More Path Tiles
Do you remember the tiles I was talking about the other day? I generated the tiles for the next couple of cases, but the number of tiles got really big really quickly. So big that I really couldn’t draw them all by hand like I did with those first few.
So I wrote a processing app which reads the data for a tileset and generates a PDF file. Here’s what the output looks like.
We talked about the square tile with one ending per side last time. It’s a really simple case, and if you cancel out the rotations, there are only these two tiles.
The hexagonal tile with one ending per side is a little bit more interesting. These are the 5 different tiles.
Once we get to two endings per side, we can have tiles which are equilateral triangles. That’s because the odd number of sides times an even number of endings per side gives us the even number of endings which we need. There are 7 distinct tiles in this case.
The square tile with two endings per side is the one that got me started on this. Here are the 35 different tiles.
Then we start to get into the big numbers. The hexagonal tiles with two endings per side yield 1,799 different tiles. Factorials get big fast, don’t they? You’ll probably want to click through, but even the highest resolution one I uploaded to flickr isn’t big enough to see much.
Then we’ve got the square tiles with three endings per side. There are 2,688 different tiles in this case. Look at them all!
The next one would be the triangle with four endings per side. I haven’t run that one yet. Like the previous two, it’s going to be culled from a set of 479,001,600 different permutations. I expect that it’ll generate a couple of thousand tiles.
The next interesting one after that is the square tile with four endings per side. I don’t think that I’m going to run that one because the search space is 20,922,789,888,000 different permutations of the 16 endings.
There are some other cases which are mildly interesting. For example the octagonal tiles with one ending per side. There are 18 of those. Of course you need to mix them with square tiles to tile the plane. The pentagonal tiles with two endings per side are also kind of interesting. The set of permutations is only 3,628,800 in that case, but I haven’t bothered running that because they don’t tile the plane well.
The “Hexagon 2” set looks like the basis for the “Entanglement” game.
How about if you offset the square tiles? Some wargames do that in lieu of hex grids.
Yes, the hex-2 set is the one that Entanglement uses. I think you’d have to play that game for quite a while to see all 1,799 tiles.
The offset tiles are an interesting idea. And here I thought I’d reached closure!
What would happen if you attached the octagonal pieces and the squares together? You would end up with an irregular piece that could be used to tessellate the plane. I think it would have more unique permutations than the octagon alone but less than a decagon, since one of the paths on the square part has to remain very short. Can your program handle that? I’d be interested to know if that results in thousands of tiles or merely hundreds.
This might also be relevant:
Looking at that information, I feel like maybe the combined Octagon and Square tiles might have less than 100 unique designs, and if not certainly less than 200.
Scratch that. I forgot for a minute that the irregular shape is going to kill any possibility of shapes that are the same when rotated. The number will probably be gigantic.
Oh, that’s a cool idea. My first guess would have been something like 105 * 3, but I don’t think it’s that simple. I’ll have to think about that.
Can anyone help me with a high res and or vector image of the 1799 hexagons?
I’d love to get hold of them for a board game prototype I’m working on 🙂