Finding the list of all closed curves in a "pair of pants" surface, having exactly 11 self-intersection points 
Given a (small) natural number, for example $k=11$, is there any way to find all closed curves in an (arbitrary) "pair of pants" surface, having exactly $k$ (topologically non-removable) self-intersection points?

My attempt: I do not have any good idea. What I am doing know is writing words, such as, $abab^{-1}...$ (where $a,b$ is generators of the fundamental group) and then counting the self-intersection points. But it is impossible write all of the words. Thanks!
 A: In order to do this you'll have to prove some kind of bound. Given a cyclically reduced word $w$ in the letters $a,a^{-1},b,b^{-1}$, if we let $L(w)$ denote the length of this word, and if we let $I(w)$ denote the self intersection number of the word, then the form of the bound should be something like
$$I(w) \ge f(L(w))
$$
where $f$ is some increasing function that is independent of $w$. Once you have that bound, you can reduce your search to finitely many possibilities by taking the first value of $L$ for which $f(L) \ge 12$.
If I had to take a wild guess, I would say that this is possible with $f$ some function of linear growth. The reason that I guess this is that there are only three isotopy classes of embedded circles on the pair of pants, represented by the three boundary circles, and also represented by the three words $a$, $b$, $b^{-1}a$. The word $w$ can be factored into a certain minimal number of subwords of the form $a$, $a^{-1}$, $b$, $b^{-1}$, $b^{-1}a$, and $a^{-1}b$, this minimal number is linear in $L(w)$, and I guess that the self-intersection number is greater than that amount, or perhaps greater than that amount minus some small constant.
