If we have the diagram that represents a transitive permutation representation of $(p,q,r_o)$ for some $p, q$ and $r_o$, we often use this diagram to get diagrams for any $r>r_o$. We can do this by adding y-arcs in each face of the diagram. The process of addung y-arcs is termed boosting.
In this diagram th vertices joined by any x-arcs(that is fixed point of x) are called free vertices. These free vertices can be used to atach new x-arcs that will "boost" the diagram.
Here we suppose that there are i consecutive free vertices on a q-gon say Q. By using x-arcs we can join them with p-i free vertices of another q-gon, Say $Q'$. As the q-gons and x-arcs both are oreinted in anticlockwise directions, we get there are i-1 vertices of Q and p-i-1 vertices of Q' fixed by xy as in this figure above.
Now I am confused in one thing that after adding a q-gon, we get $q-(i-1+p-i-1)=q-p+2$ new y-arcs in the face to which Q' was added? Thus , the length of the cucle of xy corresponding to the face is incresed by q-p+2?
As far as I understand, Given before boosting D is the diagram for $<x,y: x^p=y^q=(xy)^r=1>$ We knoe thar each of $a_1,a_2,...,a_q$ in Q gon and $b_1,..b_q$ must be fixed by $(xy)^r$.
We know than $a_1,...,a_i$ are fixed by x, and $ b_1,..,b_{p-i}$ are fixed by x, This tell us something about the cycle for xy . When we add new x-edges $a_1$ to $b_{p-i}$ and $b_1$ to $a_i$ then what happens to xy- cucles that it is increased by r+(q-p+2). How?