1 Introduction
Let be a graph with maximum vertex degree . The bend planar slope number of is the minimum number of slopes for the edge segments needed to construct a bend planar drawing of , i.e., a planar drawing where each edge is a polyline with at most bends. Since no more than two edge segments incident to the same vertex can use the same slope, is a trivial lower bound for the bend planar slope number of , irrespectively of . Besides its theoretical interest, this problem forms a natural extension of two wellestablished graph drawing models: The orthogonal [5, 15, 17, 29] and the octilinear drawing models [2, 3, 6, 25], which both have several applications, such as in VLSI and floorplanning [24, 30], and in metromaps and mapschematization [20, 26, 28]. Orthogonal drawings use only slopes for the edge segments ( and ), while octilinear drawings use no more than slopes (, , , and ); consequently, they are limited to graphs with and , respectively.
These two drawing models have been generalized to graphs with arbitrary maximum vertex degree by Keszegh et al. [22], who proved that every planar graph admits a bend planar drawing with equispaced slopes. As a witness of the tight connection between the two problems, the result by Keszegh et al. was built upon an older result for orthogonal drawings of degree planar graphs by Biedl and Kant [5]. In the same paper, Keszegh et al. also studied the bend planar slope number and showed an upper bound of and a lower bound of for this parameter. The upper bound has been recently improved, initially by Knauer and Walczak [23] to and subsequently by Angelini et al. [1] to . Angelini et al. actually proved a stronger result: Given any set of slopes, every planar graph with maximum vertex degree admits a 1bend planar drawing whose edge segments use only slopes in . Any such slope set is hence called universal for bend planar drawings. This result simultaneously establishes the bestknown upper bound on the bend planar slope number of planar graphs and the bestknown lower bound on the angular resolution of bend planar drawings, i.e., on the minimum angle between any two edge segments incident to the same vertex. Indeed, if the slopes in are equispaced, the resulting drawings have angular resolution at least .
In this paper we study slope sets that are universal for bend upward planar drawings of directed graphs (or digraphs for short). Recall that in an upward drawing of a digraph , every edge is drawn as a monotone nondecreasing curve from to . Also, admits an upward planar drawing if and only if it is a subgraph of a planar graph [12, 21]. As such drawings are common for representing planar digraphs, they have been extensively studied in the literature (see, e.g., [4, 8, 14, 17, 19]). A preliminary result for this setting is due to Di Giacomo et al. [13], who proved that every seriesparallel digraph with maximum vertex degree admits a bend upward planar drawing that uses at most slopes, and this bound on the number of slopes is worstcase optimal. Notably, their construction gives rise to drawings with optimal angular resolution (but it uses a predefined set of slopes). Upward drawings with one bend per edge and few slopes have also been studied for posets by Czyzowicz et al. [10].
Contribution. We extend the study of universal sets of slopes to upward planar drawings, and present the first constructive technique that works for all planar graphs. This technique exploits a linear ordering of the vertices of a planar digraph introduced by Gronemann [18], called bitonic ordering (see also Section 2). We show that any set of slopes containing the horizontal slope is universal for bend upward planar drawings of degree planar digraphs having a bitonic ordering (Section 3). We remark that the size of is worstcase optimal [13] and, if the slopes of are chosen to be equispaced, the angular resolution of the resulting drawing is at least (also optimal); see Fig. 0(a) for an illustration. We then extend our construction to all planar graphs by using two bends on a restricted number of edges (Section 4). More precisely, we show that, given a set of slopes containing the horizontal slope, every vertex upward planar digraph with maximum vertex degree has a bend upward planar drawing that uses only slopes in and with at most bends in total; see Fig. 0(b) for an illustration.
For space reasons some proofs are in appendix.
2 Preliminaries
We assume familiarity with common notation and definitions about graphs, drawings, and planarity (see, e.g., [11]).
An upward planar drawing of a directed simple graph (or digraph for short) is a planar drawing such that each edge of is drawn as a curve monotonically nondecreasing in the direction. An upward drawing is strict if its edge curves are monotonically increasing. A digraph is upward planar if it admits an upward planar drawing. Note that if a digraph admits an upward drawing then it also admits a strict upward drawing. A digraph is upward planar if and only if it is a subgraph of a planar graph [12]. Let be an vertex planar graph, i.e., is a plane acyclic digraph with a single source and a single sink , such that and belong to the boundary of the outer face and the edge [12]. (Other works do not explicitly require the edge to be part of , see, e.g., [18].) An ordering of is a numbering such that for each edge , it holds (which implies and ). Every planar graph has an ordering, which can be computed in time (see, e.g., [9]). If and are two adjacent vertices of such that , we say that is a successor of , and is a predecessor of . Denote by the sequence of successors of ordered according to the clockwise circular order of the edges incident to in the planar embedding of . The sequence is bitonic if there exists an integer such that ; see Fig. 1(a) for an illustration. Notice that when or , is actually a monotonic decreasing or increasing sequence. A bitonic ordering of is an ordering such that, for every vertex , is bitonic [18]. A planar graph is a bitonic graph if it admits a bitonic ordering. Deciding whether is bitonic can be done in linear time both in the fixed [18] and in the variable [7] embedding settings. If is not bitonic, every ordering of contains a forbidden configuration defined as follows. A sequence of successors of a vertex forms a forbidden configuration if there exist two indices and , with , such that and , i.e. there is a path from to and a path from to ; see Fig. 1(b).
Let be an vertex maximal plane graph with vertices , , and on the boundary of the outer face. A canonical ordering [16] of is a linear ordering of , such that for every : C1: The subgraph induced by is connected and internally triangulated, while the boundary of its outer face is a cycle containing ; C2: If , belongs to and its neighbors in form a subpath of the path obtained by removing from .
Computing takes time [16]. Also, is upward if for every edge of a digraph precedes in .
The slope of a line is the angle that a horizontal line needs to be rotated counterclockwise in order to make it overlap with . If we say that the slope of is horizontal. The slope of a segment is the slope of the line containing it. Let be a set of slopes such that . The slope set is equispaced if , for . Consider a bend planar drawing of a graph , i.e., a planar drawing in which every edge is mapped to a polyline containing at most segments. For a vertex in each slope defines two different rays that emanate from and have slope . If is horizontal these rays are called left horizontal ray and right horizontal ray. Otherwise, one of them is the top and the other one is the bottom ray of . We say that a ray of a vertex is free if there is no edge attached to through in . We also say that is outer if it is free and the first face encountered when moving from along is the outer face of . The slope number of a bend drawing is the number of distinct slopes used for the edge segments of . The bend upward planar slope number of an upward planar digraph is the minimum slope number over all bend upward planar drawings of .
3 bend Upward Planar Drawings
Let be an vertex planar graph with a bitonic ordering ; see, e.g., Fig. 2(a). We begin by describing an augmentation technique to “transform” into an upward canonical ordering of a suitable supergraph of . We start from a result by Gronemann [18], whose properties are summarized in the following lemma; see, e.g., Fig. 2(b).
Lemma 1 ([18])
Let be an vertex planar graph that admits a bitonic ordering . There exists a planar graph with an ordering such that: (i) ; (ii) and ; (iii) and are on the boundary of the outer face of ; (iv) Every vertex of with less than two predecessors in has exactly two predecessors in . Also, and are computed in time.
We call a canonical augmentation of . Observe that always contains the edges and because of (iv). We also insert the edge , which is required according to our definition of graph; this addition is always possible because and are both on the boundary of the outer face. The next lemma shows that any planar graph obtained by triangulating admits an upward canonical ordering; see, e.g., Fig. 2(c).
Lemma 2
Let be a canonical augmentation of an vertex bitonic graph . Every planar graph obtained by triangulating has the following properties: (a) it has no parallel edges; (b) is an upward canonical ordering.
Proof
Concerning Property (a), suppose for a contradiction that has two parallel edges and connecting with . Let be the cycle formed by and and let be the set of vertices distinct from and that are inside in the embedding of . is not empty, as otherwise would be a nontriangular face of . Let be the vertex with the lowest number in among those in . Since is planar (in particular and are not crossed) and has a single source, it contains a directed path from to every vertex in . Hence, it has an edge from to . Also, by assumption, there is no vertex in such that , which implies that is the only predecessor of in , a contradiction to Lemma 1(iv). Concerning Property (b), if is a canonical ordering of , then is actually an upward canonical ordering because it is also an ordering. To see that is a canonical ordering, observe first that , and are on the boundary of the outer face of by construction. Denote by the subgraph of induced by and let be the boundary of its outer face. We first prove by induction on (for ) that is connected. In the base case , is a cycle and therefore it is connected. In the case , is connected by induction and has at least two predecessors in by Lemma 1(iv), thus is connected. We now prove that each , for , is internally triangulated, which concludes the proof of condition C1 of canonical ordering. Suppose, for a contradiction, that there exists an inner face that is not a triangle. Since is triangulated, there exists a vertex , with , that is embedded inside in . Since is an ordering, there is no directed path from to any vertex of . On the other hand, either or there is a directed path from to . Both cases contradict the fact that belongs to the boundary of the outer face of . We finally show that belongs to , for . Since we already proved that is triangulated, this is enough to prove C2. By the planarity of , there is a face in such that all the neighbors of in belong to the boundary of . We claim that is the outer face of . If it was an inner face, then would be embedded inside in and, by the same argument used above, would not belong to the boundary of the outer face of .
We now show that any set of slopes that contains the horizontal slope is universal for bend upward planar drawings of bitonic graphs. The algorithm is inspired by a technique of Angelini et al. [1]. We will use important additional tools with respect to [1], such as the construction of a triangulated canonical augmentation, extra slopes to draw the edges inserted by the augmentation procedure, and different geometric invariants. Let be an vertex bitonic graph with maximum vertex degree ; see Fig. 2(a). The algorithm first computes a triangulated canonical augmentation of ; see Figs. 2(b)–2(c). We call dummy edges all edges that are in but not in and real edges the edges in that are also in . By Lemma 2, admits an upward canonical ordering , where is an ordering such that each vertex distinct from and has at least two predecessors. Let be any set of slopes, which we call real slopes. Let be the smallest angle between two slopes in and let be the maximum number of dummy edges incident to a vertex of . For each slope (, we add dummy slopes such that , for . Hence, there are dummy slopes between any two consecutive real slopes. We will use the real slopes for the real edges and the dummy slopes for the dummy ones.
Let be the subgraph of induced by . The algorithm constructs the drawing by adding the vertices according to . More precisely, it computes a drawing of the digraph obtained from by removing the dummy edges and , which exist by construction, and if it exists. Let be the boundary of the outer face of , and let be the path obtained by removing from . For a vertex of , we denote by (resp. ) the number of real (resp. dummy) edges incident to that are not in and by (resp. ) the th outer real top ray in encountered in clockwise (resp. counterclockwise) order around starting from the left (resp. right) horizontal ray. For dummy top rays, we define analogously and .
satisfies the following invariants:
 I1

is a bend upward planar drawing whose real edges use only slopes in .
 I2

Every edge of contains a horizontal segment.
 I3

Every vertex of has at least outer real top rays; see Fig. 3(a).
 I4

Every vertex of has at least outer dummy top rays between and (resp. and ), including (resp. ); see Fig. 3(b).
 I5

Let be any horizontal line and let and be any two intersection points between and the polyline representing in ; walking along from left to right, and are encountered in the same order as when walking along from to ; see Fig. 3(b).
The last vertex is added to in a slightly different way and the resulting drawing will satisfy I1. The next two lemmas state important properties of any bend upward planar drawing satisfying I1–I5. Similar lemmas are proven in [1, Lemmas 2 and 3], but for drawings that satisfy different invariants.
Lemma 3
Let be a drawing of that satisfies Invariants I1–I5. Let be any edge of such that is encountered before along when going from to , and let be a positive number. There exists a drawing of that satisfies Invariants I1–I5 and such that: (i) the horizontal distance between and is increased by ; (ii) the horizontal distance between any two other consecutive vertices along is the same as in .
Lemma 4
Let be a drawing of that satisfies Invariants I1–I5. Let be a vertex of , and let be any outer top ray of that crosses an edge of in . There exists a drawing of that satisfies Invariants I1–I5 in which does not cross any edge of .
We now describe our drawing algorithm starting with the computation of . We aim at drawing both and horizontally aligned between and . Note that is the source of , and, by the definition of a canonical augmentation, is adjacent to both and , while is adjacent to and to at least one of and . We remove the dummy edges and , and the dummy edge if it exists. The resulting graph is either the path or the path , which we draw along a horizontal segment.
Lemma 5
Drawing satisfies Invariants I1–I5.
Assume now that we have constructed drawing of satisfying I1–I5 . Let be the neighbors of the next vertex along . Let be either , if is real, or , if is dummy. Symmetrically, let be either , if is real, or , if is dummy. Let (for ) be any outer real (resp. dummy) top ray emanating from if is real (resp. dummy). By I3 all such top rays exist and by Lemma 4 we can assume that none of them crosses .
Let be a horizontal line above the topmost point of . Let be the intersection point of and . We can assume that, for , is to the left of . If this is not the case, we can increase the distance between and so to guarantee that and appear in the desired order along ; this can be done by applying Lemma 3 with respect to each edge for a suitable choice of ; see Figs. 4(a)4(b) for an illustration. We will place above using bottom rays of for the segments of the edges () incident to such that: (i) () is real (resp. dummy) if is real (resp. dummy); (ii) precedes in the counterclockwise order around starting from . This choice is possible for the real rays because has real bottom rays and it has at least one incident real edge not in (otherwise it would be a sink of , which is not possible because ). Concerning the dummy rays, we have at most dummy edges incident to and dummy bottom rays between any two consecutive real rays. Consider the ray and choose a point to the right of and above such that placing on guarantees that , where and are the intersection points of the rays with the line (see Fig. 4(c)). Observe that for a sufficiently large coordinate, point can always be found. We now apply Lemma 3 to each of the edges , , , , in this order, choosing so that each is translated to (for ). We finally apply again the same procedure to so that the intersection point between and the horizontal line passing through is to the right of (see Fig. 4(d)). After this translation procedure, we can draw the edge (resp. ) with a bend at the intersection point between (resp. ) and and therefore using the slope of (resp. ) and the horizontal slope (see Fig. 4(e)). The edges () are drawn with a bend point at and therefore using the slopes of and .
Lemma 6
Drawing , for , satisfies Invariants I1–I5.
Proof
The proof is by induction on . satisfies Invariants I1–I5 by Lemma 5 when , and by induction when .
Proof of I1. By construction, each () is drawn as a chain of at most two segments that use real and dummy slopes. In particular, if is real, then it uses real slopes, i.e., slopes in . By the choice of , the bend point of has coordinate strictly greater than that of and smaller than or equal to that of . Since each is oriented from to (as is an upward canonical ordering), the drawing is upward. Concerning planarity, we first observe that is planar and it remains planar each time we apply Lemma 3. Also, by Lemma 4 each () does not intersect (except at ). Further, the order of the bend points along guarantees that the edges incident to do not cross each other.
Proof of I2. The only edges of that are not in are and . For both these edges the segment incident to is horizontal by construction.
Proof of I3. For each vertex of distinct from , and , I3 holds by induction. Invariant I3 also holds for because (as otherwise would be a source of , which is not possible because ) and all the real top rays of , which are , are outer. Consider now vertex (a symmetric argument applies to ). If is real, then ; in this case and therefore all the other outer real top rays of in remain outer in . If is dummy, then ; in this case and therefore all the outer real top rays of in remain outer in .
Proof of I4. For each vertex of distinct from , and , I4 holds by induction. I4 also holds for because and there are dummy top rays between and including (all the top rays of are outer). Analogously, there are outer dummy top rays between and including . Consider now (a symmetric argument applies to ). If is real, then ; in this case and there are outer dummy top rays between and including (namely, all those between and ). If is dummy, then ; in this case and therefore all the other outer dummy top rays of , which by induction were between and , remain outer in .
Proof of I5. Notice that the various applications of Lemma 3 to preserve I5. Let and be any two intersection points between a horizontal line and the polyline representing in , with to the left of along . If and belong to , I5 holds by induction. If both and belong to the path , I5 holds by construction. If belongs to and belongs to , then belongs to the subpath of that goes from to because the subpath from to is completely to the right of , hence I5 holds also in this case. If belongs to and belongs to , the proof is symmetric.
Lemma 7
has a bend upward planar drawing using only slopes in .
Proof
By Lemma 6, drawing satisfies Invariant I1–I5. We explain how to add the last vertex to obtain a drawing that satisfies Invariant I1. Let be the predecessors of on . Notice that, in this case and . Vertex is added to the drawing similarly to all the other vertices added in the previous steps of the algorithm. The only difference is that the number of real incoming edges incident to in can be up to . If this is the case, since the real bottom rays are , they are not enough to draw all the real edges incident to . Let be the smallest index such that is a real edge. We ignore all the dummy edges , for , and apply the construction used in the previous steps considering only as predecessors of (notice that such predecessors are at least two because has at least two incident real edges). By ignoring these dummy edges, the segment of the real edge incident to will be drawn using the left horizontal slope. Denote by the resulting drawing. As in the proof of Lemma 6, we can prove that I1 holds for and therefore is a bend upward planar drawing whose real edges use only slopes in . The drawing of is obtained from by removing all its dummy edges and the two dummy vertices and .
Lemma 8
Drawing can be computed in time.
Lemmas 7 and 8 are summarized by Theorem 3.1. Corollary 1 is a consequence of Theorem 3.1 and of a result in [13].
Theorem 3.1
Let be any set of slopes including the horizontal slope and let be an vertex bitonic planar graph with maximum vertex degree . Graph has a bend upward planar drawing using only slopes in , which can be computed in time.
Corollary 1
Every bitonic graph with maximum vertex degree has 1bend upward planar slope number at most , which is worstcase optimal.
If is equispaced, Theorem 3.1 implies a lower bound of on the angular resolution of the computed drawing, which is worstcase optimal [13]. Also, Theorem 3.1 can be extended to planar graphs with , as any such digraph can be made bitonic by only rerouting the edge .
Theorem 3.2
Every planar graph with maximum vertex degree has bend upward planar slope number at most .
We conclude with the observation that an upward drawing constructed by the algorithm of Theorem 3.1 can be transformed into a strict upward drawing that uses slopes rather than . It suffices to replace every horizontal segment oriented from its leftmost (rightmost) endpoint to its rightmost (leftmost) one with a segment having slope (), for a sufficiently small value of .
4 bend Upward Planar Drawings
We now extend the result of Theorem 3.1 to nonbitonic planar graphs. By adapting a technique of Keszegh et al. [22], one can construct bend upward planar drawings of planar graphs using at most slopes. We improve upon this result in two ways: (i) The technique in [22] may lead to drawings with bends in total, while we prove that bends suffice; (ii) It uses a fixed set of slopes (and it is not immediately clear whether it can work with any set of slopes), while we show that any set of slopes with the horizontal one is universal.
Let be an vertex nonbitonic planar graph. All forbidden configurations of can be removed in linear time by subdividing at most edges of [18]. Let be the resulting bitonic graph, called a bitonic subdivision of . Let be a directed path of obtained by subdividing the edge of with the dummy vertex . We call the lower stub, and the upper stub of . We can prove the existence of an augmentation technique similar to that of Lemma 1, but with an additional property on the upper stubs.
Lemma 9
Let be an vertex planar graph that is not bitonic. Let be an vertex bitonic subdivision of , with a bitonic ordering . There exists a planar graph with an ordering such that: (i) ; (ii) and ; (iii) and are on the boundary of the outer face of ; (iv) Every vertex of with less than two predecessors in has exactly two predecessors in . (v) There is no vertex in such that its leftmost or its rightmost incoming edge is an upper stub. Also, and are computed in time.
Theorem 4.1
Let be any set of slopes including the horizontal slope and let be an vertex planar graph with maximum vertex degree . Graph has a bend upward planar drawing using only slopes in , which has at most bends in total and which can be computed in time.
Proof
We compute a triangulated canonical augmentation of by (1) applying Lemma 9 and (2) triangulating the resulting digraph. By Lemma 2, has an upward canonical ordering . The algorithm of Theorem 3.1 to would lead to a bend drawing of (by interpreting every subdivision vertex as a bend). We explain how to modify it to construct a drawing of with at most bends per edge and bends in total. Let the next vertex to be added according to and let its neighbors in . Suppose that is a dummy vertex and that is an upper stub. To save one bend along the edge subdivided by , we draw without bends. By Lemma 9 (v), we have that . The ray used to draw the segment of incident to can be any outer real top ray; we choose the ray with same slope as the real bottom ray used to draw the segment of incident to . This is possible because all real top rays of are outer (since is the only real outgoing edge of ). Hence, edge has no bends. The drawing of is obtained from by removing dummy edges and replacing dummy vertices (except and , which are removed) with bends. Since the upper stubs of subdivided edges has bends, each edge of has at most bends. Let and be the number of edges drawn with and bends, respectively; we have and . Thus the total number of bends is at most . Finally, can be computed in time (Lemma 9) and the modified drawing algorithm still runs in linear time.
A planar graph with a source/sink of degree requires at least slopes in any upward planar drawing; thus the gap with Theorem 4.1 is one unit. Similarly to Theorem 3.1, Theorem 4.1 implies a lower bound of on the angular resolution of ; an upper bound of can be proven with the same digraph used for the lower bound on the slope number. Finally, Theorem 4.2 extends the result of Theorem 4.1 to every upward planar graph using an additional slope.
Theorem 4.2
Let be any set of slopes including the horizontal slope and let be an vertex upward planar graph with maximum vertex degree . Graph has a bend upward planar drawing using only slopes in .
5 Open Problems
Acknowledgments. Research partially supported by project: “Algoritmi e sistemi di analisi visuale di reti complesse e di grandi dimensioni  Ricerca di Base 2018, Dipartimento di Ingegneria, Università degli Studi di Perugia”.
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Appendix
Appendix 0.A Missing proofs of Section 3
Lemma 3
Let be a drawing of that satisfies Invariants I1–I5. Let be any edge of such that is encountered before along when going from to , and let be a positive number. There exists a drawing of that satisfies Invariants I1–I5 and such that: (i) the horizontal distance between and is increased by ; (ii) the horizontal distance between any two other consecutive vertices along is the same as in .
Proof
We prove by induction on that there exists a cut such that: (i) the vertices of the subpath of from to belong to , while the vertices of the subpath of from to belong to ; (ii) every edge that crosses the cut has a horizontal segment (see Figure 5(a)).
This is trivially true for each edge in . Assume that it is true for by induction (). Let be the neighbors of