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.