Drawing Planar Graphs with Many Collinear Vertices

Consider the following problem: Given a planar graph $G$, what is the maximum number $p$ such that $G$ has a planar straight-line drawing with $p$ collinear vertices? This problem resides at the core of several graph drawing problems, including universal point subsets, untangling, and column planarity. The following results are known for it: Every $n$-vertex planar graph has a planar straight-line drawing with $\Omega(\sqrt{n})$ collinear vertices; for every $n$, there is an $n$-vertex planar graph whose every planar straight-line drawing has $O(n^\sigma)$ collinear vertices, where $\sigma<0.986$; every $n$-vertex planar graph of treewidth at most two has a planar straight-line drawing with $\Theta(n)$ collinear vertices. We extend the linear bound to planar graphs of treewidth at most three and to triconnected cubic planar graphs. This (partially) answers two open problems posed by Ravsky and Verbitsky [\emph{WG}~2011:295--306]. Similar results are not possible for all bounded treewidth planar graphs or for all bounded degree planar graphs. For planar graphs of treewidth at most three, our results also imply asymptotically tight bounds for all of the other above mentioned graph drawing problems.


Introduction
A set S ⊆ V (G) of vertices in a planar graph G is a collinear set if G has a planar straightline drawing where all the vertices in S are collinear.Ravsky and Verbitsky [21] consider the problem of determining the maximum cardinality of collinear sets in planar graphs.A stronger notion is dened as follows: a set R ⊆ V (G) is a free collinear set if a total order < R of R exists such that, given any set of |R| points on a line , the graph G has a planar straight-line drawing where the vertices in R are mapped to the given points and their order on matches the order < R .Free collinear sets were rst used (although not named) by Bose et al. [4]; also, they were called free sets by Ravsky and Verbitsky [21].Clearly, every free collinear set is also a collinear set.In addition to this obvious relationship to collinear sets, free collinear sets have connections to other graph drawing problems, as will be discussed later in this introduction.

Journal of Computational Geometry jocg.org
Based on the results in [4], Dujmovi¢ [9] showed that every n-vertex planar graph has a free collinear set of cardinality at least n/2.A natural question to consider would be whether a linear bound is possible for all planar graphs.Ravsky and Verbitsky [21] provided a negative answer to that question.In particular, they observed that if a planar triangulation has a large collinear set, then its dual graph has a cycle of proportional length.
Since there are m-vertex triconnected cubic planar graphs whose longest cycle has length O(m σ ) [13], it follows that there are n-vertex planar graphs in which the cardinality of every collinear set is O(n σ ).Here σ is a known graph-theoretic constant called shortness exponent, for which the best known upper bound is σ < 0.986 [13].
In addition to the natural open problem of closing the gap between the Ω(n 0.5 ) and O(n σ ) bounds for general n-vertex planar graphs, these results raise the question of which classes of planar graphs have (free) collinear sets of linear cardinality.Goaoc et al. [11] proved (implicitly) that n-vertex outerplanar graphs have free collinear sets of cardinality (n + 1)/2; this result was explicitly stated and proved by Ravsky and Verbitsky [21].Ravsky and Verbitsky [21] also considerably strengthened that result by proving that all n-vertex planar graphs of treewidth at most two have free collinear sets of cardinality n/30; they also asked for other classes of graphs with (free) collinear sets of linear cardinality, calling special attention to planar graphs of bounded treewidth and to planar graphs of bounded degree.
In this paper we prove the following results: 1. every n-vertex planar graph of treewidth at most three has a free collinear set with cardinality n−3 8 ; 2. every n-vertex triconnected cubic planar graph has a collinear set with cardinality n 4 ; and 3. every planar graph of treewidth k has a collinear set with cardinality Ω(k 2 ).
Our rst result generalizes the previous result on planar graphs of treewidth at most two [21].As noted by Ravsky and Verbitsky in the full version of their paper [22,Corollary 3.5], there are n-vertex planar graphs of treewidth at most 8 whose largest collinear set has cardinality o(n).In order to obtain that, the authors show a construction relying on the dual of the Barnette-Bosák-Lederberg's non-Hamiltonian cubic triconnected planar graph.We observe that the Tutte's graph, which is also non-Hamiltonian, cubic, planar, and triconnected, has a dual graph that has treewidth 5. Hence, by exploiting the same construction proposed by Ravsky and Verbitsky, however by relying on the dual of the Tutte's graph rather than on the dual of the Barnette-Bosák-Lederberg's graph, we get that the sub-linear upper bound for the size of the largest collinear set holds true for planar graphs of treewidth at most 5. Thus, our rst result leaves k = 4 as the only remaining open case for the question of whether planar graphs of treewidth at most k admit (free) collinear sets with linear cardinality.
Our second result provides the rst linear lower bound on the cardinality of collinear sets for a fairly wide class of bounded-degree planar graphs.The result cannot be extended to all bounded-degree planar graphs.In particular it cannot be extended to planar graphs of degree at most 7, since there exist n-vertex planar triangulations of maximum degree 7 whose dual graph has a longest cycle of length o(n) [18].
Journal of Computational Geometry jocg.orgFinally, our third result improves the Ω( √ n) bound on the cardinality of collinear sets in general planar graphs for all planar graphs whose treewidth is ω( 4 √ n).
We now discuss applications of our results to other graph drawing problems.Since our rst result gives free collinear sets, its consequences are broader.
A column planar set in a planar graph G is a set Q ⊆ V (G) satisfying the following property: there exists a function γ : Q → R such that, for any function λ : Q → R, there exists a planar straight-line drawing of G in which each vertex v ∈ Q is mapped to point (γ(v), λ(v)).Column planar sets were dened by Barba et al. [2] motivated by applications to partial simultaneous geometric embeddings. 1 They proved that n-vertex trees have column planar sets of size 14n/17.The lower bounds in all our three results carry over to the size of column planar sets for the corresponding graph classes.
A universal point subset for the n-vertex planar graphs is a set P of k ≤ n points in the plane such that, for every n-vertex planar graph G, there exists a planar straight-line drawing of G in which k vertices are placed at the k points in P .Universal point subsets were introduced by Angelini et al. [1].Every set of n points in general position is a universal point subset for the n-vertex outerplanar graphs [12,3,6] and every set of n/2 points in the plane is a universal point subset for the n-vertex planar graphs [9].As a corollary of our rst result, we obtain that every set of n− 3 8 points in the plane is a universal point subset for the n-vertex planar graphs of treewidth at most three.
Given a straight-line drawing of a planar graph, possibly with crossings, to untangle it means to assign new locations to some of its vertices so that the resulting straight-line drawing is planar.The goal is to do so while keeping xed (i.e., not changing the location of ) as many vertices as possible.Several papers have studied the untangling problem [19,5,8,4,11,16,21].It is known that general n-vertex planar graphs can be untangled while keeping Ω(n 0.25 ) vertices xed [4] and that there are n-vertex planar graphs that cannot be untangled while keeping Ω(n 0.4948 ) vertices xed [5].Asymptotically tight bounds are known for paths [8], trees [11], outerplanar graphs [11], and planar graphs of treewidth two [21].
As a corollary of our rst result, we obtain that every n-vertex planar graph of treewidth at most three can be untangled while keeping Ω( √ n) vertices xed.This bound is the best possible, as there are forests of stars that cannot be untangled while keeping ω( √ n) vertices xed [4].Our result generalizes previous results on trees, outerplanar graphs and planar graphs of treewidth at most two.
G as if it is drawn according to its plane embedding; also, when we talk about a planar drawing of G, we always mean that it respects the plane embedding of G.The interior of G is the closure of the union of the internal faces of G.We associate with a subgraph H of G the plane embedding obtained from the one of G by deleting vertices and edges not in H.
The degree of a vertex v in a graph G is denoted by δ G (v).A graph is cubic (subcubic) if every vertex has degree 3 (resp.at most 3).Let G be a graph and U ⊆ V (G).Let G − U be the graph obtained from G by removing the vertices in U and their incident edges.The subgraph of G induced by U has U as vertex set and has an edge e ∈ E(G) if and only if both its end-vertices are in Let G be a connected graph.A cut-vertex is a vertex whose removal disconnects G.If G has no cut-vertex and it is not a single edge, then it is biconnected.A biconnected component of G is a maximal (with respect to both vertices and edges) biconnected subgraph of G. Let G be a biconnected graph.A separation pair is a pair of vertices whose removal disconnects G.If G has no separation pair, then it is triconnected.Given a separation pair {a, b} in a biconnected graph G, an {a, b}-component is either a trivial {a, b}-component the edge (a, b) or a non-trivial {a, b}-component a subgraph of G induced by a, b, and the vertices of a connected component of G − {a, b}.

From a Geometric to a Topological Problem
In this section we show that the problem of determining a large collinear set in a planar graph, which is geometric by denition, can be transformed into a purely topological problem.This result may be useful to obtain bounds for the size of collinear sets in classes of planar graphs dierent from the ones we studied in this paper.
A curve is simple if it does not self-intersect.Given a planar drawing Γ of a plane graph G, we say that an open simple curve λ is good for Γ if, for each edge e of G, the curve λ either entirely contains e or has at most one point in common with e (if λ passes through an end-vertex of e, that counts as a common point).Clearly, the existence of a good curve passing through a certain sequence of vertices, edges, and faces of G does not depend on the actual drawing Γ, but only on the plane embedding of G.For this reason we often talk about the existence of good curves in plane graphs, rather than in their planar drawings.
Let R G,λ be the only unbounded region of the plane dened by G and λ.The curve λ is proper if both its end-points are incident to R G,λ .We have the following.Theorem 1.A plane graph G has a planar straight-line drawing with χ collinear vertices if and only if G has a proper good curve that passes through χ vertices of G.
Proof.For the necessity, assume that G has a planar straight-line drawing Γ with χ vertices lying on a common line .We transform into a straight-line segment λ by cutting o two disjoint half-lines of in the outer face of G.This immediately implies that λ is proper.Further, λ passes through χ vertices of G since does.Finally, if an edge e has two common points with λ, then λ entirely contains it, since λ and e are straight-line segments in Γ.
For the suciency, assume that G has a proper good curve λ passing through χ of its vertices; see Fig. 1 Represent C 1 as a convex polygon Q 1 whose all vertices, except for d 1 , lie along a horizontal line , with a to the left of b and d 1 above ; see Fig. 1(c).The graph G 1 is triconnected, as it contains no edge between any two non-consecutive vertices of its only non-triangular face. 2 Thus, there exists a planar straight-line drawing of G 1 in which C 1 2 This implication can be proved as follows.Suppose that G1 is not triconnected; we prove that there exists an edge between two non-consecutive vertices along C1.Augment G1 to a graph G + 1 by adding, in the outer face of G1, a vertex v * and edges between v * and all the vertices of C1.Since the outer face of G1 was its only non-triangular face, we have that G + 1 is a maximal planar graph.It follows that G + 1 is 3-connected and every 3-cut {u, v, z} of G + 1 induces a cycle (u, v, z) [14].Now, for any vertex 2-cut {s, t} of G1, we have that {s, t, v * } is a vertex 3-cut of G + 1 , hence (s, t, v * ) is a cycle of G + 1 ; it follows that the edge (s, t) belongs to G1 and that s and t are vertices of C1.Finally, the fact that s and t are not consecutive along C1 follows from the fact that the cycle (s, t, v * ) contains vertices of G + 1 on both sides, given that {s, t, v * } is a vertex Journal of Computational Geometry jocg.org is represented by Q 1 [24].Analogously, represent C 2 as a convex polygon Q 2 whose all vertices, except for d 2 , lie at the same points as in Q 1 , with d 2 below .Construct a planar straight-line drawing of G 2 in which C 2 is represented by Q 2 .
Removing the dummy vertices and edges results in a planar drawing Γ of the original graph G in which each edge e is a y-monotone curve; see Fig. 1(d).In particular, the fact that λ crosses e at most once ensures that e is either a straight-line segment or is composed of two straight-line segments that are one below and one above and that share an end-point on .As proved in [10,20], there exists a planar straight-line drawing Γ of G in which the y-coordinate of each vertex is the same as in Γ.Since λ passes through χ vertices of G, we have that χ vertices of G lie along in Γ .Theorem 1 can be stated for planar graphs without a given plane embedding as follows: A planar graph has a collinear set with χ vertices if and only if it admits a plane embedding for which a proper good curve can be drawn that passes through χ of its vertices.
While this version of Theorem 1 might be more general, it is less useful for us, so we preferred explicitly stating its version for plane graphs.

Planar Graphs with Treewidth at most Three
In this section we prove the following theorem.
Theorem 2. Every n-vertex plane graph of treewidth at most three admits a planar straightline drawing with at least n−3 8 collinear vertices.
For technical reasons, we regard a plane cycle with three vertices as a plane 3-tree.Then every plane graph G with n ≥ 3 vertices and treewidth at most three can be augmented with dummy edges to a plane 3-tree G [17] which is a plane triangulation.A planar straightline drawing of G with n−3 8 collinear vertices can be obtained from a planar straight-line drawing of G with n−3 8 collinear vertices by removing the inserted dummy edges.Thus for the remainder of this section, we assume that G is a plane 3-tree.
By Theorem 1 it suces to prove that G admits a proper good curve passing through at least n−3 8 vertices of G. Let u, v, and z be the external vertices of G.If n = 3, then G does not contain any internal vertex and we say that it is hollow.If G is not hollow, let w be the unique internal vertex of G adjacent to all of u, v, and z; we say that w is the central vertex of G. Let G 1 , G 2 , and G 3 be the plane 3-trees which are the subgraphs of G whose outer faces are delimited by the cycles (u, v, w), (u, z, w), and (v, z, w).We call G 1 , G 2 , and G 3 children of G and children of w.
We associate to each internal vertex x of G a plane 3-tree G(x), which is a subgraph of G, as follows.We associate G to w and we recursively associate plane 3-trees to the internal vertices of the children G 1 , G 2 , and G 3 of G.Note that x is the central vertex of the plane 3-tree G(x) associated to it.
We now introduce a classication of the internal vertices of G; see Fig. 2(a).Consider an internal vertex x of G.We say that x is of type A, B, C, or D if, respectively, 3, 2, 1, Journal of Computational Geometry jocg.org or 0 of the children of G(x) are hollow.Let a(G), b(G), c(G), and d(G) be the number of internal vertices of G of type A, B, C, and D, respectively.Let m = n − 3 be the number of internal vertices of G.In the following we present an algorithm that constructs three proper good curves λ u (G), λ v (G), and λ z (G) lying in the interior of G.For every edge (x, y) of G, let p xy be an arbitrary internal point of (x, y).The end-points of λ u (G) are p uv and p uz , the end-points of λ v (G) are p uv and p vz , and the end-points of λ z (G) are p uz and p vz .Although each of λ u (G), λ v (G), and λ z (G) is a good curve, any two of these curves might cross each other arbitrarily and might pass through the same vertices of G.Each of these curves passes through all the internal vertices of G of type A, through no vertex of type C or D, and through some vertices of type B. We will prove that the total number of internal vertices of G through which the curves λ u (G), λ v (G), and λ z (G) pass is at least 3m 8 , hence one of them passes through at least m 8 internal vertices of G.The curves λ u (G), λ v (G), and λ z (G) are constructed by induction on m.In the base case we have m ≤ 1; refer to Fig. 2(b).If m = 0, then λ u (G) starts at p uv , traverses the internal face (u, v, z) of G, and ends at p uz .The curves λ v (G) and λ z (G) are constructed analogously.If m = 1, then λ u (G) starts at p uv , traverses the internal face (u, v, w) of G, passes through the central vertex w of G, traverses the internal face (u, z, w) of G, and ends at p uz .The curves λ v (G) and λ z (G) are constructed analogously.
In the inductive case we have m > 1 and the central vertex w of G is of one of types BD.We outline our strategy for constructing λ u (G), λ v (G), and λ z (G).If w is of type C or D, then proper good curves are inductively constructed for the children of G and composed to obtain λ u (G), λ v (G), and λ z (G).If w is of type B, then a sequence of vertices of type B is considered; this sequence is called a B-chain.The B-chain starts at w 1 = w; while the only child H i of the last vertex w i in the sequence has a central vertex w i+1 of type B, the sequence is enriched with w i+1 ; once w i+1 is not of type B, induction is applied on H i , and the three curves obtained by induction are composed with curves passing through vertices of the B-chain to get λ u (G), λ v (G), and λ z (G).
We now detail our construction.Assume rst that w is of type C or D. Refer to Journal of Computational Geometry jocg.orgexternal vertices of H 1 Table 1: Denition of P u , P v , and P z depending on the external vertices of H 1 .Fig. 2(c).Inductively construct curves λ u (G 1 ), λ v (G 1 ), and λ w (G 1 ) for G 1 , curves λ u (G 2 ), λ z (G 2 ), and λ w (G 2 ) for G 2 , and curves λ v (G 3 ), λ z (G 3 ), and λ w (G 3 ) for G 3 .Let Next, consider the case in which w is of type B. In order to describe how to construct curves λ u (G), λ v (G), and λ z (G), we need to further explore the structure of G.
Let H 0 = G, let w 1 = w, and let H 1 be the only non-hollow child of G.We dene three paths P u , P v , and P z as described in Table 1, depending on which among u, v, z, and w are the external vertices of H 1 .Now suppose that, for some i ≥ 1, a sequence w 1 , . . ., w i of vertices of type B, a sequence H 0 , H 1 , . . ., H i of plane 3-trees, and three paths P u , P v , and P z (possibly single vertices or edges) have been dened so that the following properties hold true.Let u , v , and z be the three external vertices of H i .Let K i be the subgraph of G induced by u, v, z, u , v , z , and by the vertices that lie inside the cycle (u, v, z) and outside the cycle (u , v , z ).
(1) For 1 ≤ j ≤ i, the vertex w j is the central vertex of H j−1 and H j is the only non-hollow child of H j−1 ; (2) P u , P v , and P z are vertex-disjoint and each of them is induced in G; (3) P u , P v , and P z connect u, v, and z with u , v , and z , respectively; (4) every vertex of K i belongs to P u , P v , or P z ; and (5) every edge of K i either belongs to P u , P v , or P z , or connects two vertices on distinct paths among P u , P v , and P z .
Properties (1)(5) are indeed satised with i = 1.Consider the central vertex of H i and denote it by w i+1 .
If w i+1 is of type B, then let H i+1 be the only non-hollow child of H i .If the cycle (v , z , w i+1 ) delimits the outer face of H i+1 , then add the edge (u , w i+1 ) to P u and leave P v and P z unaltered.The cases in which the cycles (u , z , w i+1 ) or (u , v , w i+1 ) delimit the outer face of H i+1 can be dealt with analogously.Properties (1)(5) are clearly satised by the described construction.In particular, every vertex of K i+1 belongs to P u , P v , or P z , given that (i) every vertex of K i belongs to P u , P v , or P z , (ii) w i+1 belongs to P u , and (iii) the children of H i dierent from H i+1 are hollow.Further, every edge of K i+1 either belongs to P u , P v , or P z , or connects two vertices on distinct paths among P u , P v , and P z , given Journal of Computational Geometry jocg.org that (i) every edge of K i either belongs to P u , P v , or P z , or connects two vertices on distinct paths among P u , P v , and P z , (ii) the edge (u , w i+1 ) belongs to P u , (iii) the edge (v , w i+1 ) connects a vertex of P v with a vertex of P u , (iv) the edge (z , w i+1 ) connects a vertex of P z with a vertex of P u , and (v) the children of H i dierent from H i+1 are hollow.
If w i+1 is not of type B, then we call the sequence w 1 , . . ., w i a B-chain of G; note that all of w 1 , . . ., w i are of type B. For the sake of the simplicity of notation, let H = H i .We label the vertices of P u , P v , and P z as follows.
• P u = (u = u 1 , u 2 , . . ., u U = u ), where U is the number of vertices of P u ; , where V is the number of vertices of P v ; and where Z is the number of vertices of P z .
We also dene the following cycles.
None of the cycles C uv , C uz , and C vz contains a vertex in its interior; further, every edge in the interior of C uv , C uz , or C vz connects two vertices on distinct paths among P u , P v , and P z .Indeed, every vertex or edge in the interior of C uv , C uz , or C vz is an internal vertex or an internal edge of K i , respectively; then the two statements follow from Properties (4) and (5).
We are going to use the following (a similar lemma can be stated for C uz and C vz ).Lemma 1.Let p 1 and p 2 be two points on the boundary of C uv , possibly coinciding with vertices of C uv , and not both on the same edge of G.There exists a good curve connecting p 1 and p 2 , lying inside C uv , except at its end-points, and intersecting every edge of G inside C uv at most once.
Proof.The lemma has a simple geometric proof.Represent C uv as a strictly-convex polygon and draw the edges of G inside C uv as straight-line segments.Then the straight-line segment p 1 p 2 is a good curve satisfying the requirements of the lemma.
We now describe how to construct curves λ u (G), λ v (G), and λ z (G).First, inductively construct curves λ u (H), λ v (H), and λ z (H) for H.The construction of λ u (G), λ v (G), and λ z (G) varies based on how many among P u , P v , and P z are single vertices.Observe that not all of P u , P v , and P z are single vertices, as w 1 / ∈ {u, v, z}.
Suppose rst that none of P u , P v , and P z is a single vertex, as in Fig. 3(a).We describe how to construct λ u (G), as the construction of λ v (G) and λ z (G) is analogous.

Journal of Computational Geometry
(a) None of P u , P v , and P z is a single vertex.(b) P z is a single vertex while P u and P v are not.(c) P u and P v are single vertices while P z is not.
u .The curve λ 0 u lies inside C uz and connects p uz with z 2 , which is internal to P z since Z > 2; the curve λ 1 u coincides with the path (z 2 , . . ., z Z−1 ) (the path consists of a single vertex if Z = 3); the curve λ 2 • If Z = 2, then λ u consists of curves λ 1 u , . . ., λ 4 u .The curve λ 1 u lies inside C uz and connects p uz with p zz ; the curve λ 2 u lies inside C vz and connects p zz with p v z ; the curves λ 3 u and λ 4 u are dened as in the case Z > 2. The curves λ 1 u , λ 2 u , and λ 4 u are constructed as in Lemma 1.
Suppose next that one of P u , P v , and P z , say P z , is a single vertex, as in Fig. 3(b).We describe how to construct λ u (G) and λ z (G); the construction of λ v (G) is analogous to the one of λ u (G).The curve λ z (G) consists of curves λ 0 z , λ 1 z , λ 2 z .The curve λ 0 z lies inside C uz and connects p uz with p u z ; the curve λ 1 z coincides with λ z (H); the curve λ 2 z lies inside C vz and connects p v z with p vz .The curves λ 0 z and λ 2 z are constructed as in Lemma 1.The curve λ u (G) is constructed as follows.
u .The curve λ 0 u lies inside C uv and connects p uv with v 2 , which is internal to P v since V > 2; the curve λ 1 u coincides with the path (v 2 , . . ., v V −1 ) (the path consists of a single vertex if V = 3); the curve λ 2 u lies inside C uv and connects v V −1 with p u v ; the curve λ 3 u coincides with λ u (H); nally, λ 4  u coincides with λ 0 z .The curves λ 0 u , λ 2 u , and λ 4 u are constructed as in Lemma 1.
The curve λ 0 u lies inside C uv and connects p uv with p u v ; the curve λ 1 u coincides with λ u (H); the curve λ 2 u coincides with λ 0 z .The curves λ 0 u and λ 2 u are constructed as in Lemma 1.
Suppose nally that two of P u , P v , and P z , say P u and P v , are single vertices, as in Fig. 3(c).We describe how to construct λ u (G) and λ z (G); the construction of λ v (G) is Journal of Computational Geometry jocg.organalogous to the one of λ u (G).The curve λ z (G) consists of curves λ 0 z , λ • If Z > 2, then λ u (G) consists of curves λ 0 u , . . ., λ 3 u .The curve λ 0 u lies inside C uz and connects p uz with z 2 , which is internal to P z since Z > 2; the curve λ 1 u coincides with the path (z 2 , . . ., z Z−1 ) (the path consists of a single vertex if Z = 3); the curve λ 2 u lies inside C vz and connects z Z−1 with p vz ; nally, the curve λ 3 u coincides with λ v (H).The curves λ 0 u and λ 2 u are constructed as in Lemma 1.
The curve λ 0 u lies inside C uz and connects p uz with p zz ; the curve λ 1 u lies inside C vz and connects p zz with p vz ; the curve λ 2 u coincides with λ v (H).The curves λ 0 u and λ 1 u are constructed as in Lemma 1.
This completes the construction of λ u (G), λ v (G), and λ z (G).Since these curves lie in the interior of G and since their end-points are incident to the outer face of G, they are proper.We now prove that they are good and pass through many vertices of G. Lemma 2. The curves λ u (G), λ v (G), and λ z (G) are good.
Proof.We prove that λ u (G) is good by induction on m; the proof for λ If w is of type C or D, then λ u (G) is composed of the three curves λ v (G 1 ), λ w (G 3 ), and λ z (G 2 ), each of which is good by induction.By construction, λ u (G) intersects the edges (u, v), (v, w), (z, w), and (u, z) at the points p uv , p vw , p zw , and p uz , respectively, and does not intersect the edges (v, z) and (u, w) at all.Consider any edge e internal to G 1 .The curves λ w (G 3 ) and λ z (G 2 ) have no intersection with the region of the plane inside the cycle (u, v, w); further, λ u (G) does not pass through u, v, or w.Hence, λ u (G) contains e or intersects e at most once, given that λ v (G 1 ) is good.Analogously, for every internal edge e of G 2 and G 3 , we have that λ u (G) contains e or intersects e at most once.
Assume now that w is of type B. We prove that, for every edge e of G, the curve λ u (G) either contains e or intersects e at most once.
• Consider an edge e internal to H.The curves that compose λ u (G) and that lie inside C uv , C uz , or C vz , or that coincide with a subpath of P v or P z have no intersection with the region of the plane inside the cycle (u , v , z ); further, λ u (G) does not pass through u , v , or z .Hence, λ u (G) contains e or intersects e at most once, given that λ u (H), λ v (H), and λ z (H) are good.
• Consider an edge e = (v j , v j+1 ) ∈ P v (the argument for the edges in P z is analogous).
If λ u (G) has no intersection with P v , then it has no intersection with e.If λ u (G) Journal of Computational Geometry jocg.orgintersects P v and V > 2, then it contains e (if 2 ≤ j ≤ V − 2), or it intersects e only at v j+1 (if j = 1), or it intersects e only at v j (if j = V − 1).Finally, if λ u (G) intersects P v and V = 2, then λ u (G) properly crosses e at p vv .
• We prove that λ u (G) intersects the edges inside C uv at most once (the argument for the edges inside C uz or C vz is analogous).Recall that, since P u and P v are induced, every edge inside C uv connects a vertex of P u and a vertex of P v .Assume that λ u (G) contains a curve λ 0 u inside C uv that connects p uv with v 2 , a curve λ 1 u that coincides with the path (v 2 , . . ., v V −1 ), and a curve λ 2 u inside C uv that connects v V −1 with p u v , as in Fig. 3(b); all the other cases are simpler to handle.
Consider any edge e incident to v 1 inside C uv .The curve λ 0 u intersects e once in fact the end-points of λ 0 u alternate with those of e along C uv , hence λ 0 u intersects e; moreover, λ 0 u and e do not intersect more than once by Lemma 1.The path (v 2 , . . ., v V −1 ), and hence the curve λ 1 u that coincides with it, has no intersection with e, since the end-vertices of e are not in v 2 , . . ., v V −1 .Further, the curve λ 2 u has no intersection with e in fact the end-points of λ 2 u do not alternate with those of e along C uv , hence if λ 2 u and e intersected, then they would intersect at least twice, which is not possible by Lemma 1.Thus, λ u (G) intersects e once.
Analogously, every edge e incident to v V inside C uv has no intersection with λ 0 u , no intersection with λ 1 u , and one intersection with λ 2 u , hence λ u (G) intersects e once.
Finally, consider any edge e incident to v j , with 2 ≤ j ≤ V − 1.The curves λ 0 u and λ 2 u have no intersection with e in fact the end-points of each of these curves do not alternate with those of e along C uv , hence each of these curves does not intersect e by Lemma 1. Further, λ 1 u contains an end-vertex of e and thus it intersects e once.It follows that λ u (G) intersects e once.This concludes the proof of the lemma.
We introduce three parameters.Let s(G) be the total number of vertices of G through which the curves λ u (G), λ v (G), and λ z (G) pass, counting each vertex with a multiplicity equal to the number of curves that pass through it.Further, let x(G) be the number of internal vertices of type B none of λ u (G), λ v (G), and λ z (G) passes through.Finally, let h(G) be the number of B-chains of G.We have the following inequalities.Lemma 3. The following hold true if m ≥ 1:     Suppose rst that w is of type C. Also, suppose that G 1 and G 2 have internal vertices; the other cases are analogous.Since w is of type C, we have h Finally, suppose that w is of type B. Then w 1 = w is the rst vertex of a B-chain w 1 , . . ., w i of G. Recall that H is the only plane 3-tree child of w i that has internal vertices.Let x be the central vertex of H.By the maximality of w 1 , . . ., w i , we have that x is not of type B, hence If x is of type C, then let L 1 and L 2 be the children of H containing internal vertices.We have h Finally, if x is of type D, then let L 1 , L 2 , and L 3 be the children of H.We have the second inequality holds by induction.(4) x(G) ≤ b(G).This inequality follows from the fact that x(G) is the number of vertices of type B of G none of λ u (G), λ v (G), and λ z (G) passes through, hence this number cannot be larger than the number of vertices of type B of G.
(5) x(G) ≤ 3h(G).Every internal vertex of G of type B belongs to a B-chain of G. Further, for every B-chain w 1 , w 2 , . . ., w i of G, the curves λ u (G), λ v (G), and λ z (G) pass through all of w 1 , w 2 , . . ., w i , except for at most three vertices u = u U , v = v V , and z = z Z (note that, in the description of the construction of λ u (G), λ v (G), and λ z (G) if w is of type B, the vertices u, v, and z are not among w 1 , w 2 , . . ., w i ).Thus, the number x(G) of vertices of type B none of λ u (G), λ v (G), and λ z (G) passes through is at most three times the number h(G) of B-chains of G. (6) Suppose rst that w is of type C. Also, suppose that G 1 and G 2 have internal vertices; the other cases are analogous.Since w is of type C, we have a the second inequality follows by induction.
Suppose nally that w is of type B. Then w 1 = w is the rst vertex of a B-chain w 1 , . . ., w i of G and H is the only plane 3-tree child of w i that has internal vertices.Every internal vertex of G of type A is internal to H, hence a(G) = a(H).Every internal vertex of G of type B is either an internal vertex of H of type B, or is one among w 1 , . . ., w i ; hence b(G) = b(H) + i.Since λ u (G), λ v (G), and λ z (G) contain all of λ u (H), λ v (H), and λ z (H), we have that s(G) is greater than or equal to s(H) plus the number of vertices among w 1 , . . ., w i through which the curves λ u (G), λ v (G), and λ z (G) pass.For the same reason, x(G) is equal to x(H) plus the number of vertices among w 1 , . . ., w i none of λ u (G), λ v (G), and λ z (G) passes through.By construction, λ u (G), λ v (G), and λ z (G) do not pass through at most three vertices among w 1 , . . ., w i , hence x(G) ≤ x(H)+3 and s(G) ≥ s(H)+i−3.Thus, we have s(G) ≥ s(H)+i−3 ≥ 3a(H)+b(H)−x(H)+i−3 = 3a(H)+(b(H)+i)−(x(H)+3) ≥ 3a(G) + b(G) − x(G); the second inequality follows by induction.
Lemma 3 can be used to prove that one of λ u (G), λ v (G), and λ z (G) passes through many vertices of G. Let k be a parameter to be determined later.
If a(G) < km, by ( 1) and ( 6 Again by (2) and by the assumption a(G) < km we get c(G) We now prove that not only there is a planar straight-line drawing of a plane 3-tree with n−3 8 collinear vertices, but the geometric placement of these vertices can be arbitrarily prescribed, as long as it satises an ordering constraint.Since every plane graph of treewidth at most three is a subgraph of a plane 3-tree [17], this implies that every plane graph of treewidth at most three has a free collinear set with n−3 8 vertices.
Theorem 3. Every collinear set in a plane 3-tree is also a free collinear set.
Let G be an n-vertex plane 3-tree with external vertices u, v, and z in this counterclockwise order along the cycle (u, v, z).Consider any planar straight-line drawing Ψ of G and a horizontal line .Label each vertex of G as ↑, ↓, or = according to whether it lies above, below, or on , respectively; let S be the set of vertices labeled =.Let E be the set of edges of G that properly cross in Ψ; thus, the edges in E have one end-vertex labeled ↑ and one end-vertex labeled ↓.Let < Ψ be the total ordering of S ∪ E corresponding to the left-to-right order in which the vertices in S and the crossing points between the edges in E and appear along in Ψ.
Let X be any set of |S|+|E | distinct points on .Each element in S ∪E is associated with a point in X: The i-th element of S ∪ E , where the elements in S ∪ E are ordered according to < Ψ , is associated with the i-th point of X, where the points in X are in leftto-right order along .Let X S and X E be the subsets of the points in X associated to the vertices in S and to the edges in E , respectively; also, let q x be the point in X associated with a vertex x ∈ S and by q xy the point in X associated with an edge (x, y) ∈ E .
We have the following lemma, which implies Theorem 3. Lemma 4.There exists a planar straight-line drawing Γ of G such that: Journal of Computational Geometry jocg.org(1) Γ respects the labeling every vertex labeled ↑, ↓, or = is above, below, or on , respectively; and (2) Γ respects the ordering every vertex in S is placed at its associated point in X S and every edge in E crosses at its associated point in X E .
Proof.The proof is by induction on n and relies on a stronger inductive hypothesis, namely that the drawing Γ can be constructed even if the cycle (u, v, z) delimiting the outer face of G is represented in Γ by an arbitrarily prescribed triangle ∆ satisfying the following properties: (i) the vertices p u , p v , and p z of ∆ representing u, v, and z appear in this counter-clockwise order along ∆; (ii) ∆ respects the labeling each of u, v, and z is above, below, or on if it has label ↑, ↓, or =, respectively; and (iii) ∆ respects the ordering every vertex in {u, v, z} ∩ S lies at its associated point in X S and every edge in {(u, v), (u, z), (v, z)} ∩ E crosses at its associated point in X E .
In the base case n = 3.Let ∆ be any planar straight-line drawing of cycle (u, v, z) satisfying properties (i)(iii).Dene Γ = ∆; then Γ is a planar straight-line drawing of G that respects the labeling and the ordering since ∆ satises properties (i)(iii).Now assume that n > 3; let w be the central vertex of G, and let G 1 , G 2 , and G 3 be its children, whose outer faces are delimited by the cycles (u, v, w), (u, z, w), and (v, z, w), respectively.We distinguish some cases according to the labeling of u, v, z, and w.In every case we draw w at a point p w and we draw straight-line segments from p w to p u , p v , and p z , obtaining triangles ∆ 1 = (p u , p v , p w ), ∆ 2 = (p u , p z , p w ), and ∆ 3 = (p v , p z , p w ).We then use induction to construct planar straight-line drawings of G 1 , G 2 , and G 3 in which the cycles (u, v, w), (u, z, w), and (v, z, w) delimiting their outer faces are represented by ∆ 1 , ∆ 2 , and ∆ 3 , respectively.Thus, we only need to ensure that each of ∆ 1 , ∆ 2 , and ∆ 3 satises properties (i)(iii).In particular, property (i) is satised as long as p w is in the interior of ∆; property (ii) is satised as long as p w respects the labeling; and property (iii) is satised as long as p w = q w , if w ∈ S, and each edge in {(u, w), (v, w), (z, w)} ∩ E crosses at its associated point, if w / ∈ S.
If all of u, v, and z have labels in the set {↑, =}, then all the internal vertices of G have label ↑, by the planarity of Ψ, and the interior of ∆ is above .Let p w be any point in the interior of ∆ (ensuring properties (i)(ii) for ∆ 1 , ∆ 2 , and ∆ 3 ).Also, w / ∈ S and (u, w), (v, w), (z, w) / ∈ E , thus property (iii) is satised for ∆ 1 , ∆ 2 , and ∆ 3 .
The case in which all of u, v, and z have labels in the set {↓, =} is symmetric.If none of these cases applies, we can assume w.l.o.g. that u has label ↑ and v has label ↓.
• Suppose that z has label =.Since u has label ↑, v has label ↓, and (u, v, z) has this counter-clockwise orientation in G, the edge (u, v) and the vertex z are respectively the rst and the last element in S ∪ E according to < Ψ .Since ∆ satises properties Journal of Computational Geometry jocg.org(i)(iii), the points q uv and q z are respectively the leftmost and the rightmost point in X; hence all the points in X − {q uv , q z } are in the interior of ∆.If w has label =, as in Fig. 4(a), then w is the last but one element in S ∪ E according to < Ψ , by the planarity of Ψ (note that the edge (w, z) lies on ).Since ∆ satises (i)(iii), the point q w is the rightmost point in X − {q z }.Let p w = q w (ensuring properties (i)(ii) for ∆ 1 , ∆ 2 , and ∆ 3 ).Then w is at q w and (u, w), (v, w), (z, w) / ∈ E (ensuring property (iii) for ∆ 1 , ∆ 2 , and ∆ 3 ).
If w has label ↑, as in Fig. 4(b), then the edge (v, w) comes after the edge (u, v) and before the vertex z in S ∪ E according to < Ψ , since (v, w) is an internal edge of G and Ψ is planar.Since ∆ satises (i)(iii), the point q vw is between q uv and q z on .Draw a half-line h starting at v through q vw and let p w be any point in the interior of ∆ (ensuring property (i) for ∆ 1 , ∆ 2 , and ∆ 3 ) after q vw on h (ensuring property (ii) for ∆ 1 , ∆ 2 , and ∆ 3 ).Then w / ∈ S, (u, w), (z, w) / ∈ E , and the crossing point between (v, w) and is q vw (ensuring property (iii) for ∆ 1 , ∆ 2 , and ∆ 3 ).
The case in which w has label ↓ is symmetric to the previous one.
• Assume now that z has label ↑.Since u and z have label ↑, since v has label ↓, and since (u, v, z) has this counter-clockwise orientation in G, the edges (u, v) and (v, z) are respectively the rst and the last element in S ∪ E according to < Ψ .Since ∆ satises properties (i)(iii), the points q uv and q vz are respectively the leftmost and the rightmost point in X; thus all the points in X − {q uv , q vz } are in the interior of ∆.
If w has label =, as in Fig. 4(c), then w comes after the edge (u, v) and before the edge (v, z) in S ∪ E according to < Ψ , since w is an internal vertex of G and Ψ is planar.Since ∆ satises (i)(iii), q w is between q uv and q vz on .Let p w = q w (ensuring properties (i)(ii) for ∆ 1 , ∆ 2 , and ∆ 3 ).Then w is at q w and (u, w), (v, w), (z, w) / ∈ E (ensuring property (iii) for ∆ 1 , ∆ 2 , and ∆ 3 ).
If w has label ↑, as in Fig. 4(d), then the edge (v, w) comes after the edge (u, v) and before the edge (v, z) in S ∪ E according to < Ψ , since (v, w) is an internal edge of G and Ψ is planar.Since ∆ satises (i)(iii), the point q vw is between q uv and q vz on .Draw a half-line h starting at v through q vw and let p w be any point in the interior of ∆ (ensuring property (i) for ∆ 1 , ∆ 2 , and ∆ 3 ) after q vw on h (ensuring property (ii) for ∆ 1 , ∆ 2 , and ∆ 3 ).Then w / ∈ S, (u, w), (z, w) / ∈ E , and the crossing point between (v, w) and is q vw (ensuring property (iii) for ∆ 1 , ∆ 2 , and ∆ 3 ).
If w has label ↓, as in Fig. 4(e), then edges (u, v), (u, w), (w, z), and (v, z) come in this order in S ∪ E according to < Ψ , since (u, w) and (w, z) are internal edges of G and Ψ is planar.Since ∆ satises (i)(iii), q uv , q uw , q wz , q vz appear in this left-to-right order on .Let p w be the intersection point between the line through u and q uw and the line through z and q wz (ensuring property (ii) for ∆ 1 ∆ 2 , and ∆ 3 ); note that p w is in the interior of ∆ (ensuring property (i) for ∆ 1 , ∆ 2 , and ∆ 3 ).Then w / ∈ S, (v, w) / ∈ E , the crossing point between (v, w) and is q vw , and the crossing point between (w, z) and is q wz (ensuring property (iii) for ∆ 1 , ∆ 2 , and ∆ 3 ).
• The case in which z has label ↓ is symmetric to the previous one.
This concludes the proof of the lemma.

Triconnected Cubic Planar Graphs
In this section we prove the following theorem.By Theorem 1 it suces to prove that, for every n-vertex triconnected cubic plane graph G, there exists a proper good curve λ passing through at least n 4 vertices of G. Our proof will be by induction on n; Lemma 5 below states the inductive hypothesis satised by λ.In order to split the graph into subgraphs on which induction can be applied, we will use a structural decomposition that is derived from a paper by Chen and Yu [7], who proved that every n-vertex triconnected planar graph contains a cycle passing through Ω(n log 3 2 ) vertices.This decomposition applies to a graph class, called strong circuit graph in [7], wider than triconnected cubic plane graphs.We will introduce the concept of wellformed quadruple in order to point out some properties of the graphs in this class.In particular, the inductive hypothesis will need to handle carefully a set (denoted by X below) of degree-2 vertices of the graph, which have neighbors that are not in the graph at the current level of the induction; since λ might pass through these neighbors, it has to avoid the vertices in X, in order to be good.Special conditions will be ensured for two vertices, denoted by u and v below, which work as a link to the rest of the graph.
We introduce some denitions.Consider a biconnected plane graph G. Given two external vertices u and v of G, let τ uv (G) (β uv (G)) be the path composed of the vertices and edges encountered when walking along the boundary of the outer face of G in clockwise (resp.counter-clockwise) direction from u to v.An intersection point between an open curve λ and β uv (G) (or τ uv (G)) is a point p belonging to both λ and β uv (G) (resp.τ uv (G)) such Journal of Computational Geometry jocg.orgthat, for any > 0, the part of λ in the disk centered at p with radius contains a point not in β uv (G) (resp.τ uv (G)).If the end-vertices of λ are in β uv (G) (or τ uv (G)), then we regard them as intersection points.An intersection point p between λ and β uv (G) (or τ uv (G)) is proper if, for any > 0, the part of λ in the disk centered at p with radius contains a point in the outer face of G.
Our proof of the existence of a proper good curve passing through n 4 vertices of G is by induction on n.In order to make the induction work, we deal with the following setting.A quadruple (G, u, v, X) is well-formed if it satises the following properties.
(a) G is a biconnected subcubic plane graph; (b) u and v are two distinct external vertices of G; (d) if the edge (u, v) exists, then it coincides with τ uv (G); (e) for every separation pair {a, b} of G we have that a and b are external vertices of G and at least one of them is an internal vertex of β uv (G); further, every non-trivial {a, b}-component of G contains an external vertex of G dierent from a and b; and (f ) X = (x 1 , . . ., x m ) is a (possibly empty) sequence of degree-2 vertices of G in β uv (G), dierent from u and v, and in this order along β uv (G) from u to v.
We have the following main lemma (refer to Fig. 5).Lemma 5. Let (G, u, v, X) be a well-formed quadruple.There exists a proper good curve λ such that: (1) λ starts at u, does not pass through v, and ends at a point z of β uv (G); (2) z is in the part of β uv (G) between x m and v (if X = ∅, this condition is vacuous); (3) let u = p 1 , p 2 , . . ., p = z be the intersection points between λ and β uv (G), ordered as they occur along λ; we have that u = p 1 , p 2 , . . ., p = z, v appear in this order along β uv (G) (note that z is the last intersection between λ and β uv (G)); Journal of Computational Geometry jocg.org(4) λ passes through no vertex in X and all the vertices in X are incident to the unbounded region R G,λ of the plane dened by G and λ; in particular, if p i , x j and p i+1 come in this order along β uv (G), then the part of λ between p i and p i+1 lies in the interior of G; (5) λ and τ uv (G) have no proper intersection point; and (6) let L λ and N λ be the subsets of vertices in V (G) − X the curve λ passes through and does not pass through, respectively; each vertex in N λ can be charged to a vertex in L λ so that each vertex in L λ is charged with at most 3 vertices and u is charged with at most 1 vertex.
Before proceeding with the proof of Lemma 5, we state and prove an auxiliary lemma that will be used repeatedly in the remainder of the section.Lemma 6.Let (G, u, v, X) be a well-formed quadruple and let {a, b} be a separation pair of G with a, b ∈ β uv (G).The {a, b}-component G ab of G containing β ab (G) either coincides with β ab (G) or consists of (see Fig. 6): • a path P 0 = (a, . . ., u 1 ) (possibly a single vertex a = u 1 ); • for i = 1, . . ., k − 1, a path P i = (v i , . . ., u i+1 ), where u i+1 = v i ; and Proof.If G contained more than two non-trivial {a, b}-components, then one of them would not contain any external vertex of G dierent from a and b, a contradiction to Property (e) of (G, u, v, X).Thus, G contains two non-trivial {a, b}-components, one of which is G ab .Possibly, G contains a trivial {a, b}-component which is an internal edge (a, b) of G.The statement is proved by induction on the size of G ab .
In the base case, G ab is a path between a and b or is a biconnected graph.In the former case, G ab coincides with β ab (G) and the statement of the lemma follows.In the latter case, the statement of the lemma follows with k = 1, G 1 = G ab , P 0 = a, and P k = b, as long as (G ab , a, b, X ab ) is a well-formed quadruple, where X ab = X ∩ V (G ab ).We now prove that this is indeed the case.
Journal of Computational Geometry jocg.org • Property (a): G ab is biconnected by hypothesis and subcubic since G is subcubic.
• Property (b): a and b are external vertices of G ab as they are external vertices of G.
• Property (c): the degree of a and b in G ab is at least 2, by the biconnectivity of G ab , and at most 2, since G is subcubic and since a and b have a neighbor in the non-trivial {a, b}-component of G dierent from G ab .
• Property (d): if the edge (a, b) exists in G, then it forms a trivial {a, b}-component and it does not belong to G ab , hence the property is trivially satised.
• Property (e): consider any separation pair {a , b } of G ab .If G ab contained more than two non-trivial {a , b }-components, as in Fig. 7(a), then one of them would be a nontrivial {a , b }-component of G that contains no external vertex of G dierent from a and b , a contradiction to Property (e) of (G, u, v, X).For the induction, we distinguish three cases.
Journal of Computational Geometry jocg.org In the rst case a has a unique neighbor a in G ab .Then a is an internal vertex of β uv (G).Since we are not in the base case, G ab is not a simple path with two edges; hence, {a , b} is a separation pair of G satisfying the conditions of the lemma.Let G a b be the {a , b}component of G containing β a b (G).Then G ab consists of G a b together with the vertex a and the edge (a, a ) and induction applies to G a b .If G a b coincides with β a b (G), then G ab coincides with β ab (G), contradicting the fact that we are not in the base case.Hence, G a b consists of: (i) a path P 0 = (a , . . ., u 1 ); (ii) for i = 1, . . ., k with k ≥ 1, a biconnected component G i of G a b that contains vertices u i and v i and such that (G i , u i , v i , X i ) is a well-formed quadruple; (iii) for i = 1, . . ., k − 1, a path P i = (v i , . . ., u i+1 ), where u i+1 = v i ; and (iv) a path P k = (v k , . . ., b).Then G ab is composed of: (i) the path (a, a ) ∪ P 0 ; (ii) for i = 1, . . ., k, the biconnected component G i of G ab ; (iii) for i = 1, . . ., k − 1, the path P i ; and (iv) the path P k .
The second case, in which b has a unique neighbor in G ab , is symmetric to the rst one.
In the third case, the degree of both a and b in G ab is greater than 1.Let G 1 be the biconnected component of G ab containing a. Let H be the subgraph of G ab induced by the vertices with at least one incident edge not in G 1 .We prove the following claim: b / ∈ V (G 1 ), and H and G 1 share a single vertex a = b, which is an internal vertex of β uv (G).
Assume, for a contradiction, that b ∈ V (G 1 ).Then G ab is biconnected.Indeed, if G 1 contains a cut-vertex of G ab , then this cut-vertex is also a cut-vertex of G, since {a, b} is a separation pair of G and G ab is an {a, b}-component of G; however, by Property (a) of (G, u, v, X) the graph G is biconnected.By the biconnectivity of G ab and the maximality of G 1 we have G 1 = G ab ; hence, we are in the base case, a contradiction.
Every G 1 ∪{b}-bridge of G ab has exactly one attachment in G 1 .Indeed, if a G 1 ∪{b}bridge B of G ab had at least two attachments b 1 and b 2 in G 1 , then there would exist a path Q between b 1 and b 2 in B, given that B is connected; however, the edges and the internal vertices of Q would not be in G 1 , hence the union of B and Q would be a biconnected subgraph of G ab containing G 1 as a proper subgraph, thus contradicting the maximality of G 1 .An analogous proof shows that there is exactly one G 1 ∪ {b}-bridge H. Let a be the only attachment of H in G 1 .Note that δ H (a ) = 1, as δ G 1 (a ) ≥ 2 since G 1 is biconnected.By the planarity of G, we have that a is incident to the outer face of G 1 , since a and b are both incident to the outer face of G. Since a is the only attachment of H in G 1 , it follows that a is an internal vertex of β uv (G).This concludes the proof of the claim.
By the claim and since G 1 and H are not single edges, given that the degree of both a and b in G ab is greater than 1, it follows that {a, a } and {a , b} are separation pairs of G satisfying the statement of the lemma, hence induction applies to G 1 and H.In particular, (G 1 , u 1 , v 1 , X 1 ) is a well-formed quadruple, with X 1 = X ∩ V (G 1 ), u 1 = a and v 1 = a .Further, H consists of: (i) for i = 1, . . ., k − 1 with k ≥ 2, a path P i = (v i , . . ., u i+1 ) where u i+1 = v i ; note that P 1 = (v 1 = a , . . ., u 2 ) satises u 2 = a since δ H (a ) = 1; (ii) for i = 2, . . ., k, a biconnected component G i of H containing vertices u i and v i (with v k = b) and such that (G i , u i , v i , X i ) is a well-formed quadruple, with X i = X ∩ V (G i ).Then G ab is composed of: (i) a path P 0 = (a); (ii) for i = 1, . . ., k with k ≥ 1, a biconnected component G i that contains vertices u i and v i and such that (G i , u i , v i , X i ) is a well-formed quadruple; Journal of Computational Geometry jocg.org(iii) for i = 1, . . ., k−1, a path P i = (v i , . . ., u i+1 ), where u i+1 = v i ; and (iv) a path P k = (b).
This concludes the proof of the lemma.
We are now ready to prove Lemma 5.The proof is by induction on the size of G.
Base case: G is a simple cycle.Refer to Fig. 8.If u and v were not adjacent, then {u, v} would be a separation pair none of whose vertices is internal to β uv (G), contradicting Property (e) of (G, u, v, X).Thus, the edge (u, v) exists and coincides with τ uv (G) by Property (d).We now construct a proper good curve λ.The curve λ starts at u; it then passes through all the vertices in V (G) − (X ∪ {v}) in the order in which they appear along β uv (G) from u to v; in particular, if two vertices in V (G) − (X ∪ {v}) are consecutive in β uv (G), then λ contains the edge between them.If the neighbor v of v in β uv (G) is not in X, then λ ends at v , otherwise λ ends at a point z in the interior of the edge (v, v ).Charge v to u and note that v is the only vertex in V (G) − X that is not on λ.It is easy to see that λ is a proper good curve satisfying Properties (1) (6).Next we describe the inductive cases.In the description of each inductive case, we implicitly assume that none of the previously described cases applies.
Case 1: The edge (u, v) exists.Refer to Fig. 9.By Property (d) of (G, u, v, X) the edge (u, v) coincides with τ uv (G).By Property (c), the vertex v has two neighbors; let v be the one dierent from u. Since G is not a simple cycle with length three, {u, v } is a separation pair of G to which Lemma 6 applies.If the {u, v }-component of G containing β uv (G) coincided with β uv (G), then G would be a simple cycle, a contradiction to the fact that we are not in the base case.Hence, the graph G obtained from G by removing the edge (u, v) consists of: (i) a path P 0 = (u, . . ., u 1 ); (ii) for i = 1, . . ., k with k ≥ 1, a biconnected component G i of G that contains vertices u i and v i and such that (G i , u i , v i , X i ) is a wellformed quadruple, where X i = X ∩V (G i ); (iii) for i = 1, . . ., k−1, a path P i = (v i , . . ., u i+1 ), where u i+1 = v i ; and (iv) a path P k = (v k , . . ., v).Inductively construct a curve λ i satisfying the properties of Lemma 5 for each quadruple (G i , u i , v i , X i ).We construct a proper good curve λ for (G, u, v, X) as follows.• The curve λ starts at u.
Journal of Computational Geometry jocg.org if there was a single H ∪ {v}-bridge B i , then y i would be a cut-vertex of G, whereas G is biconnected.This and δ G (v) = 2 imply that G has two H ∪ {v}-bridges B 1 and B 2 .
Finally, one of y 1 and y 2 , say y 1 , belongs to τ uv (G), while the other one, say y 2 , belongs to β uv (G).Hence, if B 1 was not a trivial H ∪ {v}-bridge, then {y 1 , v} would be a separation pair none of whose vertices is internal to β uv (G), which would contradict Property (e) of (G, u, v, X).This concludes the proof of the claim.
By Claim 1 the graph G is composed of three subgraphs: a biconnected graph H, an edge B 1 = (y 1 , v), and a graph B 2 , where H and B 1 share the vertex y 1 , H and B 2 share the vertex y 2 , and B 1 and B 2 share the vertex v. Before proceeding with the case distinction, we argue about the structure of H. Let X = {y 2 } ∪ (X ∩ V (H)).We have the following.Claim 2. (H, u, y 1 , X ) is a well-formed quadruple.
Next, we discuss Property (e).Consider any separation pair {a, b} of H. First, if a was not an external vertex of H, then {a, b} would also be a separation pair of G such that a is not an external vertex of G; this would contradict Property (e) of (G, u, v, X).Second, if both a and b were in τ uy 1 (H), then {a, b} would be a separation pair of G whose vertices are both in τ uv (G), given that τ uy 1 (H) ⊂ τ uv (G); again, this would contradict Property (e) of (G, u, v, X).Third, if an {a, b}-component H ab of H contained no external vertex of H dierent from a and b, then H ab would also be an {a, b}-component of G containing no external vertex of G dierent from a and b, again contradicting Property (e) of (G, u, v, X).
Finally, we deal with Property (f ).The vertices in X ∩ X have degree 2 in H since they have degree 2 in G; further, they are internal to β uy 1 (H) since they are internal to β uv (G).Further, we have that δ H (y 2 ) = 2.In order to prove the previous statement, note that δ H (y 2 ) ≥ 2 since H is biconnected; further, δ H (y 2 ) < δ G (y 2 ) since y 2 has a neighbor in B 2 not in H; nally, δ G (y 2 ) ≤ 3 since G is a subcubic graph.Also, y 2 is an internal vertex of β uy 1 (H), since it is an internal vertex of β uv (G) and is in H.This concludes the proof of the claim.
Case 2: B 2 contains a vertex not in X ∪ {v, y 2 }.Refer to Fig. 10.The curve λ will be composed of three curves λ 1 , λ 2 , and λ 3 .Journal of Computational Geometry jocg.org The curve λ starts at u.By Claim 2, a curve λ 1 satisfying the properties of Lemma 5 can be inductively constructed for (H, u, y 1 , X ).Notice that y 2 ∈ X , thus λ 1 terminates at a point z 0 in β y 2 y 1 (H), by Property (2) of λ 1 .
The curve λ 2 lies in the internal face f of G incident to the edge (y 1 , v) and connects z 0 with a vertex u in B 2 determined as follows.Traverse β uv (G) from y 2 to v and let u = y 2 be the rst encountered vertex not in X.By Property (f ) of (G, u, v, X), every vertex in X ∩ V (B 2 ) has degree 2 in G and in B 2 ; also, δ B 2 (y 2 ) = δ B 2 (v) = 1.If all the internal vertices of β y 2 v (G) belong to X, then B 2 is a path whose internal vertices are in X, a contradiction to the hypothesis of Case 2. Hence, u = v, β y 2 u (G) is induced in B 2 , u is incident to f , and the interior of λ 2 crosses no edge of G.It is vital here that λ 1 satises Properties (3)(5), ensuring that y 2 is not on λ 1 and that the edge incident to y 2 in B 2 is in R G,λ 1 .Thus, if such an edge is (y 2 , u ), still λ intersects it only once.
The curve λ 3 connects u with a point z / ∈ {y 2 , v} on β y 2 v (G).Note that {y 2 , v} is a separation pair of G, since by hypothesis B 2 is not an edge; further, y 2 and v both belong to β uv (G).Hence Lemma 6 applies and the curve λ 3 is constructed as in Case 1, with the part between y 2 and u removed from the construction.
The curve λ satises Properties (1)(5) of Lemma 5. We determine inductively the charge of the vertices in (N λ ∩ V (H)) − {y 2 } to the vertices in L λ ∩ V (H), and the charge of the vertices in N λ in each biconnected component G i of B 2 to the vertices in L λ ∩ V (G i ).The only vertices in N λ that have not yet been charged to vertices in L λ are y 2 and v; charge them to u .Then u is charged with at most 1 vertex of H; every vertex in L λ − {u, u } is charged with at most 3 vertices if it is in H or in a biconnected component of B 2 , or with no vertex otherwise; nally, u is charged with y 2 , v, and with no other vertex if δ G (u ) = 2 or with at most 1 other vertex if δ G (u ) = 3; indeed, in the latter case u = u 1 is such that induction is applied on a quadruple (G 1 , u 1 , v 1 , X 1 ).Thus, Property (6) is satised by the constructed charging scheme.
If Case 2 does not apply, then B 2 is a path between y 2 and v whose internal vertices are in X.In order to proceed with the case distinction, we explore the structure of H.
Case 3: The edge (u, y 1 ) exists.By Claim 2, (H, u, y 1 , X ) is a well-formed quadruple, thus by Property (d) the edge (u, y 1 ) coincides with τ uy 1 (H).Let y be the unique neighbor of y 1 in β uy 1 (H).If every vertex of H dierent from u and y 1 is in X (as in Fig. 11(a)), then λ consists Let v i,j be the vertices of H, with 1 ≤ i, j ≤ g, where v i,j and v i ,j are adjacent in H if and only if |i − i | + |j − j | = 1.Let G i,j be the connected subgraph of G represented by v i,j in H.Note that, since H is a minor of G, each vertex v i,j of H corresponds to a connected subgraph G i,j of G such that, starting from G and contracting all the edges in the graphs G i,j , the graph H is obtained; then v i,j is the vertex obtained by contracting all the edges of G i,j .By the planarity of G, every edge of G that is incident to a vertex in G i,j , for some 2 ≤ i, j ≤ g − 1, has its other end-vertex in a graph G i ,j such that |i − i | ≤ 1 and |j − j | ≤ 1. (The previous statement might not be true for an edge that is incident to a vertex in G i,j with i = 1, i = g, j = 1, or j = g.) Refer to Fig. 14(a).For every edge (v i,j , v i+1,j ) of H, arbitrarily choose an edge e i,j connecting a vertex in G i,j and a vertex in G i+1,j as the reference edge for the edge (v i,j , v i+1,j ) of H.Such an edge exists since H is a minor of G. Reference edges e i,j for the edges i,j , v i,j+1 ) of H are dened analogously. (b) Figure 14: (a) Cells, boundaries, and references edges.The cell C i,j is green.The graphs G i,j , G i+1,j , G i,j+1 , and G i+1,j+1 are surrounded by violet curves; their interior is gray.The references edges are red and thick.The right-top boundary of G i,j is blue.(b) The construction of λ (represented as a thick orange line).Large disks represent the graphs G i,j such that λ passes through vertices of G i,j .Small circles represent the graphs G i,j such that λ does not pass through any vertex of G i,j .White squares represent the intersections between λ and reference edges.
For every pair of indices 1 ≤ i, j ≤ g − 1, we call right-top boundary of G i,j the walk that starts at the end-vertex of e i,j in G i,j , traverses the boundary of the outer face of G i,j in clockwise direction and ends at the end-vertex of e i,j in G i,j .The right-bottom boundary of G i,j (for every 1 ≤ i ≤ g − 1 and 2 ≤ j ≤ g), the left-top boundary of G i,j (for every 2 ≤ i ≤ g and 1 ≤ j ≤ g − 1), and the left-bottom boundary of G i,j (for every 2 ≤ i, j ≤ g) are dened analogously.
For each 1 ≤ i, j ≤ g − 1, we dene the cell C i,j as the bounded closed region of the plane that is delimited by (in clockwise order along the boundary of the region): the righttop boundary of G i,j , the edge e i,j , the right-bottom boundary of G i,j+1 , the edge e i,j+1 , the left-bottom boundary of G i+1,j+1 , the edge e i+1,j , the left-top boundary of G i+1,j , and Journal of Computational Geometry jocg.orgsince the region of the plane dened by each cell is connected and hence so is D * ); draw γ as P plus two curves connecting f p and f q with p(γ) and q(γ), respectively.Also, γ intersects each edge of G at most once, since P does.Finally, γ lies in the interior of R(γ), except at points p(γ) and q(γ).Thus, γ satises the required properties.p(γ) q(γ) f q f p p(γ) q(γ) , assume that γ connects two points p(γ) and q(γ) respectively belonging to the interior of e i,j−1 and e i+2,j ; the case in which p(γ) and q(γ) respectively belong to the interior of e i,j and e i+2,j−1 is analogous.The curve γ is composed of three curves, namely: (1) a curve γ 1 that connects p(γ) and a vertex v p on the left-bottom boundary of G i+1,j , and that lies in the interior of C i,j−1 , except at p(γ) and v p ; (2) a curve γ 2 that connects v p and a vertex v q on the right-top boundary of G i+1,j , and that is an induced path in G i+1,j ; and (3) a curve γ 3 that connects v q and q(γ), and that lies in the interior of C i+1,j , except at v q and q(γ).The curve γ 2 might degenerate to be a single point if v p = v q .
We start with γ 2 .Consider a path P in G i+1,j which is a shortest path connecting a vertex on the left-bottom boundary of G i+1,j and a vertex on the right-top boundary of G i+1,j .Let v p and v q be the end-vertices of P .Note that, possibly, v p = v q .Such a path P always exists since G i+1,j is connected; also, P has no internal vertex incident to the left-bottom boundary or to the right-top boundary of G i+1,j , as otherwise there would exist a path shorter than P between a vertex on the left-bottom boundary of G i+1,j and a vertex on the right-top boundary of G i+1,j .Draw γ 2 as P .
In order to draw γ 1 (the curve γ 3 is drawn similarly), draw the dual graph D of G so that each edge of D only intersects its dual edge; restrict D to the vertices and edges in the interior of C i,j−1 , obtaining a graph D * ; nd a shortest path P p in D * connecting the vertex f p of D * incident to the reference edge to which p(γ) belongs and a vertex representing a Journal of Computational Geometry jocg.orgface of G incident to v p .Let f p be the second end-vertex of such a path; draw γ 1 as P plus two curves connecting f p and f p with p(γ) and v p , respectively.
The curve γ has no intersections with the boundary of R(γ) other than at p(γ) and q(γ).We now prove that γ intersects each edge in R(γ) at most once.First, γ intersects each edge of G i+1,j at most once, since γ 2 is a shortest path in G i+1,j and since γ 1 and γ 3 have no intersections with the edges of G i+1,j , except at v p and v q .Second, γ intersects each edge in C i,j−1 at most once, since P p does, since γ 1 does not cross any edge incident to v p (given that P p is a shortest path between f p and any face incident to v p ), and since γ 2 and γ 3 do not intersect edges in C i,j−1 other than at v p (given that P does not contain vertex incident to the left-bottom boundary of G i+1,j other than v p ); similarly, γ intersects each edge in C i+1,j at most once.Third, γ intersects each edge in C i+1,j−1 at most once, namely at its possible end-vertex in G i+1,j ; similarly, γ intersects each edge in C i,j at most once.Thus, γ satises the required properties.This concludes the proof of Theorem 5.

Implications for other graph drawing problems
In this section, we present a number of corollaries of our results to other graph drawing problems.The following lemma is one of the key tools to establish these connections.For the sake of completeness we explicitly state it here (in a more readily applicable form than the original, see [4, Lemma 1]).Lemma 8. [4] Let G be a planar graph that has a planar straight-line drawing Γ in which a (collinear) set S ⊆ V (G) of vertices lie on the x-axis.Then, for an arbitrary assignment of y-coordinates to the vertices in S, there exists a planar straight-line drawing Γ of G such that each vertex in S has the same x-coordinate as in Γ and has the assigned y-coordinate.
The above lemma immediately implies the following.Lemma 9. [4] Let G be a planar graph, R ⊆ V (G) be a free collinear set, and < R be the total order associated with R. Consider any assignment of xand y-coordinates to the vertices in R such that the assigned x-coordinates are all distinct and the order by increasing x-coordinates of the vertices in R is < R (or its reversal).Then there exists a planar straightline drawing of G such that each vertex in R has the assigned xand y-coordinates.
We rst apply Lemma 9 to obtain an optimal bound (up to a multiplicative constant) on the size of universal point subsets for planar graphs of treewidth at most three.points in the plane is a universal point subset for all n-vertex plane graphs of treewidth at most three.
Proof.If necessary, rotate the Cartesian axes so that no two points in P have the same x-coordinate.By Theorems 2 and 3 every n-vertex plane graph G of treewidth at most three has a free collinear set R of cardinality |P |.Let < R be the total order associated with R. Since no two points in P have the same x-coordinate, there exists a bijective mapping

Figure 1 :
Figure 1: (a) A proper good curve λ (orange) for a plane graph G (black).(b) Augmentation of G with dummy vertices and edges.(c) A planar straight-line drawing of the augmented graph G.(d) Planar polyline (top) and straight-line (bottom) drawings of the original graph G.

4 u
u lies inside C vz and connects z Z−1 with p v z ; the curve λ 3 u coincides with λ v (H); nally, λ lies inside C uv and connects p u v with p uv .The curves λ 0 u , λ 2 u , and λ 4 u are constructed as in Lemma 1.

( 2 )
a(G) = c(G) + 2d(G) + 1.We use induction on m.If m = 1, then the statement is easily proved, as then the only internal vertex w of G is of type A, hence a(G) = 1 and c(G) = d(G) = 0.If m > 1, then the central vertex w of G is of one of types BD.Suppose rst that w is of type B. Also, suppose that G 1 has internal vertices; the other cases are analogous.Since w is of type B, we have a(G) = a(G 1 ), c(G) = c(G 1 ), and d(G) = d(G 1 ).Hence, a(G) = a(G 1 ) = c(G 1 ) + 2d(G 1 ) + 1 = c(G) + 2d(G) + 1;the second equality holds by induction.Suppose next that w is of type C. Also, suppose that G 1 and G 2 have internal vertices; the other cases are analogous.Since w is of type C, we have a the second equality holds by induction.Suppose nally that w is of type D. Then we have a the second equality holds by induction.

( 3 )
h(G) ≤ 2c(G) + 3d(G) + 1.We use induction on m.If m = 1, then the only internal vertex w of G is of type A, hence h(G) = 0 < 1 = 2c(G) + 3d(G) + 1.If m > 1, then the central vertex w of G is of one of types BD.
the second inequality holds by induction.
the second inequality follows by induction.Suppose next that w is of type D. Then we have a

m 8 .−3 8
It follows that one of λ u (G), λ v (G), and λ z (G) is a proper good curve passing through ninternal vertices of G.This concludes the proof of Theorem 2.

Figure 4 :
Figure 4: Cases for the proof of Lemma 4. The line is orange, the points in X S are green, and the points in X E are purple.(a) z and w have label =; (b) z has label = and w has label ↑; (c) z has label ↑ and w has label =; (d) z and w have label ↑; and (e) z has label ↑ and w has label ↓.

Theorem 4 .
Every n-vertex triconnected cubic plane graph admits a planar straight-line drawing with at least n 4 collinear vertices.

Figure 5 :
Figure 5: Illustration for the statement of Lemma 5.The gray region represents the interior of G.The curve λ is orange, the vertices in X are red squares, the intersection points between λ and β uv (G) are green circles, and the vertices u and v are black disks.

Figure 7 :
Figure 7: (a) G ab contains more than two non-trivial {a , b }-components.(b) G ab does not contain external vertices of G ab .(c) a and b both belong to τ ab (G ab ).

Figure 8 :
Figure 8: Base case for the proof of Lemma 5.

Figure 10 :
Figure 10: Case of the proof of Lemma 5.

Figure 11 :
Figure 11: Case 3 of the proof of Lemma 5. (a) Every vertex of H dierent from u and y 1 is in X .(b) H contains a vertex not in X ∪ {u, y 1 }.

Figure 15 :
Figure 15: (a) Drawing a curve γ of Type A. The region R(γ) is pink.The graph D * has vertices represented by white circles; the edges of D * in P are thick orange lines, while the edges of D * not in P are dashed black lines.(b) Drawing a curve γ of Type C. The region R(γ) is turquoise.The internal vertices of the path P in G i+1,j are black disks if they belong to the boundary of G i+1,j , or orange circles if they are internal vertices of G i+1,j .

Corollary 1 .
Every set P of at most n−3

8
then let H ∪ {v} be the subgraph of G composed of H and of the isolated vertex v.An H-bridge B is either a trivial H-bridge an edge of G not in H with both end-vertices in H or a non-trivial H-bridge a connected component of G − V (H) together with the edges from that component to H; the vertices in 1 z , λ 2 z .The curve λ 0 z lies inside C uz and connects p uz with p uz ; the curve λ 1 z coincides with λ z (H); the curve λ 2 z lies inside C vz and connects p vz with p vz .The curves λ 0 z and λ 2 z are constructed as in Lemma 1.The curve λ u (G) is constructed as follows.