Planar Confluent Orthogonal Drawings of 4-Modal Digraphs

In a planar confluent orthogonal drawing (PCOD) of a directed graph (digraph) vertices are drawn as points in the plane and edges as orthogonal polylines starting with a vertical segment and ending with a horizontal segment. Edges may overlap in their first or last segment, but must not intersect otherwise. PCODs can be seen as a directed variant of Kandinsky drawings or as planar L-drawings of subdivisions of digraphs. The maximum number of subdivision vertices in an edge is then the split complexity. A PCOD is upward if each edge is drawn with monotonically increasing y-coordinates and quasi-upward if no edge starts with decreasing y-coordinates. We study the split complexity of PCODs and (quasi-)upward PCODs for various classes of graphs.


Introduction
We consider plane digraphs, i.e., planar directed graphs with a fixed planar embedding and a fixed outer face. Directions of edges in node-link diagrams are usually indicated by arrow heads. Since this might cause clutter at vertices with high indegree, Angelini et al. [4] proposed L-drawings in which each edge is drawn with a 1-bend orthogonal polyline starting with a vertical segment at the tail. A plane digraph can only have an L-drawing without crossings if it is 4-modal, where a plane digraph is k-modal if in the cyclic order around a vertex there are at most k pairs of consecutive edges that are neither both incoming nor both outgoing. However, not every 4-modal digraph admits a planar L-drawing. This motivates to extend the model to drawings with more than one bend per edge.
In a planar confluent orthogonal drawing (PCOD) of a digraph, vertices are represented as points in the plane with distinct x-and y-coordinates and each edge is represented as an orthogonal polyline starting with a vertical segment at the tail and ending with a horizontal segment at the head. Distinct edges may overlap in a first or last segment, but must not intersect otherwise. For better readability bends have distinct coordinates and are drawn with rounded corners. A plane digraph has a PCOD if and only if it is 4-modal.
The work of Sabine Cornelsen was funded by the German Research Foundation DFG -Project-ID 50974019 -TRR 161 (B06).  A PCOD of a digraph G corresponds to a planar L-drawing of a subdivision of G. The number of subdivision vertices on an edge is its split complexity. See the red encircled vertex in Fig. 1b for the subdivision vertex. Since each edge starts with a vertical segment and ends with a horizontal segment, the number of bends on an edge is odd. An edge with split complexity k has 2k + 1 bends. The split complexity of a PCOD is the maximum split complexity of any edge. The PCOD in Fig. 1b has split complexity one. A planar L-drawing [4,17] is a PCOD of split complexity zero. If the embedding is not fixed, then it is NPcomplete to decide whether a digraph admits a planar L-drawing [17]. Every 2-modal digraph without 2-cycles has a planar L-drawing [3].
A PCOD of a digraph corresponds to a Kandinsky drawing [20] of the underlying undirected graph with the only difference that edges partially overlap instead of being drawn in parallel with a small gap. See Fig. 1c. While every simple planar graph has a Kandinsky drawing with one bend per edge [16], deciding whether a multigraph has a Kandinsky drawing with one bend per edge [16] or finding the minimum number of bends in a Kandinsky drawing of a plane graph [14] is NP-hard. For the bend-minimization problem in the Kandinsky model there are 2-approximization algorithms [19,5] and heuristics [7].
Among the results for orthogonal drawings of undirected graphs where edges must not overlap, we mention three: With one exception, every plane graph of maximum degree four admits an orthogonal drawing with at most two bends per edge [11]. In a bend-minimum drawing, however, there might have to be an edge with a linear number of bends [26]. An orthogonal drawing with the minimum number of bends can be computed by means of a min-cost flow approach [25] even if an upper bound on the number of bends per edge must be respected.
A PCOD is upward if each edge is drawn with monotonically increasing ycoordinates. A digraph is upward-planar if and only if it has an upward PCOD. A plane st-graph, i.e., a plane acyclic digraph with a single sink and a single source, both on the outer face, is always upward-planar; moreover, it has an upwardplanar L-drawing if and only if it admits a so-called bitonic st-ordering [17]. Since it suffices to subdivide the edges of a plane st-graph at most once in order to obtain a digraph that admits a bitonic st-ordering [21,1], it follows that every plane st-graph admits an upward PCOD with split complexity one. Moreover, the minimum number of bends in an upward PCOD of a plane st-graph can be determined in linear time. In general, a digraph admits an upward-planar L-drawing, if and only if it is a subgraph of a plane st-graph admitting a bitonic st-ordering [2]. Not every 2-modal tree admits an upward-planar L-drawing [2].
In a quasi-upward-planar drawing [8] edges must be strictly monotonically increasing in y-direction in a small vicinity around the end vertices. A digraph has a 2-modal embedding if and only if it admits a quasi-upward-planar drawing. Every 2-modal graph without 2-cycles admits a quasi-upward planar drawing with at most two bends per edge and the curve complexity in such drawings can be minimized utilizing a min-cost flow approach [13]. We call a PCOD quasiupward if no edge starts with decreasing y-coordinates.
Our Contribution. We show that PCODs of 4-modal trees have split complexity zero (Theorem 2), split complexity two is sufficient (Theorem 4) and sometimes necessary (Theorem 3) for PCODs of 4-modal digraphs with parallel edges or loops, while split complexity one suffices for 4-modal irreducible triangulations (Theorem 5), i.e., internally triangulated 4-connected graphs with an outer face of degree 4. Split complexity one also suffices for upward PCODs of upwardplane digraphs (Theorem 6) and for quasi-upward PCODs of 2-modal digraphs without 2-cycles (Theorem 7). Using an ILP, we conducted experiments that suggest that every simple 4-modal digraph without separating 2-cycles admits a PCOD with split complexity one (Sect. 8). Constant split complexity is not to be expected for bend-minimum PCODs (Theorem 1).

Preliminaries
Two consecutive incident edges of a vertex v are a switch if both edges are incoming or both outgoing edges of v. The drawing of a PCOD is determined by the coordinates of the vertices and the coordinates of every second bend of an edge. We call a bend independent if it is the second, fourth, etc. bend of an edge. Considering a PCOD as an L-drawing of a subdivision, the independent bends correspond to the subdivision vertices. The split complexity of an edge is the number of its independent bends. The total number of bends equals the number of edges plus twice the number of independent bends. The top, left, bottom, and right side of a vertex is its North, West, South, and East port, respectively.

Confluent Orthogonal Representation
Let Γ be a PCOD of a plane digraph G. We call a bend covered if it is contained in the drawing of another edge. We associate an orthogonal drawing of a plane graph G Γ with Γ as follows [3]: Replace every covered bend in Γ by a dummy vertex. See Figs. 1d and 2b. A zig-zag is a pair of uncovered bends on an edge, one with a left turn, and one with a right turn. E.g., on the edge (u, v) in Fig. 2a there is a zig-zag, while on the edge (u, w) there is both a left and a right turn, but the left turn is covered, so there is no zig-zag. Since the number of bends in an orthogonal drawing can always be reduced by eliminating zig-zags, we will also do so in PCODs (see Fig. 2c) and, thus, the ordering of left-and rightturns at uncovered bends of an edge will not matter. Since planar (confluent) orthogonal drawings can be stretched independently in x-and y-directions, it is algorithmically often easier not to work with actual x-and y-coordinates, but rather with the shape of the faces in terms of bends on the edges and angles at the vertices. See also [25,20].
A confluent orthogonal representation R of a plane digraph G = (V, E) is a set of circular lists H(f ), one for each face f of G. The elements of H(f ) are tuples r = (e, v, a, s, b) associated with the edges e incident to f in counterclockwise order. (a) v is the end vertex of e traversed immediately before e. (b) a ∈ {0, π 2 , π, 3π 2 , 2π} is the angle at v between e and its predecessor on f . It is a multiple of π if and only if it describes an angle at a switch. (c) s is the number of left turns (when traversing e starting from v) at bends in e. ( This ensures that covered bends are counted by s and that covered bends adjacent to the head or the tail of an edge must be s1 t1 s2 From a Representation to a PCOD. In order to construct a PCOD from a feasible confluent orthogonal representation R of a plane digraph G, we transform G into a graph G R of maximum degree 4 and a feasible orthogonal representation R without 0 or 2π angles. Using compaction for orthogonal representations [25] on G R then yields a PCOD or a π/2-rotation of a PCOD in linear time. The idea for the construction of G R is analogous to the construction of G Γ from a PCOD Γ and is as follows: Consider a vertex v ∈ V and let e 1 , . . . , e k be a maximum sequence of consecutive edges around v with 0 angles. Let The representation R is updated accordingly.

Some Initial Results
Theorem 1. There is a family G k , k > 0 of 4-modal digraphs with 14k − 3 vertices and 16k − 4 edges such that in any bend-minimum PCOD of G k there is an edge with split complexity at least k + 2.
Proof. Consider the digraphs G k indicated in Fig. 3a. Let e be the red dashed edge. Let P k be the path s 1 , x 1 , t 1 , y 1 , . . . , s k , x k , t k of length 4k − 2 in G k that is drawn vertically in Fig. 3a. Consider a planar L-drawing of G k − e in which all edges of P k (traversed from s 1 to t k ) bend to the left and the edge incident to s 1 is to the top of s 1 . Such a drawing for G 2 is indicated in Fig. 3b. Since all vertices of P k are 4-modal this uniquely determines the drawing of P k and also of the transitive edges of P k . In order to preserve the embedding, e can only be inserted into the drawing with split complexity at least k + 2.
Consider a PCOD of G k with fewer bends on e. Since all vertices are 4-modal, the rotation of the cycle C composed of P k and e can only be maintained, if the number of bends on at least one edge of P k , say (s i , x i ), is increased. But then we also must increment the number of bends on an edge (s i , t i ) to maintain the rotation of the face bounded by the edges (s i , t i ), (s i , x i ), (x i , t i ). Thus, for each independent bend less on e the total number of bends increases by at least 2.
Even though not every 2-modal tree has an upward-planar L-drawing [2], every 4-modal tree has a planar L-drawing, despite its fixed embedding.
Theorem 2. Every 4-modal tree has a PCOD with split complexity zero. Moreover, such a drawing can be constructed in linear time.
Proof. Let T be a 4-modal tree and let v be a leaf of T . We show by induction on the number m of edges that we can draw T as a PCOD Γ α with split complexity zero such that v is in the corner α (lower left ( ), lower right ( r), upper left (u ), upper right (ur)) of the bounding box of Γ α . We give the details for Γ ; the other cases are analogous. If m = 1, draw v at (0, 0) and its neighbor at (1, 1).
If m > 1, let the neighbor of v be v , and let the connected components of T − v be v, T 1 , . . . , T k in clockwise order around v . See Fig. 4. Let T 0 be the subtree consisting of v only. Each tree T i +v , i = 0, . . . , k has at most m−1 edges and the leaf v ; therefore, by the inductive hypothesis, we can construct PCODs Γ α i , α ∈ { , r, u , ur} of T i + v with v in the respective corner of the bounding box. W.l.o.g. let v be the tail of the edge connecting v and v , see Fig. 4a. Let 1 ≤ a ≤ b ≤ c ≤ d ≤ k such that T d+1 , . . . , T k , T 0 , T 1 , . . . , T a and T b+1 , . . . , T c are connected to v by an incoming edge and T a+1 , . . . , T b and T c+1 , . . . , T d by outgoing edges. Choose Γ ur 0 , Γ r 1 , . . . , Γ r a , Γ a+1 , . . . , Γ c , Γ ur c+1 , . . . , Γ ur k for T i +v , i = 1, . . . , k. Finally, merge the drawings of the subtrees at v .
In order to compute a confluent orthogonal representation, using dynamic programming, only O(deg(v )) steps are required for each vertex v . Thus, the total time complexity is linear.

Multi-Graphs
Theorem 3. There are 4-modal multigraphs that need split complexity at least two in any PCOD.
v   Proof. Consider the digraph containing a loop in Fig. 5a or the digraph containing two parallel edges in Fig. 5b. The incident 4-modal vertices on the one hand and the rotation of the outer face on the other hand, imply that the loop and one of the two parallel edges, respectively, must have split complexity two. In the case of the loop, the angle at the vertex is convex. Since the rotation of the outer face is −4, it follows that five concave bends on the loop are needed. In the case of two parallel edges, the angles at the vertices in the outer face are zero. Thus, the two edges together must have eight concave bends. Since each edge has an odd number of bends, there must be an edge with five bends.
Theorem 4. Every 4-modal multigraph has a PCOD with split complexity at most two. Moreover, such a drawing can be computed in linear time.
Proof. The approach is inspired by [11]. Subdivide each loop. Let the resulting digraph be G. Then make the digraph biconnected maintaining its 4-modality [3]. Now compute in linear time [15] an st-ordering v 1 , . . . , v n of this biconnected graph G (without taking into account the direction of the edges). Iteratively add the vertices with increasing y-coordinates in the order of the st-ordering, maintaining a column for each edge that has exactly one end vertex drawn.
Let v k be a vertex. An edge e incident to v k is incident to v k from below if e has an end vertex that is before v k in the st-ordering. Let e 1 , . . . , e j be the sequence of edges incident to v k from below as they appear from left to right. Since v k is 4-modal, e 1 , . . . , e j can be divided into at most five subsequences of edges consisting only of incoming (−) or only of outgoing (+) edges of v k . Depending on the arrangement of these subsequences, we assign the bends around v k . E.g., consider the Case +− in Fig. 6b, i.e., among the edges incident to v k from below, there are first some outgoing edges, followed by some incoming edges. All outgoing edges from below are attached to the South port, while all incoming edges from below are attached to the East port. Mind that all outgoing edges except one need two bends near v k . Consider now the edges incident to v k to later vertices in the st-ordering. By 4-modality, there can be at most some incoming edges, followed by some outgoing, some incoming and again some outgoing edges in counter-clockwise order around v k . We attach them to the East, North, West, and South port of v k , respectively. The edges from below determine the position of v k . In our case, v k is drawn above one of the edges attached to its South port. See Fig. 6 for the routing of the edges from below and the possible edges to later end vertices in the other cases. For v 1 , we choose the assignment according to Fig. 6a or Fig. 6f. After all vertices are placed, we remove edges that are in G but not in G. If x is a vertex that was inserted into a loop, we reroute the two incident edges near x such that the incoming edge of x has exactly one bend near x and the outgoing edge has no bend near x. Finally, we eliminate zig-zags.
By the st-ordering, the columns of the edges incident to v k from below are consecutive among the edges with exactly one end vertex drawn [11]. This implies planarity if the columns for the new edges are inserted directly next to v k . For each edge e, there are at most two bends near the tail of e and at most three bends near the head of e. Consider now a 2-cycle (v, x), (x, v) replacing a loop at v. Since the subdivision vertex x is incident to exactly one incoming and one outgoing edge, it follows that near x there is no bend on (x, v) and one bend on (v, x). If (x, v) does not have three bends near v then in total there are at most six bends on the loop, namely the four bends near v plus one bend near x plus the bend on x. Since the number of bends on an edge must be odd, there are only five. Consider now the case that (x, v) has three bends near v (Figs. 6f  and 6j). If in addition (v, x) has two bends near v, then there are seven bends on the loop. However, in this case, there is a zig-zag on (v, x) formed by the bend near x and the second bend near v. Thus, after eliminating zig-zags, the split complexity is at most two.

Irreducible Triangulations
An irreducible triangulation is an internally triangulated graph with an outer face of degree four that does not contain any separating triangles. We prove that every 4-modal digraph whose underlying undirected graph is an irreducible triangulation has a PCOD with split complexity at most one. Motivated by the approaches in [3,12], we use rectangular duals, a contact representation of an irreducible triangulation G = (V, E) with the following properties. The vertices v ∈ V are represented by internally disjoint axis-parallel rectangles R(v). Two rectangles touch if and only if the respective vertices are adjacent in G. Moreover, no four rectangles representing a vertex meet at the same point and v∈V R(v) is a rectangle. See the rectangles in Fig. 7a. A rectangular dual for an irreducible triangulation can be computed in linear time [9,10,22,23].
Given a rectangular dual, we perturb the coordinate system such that in each rectangle the axes correspond to the diagonals. The perturbed x-axis is the diagonal containing the bottommost-leftmost point of the rectangle, the other diagonal is the perturbed y-axis. A perturbed orthogonal polyline is a polyline such that in each rectangle the segments are parallel to one of the axes. A bend of a perturbed orthogonal polyline at the boundary of two rectangles is a real bend if among the two incident segments one is parallel to a perturbed x-axis and the other parallel to a perturbed y-axis. Bends inside a rectangle are always real. In a perturbed PCOD each vertex v is drawn at the center of R(v). An edge (u, v) is a perturbed orthogonal polyline in R(u) ∪ R(v) between u and v starting with a segment on the perturbed y-axis in R(u) and ending with a segment on the perturbed x-axis in R(v). The drawing of (u, v) must have at least one bend in the interior of both R(u) and R(v) and must cross the boundary of R(u) and R(v) exactly once. Distinct edges may overlap in a first or last segment, but must not intersect otherwise. No two bends have the same coordinates. See   Fig. 7a. The North port of v is the port above and to the left of the center of R v . TheWest, South, and East ports are the other ports in counter-clockwise order.
In analogy to the arguments in [3], we obtain that a perturbed PCOD yields a confluent orthogonal representation where the number s of left turns counts only real bends. The next theorem says that we can derive a PCOD with split complexity one from a suitable perturbed PCOD after zig-zag elimination. See Fig. 7b.
Theorem 5. Every 4-modal irreducible triangulation has a PCOD with split complexity at most one; and such a drawing can be computed in linear time.
Proof. Let G be an irreducible triangulation. We construct a rectangular dual for G ignoring edge directions. Routing the edges inside any rectangle independently, we then construct a perturbed PCOD that yields a confluent orthogonal representation with split complexity at most one after zig-zag elimination.
Let v be a vertex of G. For a side s of R(v) let u i , i = 1, . . . , k be the adjacent vertices of v in counter-clockwise order such that s and R(u i ) intersect in more than a point. Let e i be the edge between v and u i . Consider the division of e 1 , . . . , e k into mono-directed classes, i.e., maximal subsequences such that any two edges in a subsequence are either both incoming or both outgoing edges. Let the modality mod(s) of s be the number of these subsequences. Since G is 4-modal we have mod(s) ≤ 5. Assume now that s is a side of R(v) with maximal modality. Assume without loss of generality that s is the right side of R(v). mod(s) = 5. If e 1 is an outgoing edge of v, assign the mono-directed classes of edges crossing s from bottom to top in this order to (i) the North port bending three times to the left, (ii) to the West port bending twice to the left, (iii) to the South port bending once to the left, (iv) to the East port bending once to the right, and (v) to the North port bending twice to the right. Route the edges as indicated in blue in Fig. 8a to s. By adding zig-zags, it is always possible to route an edge e i between v and u i in such a way that the parts of e i in R(v) and R(u i ) meet in s. See Fig. 8b. Edges crossing other sides of R v are all outgoing edges of v and are assigned to the North port, bending once or twice in the direction of the side where they leave R v . See the purple edges in Fig. 8a. If e 1 is an incoming edge, start analogously with the West port. See Fig. 8c. mod(s) ∈ {1, . . . , 4}. The assignment of edges to ports and the routing of the edges are contained in the drawing of the case mod(s) = 5. See the blue edges in the second and third row in Fig. 8. We make again sure that an edge to a side of R(v) with modality one has at most two bends in the interior of R(v). In order to do so, we have to take special care if mod(s) = 4 and the bottommost edge is an outgoing edge of v. Let s t and s b be the top and bottom side, respectively, of R(v). If v is incident to an outgoing edge crossing s t , we opt for the variant in Fig. 8e and otherwise (including the case that v is incident to an incoming edge crossing s b ) for the variant in Fig. 8g. No particular care has to be taken if mod(s) = 2 (Figs. 8l and 8m). Let now e be an edge between two vertices u and v. We consider the number b e (u) and b e (v) of bends on e in R(u) and R(v), respectively, after eliminating zig-zags. We assume without loss of generality that b e (u) ≤ b e (v). Recall that then 1 ≤ b e (u) ≤ b e (v) ≤ 3. Let s u and s v be the sides of R(u) and R(v), respectively that contain the intersection s of R(u) and R(v).
We have to show that up to zig-zags there are in total at most three bends on e. This is clear if b e (u) = b e (v) = 1. Assume now that b e (v) ≥ 2. Since the number of real bends on e is odd it follows that the bend on s is real if and only if b e (u) + b e (v) is even. In this case the bend on s bends in opposite direction as the next bend in R(u) and R(v) (otherwise e does not cross s). Since b e (v) ≥ 2, the real bend of e on s and the next bend of e in R(v) form a zig-zag and can be eliminated. See Fig. 9a. Thus, if b e (u) + b e (v) ≤ 4 then there are at most three bends on e after zig-zag elimination.
It remains to consider the case that b e (v) = 3 and b e (u) ≥ 2. This implies mod(s v ) > 1 and e is in the first or last mono-directed class among the edges crossing s v . We assume without loss of generality that s v is the right side of R v and that e is in the bottommost mono-directed class. See Fig. 9b. It follows that e is an outgoing edge of v and, thus, an incoming edge of u. Since b e (u) ≥ 2 it follows that e is attached to the East port of u. Assume first that b e (u) = 2. Then the bends of e in R(u) are in opposite direction as the bends of e in R(v). Thus, there is at least one zig-zag consisting of a bend in R(u) and a bend in R(v). After eliminating this zig-zag there are only three bends left.
Assume now that b e (u) = 3. This is only possible if mod(s u ) ≥ 2. Hence, R(u) is the topmost or bottommost rectangle incident to the right of R(v). Since e is in the bottommost class with respect to s v , it must be the bottommost one. Thus, R(v) is the topmost neighbor to the left of R(u). Moreover, since mod(s u ) ≥ 2 there must be a port of u other than the East port that contains an edge e (red edge in Fig. 9c) crossing s u . But e would have to bend at least four times in the interior of R(u), which never happens according to our construction.

(Quasi-)Upward-Planar Drawings
Theorem 6. Every upward-plane digraph admits an upward PCOD with split complexity at most one. Moreover, for plane st-graphs both the split complexity and the total number of bends can be minimized simultaneously in linear time.
Proof. Let G be an upward-plane digraph. Then G can be augmented to a plane st-graph by adding edges [6]. Subdividing each edge once yields a plane st-graph with a bitonic st-ordering [1] and, thus, with an upward-planar L-drawing [17]. This corresponds to an upward PCOD of split complexity one for G.
If G is a plane st-graph it can be decided in linear time whether G has an upward-planar L-drawing [17], and thus, an upward PCOD of split complexity zero. Otherwise, the minimum number of edges that has to be subdivided in order to obtain a digraph that has a bitonic st-ordering can be computed in linear time [1]. Thus, a PCOD with the minimum number of bends among all upward PCODs of G with split complexity one can be computed in linear time. Observe that the total number of bends cannot be reduced by increasing the split complexity, since the subdivision of edges is only performed in order to break one of the transitive edges in a valley. the South port of v, we reroute them such that they are attached to the North port. Consider first the case that at least one among the East or the West portsay the East port -of v does not contain edges. Then we can reroute the edges as indicated in Fig. 10b. Edges attached to the South port of v in Γ are now attached to the North port of v and get two additional bends near their tail.
Assume now that both the East and the West port of v contain edges. Then, by 2-modality, no edge is attached to the North port of v. Those edges incident to the East port that bend upward are reattached to the West port without adding any additional bend. This is possible since these edges have already been rerouted near their other end vertex and two new bends have been inserted. So there are enough bends for the bend-or-end property. The rotation of the faces and the angular sum around the vertices are also maintained. The edges incident to the East port bending downward are rerouted from their original drawing to the West port with two new bends. See Fig. 10d. Now we can reroute the edges attached to the South port as in the first case.
An edge attached to the South port in Γ gets at most two new bends near its tail; an edge attached to the North port at most two new bends near its head. Thus, in the end each edge has at most three bends, i.e., split complexity 1.

Experiments using an ILP
Based on the definition of confluent orthogonal representations and the fact that each 4-modal multigraph has a PCOD with split complexity at most two, we developed an ILP to compute PCODs with minimum split complexity for 4-modal graphs. See the appendix. Since each simple 4-modal graph without 2-cycles can be extended to a triangulated 4-modal graph [3], we first sampled several thousand upward-planar triangulations for various numbers n ≤ 500 of vertices with two different methods: sampling (a) undirected triangulations uniformly at random [24] orienting the edges according to an st-ordering [15] and (b) with an OGDF method [18]. Then we flipped the direction of each edge with probability 0.5 maintaining 4-modality. Finally, we added as many 2-cycles as 4-modality allowed. The resulting digraphs contained ( 3 4 ± 1 4 )n separating triangles, roughly n 2-cycles, but no separating 2-cycles. All digraphs had curve complexity one.
We examined the split complexity of PCODs of various graph classes. In particular, we have shown that every 4-modal digraph admits a PCOD with split complexity two even if it contains loops and parallel edges and that split complexity two is sometimes necessary. For simple digraphs, we made a first step, by proving that every 4-modal irreducible triangulation admits a PCOD with split complexity one. It still remains open whether split complexity one suffices for all simple 4-modal digraphs. Experiments suggest that this could very well be true. It would also be interesting to know whether the minimum split complexity or the minimum number of bends in a PCOD or a (quasi-)upward PCOD can be efficiently determined in the case of a given 4-modal, 2-modal, or upward-planar embedding, respectively, as well as in the case when no embedding is given.