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Comment by throwaway000002

10 years ago

Brilliant!

Don't really understand the technique, but would like your thoughts on if it's possible to give a Penrose tile set as a seed and see if aperiodic order is generated.

Lovin' it!

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throwaway000002

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ExUtumno  10 years ago

Thanks!

I'm not sure, but I think that Penrose tilesets are what I call "easy": you can't run into a situation where you can't place a new tile. It would be great if someone here could confirm or deny this.

So if this is the case, then Penrose tilesets are not interesting to WFC, because you can produce arbitrary tilings with much simpler algorithms.

Right now though WFC is only working with square tiles, but it's not hard to generalize it to arbitrary shapes. Paul F. Harrison made a tiling program that supports hex tiles: http://logarithmic.net/pfh/ghost-diagrams See also the relevant paragraph in the readme (just search the word "easy").

  • derefr  10 years ago

    Penrose tiles are "easy" on an unbounded canvas, but I'm pretty sure they're a 100%-probable "contradiction" (because they're aperiodic) on a bounded toroidal canvas.

  • hackpert  10 years ago

    I think you're right about Penrose tiles being easy. One could, however, potentially use this with polygonal tiles to find their Heesch numbers and maybe even make some advancements in solving some of the many unsolved problems associated with Heesch tiling.

  • DennisP  10 years ago

    Maybe it could be interesting to place Penrose tiles with the simple algorithm, but color them with your algorithm just like you're currently coloring squares.

    • ExUtumno  10 years ago

      So basically make a not-easy tileset with the shapes of Penrose tiles. Yes, this could be interesting.