4 ms·

I'd rather get these tiles for paving in my house floor.

• amelius 11y ago
  These tiles have nice properties, but it seems the property that you can cover an arbitrary shape with them is missing (even if allowing cropping of the result).
  • gjm11 11y ago
    If you allow cropping, you can always do this. Suppose the tiles are T1, ..., Tn. The "self-tiling" property means you can tile each Tk with smaller copies of T1, ..., Tn. Or, equivalently, you can tile a larger copy of each Tk with copies of T1, ..., Tn. So: pick one of the tiles. Tile it with smaller copies of the tiles. Tile each of those with still-smaller copies. Tile each of those with still-smaller copies. Etc. Now pick a region in the original tile that's the same shape as the thing you're trying to cover. This process covers (the whole original tile, and hence in particular) that region with finer and finer copies of the tiles.
    • amelius 11y ago
      Yes, it seems simple, but I'm not sure if this proof is sufficiently rigorous that it allows for mathematically pathological tiles.
      • gjm11 11y ago
        I don't think it requires anything more than that at least one tile has nonempty interior.
• madcaptenor 11y ago
  I was thinking the same. I've got a couple bathrooms that need updating.