Disconnectivity and Relative Positions in Simultaneous Embeddings

The problem Simultaneous Embedding with Fixed Edges (SEFE) asks for two planar graph $G^1 = (V^1, E^1)$ and $G^2 = (V^2, E^2)$ sharing a common subgraph $G = G^1 \cap G^2$ whether they admit planar drawings such that the common graph is drawn the same in both. Previous results on this problem require $G$, $G^1$ and $G^2$ to be connected. This paper is a first step towards solving instances where these graphs are disconnected. First, we show that an instance of the general SEFE-problem can be reduced in linear time to an equivalent instance where $V^1 = V^2$ and $G^1$ and $G^2$ are connected. This shows that it can be assumed without loss of generality that both input graphs are connected. Second, we consider instances where $G$ is disconnected. We show that SEFE can be solved in linear time if $G$ is a family of disjoint cycles by introducing the CC-tree, which represents all simultaneous embeddings. We extend these results (including the CC-tree) to the case where $G$ consists of arbitrary connected components, each with a fixed embedding. Note that previous results require $G$ to be connected and thus do not need to care about relative positions of connected components. By contrast, we assume the embedding of each connected component to be fixed and thus focus on these relative positions. As SEFE requires to deal with both, embeddings of connected components and their relative positions, this complements previous work.