About a year ago I made this topic, which never really got off the ground.

For large cubes, there are 2 types:
Even (no fixed center piece)
Odd (fixed center piece)

However, for large Pyraminxes and FTOs, you can classify them based on the center into 3 groups:

No Center (self-explanatory)
Regular Center (the center triangle points in the same direction as the whole face)
Inverted Center (the center triangle points in the opposite direction as the whole face)

Starting from the Master Pyraminx or FTO (4 layers) because it's the first with a center, that means that we don't hit repetition until the 7-layer version. But does that mean that all three solve types are unique in some way (like how even cubes are distinct from odd ones because they can have parity problems)? And, once we do hit repetition (7, 8, and 9 layer versions), is it just "more of the same" like cubes, or are there other factors?

(Brandon, I'd love to read what you were going to write in that other topic, but didn't... )

Hey Jared,

I didn't write about this in part because I couldn't figure out the exact properties of each puzzle as you go bigger and bigger. I'm hindered in part because Geltinbrain's 5.1.X series doesn't go very high.

I have one of Timur's Royal Pyraminxs now which should help some. Even identifying all of the piece types is somewhat hard. As you have pointed out there is some variety to the center triangle types and there is some variety to the edge pieces too. For Pyraminxs I have a really hard time doing it in my head. Being able to turn the puzzle really helps me work through each pieces behavior.

You don't ever run into any parity issues with any sized Pyraminx because every turn is modulo 3 so the whole puzzle stays in an even permutation at all times.

I will try again to take a stab at this

Have you tried Ultimate Magic Cube?

Edit: The 6x6 and 7x7 cubes DO add another piece type (centers which are neither + or X), but that's mostly only important in super-solves. Perhaps the 7, 8, and 9 layer Pyraminxes and FTOs also have this property, and 10 is where they start getting redundant...

Alright I figured this out.

So we're talking about N-pyraminx where N is the number of layers. The original Pyraminx is N = 3. The Master Pyraminx is N = 4, and so on. I'm not going to bother with N = 1 or 2.

Broadly speaking a N-pyraminx has the following part types and counts:

• 4 trivial tips (3 orientations each)
• 4 tip bases (3 orientations each)
• 6 middle-edges (2 orientations each)
• Some number * 12 of edge wings (no orientation)
• 0 or 4 centers (3 orientations each but orientation not visible)
• Some number * 12 of wing-centers (no orientation)

So lets tackle the easy parts first. The trivial tips and tip pases are fixed at 4 each.

There are two types of edges. "outer edges" and "inner edges". Outer-edges are easy to make a pure [1,1] commutator 3-cycle for and inner-edges can be cycled along with an outer-edge with [3,1] commutator 3-cycles. If the N for the N-pyraminx is odd then the middle-edge will be an outer-edge and if it is even it'll be a inner-edge. The difference between the edge types is superficial.

There are always 6 middle-edges and the number of edge-wings is (N - 3) * 12.

There are either 0 or 4 centers. If N % 3 == 0 then there are no centers and otherwise there is 1 center for each face for a total of 4.

For the number of wing-centers, the number is simple to calculate but figuring it out can be tricky. Each center-wing appears on a face 3 times so you could try to count out how many there are and find a formula. I have highlighted them on the N-pyraminx where N in {5, ... 10}:
If you made a table it would be:
3: 0
4: 0
5: 1
6: 3
7: 5
8: 8
9: 12
10: 16
11: 21
[...]

The trick is not to try to find the pattern of highlighted pieces to derive the formula. Instead observe that the total number of pieces in the cent for a face is (N - 3) ^ 2. A piece in the center either appears 3 times because it is a center-wing or it appears once because it is a center. That means the number of center-wings in the (total number of center pieces - the center if it's there) / 3.

So for a 7-pyraminx there are ((7 - 3)^2 - 1) / 3 = 5 center-wings. For a 9-pyraminx there are ((9 - 3) ^ 2 - 0) / 3 = 12 center-wings.


So if you will permit me the modulus operator (%) and 0^0 == 1 then...

The generalized N-pyraminx distinct position formula is:
npyra(n) = ((4! / 2) * (3^4)^2 * (6! / 2) * (2^6 / 2) * (12! / 2)^(n - 3) * (((4! / 2) * (1 - (0^(n % 3)))) + (0^(n % 3))) * (12! / ((3!)^4))^(((n - 3)^2 - (1 - (0^(n % 3)))) / 3)) / 12

Which produces:
? npyra(3)
% = 75582720

? npyra(4)
% = 217225462874112000

? npyra(5)
% = 19228688422470957567836160000000

? npyra(6)
% = 52425105093745505455227317179687895040000000000000

? npyra(7)
% = 20582183665033346731252760185588627707250626464317440000000000000000000
[...]
npyra(17) = 1.48 * 10^488
[...]
npyra(100) = 1.43 * 10^18282


As for solving, you can always solve centers-out with a combination of intuition, [1,1], and [3,1] commutators. Once you've solved the centers, reduce edges by first grouping all of the inner-edge-wings to the middle-edge and then all of the outer-edge wings. Then solve the very shallow reduced H-M Pyramid. You could end up with a 3-cycle in the center-groups and the end but the fix is simple.

Wow! Thanks so much! I was only hoping for a bit on solving, but you went way past that.

So, unlike cubes, when it comes to solving, there really isn't a general difference between even/odd or k/k+1/k+2... what a bummer!

So, now I'm curious as to how this could be related to FTOs. If you bandage an order-3 FTO right, you get an Octaminx, which is a truncated Pyraminx. In fact, the classic FTO is simply 2 Octaminxes in one puzzle, as can be seen by the circle version (GB 4.1.11) - the circles on each of the two separate sets of faces show the underlying Octaminxes, so you have to solve both of them in addition to the FTO itself.

So, how do higher-order FTOs relate to higher-order Pyraminxes, if they do?

I dunno, I haven't really thought about it. Obviously the face patterns are roughly the same but on the FTO there are no inner-edges, those are two discrete center triangle pieces.

Also, the FTOs have center pieces in two orbits. FTOs can probably be thought of as (mostly) two disjoint pyraminxes. I might take a stab at a generalized FTO analysis since it should follow from most of my analysis of the Pyraminx family.

I really owe you a drink or something for all this.

Alright I did the big FTOs too.

Observe the centers as they get bigger:
Note that the non-outer-layer centers grow at the same rate as the Pyraminx analysis above. Also there ale always 3 centers on the outside and the number of triangle centers on the edge grow linearly with N.

For the edges, there is no difference between middle edges and edge wings. Edge-wings are actually two different sets of pieces (in different orbits). They don't have any orientation.

There are 12 orientations available to the puzzle rather than 24 due to twist restrictions on the corners and edges.

The center pieces are in two orbits. That is, 4 faces are in one orbit and 4 faces are in the other. This is basically like two Pyraminx puzzles that have been fused.

Taking all of that into account...

The generalized N-FTO formula is:
nfto(n) = ((6! / 2) * (2^6 / 2) * (((4! / 2)^2 * (1 - (0^(n % 3)))) + (0^(n % 3))) * ((12! / ((3!)^4))^2)^((((n - 3)^2 - (1 - (0^(n % 3)))) / 3) + 1 + (n - 3)) * (12! / 2)^(n - 2)) / 12

Here are some interesting values:
nfto(3) = 31408133379194880000000
nfto(4) = 147970626354794338756025552732160000000000000
nfto(5) = 661317980389074634392982723584033456560017419062476800000000000000000000000
nfto(6) = 2.804 * 10^114
nfto(7) = 2.465 * 10^158
[...]
nfto(17) = 4.649 * 10^1021
[...]
nfto(100) = 2.530 * 10^36838

Solving-wise, first reduce the centers using [3,1] commutators. Be sure to reduce the center groups into an even permutation. Then form edge-groups with [3,1] commutators (paring the edge wings up to each other or to a middle edge if there is one). Then complete the edge groups by cycling the triangles between the edge wings with [3,1] commutators.

Then solved the reduced, shallow-cut FTO.

The [3,1] commutators needed for the various piece types are trivial to extend into bigger and bigger N-FTOs.

The "hardest" N-FTOs are the ones where N is divisible by 3 because they don't have the middle center pieces so it's possible to reduce the centers into an odd permutation.

Thanks!!!

So, by "odd permutation" you mean a situation kind of like a 4x4's centers being out of order, right? So, if you reduce the centers into the right spots, you can avoid this? Or do you mean that the order-6 FTO (for example) can have parity problems towards the end?

I mean you could end up with two center groups swapped. You want to make sure you reduce them to positions where this won't happen.

The 4x4x4 centers example is not so great because once you reduce to 3x3x3 centers you can't permute the centers relative to each other at all.