Valerie

i realized i never drew nerevars face

uwa... foul murder...

req'd by @psyanidemilk

for the life of me i can't figure out what a sports lorry is

So my current project of n-dimensional chess has involved perhaps, some of the most interesting combinatorics I've ever handled. Let's talk about them! It's a perfect showcase of what I think are the 3 most important parts of math 1.> Start with simple, easy examples 2.> Break large problems into smaller parts 3.> Shift your perspective and abstract problems in a way that they are easier to deal with, while retaining a way to shift back once you've solved it to reapply what you've done. So, let's first start with the problem. Here's a piece in 5D chess known as a unicorn. It can move along any triagonal, which means it has to pick 3 out of the 5 dimensions available. in 4D time travel chess, it actually gets to move in 6 dimensional space, but it was originally introduced in space chess, a 3D variant. My point here is that n-dimensional chess means that we have to be able to list these kinds of moves for a lot of different dimensionalities, and hard coding them quite simply isnt an option. So, we turn to combinatorics!

We'll start with point 1.> by starting with our easiest example. Let's take a look at a bishop's movement in 2D space first to get a good baseline. The directions a bishop can travel along in 2D space are as follows

Now I purposely arranged this in a way that if we replace negative numbers with 0, we get

And hey! That's just counting in binary! Which makes sense, because we only have two choices, positive and negative. Once we move up to 3D space, we can actually re-use this tactic as well. Let's list all the possible combinations of directions in 3D space first, without worrying about the sign. This is what I mean in part 2.> by breaking complex problems into smaller ones, in this case breaking the movement into directions and signs.

Basically, now we can take each one of these 3 possibilities and just apply our same trick by ignoring whichever part the 0 is in. And this works for triagonals and quadragonals in higher dimensional space as well. This makes iterating through signs the easier part Of course, that means the other half is going to be the hard part. Now, with diagonals it isn't too bad. Lets look at diagonals in 5D space. Get ready because this one is a bit more of a jump

Basically what I'm doing here is I'm starting with each 1 as far left as possible, and sliding them over repeatably. not the best explanation, but its doesn't matter because this method doesn't extend well to triagonals very well (choosing 3 directions at once). Instead, we take a shift of perspective, demonstrating point 3.> Instead of representing this as a series of 1's and zeros, we simply represent the *position* of each 1. For example

We have a 1 in the 0th and 3rd locations, so we can simply represent this as just (0, 3). Suddenly, instead of trying to figure out how many ways there are to arrange 1's and 0's, we are asking how many ways can we choose 2 numbers out of 0-4. Now, there is a slight catch in that (0, 3) and (3, 0) are the same thing, so when counting them order doesn't matter, making this a simple combination problem. However we don't just want to count them, we also want to iterate over them. Lets try to generate all possible triagaonals within 7D space. That means we want to choose 3 numbers ranging from 0-6. We start with our easiest one

In order to avoid repeats (no pun intended) we maintain a strict order that is always increasing. Each number should always be larger than the previous one. To iterate through new combinations, we simply increase the last number until we can't anymore.

Now at this point, we can't increase the last number any more, so we increase the second to last and start over

And since the number after 2 has to be larger than 2, the smallest number we can roll over to is 3. From here we continue. Basically, it turns this complicated problem into a type of counting game. Just like with regular numbers, we increase the last one until it can't go anymore, then we increase the next one over and restart, just instead of restarting to 0, we restart to the smallest allowed number that keeps things in order.

And at this point, we now have to increase our first number. I'll go ahead and finish this up despite it being pretty long, so that we can make sure we have all the possibilities

Now if we do some quick calculations for how many combinations we have of 3 numbers out of 7 total we get 3 choose 7 = 7! / ( 3! * (7-3)! ) = 7! / ( 3! * 4! ) = (7 * 6 * 5) / (3 * 2 * 1) = (7 * 5) = 35

Which is exactly how much we counted to! There are a few reasons we know there aren't any repeats but this has gotten long enough, and I can't quite articulate them off the top of my head. From here we convert back into 1's and 0's, then we count up in binary for each one of them to get the sign permutations. To wrap things up, I'll finish up by demonstrating all those sign permutations for one of these direction combinations. As stated in point 3.> we can translate back from our abstraction back to the original problem, so we can apply the smaller part that is generating the sign permutations.

There we go. A little bit of a mess, but that's it! And that's how I generate arbitrary m-agonals in arbitrary n-dimensional space for my chess engine. Finally, here's a picture of a queen's movement allowed to move along any agonal up to heptagonals (seven!) on a 7D board generated in a precursor to my current iteration of nDimensional chess

Y'know, not that anyone would ever play that lol. Here's a bishop on the same board, which is (somewhat) more... manageable lol.