Simultaneous Embeddings with Few Bends and Crossings

A simultaneous embedding with fixed edges (SEFE) of two planar graphs $R$ and $B$ is a pair of plane drawings of $R$ and $B$ that coincide when restricted to the common vertices and edges of $R$ and $B$. We show that whenever $R$ and $B$ admit a SEFE, they also admit a SEFE in which every edge is a polygonal curve with few bends and every pair of edges has few crossings. Specifically: (1) if $R$ and $B$ are trees then one bend per edge and four crossings per edge pair suffice (and one bend per edge is sometimes necessary), (2) if $R$ is a planar graph and $B$ is a tree then six bends per edge and eight crossings per edge pair suffice, and (3) if $R$ and $B$ are planar graphs then six bends per edge and sixteen crossings per edge pair suffice. Our results improve on a paper by Grilli et al. (GD'14), which proves that nine bends per edge suffice, and on a paper by Chan et al. (GD'14), which proves that twenty-four crossings per edge pair suffice.Comment: Full version of the paper "Simultaneous Embeddings with Few Bends and Crossings" accepted at GD '1

Relating Graph Thickness to Planar Layers and Bend Complexity

The thickness of a graph $G=(V,E)$ with $n$ vertices is the minimum number of planar subgraphs of $G$ whose union is $G$. A polyline drawing of $G$ in $\mathbb{R}^2$ is a drawing $\Gamma$ of $G$, where each vertex is mapped to a point and each edge is mapped to a polygonal chain. Bend and layer complexities are two important aesthetics of such a drawing. The bend complexity of $\Gamma$ is the maximum number of bends per edge in $\Gamma$, and the layer complexity of $\Gamma$ is the minimum integer $r$ such that the set of polygonal chains in $\Gamma$ can be partitioned into $r$ disjoint sets, where each set corresponds to a planar polyline drawing. Let $G$ be a graph of thickness $t$. By F\'{a}ry's theorem, if $t=1$, then $G$ can be drawn on a single layer with bend complexity $0$. A few extensions to higher thickness are known, e.g., if $t=2$ (resp., $t>2$), then $G$ can be drawn on $t$ layers with bend complexity 2 (resp., $3n+O(1)$). However, allowing a higher number of layers may reduce the bend complexity, e.g., complete graphs require $\Theta(n)$ layers to be drawn using 0 bends per edge. In this paper we present an elegant extension of F\'{a}ry's theorem to draw graphs of thickness $t>2$. We first prove that thickness-$t$ graphs can be drawn on $t$ layers with $2.25n+O(1)$ bends per edge. We then develop another technique to draw thickness-$t$ graphs on $t$ layers with bend complexity, i.e., $O(\sqrt{2}^{t} \cdot n^{1-(1/\beta)})$, where $\beta = 2^{\lceil (t-2)/2 \rceil }$. Previously, the bend complexity was not known to be sublinear for $t>2$. Finally, we show that graphs with linear arboricity $k$ can be drawn on $k$ layers with bend complexity $\frac{3(k-1)n}{(4k-2)}$.Comment: A preliminary version appeared at the 43rd International Colloquium on Automata, Languages and Programming (ICALP 2016

On Visibility Representations of Non-planar Graphs

A rectangle visibility representation (RVR) of a graph consists of an assignment of axis-aligned rectangles to vertices such that for every edge there exists a horizontal or vertical line of sight between the rectangles assigned to its endpoints. Testing whether a graph has an RVR is known to be NP-hard. In this paper, we study the problem of finding an RVR under the assumption that an embedding in the plane of the input graph is fixed and we are looking for an RVR that reflects this embedding. We show that in this case the problem can be solved in polynomial time for general embedded graphs and in linear time for 1-plane graphs (i.e., embedded graphs having at most one crossing per edge). The linear time algorithm uses a precise list of forbidden configurations, which extends the set known for straight-line drawings of 1-plane graphs. These forbidden configurations can be tested for in linear time, and so in linear time we can test whether a 1-plane graph has an RVR and either compute such a representation or report a negative witness. Finally, we discuss some extensions of our study to the case when the embedding is not fixed but the RVR can have at most one crossing per edge

Compact Drawings of 1-Planar Graphs with Right-Angle Crossings and Few Bends

We study the following classes of beyond-planar graphs: 1-planar, IC-planar, and NIC-planar graphs. These are the graphs that admit a 1-planar, IC-planar, and NIC-planar drawing, respectively. A drawing of a graph is 1-planar if every edge is crossed at most once. A 1-planar drawing is IC-planar if no two pairs of crossing edges share a vertex. A 1-planar drawing is NIC-planar if no two pairs of crossing edges share two vertices. We study the relations of these beyond-planar graph classes (beyond-planar graphs is a collective term for the primary attempts to generalize the planar graphs) to right-angle crossing (RAC) graphs that admit compact drawings on the grid with few bends. We present four drawing algorithms that preserve the given embeddings. First, we show that every $n$-vertex NIC-planar graph admits a NIC-planar RAC drawing with at most one bend per edge on a grid of size $O(n) \times O(n)$. Then, we show that every $n$-vertex 1-planar graph admits a 1-planar RAC drawing with at most two bends per edge on a grid of size $O(n^3) \times O(n^3)$. Finally, we make two known algorithms embedding-preserving; for drawing 1-planar RAC graphs with at most one bend per edge and for drawing IC-planar RAC graphs straight-line

Simultaneous Orthogonal Planarity

We introduce and study the $\textit{OrthoSEFE}-k$ problem: Given $k$ planar graphs each with maximum degree 4 and the same vertex set, do they admit an OrthoSEFE, that is, is there an assignment of the vertices to grid points and of the edges to paths on the grid such that the same edges in distinct graphs are assigned the same path and such that the assignment induces a planar orthogonal drawing of each of the $k$ graphs? We show that the problem is NP-complete for $k \geq 3$ even if the shared graph is a Hamiltonian cycle and has sunflower intersection and for $k \geq 2$ even if the shared graph consists of a cycle and of isolated vertices. Whereas the problem is polynomial-time solvable for $k=2$ when the union graph has maximum degree five and the shared graph is biconnected. Further, when the shared graph is biconnected and has sunflower intersection, we show that every positive instance has an OrthoSEFE with at most three bends per edge.Comment: Appears in the Proceedings of the 24th International Symposium on Graph Drawing and Network Visualization (GD 2016