing of G such that each vertex of G is mapped to a distinct point in the Euclidean plane, and each edge is mapped to a polygonal chain between its endpoints.Let L k = {l 1 , l 2 , . . ., l k } be a set of k horizontal lines such that for each i ≤ k, line l i passes through the point (0, i).A polyline drawing of G is called a polyline drawing on L k if the vertices and bends of the drawing lie on the lines of L k .The height of such a drawing is k, i.e., the number of parallel horizontal lines used by the drawing.Such a drawing is also referred to as a k-layer drawing in the literature [21,25].
Let Γ be a polyline drawing of G.We call Γ a t-bend polyline drawing if each of its edges has at most t bends.Thus a 0-bend polyline drawing is also known as a straight-line drawing.G is called a planar triangulation if every face of G is bounded by a cycle of three vertices.Figure 1(a) shows a planar graph G, and  Drawing planar graphs on a small integer grid is an active research area in graph drawing [4,9,17,24,15], which is motivated by the need of compact layout of VLSI circuits and visualization of software architecture.In visualization applications, the constraint on area is imposed naturally by the size of the display screen.For VLSI circuit layout, compact drawings reduce the microchip area.Minimizing area often requires the edges to have bends.Since simultaneously optimizing the width and height of the drawing is very challenging, researchers have also focused their attention on optimizing one dimension of the drawing [7,18,21,25], while the other dimension is unbounded.
In this paper we develop new techniques that can produce drawings with small height.We distinguish between the terms 'plane' and 'planar'.A plane graph is a planar graph with a fixed combinatorial embedding and a specified outer face.While drawing a planar graph, we allow the output to represent any planar embedding of the graph.On the other hand, while drawing a plane graph, the output is further constrained to respect the input embedding.
Related Work: State-of-the-art algorithms that compute straight-line drawings of n-vertex plane graphs on an (O(n) × 2n/3)-size grid imply an upper bound of 2n/3 on the height of straight-line drawings [6,7].This bound is tight for plane graphs, i.e., there exist n-vertex plane graphs such as plane nested triangles graphs and some plane 3-trees that require a height of 2n/3 in any of their straight-line drawings [12,22].Recall that an n-vertex nested triangles graph is a plane graph formed by a sequence of n/3 vertex disjoint cycles, , where for each i ∈ {2, . . ., n/3}, cycle C i contains the cycles C 1 , . . ., C i−1 in its interior, and a set of edges that connect each vertex of C i to a distinct vertex in C i−1 .Besides, a plane 3-tree is a triangulated plane graph that can be constructed by starting with a triangle, and then repeatedly adding a vertex to some inner face of the current graph and triangulating that face.
The 2n/3 upper bound on the height is also the currently best known bound for polyline drawings, even for planar graphs, i.e., when we are allowed to choose a suitable embedding for the output drawing.In the variable embedding setting, Frati and Patrignani [17] showed that every n-vertex nested triangles graph can be drawn with height at most n/3 + O(1), which is significantly smaller than the lower bound of 2n/3 in the fixed embedding setting.Zhou et al. [28] showed that series-parallel graphs can be drawn with 0.3941n 2 area, and hence with height 0.628n < 2n/3.Similarly, Hossain et al. [18] showed that an universal set of n/2 horizontal lines can support all n-vertex planar 3-trees, i.e., every planar 3-tree admits a drawing with height at most n/2.They also showed that 4n/9 lines suffice for some subclasses of planar 3-trees, and asked whether 4n/9 is indeed an upper bound for planar 3-trees.
In the context of optimization, Dujmović et al. [13] gave fixed-parametertractable (FPT) algorithms, parameterized by pathwidth, to decide whether a planar graph admits a straight-line drawing on k horizontal lines.Drawings with minimum number of parallel lines have been achieved for trees [21].Recently, Biedl [3] gave an algorithm to approximate the height of straight-line drawings of 2-connected outer planar graphs within a factor of 4. Several researchers have attempted to characterize planar graphs that can be drawn on few parallel lines [8,16,26].

Contributions:
In this paper we show that every n-vertex planar graph with maximum degree ∆, having an edge separator of size λ, admits a drawing with height 4n/9 + O(λ), which is better than the previously best known bound of 2n/3 for any λ ∈ o(n).This result is an outcome of a new application of the planar separator theorem [10].The resulting drawing is not a grid drawing, i.e., the vertices and bends are not restricted to lie on integer grid points, and it is not obvious whether our technique can be immediately adapted to improve the current best 8 9 n 2 -area upper bound [6] on the grid drawings of planar graphs.
However, the techniques developed in this paper have the potential to provide powerful tools for computing compact drawings for planar triangulations in the variable embedding setting.
If the input graphs are restricted to planar 3-trees, then we can improve the JGAA, 0(0) 0-0 (0) 3 upper bound to 4n/9 + O(1), which settles the question of Hossain et al. [18] affirmatively.Furthermore, the drawing we construct in this case is a straightline drawing.

Preliminary Definitions and Results
Let G be an n-vertex plane graph.G is called connected if there exists a path between every pair of vertices in G.We call G a k-connected graph, where Let P k , for some k ∈ {3, 4, . . ., n}, be the path w 1 (= v 1 ), . . ., w l , v k (= w l+1 ), w r , . . ., w t (= v 2 ).The edges (w l , v k ) and (v k , w r ) are the l-edge and r-edge of v k , respectively.The other edges incident to v k in G k are called the m-edges.
For example, in Figure 2(c), the edges (v 6 , v 1 ), (v 6 , v 4 ), and (v 5 , v 6 ) are the l-, rand m-edges of v 6 , respectively.Let E m be the set of all m-edges in G. Then the graph T vn induced by the edges in E m is a tree with root v n .Similarly, the graph and the graph T v2 induced by all r-edges except (v 2 , v n ) is a tree rooted at v 2 .
These three trees form the Schnyder realizer [24] of G, e.g., see Figure 2(a).Lemma 1 (Bonichon et al. [5]) The total number of leaves in all the trees in any Schnyder realizer of an n-vertex triangulation is at most 2n − 5.
Let G be a planar graph and let Γ be a straight-line drawing on k parallel lines.By l(v), where v is a vertex of G, we denote the horizontal line in Γ that passes through v.We now have the following lemma that bounds the height of a straight-line drawing in terms of the number of leaves in a Schnyder tree.
Although the lemma can be derived from known straight-line [6] and polyline drawing algorithms [4], we include a proof for completeness.
Lemma 2 Let G be an n-vertex plane triangulation and let v 1 , v n , v 2 be the outer vertices of G in clockwise order on the outer face.Assume that T vn has at most p leaves.Then for any placement of v n on line l 1 or l p+2 , there exists a straight-line drawing Γ of G on L p+2 such that v 2 and v 1 lie on lines l p+2 and l 1 , respectively.Symmetrically, there exists a straight-line drawing Γ of G on L p+2 such that v 1 and v 2 lie on lines l p+2 and l 1 , respectively.
Proof: We construct Γ by a variant of the shift algorithm [9].The case when G has n = 3 vertices is straightforward, and hence we assume that n > 3. The construction of Γ is incremental.We start with the drawing of G 3 and then add the other vertices in the canonical order corresponding to T vn .Let Γ 3 be the drawing of G 3 on L 3 , where v 1 and v 2 are placed on l 1 and l 3 , respectively, along a vertical line, and v 3 is placed on l 2 to the left of edge (v 1 , v 2 ), e.g., see Figure 3(b).We now add the vertices v i , where 3 < i < n, maintaining the following invariants: (a) P i is drawn as a strictly y-monotone polygonal chain.
JGAA, 0(0) 0-0 (0) 5 Observe that Γ 3 maintains all the above invariants.We now assume that i > 3 and for all j < i, Γ j maintains the above invariants, and consider the insertion of v i .Let w p , . . ., w q be the neighbors of v i on P i−1 .If q − p ≥ 2, then v i is a non-leaf vertex in T vn .In this case we place v i on l(w q−1 ) and add the edges (v i , w), where w ∈ {w p , . . ., w q }.Since P i−1 is strictly y-monotone, we can place v i sufficiently far from w q−1 to the left such that the edges (v i , w) do not create any edge crossing, and P i is strictly y-monotone in Γ i .Figures 3(d)-(e) illustrate such a scenario.Since the number of leaves in v 3 , . . ., v i is same as the number of leaves in v 3 , . . ., v i−1 , Invariants (a)-(c) hold in Γ i .
In the remaining case, q − p = 1, i.e., v i is a leaf in T vn .Here we shift the vertices w q , . . ., w t (= v 2 ) and their descendants in T vn above by one unit from their current positions.Such a shift does not create edge crossings [9].
Figures 3(b)-(c),(f) illustrate such a scenario.We then place v i on l(w q ) − 1 sufficiently far to the left such that the edges (v i , w p ) and (v i , w q ) do not create any edge crossing, and P i is strictly y-monotone in Γ i .Since the number of leaves in v 3 , . . ., v i is one more than the number of leaves in v 3 , . . ., v i−1 , Invariants (a)-(c) hold in Γ i .
Since P n−1 is strictly y-monotone in Γ n−1 , there exists a point c on l 1 (similarly, on l p+2 ) which is visible to all the vertices on P n−1 .We place v n at c, and draw the edges incident to it, which completes the drawing of G.
Chrobak and Nakano [7] showed that every planar graph admits a straight-line drawing with height 2n/3.We now observe some properties of Chrobak and Nakano's algorithm [7].Let G be a plane triangulation with n vertices and let x, y be two prescribed outer vertices of G in clockwise order on the outer face of G. Let Γ be the drawing of G produced by the Algorithm of Chrobak and Nakano [7].Then Γ has the following properties: (CN 1 ) Γ is a drawing on L q , where q ≤ 2n/3.
(CN 2 ) For the vertices x and y, we have l(x) = l 1 and l(y) = l q in Γ.The remaining outer vertex z lies on either l 1 or l q .
Note that the placement of z cannot be prescribed to the algorithm, i.e., the algorithm may produce a drawing where l(x) = l 1 , l(y) = l q and l(z) = l 1 , however, this does not imply that there exists another drawing where l(x) = l 1 , l(y) = l q and l(z) = l q .We end this section with the following lemma.
Lemma 3 Let G be a plane graph and let Γ be a straight-line drawing of G on a set L k of k horizontal lines, where the lines are not necessarily equally spaced.
Then there exists a straight-line drawing Γ of G on a set of k horizontal lines that are equally spaced.Furthermore, for every i ∈ {1, 2, . . ., k}, the left to right order of the vertices on the ith line in Γ coincides with that of Γ .In fact, this construction is inspired by the technique for computing visibility representation of planar graphs, as described in [27,1].
We now adjust the length of the vertical edges so that the layers in D become equally spaced.Biedl [2] showed that such a drawing D can be transformed to the required straight-line drawing Γ , where for every i ∈ {1, 2, . . ., k}, the left to right order of the vertices on the ith line in D coincides with that of Γ .
In the following sections we describe our drawing algorithms.For simplicity we often omit the floor and ceiling functions while defining different parameters of the algorithms.One can describe a more careful computation using proper floor and ceiling functions, but that does not affect the asymptotic results discussed in this paper.

Drawing Triangulations with Small Height
Every planar triangulation has a simple cycle separator of size O( √ n) [11].In the preliminary version of this paper [14], we used this result to prove that every n-vertex planar graph with maximum degree ∆ ∈ o( √ n) admits a 4-bend polyline drawing with height at most 4n/9 + o(n).In this section we use edge separator, and prove that every planar graph with ∆ ∈ o(n) can be drawn with 3 bends per edge and height at most 4n/9 + o(n).
We first present an overview of our algorithm, and then describe the algorithmic details.

Algorithm Overview
Let G = (V, E) be an n-vertex planar graph, where n ≥ 9, and let Γ be a planar drawing of G on the Euclidean plane.Without loss of generality assume that G is a planar triangulation.Let M ⊆ E be an edge separator of G such that the corresponding edges in the dual graph G * form a simple cycle (respectively, V i ⊆ V ) be the vertices that lie outside (respectively, inside) of C * .Diks et al. [10] proved that there always exists such an edge separator illustrate a planar triangulation G and an edge separator of G. Since G is a planar triangulation, there must be an outer vertex q on G i or G o such that q is incident to two or more edges of M .Without loss of generality assume that q lies on G i , e.g., see vertex v

Algorithm Details
Let G be the embedding obtained from G by choosing q, b, c as the outer face.
We We now use the algorithm of Chrobak and Nakano [7] to compute a straight- We now construct a graph G i from G i , as follows.Observe that the vertex a is an outer vertex of G i , which appears immediately after q while walking on the outer face of G i .We add a vertex w d on the outer face of G i , and make it adjacent to q and a.We now add another vertex w i on the outer face, and make it adjacent to w d and q such that the cycle w i , q, w d becomes the boundary of the outer face, e.g., see Figure 5(b).
If w o lies in l x in Γ o , then we make w i adjacent to all the outer vertices of G i .Otherwise, we make w d adjacent to all the outer vertices of G i .We remove any resulting multi-edges by subdividing each corresponding inner edge with a dummy vertex, and then by triangulating the resulting graph.To construct a drawing of G , we merge the drawings of G o and G i .
Γ o Γ i the outer vertices of G o and G i , respectively.Let i be a vertical line to the right of segment w d w i in Γ i such that all the other vertices of Γ i are in the right halfplane of i .Furthermore, i must be close enough such that all the intersection points with the edges incident to w i lie in between the horizontal line l(w i ) and the horizontal line immediately below l(w i ).For each intersection point, we insert a division vertex at that point and create a horizontal line through that vertex.We then delete vertex w i from Γ i , but not the division vertices.Let the resulting drawing be D. Since every planar triangulation with maximum degree ∆ has an edge separator of size O( √ ∆n) [10], we obtain the following corollary.
Corollary 1 Every n-vertex planar triangulation with maximum degree o(n) admits a polyline drawing with height at most 4n/9 + o(n).
Pach and Tóth [23] showed that polyline drawings can be transformed into straight-line drawings while preserving the height if the polyline drawing is monotone, i.e., if every edge in the polyline drawing is drawn as a y-monotone curve.Unfortunately, our algorithm does not necessarily produce monotone drawings.

Drawing Planar 3-Trees with Small Height
In this section we examine straight-line drawings of planar 3-trees.We first introduce a few more definitions and recall some known results.Afterwards, we describe the algorithm details.

Technical Background
Let G be an n-vertex planar 3-tree and let Γ be a straight-line drawing of G.
Then Γ can be constructed by starting with a triangle, which corresponds to the outer face of Γ, and then iteratively inserting the other vertices into the inner faces and triangulating the resulting graph.Let a, b, c be the outer vertices of Γ in clockwise order.If n > 3, then Γ has a unique vertex p that is incident to all the outer vertices.This vertex p is called the representative vertex of G.
For any cycle i, j, k in G, let G ijk be the subgraph induced by the vertices i, j, k and the vertices lying inside the cycle.Let G * ijk be the number of vertices in G ijk .The following two lemmas describe some known results.
Lemma 4 (Mondal et al. [22]) Let G be a plane 3-tree and let i, j, k be a cycle of three vertices in G. Then G ijk is a plane 3-tree.Let G be a plane 3-tree and let a, b, c be the outer vertices of G. Assume that G has a drawing Γ on L k , where a, b lie on lines l 1 , l k , respectively, and c lies on line l i , where 1 ≤ i ≤ k.Then the following properties hold for Γ [18].
Reshape.Let p, q and r be three distinct non-collinear points on lines l 1 , l k and l i , respectively.Then G has a drawing Γ on L k such that the outer face of Γ coincides with triangle pqr (e.g., Figures 7(a)-(b)).
Stretch.For any integer t ≥ k, G admits a drawing Γ on L t such that a, b, c lie on l 1 , l t , l i , respectively (e.g., Figure 7(c)).Lemma 6 Let G be an n-vertex plane 3-tree with outer vertices a, b, c in clockwise order.Let w 1 , . . ., w k (= p), w k+1 (= q), . . ., w t (= c) be the maximal path P such that each vertex on P is adjacent to both a and b (e.g., see Figure 8).Assume that n = n + 3, and and max ∀i>k+1 {G * awiwi−1 , G * bwiwi−1 } ≤ 4n /9, then G admits a drawing with height at most 4n /9 + 4.
Proof: To construct the required drawing of G, we distinguish two cases depending on whether leaf(T p,G abp ) ≤ x or not.Let H be the subgraph of G induced by the vertices {a, b} ∪ {w k , . . ., w t }.In each case, we first construct a drawing of H on L x+4 , and then extend it to compute the required drawing using Lemmas 2-5.
Case 1 (leaf(T p,Gabp ) ≤ x).Since G * bqp ≤n /2, by Lemma 1, one of the trees in the Schnyder realizer of G bqp has at most n /3 ≤ x leaves.We now draw G abq considering the following scenarios.
Case 1A (leaf(T p,Gbqp ) ≤ x).We refer the reader to Figures 9(a such that a, p, q lie on l 1 , l x+2 , l 1 , respectively, and by the Reshape property we can merge these drawings to obtain a drawing of G abq on L x+3 .Case 1B (leaf(T q,Gbqp ) ≤ x).We refer the reader to Figures 9(a)-(b).
By Lemma 2 and the Stretch condition, G abp admits a drawing Γ abp on JGAA, 0(0) 0-0 (0) 13  L x+2 such that the vertices a, b, p lie on l 1 , l x+2 , l 1 , respectively.Similarly, G bqp admits a drawing Γ bpq on L x+2 such that the vertices p, b, q lie on l 1 , l x+2 , l x+2 , respectively.By Lemma 5, G apq admits a drawing Γ apq on L (n +2)/3 such that a, p, q lie on l 1 , l 1 , l (n +2)/3 , respectively.By Stretch, we modify Γ apq such that a, p, q lie on l 1 , l 1 , l x+2 , respectively.Finally, by Stretch and Reshape we can merge these drawings to obtain a drawing of G abq on L x+3 .Figures 9(c)-(d) show an illustration.
Case 1C (leaf(T b,Gbqp ) ≤ x).The drawing of this case is similar to Case 1B.The only difference is that we use T b,G bqp while drawing G bqp .
Observe that each of the Cases 1A-1C produces a drawing of G abq such that a, b lie on l 1 , l x+3 , respectively, and q lies on either l 1 or l x+3 .We use the Stretch operation to modify the drawing such that a, b lie on l 1 , l x+4 , respectively, and q lies on either l 2 or l x+3 .Specifically, if q is on l x+3 , then we push b to l l+4 .
Otherwise, q is on l 1 , and in this case we push a to l 0 , and then shift the drawing up by one layer to move a back to l 1 .
If q lies on l x+3 , then we place the vertices w k+1 , . . ., w t (= c) on l 2 and l x+3 alternatively, as shown in Figure 10(a).Similarly, if q lies on l 2 , then we draw the path w k+1 , . . ., w t (= c) in a zigzag fashion, placing the vertices on l x+3 and l 2 alternatively such that each vertex is visible to both a and b.For each If leaf(T p,G bpq ) > x, then either leaf(T b,G bpq ) ≤ x or leaf(T q,G bpq ) ≤ x.
In this case we draw G abq on L x+3 , where a, b, p, q lie on l 1 , l x+3 , l 2 , l x+3 , respectively, as in Figure 10   JGAA, 0(0) 0-0 (0) 15 Each of the Cases 2A-2C produces a drawing of G abq such that a, b lies on l 1 , l x+3 , respectively, and q lies on either l 1 or l x+3 .Hence we can extend these drawings to draw G as in Case 1.

Drawing Algorithm
We are now ready to describe our algorithm.

Decomposition.
Let G be an n-vertex plane 3-tree with the outer vertices a, b, c and the representative vertex p.A tree spanning the inner vertices of G is called the representative tree T if it satisfies the following conditions [22]: -The plane 3-tree G i , where 1 ≤ i ≤ 3, is determined by the representative tree rooted at the ith child of v, and thus contains at most r/2 + 3 = (n − 3)/2 + 3 = (n + 3)/2 vertices.
-The plane 3-tree G 4 is obtained by deleting v and the vertices from G that are descendent of v in T , and contains at most (n + 3)/2 vertices.

Drawing Technique.
Without loss generality assume that

Case 1 (G
then G admits a drawing on L x+4 by Lemma 6.We may thus assume that there exists some j > k + 1 such that either G * awj wj−1 > x or G * bwj wj−1 > x.
We first show that G abq can be drawn on L x+3 in two ways: One drawing Γ 1 contains the vertices a, b, q on l 1 , l x+3 , l 2 , respectively, and the other drawing Γ 2 contains a, b, q on l 1 , l x+3 , l x+2 , respectively.We then extend these drawings to obtain the required drawing of G. Consider the following scenarios depending on whether G * 1 ≤ x or not. - Here we draw the subgraph G induced by the vertices a, b, p, q such that they lie on l 1 , l x+3 , l x+2 , l 2 , respectively.Since can be drawn inside their corresponding triangles, which corresponds to Γ 1 .Similarly, we can find another drawing Γ 2 of G abq , where the vertices a, b, p, q lie on l 1 , l x+3 , l 2 , l x+2 , respectively. - , we can use Chrobak and Nakano's algorithm [7] and Stretch operation to draw G 1 such that that a, b lie on l 1 , l n /3+1 , respectively, and p lies either on l 2 or l n /3 .First consider the case when p lies on l n /3 .We then use the Stretch condition to push b to l x+3 .To construct Γ 1 , we place q on l 2 , and to construct Γ 2 , we place q on l x+2 .Since G * 3 ≤ G * 2 ≤ n /9, for each placement of q, we can draw G 2 and G 3 using Lemma 5 inside their corresponding triangles.
The case when p lies on l 2 is handled symmetrically, i.e., first by pushing a downward using Stretch operation so that the drawing spans (x + 3) horizontal lines, then shifting the drawing upward such that a comes back to l 1 , and finally placing the vertex q on l 2 (for Γ 1 ) and l x+2 (for Γ 2 ) .
We now show how to extend the drawing of G abq to compute the drawing of G.
-Assume that G * awj wj−1 > x.Shift b to l x+4 , and draw the path w k+1 , . . ., w j−1 in a zigzag fashion, placing the vertices on l 2 and l x+3 alternatively, such JGAA, 0(0) 0-0 (0) 17 that l(w k+1 ) = l(w k+2 ), and each vertex is visible to both a and b.Choose Γ 1 or Γ 2 such that the edge (a, w j−1 ) spans at least x + 3 lines.We now draw G awj wj−1 using Chrobak and Nakano's algorithm [7].Since -Assume that G * bwj wj−1 > x.The drawing in this case is similar to the case when G * awj wj−1 > x.The only difference is that while drawing the path w k+1 , . . ., w j−1 , we choose Γ 1 or Γ 2 such that the edge (b, w j−1 ) spans at least x + 3 lines.The following theorem summarizes the result of this section.

Conclusion
In this paper we have shown that every n-vertex planar graph with maximum degree ∆, having an edge separator of size λ, admits a polyline drawing with height 4n/9 + O(λ), which is 4n/9 + o(n) for any planar graph with ∆ ∈ o(n).
While restricted to n-vertex planar 3-trees, we compute straight-line drawings with height at most 4n/9 + O(1).In some cases the width of the drawings that we compute for plane 3-trees may be exponentially large over n.Hence it would be interesting to find drawing algorithms that can produce drawings with the same height as ours, but bound the width as a polynomial function of n.
Several natural open question follows.
-Does every n-vertex planar triangulation admit a straight-line drawing with height at most 4n/9 + O(1)?
-What is the minimum constant c such that every n-vertex planar 3-tree admits a straight-line (or polyline) drawing with height at most cn?
-Does a lower bound on the height for straight-line drawings of triangulations determine a lower bound also for their polyline drawings?
Recently, Biedl [2] has examined height-preserving transformations of planar graph drawings, which shed some light on the last open question.

Figure 1 (
Figure 1(b) illustrates a 1-bend polyline drawing of G on L 8 .

Figure 1 :
Figure 1: (a) A triangulation G. (b) A polyline drawing of G with height 8.
if the removal of fewer than k vertices does not disconnect the graph.A plane graph delimits the plane into topologically connected regions called faces.The bounded regions are called the inner faces and the unbounded region is called the outer face of G.The vertices on the boundary of the outer face are called the outer vertices, and the remaining vertices are called the inner vertices of G.If every face of G (including the outer face) is a cycle of length three, then we call G a triangulation, or a maximal planar graph.G is called an internally triangulated graph if every face except the outer face is a cycle of length three.Let G = (V, E) be an n-vertex triangulated plane graph.A simple cycle C in G is called a cycle separator if the interior and the exterior of C each contains at most 2n/3 vertices.An edge separator of G is a subset of edges M of G such that the graph G = (V, E \ M ) consists of two induced subgraphs, each containing at most 2n/3 vertices.Every planar graph with maximum degree ∆ admits an edge separator of size 2 √ 2∆n, where the corresponding edges in the dual graph form a simple cycle [10].Let v 1 , v n and v 2 be the outer vertices of G in clockwise order on the outer face.Let σ = (v 1 , v 2 , ..., v n ) be an ordering of all vertices of G.By G k , 2 ≤ k ≤ n, we denote the subgraph of G induced by v 1 , v 2 , . . ., v k .For each G k , the notation P k denotes the path (while walking clockwise) on the outer face of G k that starts at v 1 and ends at v 2 .We call σ a canonical ordering of G with respect to the outer edge (v 1 , v 2 ) if for each k, 3 ≤ k ≤ n, the following conditions are satisfied [9]: (a) G k is 2-connected and internally triangulated.(b) If k ≤ n, then v k is an outer vertex of G k and the neighbors of v k in G k−1 are consecutive on P k−1 .

Figure 2 :
Figure 2: (a) A plane triangulation G with a canonical ordering.The associated realizer, where the l-, r-and m-edges are shown in dashed, bold-solid, and thinsolid lines, respectively.(b) T v1 .(c) Neighbors of v 6 in G 6 .(d)-(g) Illustrating Lemma 3.

Figure 3 :
Figure 3: (a) A plane triangulation G with a canonical ordering of its vertices.(b)-(f) Illustration for drawing Γ i .
Proof: A flat visibility drawing of G on L k maps each vertex of G to a distinct horizontal interval on some horizontal line of L k , and each edge of G to a horizontal or vertical line segment between the corresponding intervals.Given a straight-line drawing Γ of G on L k , it is straightforward to transform Γ into a flat visibility drawing D on L k such that for every i ∈ {1, 2, . . ., k}, the left to right order of the vertices on the ith line in Γ coincides with that of D, and for every vertex v in D, the clockwise ordering of the edges around v coincides with the ordering in Γ.One way to construct such a drawing D is to direct the edges of Γ from bottom to top, and then draw the directed paths in a depth-first search order from left to right.Figures 2(d)-(g) illustrate such a construction.
be the subgraphs of G induced by the vertices of V i and V o , respectively.Since n ≥ 9, each of G i and G o contains at least 3 vertices.
5 in Figure4(c).Let a, b, c be three consecutive neighbors of q in G in counter clockwise order such that a ∈ V i and {b, c} ⊆ V o .We take an embedding G of G with q, b, c as the outer face, as shown in Figure4(d) with q = v 5 , a = v 3 , b = v 2 , and c = v 11 .Consequently, G o and G i lie on the outer face of each other, as illustrated in Figures 4(d)-(e).We first draw G o and G i separately with small height, and then merge these drawings to compute the final output.The drawings of G o and G i are placed side by side.Consequently, the height of the final output can be expressed in terms of the maximum height of the drawings of G o and G i , and hence the area of the final drawing becomes small.
first construct a graph G o from G o by adding a vertex w o on the outer face of G o , and making w o adjacent to all the outer vertices of G o such that the edge (b, c) remains as an outer edge.We remove any resulting multi-edges by subdividing each corresponding inner edge with a dummy vertex, and then by triangulating the resulting graph.Note that we do not need to add dummy vertices on the outer edges.Figure 5(a) illustrates an example of G o , where the dummy vertex d removes the multi-edges between v 7 and w o .Since there are O( √ ∆n) edges in M , the number of vertices in G o is at most 2n/3 + O( √ ∆n).

Figure 4 :
Figure 4: (a) A planar triangulation.(b) An edge separator M of G, and the corresponding simple cycle in the dual graph.The edges of M and C * are shown in thin and thick gray, respectively.(c) G o and G i are shaded in light-gray and dark-gray, respectively.(d)-(e) Choosing a suitable embedding G .

Figure 5
(b) illustrates an example of G i , where d is a dummy vertex.Since there are O( √ ∆n) edges in M , the number of vertices in G i is at most 2n/3 + O( √ ∆n).We now use the algorithm of Chrobak and Nakano[7] to compute a straightline drawing Γ i of G i with height y = 4n/9 + O( √ ∆n) such that w d , w i lie on l 1 , l y , respectively, and the segment w d w i is vertical.Assume without loss of generality that all the vertices of G i are in the right half-plane of the line determined by w d , w i .

Figures
Figures 6(c)-(d) illustrate this scenario.By Lemma 3, we can modify Γ i such

Figure 6 (Theorem 1
e) shows a schematic representation of D. Since the division vertices correspond to the bends, each edge may contain at most four bends (one bend inside Γ o , one bend inside Γ i , and two bends to merge the drawings Γ i and Γ o ).Since there are at most O( √ ∆n) edges that may have bends, the number of bends is at most O( √ ∆n) in total.Note that for every edge containing four bends, two of the bends correspond to w o and w i , and they are adjacent one the same horizontal line in the final drawing.Therefore, we can now transform D into a flat-visibility drawing, where the adjacent pair of bends correspond to a single vertex, and then transform the flat-visibility drawing back into a polyline drawing (similar to the proof of Lemma 3), where the bends that correspond to w o and w i are merged to a single bend.Consequently, the number of bends per edge reduces to 3. The following theorem summarizes the result of this section.Let G be an n-vertex planar graph.If G contains a simple cycle separator of size λ, then G admits a 3-bend polyline drawing with height 4n/9 + O(λ) and at most O(λ) bends in total.

Lemma 5 (
Hossain et al.[18]) Let G be an n-vertex plane 3-tree with the outer vertices a, b, c in clockwise order.Let D be a drawing of the outer cycle a, b, c on L n , where the vertices lie on l 1 , l k and l i with k ≤ n and i ∈ {l 1 , l 2 , l n , l n−1 }.Then G admits a straight-line drawing Γ on L k , where the outer cycle of Γ coincides with D.

For
any triangulation H with the outer vertices a, b, c, let T a,H , T b,H , T c,H be the Schnyder trees rooted at a, b, c, respectively.By leaf(T ) we denote the number of leaves in T .The following lemma establishes a sufficient condition for a plane 3-tree G to have a straight-line drawing with height at most 4(n+3)/9+4.

3 Figure 8 :
Figure 8: Illustration for Lemma 6, where the graph G abp is in shaded region.
)-(b).By Lemma 2 and the Stretch condition, G abp admits a drawing Γ abp on L x+2 such that the vertices a, b, p lie on l 1 , l x+2 , l x+2 , respectively.Similarly, since leaf(T p,G bqp ) ≤ x, by Lemma 2 G bqp admits a drawing Γ bpq on L x+2 such that the vertices q, b, p lie on l 1 , l x+2 , l x+2 , respectively, as shown in Figure 9(a).By the Stretch property, Γ abp can be extended to a drawing Γ abp on L x+3 , where a, b, p lie on l 1 , l x+3 , l x+2 , respectively.Similarly, Γ bqp can be extended to a drawing Γ bqp on L x+3 , where q, b, p lie on l 1 , l x+3 , l x+2 , respectively.Since G * apq ≤ (n + 2)/3, by Lemma 5 and the Stretch condition, G apq admits a drawing Γ apq on L (n +2)/3 .Finally, by the Stretch property Γ apq can be extended to a drawing Γ apq on L x+2
(c).Specifically, we use Lemma 2 to draw G bpq .Since leaf(T a,G abp ) ≤ x, we use Lemma 2 to draw G abp , and since G * apq ≤ (n + 2)/3, we draw G apq using Lemma 5. Finally, we use Stretch and Reshape to merge these drawings.Case 2B (leaf(T a,Gabp ) > x and leaf(T b,Gabp ) ≤ n /9).If leaf(T p,G bpq ) ≤ n /3, then we first draw G bpq using Lemma 2 such that b, p, q lie on l n /3+2 , l n /3+2 , l 1 , respectively, and then use the Stretch condition to shift b to l x+3 .By Lemma 2 and the Stretch condition, there exists a drawing of G abp on L x+3 with a, b, p lying on l 1 , l x+3 , l n /3+2 , respectively.Since G * apq ≤ (n + 2)/3, we can draw G apq using Lemma 5 inside triangle apq.

Figure 10 (
Figure 10(d) illustrates the scenario after applying Stretch and Reshape.If leaf(T p,G bpq ) > n /3, then by Lemma 1 either leaf(T b,G bpq ) ≤ n /3 − 2 or leaf(T q,G bpq ) ≤ n /3 − 2. Hence we can use Lemma 2 and the Stretch condition to draw G bpq such that b, p, q lie on l x+3 , l n /9+2 , l x+3 , respectively.On the other hand, we use Lemma 2 to draw G abp such that a, b, p lie on l 1 , l n /9+2 , l n /9+2 , respectively, and then use the Stretch condition to move b to l x+3 .Since G * apq ≤ (n + 2)/3, we can draw G apq using Lemma 5 inside triangle apq.Figure 10(e) illustrates the scenario

Figure 10 (
e) illustrates the scenario after applying Stretch and Reshape.Case 2C (leaf(T a,Gabp ) ≤ n /9 and leaf(T b,Gabp ) > x).The drawing in this case is analogous to Case 2B.The only difference is that we use T a,G abp while drawing G abp .
(a) If n = 3, then T is empty.(b) If n = 4, then T consists of a single vertex.(c) If n > 4, then the root p of T is the representative vertex of G and the subtrees rooted at the three clockwise ordered children p 1 , p 2 and p 3 of p in T are the representative trees of G abp , G bcp and G cap , respectively.Recall that every r-vertex tree T has a vertex v such that the connected components of T \ v are all of size at most r/2[19].Such a vertex v in T corresponds to a decomposition of G into four smaller plane 3-trees G 1 , G 2 , G 3 , and G 4 , as follows.
to the outer face of G, then let (a, b) be the corresponding outer edge.Otherwise, G 1 does not have any edge incident to the outer face of G.In this case there exists an inner face f in G that is incident to G 1 , but does not belong to G 1 .We choose f as the outer face of G, and now we have an edge (a, b) of G 1 that is incident to the outer face of G. Let P =(w 1 , . . ., w k (= p), w k+1 (= q), . . ., w t ) be the maximal path in G such that each vertex on P is adjacent to both a and b, where {a, b, p}, {a, p, q}, {b, q, p} are the outer vertices of G 1 , G 2 , G 3 , respectively, e.g., see Figure11.Assume that n = n + 3 and x = 4n /9.We draw G on L x+4 by distinguishing two cases depending on whether G * 4 > x or not.

x < G * awj wj− 1 ≤ n / 2 ,
we can draw G awj wj−1 on at most n /3 parallel lines.By the Stretch and Reshape conditions, we merge this drawing with the current drawing such that w j lies on either l x+3 or l n /9+2 .Since G * bwj wj−1 ≤ n /9, we can draw G bwj wj−1 inside its corresponding triangle using Lemma 5. Since max ∀i>j {G * awiwi−1 , G * bwiwi−1 } ≤ n /9, it is straightforward to extend the current drawing to a drawing of G on x + 4 parallel lines by continuing the path w j , . . ., w t in the zigzag fashion.
Let o be a vertical line to the left of w o in Γ o such that all the other vertices of Γ o are in the left half-plane of o .Furthermore, o must be close enough such that all the intersection points with the edges incident to w o lie in between l(w o ) and l(w o ) − 1.For each intersection point, we insert a division vertex at that point and create a horizontal line through that vertex.Delete vertex w o , but not the division vertices.Finally, by Lemma 3, we can modify Γ o such that the horizontal lines are equally spaced.Note that Γ o is a drawing on at most