Let
be the subsequence
of
consisting of those elements
incident to gray edges that occur in unoriented components
of
.
Order
on a circle CR such that
follows
for
and
follows
.
Let M be an unoriented connected component in
.
Let
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
denote the number of hurdles in a permutation
.
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
by
,
and denote the change in
by
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
has an odd
number of hurdles, all of which are superhurdles. We call
a
fortress in such a case, and write
;
otherwise we write
.
Note that in this case, one extra reversal is required
to sort
.