Sorry, title probably not great, but I can't think of a better one. Here's my question. Suppose I have a $2n$ gon, with the edges identified in pairs, and no neighboring edges identified. Suppose also that I have have already done the standard thing and cut and pasted to eliminate "unnecessary" edge pairs, in other words I have the fundamental domain for surface of genus $n/2$. Now of course if I allow myself to cut and paste between any vertices I can rearrange this into the standard fundamental polygon for the surface. But suppose instead that before I allow myself to cut and paste to do this, I triangulate the polygon by connecting vertices, and then only allow myself to cut and paste respecting this triangulation. I can no longer necessarily get to the standard polygon, but has anyone seen anything about how close I can get in this case? In particular, I guess the question is how low can I make the total interleaving, say can I guarantee that no two identified sides will have more than N edges between them? I played with Mathematica about this a while ago, with $n=4$ and $n=6$, and it seemed that I could make sure that no paired edges ever had more than 2 others between them, but doubt that really means anything.

Thanks.

Here is an attempt to make clearer what I'm interested in and what I mean I agree with Lee that the question is probably confusing, I found it very difficult to explain without being able to draw a good picture and move things around. Let me try again. I am referring to the standard polygonal proof, and yes of course you might have to do a series of such cuts to get the final "standard" form. As far as what "no two sides will have more than N edges between them" means, think of starting at one edge and walking around the polygon, counting edges until you hit the edge identified with the edge you started on. What I am interested in is trying to minimize the maximum such count, in other words do this for each edge pair and for every possible rearrangement of the polygon, and then find the minimum of the maximum value. What this will depend on is the original triangulation. If you have no triangulation and allow for any cuts to rearrange, then of course the min will be $N=1$ since you will end up with $ABA^{-1}B^{-1}...$, but what I am interested in is if in an $n$ ton you start by triangulating with $n-3$ cuts connecting pairs of vertices and you are only allowed to use those to rearrange. Hope this makes it clearer.