Let $S$ be a Riemann Surface of genus 2. Is there a picture in the literature for a regular tiling of $S$ by 12 heptagons (where three heptagons meet at each vertex). Also, apart from the obvious restriction given by the Euler characteristic $2-2g=f-nf/2+nf/v$ (where $g$ is the genus, $f$ is the number of faces, $v$ is the number of faces meeting at each vertex) are there any obstructions for the existence of a tiling of a surface of genus $g$ by $n$-gons (where the same number $v$ of $n$-gons meet at each vertex).

I know that such a tiling exists (for a surface of genus 2 by heptagons), but I am unable to make a drawing. The construction goes like this (thanks to Mladen Bestvina): Start with a sphere. View this as a octahedron so that you have 8 triangles. 
Pick a hexagon (start from the top go down, left, down come back up from the opposite side)
We fatten up these 6 sides to look like circles with one ray coming from the middle. You should think of these as the quotient of 2 triangles attached along one edge <|> via rotation by 180 degrees (so you get two edges meeting in a point A and 2 edges plus a ray meeting in the other point B). We call the center of the circle C so that the ray goes from C to B.  
You have to view the point B_i (of the i-th side, i=1,2,3,4,5,6) as being attached to A_{i+1} for i=1,2,3,4,5,6 (of course mod 6) so each of these vertices has valence 5+2=7.
Now you take the double cover with fixed points C_1....C_6  and you get a surface of genus 3 with 28 triangles (28=2x8+2x6 where 8 is the number of triangles and 6 are the circles with a ray that unwrap in to two triangles). The vertices still have valence 7 (because the cover is etale here). Take the dual tiling and you are done!

This is just a curiosity.  I tried asking the first part of this on stackexchange with no luck.
I am also aware of the beautiful pictures of the famous tiling of a surface of genus 3 by 24 heptagons but you can not use that to obtain the tiling of a genus two surface in an obvious way)