The Overlap Graph

Two intervals on the real line overlap if their intersection is nonempty but neither one of them properly contains the other. An interval overlap graph is a graph G(V,E), for which there is an assignment of an interval to each vertex such that two vertices are adjacent if and only if the corresponding intervals overlap. For a permutation $\pi$, we associate with a gray edge $(\pi_i,\pi_j)$ the interval [i,j]. The overlap graph of a permutation $\pi$, denoted $OV(\pi )$, is the interval overlap graph of the gray edges of $B(\pi )$. Namely, the vertex set of $OV(\pi )$ is the set of gray edges in $B(\pi )$, and two vertices are connected if the intervals associated with their gray edges overlap. We shall identify a vertex in $OV(\pi )$ with the edge it represents and with its interval in the representation. Thus, the endpoints of a gray edge are actually the endpoints of the interval representing the corresponding vertex in $OV(\pi )$. A connected component of $OV(\pi )$ that contains an oriented edge is called an oriented component, otherwise, it is called an unoriented component. Figure 11.9(c) shows the interval overlap graph for $\pi = (4,-3,1,-5,-2,7,6)$. It has only one oriented component. Figure 11.10(b) shows the overlap graph of the permutation $\pi '=(4,-3,1,2,5,7,6)$, which has two connected components, one oriented and the other unoriented.

  

Figure: a) The breakpoint graph of $\pi '=(4,-3,1,2,5,7,6)$. $\pi '$ was obtained from $\pi$ of figure 11.9 by the reversal $\rho (7,10)$; or, equivalently, by the reversal defined by the gray edge (2,3). b) The overlap graph of $\pi '$.

Lemma 11.9   The reversal acting on a gray edge flips the orientation of all edges overlapping it, leaving all other edges unchanged.

We shall see that any connected component which is oriented can be transformed by a series of reversals to a set of trivial connected components that correspond to the identity permutation. The unoriented connected components impose a problem for us since we cannot split any of their cycles, nor delete any of their breakpoints by applying a single reversal. In some cases we can eliminate unoriented components. This is done either by applying a reversal that does not increase the number of cycles, but rather transforms some of the edges to oriented edges, or by applying a reversal that merges two or more unoriented connected components into one oriented component. The above idea for eliminating unoriented components allows a characterization of the unoriented components, on which we have to spend an extra reversal operation. We denote these components as hurdles. A more accurate description follows.