A strip tiling tiles a strip between 2 right angles of some fixed distance. A finite set of finite tiles is considered here.

Let $StripT$ be the set of lengths of strips that can be tiled. $$\left({l_1,l_2}\right)\in StripT \Rightarrow l_1+l_2\in StripT$$

An angle tiling tiles a sector of some fixed angle.

Let $AngleT$ be the set of lengths of strips that can be tiled. $$\left[{\left({\theta_1,\theta_2}\right)\in AngleT \land \theta_1+\theta_2\leq 2\pi}\right] \Rightarrow \theta_1+\theta_2\in AngleT$$

\section{Bootstrapped Tilings}

Let the tuples of sets $\left({Lengths_1,Angles_1}\right)$ and $\left({Lengths_2,Angles_2}\right)$ exist such that $Lengths_1\cap Lengths_2$ and $Angles_1\cap Angles_2$ are both empty (showing disjoint sets).

Suppose the following statement is true: $$\left[{Lengths_1\subset StripT\land Angles_1\subset AngleT}\right] \Leftrightarrow \left[{Lengths_2\subset StripT\land Angles_2\subset AngleT}\right] $$ Then fractal tilings can be made to bootstrap tilings of these strip lengths and sector angles. I would like an example nontrivial bootstrapped tiling or proof that such bootstrapping is impossible.

Such a bootstrapped tiling would likely use strip lengths or sector angles not in the sets in the tuples. Example:

\section{Regular Polygon Tilings}

The simplest tilings use regular polygons. A regular polygon with $n>2$ sides and angle $\theta<\frac{pi}{2}-\frac{pi}{n}$ has edge length $2\text{ArcCosh}\left({\frac{\cos\left({\frac{\pi}{n}}\right)}{\sin\left({\frac{\theta}{2}}\right)}}\right)$.

Some of these tilings neither fill a sector nor a strip.

To tile a sector, set $\theta=\frac{\pi}{k},k\in\mathbb{Z}^+$. This gives $$\left\{{\frac{\pi}{k} \mid k\in\mathbb{Z}\land k>1}\right\}\subset AngleT$$ and via addition: $$\left\{{q2\pi \mid q\in\mathbb{Q} \land 0<q\leq 1}\right\}\subset AngleT$$

To tile a strip, set $\theta=\frac{\pi}{2k},k\in\mathbb{Z}^+$. This gives $$\left\{{2\text{ArcCosh}\left({ \frac{\cos\left({\frac{\pi}{n}}\right)} {\sin\left({\frac{\pi}{4k}}\right)}}\right) \mid \left({n,k}\right)\subset\mathbb{Z}\land k>1 \land n>2}\right\}\subset LengthT$$


\section{Simple bootstrapping from known tilable angle}

Let $\theta\in AngleT\land k\in\mathbb{Z}^+$. Consider a pentagon with angles $\left({\pi-\theta,\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2}}\right)$. Start tiling a strip length $l_1$. Let the pentagon have sides $\left({l_1,l_2,l_3,l_4,l_5}\right)$. The angle $\pi-\theta$ touches the vertex touching the sides $l_4,l_5$. Let $l_3=k*l_1+l\land l\in StripT$. $l_1$ bootstraps as a fractal. Todo: express $l_1$ in terms of $\theta,l$.

It would be nice to see lengths and sides which simultaneously bootstrap off of each other rather than requiring a known tiling for the angle.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.