On a Tree and a Path with no Geometric Simultaneous Embedding

Two graphs $G_1=(V,E_1)$ and $G_2=(V,E_2)$ admit a geometric simultaneous embedding if there exists a set of points P and a bijection M: P -> V that induce planar straight-line embeddings both for $G_1$ and for $G_2$. While it is known that two caterpillars always admit a geometric simultaneous embedding and that two trees not always admit one, the question about a tree and a path is still open and is often regarded as the most prominent open problem in this area. We answer this question in the negative by providing a counterexample. Additionally, since the counterexample uses disjoint edge sets for the two graphs, we also negatively answer another open question, that is, whether it is possible to simultaneously embed two edge-disjoint trees. As a final result, we study the same problem when some constraints on the tree are imposed. Namely, we show that a tree of depth 2 and a path always admit a geometric simultaneous embedding. In fact, such a strong constraint is not so far from closing the gap with the instances not admitting any solution, as the tree used in our counterexample has depth 4.Comment: 42 pages, 33 figure

The Complexity of Simultaneous Geometric Graph Embedding

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 $2^{2^{\Omega (\sqrt{n})}}$

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