Rose V: Square Wrap-Up and Isometric!

We’ve made a lot of progress on the Rose problems! Last time, we proved that given a rule of the form “N colored points in a row is not allowed”, we have optimal colorings for all square-grid spaces for as many dimensions as we want! That may sound pretty specific, but let’s take a broader look at just how many cases that covers.

I’m imagining sorting all possible rules into three different categories. They are: rules that prohibit only colored points, like “Three colored points in a row are not allowed”, rules that prohibit only uncolored points, like “Three uncolored points in a row are not allowed”, and rules that have some combination of those rules, like “an uncolored point between two colored points is not allowed”. Sidenote: If you remember from the Rose II definition, I decided to phrase all rules as negative, as “XYZ is not allowed”, because positive rules, like “you must have a colored point between two uncolored points” vanish as we go to infinity, or can be rephrased as negative rules.

If you think about these three categories, you’ll realize that we just solved the first one, for square grids of N dimensions, and the latter two are trivial: color the whole space and you’ll get an optimal coloring that won’t ever need an uncolored point for the rules to apply to.

What I’m saying is, give me a square grid of any dimension and any single 1D rule you can think of, and I can find and prove what the optimal coloring is!

But before we get to the more complicated “multiple rule” cases, let’s look at some alternatives to regular integer spaces. Namely, the isometric space!

This is a beautiful, interesting space. The purple triangles represent the points, and each point has six neighbors! The 2D square grid has two “directions” in which rules can be broken and adds a direction whenever you add a dimension. The 2D isometric grid has three directions from the get-go: one depicted horizontally here, and two on the 60° and 120° diagonals. And it’s even more interesting in other dimensions – I can sort of visualize a 3D isometric space, but it’s not clear to me if it’s well-defined or even exists in higher dimensions!

Let’s jump right in, as usual. For square grids, the easiest, quickest rule has been “Two colored points in a row are not allowed”. Let’s try our techniques on this new space.

And somehow, alternating colored points on the first row sort of… doesn’t work. Doing it means the entire next row must be uncolored. We can repeat this process on the next few rows to get a total of ¼ of the space colored. We can tell that it’s ¼ because of the trick we used for the square grids – find a tile that repeats. The red outlined parallelogram can be used to tile the whole pattern, and it’s ¼ colored, so the space is ¼ colored.

Then, I tried spacing them out more evenly, to get this configuration. It gets us 1/3 of the space colored, using the red outlined parallelogram.

After I found that one, I tried and tried and couldn’t find one that worked better. Maybe this is the best we can do in the isometric space. So then I went about trying to prove it when I hit upon this tile.

A simple, triangular tile, which you can alternate to cover every point in the space. But under our rules, the most colored that this tile can be is 1/3. If we color two points, that means there are two adjacent colored points, and that’s not allowed. And because this tile can tile the entire space, that means that the most colored the space can be is 1/3 colored!

So, we’ve proven that the 1/3 colored space above is the best you can do. We’ve done this by coming up with a space that is 1/3 colored, making 1/3 the lower bound, and using the pigeonhole principle to show that there can’t be anything better. If someone tells me they have a better fraction, I can break the space into these triangles with three points in them, and show that their coloring must have at least one triangle that has more than one colored point!

Next time we’ll look at some other rules for the isometric space, which is already looking much more complex than the square grids we’ve solved! With the rule we solved for today, we got ½ of the square grid colored easily!