Construct a set of local rules in plane such that all permitted tilings are non periodic.(there exists at least one tiling)

Question

This is the problem given to me on a practice.

I imagine the problem is understandable but if there is anything unexplained, please tell me.

I have tried various things to solve this problem but all of them lead to a dead-end, and so far I don't think I have made any good progress to share.

Here are some definitions I think would be helpful to write.

"local rules: a set of $n.n$ squares which are forbidden in the tiling ($n$ is not a constant and can be any positive integer greater than $1$)",

"non periodic : means there is not a non_zero vector such that after adding the tiling to that vector we reach the same tiling",and the plane we are tiling the plane using $1.1$ black or white squares

Answer 1

You can use a set of aperiodic Wang tiles as a starting point.

Wang tiles are normally depicted with colours, but you can choose to represent each tile by almost any n by n pattern of black and white cells, as long as they are unique enough.

Then enumerate all ways that you are allowed to make a patch of four Wang tiles which form a 2n by 2n square of black and white cells. List all n by n sub-squares of these allowed patches, giving you all the posssible n by n patterns that you will allow in the tiling as a whole. Then disallow all other n by n patterns.

This shows it can be done, though the list of local rules will be very long.

Answer 2

Allow only "pixelated sections" of Penrose tilings.

To elaborate: Let us first assign integer coordinates to all parallelogram vertices in the above image:

• Assign $(0,0)$ to the centre (where five "thick" parallelograms meet)
• Assign $(-3,0)$ to its left neighbour and from there assign $(-1,-2)$, $(2,-1)$, $(2,1)$, $(-1,2)$ to its other four neighbours counter-clockwise.
• From this, you can extend assigning integer coordinates to parallelogram vertices by using the parallelogram rule, i.e., that we demand $ x_2-x_1=x_3-x_4$ and $y_2-y_1=y_3-y_4$ for a parallelogram with vertices $(x_1,y_1),\ldots,(x_4,y_4)$.

Now turn this into a tiling of the square latice by black/white tiles, namely make the tile at $(x,y)$ black iff $(x,y)$ is one of the infinitely many vertices labelled above. This is an aperiodic tiling of the square lattice.

Now forbid all $6\times6$ patterns that do not appera in this aperiodic tiling. (Or put differently: Allow only such $6\times 6$ patterns that do occur). It requires checking, but it seems quite clear that all lattice tilings that use only allowed patterns must correspond to a Penrose tiling and therefore are aperiodic.