# A brickwork fractal

On my way to work in Yaletown, I walk along some sidewalks with interesting brick patterns. The three above happen to satisfy a nice mathematical property: the bricks are arranged so that no four corners meet at the same place.

As my relative Oliver Linton has pointed out, paving bricks come in a variety of shapes and sizes, which allows for many more beautiful tilings. For example, the addition of 2×2 square tiles make it possible to construct rectangular tilings that fit together to tesselate the plane while preserving the four-corner rule.

This is a very exciting observation — at least, to anybody who likes recursion! We might be able to use the same trick to construct an infinite sequence of increasingly intricate tilings that converge to a self-similar “fractal tiling”. The simplest non-trivial example I could find involves a set of 5×5, 10×10, and two 5×10 rectangular tilings.

Starting with any of these four layouts, we can replace each of the 1×1, 2×2, and 1×2 bricks with a corresponding 5×5, 10×10, or 5×10 rectangular tiling in the correct orientation. (This will produce a few four-corner intersections, but we can fix these by merging adjacent pairs of 1×2 bricks.)

Repeatedly performing this operation gives an infinite sequence of tilings, but can we say they converge to anything? A tiling *T* can be identified with its outline *∂T* (i.e. the set of points on boundaries between two or more tiles). Note that if a point is in *∂T _{i}*, then it will be in every subsequent

*∂T*unless it is one of the few bricks merged in

_{j}*∂T*. So we might sensibly define the limiting object of the tiling sequence

_{i+1}*T_i*as the union

*⋃*. This self-similar dense path-connected set satisfies the topological equivalent of the “four corners rule” — a pretty interesting list of mathematical properties!

_{i}(∂T_{i}⋂∂T_{i+1})The same strategy could be applied to other sets of (2n+1)×(2n+1), (4n+2)×(4n+2), and (2n+1)×(4n+2) tiles with similar boundaries. What’s the prettiest brickwork fractal you can find?