Journal of Graph Algorithms and Applications Universal Line-sets for Drawing Planar 3-trees 60 Hossain Et Al. Universal Line-sets for Drawing Planar 3-trees

A set S of lines is universal for drawing planar graphs with n vertices if every planar graph G with n vertices can be drawn on S such that each vertex of G is drawn as a point on a line of S and each edge is drawn as a straight-line segment without any edge crossing. It is known that ⌊ 2(n−1) 3 ⌋ parallel lines are universal for any planar graph with n vertices. In this paper we show that a set of ⌊ n+3 2 ⌋ parallel lines or a set of ⌈ n+3 4 ⌉ concentric circles are universal for drawing planar 3-trees with n vertices. In both cases we give linear-time algorithms to find such drawings. A by-product of our algorithm is the generalization of the known bijection between plane 3-trees and rooted full ternary trees to the bijection between planar 3-trees and unrooted full ternary trees. We also identify some subclasses of planar 3-trees whose drawings are supported by fewer than ⌊ n+3 2 ⌋ parallel lines.


Introduction
Many researchers in the graph drawing community have concentrated their attention on drawing graphs on point-sets [3,8,15] and on line-sets [7,10,13] due to strong theoretical and practical motivation for such drawings (e.g., computing small-width VLSI layout, approximating pathwidth and data visualization on small form factor).A set S of lines supports a drawing of a planar graph G if G has a planar drawing, where each vertex is drawn as a point on a line in S and each edge is drawn as a straight line segment.We say G has a drawing on S if S supports a drawing of G.A set of lines that supports the drawing of all n-vertex graphs in some class is called universal for that class.In this paper we study the problem of finding universal line sets of smaller size for planar graphs.Given a plane graph G with n vertices, Chrobak and Nakano [5] gave an algorithm to compute a drawing of G on a ⌊ 2(n−1) 3 ⌋ × 4⌊ 2(n−1) 3 ⌋ grid.This implies that ⌊ 2(n−1) 3 ⌋ parallel lines are universal for any planar graph with n vertices.Note that a plane graph is a planar graph with a fixed planar embedding.
Recently, several researchers have studied a labeled version of the problem where both the lines in the point set S and vertices of G are labeled from 1 to n and each vertex is drawn on its associated line.Estrella-Balderrama et al. [10] showed that no set of n parallel lines supports all n-vertex planar graphs when each vertex is drawn as a point on its associated line.Dujmović et al. [7] showed that there exists a set of n lines in general position that does not support all n-vertex planar graphs.An unlabeled version of the problem has appeared in the literature as "layered drawing."A layered drawing of a plane graph G is a planar drawing of G, where the vertices are drawn on a set of horizontal lines called layers and the edges are drawn as straight line segments.Finding a layered drawing of a graph on the minimum number of layers is a challenging task.Dujmović et al. [9] gave a parametrized algorithm to check whether a given planar graph with n vertices admits a layered drawing on h layers or not.Mondal et al. [14] gave an O(n 5 )-time algorithm to compute a layered drawing of a "plane 3-tree" G, where the number of layers is minimum over all possible layered drawings of G.
In this paper we consider the problem of finding a universal line set of smaller size for drawing "planar 3-trees."A planar 3-tree G n with n ≥ 3 vertices is a planar graph for which the following two conditions, (a) and (b) hold: (a) G n is a maximal planar graph; (b) if n > 3, then G n has a vertex whose deletion gives a planar 3-tree G n−1 .Many researchers have shown their interest on planar 3-trees for a long time for their beautiful combinatorial properties which have applications in computational geometry [1,2,6,14,18].In this paper we show that a set of ⌊ n+3 2 ⌋ parallel lines and a set of ⌈ n+3 4 ⌉ concentric circles are universal for planar 3-trees with n vertices.In both cases we give lineartime algorithms to find such drawings.A by-product of our algorithm is the generalization of the known bijection between plane 3-trees and rooted ternary trees to the bijection between planar 3-trees and unrooted full ternary trees.We also identify some subclasses of planar 3-trees whose drawings are supported by fewer than ⌊ n+3 2 ⌋ parallel lines.Let G be a plane 3-tree, i.e., a planar 3-tree with a fixed planar embedding.Clearly the outer face of G is a triangle, and let a, b and c be the three outer vertices of G.There is a vertex p in G, which is the common neighbor of a, b and c.The vertex p is called the representative vertex of G [14].The vertex p along with the three outer vertices of G divides the interior region of G into three new regions.It is known that the subgraphs G 1 , G 2 and G 3 enclosed by those three regions are also plane 3-trees [14].G can be represented by a representative tree whose root is the representative vertex p of G and the subtrees rooted at the children of p are the representative trees of G 1 , G 2 and G 3 .Figure 1(b) illustrates the representative tree of the plane 3-tree in Figure 1(a).The depth ρ of a plane 3-tree is the number of vertices that lie on the longest path from the root to a leaf in its representative tree.
We now give an outline of our idea for drawing a planar 3-tree G on ρ + 2 parallel lines.One can observe that the depth of different embeddings of a planar 3-tree may differ.Figures 1(a) and (c) illustrate two different planar embeddings of the same planar 3-tree, with depths 3 and 4, respectively.We thus find an embedding of the planar 3-tree with the minimum depth ρ ′ , and find a drawing on ρ ′ + 2 parallel lines.We show that ρ ′ is at most 2 ⌋ parallel lines support a drawing of a planar 3-tree with n vertices.The rest of the paper is organized as follows.Section 2 describes some of the definitions that we have used in our paper.Section 3 deals with drawing plane 3-trees on parallel lines and concentric circles.In section 4 we obtain our bound on universal line set and universal circle set for planar 3-trees, and in Section 5 we consider drawings of some subclasses of planar 3-trees.Finally, Section 6 concludes our paper with discussions.A preliminary version of this paper has been presented at the 6th International Workshop on Algorithms and Computation (WALCOM 2012) [12].

Preliminaries
In this section we introduce some definitions and known properties of plane 3-trees.For the graph theoretic definitions not described here, see [17].
A graph is planar if it can be embedded in the plane without edge crossing except at the vertices where the edges are incident.A plane graph is a planar graph with a fixed planar embedding.A plane graph divides the plane into some connected regions called the faces.The unbounded region is called the outer face and all the other faces are called the inner faces.The vertices on the outer face are called the outer vertices and all the other vertices are called inner vertices.If all the faces of a plane graph G are triangles, then G is called a triangulated plane graph.We denote by C o (G) the contour outer face of G.For a cycle C in a plane graph G, we denote by G(C) the plane subgraph of G inside C (including C).A maximal planar graph is one to which no edge can be added without losing planarity.Thus in any embedding of a maximal planar graph G with n ≥ 3 vertices, the boundary of every face of G is a triangle, and hence an embedding of a maximal planar graph is often called a triangulated plane graph.
Let G be a plane 3-tree.By a triangle C xyz of G we denote a cycle C of three vertices, where x, y, z are the vertices on the boundary of C in anticlockwise order.The following result is known on plane 3-trees [14].
Lemma 1 [14] Let G be a plane 3-tree of n ≥ 3 vertices and let C be any triangle of G. Then the subgraph G(C) is a plane 3-tree.
Let p be the representative vertex and a, b, c be the outer vertices of G in anticlockwise order.The vertex p, along with the three outer vertices a, b and c, form three triangles C abp , C bcp and C cap .We call these triangles the nested triangles around p.
We now define the representative tree of a plane 3-tree G of n > 3 vertices as an ordered rooted tree T satisfying the following two conditions (a) and (b).
(a) if n = 4, then T is a single vertex, which is the representative vertex of G.
(b) if n > 4, then the root p of T is the representative vertex of G and the subtrees rooted at the three anticlockwise ordered children q 1 , q 2 and q 3 of p in T are the representative trees of G(C 1 ), G(C 2 ) and G(C 3 ), respectively, where C 1 , C 2 and C 3 are the nested triangles around p in anticlockwise order.
Figure 1(b) illustrates the representative tree T of the plane 3-tree G of Figure 1(a).We define the depth ρ of G as the number of vertices that lie on the longest path from the root to a leaf in its representative tree.The following lemma describes a property of a representative tree.Lemma 2 ( [14]) Let G be a plane 3-tree and let T be its representative tree.Every vertex v in T corresponds to a unique cycle C of three vertices in G such that G(C) is a plane 3-tree with representative vertex v.Moreover, the subtree rooted at v in T is the representative tree of G(C).
Let a, b and c be the three outer vertices of a plane 3-tree G.We denote by △abc the drawing of the outer face of G as a triangle.A line or arc l crosses JGAA, 17(2) 59-79 (2013) 63 a triangle △abc if there exists at least one point p on l in the proper interior of the triangle △abc.A line or arc l touches the triangle △abc if it does not cross the triangle △abc and at least one point among a, b, c lies on l.
The center of a tree T is either a single node or an edge, which is obtained by repeatedly deleting all the nodes of degree one, until a single node or an edge is left.Let p and q be two vertices of T .By d T (p, q) we denote the distance, i.e., the length of the unique path, between p and q in T .Two trees T and T ′ are isomorphic if there exists a bijective mapping φ from the vertices of T to the vertices of T ′ such that two vertices u and v are adjacent in T if and only if φ(u) and φ(v) are adjacent in T ′ .
Given a plane graph G with n vertices, Chrobak and Nakano [5] gave an algorithm to compute a straight-line drawing of G on a ⌊ 2(n−1) 3 ⌋ × 4⌊ 2(n−1) 3 ⌋ grid.We now observe some properties of their drawing algorithm.Let Γ be a triangulated plane graph with n vertices and let x, y be two arbitrary outer vertices of Γ in anticlockwise order.Let D be the drawing of Γ produced by the algorithm of Chrobak and Nakano [5].Then D has the following properties.
(CN2) Vertex x and vertex y lie on lines l 0 and l q in D, respectively.The remaining outer vertex lies on either line l 0 or l q .
3 Drawings on Parallel Lines and Concentric Circles In this section we prove that any plane 3-tree of depth ρ has a drawing on ρ + 2 parallel lines.We first need the following lemma.
Lemma 3 Let a, b, and c be the three vertices of the outer face C o (G) of a plane 3-tree G, and let v be the representative vertex of G. Let △abc be a drawing of C o (G) on a set of k + 2 parallel lines, for some positive integer k, such that two of the vertices among a, b, c lie on the same or consecutive lines.Assume that k parallel lines l 1 , l 2 , ..., l k cross △abc.Then there exists a line l x , 1 ≤ x ≤ k such that we can place vertex v on line l x interior to △abc, where at least k − 1 parallel lines cross each of the triangles △abv, △bcv and △acv.
Proof: Without loss of generality assume that a is a top-most and c is the bottom-most vertices in the △abc, i.e., vertex a and c lie on the lines l 0 and l k+1 , respectively.We now consider the following four cases according to the positions of the vertex b.We now have the following lemma.
Lemma 4 Every plane 3-tree G with depth ρ has a drawing on ρ + 2 parallel lines.
Proof: We prove a stronger claim as follows: Given a drawing D of the outer face of G on ρ + 2 lines such that two of its outer vertices lie on the same or consecutive lines, we can extend the given drawing to a drawing D ′ of G such that D ′ is also a drawing on ρ + 2 lines.
The case when ρ = 0 is straightforward, since in this case G is a triangle and any given drawing D of the outer face of G on two lines is itself a drawing of G.We may thus assume that ρ > 0 and the claim holds for any plane 3-tree of depth ρ ′ , where ρ ′ < ρ.
Let G be a plane 3-tree of depth ρ and let a, b and c be the three outer vertices of G in anticlockwise order.Let p be the representative vertex of G.We draw C o (G) on ρ + 2 parallel lines by drawing the outer vertex a on Line l 0 , and the other two outer vertices b and c on Line l ρ or on Lines l ρ and l ρ+1 , respectively.According to Lemma 3, there is a line l x , 1 ≤ x ≤ ρ + 1 such that the placement of p on line l x inside △abc ensures that the triangles △abp, △acp and △cbp are crossed by at least ρ − 1 parallel lines.
We place p on l x inside △abc.By Lemma 1, G(C abp ), G(C bcp ) and G(C cap ) are plane 3-trees.Observe that the depth of each of these plane 3-trees is at most ρ − 1.By induction hypothesis, each of these plane 3-trees admits a drawing on ρ + 1 parallel lines inside the triangles △abp, △bcp and △cap, respectively.
Based on the proof of Lemma 4, one can easily develop an O(n)-time algorithm for finding a drawing of a plane 3-tree G of n vertices on ρ + 2 parallel lines, where ρ is depth of G. Thus the following theorem holds.
Theorem 1 Let G be a plane 3-tree of n vertices.Then one can find a drawing of G on ρ + 2 parallel lines in O(n) time, where ρ is the depth of G.
We now consider the problem of drawing a plane 3-tree on a concentric circle set.Since a set of ρ+2 parallel lines can be formed with ⌈ ρ+2 2 ⌉ infinite concentric circles, each of which contributes two parallel lines, every plane 3-tree admits a drawing on ⌈ ρ+2 2 ⌉ concentric circles.We can observe that Lemma 3 holds even if we consider a set C of non-crossing concentric circular arcs1 of finite radii instead of a set of parallel lines, and hence we have the following corollary.
Corollary 1 Let G be a plane 3-tree of depth ρ.Then G has a drawing on ⌈ ρ+2 2 ⌉ concentric circles.Furthermore, such a drawing can be found in lineartime.

Universal Line Sets for Drawing Planar 3-Trees
In this section we give an algorithm to find an embedding of a planar 3-tree with minimum depth and prove the ⌊ n+3 2 ⌋ upper bound on the size of the universal line set for planar 3-trees.For any planar 3-tree the following fact holds.
Fact 1 Let G be a planar 3-tree and let Γ and Γ ′ be two planar embeddings of G. Then any face in Γ is a face in Γ ′ and vice versa.
We call a triangle, i.e., a cycle of three vertices, in a planar 3-tree G a facial triangle if it appears as a face boundary in a planar embedding of G.
Let G be a planar 3-tree of n vertices and let Γ be a planar embedding of G (i.e., Γ is a plane 3-tree).We now define a tree structure that contains the faces of Γ as its leaves.Later, we will prove that such tree structures that correspond to different planar embeddings of G are isomorphic, and consequently, we will be able to find a minimum depth embedding G examining only a single tree structure.A face-representative tree of Γ is an ordered rooted tree T f that satisfies the following conditions.(ii) The subtrees rooted at the three anticlockwise ordered children q 1 , q 2 and q 3 of p in T f are the face-representative trees of Γ(C 1 ), Γ(C 2 ) and Γ(C 3 ), respectively, where C 1 , C 2 and C 3 are the three nested triangles around p in anticlockwise order.Figure 2 illustrates a face-representative tree of a plane 3-tree where black nodes are vertex-nodes and white nodes are face-nodes.Observe that every internal node in a face-representative tree has exactly four neighbors.We call such a tree an unrooted full ternary tree.A face-representative tree has 2n − 4 face-nodes and n − 3 vertex-nodes.Deletion of the face-nodes from the facerepresentative tree yields the representative tree of Γ.
A rooted tree is semi-labeled if its internal vertices are unlabeled and the leaves are labeled.Two semi-labeled trees are isomorphic at root, if we can assign labels to the unlabeled nodes such that the trees become identical and the labels of the two roots are the same.It is easy to see that if two semi-labeled trees are isomorphic at root, then they are isomorphic.The unordered rooted tree obtained by deleting the labels of the internal nodes of a face-representative tree is a semi-labeled face-representative tree.Let T 1 and T 2 be two semi-labeled face representative trees of two different embeddings of a planar 3-tree G.If f is a facial triangle in G, then there is a face-node corresponding to f in T 1 and in T 2 , by Fact 1.For convenience, we often denote each of these face-nodes as f .
We now prove that the face-representative trees obtained from different embeddings of a planar 3-tree are isomorphic.In fact, we have a stronger claim in the following lemma.
Lemma 5 Let G be a planar 3-tree and let Γ ′ , Γ ′′ be two different planar embeddings of G. Let f be a facial triangle in G, and let T ′ and T ′′ be the semi-labeled face-representative trees obtained from the face-representative trees of Γ ′ and Γ ′′ , respectively, by choosing f as their roots.Then T ′ and T ′′ are isomorphic at f .Proof: We employ induction on the number of vertices n.The case when n ≤ 4 is straightforward.We thus assume that n > 4 and the claim holds for all planar 3-trees of less than n vertices.Let the outer face of Γ ′ and Γ ′′ be C abc and C xyz , respectively.Let the representative vertex of Γ ′ be v.Then C xyz is a face 1 and Γ ′ 2 , respectively.Let f be a facial triangle in G, which is determined by the vertices a, c and v.By induction hypothesis, the semi-labeled face-representative trees, which are obtained from T 1 and T 2 by choosing f as their roots, are isomorphic at f .Similarly, the semi-labeled face-representative trees T ′ f1 and T ′′ f1 of Γ ′ (C bcv ) and Γ ′′ (C cbv ) rooted at f 1 are isomorphic at f 1 , where f 1 is the face determined by the vertices b, c and v.The semi-labeled face-representative trees T ′ f2 and T ′′ f2 of Γ ′ (C abv ) and Γ ′′ (C bav ) rooted at f 2 are isomorphic at f 2 , where f 2 is the face determined by the vertices a, b and v. Let T f3 be the face-representative tree of a plane 3-tree of exactly three vertices.Assign the label "abc" to f 3 .
We now connect a copy of T ′ f1 , T ′ f2 and T f3 with T 1 by adding edges (f, f 1 ), (f, f 2 ) and (f, f 3 ).Remove the label of f and the label of f i , i ∈ {1, 2}, if T ′ fi consists of at least two vertices.Let X be the resulting semi-labeled tree.It is now straightforward to observe that the two trees, which are obtained from X and T ′ by choosing f 3 as their roots, are isomorphic at f 3 .
Similarly, we connect a copy of T ′′ f1 , T ′′ f2 and T f3 to T 2 by adding edges (f, f 1 ), (f, f 2 ) and (f, f 3 ).We then remove the label of f and the label of fi consists of at least two vertices.Let Y be the resulting semi-labeled tree.It is now straightforward to observe that the trees, which are obtained from Y and T ′′ by choosing f 3 as their roots, are isomorphic.
According to the construction, X and Y are isomorphic at f 3 .Therefore, to complete the proof, we show that for any facial triangle f ′ in G, f ′ = f 3 , the trees X ′ and Y ′ rooted at f ′ , which are obtained respectively from X and Y , are isomorphic at f ′ .Suppose for a contradiction that X ′ is not isomorphic to Y ′ at f ′ .Since X and Y are isomorphic at f 3 , the unlabeled vertices of X and Y can be labeled such that X and Y become identical.Such a labeling can determine an isomorphism for X ′ and Y ′ at f ′ , a contradiction.
Observe that if two semi-labeled face representative trees are isomorphic at their root, then they are isomorphic.Therefore, we have the following corollary.
Corollary 2 Let G be a planar 3-tree and let Γ ′ , Γ ′′ be two different planar embeddings of G. Let T ′ and T ′′ be the semi-labeled face representative trees of Γ ′ and Γ ′′ , respectively.Then T ′ and T ′′ are isomorphic.
Let G be a planar 3-tree of n vertices.Since the semi-labeled face-representative trees obtained from different planar embeddings of G are isomorphic, we can choose any leaf of a face-representative tree T f to obtain another semi-labeled face-representative tree that corresponds to a different planar embedding of G. Observe that T f has 2n − 4 face-nodes and let x be a face-node in T f such that the depth of the tree T x obtained from T f by choosing x as the root is minimum over all the 2n − 4 possible choices for x.Recall that deletion of the face-nodes from the face-representative tree yields the representative tree of the corresponding embedding.Therefore, deletion of the face-nodes from T x gives us a representative tree with minimum depth, which in turn corresponds to a minimum-depth embedding of G.The following fact states that x is the nearest face-node from the center of T f .Fact 2 Let T f be a face-representative tree and let x be a face-node of T f such that the distance between x and the center of T f is minimum over all the face nodes of T f .Then the depth of the tree obtained from T f by choosing a face-node as the root is greater than or equal to the depth of the tree obtained from T f by choosing x as the root.
Proof: First assume that center of T f is an edge (u, v).Deletion of the edge (u, v) from T f yields two connected components T u and T v that contain u and v, respectively.Let p and q be two vertices of some tree T .Then by d T (p, q), we denote the distance (i.e., length of the unique path) between p and q in T .Let Let y be a leaf such that the distance between y and the center of T f is maximum over all nodes in T f .Let be the depth of the tree obtained from T f by choosing x as the root.Since the center of T f is an edge, D x = k + k ′ + 1, which is independent of the position of x in T i , i ∈ {u, v}.
Let z be a face-node in T f , where l = min{d T f (z, u), d T f (z, v)} and l > k.If there is no such z, then we are done.Otherwise, suppose for a contradiction that the depth of the tree obtained from T f by choosing z as the root is D z , where D z < D x .Observe that D z = l + k ′ + 1, which is independent of the position of z in T i .Since l > k, therefore D z > D x , a contradiction.Now assume that the center is a vertex v. Deletion of the vertex u from T f yields four connected components T 1 , T 2 , T 3 and T 4 .Let k = d T f (x, v).Let y be a leaf such that the distance between y and the center of T f is maximum over all nodes in T f .Let k ′ = d T f (y, v).Assume that y is in some T j , 1 ≤ j ≤ 4. Then there is a leaf y ′ not in T j such that d T f (y ′ , v) = k ′ .Let the depth of the tree obtained from T f by choosing x as the root be D x .Observe that D x = k + k ′ , which is independent of the position of x in T j .
Let z be a face-node in T f , where l = d T f (z, v) and l > k.If there is no such z, then we are done.Otherwise, suppose for a contradiction that the depth of the tree obtained from T f by choosing z as the root is D z , where The center of a tree is either a single node or an edge, and it is straightforward to find the center of T f in O(n) time by repeatedly deleting the nodes of degree one, until a single node or an edge is left.We then do a breath-first search to select a nearest node x, which also takes O(n) time.Then by Fact 2, the planar embedding of G that corresponds to the face-representative tree obtained by choosing x as the root is the minimum-depth embedding of G. Thus the following lemma holds.
Lemma 6 Let G be a planar 3-tree.An embedding Γ of G with the minimum depth can be found in linear time.
We now have the following lemma on the bound of minimum-depth.

Lemma 7
The depth of a minimum-depth embedding Γ of a planar 3-tree G with n vertices is at most ⌊ n−3 2 ⌋ + 1.
Proof: Let T x be the face-representative tree of Γ, where x is the root of T x .Since Γ is a minimum-depth embedding, the length k of the unique path between x and the center of T x is minimum over all the face nodes of T x .Let O be the center of T x , which may be a node or an edge of T x .Every internal node of T x has exactly four neighbors and x is a nearest node from O. Let T be the representative tree of Γ.Note that T is obtained by deleting all face-nodes from T x .T contains exactly n − 3 vertices and the distance from the root of T to O is k − 1.Since every internal node of T x has exactly four neighbors and x is a nearest node from the center, the depth of the representative tree T obtained by deleting all the face-nodes from T x is at most ⌊ (n−3) 2 ⌋ + 1, when k = 1.We thus assume that k > 1.Since every internal node of T x is a node of degree four and no leaf node of T x is within the distance k − 1 from O, every node of T within the distance k − 2 from O is a node of degree four.The number of nodes in T which are within the distance k − 1 from the center is  path in T , these nodes can contribute at most We now use Theorem 1 and Corollary 1 to obtain the upper bounds on the sizes of universal line set and universal circle set for planar 3-trees, as in the following theorem.
2 ⌋ parallel lines and a set of ⌈ n+3 4 ⌉ concentric circles are universal for planar 3-trees with n vertices.

Bounds for Special Classes of Planar 3-Trees
In this section we categorize planar 3-trees into three types: Type 0, Type 1 and Type 2. We prove that every planar 3-tree of Type 0 and Type 1 can be embedded on ⌈ (n+3) 3 ⌉ and ⌊4n/9⌋ parallel lines, respectively.We conjecture that every planar 3-tree of Type 2 admits an embedding on ⌊4n/9⌋ parallel lines.
Let T be a rooted tree with n vertices.Then there exists a vertex v in T such that the number of vertices in the subtree rooted at v is more than 2n/3 and the number of vertices in each of the subtrees rooted at the children of v is at most 2n/3 [16, Theorem 9.1].If T is a representative tree of some plane 3-tree, then T must have such a vertex.Consequently, we can use Lemma 2 to prove the following lemma.
Lemma 8 Let Γ be a plane 3-tree.Then there exists a triangle C in Γ satisfying the following.Let r be the representative vertex of Γ(C) and let C 1 , C 2 , C 3 be the three nested triangles around r. Then the number of inner vertices in Γ(C) is more than 2(n − 3)/3 and the number of inner vertices in each We call C a heavy triangle of Γ. Observe that for any heavy triangle C of Γ, one of the following properties hold.Let G be a planar 3-tree.If G admits a plane embedding that contains a heavy triangle satisfying Property (a), then we call G a planar 3-tree of Type 0. If G is not a planar 3-tree of Type 0, but admits a plane embedding that contains a heavy triangle satisfying Property (b), then we call G a planar 3-tree of Type 1.If G is not a planar 3-tree of Type 0 or Type 1, but admits a plane embedding that contains a heavy triangle satisfying Property (c), then we call G a planar 3-tree of Type 2. We now have the following lemma.
Lemma 9 Given a planar 3-tree G, one can determine whether it is of Type 0, Type 1 or Type 2 in linear time.
Proof: Let Γ be any plane embedding of G and let T be its semi-labeled face representative tree.Let v be any vertex-node in T .By N v we denote the number of vertex-nodes in the subtree rooted at v and by N (T ) we denote the number of vertex-nodes in tree T .We now do a postorder traversal on T to find the number N v for each vertex-node v in T .While traversing T , at each vertex-node v we perform a Type Test, as described below.Throughout the computation we maintain five variables t 1 , t 2 , . . ., t 5 and the value stored in t 1 after the end of the computation indicates the type of G. Initially, we set t 1 = 2.
Type Test: Let P, Q, R and S be the subtrees obtained by deleting the vertex v from T .For each subset {I, J, K} ⊂ {P, Q, R, S} of three subtrees, we do the following.
, then we set t = 0 and stop the tree traversal.
Observe that at each step of the traversal we perform only a constant number of tests and numbers N (.) can be computed in constant time with the help of N v values using the knowledge of the total number of vertex-nodes in T .Therefore, the traversal can be performed in linear time.
In the following we prove that the value stored in t corresponds to the type of G, and such a Type t embedding of G can be constructed in linear time.
Without loss of generality we assume that G is a Type 0 plane 3-tree.The proof for the case when G is a Type 1 or a Type 2 plane 3-tree is similar.
By definition, G has a planar embedding Γ ′ such that for some vertex v ′ of its face-representative tree, the condition of Test (A) holds.Observe that by Corollary 2, the face-representative trees obtained from different embeddings of G are isomorphic.Therefore, there exists a vertex v in T and three subtrees {I, J, K} ⊂ {P, Q, R, S}, such that N (I) + N (J) + N (K) > 2(n − 3)/3 and each of N (I), N (J), N (K) is at most (n − 3)/3.Consequently, t must be 0.
We now compute a Type 0 embedding of G. Let T ′′ = {P, Q, R, S} \ {I, J, K}, and let f be a leaf of T ′′ .We claim that the embedding of G taking f as the outer face gives us a Type 0 embedding of G.We distinguish two cases depending on whether v is an ancestor of f or not.
Consider first the case when v is an ancestor of f in T , as shown in  The case when f is a descendant of v in T is simpler.By Lemma 2, v corresponds to a unique cycle C xyz of three vertices x, y, z such that Γ(C xyz ) is a plane 3-tree with representative vertex v. Observe that I, J, K correspond to the three nested triangles around v in Γ(C xyz ).Consequently, xyz is the required heavy triangle and Γ itself is a Type 0 embedding of G.
Before proving the upper bounds for planar 3-trees of Type 0 and Type 1, we need to explain some properties of drawings on line set and some properties of the drawing algorithm of Chrobak and Nakano [5].
Fact 3 Let G be a plane 3-tree and let x, y, z be the outer vertices of G. Assume that G has a drawing D on k parallel lines, where x lies on line l 0 , y lies on line l k−1 and z lies on line l i , 0 ≤ i ≤ k − 1.
(a) Let p, q and r be three non-collinear points on lines l 0 , l k−1 and l i , respectively.Then G has a drawing D ′ on k parallel lines, where the vertices x, y, z lie on points p, q, r, respectively, and for each vertex u, if u lies on line l in D then u lies on line l in D ′ .Moreover, D ′ respects the combinatorial planar embedding determined by D. If (x, y) is an outer edge of Γ, then redefine Γ as Γ ′ .Otherwise, consider an embedding Γ ′ of G such that (x, y) is an outer edge of Γ ′ and the embeddings of Γ ′ (C xyz ) and Γ(C xyz ) are the same.Observe that any embedding of G taking a face xyv of G as the outer face, where v is not a vertex of Γ(C xyz ), will suffice.An example is illustrated in Figure 7.
Since the number of inner vertices in each of the plane 3-trees Γ ′ (C xyw ), Γ ′ (C yzw ) and Γ ′ (C zxw ) is at most (n − 3)/3, the depth of the representative tree of Γ ′ (C xyz ) is at most (n − 3)/3 + 1.It is now tempting to claim that the depth of the representative tree of Γ ′ is bounded by (n − 3)/3 + 2 and we can produce a drawing on (n − 3)/3 + 4 parallel lines by Theorem 1.However, z might not be the representative vertex in Γ ′ .Therefore, the depth of the representative tree of Γ ′ may be very large and hence we compute the drawing in a different technique as described below.
Let t 0 (= z), t 1 , t 2 , . . ., t q (= v) be all the vertices of Γ ′ such that no t i is interior to Γ ′ (C xyz ) and each t i , 0 ≤ i ≤ q is adjacent to both x and y, and for each j, 0 ≤ j < q, vertex t j is interior to the triangle xyt j+1 .We claim that t 0 (= z), t 1 , t 2 , . . ., t q (= v) is a path in Γ ′ .Otherwise, assume that t j and t j+1 are not adjacent.By Lemma 1, Γ ′ (C xytj+1 ) is a plane 3-tree.Let t ′ j be the representative vertex of Γ ′ (C xytj+1 ) which is adjacent to both x and y.If t ′ j does not coincide with t j , then j ′ > j + 1, a contradiction to the assumption that t ′ j is the representative vertex of Γ ′ (C xytj+1 ).We now draw Γ ′ on k = ⌈ (n+3) 3 ⌉ parallel lines.Place the vertices x and y on lines l 0 and l k−1 , respectively, with the same x-coordinate.Place the vertices t 0 (= z), t 1 , t 2 , . . ., t q (= v) on lines l 1 and l k−2 alternatively with increasing x coordinates such that the triangles xyt i can be drawn maintaining their nesting order avoiding edge crossings.Then add the edges between t j and t j+1 .See Figure 8 (a).Let the resulting drawing be D. Since Γ ′ (C xyz ) contains more than 2(n−3)/3 inner vertices, each plane 3-tree Γ ′ (C xtj tj+1 ) and Γ ′ (C ytj tj+1 ) contains less than (n − 3)/3 vertices.Consequently, the depth of the representative tree of each plane 3-tree Γ ′ (C xtj tj+1 ) and Γ ′ (C ytj tj+1 ) is at most (n − 3)/3.Since Case 1, we can show that t 0 , t 1 , t 2 , . . ., t q (= v) is a path in Γ ′ .Now place the vertices t 1 , t 2 , . . ., t q (= v) on lines l 1 and l k−2 alternatively with increasing x coordinates such that the triangles xyt i can be drawn maintaining their nesting order avoiding edge crossings.Then add the edges between t j and t j+1 .See Figure 8 (b).Let the resulting drawing be D. Observe that each of the plane 3-trees Γ ′ (C xtj tj+1 ) and Γ ′ (C ytj tj+1 ) contains at most (n − 3)/3 inner vertices.Therefore, the depth of the representative tree of each of those plane 3-trees is at most (n − 3)/3.On the other hand, each of the triangles xt j t j+1 and yt j t j+1 contains two vertices on consecutive lines and is crossed by more than (n − 3)/3 parallel lines.Consequently, we can draw the plane 3-trees Γ ′ (C xtj tj+1 ) and Γ ′ (C ytj tj+1 ) on k − 1 lines and then insert those drawings into the corresponding △xt j t j+1 and △yt j t j+1 in D using Property (a) of Fact 3.
The technique we used to draw Type 1 plane 3-trees, uses the property (CN2), i.e., Γ(C xyw ) admits a drawing such that w lies either on line l k−2 , or l 1 ; as shown in Figure 8(b) and Figure 9, respectively.If we could choose the position of w on l 1 or on l k−2 arbitrarily, then we could find a drawing of Type 2 plane 3-trees on k = 4n/9+1 parallel lines, as follows.Without loss of generality assume that each of Γ(C xyw ) and Γ(C wxt1 ) contains more than (n − 3)/3 inner vertices.Observe that if we could choose the position of w arbitrarily, then we could compute the drawings of Γ(C xyw ) and Γ(C wxt1 ) inside triangles xyw and wxt 1 , respectively.However, it seems very difficult to modify Chrobak and Nakano's algorithm [5] to compute a straight-line drawing respecting a given position for w.Consequently, it would be interesting to examine whether the upper bound of ⌈(n + 3)/2⌉, as proved in Theorem 2, is tight.

Conclusion
Let n be a positive integer multiple of six, then there exists a planar 3-tree with n vertices requiring at least n/3 parallel lines in any of its drawing on parallel lines [11].On the other hand, we have proved that ⌊ n+3 2 ⌋ parallel lines are universal for planar 3-trees with n vertices.It would be interesting to close the gap between the upper bound and the lower bound of universal line set for planar 3-trees.Finding a universal line set of smaller size for drawing planar 3-trees where the lines are not always parallel is left as an open problem.
Open Problem 1.What is the smallest constant c such that every planar 3-tree with n vertices admits a drawing on cn parallel straight lines?JGAA, 17(2) 59-79 (2013) 77 Observe that we tried to find straight-line drawings with small height.Although in Section 5 we use an algorithm of Chrobak and Nakano [5] that can produce O(n 2 )-area grid drawings, the drawings we produce can have exponential width because of the scenario depicted in Figure 9.One can decide whether a planar 3-tree admits a straight-line grid drawing on a given area [14], but the only upper bound known is O(8n 2 /9) area, which is implied by the algorithm of Brandenburg [4] that can compute an O(8n 2 /9)-area straight-line grid drawing for arbitrary planar graphs.Since the lower bound on the area requirement of straight-line grid drawings of plane 3-trees is Ω(n 2 ), we ask the following question.
Open Problem 2. What is the smallest constant c such that every planar 3-tree with n vertices admits a straight-line grid drawing on cn 2 area?

JGAA, 17 Figure 1 :
Figure 1: (a) A plane 3-tree G, (b) representative tree T of G, (c) another embedding G ′ of G and (d) representative tree T of G ′ .

Case 1 : 3 : 4 :
Vertex b lies on the line l k+1 .In this case, vertices b and c lie on the same line l k+1 .If we place the representative vertex v on the line l 1 inside the △abc, then k, k − 1 and k lines cross the triangles △abv, △bcv and △acv, respectively.Case 2: Vertex b lies on the line l 0 .In this case, vertices b and a lie on the same line l 0 .If we draw v on the line l k inside the △abc, then k − 1, k and k lines cross the triangles △abv, △bcv and △acv, respectively.Case Vertex b lies on the line l 1 .In this case, vertices a and b lie on consecutive lines.If we draw v on the line l k inside the △abc, then k − 1, k − 1 and k lines cross the triangles △abv, △bcv and △acv, respectively.Case Vertex b lies on the line l k .In this case, vertices b and c lie on consecutive lines.If we draw v on the line l 1 inside the △abc, then k − 1, k − 1 and k lines cross the triangles △abv, △bcv and △acv, respectively.
(a) If n = 3, then T f is a single face-node.(b)If n > 3, then any vertex in T f is either a vertex-node, which corresponds to a vertex of Γ or a face-node, which corresponds to a face of Γ.Moreover, the following (i)-(ii) hold.(i)The root is a face-node that corresponds to the outer face of Γ. Root has only one child which is the representative vertex p of Γ.Every vertex-node has exactly three children.Every face-node other than the root is a leaf in T f .

Figure 2 :
Figure 2: (a)A plane 3-tree Γ and (b) the face-representative tree T f of Γ.
Figure 4 illustrates an example, where the nodes within the distance k−1 from the center lie in the shaded region.The number of nodes of T which are not counted within the distance k − 1 is n − 3 − 2 • 3 k−1 + 1.Since O is on the middle of a longest
. Let w be the neighbor of v that belongs to T ′′ .By Lemma 2, w corresponds to a unique cycle C xyz of three vertices x, y, z such that Γ(C xyz ) is a plane 3-tree with representative vertex w.We delete all the inner vertices of Γ(C xyz ) from Γ. Let Γ ′ be the resulting embedding, as shown in Figures5(b)-(c).Take another embedding Γ ′′ of Γ ′ with xyz as the outer face, as shown in Figure5(d).It is now straightforward to observe that I, J, K correspond to the three nested triangles around the representative vertex of Γ ′′ , and hence xyz is the required heavy triangle.We now extend Γ ′′ to a Type 0 embedding of G as follows.First take an embedding Γ(C xyz ) ′ of Γ(C xyz ) with f as the outer face.Then insert Γ ′′ into the face xyz of Γ(C xyz ) ′ such that the outer face of Γ ′′ coincide with the face xyz creating an embedding of G.

Figure 5 (
e) illustrates such a scenario.

Figure 9 :
Figure 9: Illustration for a possible drawing of a Type 2 Plane 3-tree.