I have a question about the pyraminx twisty puzzle. I’ve included a bit of background to make sure we’re all using the same words.

(background)

To the initiated, a pyraminx is a triangular-looking rubiks-cube-like puzzle that, it turns out, is actually quite a bit easier than the rubiks cube, or even the 2×2 cube.

I’ve been trying to understand the “math” of the pyraminx; despite it being a fairly common puzzle, there doesn’t seem to be as much written material about it, maybe because it’s not too hard to solve by hand, and because on a computer, the number of states is small enough that you can solve by brute-force without thinking too much about the structure of the thing.

But what I’m interested in is the number of solvable configurations. According to wikipedia (https://en.wikipedia.org/wiki/Pyraminx#Description) the right answer is the product of a few factors, which I guess you can pick independently:

• $3^4$ — The four “trivial tips” can be in any rotated state and have nothing to do with anything else in the puzzle.
• $3^4$ — The four “axials” (wikipedia calls them that; basically the things that aren’t edges or tips) can be rotated in 3 different configurations, independently.
• $6! / 2$ — The six edge pieces’ positions can be permuted more or less as desired, except for a parity concern; the permutation needs to be even. You can see this from the fact that the twists all move the edges in a 3-cycle, which is an even permutation, so there’s just no way to do an odd permutation. The fact that you can get every even permutation isn’t totally obvious but it’s fine, it’s just a consideration of cases.
• $2^6 / 2$ — The six edge pieces’ orientations can be picked more or less arbitrarily, except you can’t flip just one edge. You can flip two adjacent edges — R U' R' U R' L R L' — and it’s not hard to see that you can then flip any pair of edges by composing this, which means you can flip any even number of edges.

Anyway so all of this is just establishing terminology to get to my actual question —

How can I prove that “one edge flip” is unreachable, without using a computer and saying “brute force says so?”

To do this with the rubiks cube what you need to do is define a notion of an orientation of an edge that makes sense even when the edge is not in the right position — basically you say it’s in the right orientation if, when you get it back to position without using any U or D moves, it’s in the right orientation (you can also define it in a few other slightly-inequivalent ways but the math works out basically the same). Then you can make some arguments about what all the moves do to the orientations and it’s all set.

What I don’t know then is how to define orientation for an edge piece when it’s out of position. If I could do that, then I assume I could do the same thing as I did for the rubiks cube, but I couldn’t come up with anything that worked at all. It’s easy to see that you have to use three different rotation types to get a piece back to where it started, but flipped over; so you could say a piece’s orientation is whatever orientation it has when you get it home using only two rotation types. But that doesn’t work; sometimes there are multiple ways to get it home, with different results. You could also try just saying you have to use a certain two rotation types (say, only by R and U), but that’s not transitive; you can’t move every edge to every place this way, so it doesn’t really work.

Anyway I feel like somebody must have already come up with something, but I couldn’t find anything by searching. Thanks in advance if anybody can invent anything or has a link.