3D Visibility Representations of 1-planar Graphs

Abstract

We prove that every 1-planar graph G has a z-parallel visibility representation, i.e., a 3D visibility representation in which the vertices are isothetic disjoint rectangles parallel to the xy-plane, and the edges are unobstructed z-parallel visibilities between pairs of rectangles. In addition, the constructed representation is such that there is a plane that intersects all the rectangles, and this intersection defines a bar 1-visibility representation of G.

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., [23, 26,27,28]), 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 k-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., [9, 18, 24]). A bar k-visibility representation is a bar visibility representation in which each visibility intersects at most k bars (see, e.g., [7, 8, 13]).

Extensions of visibility representations to 3D have also been studied. Of particular interest for us are z -parallel visibility representations (ZPRs), in which the vertices of the graph are isothetic disjoint rectangles parallel to the xy-plane, and the edges are visibilities parallel to the z-axis. Bose et al. [6] proved that $K_{22}$ admits a ZPR, while $K_{56}$ does not. Štola [25] reduced this gap by showing that $K_{51}$ does not admit any ZPR. If the rectangles are restricted to unit squares, then $K_7$ is the largest representable complete graph [14]. Other 3D visibility models are box visibility representations [15], and 2.5D box visibility representations [2].

In this paper we study 3D visibility representations of 1-planar graphs. We recall that a graph is 1-planar if it can be drawn with at most one crossing per edge (see, e.g., [5, 19, 22]). The 1-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., [17, 21]). Brandeburg [7] and Evans et al. [13] proved that every 1-planar graph admits a bar 1-visibility representation. Later, Biedl et al. [4] proved that a 1-plane graph (i.e., an embedded 1-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 1-planar graphs can be realized, regardless of their 1-planar embedding. On the other hand, every 1-planar graph can be represented with vertices that are orthogonal polygons with several reflex corners [12]. Our goal is to represent 1-planar graphs with vertices drawn as rectangles (rather than more complex polygons) by exploiting the third dimension. We prove that every 1-planar graph G has a ZPR $\gamma$. In addition, $\gamma$ is 1-visible, i.e., there is a plane that is orthogonal to the rectangles of $\gamma$ and such that its intersection with $\gamma$ defines a bar 1-visibility representation of G (see Sect. 2 for formal definitions).

Our main contribution is summarized by the following theorem.

Theorem 1

Every 1-planar graph G with n vertices admits a 1-visible ZPR $\gamma$ in $O(n^3)$ volume. Also, if a 1-planar embedding of G is given as part of the input, then $\gamma$ can be computed in O(n) time.

An embedding is needed, as recognizing 1-planar graphs is NP-complete [16, 20]. An example of a 1-visible ZPR is shown in Fig. 1. We also remark that, as pointed out by Kobourov et al. in a recent survey [19], very little is known on 3D representations of 1-planar graphs, and our result sheds some light on this problem.

Fig. 1.

(a) A 1-planar graph G. (b) The intersection of a 1-visible ZPR $\gamma$ of G with the plane $Y=0$; the red (bold) visibilities traverse a bar. (c) The projection to the yz-plane of $\gamma$ (only the red visibilities are shown). (Color figure online)

From a high-level perspective, to prove Theorem 1 (see Sect. 3) we start by constructing a bar 1-visibility representation $\gamma _1$ of G, which is then used as the intersection of the ZPR $\gamma$ with the plane $Y=0$ (see, e.g., Fig. 1b). In particular, we transform each bar b of $\gamma _1$ into a rectangle $R_b$ by computing the y-coordinates of its top and bottom sides, so that each visibility in $\gamma _1$ that traverses a bar b can be represented as a visibility in $\gamma$ that passes above or below $R_b$ (see, e.g., Fig. 1c). This is done by using two suitable acyclic orientations of the edges of G.

For reasons of space some proofs and technicalities have been omitted and can be found in [1].

2 Preliminaries and Definitions

We assume familiarity with the concepts of planar drawings and planar embeddings, see, e.g., [10]. 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 1-plane graph is a 1-planar graph with a 1-planar embedding, i.e., an embedding where each edge is incident to at most one dummy vertex. A kite is a 1-plane graph isomorphic to $K_4$ 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 1-plane graph G and a kite $K=\{a,b,c,d\}$, with $K \subseteq G$, kite K is empty if it contains no vertex of G inside the 4-cycle $\langle a,b,c,d \rangle$.

A (partial) orientation $\mathcal O$ of a graph G is an assignment of directions to (a subset of) the edges of G. The graph obtained by orienting the edges of G according to $\mathcal O$ is the directed (or mixed) graph $G_{\mathcal O}$. A planar st-(multi)graph G is a plane acyclic directed (multi)graph with a single source s and a single sink t, with both s and t on its outer face [11]. The sets of incoming and outgoing edges incident to each vertex v of G are bimodal, i.e., they are contiguous in the cyclic ordering of the edges at v. Each face f of G is bounded by two directed paths with a common origin and destination, called the left path and right path of f. Face f 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 u to vertex v, u precedes v in the ordering.

A set $\mathcal R$ of disjoint rectangles in $\mathbb {R}^3$ is z -parallel, if each rectangle has its sides parallel to the x- and y-axis. Two rectangles of $\mathcal R$ are visible if and only if they contain the ends of a closed cylinder C of radius $\varepsilon >0$ parallel to the z-axis and orthogonal to the xy-plane, and that does not intersect any other rectangle.

Definition 1

A z -parallel visibility representation (ZPR) $\gamma$ of a graph G maps the set of vertices of G to a z-parallel set of disjoint rectangles, such that for each edge of G the two corresponding rectangles are visible¹. If there is a plane that is orthogonal to the rectangles of $\gamma$ and such that its intersection with $\gamma$ defines a bar k-visibility representation of G, then $\gamma$ is a k -visible ZPR.

3 Proof of Theorem 1

Let $G=(V,E)$ be a 1-plane graph with n vertices. To prove Theorem 1, we present a linear-time algorithm that takes G as input and computes a 1-visible ZPR of G in cubic volume. The algorithm works in three steps, described in the following.

Step 1. We compute a bar 1-visibility representation $\gamma _1$ of G by applying Brandenburg’s linear-time algorithm [7], which produces a representation with integer coordinates on a grid of size $O(n^2)$. This algorithm consists of the following steps. (a) A 1-plane multigraph $G'=(V,E'\supseteq E)$ is computed from G such that: The four end-vertices of each pair of crossing edges of $G'$ induce an empty kite; no edge can be added to $G'$ without introducing crossings; if two vertices are connected by a set of $k>1$ parallel edges, then all of them are uncrossed and non-homotopic. We remark that the embedding of $G'$ may differ from the one of G due to the rerouting of some edges. (b) Let P be the plane multigraph obtained from $G'$ by removing all pairs of crossing edges. Let $\mathcal O$ be an orientation of P such that $P_{\mathcal O}$ is a planar st-multigraph. Then the algorithm by Tamassia and Tollis [26] is applied to compute a bar visibility representation of $P_{\mathcal O}$. (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 $b_v$ of $\gamma _1$ to a preliminary rectangle $R_v$. We assume that $\gamma _1$ lies on the xz-plane and that the bars are parallel to the x-axis. Let z(v) be the z-coordinate of $b_v$ and let $x_L(v)$ and $x_R(v)$ be the x-coordinates of the left and right endpoints of $b_v$, respectively. The rectangle $R_v$ lies on the plane parallel to the xy-plane with equation $Z=z(v)$. Also, its left and right sides have x-coordinates equal to $x_L(v)$ and $x_R(v)$, respectively. It remains to compute the y-coordinates of the top and bottom sides of $R_v$. We preliminarily set the y-coordinates of the bottom sides and of the top sides of all the rectangles to $-1$ and $+1$, respectively. All the visibilities of $\gamma _1$ that do not traverse any bar can be replaced with cylinders of radius $\varepsilon < \frac{1}{2}$. Let $P'$ be the subgraph of $G'$ induced by all such visibilities, and let $\gamma _2$ be the resulting ZPR. The next lemma follows.

Lemma 1

$\gamma _2$ is a ZPR of $P'$.

Step 3. To realize the remaining visibilities of $\gamma _1$, we modify the y-coordinates of the rectangles. The idea is to define two partial orientations of the edges of P, denoted by $\mathcal O_1$ and $\mathcal O_2$, to assign the final y-coordinates of the top sides and of the bottom sides of the rectangles, respectively. In particular, an edge oriented from u to v in $\mathcal O_1$ ($\mathcal O_2$) encodes that the top side (bottom side) of $R_u$ will have y-coordinate greater (smaller) than the one of $R_v$. The orientations are such that if two vertices u and v see each other through a third vertex w in $\gamma _1$, then their top (bottom) sides both have larger (smaller) y-coordinate than the one of w. Hence, both $\mathcal O_1$ and $\mathcal O_2$ are defined based on $\gamma _1$, using the following three rules.

Let $f=\{o,u,v,d\}$ be a face of $P_{\mathcal O}$ (and hence of P) such that $\{o,u,v,d\}$ are part of an empty kite of $G'$. In what follows we assume that o is the origin and d is the destination of the face. We borrow some terminology from [7], refer to Fig. 2 (the black thin edges only). If the left (resp., right) path of f is composed of the single edge (o, d), then f is called a right wing (resp., left wing). If both the left path and the right path of f consist of two edges, then f is a diamond.

Fig. 2.

(a)-(b) A right wing. (c)-(d) A left wing. (e)-(f) A diamond.

(R.1) If f is a right wing, we may assume that $b_v$ is above $b_u$. Consider the restriction of $\gamma _1$ with respect to $\{o,u,v,d\}$. Either the visibility between $b_u$ and $b_d$ traverses $b_v$ (as in Fig. 2b), or the visibility between $b_o$ and $b_v$ traverses $b_u$. In both cases we only orient edges in $\mathcal O_1$. In the first case we orient (u, v) from u to v and (v, d) from d to v (see the green bold edges in Fig. 2a). In the second case we orient (o, u) from o to u and (u, v) from v to u. (R.2) If f is a left wing, we may assume that $b_v$ is above $b_u$. As for a right wing, either the visibility between $b_u$ and $b_d$ traverses $b_v$ (as in Fig. 2d), or the visibility between $b_o$ and $b_v$ traverses $b_u$. We orient the edges as for a right wing, but we only consider $\mathcal O_2$ (see, e.g., the blue bold edges in Fig. 2c). (R.3) If f is a diamond, we may assume that $b_u$ is to the left of $b_v$. Either the visibility between $b_o$ and $b_d$ traverses $b_v$, or the visibility between $b_o$ and $b_d$ traverses $b_u$. In the first case we orient (o, v) from o to v and (v, d) from d to v in $\mathcal O_1$ (see the green bold edges in Fig. 2e). In the second case we orient (o, u) from o to u and (u, d) from d to u in $\mathcal O_2$.

By applying the above three rules for all left and right wings, and for all diamonds of $P_{\mathcal O}$, we obtain $\mathcal O_1$ and $\mathcal O_2$. 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 $\mathcal O_1$ (resp., $\mathcal O_2$) 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 $P_{\mathcal O_1}$ and $P_{\mathcal O_2}$ are acyclic, i.e., they have no oriented cycles.

Lemma 2

Both $P_{\mathcal O_1}$ and $P_{\mathcal O_2}$ are acyclic.

Sketch of proof

We prove that $P_{\mathcal O_1}$ is acyclic. The argument for $P_{\mathcal O_2}$ is symmetric. Suppose, for a contradiction, that $P_{\mathcal O_1}$ contains a directed cycle $C =\langle e_1,e_2,\dots ,e_c \rangle$, as shown in Fig. 3a. First, note that $c>2$. If $c=2$, there are two non-homotopic parallel edges that are both part of the right path of a right wing or of a diamond in $P_\mathcal O$. But this is impossible since each pair of crossing edges in $G'$ forms an empty kite. Some edges of C have opposite orientations in $\mathcal O$ and $\mathcal O_1$, since $\mathcal O$ is acyclic. In particular, there is at least a non-empty maximal subsequence $S = \langle e_i,e_{i+1}, \dots , e_j \rangle$ of C with this property. We distinguish two cases, whether C is oriented clockwise or counter-clockwise in a closed walk along its boundary. Let a and b be the origin of $e_i$ and the destination of $e_j$, respectively. Note that there is a directed path from b to a in $P_{\mathcal O}$ (and from a to b in $P_{\mathcal O_1}$).

Case 1. Refer to Fig. 3b. Since $e_j$ is oriented in $P_{\mathcal O_1}$, it belongs to the right path of a right wing or of a diamond f of $P_{\mathcal O}$ by R.1 and R.3. Also, b is the origin of f, as otherwise b would have an incoming edge between $e_j$ and $e_{j+1}$ in counterclockwise order from $e_j$, which violates the bimodality of the edges around b or the fact that the source s of $P_{\mathcal O}$ is on the outer face. But then the orientation of $e_j$ in $\mathcal O_1$ contradicts R.1 or R.3.

Case 2. This case can be handled similarly by observing that a is the destination of a face f having $e_{i-1}$ in its right path.     $\square$

Fig. 3.

Illustration for the proof of Lemma 2. In black (thin) we show the orientation of the edges according to $\mathcal O$, while in green (bold) according to $\mathcal O_1$. (Color figure online)

For each maximal subsequence of the edges of $P_{\mathcal O_1}$ 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 $P_{\mathcal O_1}$, denoted by $\sigma _1$. Set the y-coordinate of the top side of the rectangle representing the i-th vertex in $\sigma _1$ equal to $n-i+1$. Apply a symmetric procedure for $P_{\mathcal O_2}$, by computing a total ordering $\sigma _2$, and by setting the y-coordinate of the bottom side of the rectangle representing the i-th vertex in $\sigma _2$ equal to $i-n-1$. This concludes the construction of $\gamma$ (possible dummy edges inserted by the augmentation procedure of Step 1(a) are simply ignored in $\gamma$). The correctness of $\gamma$ easily follows.

Lemma 3

$\gamma$ is a 1-visible ZPR of G.

Since $\gamma _1$ takes $O(n^2)$ area, and each rectangle of $\gamma$ has height at most 2n, it follows that $\gamma$ takes $O(n^3)$ volume. Also, each step of the algorithm can be performed in linear time. This concludes the proof of Theorem 1.

4 Open Problems

Our research suggests interesting research directions, such as: (i) The algorithm in [7] can be adjusted to compute bar 1-visibility representations of optimal 2-planar graphs [3] (i.e., 2-planar graphs with maximum density), and our construction can be also modified to obtain 1-visible ZPRs for these graphs. Does every 2-planar graph admit a 1-visible ZPR? (ii) Can we generalize our result so to prove that every graph admitting a bar 1-visibility representation also admits a 1-visible ZPR? (iii) Our algorithm computes ZPRs in which all the rectangles are intersected by the plane $Y=0$. Can this plane contain all bottom sides of the rectangles? If this is not possible, we wonder if every 1-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).