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 along CR. Let $h(\pi)$ denote the number of hurdles in a permutation $\pi$.

 

Therefore, from lemma 10.7 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 denote the change in $h(\pi)$ by $\Delta h$ when a reversal is applied, we conclude that:

 

The situation described in theorem 10.8 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 $\pi$.