For two planar graph $G^{\textcircled1}$ = ($V^{\textcircled1}$, $E^{\textcircled1}$) and $G^{\textcircled2}$ = ($V^{\textcircled2}$, $E^{\textcircled2}$) sharing a common subgraph G = $G^{\textcircled1}$ ∩ $G^{\textcircled2}$ the problem Simultaneous Embedding with Fixed Edges (SEFE) asks whether they admit planar drawings such that the common graph is drawn the same. Previous algorithms only work for cases where G is connected, and hence do not need to handle relative positions of connected components. We consider the problem where G, $G^{\textcircled1}$ and $G^{\textcircled2}$ are not necessarily connected.#R##N##R##N#First, we show that a general instance of SEFE can be reduced in linear time to an equivalent instance where $V^{\textcircled1}$ = $V^{\textcircled2}$ and $G^{\textcircled1}$ and $G^{\textcircled2}$ are connected. Second, for the case where G consists of disjoint cycles, we introduce the CC-tree which represents all embeddings of G that extend to planar embeddings of $G^{\textcircled1}$. We show that CC-trees can be computed in linear time, and that their intersection is again a CC-tree. This yields a linear-time algorithm for SEFE if all k input graphs (possibly k>2) pairwise share the same set of disjoint cycles. These results, including the CC-tree, extend to the case where G consists of arbitrary connected components, each with a fixed embedding. Then the running time is O(n2).
['Thomas Bläsius', 'Ignaz Rutter']
Disconnectivity and relative positions in simultaneous embeddings