Hurdles

Let $\pi_{i_1},\pi_{i_2},\ldots,\pi_{i_k}$ be the subsequence of $0,\pi_1,\ldots,\pi_n,n+1$ consisting of those elements incident to gray edges that occur in unoriented components of $OV(\pi )$. Order $\pi_{i_1},\pi_{i_2},\ldots,\pi_{i_k}$ on a circle CR such that $\pi_{i_j}$ follows $\pi_{i_{j-1}}$ for $2\le j\le k$ and $\pi_{i_1}$ follows $\pi_{i_k}$. Let M be an unoriented connected component in $OV(\pi )$. Let $E(M)\subset \{\pi_{i_1},\pi_{i_2},\ldots,\pi_{i_k}\}$ be the set of endpoints of the edges in M. An unoriented component M is a hurdle if the elements of E(M) occur consecutively on CR. Let $h(\pi)$ denote the number of hurdles in a permutation $\pi$.

Lemma 11.10   Unoriented connected components cannot be resolved (transformed into the identity permutation) only by good moves.

Therefore, from lemma 11.10 we see that hurdles are unoriented connected components that cannot be solved by good moves. We can still make either a profitless move on a hurdle, that can possibly change some unoriented edges into oriented ones, or make a bad move, joining cycles from different hurdles, thus merging them and flipping the orientation of many edges and components on the way. For some hurdles, called superhurdles, there exist another unoriented component, which upon deletion of the superhurdle by a profitless move, becomes a hurdle itself. Furthermore, note that when merging two hurdles which are not consecutive along CR, no new hurdles are formed. Therefore, if we denote the number of hurdles in $B(\pi )$ by $h(\pi)$, and its change by $\Delta h$, we conclude that:

Theorem 11.11   Unless $\pi$ has exactly three hurdles, all of which are superhurdles, there exist a reversal for which $\Delta b - \Delta c + \Delta h=-1$. Thus the minimum number of reversals required to sort $\pi$ is $b(\pi)-c(\pi) + h(\pi)$.

Proof:A hurdle is destroyed by a profitless move, or at most two are destroyed (merged) by a bad move. In either cases at least one reversal is required to eliminate each hurdle. The argument above shows that it is possible to do so without generating any new hurdles along the way. The situation described in theorem 11.11 is bound to occur sometime during the sorting process if $\pi$ has an odd number of hurdles, all of which are superhurdles. We call $\pi$ a fortress in such a case, and write $f(\pi)=1$; otherwise we write $f(\pi)=0$. Note that in this case, one extra reversal is required to sort $\pi$.

Theorem 11.12    [13]. If $\pi$ is a signed permutation, then $d(\pi)=b(\pi)-c(\pi)+h(\pi)+f(\pi)$.