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.


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 2x2 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.

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    $\begingroup$ related $\endgroup$
    – msh210
    Commented Dec 11, 2023 at 3:43
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    $\begingroup$ I think this might be more suitable on the math SE. Abstractly the situation looks similar to the regular Rubik's cube. There is a collection of possible states that has two distinct components where only states within one component can be transformed into each other by the regular moves. $\endgroup$
    – quarague
    Commented Dec 11, 2023 at 8:11
  • $\begingroup$ I figured if it's already a known problem then people here are more likely to know than on math.se, since pyraminx is an established puzzle. If it needs to be invented, I agree math.se is a better fit. $\endgroup$ Commented Dec 11, 2023 at 17:16
  • $\begingroup$ In any case I'll give it another day or two before I cross-post, I don't want to spam. But thanks for the redirect. $\endgroup$ Commented Dec 11, 2023 at 17:22

1 Answer 1


Sorry for all the edits. I keep looking back at this and understanding it better and better. Here is my (more) concise answer:

Step 0: Get on the same page.

If you orient the Pyraminx with a flat side down and one of the bottom three points facing away from you, this is the standard orientation of a Pyraminx, and what I'm going to be working from. The color scheme I'm using is Blue down, Green Front, Yellow in the back left, and Red in the front right. The possible turns are U, R, L and B, and we are completely ignoring the tips. I use the term "sticker" to talk about one colored side of the edge piece. If you have a stickerless pyraminx, just imagine it has stickers.

Step 1: Decide what is oriented correctly.

All 6 edges are either in the bottom layer (one sticker facing straight down and touching the table if you set down the puzzle) or the top layer. In the bottom layer, the edge stickers are facing either DOWN or UP. In the top layer, the edge stickers are facing either clockwise (CW) or counterclockwise (CC).

In my method, DOWN and CW are considered the same (therefore UP and CC are also the same). So, for an edge piece that belongs in the bottom row, it is oriented correctly if the sticker that is supposed to be facing DOWN is facing DOWN or CW. With our current colors and orientation, this means all blue stickers should be facing DOWN or CW. For edge pieces in the top row, the stickers that, in their solved state will face CW should be facing either CW or DOWN. On our pyraminx this means the Yellow in Yellow/Green, the Red in Red/Yellow, and the Green in Green/Red should all be facing either DOWN or CW.

Step 2: How Moves flip edges

With the above definitions, U moves won't affect the orientation, because all three edge pieces will retain their CC/CW orientation.

R, L and B moves will each flip 2 edges as described here:

  • If the edge moves from one bottom layer space directly across to another bottom layer space, it will be flipped (the DOWN side is now facing UP)
  • If the edge moves from bottom to top in a CC move such as L', R' or B', it will be flipped, because the sticker that was DOWN is now facing CC.
  • If the edge moves from top to bottom in a CW move such as R, L or B, it will be flipped because the sticker that was facing CW is now facing UP
  • The remaining edge piece moved on any of these moves (the one that moves from top to bottom on a CC move or bottom to top on a CW move) will not be flipped.

Using the terminology from the 3x3 rule you used above: "A piece is in the right orientation if, when you get it back into position without moving it across the bottom, up on a CC move, or down on a CW move, it's in the right orientation."

You can see my original answer below for an example of this theory in action.

In the end, this proves that only an even number of edges can be flipped because every move you can make on the puzzle (including tips) flips an even number of edges (0 or 2). I admit that I developed this theory by starting with the assumption that it was impossible to flip only one edge, and using it to prove that statement seems circular. However, as I thought through it, and noticed the CW/DOWN vs CC/UP rule, I was able to start over with only that, and still came to the same conclusion.

Pictures First: The regular look with some edge stickers labeled: The regular look with some edge stickers labeled

Next: What an L' would look like: An L Prime

Last, what an L move would look like (from solved): An L move

Original answer:

Ok, this is arbitrary and a little complex, so bear with me (Edit, it ends up being rather simple, and even intuitive after you work with it for a while... see my edit at the end). And remember that the 3x3 edge orientation definition is a little arbitrary, too (we could say R and L are the flipping moves instead of U and D, and get the same math).

We have to first define which moves flip which edges. On a 3x3, that worked out very nicely, but not so on a Pyraminx, which only has 4 "faces" to turn, and each one moves 3 edges. So, each turn must flip either 0 or 2 edges. So, in addition to defining which turns are "flipping turns," we also need to know which 2 edges are considered flipped and which one stays the same. Assuming that an R move flips two edges, it must be set up so that opposing moves (R and R') flip the same pieces, and that R3 will flip all three edges twice.

The simple edge flipping definition on a 3x3 is only possible because you are defining a static orientation of the puzzle (you never rotate the whole puzzle so that up is always up). In my solution, the same is required. We define the U, R, L and B "faces" and always keep them in their relative positions. Furthermore, I still think in 3x3, so when I say the R "face" I mean the part of the puzzle that rotates when executing an R or R' move. So the "faces" I'm talking about are actually pyramids in shape rather than flat faces.

Starting from those assumptions, I came up with this:

  • U turns are non-flipping. No edges become flipped by any U moves.
  • R, L and B moves all flip two pieces each.
    • R', L' and B' DO NOT flip the edge that moves DOWN into the bottom layer, but DO flip the other two
    • R, L and B DO NOT flip the edge that moves UP into the top layer, but DO flip the other two pieces.

After some basic turning, it looks like this always comes out correct.

Let's apply this to your flipping algorithm. With Blue on the bottom, Green in front, yellow in the back left and red in the back right, here is what we have:

  • R - This move affects the RG, BR and GB edges. GB moves up and is not flipped, but RG and BR are now flipped.
  • U - Nothing is flipped on U turns
  • R' - This time, the move affects RG, BR and RY, with RY moving down and not getting flipped. RG and BR get flipped, returning them to their original state. Now no pieces are flipped.
  • U' - No flips
  • R' - Affects BR, GB and RY flipping BR and RY
  • L - Affects GY, BY and GB, flipping GY and GB (So that BR, RY, GY and GB are all flipped)
  • R - Affects GY, BR, and RY, effectively unflipping BR and RY. (GY and GB are flipped)
  • L' - Affects BY, GB and RY. Leaves BY alone, unflips GB and flips RY.

At the end, my rules show that GY and RY ought to be flipped. If you finish with one more U turn, you can see that GR is in its spot and is not flipped, while GY and RY in their spots are flipped. I also tested this with my flipping alg (R' L R L' U L' U' L ), and it held true.

The way to disprove this would be to move a piece around in such a way that it scores an even number of flips by my definition, but winds up flipped within its correct spot. I didn't see a way to do this, but I'm also about at the end of my brain's capacity to keep track of all these edges. My next step would be to write a program that can simulate a pyraminx and run tests on that. But the last one I did for the 3x3 took me FOR. EV. ER, so I'm not going to do that just now.

EDIT: The more I think about this, the more sense it makes. If you take an edge piece in the top, rotating it around to a different slot wouldn't change its orientation.. Same with the bottom. If it's in the bottom front slot and you bring it to the top with R (no flip), then U to move it to the left, and L' to put it back in the bottom, no flip has occurred, and the sticker that was facing down is still facing down. So regardless of which bottom slot it's in, the same sticker is always facing down unless it's a flipped edge. Bringing it up with a counterclockwise turn, bringing it down with a clockwise turn, or turning it across the bottom are the moves that make it change its orientation.

Since each edge in the bottom layer has one sticker down and one sticker up, and each edge in the top layer has one sticker facing clockwise and one facing counterclockwise (cc), the "Correct" orientation of each piece is "down/cw sticker facing down or clockwise." That would be (on the color scheme I defined above) the Green of Red/Green (so the G comes first in GR), Red in RY, Yellow in YG, and then all blue faces down, so BG, BY, BR. If a Blue sticker is facing up or clockwise, it's flipped. If the G in GR is up or clockwise, it's flipped, and so on. Aside from memorizing the three GR/RY/YG, this means checking edge orientation visually is pretty trivial! You could easily spot a twisted edge on a scrambled pyraminx just by looking at it!

I really think it's the answer. Like you, I haven't heard of an accepted standard for this, but I feel like mine is as good as any.

  • $\begingroup$ This is great, thank you. I knew the trick would be to make some kind of arbitrary decision about which move(s) are "special" somehow, but I couldn't quite get something that worked. I'll be sure to accept once I have time to go through the details (after work). $\endgroup$ Commented Dec 12, 2023 at 13:23
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    $\begingroup$ Yes, beautiful, it all checks out. I appreciate your level of detail. $\endgroup$ Commented Dec 12, 2023 at 21:51

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