Speedsolving Posts: 2D State Diagrams

The posts here are from January 2009 and can be found in this topic: http://www.speedsolving.com/forum/showthread.php?8628

This is a topic on puzzle theory - specifically, it deals with the theory of symmetrical twisty puzzles. There's no real "theory" section in this forum but it doesn't belong in Off-Topic, so I've put it here so more people will see it. (This does have applications to speedcubing, in terms of solving gelatinbrain puzzles.)

So, as some of you may know, symmetrical twisty puzzles can be described by talking about the shape and the number of separate types of turns in each of the three basic categories (face turns, edge turns, and vertex turns) which are named around where the axis (which the pieces rotate around) points. A 3x3x3, for instance, is a cube with one type of turn (in the face turn category); in my categorization scheme that would make it type Cf (C for cube, f for face turn). The only other type-Cf puzzle is the 2x2x2, which has a deeper turn that just happens to line up with the turn on the opposite face. (The 4x4x4 and 5x5x5 have two face turn types, so they are type Cff. Similarly the 6x6x6 and 7x7x7 are type Cfff.) Note that I'm considering a 3x3 with slightly shallower or deeper cuts to be the same, because it has the same number and types of pieces and the exact same solution, so the only difference is in the shape of the pieces.

If we want to talk about symmetrical puzzles with only one type of turn, it's relatively simple, because we can describe the depth of the turn as a single number. But if we have two types of turns, we need two numbers: for instance, on a 5x5x5, if we imagine a turn of depth 0 to turn nothing and a turn of depth 1 to turn the entire cube, the two turns have depths 1/5 and 2/5. But it's hard to imagine exactly what numbers will give a certain puzzle, and which numbers will give a different one. So the question is this: how many possible puzzles are there of a given type, and, more importantly, how can we understand where one puzzle begins and another ends?

My answer to this is the 2D State Diagram. This particular one describes the Tvv-type polyhedra (that is, tetrahedra with two vertex turn types - note that a vertex turn and a face turn are the same thing on a tetrahedron). The two axes correspond to the two depths of turn, and each one goes from 0 (no part of the puzzle is turned) to 1 (the entire puzzle is turned).

So what are the lines in the diagram? Each one represents some kind of puzzle which only exists for a very specific set of turn depths. The diagonal line from the top-left to bottom-right represents the degenerate case where the two depths are the same (so, type Tv puzzles with only one type of cut). Each of those big spaces between the lines, however, is a specific puzzle, which remains the same anywhere in that space. There are also puzzles on the lines themselves, and each place where lines intersect is another puzzle, albeit a very specific one because even a small change in either of the two depths will make it different. The Pyraminx, for example, is located at the intersection of a horizontal or vertical 2/3 line, and the diagonal line from bottom left to top right.

Finally, I have prepared an interactive simulation (!) of this state diagram, so that you can play with it to see how it works out. It uses the GeoGebra geometry program, which I'm really fond of. You can download it from my website. To use it, just move the red point around inside the square (use the white pointer symbol to move a point around), and watch how the puzzle changes when you cross or move along lines.

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I wrote up the state diagram for type Dff puzzles (dodecahedra with two different depths of face turns). The Gigaminx is from this category. Note that this category is MUCH more complicated than the Tvv type I showed in the first post - I'm not completely sure of the count, but I think there are a whopping 98 puzzles of this type!!! I'm really glad I made an interactive simulation, because there is no way I'd ever want to draw all of them out by hand. You can also see that there are 7 puzzles along the diagonal (these are the Df type puzzles).

In the image, I have marked a few lines as green. Those lines are distinctive because, although they separate different puzzles, the puzzles on the lines themselves are equivalent to one of the puzzles on one side of the line. For instance the Megaminx and Supernova, although they look like they might different puzzles, function exactly the same way.The full way to count the number of puzzles is:
- Any open space is a separate puzzle, although you have to make sure to only count puzzles on one side of the main diagonal (for everything here).
- Any line segment (i.e. a line between two intersections) is a separate puzzle, except (i) if the line segment is green; (ii) if the line segment is on the main diagonal; (iii) if the line segment is on the top or left. The second and third cases remove puzzles with essentially only one depth of turn, and the first removes a puzzle that looks like a different puzzle but isn't.
- Any intersection of lines is a separate puzzle, as long as (i) it isn't on the the top or left or the main diagonal, and (ii) it has at least two black lines crossing it.
The Tvv case gives 11+10+2 = 23 puzzles like this (I think). Note that, although I haven't added it in yet, the halfway lines and the bottom-left-to-top-right diagonal in the Tvv diagram should also be green. It's only important to know what lines are green and what are not when you are counting puzzles, though.

Finally, there is a GeoGebra simulation of this, although I have made the two-dimensional slider much larger to compensate for the complexity. That way you can really see what lines you are crossing.

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For each polyhedron there are actually six types of second-order puzzles: the familiar vv, ee, and ff, but also fv, ev, and ef. The first three use only one type of cut, so they're symmetrical, but the other three use two different types of cuts and have much more complicated drawings. As an example I've made one for the Cfv group. There are around 50 puzzles in this group, most of which have probably never been seen before. I hope to eventually make an interactive simulation of all 5*6 categories of second-order regular twisty puzzles. I imagine there will be over 1000 puzzles in total!

Speedsolving Posts: Parity Errors

I'm going to be posting some of my longer posts on speedsolving.com here. I think I've written some very useful things down there, but it's a kinda big forum (and I've made a lot of posts!) so it's easy for stuff to get lost. The plan is to copy some of them here, with a link to the original post (for those who want to view the thread). Here goes...

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From http://www.speedsolving.com/forum/showthread.php?p=852756:

In speedcubing there are two things we think of as parity: "reduction parity" and mathematical parity. Some things we call parity fall into both categories, but others only fall into one of them, and you will often see people disagree on whether certain things count as parity, because they disagree that both of these two definitions are valid.

Reduction parity occurs when you try to reduce the puzzle so it can be solved by a constrained set of moves, putting it into some subset of the positions. However, you can often reach a position which seems like it is in your subset, but which is actually not, and to solve the puzzle you have to briefly go outside your constrained set of moves to bring the puzzle back into the subset you want. Typically the number of positions you can encounter is some small multiple of the number of positions you expect. The obvious example is PLL parity in 4x4x4: all the centers and edges are properly paired, so you expect to be able to finish the puzzle with only outer layer turns, but this isn't quite possible. OLL parity falls under this definition too (so the reduced 4x4x4 has four times as many positions as you would expect). Square-1 parity also falls under this definition - your constrained set of moves are any moves that keep the puzzle in cubeshape. BLD parity is not of this type (the solver has not reduced the puzzle).

Mathematical parity is based on the idea of the mathematical definition of an odd permutation. Basically, at least one orbit of pieces has an odd permutation, and thus cannot be solved with just 3-cycles. You see this type of parity crop up in blindfolded solves, because blindfolded solvers attempt to solve most or all of the puzzle with 3-cycles and thus an odd permutation is very noticeable. OLL parity on 4x4x4 and Square-1 parity can also be thought of as this way, as they originate from some kind of 2-cycle. PLL parity on 4x4x4 is not of this type (it can be solved with 3-cycles).

Advanced Counting: Burnside's Lemma

Sometimes the theory we have discussed so far is not enough to calculate the number of positions of a puzzle. Fortunately there is a theorem from group theory that will help us with this. It is called Burnside's Lemma (although he was not the first to discover or prove it):

Theorem: Suppose we have a finite set X, and a finite group G of permutations of X. For each element g in G write |Fix(g)| for the number of elements of X that are not changed by g, and write |G| for the number of elements of G. Then the number of orbits of X is equal to 1/|G| Σ_g∈G |Fix(g)|.

What does this mean, though? Well, when we are counting the number of positions of a puzzle, we want the number of positions regardless of orientation - you shouldn't be able to get from one position to another just by rotating the puzzle, but we also don't want to miss any. The way we do this is by setting X to the set of every possible position in every possible orientation and G the possible ways to rotate the cube. The number of elements in X is relatively easy to calculate (it's like finding the number of positions of a puzzle, except that you don't fix anything; for example the size of X for the 2x2 cube would be 8! 3^7), and the number of elements in G can mostly be looked up from a small table, since it is 12 for a puzzle with tetrahedral symmetry, 24 for puzzle with the symmetry of a cube or octahedron, and 60 for one with the symmetry of a dodecahedron or icosahedron. Now, the number of orbits is simply the number of positions of a puzzle regardless of orientation, which is exactly what we want - so as long as we can compute the Fix(g) values, we know exactly how many positions a puzzle has, even if the puzzle has repeated pieces.

One really important thing to notice is that, for any cycle in a permutation in G, every piece on the cycle has to look the same. This is pretty important because it tells us that if we have, say, three of each piece, it is impossible for g to fix any positions if g contains a 4-cycle (or bigger), because there are no sets of 4 of the same piece. Also, if g contains only 3-cycles, and the number of duplicate pieces in each set is not a multiple of 3, g cannot fix any positions.

I want to do two examples to show how this works. First I'll do the very simple example of a cube with three red faces and three blue faces which can be switched any way we want. We already saw that our basic approximation gave something like .83 when the actual answer was 2. Let's use Burnside's Lemma. The size of X is 6 choose 3 or 20, and the size of G is 24 because the puzzle is cubical. Let's calculate the |Fix(g)|'s: if g is the permutation that does nothing, 20 positions remain the same; if g is a rotation around a corner (there are 8 of these), 2 positions are fixed; if g is a 90-degree rotation around a face (there are 6 of these), there are no fixed positions; if g is a 180-degree rotation around a face (there are 3 of these), there are 4 fixed positions; and if g is a flip around an edge (there are 6 of these), there are no fixed positions. So the total number of positions is 1/24 (20 + 8*2 + 0 + 3*4 + 0) = 1/24 (48) = 2. It's exact!

Now let's do a very hard example: the (non-jumbled) Little Chop or 24-cube, which has 24 pieces, with 4 each of 6 colors and no orientation. The size of X is 24!/(4!^6) and the size of G is 24 again. Calculate the values of |Fix(g)|: if g does nothing, 24!/(4!^6) positions remain the same; if g rotates around a corner, we create 8 3-cycles so no positions can be fixed; if g is a 90-degree rotation around a face, there are 6 4-cycles so 6! positions are fixed; if g is a 180-degree rotation around a face, there are 12 2-cycles so 12!/(2!^6) positions are fixed; and if g is a flip around an edge, there are again 12 2-cycles so 12!/(2!^6) positions are fixed. The total number of positions is 1/24 (24!/(4!^6) + 6*6! + 3*12!/(2!^6) + 6*12!/(2!^6)), which works out to 135,277,941,853,080. We can't really verify this, but if we have done our calculations right it is the correct value. Note that with our basic computations, fixing one of the 4 pieces of a specific color, we would get an approximate value of 1/4 (23!/(3! * 4!^5)) = 135,277,939,046,250, which is pretty close (but not exact).

Twisty Polyhedron Categorization Project

There are a lot of twisty puzzles out there, and they can be categorized in a multitude of ways. This project, the TPCP, is an attempt to categorize all of the most basic twisty puzzles into a list, ideally giving each one an identification code, an image, the gelatinbrain number or common name if applicable, a list of the types of pieces and their orbits, and a basic solution. I do not expect to complete all of it but I hope to get a good chunk of it done.

The most basic twisty puzzles, as I define them, satisfy the following characteristics:
- They are based around a regular polyhedron.
- There is exactly one type of axis around which turns can be made.
- All possible turn axes are represented, and they all have the same depth. Thus the puzzle is symmetrical and remains mathematically unchanged under a rotation.
- Each face of the solved puzzle has one unique solid color and the goal is to bring the puzzle into a state where each face has one solid color.
- After any turn, the puzzle is still in a regular-polyhedral form.

The foundation for the TPCP is very mathematical. Each puzzle belongs to one type, which is described by a capital letter denoting the polyhedron (T, C, O, D, and I for tetrahedron, cube, octahedron, dodecahedron, and icosahedron, respectively) and a lowercase letter denoting the type of axis (f, e, and v for face, edge, and vertex turns, respectively). Thus for example the 3x3x3 cube belongs to type Cf. Within the types, each puzzle gets a natural number, starting with puzzle 1 for the shallowest cut and increasing for deeper cuts. Puzzles with different-depth cuts which function in the same way are considered to be the same. Even trivial puzzles will be considered, except for puzzles with a cut depth of 0 where turns do not move any pieces and thus no scrambling at all is possible. In this case the 3x3x3 cube would be Cf1.

For standardization of images I have chosen to use an image size given by the downloaded Gelatinbrain applet: 312x312 pixels. Of course I have used the lossless .png format so the images should not take too much time to load. Images for puzzles which are not represented in the downloadable applet have been created by me.

Note that in this project if a set of pieces is described as "trivial" the pieces cannot be permuted relative to each other but do have an orientation.

Although it is not addressed in this project, it is possible to use this categorization system to create types for higher-order puzzles (which I call compound types), and I may investigate some of these series in the future. The type of, for example, a dodecahedral puzzle with two types of edge turn and one type of vertex turns would be Deev, and the Super-X would belong to type Cfv. Although the numbers within the types cannot be as unambiguous as in the TPCP puzzles, we can still assign each puzzle in a compound type a number, with the expectation that someone wishing to do research in this topic would look up a puzzle's number in an online list.

With that said, let's start to classify some of the types.

Number of Positions of Generalized Twisty Polyhedra

I wrote this on 3/30/08.

The goal of this article is to determine the number of positions of various twisty polyhedra. I will first show an algorithmic theory where we can find the number of positions in well-behaved twisty polyhedra (that is, puzzles which are symmetrical and have no move constraints based on the position) without difficulty, and then use it to find the number of positions in a few infinite families of theoretical twisty polyhedra, and a few examples of how to modify this theory for trickier puzzles.

Theory of Positions

Before we start determining the number of positions of a puzzle we must fix the orientation somehow. This is necessary since simply turning a puzzle around without doing moves on it will not change its position. The normal way to do this is to fix one specific piece, or a set of pieces which are fixed relative to one another. If this is impossible we can divide the number of positions we calculate later by the number of possible positions of a solved puzzle (most usually 12 for a puzzle with tetrahedral symmetry, 24 for a puzzle with cubical or octahedral symmetry, or 60 for a puzzle with icosahedral or dodecahedral symmetry) but it is not certain that this will give us the correct result. Note, however, that if we have a few choices for what piece to fix, you will get the same answer no matter which piece you choose.

After this the first step to finding the number of positions of a twisty puzzle is to determine the orbits. An orbit is the set of positions that a given non-fixed piece can be in, and orbits containing the same positions are considered the same. In a 3x3x3 cube we will typically fix the orientation by forcing the centers to be in specific places; then there are two orbits, one for the 12 edges and one for the 8 centers. Note that there can be more than one orbit containing similar-looking pieces; for example on the 3x3x3 square group (where only half turns can be used) there are in fact three edge orbits and two corner orbits.

In general we will proceed by finding the number of positions in each orbit and multiplying them together. Because of parity concerns, the best way to do this is to order the orbits in some way and consider them one by one, assuming that while considering one orbit all pieces in the previously considered orbits are solved. This way, when two orbits have the same parity, we see that one does not obey a parity constraint and the other does, so we correctly divide by 2 only once.

So how many positions are in a single orbit? This is generally an easy problem, as it can be directly calculated with a little casework. First we consider the permutations. If there are n pieces, we start with n!; if there are n pieces each of k types, as in the centers of a higher-order Rubik's Cube, we start with (nk)!/n!^k. If the pieces of the orbit have to obey a (permutation) parity constraint then we divide by 2; note that this is only possible in the first type because if we have indistinguishable pieces we can always make a 2-cycle of same-colored pieces to cancel out a wrong parity. Next we consider the orientation of the pieces. If the pieces cannot be oriented differently when they are in a given position we do not have to consider orientation and we can just multiply by 1. Otherwise, if we let k be the number of possible orientations one piece can have if it is a given position and n be the number of pieces then we should multiply by k^n. If the pieces of the orbit obey an orientation parity, we divide by its order, which is generally 2 or 3 but can be higher.

Let's work through an example to illustrate the technique: the 3x3x3 cube. As I already mentioned, if we fix the orientation by fixing the centers' positions, we have two orbits: edges and corners. First we'll consider edges: there are 12 pieces (12!) with no parity (since corners are unknown), and each has an orientation of order 2 (2^12) with an orientation parity (/2), so the total number of possibilities here is 12! * 2^11. Then we consider corners: there are 8 pieces (8!) with parity (/2, since edges are fixed), and each has an orientation of order 3 (3^8) with an orientation parity (/3), so the total number of possibilties is 8!/2 * 3^7. Thus the total number of positions is (12! * 2^11)(8!/2 * 3^7) = 43252003274489856000, as I'm sure you already know.

Number of Positions of Infinite Families of Twisty Polyhedra

The most important concept here is that of layers. It is difficult to put into words, but the number of layers of a puzzle is basically the number of parallel, twistable slices (not counting slices on the opposite side of the puzzle's center). Puzzles with one layer include the 2x2x2, 3x3x3, Megaminx, and Dino Cube; puzzles with two layers include the 4x4x4, 5x5x5, Gigaminx, and Lattice Cube. A family of puzzles is basically a set of puzzles that function similarly (have the same axes, the same fixed pieces, the same shape, and so on), but which differ in the number of layers, a constant which I will call n throughout the rest of the article. I don't count 'trivial tips' (twistable elements of a puzzle which do not affect any other elements, are not affected by any other elements, and can always be solved in one move) as layers, because they do not share any pieces with the rest of the puzzle; thus, although most families only have twistable puzzles for n >= 1, there are some families where the case for n = 0 has more than 1 possibility.

There are many families of puzzles; however, most families have only had one puzzle constructed or even considered, and there are other families (such as stacked pucks and 2x2x(2n+1) cuboids) where adding one layer simply multiplies the number of combinations by a fixed amount. In this article I will only consider the number of combinations for families of which at least two puzzles have been constructed or designed. Of course, these are the most popular families to consider. All numbers I give have been calculated using the formula given here.

Odd NxNxN Cubes: Just like the 3x3x3, we will hold the centers constant. Then our orbits are: 12 edges with 2 orientations each (12! * 2^11), 8 corners with 3 orientations each (8!/2 * 3^7), n-1 orbits of 24 edge wings ((24!)^(n-1)), and n^2-n different orbits of 24 centers, all of which are made up of 4 pieces each of 6 types ((24!/4!^6)^(n^2-n)). Thus the total number of positions is 12! * 8! * 24!^(n^2-1) * 4!^(-6n^2+6n) * 2^10 * 3^7. For this family the existing puzzles are n=1 (3x3x3, with 4.325 * 10^19 positions), n=2 (5x5x5, with 2.829 * 10^74 positions), and n=3 (7x7x7, with 1.950 * 10^160 positions).

Odd NxNxN Supercubes: This is like the odd NxNxN cubes, except that there are much more parities since the center pieces have no duplicates, and also the centers have orientation. Again keep the centers fixed, then the orbits are 6 centers with 4 orientations each (1 * 4^6), 12 edges with 2 orientations each (12!/2 * 2^11), 8 corners with 3 orientations each (8!/2 * 3^7), n-1 orbits of 24 edge wings ((24!)^(n-1)), and n^2-n orbits of 24 centers ((24!/2)^(n^2-n)). So the total number of positions is 12! * 8! * 2^(-n^2+n+21) * 3^7 * 24!^(n^2-1). The puzzles that have been made so far are n=1 (3x3x3 Supercube, with 8.858 * 10^22 positions) and n=2 (5x5x5 Supercube, with 5.289 * 10^93 positions).

Even NxNxN Cubes: We can't hold centers constant, because there are no centers, but if we fix one specific corner we will always be guaranteed a fixed orientation. So our orbits are: 7 corners with 3 orientations each (7! * 3^6), n-1 orbits of 24 edge wings ((24!)^(n-1)), and n^2-2n+1 orbits of 24 centers, each of 4 pieces in 6 colors ((24!/4!^6)^(n^2-2n+1)). The total number of positions here is 7! * 24!^(n^2-n) * 4!^(-6n^2+12n-6) * 3^6. The puzzles in this family that have been constructed are n=1 (2x2x2, with 3.674 * 10^6 positions), n=2 (4x4x4, with 7.401 * 10^45 positions), and n=3 (6x6x6, with 1.572 * 10^116 positions).

Even NxNxN Supercubes: This is very much like the odd NxNxN supercubes. Keep a corner fixed, and our orbits become 7 corners with 3 orientations each (7! * 3^6), n-1 orbits of 24 edge wings ((24!)^(n-1)), and n^2-2n+1 orbits of 24 centers, all with parity ((24!/2)^(n^2-2n+1)). So the total number of positions is 7! * 24!^(n^2-n) * 3^6 * 2^(-n^2+2n-1). The existing puzzles are n=1 (2x2x2, with 3.674 * 10^6 positions) and n=2 (4x4x4 Supercube, with 7.072 * 10^53 positions).

Megaminx Family: These puzzles work just like the odd NxNxN cubes, except with larger numbers of pieces in each orbit. However, in these puzzles all turns create 5-cycles, so every orbit has a permutation parity. There are 30 edges with 2 orientations each (30!/2 * 2^29), 20 corners with 3 orientations each (20!/2 * 3^19), n-1 orbits of 60 edge wings ((60!/2)^(n-1)), and n^2-n orbits of 60 centers, in 12 colors of 5 pieces each ((60!/5!^12)^(n^2-n)). So the total number of positions is 30! * 20! * 60!^(n^2-1) * 5!^(-12n^2+12n) * 2^(28-n) * 3^19. The existing puzzles are n=1 (Megaminx, with 1.007 * 10^68 positions) and n=2 (Gigaminx, with 3.648 * 10^263 positions).

Pyraminx Family: These puzzles are slightly more tricky to calculate since some of them have middle centers (where there are only 4 pieces instead of 12) and some do not. I am not including the trivial tips in the calculation because they are not an inherent part of the puzzle, but of course to put them back in we can just multiply by 3^4. In this puzzle we are keeping the corners in the same position, although their orientation can change. Note that, just as in the Megaminx family, all turns create even permutations, so every orbit has permutation parity. In this family our orbits are 4 corners with 3 orientations each (1 * 3^4), 6 middle edges with 2 orientations each (6!/2 * 2^5), n-1 orbits of 12 edge wings ((12!/2)^(n-1)), a total of floor((n-1)^2/3) orbits of 12 centers, in 3 centers for each of 4 colors ((12!/3!^4)^(floor((n-1)^2/3))), and ((n-1)^2 mod 3) orbits of 4 centers ((4!/2)^((n-1)^2 mod 3)). Thus the number of positions is 6! * 2^(5-n-((n-1)^2 mod 3)) * 3^4 * 12!^(n-1+floor((n-1)^2/3)) * 3!^(-4floor((n-1)^2/3)) * 4!^((n-1)^2 mod 3). The existing puzzles are n=1 (Tetraminx, with 9.331 * 10^5 positions), n=1 with tips (Pyraminx, with 7.558 * 10^7 positions), and n=2 with tips (Master Pyraminx, with 2.172 * 10^17 positions).

Magic Octahedron Family:This family works just like the Pyraminx family, but now there are parities involved with some of the edges since a turn creates edge 4-cycles. Again, we want to fix the corners, and we're not going to fix the tips. The orbits are 6 corners with 4 orientations each (1 * 4^6), n-1 sets of 24 edge wings ((24!)^(n-1)), one orbit of middle edges (12!/2 * 2^11), a total of floor((n-1)^2/3) orbits of 24 centers, 3 each in 8 colors ((24!/3!^8)^(floor((n-1)^2/3)), and ((n-1)^2 mod 3) orbits of 8 centers ((8!/2)^((n-1)^2 mod 3)). So the number of positions is 2^(22-((n-1)^2 mod 3)) * 24!^(n-1+floor((n-1)^2/3)) * 12! * 3!^(-8floor((n-1)^2/3)) * 8!^((n-1)^2 mod 3). The existing puzzles are n=1 (Magic Jewel, with 2.009 * 10^15 positions), n=1 with tips (Magic Octahedron, with 8.229 * 10^18 positions), and n=2 with tips (Master Octahedron, with 1.029 * 10^47 positions).

Trajber's Octahedron Family: Although the Trajber's Octahedron functions in an analogous way to an odd NxNxN cube, the pieces are colored differently, which means that a different method is required when solving it, and it has a different number of positions. Thus it makes sense to consider it a separate puzzle. Note that octahedra made from even NxNxN cubes are entirely different, in that they have many solutions and no piece has multiple stickers. We will hold the corners constant, although allowing them to rotate. Thus our orbits are 6 corners with 4 orientations each (1 * 4^6), 12 middle edges with two orientations each (12!/2 * 2^11), n-1 orbits of 24 edge wings ((24!)^(n-1)), n^2-n outer center orbits, with 3 pieces each of 8 colors ((24!/3!^8)^(n^2-n)), and 8 inner centers (8!/2). Thus the total number of positions is 12! * 8! * 2^21 * 24!^(n^2-1) * 3!^(-8n^2+8n). The existing puzzles are n=1 (Trajber's Octahedron, with 4.050 * 10^19 positions) and n=2 (5x5x5 Trajber's Octahedron, with 3.429 * 10^78 positions).

Dino Cube Family: This is an interesting set of puzzles to think about, although it is easy to solve, since you basically just have a large number of orbits that are all solved like a dino cube. Note that a true n-layer Dino Cube will have 2n(n+1) stickers on each face, and also has two solutions - if you solve it to a mirrored color scheme you will not encounter any problems. This doesn't change the number of combinations, but it does make the puzzle slightly 'easier'. There are n(n+1)/2 orbits on this puzzle, each of which must have an even permutation because all turns create 3-cycles, and no pieces have orientation since they are in a given orientation whenever they are in a given position. We will fix one piece in the centermost orbit (11!/2) and then keep the other ones free ((12!/2)^((n^2+n)/2-1)), so the total number of positions is 11! * 12!^((n^2+n-2)/2) * 2^((-n^2-n)/2). The existing puzzles are n=1 (Dino Cube, with 1.996 * 10^7 positions) and n=2 (Lattice Cube, with 1.145 * 10^24 positions).

Number of Positions of Some Other Interesting Twisty Polyhedra

We can use these techniques for basically any twisty polyhedron, but in some specific cases we need to use a trick to find the correct answer, because if we are not careful we will miss an important detail about the puzzle. I'll illustrate each one with a puzzle. I will make a large table with the number of positions for each of a number of existing puzzles in another article, so if you are interested in the number of positions of a specific puzzle that isn't mentioned here it should be on that list.

Siamese Cube: The Siamese Cube is composed of two separate puzzles which do not share pieces. Since they are the same, you can find the number of positions by finding the number of positions for one of those puzzles and squaring it. Fix the block and the number of positions is ((24)(120 * 3^5)(11!/2 * 2^10))^2 or 2.046 * 10^32.

Helicopter Cube: The tricky thing about this cube is that (in its non-jumbleable form) it actually has four orbits of centers, which each contains one center of each color, and at least one orbit can have an odd permutation as long as the permutation of all of the orbits combined is even (since each turn flips parity on two orbits but doesn't change the overall parity). So if we fix a corner the number of positions ends up as (7! * 3^6)(6!)(6!)(6!)(6!/2) or 4.937 * 10^17.

Alexander's Star: In the standard coloring scheme each piece has a duplicate. So we want to choose one pair of pieces and fix one. There are two identical pieces, though, so we can fix each position exactly two ways, so we have to divide the result by 2. But note that there still may be positions which look the same in more than one of these states (which we assume to be different), which would mean that the result we calculate is actually too small. Generally there are a very small number of these, not enough to influence the calculated result if we write it scientific notation, but it is important to realize that if we wrote out the number in full it would almost certainly be incorrect. The number of positions is approximately (29*(28!/2!^14) * 2^28)/2 or 7.243 * 10^34.

Here is another way to look at this phenomenon: consider a "puzzle" with six tiles arranged like the faces of a cube, where you can switch any two tiles at any time. Suppose we color these tiles so there are three red ones and three blue ones. Now intuition tells us there are exactly 2 positions; our calculation, on the other hand, says that we can fix one of them in a specific orientation and get (5!/(2!*3!))/3*4 = 120/144 or 0.833... positions. Of course this is a contrived result, with many more identical positions than a real puzzle would have, but it does show how the calculated number is too small.