How do you calculate the smallest cycle possible for a given tile shape? If you connect together a bunch of regular hexagons, they neatly fit together (as everyone knows), each tile having six neighbors.  Making a graph of the connectivity of these tiles, you can see that there are cycles all over the place.  The smallest cycles in this graph have three nodes.
$f(\text{regular hexagon}) = 3$

If you use regular pentagons, they don't fully tile the plane, but they do still form cycles.  The smallest cycles have 6 nodes.
$f(\text{regular pentagon}) = 6$

If you use regular heptagons, the smallest cycle also seems to require 6 tiles.
$f(\text{regular heptagon}) = 6$

And it's the same with nonagons, 6 tiles to a cycle:
$f(\text{regular nonagon}) = 6$

Is there a general way to calculate the minimum number of tiles needed to form a cycle?
$f(\text{polygon}) = ?$
Ideally, I'd love to find a method that worked for any equilateral non-intersecting polygon, but even a method that works for any regular polygon would be great.
Sorry for the crudeness of the drawings.
 A: I did some experiments for regular polygons, for up to $n=24$ (click to enlarge):

These experiments suggest that


*

*If $6\vert n$ you get a $3$-cycle

*Else if $2\vert n$ you get a $4$-cycle

*Else you get a $6$-cycle


That you can get the $3$-cycle for anything that has edges aligned as a hexagon has them is pretty obvious. On the other hand, for reasons of symmetry any arrangement of $3$ regular $n$-gons has to be symmetric under $120°$ rotation, so you have to have those hexagon-aligned edges else it won't work.
It's also easy to see that you can get a $4$-cycle if $4\vert n$, since that's the obvious arrangement for squares. The other $4$-cycles are more tricky. The most systematic (and reasonably symmetric) choice there would be a configuration where two polygons have opposite vertices aligned with one axis, while the other two polygons have an axis perpendicular to the first passing through the centers of two opposite edges. Something like this:

Right now I have no obvious answer as to why you can't get $4$-cycles for odd $n$, nor why $6$-cycles are always possible. Looking at the angles it's easy to see that the length of the cycle has to be even if $n$ is odd, so once the $4$-cycles are ruled out for those cases, the $5$-cycles are out of the question and the $6$-cycles will provide the solution. Perhaps others can build more detailed answers with the help of my pictures.
