# How to calculate the number of different tiles in a tessellation? [closed]

For example, in the $$4.8.8$$ tiling, are there roughly as many octagons as squares? How would I find that out for an arbitrary tiling?
Edit: Ok yes, obviously there are infinitely many of each. What I mean is for any arbitrary section of that infinite tiling, what is the ratio of tiles? Obviously it won't be the same for exactly every section but I assume there would some kind of rough ratio.

• It is not at all clear what you are asking. For instance, as I would think that the literal answer to your question is Yes, because there are infinitely many octagons and infinitely many squares... but then I would guess that this answer would not be satisfactory to you. I suggest you hit the edit button and clarify what you are asking. Commented Jul 11 at 20:38

If the tiling is periodic, then you should find a fundamental domain of that tiling. A fundamental domain is a patch of tiles $$\mathcal{P} = \{T_1, \ldots, T_k\}$$ so that there exists a full-rank lattice $$L\subset \mathbb{R}^d$$ such that
• $$\bigcup_{t \in L} \mathcal{P}+t = \mathbb{R}^d$$ (periodically covers), and
• for all $$t,t'\in L$$, if $$(\mathcal{P}+t) \cap (\mathcal{P}+t') \neq \emptyset$$, then $$t=t'$$ (no non-trivial overlaps).
The frequency of tiles in a fundamental domain will then be the asymptotic frequency for large patches of your tiling. That is, for all $$\epsilon >0$$, there exists an $$R>0$$ so that all patches (that are not too weird) with radius at least $$R$$ will have frequencies within epsilon of the frequencies of tiles in a fundamental domain.