Windrose Planarity: Embedding Graphs with Direction-Constrained Edges

TL;DR

This paper introduces the concept of windrose planarity for embedding graphs with direction constraints, providing a polynomial-time algorithm for testing such embeddings and methods for constructing drawings with limited bends.

Contribution

It presents the first polynomial-time algorithm for windrose planarity testing given a fixed embedding and offers construction techniques for drawings with bounded bends and grid size.

Findings

Polynomial-time algorithm for windrose planarity testing

Construction of windrose-planar drawings with at most one bend per edge

Drawings can be embedded on a polynomial-sized grid

Abstract

Given a planar graph $G$ and a partition of the neighbors of each vertex $v$ in four sets $UR(v)$, $UL(v)$, $DL(v)$, and $DR(v)$, the problem Windrose Planarity asks to decide whether $G$ admits a windrose-planar drawing, that is, a planar drawing in which (i) each neighbor $u \in UR(v)$ is above and to the right of $v$, (ii) each neighbor $u \in UL(v)$ is above and to the left of $v$, (iii) each neighbor $u \in DL(v)$ is below and to the left of $v$, (iv) each neighbor $u \in DR(v)$ is below and to the right of $v$, and (v) edges are represented by curves that are monotone with respect to each axis. By exploiting both the horizontal and the vertical relationship among vertices, windrose-planar drawings allow to simultaneously visualize two partial orders defined by means of the edges of the graph. Although the problem is NP-hard in the general case, we give a polynomial-time algorithm for testing whether there exists a windrose-planar drawing that respects a given combinatorial embedding. This algorithm is based on a characterization of the plane triangulations admitting a windrose-planar drawing. Furthermore, for any embedded graph with $n$ vertices that has a windrose-planar drawing, we can construct one with at most one bend per edge and with at most $2n-5$ bends in total, which lies on the $3n \times 3n$ grid. The latter result contrasts with the fact that straight-line windrose-planar drawings may require exponential area.