Understanding the proof of solvable conjugacy problem for hyperbolic groups. https://www.math.ucdavis.edu/~kapovich/280-2009/hyplectures_papasoglu.pdf
https://courses.maths.ox.ac.uk/node/view_material/48431
In the first link the theorem I am talking about is on page $29$, Theorem $3.18$. In the second link the theorem is on page $63$, Lemma $6.2$.
We want to prove that that the conjugacy problem is solvable for $\delta$- Hyperbolic groups. To do that we obtain a bound on $|x|$ where $g_1 = xg_2 x^{-1}$ for $g_1,g_2 \in G$. The argument starts in both cases by claiming that if $x$ is a minimal $x$ such that $g_1 = xg_2 x^{-1}$ with $x= y_1...y_n$, and all $y_i$ are generators, and we let $x_i = y_1...y_i$, then $|x_ig_1x_i^{-1}| \leq 2\delta +|g_1|$ for $|g_1| < i \leq n-|g_2|$. I do not understand where this comes from. It probably comes from using $\delta$ thinness on suitably chosen triangles as hinted in the first link, but I do not understand how we can bound $|x_i|$ at all. The range for $i$ also remains a mystery to me. Further clarifications on this proof would be appreciated as maybe then I would be able to complete the proof.
My understanding of $\delta$ thinness here, is that given any triangle and given any point on one side, there is a point on the other edges that's at least '$\delta$ close the first point. I'm not sure how that translates in terms of words and generators on the Cayleh graph.
 A: Consider the geodesic quadrangle in the Cayley graph with two "vertical sides" and two horizontal sides labelled $x, g_1, x^{-1}, g_2^{-1}$, the ("horizontal") sides $g_1,g_2$ are much shorter than the "vertical" sides labelled by $x$ because we assume, by contradiction, that there is no algorithm to find $x$ given $g_1,g_2$. We can also, as you  noted, assume that $x$ is the shortest possible. Then each side is in a union of $2\delta$-neighborhoods of the other three sides (divide the quadrangle by a diagonal). The intersections of $2\delta$ neighborhoods of the short sides with the left vertical side are small. Therefore a large portion of the left side is in a $2\delta$-neighborhood of the right vertical side. That means for  most $i$ $x_ig_1x_i^{-1}$ has length at most $d=2\delta(1+|g_1|+|g_2|)$. Here $x_i$ is  the suffix of $x$ of length $i$. The length of $x$ can be assumed to be $\ge \exp(d)$, so for some $i<j$ we have $x_ig_1x_i=x_jg_1x_j$. But that implies, we can cut the subword  between $x_i$ and $x_j$ from $x$ and still get a (shorter) conjugator $x'$, a contradiction.
