Drawing Subcubic 1-Planar Graphs with Few Bends, Few Slopes, and Large Angles

We show that the 1-planar slope number of 3-connected cubic 1-planar graphs is at most 4 when edges are drawn as polygonal curves with at most 1 bend each. This bound is obtained by drawings whose vertex and crossing resolution is at least $\pi/4$. On the other hand, if the embedding is fixed, then there is a 3-connected cubic 1-planar graph that needs 3 slopes when drawn with at most 1 bend per edge. We also show that 2 slopes always suffice for 1-planar drawings of subcubic 1-planar graphs with at most 2 bends per edge. This bound is obtained with vertex resolution $\pi/2$ and the drawing is RAC (crossing resolution $\pi/2$). Finally, we prove lower bounds for the slope number of straight-line 1-planar drawings in terms of number of vertices and maximum degree.


Introduction
A graph is 1-planar if it can be drawn in the plane such that each edge is crossed at most once.The notion of 1-planarity naturally extends planarity and received considerable attention since its first introduction by Ringel in 1965 [33], as witnessed by recent surveys [14,27].Despite the efforts made in the study of 1-planar graphs, only few results are known concerning their geometric representations (see, e.g., [1,4,7,11]).In this paper, we study the existence of 1-planar drawings that simultaneously satisfy the following properties: edges are polylines using few bends and few distinct slopes for their segments, edge crossings occur at large angles, and pairs of edges incident to the same vertex form large angles.For example, Fig. 1d shows a 1-bend drawing of a 1-planar graph (i.e., a drawing in which each edge is a polyline with at most one bend) using 4 distinct slopes, such that edge crossings form angles at least π/4, and the angles formed by edges incident to the same vertex are at least π/ 4. In what follows, we briefly recall known results concerning the problems of computing polyline drawings with few bends and few slopes or with few bends and large angles.
Related work.The k-bend (planar) slope number of a (planar) graph G with maximum vertex degree ∆ is the minimum number of distinct edge slopes needed to compute a (planar) drawing of G such that each edge is a polyline with at most k bends.When k = 0, this parameter is simply known as the (planar) slope arXiv:1808.08496v1[cs.CG] 26 Aug 2018 number of G. Clearly, if G has maximum vertex degree ∆, at least ∆/2 slopes are needed for any k.While there exist non-planar graphs with ∆ ≥ 5 whose slope number is unbounded with respect to ∆ [3,32], Keszegh et al. [25] proved that the planar slope number is bounded by 2 O(∆) .Several authors improved this bound for subfamilies of planar graphs (see, e.g., [22,26,28]).
Of particular interest for us is the k-bend 1-planar slope number of 1-planar graphs, i.e., the minimum number of distinct edge slopes needed to compute a 1-planar drawing of a 1-planar graph such that each edge is a polyline with at most k ≥ 0 bends.Di Giacomo et al. [12] proved an O(∆) upper bound for the 1-planar slope number (k = 0) of outer 1-planar graphs, i.e., graphs that can be drawn 1-planar with all vertices on the external boundary.
Finally, the vertex resolution and the crossing resolution of a drawing are defined as the minimum angle between two consecutive segments incident to the same vertex or crossing, respectively (see, e.g., [17,20,30]).A drawing is RAC (right-angle crossing) if its crossing resolution is π/2.Eades and Liotta proved that 1-planar graphs may not have straight-line RAC drawings [18], while Chaplick et al. [8] and Bekos et al. [4] proved that every 1-planar graph has a 1-bend RAC drawing that preserves the embedding.
Our contribution.We prove upper and lower bounds on the k-bend 1-planar slope number of 1-planar graphs, when k ∈ {0, 1, 2}.Our results are based on techniques that lead to drawings with large vertex and crossing resolution.
In Section 3, we prove that every 3-connected cubic 1-planar graph admits a 1-bend 1-planar drawing that uses at most 4 distinct slopes and has both vertex and crossing resolution π/4.In Section 4, we show that every subcubic 1-planar graph admits a 2-bend 1-planar drawing that uses at most 2 distinct slopes and has both vertex and crossing resolution π/2.These bounds on the number of slopes and on the vertex/crossing resolution are clearly worst-case optimal.In Section 5.1, we give a 3-connected cubic 1-plane graph for which any embeddingpreserving 1-bend drawing uses at least 3 distinct slopes.The lower bound holds even if we are allowed to change the outer face.In Section 5.2, we present 2-connected subcubic 1-plane graphs with n vertices such that any embeddingpreserving straight-line drawing uses Ω(n) distinct slopes, and 3-connected 1plane graphs with maximum degree ∆ ≥ 3 such that any embedding-preserving straight-line drawing uses at least 9(∆ − 1) distinct slopes, which implies that at least 18 slopes are needed if ∆ = 3. Preliminaries can be found in Section 2, while open problems are in Section 6.

Preliminaries
We only consider simple graphs with neither self-loops nor multiple edges.A drawing Γ of a graph G maps each vertex of G to a point of the plane and each edge to a simple open Jordan curve between its endpoints.We always refer to simple drawings where two edges can share at most one point, which is either a common endpoint or a proper intersection.A drawing divides the plane into topologically connected regions, called faces; the infinite region is called the outer face.For a planar (i.e., crossing-free) drawing, the boundary of a face consists of vertices and edges, while for a non-planar drawing the boundary of a face may also contain crossings and parts of edges.An embedding of a graph G is an equivalence class of drawings of G that define the same set of faces and the same outer face.A (1-)plane graph is a graph with a fixed (1-)planar embedding.Given a 1-plane graph G, the planarization G * of G is the plane graph obtained by replacing each crossing of G with a dummy vertex.To avoid confusion, the vertices of G * that are not dummy are called real.Moreover, we call fragments the edges of G * that are incident to a dummy vertex.The next lemma will be used in the following and can be of independent interest, as it extends a similar result by Fabrici and Madaras [19].The proof is given in Appendix A.
Lemma 1.Let G = (V, E) be a 1-plane graph and let G * be its planarization.We can re-embed G such that each edge is still crossed at most once and (i) no cutvertex of G * is a dummy vertex, and (ii A drawing Γ is straight-line if all its edges are mapped to segments, or it is k-bend if each edge is mapped to a chain of segments with at most k > 0 bends.The slope of an edge segment of Γ is the slope of the line containing this segment.For convenience, we measure the slopes by their angle with respect to the x-axis.Let S = {α 1 , . . ., α t } be a set of t distinct slopes.The slope number of a k-bend drawing Γ is the number of distinct slopes used for the edge segments of Γ .An edge segment of Γ uses the north (N) port (south (S) port) of a vertex v if it has slope π/2 and v is its bottommost (topmost) endpoint.We can define analogously the west (W) and east (E) ports with respect to the slope 0, the north-west (NW) and south-east (SE) ports with respect to slope 3π/4, and the south-west (SW) and north-east (NE) ports with respect to slope π/4.Any such port is free for v if there is no edge that attaches to v by using it.
We will use a decomposition technique called canonical ordering [24].Let G = (V, E) be a 3-connected plane graph.Let δ = {V 1 , . . ., V K } be an ordered Each G i is 2-connected and internally 3-connected, that is, removing any two interior vertices of G i does not disconnect it.(v) For each i ∈ {2, . . ., K − 1}, one of the following conditions holds: (a) V i is a singleton v i that lies on C i and has at least one neighbor in In particular, every singleton has at least two predecessors and at least one successor, while every vertex in a chain has either zero or one predecessor and at least one successor.Kant [24] proved that a canonical ordering of G always exists and can be computed in O(n) time; the technique in [24] is such that one can arbitrarily choose two adjacent vertices u and w on the outer face so that u = v 1 and w = v 2 in the computed canonical ordering.
An n-vertex planar st-graph G = (V, E) is a plane acyclic directed graph with a single source s and a single sink t, both on the outer face [10].An st-ordering of G is a numbering σ : V → {1, 2, . . ., n} such that for each edge (u, v) ∈ E, it holds σ(u) < σ(v) (thus σ(s) = 1 and σ(t) = n).For an st-graph, an st-ordering can be computed in O(n) time (see, e.g., [9]) and every biconnected undirected graph can be oriented to become a planar st-graph (also in linear time).

1-bend Drawings of 3-connected cubic 1-planar graphs
Let G be a 3-connected 1-plane cubic graph, and let G * be its planarization.We can assume that G * is 3-connected (else we can re-embed G by Lemma 1).We choose as outer face of G a face containing an edge (v 1 , v 2 ) whose vertices are both real (see Fig. 1a).Such a face exists: If G has n vertices, then G * has fewer than 3n/4 dummy vertices because G is subcubic.Hence we find a face in G * with more real than dummy vertices and hence with two consecutive real vertices.Let δ = {V 1 , . . ., V K } be a canonical ordering of G * , let G i be the graph obtained by adding the first i sets of δ and let C i be the outer face of G i .
Note that a real vertex v of G i can have at most one successor w in some set V j with j > i.We call w an L-successor (resp., R-successor ) of v if v is the leftmost (resp., rightmost) neighbor of V j on C i .Similarly, a dummy vertex x of G i can have at most two successors in some sets V j and V l with l ≥ j > i.
In both cases, a vertex v of G i having a successor in some set V j with j > i is called attachable.We call v L-attachable (resp., R-attachable) if v is attachable and has no L-successor (resp., R-successor) in G i .We will draw an upward edge at u with slope π/4 (resp., 3π/4) only if it is L-attachable (resp., R-attachable).Let u and v be two vertices of C i , for i > 1. Denote by P i (u, v) the path of C i having u and v as endpoints and that does not contain (v 1 , v 2 ).Vertices u and v are consecutive if they are both attachable and if P i (u, v) does not contain any other attachable vertex.Given two consecutive vertices u and v of C i and an edge e of C i , a uv-cut of G i with respect to e is a set of edges of G i that contains both e and (v 1 , v 2 ) and whose removal disconnects G i into two components, one containing u and one containing v (see Fig. 1c).We say that u and v are L-consecutive (resp., R-consecutive) if they are consecutive, u lies to the left (resp., right) of v on C i , and u is L-attachable (resp., R-attachable).
We construct an embedding-preserving drawing Γ i of G i , for i = 2, . . ., K, by adding one by one the sets of δ.A drawing Γ i of G i is valid, if: P1 It uses only slopes in the set {0, π 4 , π 2 , 3π 4 }; P2 It is a 1-bend drawing such that the union of any two edge fragments that correspond to the same edge in G is drawn with (at most) one bend in total.A valid drawing Γ K of G K will coincide with the desired drawing of G, after replacing dummy vertices with crossing points.Construction of Γ 2 .We begin by showing how to draw G 2 .We distinguish two cases, based on whether V 2 is a singleton or a chain, as illustrated in Fig. 2. Construction of Γ i , for 2 < i < K.We now show how to compute a valid drawing of G i , for i = 3, . . ., K − 1, by incrementally adding the sets of δ.
We aim at constructing a valid drawing Γ i that is also stretchable, i.e., that satisfies the following two more properties; see Fig. 3.These two properties will be useful to prove Lemma 2, which defines a standard way of stretching a drawing by lengthening horizontal segments.P3 The edge (v 1 , v 2 ) is drawn with two segments s 1 and s 2 that meet at a point p.Segment s 1 uses the SE port of v 1 and s 2 uses the SW port of v 2 .Also, p is the lowest point of Γ i , and no other point of Γ i is contained by the two lines that contain s 1 and s 2 .P4 For every pair of consecutive vertices u and v of C i with u left of v on C i , it holds that (a) If u is L-attachable (resp., v is R-attachable), then the path P i (u, v) is such that for each vertical segment s on this path there is a horizontal segment in the subpath before s if s is traversed upwards when going from u to v (resp., from v to u); (b) if both u and v are real, then P i (u, v) contains at least one horizontal segment; and (c) for every edge e of P i (u, v) such that e contains a horizontal segment, there exists a uv-cut of G i with respect to e whose edges all contain a horizontal segment in Γ i except for (v 1 , v 2 ), and such that there exists a y-monotone curve that passes through all and only such horizontal segments and (v 1 , v 2 ).
Lemma 2. Suppose that Γ i is valid and stretchable, and let u and v be two consecutive vertices of C i .If u is L-attachable (resp., v is R-attachable), then it is possible to modify Γ i such that any half-line with slope π/4 (resp., 3π/4) that originates at u (resp., at v) and that intersects the outer face of Γ i does not intersect any edge segment with slope π/2 of P i (u, v).Also, the modified drawing is still valid and stretchable.
Proof sketch.Crossings between such half-lines and vertical segments of P i (u, v) can be solved by finding suitable uv-cuts and moving everything on the right/left side of the cut to the right/left.The full proof is given in Appendix B Let P be a set of ports of a vertex v; the symmetric set of ports P of v is the set of ports obtained by mirroring P at a vertical line through v.We say that Γ i is attachable if the following two properties also apply.P5 At any attachable real vertex v of Γ i , its N, NW, and NE ports are free.P6 Let v be an attachable dummy vertex of Γ i .If v has two successors, there are four possible cases for its two used ports, illustrated with two solid edges in Fig. 4a-d.If v has only one successor not in Γ i , there are eight possible cases for its three used ports, illustrated with two solid edges plus one dashed or one dotted edge in Fig. 4a-e (see Fig. 16 in Appendix B).Observe that Γ 2 , besides being valid, is also stretchable and attachable by construction (see also Fig. 2).Assume that G i−1 admits a valid, stretchable, and attachable drawing Γ i−1 , for some 2 ≤ i < K − 1; we show how to add the next set V i of δ so to obtain a drawing Γ i of G i that is valid, stretchable and attachable.We distinguish between the following cases.Case 1. V i is a singleton, i.e., V i = {v i }.Note that if v i is real, it has two neighbors on C i−1 , while if it is dummy, it can have either two or three neighbors on C i−1 .Let u l and u r be the first and the last neighbor of v i , respectively, when walking along C i−1 in clockwise direction from v 1 .We will call u l (resp., u r ) the leftmost predecessor (resp., rightmost predecessor ) of v i .Case 1.1.Vertex v i is real.Then, u l and u r are its only two neighbors in C i−1 .Each of u l and u r can be real or dummy.If u l (resp., u r ) is real, we draw (u l , v i ) (resp., (u r , v i )) with a single segment using the NE port of u l and the SW port of v i (resp., the NW port of u r and the SE port of v i ).If u l is dummy and has two successors not in Γ i−1 , we distinguish between the cases of Fig. 4 as shown in Fig. 5.The symmetric configuration of C3 is only used for connecting to u r .
If u l is dummy and has one successor not in Γ i−1 , we distinguish between the various cases of Fig. 4 as indicated in Fig. 6(see Fig. 16 for all cases in Appendix B).Observe that C1 requires a local reassignment of one port of u l .The edge (u r , v i ) is drawn by following a similar case analysis(depicted in Fig. 16 of Appendix B).Vertex v i is then placed at the intersection of the lines passing through the assigned ports, which always intersect by construction.In particular, the S port is only used when u l has one successor, but the same situation cannot occur when drawing (u r , v i ).Otherwise, there is a path of C i−1 from u l via its successor x on C i−1 to u r via its successor y on C i−1 .Note that x = y is possible but x = u r .Since the first edge on this path goes from a predecessor to a successor and the last edge goes from a successor to a predecessor, there has to be a vertex z without a successor on the path; but then u l and u r are not consecutive.To avoid crossings between Γ i−1 and the new edges (u l , v i ) and (u r , v i ), we apply Lemma 2 to suitably stretch the drawing.In particular, possible crossings can occur only with vertical edge segments of P i−1 (u l , u r ), because when walking along P i−1 (u l , u r ) from u l to u r we only encounter a Fig. 5: A real singleton when u l is dummy with two successors not in Γi−1 Fig. 6: Some cases for the addition of a real singleton when u l is dummy with one successor not in Γi−1 (possibly empty) set of segments with slopes in the range {3π/4, π/2, 0}, followed by a (possibly empty) set of segments with slopes in the range {π/2, π/4, 0}.Case 1.2.Vertex v i is dummy.By 1-planarity, the two or three neighbors of v i on C i−1 are all real.If v i has two neighbors, we draw (u l , v i ) and (u r , v i ) as shown in Fig. 7a, while if v i has three neighbors, we draw (u l , v i ) and (u r , v i ) as shown in Fig. 7b.Analogous to the previous case, vertex v i is placed at the intersection of the lines passing through the assigned ports, which always intersect by construction, and avoiding crossings between Γ i−1 and the new edges (u l , v i ) and (u r , v i ) by applying Lemma 2. In particular, if v i has three neighbors on C i−1 , say u l , w, and u r , by P4 there is a horizontal segment between u l and w, as well as between w and u r .Thus, Lemma 2 can be applied not only to resolve crossings, but also to find a suitable point where the two lines with slopes π/4 and 3π/4 meet along the line with slope π/2 that passes through w.Case 2. V i is a chain, i.e., V i = {v i 1 , v i 2 , . . ., v i l }.We find a point as if we had to place a vertex v whose leftmost predecessor is the leftmost predecessor of v i 1 and whose rightmost predecessor is the rightmost predecessor of v i l .We then draw the chain slightly below this point by using the same technique used to draw V 2 .Again, Lemma 2 can be applied to resolve possible crossings.
We formally prove the correctness of our algorithm in Appendix B.
Construction of Γ K .We now show how to add V K = {v n } to Γ K−1 so as to obtain a valid drawing of G K , and hence the desired drawing of G after replacing dummy vertices with crossing points.Recall that (v 1 , v n ) is an edge of G by the definition of canonical ordering.We distinguish whether v n is real or dummy; the two cases are shown in Fig. 8.Note that if v n is dummy, its four neighbors are all real and hence their N, NW, and NE ports are free by P5.If v n is real, it has three neighbors in Γ K−1 , v 1 is real by construction, and the S port can be used to attach with a dummy vertex.Finally, since Γ K−1 is attachable, we can use Lemma 2 to avoid crossings and to find a suitable point to place v n .A complete drawing is shown in Fig. 1d.The theorem follows immediately by the choice of the slopes.

2-bend drawings
Liu et al. [29] presented an algorithm to compute orthogonal drawings for planar graphs of maximum degree 4 with at most 2 bends per edge (except the octahedron, which requires 3 bends on one edge).We make use of their algorithm for biconnected graphs.The algorithm chooses two vertices s and t and computes an st-ordering of the input graph.The edges around each vertex v i , 1 ≤ i ≤ n, are assigned to the four ports as follows.If v i has only one outgoing edge, it uses the N port; if v i has two outgoing edges, they use the N and E port; if v i has three outgoing edges, they use the N, E, and W port; and if v i has four outgoing edges, they use all four ports.Symmetrically, the incoming edges of v i use the S, W, E, and N port, in this order.The edge (s, t) (if it exists) is assigned to the W port of both s and t.If deg(s) = 4, the edge (s, v 2 ) is assigned to the S port of s (otherwise the port remains free); if deg(t) = 4, the edge (t, v n−1 ) is assigned to the N port of t (otherwise the port remains free).Note that every vertex except s and t has at least one incoming and one outgoing edge; hence, the given embedding of the graph provides a unique assignment of edges to ports.Finally, they place the vertices bottom-up as prescribed by the st-ordering.The way an edge is drawn is determined completely by the port assignment, as depicted in Fig. 9.
Let G = (V, E) be a subcubic 1-plane graph.We first re-embed G according to Lemma 1.Let G * be the planarization of G after the re-embedding.Then, all vertical horizontal L-shapes C-shapes U-shapes Fig. 9: The shapes to draw edges cutvertices of G * are real vertices, and since they have maximum degree 3, there is always a bridge connecting two 2-connected components.Let G 1 , . . ., G k be the 2-connected components of G, and let G * i be the planarization of G i , 1 ≤ i ≤ k.We define the bridge decomposition tree T of G as the graph having a node for each component G i of G, and an edge (G i , G j ), for every pair G i , G j connected by a bridge in G.We root T in G 1 .For each component G i , 2 ≤ i ≤ k, let u i be the vertex of G i connected to the parent of G i in T by a bridge and let u 1 be an arbitrary vertex of G 1 .We will create a drawing Γ i for each component G i with at most 2 slopes and 2 bends such that u i lies on the outer face.
To this end, we first create a drawing Γ * i of G * i with the algorithm of Liu et al. [29] and then modify the drawing.Throughout the modifications, we will make sure that the following invariants hold for the drawing i is a planar orthogonal drawing of G * i and edges are drawn as in Fig. 9; (I2) u i lies on the outer face of Γ * i and its N port is free; (I3) every edge is y-monotone from its source to its target; (I4) every edge with 2 bends is a C-shape, there are no edges with more bends; (I5) if a C-shape ends in a dummy vertex, it uses only E ports; and (I6) if a C-shape starts in a dummy vertex, it uses only W ports.
Proof sketch.We choose t = u i and some real vertex s and use the algorithm by Liu et al. to draw G i .Since s and t are real, there are no U-shapes.Since no real vertex can have an outgoing edge at its W port or incoming edge at its E port, the invariants follow.The full proof is given in Appendix C.
We now iteratively remove the C-shapes from the drawing while maintaining the invariants.We make use of a technique similar to the stretching in Section 3. We lay an orthogonal y-monotone curve S through our drawing that intersects no vertices.Then we stretch the drawing by moving S and all features that lie right of S to the right, and stretching all points on S to horizontal segments.After this stretch, in the area between the old and the new position of S, there are only horizontal segments of edges that are intersected by S. The same operation can be defined symmetrically for an x-monotone curve that is moved upwards.Lemma 5. Every G i admits an orthogonal 2-bend drawing such that u i lies on the outer face and its N port is free.Proof sketch.We start with a drawing Γ * i of G * i that satisfies invariants (I1)-(I6), which exists by Lemma 4. By (I2), u i lies on the outer face and its N port is free.If no dummy vertex in Γ * i is incident to a C-shape, by (I4) all edges incident to dummy vertices are drawn with at most 1 bend, so the resulting drawing Γ i of G i is an orthogonal 2-bend drawing.Otherwise, there is a C-shape between a real vertex u and a dummy vertex v.We show how to eliminate this C-shape without introducing new ones while maintaining all invariants.
We prove the case that (u, v) is directed from u to v, so by (I5) it uses only E ports; the other case is symmetric.We do a case analysis based on which ports at u are free.We show one case here and the rest in Appendix C.
Case 1.The N port at u is free; see Fig. 10.Create a curve S as follows: Start at some point p slightly to the top left of u and extend it downward to infinity.Extend it from p to the right until it passes the vertical segment of (u, v) and extend it upwards to infinity.Place the curve close enough to u and (u, v) such that no vertex or bend point lies between S and the edges of u that lie right next to it.Then, stretch the drawing by moving S to the right such that u is placed below the top-right bend point of (u, v).Since S intersected a vertical segment of (u, v), this changes the edge to be drawn with 4 bends.However, now the region between u and the second bend point of (u, v) is empty and the N port of u is free, so we can make an L-shape out of (u, v) that uses the N port at u.This does not change the drawing style of any edge other than (u, v), so all the invariants are maintained and the number of C-shapes is reduced by one.
Finally, we combine the drawings Γ i to a drawing Γ of G. Recall that every cutvertex is real and two biconnected components are connected by a bridge.Let G j be a child of G i in the bridge decomposition tree.We have drawn G j with u j on the outer face and a free N port.Let v i be the neighbor of u j in G i .We choose one of its free ports, rotate and scale Γ j such that it fits into the face of that port, and connect u j and v i with a vertical or horizontal segment.Doing this for every biconnected component gives an orthogonal 2-bend drawing of G.
Theorem 2. Every subcubic 1-plane graph admits a 2-bend 1-planar drawing with at most 2 distinct slopes and both angular and crossing resolution π/2. 5 Lower bounds for 1-plane graphs

1-bend drawings of subcubic graphs
Theorem 3.There exists a subcubic 3-connected 1-plane graph such that any embedding-preserving 1-bend drawing uses at least 3 distinct slopes.The lower bound holds even if we are allowed to change the outer face.
Proof.Let G be the K 4 with a planar embedding.The outer face is a 3-cycle, which has to be drawn as a polygon Π with at least four (nonreflex) corners.Since we allow only one bend per edge, one of the corners of Π has to be a vertex of G.The vertex in the interior has to connect to this corner, however, all of its free ports lie on the outside.Thus, no drawing of G is possible.

Straight-line drawings
The full proofs for this section are given in Appendix D. Theorem 4.There exist 2-regular 2-connected 1-plane graphs with n vertices such that any embedding-preserving straight-line drawing uses Ω(n) distinct slopes.
Proof sketch.Let G k be the graph given by the cycle a 1 . . ., a k+1 , b k+1 , . . ., b 1 , a 1 and the embedding shown in Fig. 11a.Walking along the path a 1 , . . ., a k+1 , we find that the slope has to increase at every step.Lemma 6.There exist 3-regular 3-connected 1-plane graphs such that any embeddingpreserving straight-line drawing uses at least 18 distinct slopes.
Proof sketch.Consider the graph depicted in Fig. 11b.We find that the slopes of the edges (a i , b i ), (a i , c i ), (c i , d i ), (c i , e i ), (e i , d i ), (e i , a i+1 ) have to be increasing in this order for every i = 1, 2, 3.
Proof sketch.Consider the graph depicted in Fig. 11c.The degree of a i , c i , and e i is ∆.We repeat the proof of Lemma 6, but observe that the slopes of the 9(∆−3) added edges lie between the slopes of (a i , b i ), (a i , c i ), (c i , e i ), and (e i , a i+1 ).

A Omitted material from Section 2
Lemma 1.Let G = (V, E) be a 1-plane graph and let G * be its planarization.We can re-embed G such that each edge is still crossed at most once and (i) no cutvertex of G * is a dummy vertex, and (ii) if G is 3-connected, then G * is 3-connected.
Proof.We first show how to iteratively remove all cutvertices from G * that are dummy vertices.Suppose that there is a cutvertex v in G * that is a dummy vertex.Let a, b, c, d be the neighbors of v in counter-clockwise order, so the edges (a, c) and (b, d) cross in G.
First, assume that one of the four edges of v, say (v, a), is a bridge, so removing v from G * gives a connected component A that contains neither b, c, d; see Fig. 12a.We shrink A and move it to the other side of v.This eliminates the crossing between (a, c) and (b, d) from G.
We can now assume that there is no bridge at v, so removal of v divides G * into two components A and B. If a and c lie in the same component, G is disconnected as there is no path from a to b; hence, assume w.l.o.g. that a, b ∈ A and c, d ∈ B; see Fig. 12b.We flip B, reroute the edge (a, c) along (a, v) and (v, c) and reroute (b, d) along (b, v) and (v, d).This eliminates the crossing between (a, c) and (b, d) from G.
This shows the first part of the lemma.For the second part, suppose that G is 3-connected and G * has no dummy vertex as a cutvertex; otherwise, apply the first part of the lemma.Assume that there is a separation pair u, v in G * where v is a dummy vertex.Let again a, b, c, d be the neighbors of v in counter-clockwise order.
First, assume that u is a real vertex.Removal of u and v splits G * in at most four connected components.If one of these connected components contains exactly one neighbor of v, say a, there are at most two vertex-connected paths from a to c in G: the edge (a, c) and one path via u.Hence, there are two connected components A and B that contain two neighbors of v each; assume w.l.o.g. that A contains a.If A contains a and c, u is a cutvertex in G, which contradicts 3-connectivity.If A contains a and b, we flip A, reroute the edge (b, d) along (b, v) and (v, d) and reroute (a, c) along (a, v) and (v, c); see Fig. 13a.This eliminates the crossing between (a, c) and (b, d) from G. If A contains a and d, we proceed analogously.Second, assume that u is also a dummy vertex with neighbors a , b , c , d in counter-clockwise order.Removal of u and v splits G * in at most four connected components.If one of these connected components contains exactly one neighbor of v, say a, and exactly one neighbor of u, say a , then there are at most two vertex-connected paths from a to c in G: the edge (a, c) and one path via the edge (a , c ).If one of the connected components contains exactly three neighbors of v, say a, b, c, and exactly one neighbor of u, say a , then d, a is a separation pair in G, as it separates a, b, c from b , c , d .Hence, each of these connected components contains exactly two neighbors of one of v and u.Let A be a connected component and assume w.l.o.g. that it contains a and one more neighbor of v.If A contains a and c, there is some neighbor of u that is not in A, say a .Since (a, c) ∈ A and b, d / ∈ A, all paths from a to a in G have to traverse the edge (a , c ) or (b , d ), so there are at most two of them; a contradiction to 3-connectivity.If A contains a and b, we flip A and reroute the edges (a, c) and (b, d) to eliminate their crossing from G. Note that, if A contains also exactly two neighbors of u, this also eliminates the crossing between (a , c ) and (b , d ); see Fig. 13b.If A contains a and d, we proceed analogously.
Each step reduces the number of crossings in the embedding, so it terminates with an embedding that has a 3-connected planarization.

B Omitted material from Section 3
Lemma 2. Suppose that Γ i is valid and stretchable, and let u and v be two consecutive vertices of C i .If u is L-attachable (resp., v is R-attachable), then it is possible to modify Γ i such that any half-line with slope π/4 (resp., 3π/4) that originates at u (resp., at v) and that intersects the outer face of Γ i does not intersect any edge segment with slope π/2 of P i (u, v).Also, the modified drawing is still valid and stretchable.
Proof.Refer to Fig. 14.Suppose there is a half-line h that originates at u (the argument is analogous for v) with slope in the set {π/4, π/2} that intersects the outer face of Γ i .Let s be the first edge segment of P i (u, v) that is intersected by h.By the slopes of h, we have that s is drawn with slope π/2.Then, s must be traversed upwards when going from u to v in P i (u, v), and thus by P4 there is a horizontal before s.
Let e be the edge containing this horizontal segment.By P4, there is a uv-cut with respect to e and there is a y-monotone curve uv that cuts the horizontal segments of this cut.Let C u and C v be the two components defined by the uvcut, such that C u contains u and C v contains v.We shift all vertices in C v and all edges having both end-vertices in C v to the right by σ units, for some suitable σ > 0. All vertices in C u and all edges having both end-vertices in C u are not modified.Furthermore, the edges having an end-vertex in C u and the other endvertex in C v are all and only the edges of the uv-cut, and thus they all contain a horizontal segment in Γ i that can be stretched by σ units.Finally, note that (v 1 , v 2 ) is also part of the uv-cut, but it does not contain any horizontal segment; however, by P3 its two segments can be always redrawn by using the SE port of v 1 and the SW port of v 2 .For a suitable choice of σ, this operation removes the crossing between h and s.Moreover, no new edge crossing can appear in the drawing because uv intersects only the edge segments of the cut.Hence, we can repeat this procedure until all crossings between h and segments of P i (u, v) are resolved.The resulting drawing is clearly still valid and stretchable.Lemma 3. Drawing Γ K−1 is valid, stretchable, and attachable.
Proof.In all the cases used by our construction, we guaranteed the drawing be valid and attachable.Concerning stretchability, observe that P3 is guaranteed by Lemma 2. To show P4, one can use induction on i ≤ K − 1 as follows.
In the base case i = 2, we have that Γ 2 is clearly stretchable by construction.When adding V i to Γ i−1 , we have that P4 holds by induction for all pairs of vertices that are consecutive both in Γ i−1 and in Γ i , because P i−1 (u, v) = P i (u, v).Also, the vertices in V i are all attachable vertices.We distinguish the following cases.Case 1. V i is singleton.Then, v i is consecutive with either u l or the attachable vertex before u l , and with either u r or the attachable vertex w after u r .Case 1.1.v i is consecutive to u l (resp., u r ).Then, u l (resp., u r ) has degree four and hence is dummy.However, now u l (resp., u r ) is not L-attachable (resp., R-attachable) anymore, so P4 holds.Case 1.2.v i is consecutive to w (a symmetric argument applies to w ).Then, observe first that P4 holds for P i−1 (w, u l ) = P i (w, u l ).Also, observe that if there is a horizontal segment in P i (w, u l ), then P4 holds for P i (w, v i ) (even if (u l , v i ) contains a vertical segment and even if v i is real).Case 1.2.1.Both w and u l are real.Then, there is a horizontal segment in P i (w, u l ) and hence P4 holds for P i (w, v i ).Case 1.2.2.w is dummy and u l is real.Then, u l is R-attachable.Furthermore, this means that (u l , v i ) will not be drawn as a vertical segment.Hence, P4 holds for P i (w, v i ).Case 1.2.3.Both w and u l are dummy.Then, v i is real.If v i is the only successor of u l , then it is drawn with Case C3L in Fig. 15, so there is a horizontal segment on (u l , v i ).Otherwise, u l has another successor x in G i .If x is an R-successor of u l , then v i is drawn with Case C1R, C2R, C2sR, or C3sR in Fig. 16, column 2. In each case, (u l , v i ) is drawn with a horizontal segment, so P4 holds.If x is an L-successor of u l , then v i is drawn with Case C1L, C2L, C2sL, or C3L in Fig. 16, column 2. In each case, the edge (u l , v i ) goes downward from v i .
We move along P (w, v i ) from v i to w until we either arrive at an edge of a chain, at w, or at a dummy vertex y = w.Since a real vertex cannot have two successors, this walk only moves downwards.Recall that all edges of a chain are drawn with a horizontal segment, so if we arrive at such an edge, then P4 holds.
If we arrive at a dummy vertex y = w, then it has an L-successor in G i .Since y and v i are not consecutive, all of its successors are already drawn, so we are in one of the cases of Fig. 16, column 2 or 3, that has at least one L-successor.Then, either one of the edges of y on C i has a horizontal segment, so P4 holds, or the edge from y goes downwards (Cases C2L and C3L in column 2), and we can continue our walk.
If we arrive at w then the last edge we traverse was to w from an L-successor of w in G i .Hence, w is not L-attachable.Further, since our walk only went downwards from v i to w, there was no upwards vertical on this segment.Hence, P4 holds.Case 1.2.4.w is real and u l is dummy, then v i is real.We proceed exactly as in Case 1.2.3.However, since w is real and attachable, the last edge cannot be from a successor of w to w.Hence, it is either an edge of a chain, so it has a horizontal segment, or on the way we encountered a dummy vertex that has an edge with a horizontal segment.In either case, P4 holds.Case 2. V i is a chain.Note first that P4 holds for each pair of consecutive vertices u and v such that both of them are in V i , since all edges of V i contain a horizontal segment.If u ∈ V j and v ∈ V i (resp., u ∈ V i and v ∈ V j ) with j < i, , and a similar argument as for the singletons can be applied.This completes the case analysis.
Finally, we should prove that for every edge e of P i (u, v) such that e contains a horizontal segment, there exists a uv-cut of G i with respect to e whose edges all contain a horizontal segment in Γ i except for (v 1 , v 2 ), and such that there exists a y-monotone curve that passes through all and only such horizontal segments and through (v 1 , v 2 ).Again, this is true by induction for all such pairs of vertices that are consecutive both in Γ i−1 and in Γ i , because If both u ∈ V i and v ∈ V i , then V i is a chain.Let u l and u r be the two vertices on C i−1 used by V i to attach to Γ i−1 .Note that, by P4, P i−1 (u l , u r ) does not contain any horizontal segment only if it contains neither pairs of consecutive real vertices, nor vertical segments.It is not difficult to see that this situation never occurs, and hence there is at least an edge e in P i−1 (u l , u r ) such that e contains a horizontal segment, there exists a uv-cut of G i−1 with respect to e whose edges all contain a horizontal segment in Γ i−1 except for (v 1 , v 2 ), and such that there exists a y-monotone curve that passes through all and only such horizontal segments and through (v 1 , v 2 ).Now, we can add the edge (u, v) to this cut, since this edge contains a horizontal segment by construction.Also, Fig. 16: Cases for dummy vertices with two successors (u, v) is on the same face of Γ i as e and above it, so the y-monotone curve that passes through e can be suitably modified so to also pass through the horizontal segment of (u, v).
If u ∈ Γ i−1 and v ∈ V i (the symmetric case is analogous), as explained above, we have that P i (u, v) is constructed from P i−1 (u, u l ), where u l is the leftmost predecessor of v (which is either a singleton v i or the first vertex v i 1 of a chain).Then, the only edge for which we may need to prove the property is the edge (u l , v) and only if it contains a horizontal segment.If so, we can again exploit the fact that there is at least an edge e in P i−1 (u l , u r ) (where u r is the rightmost predecessor of v if V i is a singleton or of v i l if V i is a chain) for which P4 holds by induction.
Proof.By construction, G * i is biconnected.First, observe that every face in G * i contains at least two real vertices since no two dummy vertices can be adjacent.Hence, there is some face that contains the real vertex t = u i and some real vertex s.We use these two vertices to compute an st-order σ and use the algorithm of Liu et al. to draw G * i .We first show that this drawing satisfies all invariants.Invariants (I1) and (I2) are trivially satisfied.
Since s and t are real vertices, both have degree at most 3, so there are no U-shapes in the drawing; since U-shapes are the only edges that are not drawn y-monotone from its source to its target, this satisfies (I3).By construction, all edges in G * i are drawn with at most 2 bends; hence, invariant (I4) holds.Consider a dummy vertex v in G * i with neighbors a, b, c, d in clockwise order; hence, the edges (a, c) and (b, d) cross in the given embedding of G i .Assume w.l.o.g. that (v, a) uses the S port, (v, b) uses the W port, (v, c) uses the N port, and (v, d) uses the E port at v. Since there are no U-shapes in the drawing, both (v, a) and (v, c) have to be drawn as a vertical or an L-shape, so they both have at most 1 bend.
Consider now the edges (v, b) and (v, d).Both edges are drawn as a horizontal, an L-shape, or a C-shape.Recall that b and c are real vertices, so they have at most degree 3.If (v, b) starts in b, it uses the N or the E port at b, but it uses the W port at v, so it cannot be a C-shape.If (v, b) ends in b, it uses the W or the S port at b, and it can only be a C-shape if it uses the W one. Symmetrically, If (v, d) starts in d, it uses the N port or the E port at d and it can only be a C-shape if it uses the E one.If (v, d) ends in d, it uses the W or the S port at d, but it uses the E port at v, so it cannot be a C-shape.This establishes invariants (I5) and (I6) and proofs the lemma.Lemma 5. Every G i admits an orthogonal 2-bend drawing such that u i lies on the outer face and its N port is free.Proof.It remains to show the cases that the N port at u is not free.Case 2. The N port at u is used by an edge (u, w) and the W port is free.We distinguish three more cases based on the drawing style of (u, w).Case 2.1.(u, w) is a vertical edge; see Fig. 17.We create a curve S as in Case 1 except that we do not pass the vertical segment of (u, v) but extend it upwards to infinity before.We stretch the drawing by moving S to the right such that u is placed below the top-right bend point of (u, v).Now the edge (u, w) is drawn with 2 bends, but the area between u and the two bend points is empty and the W port of u is unused, so we can make an L-shape out of (u, w) that uses the W port at u. Furthermore, similar to Case 1, the region between u and the top-right bend point of (u, v) is free and now the N port of u is unused, so we can make an L-shape out of (u, v) that uses the N port at u. Case 2.2.(u, w) is an L-shape and w lies to the left of u; see Fig. 18.Assume first that w lies below v.We claim that there is no vertex in the region bounded by the vertical segment of (u, w) from the left, the first horizontal segment of (u, v) from the bottom, the vertical segment of (u, v) to the right and the y-coordinate of w from the top.Assume to the contrary that there is some vertex in this region and let x be the bottom-most one.Every vertex has at least one incoming edge, and by invariant (I3) the target vertex is not below the source vertex.Hence, there has to be an edge from some vertex y to vertex x such that y does not lie above x.
Since the E and the N port of u are already used, y cannot be u.If y lies below x, by choice of x the edge (y, x) has to intersect (u, w) or (u, v), which contradicts invariant (I1).Otherwise, y lies next to x and (y, x) is a horizontal segment.If y lies to the right (left) of x, we choose the rightmost (leftmost) vertex z such that there is a directed path from z to x that only contains horizontally drawn edges.If z lies outside the region, some edge on this path has to cross (u, v) or (u, w), which contradicts (I1).Otherwise, we repeat the argument with z; since z cannot have an incoming edge from a vertex with the same y-coordinate, it cannot have any incoming edge without a crossing.Hence, this region is empty and we can move x upwards to the same ycoordinate as w.Now (u, w) uses the W port at u and we can use Case 1 to make (u, v) an L-shape.
On the other hand, if w does not lie below v, we can use the same argument that the area described above, but bounded form the top by the y-coordinate of v, is empty.However, there has to be an edge that uses the S port of v, and it has to be y-monotone by invariant (I3), so its source has to lie below v; a contradiction.Case 2.3.(u, w) is an L-shape and w lies to the right of u; see Fig. 19.By the same argument as in Case 2.2, w has to lie below v.We can also use the exact argument to show that the region between (u, w) and (u, v) is empty.
We create a curve S as in Case 2.1.We stretch the drawing by moving S to the right such that u is placed directly below w.Because of the empty region, we can now make (u, w) a vertical edge and then use Case 2.1 to make (u, v) an L-shape.Case 3. The N port and the W port at u are used.Let (u, w) be the edge that uses the W port at u; since (u, v) is an outgoing edge and the edge at the N port has to be outgoing by invariant (I3), (u, w) is an incoming edge at u.By invariants (I3) and (I5), it has to be drawn as a horizontal segment or as an L-shape such that w lies below u.We distinguish two more cases based on the drawing style of (u, w).Case 3.1.(u, w) is an L-shape and w lies below u; see Fig. 20.We create a curve S as follows: We start at some point p slightly to the top left of w and extend it downward to infinity.Then we extend it from p to the right until it passes u and extend it upwards to infinity.We place the curve close enough to (u, w) such that no vertex or bend point lies between S and (u, w).Then, we stretch the drawing by moving S to the right such that w is placed below u.After this operation, the S port of u is free and there is no edge or vertex on the vertical segment between u and w, so we can make (u, w) a vertical edge and then use Case 2 to make (u, v) an L-shape.Case 3.2.(u, w) is a horizontal edge and w at the same y-coordinate as u; see Fig. 21.We now create an x-monotone curve S as follows: We start at some point p slightly to the top left of u and extend it leftward to infinity.Then we extend it from p to the bottom until it passes (u, w) and extend it rightwards to infinity.We place the curve close enough to (u, w) such that no vertex or bend point lies between S and (u, w).Then, we stretch the drawing by moving S upwards for a short distance.After this operation, the S port of u is free and the whole region between w and u is empty, so we can make (u, w) an L-shape and then use Case 2 to make (u, v) an L-shape.
Obviously, each of the above operations maintains all the invariants.Hence, by repeating them for every C-shape, we obtain the desired drawing of G i .

D Omitted proofs from Section 5
The following construction is the same as the one of Hong et al. [21] to prove an exponential area lower bound for straight-line drawings of 1-plane graphs, with two edges added to make the graph biconnected.and b i .Note that the edges in W i incident to c i and the edges in W i+1 incident to c i have the same (pair of) slopes.Hence, we can rotate W i+1 by π around c i and align it with W i such that the edges meeting in c i of both quadrilaterals overlap.The rotation will maintain the slopes.Thus we can draw a "tower" of disjoint copies of all W i (see Fig. 22c).For all 1 < i < k the supporting lines of a i c i and a i+1 c i+1 differ by rotation (angle < π) in the same direction.The total rotation of those edges cannot exceed π, since no a i c i can "overtake" b 2 a 2 .As a consequence, the slopes of all edges a i c i are different and thus also all the slopes of the edges a i a i+1 have to be different.Lemma 6.There exist 3-regular 3-connected 1-plane graphs such that any embeddingpreserving straight-line drawing uses at least 18 distinct slopes.
Proof.Consider the graph G depicted in Fig. 23a-b.To simplify the analysis we exploit a similar idea as in Theorem 4. Let x i be the crossing between a i c i and a i+1 e i (indices modulo 3).Fix any straight-line drawing of G and let T i be the triangle e i c i x i including the two segments e i d i and c i d i .For i = 1, 2, 3 we cut T i , rotate it by π around x i and put it back to the drawing.This leaves the slopes unchanged.By this we obtain a drawing which contains a pseudo-triangle, whose chains (a i c i d i e i a i+1 ) have 4 edges (see Fig. 23c).Further, for every chain there is an edge (e i c i ) between the second and fourth vertex cutting off d i .The edges of a pseudo-triangle have different slopes.Thus, we have 12 different slopes here.Moreover, if you traverse the edges of a pseudo-triangle in cyclic order they will be ordered by slope.Since a i b i is sandwiched between a i c i and a i e i+2 we have three more distinct slopes.Finally, we note that replacing the edges e i d i and c i d i with e i c i gives another pseudo triangle that avoids the slopes of e i d i and c i d i .As a consequence, the edges e i c i will give us three new slopes and we end up with 18 different slopes.
To obtain an infinite family of graphs, observe that we do not use the edges between b-and f -vertices in our analysis.Hence, we can subdivide the edges (b 1 , f 3 ) and (b 2 , f 1 ) several times and connect pairs of subdivision vertices.
Proof.Consider the graph depicted in Fig. 24.The degree of a i , c i and e i is ∆.
We can repeat the argument of the proof of Lemma 6.There are only two differences: (i) instead of a single edge a i b i there is a bundle of edges incident to a i .However the whole bundle lies in between a i c i and a i e i+2 and therefore the slopes of these edges are distinct.(ii) Instead of the pseudo-triangle with chains a i c i d i e i a i+1 we have now a sequence of nested chains given by the edges incident to e i and c i .All these "subchains" are contained in the triangle e i c i x i , where x i is the crossing between a i c i and a i+1 e i .Their slopes lie between the slopes of a i c i and e i a i+1 .This means that the three edge-bundles of subchains are separated by slopes.Clearly the slopes within each bundle have to be different.
Counting the slopes we see that we have the 18 slopes of the subgraph shown in Fig. 23a and then there are 9 vertices, each incident to ∆ − 3 new edges, that will need a new slope.In total we need 18 + 9(∆ − 3) = 9(∆ − 1) distinct slopes.
To obtain an infinite family of graphs, we can again subdivide (b 1 , f 3 ) and (b 2 , f 1 ) several times and connect pairs of the subdivision vertices.

Fig. 1 :
Fig. 1: (a) A 3-connected 1-plane cubic graph G; (b) a canonical ordering δ of the planarization G * of G-the real (dummy) vertices are black points (white squares); (c) the edges crossed by the dashed line are a uv-cut of G5 with respect to (u, w)-the two components have a yellow and a blue background, respectively; (d) a 1-bend 1-planar drawing with 4 slopes of G

Fig. 4 :
Fig. 4: Illustration for P6.If v has two successors not in Γi, then the edges connecting v to its two neighbors in Γi are solid.If v has one successor in Γi, then the edge between v and this successor is dashed or dotted.

Fig. 11 :
Fig. 11: The constructions for the results of Section 5

Fig. 15 :
Fig. 15: Cases for dummy vertices with one successor