The Breakpoint Graph

Let us now define a one-to-one mapping u from the set of signed permutations of order n into the set of unsigned permutations of order 2n. Let $\pi$ be a signed permutation. To obtain $u(\pi)$replace each positive element x in $\pi$ by 2x-1 and 2x, and each negative element -x by 2x and 2x-1. For any signed permutation $\pi$, let $B(\pi) = B(u(\pi))$. This description of $B(\pi )$ is equivalent to the following description: given a permutation $\pi = (\pi_1,\ldots,\pi_n)$, obtain a graph of 2n+2 vertices by replacing each positive element x in $\pi$ by 2x-1and 2x, each negative element -x by 2x and 2x-1, and augment with begin and end vertices, 0 and 2n+1. Black edges connect vertices $\pi_{2i},\pi_{2i+1}$ and gray edges connect vertices 2i and 2i+1.

We may now limit our discussion to signed permutations given that the reversals we perform on our unsigned permutation have a one-to-one correspondance to reversals in signed permutations.

Thus, sorting $\pi$ by reversals is equivalent to sorting the unsigned permutation $u(\pi)$ by even reversals. From now on we will consider the latter problem and by reversals we will always mean an even reversal.
Note that in $B(\pi )$ every vertex has exactly one black edge and one gray edge incident on it. Therefore, there is a unique decomposition of $B(\pi )$ into cycles. The edges of each cycle are alternating gray and black. Our goal is to sort our given graph $B(\pi )$ into n+1 trivial cycles.
Notice that in this formulation all reversals are one of three types:

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A reversal can act on two cycles, joining them. We call this move a bad move.

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It can act on one cycle, changing it. We call this move a profitless move.

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It can act on one cycle, splitting it. We call this a good move.

Let $b(\pi)=b(u(\pi))$ and let $c(\pi)$ be the number of cycles in $B(\pi )$.

Figure 10.10(a) shows the breakpoint graph of the permutation $\pi = (4,-3,1,-5,-2,7,6)$. It has eight breakpoints and decomposes into two alternating cycles, i.e. $b(\pi) = 8$, and $c(\pi) = 2$. The two cycles are shown in figure 10.10(b).

An alternative construction of the breakpoint graph constructs $B'(\pi)$, with vertices $0,\ldots,2n+1$, black edges $(\pi_{2i},\pi_{2i+1})$, and grey edges (2i,2i+1) for all $i=0,\ldots,n$. All vertices of $B'(\pi)$ are in disjoint cycles, but we ignore trivial cycles, i.e. cycles of length 2. Figure 10.11(a) shows the breakpoint graph of $\bar{\pi} = (4,-3,1,2,5,7,6)$, $B'(\pi)$ that has seven breakpoints and decomposes into two cycles (the trivial cycle that is composed of the vertices {2,3} is ignored). Let $c'(\pi)$ denote the number of cycles in $B'(\pi)$.

For an arbitrary reversal $\rho$ on a permutation $\pi$, define $\Delta b(\pi,\rho) = b(\pi,\rho)-b(\pi)$ and $\Delta c'(\pi,\rho) = c'(\pi,\rho)-c'(\pi)$. When the reversal $\rho$ and the permutation $\pi$ will be clear from the context we will abbreviate $\Delta b(\pi,\rho)$ to $\Delta b$ and $\Delta c'(\pi,\rho)$ to $\Delta c'$. As Bafna and Pevzner [2] observed, the following values are taken by $\Delta b$ and $\Delta c'$ depending on the types of the gaps $\rho(i,j)$ acts on (see figure 10.9):

1.

Two adjacencies: $\Delta c'=1$ and $\Delta b=2$.

2.

A breakpoint and an adjacency: $\Delta c'=0$ and $\Delta b=1$.

3.

Two breakpoints each belonging to a different cycle: $\Delta b=0$, $\Delta c'=-1$.

4.

Two breakpoints of the same cycle C:
a. $(\pi_i,\pi_{j+1})$ and $(\pi_{i-1},\pi_j)$ are gray edges: $\Delta c'=-1$, $\Delta b=-2$.
b. Exactly one of $(\pi_i,\pi_{j+1})$ and $(\pi_{i-1},\pi_j)$is a gray edge: $\Delta c'=0$, $\Delta b=-1$.
c. Neither $(\pi_i,\pi_{j+1})$ nor $(\pi_{i-1},\pi_j)$ is a gray edge, and when breaking Cat i and jvertices i-1 and j+1 end up in the same path: $\Delta b=0$, $\Delta c'=0$.
d. Neither $(\pi_i,\pi_{j+1})$ nor $(\pi_{i-1},\pi_j)$ is a gray edge, and when breaking Cat i and jvertices i-1 and j+1 end up in different paths: $\Delta b=0$, $\Delta c'=1$.

 

  

Figure: All possible cases of changes to $\Delta b$ and $\Delta c'$by applying a reversal (see section 10.4.4).

We call a reversal proper if $\Delta b-\Delta c'=-1$, i.e. it is either of type 4a, 4b, or 4d. We say that a reversal $\rho$ acts on a gray edge e if it acts on the breakpoints which correspond to the black edges incident on e. A gray edge is oriented if a reversal acting on it is proper, otherwise it is unoriented. Notice that a gray edge $(\pi_k,\pi_l)$ is oriented if and only if k+l is even. For example, the gray edge (0,1) in the graph of figure 10.10(a) is unoriented, while the gray edge (7,6) is oriented.

  

Figure: (a) The breakpoint graph, $B(\pi )$, of the permutation $\pi = (4,-3,1,-5,-2,7,6)$. Black edges are solid; gray edges are dashed; oriented edges are bold. (b) $B(\pi )$ decomposes into two disjoint alternating cycles. We disregard whether grey edges are oriented or not. (c) The overlap graph, $OV(\pi )$. Black vertices correspond to oriented edges.