If your single 1D tile can be reduced to a finite set of nonnegative integers, including 0 for definiteness, then any tiling can be converted to a periodic one. Suppose the integers are covered by some tiling. For each K, consider the set of integers >K which are covered by some tile that begins at a value <=K. This set is of bounded width, not exceeding the width of the tile. So there are at most 2^width possible shapes for the set. Somewhere in the first 2^width+1 integers, there will be a K' and K'' with congruent sets (i.e., shifted by K''-K'). The supertile defined by the shift, i.e. tiles covering up to K'' but not entirely <=K', has a set of holes on the left that matches the covered squares on the right. So the supertile makes a periodic covering of the integers. There are easy tiles which can't be 'integrated' (rationalized?), but do cover the real line, so maybe there's an irrational aperiodic disconnected 1D tile. Rich -------- Quoting Warren Smith <warren.wds@gmail.com>: <clip>
Another problem I'm interested in is this.
You've all heard of the Penrose tiling (set of two tiles which only tile aperiodically). It is a famous open problem ("Einstein" problem) whether a single 2D tile exists which only tiles aperiodically. I would like to pose the same problem in 1D where now I do not demand that the tile be a connected set. I mean: 1D tiling? Now THAT'S simple. But... I do not know the answer, even in 1D. I feel quite embarrassed about it. I mean, you have to be an idiot not to be able to solve 1D tiling.
If there is such a tile, it probably has a fractal-esque nature. For a (not quite) example Cantor's "exclude middle third" set is tiled by two copies of itself and is fractal in nature, so you could say it is a solution of this optimization problem: "minimize area subject to requirement two copies tile itself and it contains >=2 points."
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