The Complexity of Simultaneous Geometric Graph Embedding

TL;DR

This paper investigates the computational complexity of simultaneously embedding multiple planar graphs on the same vertex set, establishing its equivalence to known hard problems and proving exponential lower bounds on grid size for such embeddings.

Contribution

It provides an elementary reduction from pseudoline stretchability to the simultaneous geometric embedding problem, proving its -hardness and tight exponential lower bounds on grid size.

Findings

The problem is -hard, equivalent to several geometric recognition problems.

There exist collections of graphs that require exponential grid size for simultaneous embedding.

The lower bound on grid size is tight, matching the upper bounds.

Abstract

Given a collection of planar graphs $G_1,\dots,G_k$ on the same set $V$ of $n$ vertices, the simultaneous geometric embedding (with mapping) problem, or simply $k$-SGE, is to find a set $P$ of $n$ points in the plane and a bijection $\phi: V \to P$ such that the induced straight-line drawings of $G_1,\dots,G_k$ under $\phi$ are all plane. This problem is polynomial-time equivalent to weak rectilinear realizability of abstract topological graphs, which Kyn\v{c}l (doi:10.1007/s00454-010-9320-x) proved to be complete for $\exists\mathbb{R}$, the existential theory of the reals. Hence the problem $k$-SGE is polynomial-time equivalent to several other problems in computational geometry, such as recognizing intersection graphs of line segments or finding the rectilinear crossing number of a graph. We give an elementary reduction from the pseudoline stretchability problem to $k$-SGE, with the property that both numbers $k$ and $n$ are linear in the number of pseudolines. This implies not only the $\exists\mathbb{R}$-hardness result, but also a $2^{2^{\Omega (n)}}$ lower bound on the minimum size of a grid on which any such simultaneous embedding can be drawn. This bound is tight. Hence there exists such collections of graphs that can be simultaneously embedded, but every simultaneous drawing requires an exponential number of bits per coordinates. The best value that can be extracted from Kyn\v{c}l's proof is only $2^{2^{\Omega (\sqrt{n})}}$.