# Possible tiling patterns for equiangular hexagon with alternating side lengths

I want to make a pattern of an equiangular hexagons with side lengths of 1-5-1-5-1-5. Here are 3 examples I made:

Which different patterns exist? Are there more than the ones I show? And which one is the most space-efficient, so that it minimizes the wasted space between the hexagons?

I checked the wikipedia page on hexagonal tilings but I cannot find examples for this specific hexagon.

• I'm going to hazard a guess that you will not be able to reduce wasted space below the current, uh ... 3/49? that it currently is. Commented Oct 7, 2016 at 9:46
• @DanUznanski: I agree that the most space-efficient tiling with the shown hexagon is $46/49$ (about 93.88% efficient; would leave small empty hexagons inside each group of six hexagons) -- basically equilateral triangle tiling (as is typical for hexagonal tilings) -- but OP's example is less efficient: each group of six hexagons has an inner hole of $6/49$ ($1/49$ per hexagon), which means Stijn's example is only .. umm.. $46/(49+1) = 23/25$ = 92% efficient? Commented Oct 7, 2016 at 11:16
• What is your definition of “pattern”? Does it have to be periodic? Periodic in two independent directions? Isohedral (which your example pattern seems to be)? Isogonal (which your first example is not)? Isotoxal (which the first example isn't either)? Aligned to multiples of $60°$? Judging from your pictures I'd assume you'd want isohedral tilings.
– MvG
Commented Oct 7, 2016 at 14:04
• Both designs 1 and 2 have the same efficiency of 46/49, right? Commented Oct 7, 2016 at 14:04
• @MvG - I just need the pattern to be "tileable", I'm not sure what isohedral, isogonal and isotoxal mean. Alignment doesn't matter and it doesn't have to be periodic. Do you know of such patterns that are non-periodic or have non-60° allignment? Commented Oct 7, 2016 at 14:12

Here are the patterns I could think of, represented by their unit cells and with their densities listed beside:

The first tiling shown in the question has no code in my diagram and has density $\frac{46}{49}=0.9388$. This is not very interesting because it is a rearrangement of the A0 tiling (the second in the question), which has the same density.

From the A0 tiling we can shift the hexagons on their long edges by one to five units to create the A1 to A5 tilings. As the holes between the tiles grow, so does the density decrease:

• A1 has density $\frac{23}{26}=0.8846$
• A2 has density $\frac{46}{61}=0.7541$
• A3 has density $\frac{23}{38}=0.6053$
• A4 has density $\frac{46}{97}=0.4742$
• A5 has density $\frac{23}{62}=0.3710$

The B tiling is the last one shown in the question and has density $\frac{23}{27}=0.8519$.

Now to answer the question of densest possible tiling. When I started typing this I thought it was tiling M, which has density $\frac{23}{24}=0.9583$ and features rows of tiles that can slide over each other.

Then I realised the rows could be pushed into each other a bit more, resulting in the real winner: tiling N with density $\frac{46}{47}=0.9787$.

So if you want to tile a real wall or floor with this shape, your best bets are tilings M and N. While M is less dense, it isn't chiral like N, and its rows may make it better suited to the (usually) rectangular areas tiles are supposed to cover.

• Wow, nice, I didn't think it was possible to get this dense! Commented Oct 7, 2016 at 15:04
• @Stijn Since you found it nice, you are encouraged to click the tick next to my answer. Commented Oct 7, 2016 at 15:11
• Ah yes, I'm a bit new here. I guess your answer would also work for hexagons with another ratio than 1:5, right? Commented Oct 7, 2016 at 15:48
• @Stijn Sure, it can be extended to hexagons of other ratio. As (short side length)/(long side length) goes to 0 the tiling approaches triangular; as it goes to 1 the tiling approaches hexagonal. Commented Oct 7, 2016 at 15:52
• @DanUznanski Inkscape. The same program I use to draw ponies (see my profile). Commented Oct 8, 2016 at 1:03