Journal of Graph Algorithms and Applications Drawing Outer 1-planar Graphs with Few Slopes 708 Di Giacomo Et Al. Drawing Outer 1-planar Graphs with Few Slopes

A graph is outer 1-planar if it admits a drawing where each vertex is on the outer face and each edge is crossed by at most another edge. Outer 1-planar graphs are a superclass of the outerplanar graphs and a subclass of the planar partial 3-trees. We show that an outer 1-planar graph G of bounded degree ∆ admits an outer 1-planar straight-line drawing that uses O(∆) different slopes, which generalizes a previous result by Knauer et al. about the outerplanar slope number of outerplanar graphs [18]. We also show that O(∆ 2) slopes suffice to construct a crossing-free straight-line drawing of G; the best known upper bound on the planar slope number of planar partial 3-trees of bounded degree ∆ is O(∆ 5) as proved by Jelínek et al. [16].


Introduction
The slope number of a graph G is defined as the minimum number of distinct edge slopes required to construct a straight-line drawing of G. Minimizing the number of slopes used in a straight-line graph drawing is a desirable aesthetic requirement and an interesting theoretical problem which has received considerable attention since its first definition by Wade and Chu [21]. Let Δ be the maximum degree of a graph G and let m be the number of edges of G, clearly the slope number of G is at least Δ 2 and at most m. For non-planar graphs, there exist graphs with Δ ≥ 5 whose slope number is unbounded (with respect to Δ ) [3,19], while the slope number of graphs with Δ = 4 is unknown, and the slope number of graphs with Δ = 3 is four [18].
Concerning planar graphs, the planar slope number of a planar graph G is defined as the minimum number of distinct slopes required by any planar straight-line drawing of G (see, e.g., [9]). Keszegh, Pach and Pálvölgyi [14] prove that O (2 O(Δ ) ) is an upper bound and that 3Δ − 6 is a lower bound for the planar graphs of bounded degree Δ . The gap between upper and lower bound has been reduced for special families of planar graphs with bounded degree. Knauer, Micek and Walczak [15] prove that an outerplanar graph of bounded degree Δ ≥ 4 admits an outerplanar straight-line drawing that uses at most Δ − 1 distinct edge slopes, and this bound is tight. Jelínek et al. [13] prove that the slope number of the planar partial 3-trees of bounded degree Δ is O(Δ 5 ), while in [17] it is proved that all partial 2-trees of bounded degree Δ have O(Δ ) slope number. Di Giacomo et al. [7] show that planar graphs of bounded degree Δ ≤ 3 and at least five vertices have planar slope number four, which is worst case optimal.
The research in this paper is motivated by the following observations. The fact that the best known upper bound on the slope number is O(Δ 5 ) for planar partial 3-trees while it is O(Δ ) for partial 2-trees suggests to further investigate the planar slope number of those planar graphs whose treewidth is at most three. Also, the fact that nonplanar drawings may require a number of slopes that is unbounded in Δ while the planar slope number of planar graphs is bounded in Δ , suggests to study how many slopes may be needed to construct straight-line drawings that are "nearly-planar" in some sense, i.e. where only some types of edge crossing are allowed.
We study outer 1-planar graphs that are graphs which admit drawings where each edge is crossed at most once and each vertex is on the boundary of the outer face (see, e.g., [2,5,11]). In 2013, Auer et al. [2], and independently Hong et al. [11], presented a linear-time algorithm to test outer 1-planarity. Both algorithms produce an outer 1planar embedding of the graph if it exists. Given an outer 1-planar graph G, we define the outer 1-planar slope number of G, as the minimum number of distinct slopes required by any outer 1-planar straight-line drawing of G. We prove the following results.
1. The outer 1-planar slope number of outer 1-planar graphs with maximum degree Δ is at most 6Δ + 12 (Section 3). Since outerplanar drawings are a special case of the outer 1-planar drawings, this result extends the above mentioned upper bound on the planar slope number of outerplanar graphs [15]. 2. Outer 1-planar drawings are known to be planar graphs and they have treewidth at most three [2]. We study crossing-free straight-line drawings of outer 1-planar graphs of bounded degree Δ and show an O(Δ 2 ) upper bound to the planar slope number (Section 4). Hence, for this special family, we are able to reduce the general O(Δ 5 ) upper bound [13].
Our results are constructive and give rise to linear-time drawing algorithms. Also, it may be worth recalling that the study of the 1-planar graphs, i.e. those graphs that can be drawn with at most one crossing per edge, has received a lot of interest in the recent graph drawing literature (see, e.g., [1,4,8,10,12,16,20]).
In Section 2 we introduce preliminaries. Section 5 lists some open problems. For reasons of space some proofs are sketched or omitted.

Preliminaries and Basic Definitions
A drawing Γ of a graph G = (V, E) is a mapping of the vertices in V to points of the plane and of the edges in E to Jordan arcs connecting their corresponding endpoints but not passing through any other vertex. Also, no two edges that share an endpoint cross. Γ is a straight-line drawing if every edge is mapped to a straight-line segment. Γ is a planar drawing if no edge is crossed; it is a 1-planar drawing if each edge is crossed at most once. A planar graph is a graph that admits a planar drawing; a 1-planar graph is a graph that admits a 1-planar drawing.
A planar drawing of a graph partitions the plane into topologically connected regions, called faces. The unbounded region is called the outer face. A planar embedding of a planar graph is an equivalence class of planar drawings that define the same set of faces. The concept of planar embedding can be extended to 1-planar drawings as follows. In a . 1. Illustration of Properties 2-4. The pertinent graph of: (a) an R-node μ; (b) a P-node μ (case (ii) of Property 3);. (c) a P-node μ that is AOS with respect to s μ ; (d) An S-node ν with a child μ that is AOS with respect to s μ . Dashed edges cross in the embedding of the graph.
1-planar drawing Γ of a graph G each crossed edge is divided into two edge fragments. Also in this case, Γ partitions the plane into topologically connected regions, which we call faces. A 1-planar embedding of a 1-planar graph is an equivalence class of 1-planar drawings that define the same set of faces. An outer 1-planar drawing is a 1planar drawing with all vertices on the outer face. An outer 1-plane graph G is a graph with a given outer 1-planar embedding.
The slope s of a line is the angle that an horizontal line needs to be rotated counterclockwise in order to make it overlap with . The slope of a segment representing an edge in a straight-line drawing is the slope of the supporting line containing the segment.
Our drawing techniques use SPQR-trees, whose definition can be found in [6].
Properties of Outer 1-planar Graphs. The structural properties of outer 1-planar graphs have been studied in [2,11]. In this paragraph we derive properties that hold in the fixed outer 1-planar embedding setting and that easily follow from the results in [11]. In Section 4 we will use the same properties explaining how to adapt them to the planar embedding setting. The following property can be found as Lemma 1 in [11].
Property 1. Let G be an outer 1-plane graph. If G is triconnected, then it is isomorphic to K 4 and it has exactly one crossing.
In what follows we consider a biconnected outer 1-plane graph G and its SPQR-tree T . Let μ be a node of T , the pertinent graph G μ of μ is the subgraph of G whose SPQRtree (with respect to the reference edge e of μ) is the subtree of T rooted at μ. Notice that the edge e is not part of G μ . From now on we assume G μ to be an outer 1-plane graph using the embedding induced from G. We give the following definition [11].

Definition 1.
A node μ of T is one sided with respect to its poles s μ and t μ , or simply OS, if the edge (s μ ,t μ ) is on the outer face of G μ .
Furthermore, we consider T to be rooted at a Q-node ρ whose (only) child is denoted by ξ . In particular, we choose ρ to be associated with an edge that is not crossed and that belongs to the boundary of the outer face of G. It can be shown that such an edge always exists. This choice implies that ξ is OS by definition. The next property derives from Lemma 5 in [11] and defines the structure of the skeleton of R-nodes, see also Figure 1(a).
Property 2. Let μ be an R-node of T . Then: (i) The skeleton σ (μ) is isomorphic to K 4 and it has one crossing; (ii) The children of μ are all OS; (iii) Two children of μ are Q-nodes whose associated edges cross each other in G μ .
Observe that if μ is an R-node of T , then it is always OS. In order to handle P-nodes, we first need to define a special kind of S-nodes [11]. Definition 2. Let μ be an S-node of T . Let η be the unique child of μ having s μ as a pole, and let η be the unique child of μ having t μ as a pole. Node μ has a tail at s μ (t μ ), if η (η ) is a Q-node.
The next property derives from Lemma 6 in [11], see also Figure 1 Property 3. Let μ be an OS P-node of T . One of the following cases holds: (i) μ has two children one of which is a Q-node and the other one is OS; (ii) μ has two children and none of them is a Q-node. Then both are OS S-nodes, one of them has a tail at s μ , and the other one has a tail at t μ . Also, the two edges associated with these two tails cross each other in G; (iii) μ has three children and one of them is a Q-node. For the remaining two children case (ii) applies. Property 3 is restricted to P-nodes that are OS. However, an internal P-node μ (different from ξ ) might not have the edge (s μ ,t μ ) on the outer face of G μ [11], see also Figure 1(c) for an illustration.
Definition 3. Let μ be a P-node of T different from ξ . Node μ is almost one sided with respect to s μ (t μ ), or simply AOS with respect to s μ (t μ ), if μ has 2 ≤ k ≤ 4 children, one of them is an S-node with a tail at s μ (t μ ), and for the remaining children one of the following cases applies: (i) If k = 2, then the other child is OS; (ii) If k > 2, all and only the cases in Property 3 can apply for the remaining k − 1 children.
Let μ be AOS with respect to s μ (t μ ), then, in order to guarantee that the graph is outer 1-planar, the edge associated with the tail at s μ (t μ ) crosses another edge, represented by a Q-node ψ in T , having t μ (s μ ) as an end-vertex. This implies that in fact, μ and ψ are two children of an S-node ν in T [11] (see also Figure 1(d)). This observation will be used in Section 3 and in the next property, that is derived from Lemma 7 in [11].
Property 4. Let μ be an S-node of T . Let η 1 , η 2 ,... ,η k be the k children of μ in T , such that t η i−1 = s η i , for i = 2,... ,k. For each 1 ≤ i ≤ k, one of the following cases applies: An immediate observation from these properties is that every

The Outer 1-planar Slope Number
In this section we first present an algorithm, called BO1P-DRAWER, that takes as input a biconnected outer 1-plane graph G with maximum degree Δ , and returns a straightline drawing Γ of G that uses at most 6Δ slopes. This result is then extended to simply connected graphs with a number of slopes equal to 6Δ + 12.
A Universal Set of Slopes. We define a universal set of slopes used by algorithm BO1P-DRAWER to draw every biconnected outer 1-plane graph G with maximum degree Δ . Let α = π 2Δ and observe that 0 < α ≤ π 6 when Δ ≥ 3. We call blue slopes the set of slopes defined as b i = (i − 1)α, for i = 1, 2,...,2Δ . For each of the 2Δ blue slopes, we also define two red slopes as r − i = b i − ε and r + i = b i + ε, for i = 1, 2,...,2Δ , where the value of ε only depends on Δ . The union of the blue and red slopes defines the universal set of slopes S Δ of size 6Δ . We choose ε as follows: 1+2 tan(2α) tan (α)−2 tan(α) tan (α) . The reason of this choice will be clarified in the proof of Lemma 3. Clearly, ε depends only on Δ and it is possible to see that it is a positive value.
Algorithm Overview. Algorithm BO1P-DRAWER exploits SPQR-trees and the structural properties presented in Section 2. It takes as input a biconnected outer 1-plane graph G with maximum degree Δ and returns a straight-line drawing Γ of G that uses only slopes in S Δ . We first construct the SPQR-tree T rooted at a Q-node ρ, whose (only) child is denoted by ξ . Moreover, the edge associated with ρ is not crossed and belongs to the boundary of the outer face of G. Then we draw G by visiting T bottomup, handling ρ and ξ together as a special case. At each step we process an internal node μ of T and compute a drawing Γ μ of its pertinent graph G μ by properly combining the already computed drawings of the pertinent graphs of the children of μ. Let s μ and t μ be the poles of μ. With a slight overload of notation for the symbol Δ , we denote by Δ (s μ ) and Δ (t μ ) the degree of s μ and t μ in G μ , respectively. For each drawing Γ μ we aim at maintaining the following three invariants. I1. Γ μ is outer 1-plane with respect to the embedding of G μ . I2. Γ μ uses only slopes in S Δ . I3. Γ μ is contained in a triangle τ μ such that s μ and t μ are placed at the corners of its base. Also, β μ < (Δ (s μ ) + 1)α and γ μ < (Δ (t μ ) + 1)α, where β μ and γ μ are the internal angles of τ μ at s μ and t μ .
We now explain how to compute a drawing Γ μ of G μ , by combining the drawings Γ η 1 ,Γ η 2 ,... ,Γ η h of the pertinent graphs G η 1 , G η 2 ,... ,G η h of the children η 1 , η 2 ,... ,η h of μ. To this aim, the drawings Γ η 1 ,Γ η 2 ,... ,Γ η h are possibly manipulated. First, observe that the triangle τ η j (1 ≤ j ≤ h) can be arbitrarily scaled without modifying the slopes used in Γ η j . Furthermore, due to the symmetric choice of the blue and red slopes, if we rotate τ η j by an angle c · α, with c integer, the resulting drawing maintains invariant I2. Namely each blue slope b i , for i = 1, 2,...,2Δ , used in τ η j will be transformed in Similarly, any red slope will be transformed into another red slope. Moreover, let η 1 and η 2 be two children of μ. When we draw G η 1 and G η 2 , although they may share one or both the poles, we consider each graph to have its own copy of its poles. Then, when computing Γ μ , we say that we attach Γ η 1 to Γ η 2 if they share either two poles (this is always true when μ is a P-node) or one pole (this may happen when μ is either an S-or R-node), meaning that we may scale, shift and rotate Γ η 1 or Γ η 2 in such a way that the points representing the shared poles on the drawing coincide.
As observed in Section 2, all the internal nodes of T are OS except for some P-nodes which are AOS. Let μ be any of these P-nodes, we know that μ is one of the children of an S-node, say ν, and it shares a pole with a Q-node, denoted by η (also a children of ν). We replace μ and η in T with a new node ϕ, that, for the sake of description, is called an S * -node. Also, the children of μ become children of ϕ. If μ and η were the only two children of ν, then we also replace ν with ϕ. The pertinent graph of ϕ is It is easy to see that ϕ is OS. By means of this transformation we can consider only P-nodes that are OS. Similarly we can handle just S-nodes whose children are OS. In what follows we distinguish between S-, P-, S * -, and R-nodes different from ξ .
Proof sketch: The drawings of the pertinent graphs of the children η 1 , η 2 ,... ,η k of μ are attached to each other as shown in Figure 2(a). Clearly all invariants hold.
Proof sketch: Recall that, thanks to the definition of S * -nodes, here we need to only handle only P-nodes that are OS. By Property 3, one of the following cases applies: (i) μ has two children one of which is a Q-node and the other one is OS. (ii) μ has two children and none of them is a Q-node. Then both are OS S-nodes, one of them has a tail at s μ , and the other one has a tail at t μ . Also, the two edges associated with these two tails cross each other in G. (iii) μ has three children and one of them is a Q-node. For the remaining two children case (ii) applies. Case (i) can be easily handled as shown in Figure 2(b). Consider case (ii) and let η 1 be the child of μ that is an S-node with a tail at t μ , and η 2 be the child of μ that is an Snode with a tail at s μ . Refer to Figure 2(c). Recall that s η 1 = s η 2 = s μ and t η 1 = t η 2 = t μ . We modify the drawing Γ η 1 as follows. We first rotate Γ η 1 so that the segment s η 1 t η 1 uses the blue slope b 2 . Then we redraw the tail of η 1 using the red slope r + 2Δ = b 2Δ + ε and so that s η 1 and t η 1 are horizontally aligned. Similarly, we modify the drawing Γ η 2 . We rotate Γ η 2 so that the segment s η 2 t η 2 uses the blue slope b 2Δ and redraw the tail of η 2 using the red slope r − 2 = b 2 − ε and so that s η 2 and t η 2 are horizontally aligned. Finally, we attach Γ η 1 and Γ η 2 (possibly scaling one of them). Invariants I1. and I2. hold by construction. Also, Γ μ is contained in a triangle τ μ such that s μ and t μ are placed at the corners of its base. Moreover, we have that Δ (s μ ) = Δ (s η 1 ) + 1, and β μ = β η 1 + α < Δ (s η 1 + 1)α + α = Δ (s η 1 + 2)α = Δ (s μ + 1)α. Similarly, Δ (t μ ) = Δ (t η 2 ) + 1, and γ μ = γ η 2 + α < Δ (t η 2 + 1)α + α = Δ (t η 2 + 2)α = Δ (t μ + 1)α. Hence, Invariant I3. holds. In case (iii) we can use the same construction as in case (ii). Notice that the edge (s μ ,t μ ) can be safely drawn using the horizontal blue slope b 1 . All invariants hold. Lemma 3. Let μ be an S * -node different from ξ . Then G μ admits a straight-line drawing Γ μ that respects Invariants I1., I2. and I3.
Proof. Refer to Figure 2(d). Denote by η the child of μ that is an S-node with a tail at either s μ or t μ . Suppose that η has a tail at t μ (the case when the tail is at s μ is symmetric). Denote by ψ the child of μ that is a Q-node having t ψ = s η and s ψ = s μ as poles. Finally denote by η 1 , η 2 ,... ,η k the remaining children of μ. Recall that s η 1 = s η i = s η k and that t η 1 = t η i = t η k . If k = 1, we first rotate Γ η 1 so that the segment s η 1 t η 1 uses the blue slope b 2Δ . If k > 1, we combine the drawings Γ η 1 ,Γ η 2 ,... ,Γ η k with the same technique described for P-nodes (recall that indeed they were children of a Pnode before the creation of the S * -node), and, again, we rotate the resulting drawing so that the base of its bounding triangle uses the blue slope b 2Δ . Then we attach Γ η to Γ η 1 (after Γ η has been horizontally flipped). Also, we scale Γ η so that its tail can be redrawn by using the red slope r + 2Δ and such that t η = t μ coincides with t η 1 = t η k . Finally, we redraw the edge associated with ψ, starting from the point representing t ψ = s η , using the red slope r − 2 and stretch it enough that s ψ = s μ and t μ are horizontally aligned. See also Figure 2(d) for an illustration. Invariants I1. and I2. hold by construction. Consider now Invariant I3.. By construction Γ μ is contained in a triangle τ μ such that s μ and t μ are placed at the corners of its base. For the sake of description, in what follow we still denote by Γ η the drawing of G η minus the tail of η (i.e., minus an edge), and as τ η the surrounding triangle of Γ η . To prove the second part of Invariant I3., we should prove that the line passing through s μ with slope b 3 = 2α does not cross the drawing of Γ η , i.e., is such that Γ η is placed in the half-plane H defined by and containing the segment s μ t μ . Denote by δ x the horizontal distance between the point where s μ is drawn and the leftmost endpoint of τ η . Also, denote by h η the height of τ η . Our condition is satisfied if the following inequality holds tan (2α)δ x ≥ tan (α)δ x + h η . Let w η be the length of the base of τ η , in the worst case (the case that maximizes h η ), we have that h η = w η 2 1 tan (α) , which means that the degree of the two vertices placed as endpoints of the base of τ η is Δ . Moreover, it is possible to see that . Substituting w η in h η and h η in the above inequality we have: 2 tan (α−ε) tan (α) . With some manipulation we get: tan (α − ε) ≥ tan (α) 2 tan(2α) tan (α)−2 tan(α) tan (α)+1 . Now, since the tangent function is strictly increasing in (− π 2 , π 2 ), we have: ε ≤ α − arctan tan (α) 2 tan(2α) tan (α)−2 tan (α) tan (α)+1 . Since the value of ε has been chosen equal to the right-hand side of the above inequality, the inequality holds. Hence, β μ < 2α = (Δ (s μ ) + 1)α (since Δ (s μ ) = 1). With a symmetric argument one can prove that the line passing through t μ with slope b 2Δ −1 = (Δ −1)π Δ does not cross the drawing of Γ η . Since Δ (t μ ) = Δ (t η k ) + 1, and γ μ = γ η k + α < (Δ (t η k ) + 1)α + α = (Δ (t η k ) + 2)α = (Δ (t μ ) + 1)α, Invariant I3. holds.

Lemma 4. Let μ be an R-node different from ξ . Then G μ admits a straight-line drawing Γ μ that respects Invariants I1., I2. and I3.
Proof. Refer to Figure 2(e). Recall that, by Property 2, (i) the skeleton σ (μ) is isomorphic to K 4 and it has one crossing; (ii) the children of μ are all OS; (iii) two children of μ are Q-nodes whose associated edges cross each other in G μ . Hence, denote by η 1 , η 2 , η 3 the three children of μ whose associated virtual edges lie on the boundary of the outer face of σ (μ) with s μ = s η 1 , t η 1 = s η 2 , t η 2 = s η 3 , and t η 3 = t μ . Also, denote by η 4 and η 5 the two children of μ that are Q-nodes whose associated edges cross each other in G μ , and so that the poles of η 4 coincides with t η 1 and t η 3 , while the poles of η 5 coincides with t η 2 and s η 1 . We rotate Γ η 1 in such a way that the segment s ν 1 t ν 1 uses the blue slope b 2 . Similarly, we rotate Γ η 3 in such a way that the segment s η 3 t η 3 uses the blue slope b 2Δ . Furthermore, we scale one of the two drawings so that t η 1 and s η 3 are horizontally aligned. Moreover, we redraw the edge associated with η 4 by using the red slope r + 2Δ and we redraw the edge associated with η 5 by using the red slope r − 2 . Observe that, attaching η 4 and η 5 to η 1 and η 3 , the length of the segment t η 1 s η 3 is determined. Thus, we attach Γ η 2 so that s η 2 coincides with t η 1 and that t η 2 coincides with s η 3 .

Lemma 5.
Let ρ be the root of T and let ξ be its unique child. Graph G = G ρ ∪ G ξ admits a straight-line drawing Γ that respects Invariants I1., I2. and I3.
Proof sketch: It is possible to prove that at least one edge (s,t) of the outer face of G is not crossed. If we root T at the Q-node associated with (s,t), the root's child ξ is OS and a drawing of G ρ ∪ G ξ can be computed as in Lemmas 1, 2, 3, and 4. Lemma 6. Let G be a biconnected outer 1-plane graph with n vertices and with maximum degree Δ . G admits an outer 1-planar straight-line drawing that maintains the given outer 1-planar embedding, and that uses at most 6Δ slopes. Also, this drawing can be computed in O(n) time.
A simply connected outerplane graph can be augmented (in linear time) into a biconnected outerplane graph by adding edges so that the maximum degree is increased by at most two. This technique can be directly applied also to outer 1-plane graphs. Theorem 1. Let G be an outer 1-plane graph with n vertices and with maximum degree Δ . G admits an outer 1-planar straight-line drawing that maintains the given outer 1planar embedding, and that uses at most 6Δ + 12 slopes. Also, this drawing can be computed in O(n) time.

The Planar Slope Number
In this section we describe an algorithm, called BP-DRAWER, that computes a planar drawing of an outer 1-planar graph G, using at most 4Δ 2 − 4Δ slopes. This result is then extended to simply connected graphs with a number of slopes equal to 4Δ 2 + 12Δ + 8.
A Universal Set of Slopes. We start by defining a universal set of slopes that are used by algorithm BP-DRAWER. Let θ = π 4Δ and observe that 0 < θ ≤ π 12 when Δ ≥ 3. We call green slopes the set of slopes defined as g i = (i − 1)θ , for i = 1, 2,...,4Δ . For each green slope g i , we define Δ − 1 yellow slopes as y i, j = g i + arctan tan(g 4Δ ) tan(g 3 ) tan(g j ) with j = 3Δ ,... ,4Δ − 2. The reason of this choice will be clarified in the proof of Lemma 10. The union of the green and yellow slopes defines the universal set of slopes T Δ . It is possible to see that g i < y i, j < g i+1 , for each 1 ≤ i < 4Δ and 3Δ ≤ j ≤ 4Δ − 2.
Algorithm Overview. Algorithm BP-DRAWER takes as input a biconnected outer 1plane graph G with maximum degree Δ and returns a planar straight-line drawing Γ of G that uses only slopes in T Δ . As in Section 3 we construct the SPQR-tree T of G rooted at a Q-node associated with an edge that is not crossed and belongs to the boundary of the outer face of G in the outer 1-planar embedding of G. Then we draw G by visiting T bottom-up. At each internal node μ of T we compute a drawing Γ μ of G μ by combining the already computed drawings of the pertinent graphs of the children of μ. For each drawing Γ μ we maintain the following three invariants: Ia. Γ μ is planar. Ib. Γ μ uses only slopes in T Δ . Ic. Γ μ is contained in a triangle τ μ such that s μ and t μ are placed at the corners of its base. Also, β μ < (Δ (s μ ) − 1)θ and γ μ < (Δ (t μ ) − 1)θ , where β μ and γ μ are the internal angles of τ μ at s μ and t μ , respectively.
As in Section 3 the root ρ of T and its unique child ξ will be handled in a special way. Also, in order to construct Γ μ we may shift, scale and rotate the drawings of the pertinent graphs of the children of μ. We observe that if we rotate τ μ by an angle c · θ , with c integer, the resulting drawing maintains invariant Ib. Namely each green slope g i , for i = 1, 2,... ,4Δ , used in τ μ will be transformed in another green slope g i+c = g i + c · θ = (i − 1 + c)θ , where i + c is considered modulo 4Δ . Similarly, any yellow slope y i, j will be transformed into another yellow slope y i+c, j .
Before describing how the drawing of the pertinent graph of each node μ is obtained by combining the drawing of the pertinent graphs of its children, we observe that the structural properties described in Properties 2, 3, or 4 hold, depending on the type of μ. However, since we want to produce a planar drawing, our algorithm embeds each pertinent graph in a planar way. One of the consequence of this fact is that we no longer need to introduce S * -nodes; namely, the P-nodes that are AOS in the outer 1-planar embedding must be embedded in a planar way and therefore they do not need to be handled in a special way anymore. On the other hand, we need to distinguish between R-nodes whose poles are adjacent in G and R-nodes whose poles are not adjacent in G. For this reason we introduce R * -nodes. Let μ be an R-node; if the poles s μ and t μ of μ are adjacent in G, then the parent ν of μ is a P-node that has (at least) another child η that is a Q-node (the edge associated with η is (s μ ,t μ )). We replace μ and η in T with a new node ϕ, that, for the sake of description, is called an R * -node. Also, the children of μ become children of ϕ. If μ and η were the only two children of ν, then we also . 3. The planar drawing of the pertinent graph of: (a) a P-node with two children such that none of them is a Q-node; (b) a P-node with three children, one of which is a Q-node; (c) a P-node that is AOS in the outer 1-planar embedding of G; (d) an R-node; (e) an R * -node. (f) Illustration for the proof of Lemma 10. replace ν with ϕ. The pertinent graph of ϕ is G ϕ = G μ ∪ G η , and the reference edge of ϕ is (s μ ,t μ ). We now explain how the different types of node are handled.
The proof of next lemmas are omitted. An illustration of how Γ μ is constructed is shown in Figures 2(a) and 3.

Lemma 7.
Let μ be an S-node different from ξ . Then G μ admits a straight-line drawing Γ μ that respects Invariants Ia., Ib. and Ic. Lemma 8. Let μ be a P-node different from ξ . Then G μ admits a straight-line drawing Γ μ that respects Invariants Ia., Ib. and Ic. Lemma 9. Let μ be an R-node different from ξ . Then G μ admits a straight-line drawing Γ μ that respects Invariants Ia., Ib. and Ic. Lemma 10. Let μ be an R * -node different from ξ . Then G μ admits a straight-line drawing Γ μ that respects Invariants Ia., Ib. and Ic.
Proof. Since μ is an R * -node, it is obtained by merging an R-node μ and a Q-node representing the edge (s μ ,t μ ). By Property 2, the skeleton σ (μ ) of μ is isomorphic to K 4 and two children of μ are Q-nodes. The two edges corresponding to these Q-nodes do not share an end vertex and each one of them is incident to a distinct pole of μ. Let η 1 , η 2 , η 3 , η 4 , and η 5 be the children of μ ; we assume that η 4 and η 5 are the two Qnodes. Also, μ has a sixth child η 6 that is a Q-node corresponding to the edge (s μ ,t μ ). We assume that s μ = s η 1 = s η 4 , t μ = t η 3 = t η 5 , t η 1 = t η 2 = s η 5 , and t η 4 = s η 2 = s η 3 . We construct a drawing of G μ as follows (see Figure 3(e)). We rotate Γ η 3 so that the segment s η 3 t η 3 uses the green slope g 4Δ , and draw the edge associated with η 5 as a segment whose slope is the green slope (4Δ − Δ (t η 3 ))θ and whose length is such that s η 5 is vertically aligned with s η 3 . We rotate Γ η 2 so that the segment s η 2 t η 2 uses the green slope g 2Δ +1 = π 2 . We then attach Γ η 2 , Γ η 3 , and Γ η 5 (possibly scaling some of them). We draw the edge corresponding to η 6 with the horizontal slope g 1 and stretch it so that s η 6 = s μ belongs to the line with slope g 2 passing through s η 5 . We now rotate Γ η 1 so that the segment s η 1 t η 1 uses the green slope g 2 and attach it to Γ η 5 and Γ η 6 . Finally, the edge corresponding to η 4 is drawn as the segment s μ s η 3 . Invariant Ia. holds because the drawings Γ η 1 , Γ η 2 , Γ η 3 , Γ η 4 , Γ η 5 , and Γ η 6 do not intersect each other except at common endpoints. About this, let τ be the triangle defined by the three vertices s μ , s η 3 , and s η 5 ; it is easy to see that Γ η 2 is completely contained inside τ except for the segment s η 3 s η 5 that Γ η 2 shares with τ. Namely the angle inside τ at s η 3 is π 2 + θ , while the angle inside τ at s η 5 is at least π 4 (because the angle inside τ at s μ is θ and 2θ < π 4 ). Since β η 2 < π 4 and γ η 2 < π 4 , the triangle τ η 2 is completely inside τ except for the vertical side shared by the two triangles. Concerning Invariant Ib., we observe that Γ η 1 , Γ η 2 , Γ η 3 , Γ η 4 , and Γ η 5 are rotated by an angle that is a multiple of θ and therefore Ib. holds by construction for each of them. We now show that the slope φ of the edge corresponding to η 4 is in fact either a green slope or a yellow one (refer to Figure 3(f)). Let δ x 1 be the horizontal distance between s η 3 and t μ and let δ x 2 be the horizontal distance between s μ and s η 3 . By simple trigonometry we have δ x 1 tan(g 4Δ ) = δ x 2 tan(φ ) and δ x 1 tan(g j ) = δ x 2 tan(g 3 ), where g j is the slope of the segment representing the edge corresponding to η 5 (and therefore j = 4Δ − Δ (t η 3 )). From the two previous equations we obtain tan(φ ) = tan(g 4Δ ) tan(g 3 )
By Lemmas 7,8,9,10, and 11, we can prove the following lemma. Lemma 12. Let G be a biconnected outer 1-plane graph with n vertices and with maximum degree Δ . G admits a planar straight-line drawing that uses at most 4Δ 2 − 4Δ slopes. Also, this drawing can be computed in O(n) time.
The result above can be extended to simply connected outer 1-planar graph with the same technique described in Section 3. We obtain the following theorem. Theorem 2. Let G be an outer 1-plane graph with n vertices and with maximum degree Δ . G admits a planar straight-line drawing that uses at most 4Δ 2 + 12Δ + 8 slopes. Also, this drawing can be computed in O(n) time.

Open Problems
An interesting open problem motivated by our result of Section 3 is whether the 1planar slope number of 1-planar straight-line drawable graphs (not all 1-planar graphs admit a 1-planar straight-line drawing [12]), is bounded in Δ or not. A second problem is whether the quadratic upper bound of Section 4 is tight or not. Finally, it could be interesting to further explore trade-offs between slopes and crossings, e.g., can we draw planar partial 3-trees with o(Δ 5 ) slopes and a constant number of crossings per edge?