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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 gr

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 http://logarithmic.net/pfh/ghost-diagrams See also the relevant paragraph in the readme (just search the word "easy").

DennisP 10y 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 10y ago
  So basically make a not-easy tileset with the shapes of Penrose tiles. Yes, this could be interesting.
hackpert 10y 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.
derefr 10y 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.
pfd1986 10y ago
Agreed. The best way to find 'complexity' and perhaps aperiodicity is by using contradictory rules. source: my own quasicrystal simulations (http://www.nature.com/nmat/journal/v14/n1/extref/nmat4152-s2.html http://www.nature.com/nmat/journal/v14/n1/extref/nmat4152-s2...)
mathgenius 10y ago
You need a backtracking algorithm to do Penrose tiles, so yeah, you can get stuck.

DennisP 10y 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 10y ago
So basically make a not-easy tileset with the shapes of Penrose tiles. Yes, this could be interesting.

hackpert 10y 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.

derefr 10y 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.

pfd1986 10y ago

Agreed. The best way to find 'complexity' and perhaps aperiodicity is by using contradictory rules. source: my own quasicrystal simulations (http://www.nature.com/nmat/journal/v14/n1/extref/nmat4152-s2.html http://www.nature.com/nmat/journal/v14/n1/extref/nmat4152-s2...)

mathgenius 10y ago

You need a backtracking algorithm to do Penrose tiles, so yeah, you can get stuck.