Relating Graph Thickness to Planar Layers and Bend Complexity

Abstract

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