## 1 Introduction

Visibility representations are a classic research topic in Graph Drawing and Computational Geometry. Motivated by VLSI applications, seminal papers studied *bar visibility representations* of planar graphs (see, e.g., [22, 25, 26, 27]), in which vertices are represented as non-overlapping horizontal segments, called *bars*, and edges correspond to vertical *visibilities* connecting pairs of bars, i.e., vertical segments that do not intersect any bar other than at their endpoints.

In order to represent non-planar graphs, more recent papers investigated models in which either two visibilities are allowed to cross, or a visibility can “go through” a vertex. Two notable examples are rectangle visibility representations and bar -visibility representations. In a *rectangle visibility representation* of a graph, every vertex is represented as an axis-aligned rectangle and two vertices are connected by an edge using either a horizontal or a vertical visibility (see, e.g., [8, 17, 23]). A *bar -visibility representation* is a bar visibility representation in which each visibility intersects at most bars (see, e.g., [6, 7, 12]).

Extensions of visibility representations to 3D have also been studied. Of particular interest for us are *-parallel visibility representations (ZPRs)*, in which the vertices of the graph are isothetic disjoint rectangles parallel to the -plane, and the edges are visibilities parallel to the -axis. Bose et al. [5] proved that admits a ZPR, while does not. Štola [24] reduced this gap by showing that does not admit any ZPR. If the rectangles are restricted to unit squares, then is the largest representable complete graph [13]. Other 3D visibility models are box visibility representations [14], and 2.5D box visibility representations [1].

In this paper we study 3D visibility representations of -planar graphs. We recall that a graph is *-planar* if it can be drawn with at most one crossing per edge (see, e.g., [4, 18, 21]). The -planar graphs are among the most investigated families of “beyond planar graphs”, i.e., graphs that extend planarity by forbidding specific edge crossings configurations (see, e.g., [16, 20]). Brandeburg [6] and Evans et al. [12] proved that every -planar graph admits a bar -visibility representation. Later,
Biedl et al. [3] proved that a -plane graph (i.e., an embedded -planar graph) admits a rectangle visibility representation if and only if it does not contain any of a set of obstructions, and that not all -planar graphs can be realized, regardless of their -planar embedding. On the other hand, every -planar graph can be represented with vertices that are orthogonal polygons with several reflex corners [11].
Our goal is to represent -planar graphs with vertices drawn as rectangles (rather than more complex polygons) by exploiting the third dimension. We prove that every -planar graph has a ZPR . In addition, is *-visible*, i.e., there is a plane that is orthogonal to the rectangles of and such that its intersection with defines a bar -visibility representation of (see Section 2 for formal definitions).

Our main contribution is summarized by the following theorem.

###### Theorem 1.1

Every -planar graph with vertices admits a -visible ZPR in volume. Also, if a -planar embedding of is given as part of the input, then can be computed in time.

An embedding is needed, as recognizing -planar graphs is NP-complete [15, 19]. An example of a -visible ZPR is shown in Fig. 4. We also remark that, as pointed out by Kobourov et al. in a recent survey [18], very little is known on 3D representations of -planar graphs, and our result sheds some light on this problem.

From a high-level perspective, to prove Theorem 1.1 (see Section 3) we start by constructing a bar -visibility representation of , which is then used as the intersection of the ZPR with the plane (see, e.g., Fig. (b)b). In particular, we transform each bar of into a rectangle by computing the -coordinates of its top and bottom sides, so that each visibility in that traverses a bar can be represented as a visibility in that passes above or below (see, e.g., Fig. (c)c). This is done by using two suitable acyclic orientations of the edges of .

Some proofs and technicalities have been moved to the appendix.

## 2 Preliminaries and definitions

We assume familiarity with the concepts of planar drawings and planar embeddings, see, e.g., [9].
The *planarization* of a non-planar drawing is a planar drawing obtained by replacing every crossing with a *dummy vertex*. An *embedding* of a graph is an equivalence class of drawings whose planarized versions have the same planar embedding. A *-plane* graph is a 1-planar graph with a *-planar embedding*, i.e., an embedding where each edge is incident to at most one dummy vertex. A *kite* is a -plane graph isomorphic to in which the outer face is composed of four vertices and four crossing-free edges, while the remaining two edges cross each other. Given a -plane graph and a kite , with , kite is *empty* if it contains no vertex of inside the 4-cycle .

A *(partial) orientation* of a graph is an assignment of directions to (a subset of) the edges of . The graph obtained by orienting the edges of according to is the directed (or mixed) graph . A *planar -(multi)graph* is a plane acyclic directed (multi)graph with a single source and a single sink , with both and on its outer face [10]. The sets of incoming and outgoing edges incident to each vertex of are *bimodal*, i.e., they are contiguous in the cyclic ordering of the edges at . Each face of is bounded by two directed paths with a common origin and destination, called the *left path* and *right path* of . Face is the *left* (resp., *right*) face for all vertices on its right (resp., left) path except for the origin and for the destination. A *topological ordering* of a directed acyclic (multi)graph is a linear ordering of its vertices such that for every directed edge from vertex to vertex , precedes in the ordering.

A set of disjoint rectangles in is *-parallel*, if each rectangle has its sides parallel to the - and -axis. Two rectangles of are *visible* if and only if they contain the ends of a closed cylinder of radius parallel to the -axis and orthogonal to the -plane, and that does not intersect any other rectangle.

###### Definition 1

A *-parallel visibility representation (ZPR)* of a graph maps the set of vertices of to a -parallel set of disjoint rectangles, such that for each edge of the two corresponding rectangles are visible^{1}^{1}1Our visibility model is often called *weak*, to be distinguished with the *strong* model in which visibilities and edges are in bijection. While this distinction is irrelevant when studying complete graphs (e.g., in [5, 24]), the weak model is commonly adopted to represent sparse non-planar graphs in both 2D and 3D (see, e.g., [1, 3, 6, 11, 12]).. If there is a plane that is orthogonal to the rectangles of and such that its intersection with defines a bar -visibility representation of , then is a *-visible ZPR*.

## 3 Proof of Theorem 1.1

Let be a -plane graph with vertices. To prove Theorem 1.1, we present a linear-time algorithm that takes as input and computes a -visible ZPR of in cubic volume. The algorithm works in three steps, described in the following.

Step 1. We compute a bar -visibility representation of by applying Brandenburg’s linear-time algorithm [6], which produces a representation with integer coordinates on a grid of size . This algorithm consists of the following steps. a) A -plane multigraph is computed from such that: The four end-vertices of each pair of crossing edges of induce an empty kite; no edge can be added to without introducing crossings; if two vertices are connected by a set of parallel edges, then all of them are uncrossed and non-homotopic. We remark that the embedding of may differ from the one of due to the rerouting of some edges. b) Let be the plane multigraph obtained from by removing all pairs of crossing edges. Let be an orientation of such that is a planar -multigraph. Then the algorithm by Tamassia and Tollis [25] is applied to compute a bar visibility representation of . c) Finally, all pairs of crossing edges are reinserted through a postprocessing step that extends the length of some bars so to introduce new visibilities. The newly introduced visibilities traverse at most one bar each. In addition, each bar is traversed by at most one visibility.

Step 2. We transform each bar of to a preliminary rectangle . We assume that lies on the -plane and that the bars are parallel to the -axis. Let be the -coordinate of and let and be the -coordinates of the left and right endpoints of , respectively. The rectangle lies on the plane parallel to the -plane with equation . Also, its left and right sides have -coordinates equal to and , respectively. It remains to compute the -coordinates of the top and bottom sides of . We preliminarily set the -coordinates of the bottom sides and of the top sides of all the rectangles to and , respectively. All the visibilities of that do not traverse any bar can be replaced with cylinders of radius . Let be the subgraph of induced by all such visibilities, and let be the resulting ZPR. The next lemma follows.

###### Lemma 1

is a ZPR of .

Step 3. To realize the remaining visibilities of , we modify the -coordinates of the rectangles. The idea is to define two partial orientations of the edges of , denoted by and , to assign the final -coordinates of the top sides and of the bottom sides of the rectangles, respectively. In particular, an edge oriented from to in () encodes that the top side (bottom side) of will have -coordinate greater (smaller) than the one of . The orientations are such that if two vertices and see each other through a third vertex in , then their top (bottom) sides both have larger (smaller) -coordinate than the one of . Hence, both and are defined based on , using the following three rules.

Let be a face of (and hence of ) such that are part of an empty kite of . In what follows we assume that is the origin and is the destination of the face. We borrow some terminology from [6], refer to Fig. 11 (the black thin edges only). If the left (resp., right) path of is composed of the single edge , then is called a *right wing* (resp., *left wing*). If both the left path and the right path of consist of two edges, then is a *diamond*.
(R.1) If is a right wing, we may assume that is above . Consider the restriction of with respect to . Either the visibility between and traverses (as in Fig. (b)b), or the visibility between and traverses . In both cases we only orient edges in . In the first case we orient from to and from to (see the green bold edges in Fig. (a)a). In the second case we orient from to and from to .
(R.2) If is a left wing, we may assume that is above .
As for a right wing, either the visibility between and traverses (as in Fig. (d)d), or the visibility between and traverses . We orient the edges as for a right wing, but we only consider (see, e.g., the blue bold edges in Fig. (c)c).
(R.3) If is a diamond, we may assume that is to the left of . Either the visibility between and traverses , or the visibility between and traverses . In the first case we orient from to and from to in (see the green bold edges in Fig. (e)e). In the second case we orient from to and from to in .

By applying the above three rules for all left and right wings, and for all diamonds of , we obtain and . Note that the above procedure is correct, in the sense that no edge is assigned a direction twice. This is due to the fact that a direction in (resp., ) is assigned to an edge only if it belongs to the right (resp., left) path of a right (resp., left) wing or of a diamond. On the other hand, an edge belongs only to one right path and to one left path. In what follows, we prove that both and are acyclic, i.e., they have no oriented cycles.

###### Lemma 2

Both and are acyclic.

Sketch of proof. We prove that is acyclic. The argument for is symmetric. Suppose, for a contradiction, that contains a directed cycle , as shown in Fig. (a)a. First, note that . If , there are two non-homotopic parallel edges that are both part of the right path of a right wing or of a diamond in . But this is impossible since each pair of crossing edges in forms an empty kite.

Some edges of have opposite orientations in and , since is acyclic. In particular, there is at least a non-empty maximal subsequence of with this property. We distinguish two cases, whether is oriented clockwise or counter-clockwise in a closed walk along its boundary. Let and be the origin of and the destination of , respectively. Note that there is a directed path from to in (and from to in ).

∎

For each maximal subsequence of the edges of such that each edge is oriented and the induced subgraph is connected, compute a topological ordering. Concatenate all such topological orderings, and append at the beginning or at the end of the sequence possible vertices that are not incident to any oriented edge. This gives a total ordering of the vertices of , denoted by . Set the -coordinate of the top side of the rectangle representing the -th vertex in equal to . Apply a symmetric procedure for , by computing a total ordering , and by setting the -coordinate of the bottom side of the rectangle representing the -th vertex in equal to . This concludes the construction of (possible dummy edges inserted by the augmentation procedure of Step 1(a) are simply ignored in ). The correctness of easily follows.

###### Lemma 3

is a -visible ZPR of .

Since takes area, and each rectangle of has height at most , it follows that takes volume. Also, each step of the algorithm can be performed in linear time. This concludes the proof of Theorem 1.1.

## 4 Open problems

Our research suggests interesting research directions, such as: (i) The algorithm in [6] can be adjusted to compute bar -visibility representations of optimal -planar graphs [2] (i.e., -planar graphs with maximum density), and our construction can be also modified to obtain -visible ZPRs for these graphs. Does every -planar graph admit a -visible ZPR? (ii) Can we generalize our result so to prove that every graph admitting a bar -visibility representation also admits a -visible ZPR? (iii) Our algorithm computes ZPRs in which all the rectangles are intersected by the plane . Can this plane contain all bottom sides of the rectangles? If this is not possible, we wonder if every -planar graph admits a 2.5D-visibility representation (i.e., vertices are axis-aligned boxes whose bottom faces lie on a same plane, and visibilities are both vertical and horizontal).

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## Appendix

###### Lemma 2

Both and are acyclic.

###### Proof

We prove that is acyclic. The argument for is symmetric. Suppose, for a contradiction, that contains a directed cycle , as shown in Fig. (a)a. First, note that . If , there are two non-homotopic parallel edges that are both part of the right path of a right wing or of a diamond in . But this is impossible since each pair of crossing edges in forms an empty kite. Some edges of have opposite orientations in and , since is acyclic. In particular, there is at least a non-empty maximal subsequence of with this property. We distinguish two cases, whether is oriented clockwise or counter-clockwise in a closed walk along its boundary. Let and be the origin of and the destination of , respectively. Note that there is a directed path from to in (and from to in ).

###### Lemma 3

is a -visible ZPR of .

###### Proof

By Lemma 1 we know that realizes all the edges of whose visibilities do not cross any bar in . Note that the top sides of the first and of the last vertex of receive -coordinates and , respectively. Similarly, the bottom sides of the first and of the last vertex of receive -coordinates and , respectively. Hence all visibilities in are preserved in .

Each visibility connecting a bar to a bar and traversing a bar in can now be replaced with a cylinder of radius and -coordinate equal either to the one of the top side of plus or to the bottom side of minus . In fact, the above construction ensures the top sides of and have -coordinates greater than the one of (by at least one unit), or that the bottom sides of and have -coordinates smaller than the one of . Also, there is no rectangle that obstructs the visibility , as otherwise would be traversed by in , which is not possible.

The -visibility of is obtained by construction, being the intersection of with the plane .

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