Tiles, tiles, and tiles

Think of any board game or a 2D computer game and most likely it involves a square grid somewhere. It could be just square tiles for displaying the graphics, but in general it’s also embodied in the entire essence of the game, guiding the gameplay. But there is one other option, and, maybe, a third one too.

We want to determine all the possible ways to use regular polygons to tile the space: all polygons should cover the entire space without overlaps or without leaving gaps. How many options are there?

First, let’s assume the case where all polygons are meeting at a vertex. Let’s say \(k\) \(n\)-gons are meeting at a single point, as in this figure:

All of these angles should be equal. If we use \(\alpha\) to mean the internal angle of the \(n\)-gon, we have that all these \(k\) copies of \(\alpha\) should sum to \(360^\circ\):

\[k \alpha = 360^\circ\]

Next, given that we use \(n\)-gons, we can compute the value of \(\alpha\), by looking at one of the side in the \(n\)-gon:

The marked angles are both \(\alpha/2\) and the angle at the center of the inscribing circle is \(360^\circ/n\). Since the angles in a triangle must sum to \(180^\circ\), we have

\[\alpha = \frac{n-2}{n} 180^\circ\]

Replacing in the other formula and solving for \(n\) we find that

\[n = \frac{2k}{k-2} = 2 + \frac{4}{k-2}\]

We know that both \(n\) and \(k\) are integers, so we need to find all integer values of \(k\) for which \(n\) would be an integer. From the last fraction, these are those values for which \(k-2\) is a divisor of 4, so \(k\) could only be 3, 4, or 6. Thus, there are only 3 possible grids: the equilateral triangle one, the square one, and the regular hexagonal one.

The square one is the one we saw in all games. It is easy to draw, each square can be mapped to an image tile. It is easy to reason about. It is used (almost) everywhere. But, it limits movement to just the 4 cardinal directions, giving rise to the Manhattan distance (also known as taxicab metric, something to cover in another article). Or, we could consider movign diagonally to cost the same as moving vertically/horizontally, giving rise to a totally different metric (to be considered in the same article).

But, what if we want more movement? Enters the hexagonal grid:

Now, there are 6 directions in which we can move with the same cost. If we want, we can add “diagonal” moves too, jumping over a side of a neighboring hexagons, though I have yet to find a place where this is used. We’re starting to see games using this grid, for example the popular Catan games. In fact, I wanted to use this grid to continue the linear algebra article. But, I got sidetracked into writing this one first, as an introduction. Will do the other one in one of the other days of the month. Until then, you can read a lot about this grid in this awesome guide.

Finally, there is the triangular grid:

Movement can be done directly to just 3 neighbors. If you want to move from one triangle to the triangle opposite which shares just a corner you have to go through 2 other triangles. This amplifies the metric issue, although one could also define a metric where you’d consider that triangle to be distance 2 apart too – by considering the cost to travel over an edge to be 1, while the cost to travel over a corner to be 2. I got this idea from this article, which also tries to convince readers that the triangular grids are not that weird. I’ll have to experiment more with these.

Before we conclude, look at the center of triangles in the triangular grid. They form regular hexagons, so they create a hexagonal grid! Similarly, the centers of hexagons in the hexagonal grid form a triangular grid. This is a nice observation that both links these two grids and can also help with subdividing one grid into a smaller one: something that is trivial with the square grid, possible with the triangular one, and complex with the hexagonal one – but this is the subject of yet another article.

Just like Columbo, a careful reader might say Just one more thing…. We started with the assumption that all regular polygons are meeting at a vertex. But what if we relax this condition?

For the squares here, we got a brick pattern. But, if you squint a little, this is just the hexagonal grid, again!

We can shift the triangles in a similar way, but not the hexagons. Try it! Other non-regular polygons can also tile the plane in this shifted approach, and we can even tile the plane with tiles that are completely aperiodic, although this is totally outside of scope for this set of articles.

Next article, we return to the subject of linear algebra and look at how these grids can give some intuition there.