Images of the fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$ whose Euclidean area is large Let $S$ be a compact subset of the closure of the upper half plane. (Assume that $S$ is a (Euclidean) rectangular box, if you wish.) Let $D$ be the standard fundamental domain of $\text{SL}_2(\mathbb{Z})\backslash \mathbb{H}$. What is a simple way to construct all maps $g\in \text{SL}_2(\mathbb{Z})$ such that the Euclidean area of $g D\cap S$ is at least $\epsilon$? Is there a (good) a priori bound for the size of the entries of a matrix $g$ satisfying such a property?
(Yes, my question is motivated by the need to produce a good conference poster, but I hope that will make it more rather than less interesting.)
 A: Well, here is a simple way to construct maps, which might work well: sample points in your domain randomly, so that the area of the Voronoi region is of order of $\epsilon$ (sampling something like $2 A(S)/\epsilon$ points should work.) Then use the ``extended'' continued fraction algorithm (which gives the transformation matrices), as described in my paper "How to pick an random integer matrix". The set of all matrices you get should be more or less what you want.
A: This turns out to be an easy problem. For any $\epsilon>0$, we can cover all of $\mathbb{H}$ except for a domain U of area $\epsilon$ and its (horizontal) translates by $\mathbb{Z}$ using words on $S = \left(\begin{matrix} 0 & -1\\
1 & 0\end{matrix}\right)$ and $T = \left(\begin{matrix} 1 & 1\\ 0 & 1\end{matrix}\right)$ that involve at most $\ll \log(1/\epsilon)$ appearances of the involution $S$ (which corresponds to the map $w:z\mapsto -1/z$), and at most $\ll 1/\epsilon$ appearances of $T$ (which corresponds to the map $z\mapsto z+1$).
This is
(a) essentially optimal: $(D+n)^{-1}$ (where $D$ is the fundamental domain) has Euclidean area $\gg 1/n^2$, and $ST^{-1}ST...ST^{-1}ST(D)$ ($(\log \epsilon)$ repetitions) isn't that small either;
(b) good enough in practice, if, say, we are producing a poster: powers of $T$ come for free, whereas each appearance of $S$ corresponds to one more step in the continued fraction algorithm (the obvious one; I just checked that it's the same as the one in page 7 of the paper Igor R. kindly linked to in the above).
Sketch of proof. Write $V$ for the closure of the standard fundamental domain $D$. The union of $V+\mathbb{Z}$, $V^{-1}+\mathbb{Z}$ and $(V+1)^{-1}+\mathbb{Z}$ covers all of $\mathbb{H}$ except for a set contained in $\{x+i y: 0< y < 1/(2\sqrt{3})\}$. Let us look at the intersection of that set with $\{x+i y: |x|\leq 1/2\}$. All of its elements $z$ satisfy $|z|\leq 1/3$. Hence, in a neighbourhood of $w^{-1}(z) = -z^{-1}$, the map $w$ shrinks
everything by a factor of at least $3-\epsilon$. Now repeat.
(As for the map $z\mapsto z+1$: the map $w$ shrinks anything in $\{x + i y: y>0, n+1/2\leq x\leq n+3/2\}$ by a factor $\ll 1/n$.)
