Number of ways to choose a closed path of given length on a square lattice

Question

Also known as self-avoiding polygons, this is an unsolved problem. However, to leading order in the asymptotic limit, the number of polygons of a given perimeter scales exponentially with perimeter length (see here for example). I am curious to know if there is a simple, intuitive argument for why we expect this to be the case, as opposed to say some power law.

Answer 1

Consider taking $n$ steps on the square grid, each step being either East or North. There are $2^n$ ways of doing this.

You can create a self-avoiding polygon by connecting together four copies of this set of $n$ edges, each copy being 90 degree rotated from the previous one. You will need 4 extra edges to space them out to make sure the copies don't overlap.

So there are at least $2^n$ self-avoiding polygons of perimeter $4n+4$.

Edit:
Actually, the above procedure double-counts a few of the self-avoiding polygons. This can be fixed by using two spacer edges (8 all together), e.g. appending one East and one North step after the $n$ freely chosen steps. The conclusion doesn't really change: There are at least $2^n$ self-avoiding polygons of perimeter $4n+8$, so it grows (at least) exponentially.