Cycle lengths of the 3x3x3

First, here is a table of operations which will make the listing of algorithm cycle lengths significantly easier for reasons that will be explained afterwards:

A list of n-move algorithms Ln is said to be complete if the closure of Ln under {Cd, Ce, Co, Cr, Mxy, Mxz, Myz, P, R, Rx, Ry, Rz} is equivalent to the exhaustive list En where En is the set of all algorithms of the form M1a1M2a2...Mnan. Thus it is not necessary to analyze 18n algorithms for all n-moved algorithms under the half-turn metric.

Prioritization from greatest to least of facial turns is {R,L,U,D,F,B} and prioritization of turning amount from greatest to least is {90deg, -90deg, 180deg}, denoted M or M1, M' or M3, and M2 respectively.

Period lengths of all edges and all corners are given as well. The total period of an algorithm is the lowest common multiple of the edge and corner period values. Some algorithms such as RUR are "equivalent" to simpler algorithms by virtue of their form, but collapsing RUR to R2U is not good for a master list because RURF is **not** equivalent to R2UF and thus RUR and R2U are technically different. (In this case, RUR => RURRURRURRUR... => RUR2UR2UR2... => R2U, with a cycle length of 30. R2UF has a period of 36 but RURF has a period of 84.)

There are only a finite number of cycle lengths, corresponding to the greatest lowest common multiple of all cycle lengths for edges and corners taken together. The maximum cycle length is 1,260. By the way the pieces are put together in relation to each other on the cube, it is impossible to do *just* the following (note that O is shorthand for an Odd number and E is shorthand for an Even number):

* Change the orientation of O edges.
* Flip the orientation of a group of corners so that the combined orientation (mod 3) is not 0. (Videos purporting to show a single-corner rotation algorithm are in-jokes and beginner trolling in the cubing community.)
* Permute an odd number of E-cycles, where the cycles can be either corners, edges, or both.

These rules allow us to calculate all possible cycle lengths without even touching the cube. For the following table, nC represents a general n-cycle, ncC an n-cycle of corners, neC an n-cycle of edges, a number after an abbreviation denotes repetition (i.e. 2cC3 stands for three 2-cycles of corners), and o after an abbreviation stands for orientation (i.e. 3cCo stands for a 3-cycle of corners where orientation is relevant, so the permutation period is 3 but the orientation period is 9, so the period of 3cCo is 9).

Since there are no more than 12 edges, 11 is the highest prime divisor of any cycle length. On larger cubes with 24-member center or edge groups the maximum prime divisor is 23. For the megaminx it is 29, higher order -minx it would be 59, and for the tuttminx it's 89. Regardless of the order of the puzzle, the limiting factor on prime divisors is basic geometry. The most common algorithm length is 60, which makes sense given 60 has 2,3,6, and 12 as factors and these are an inherent part of a cube's geometry.

For a sufficiently large n x n x n cube, the maximum cycle length is the LCM of all primes underneath 24 (the maximum group size) along with some extra multiplications by 2 and 3. These extra multiplications are derived from the presence of cycles such as 2C + 8eCo, one to construct a permutative-orientative 16 cycle. On cubes larger than the 3 x 3 x 3 there are 24 edges not 12 so one can make a 2C + 16eCo or 32-cycle. All cubes have the same number of corners, so 9 is the maximum power of 3 that can be in a cycle length. 32•9•5•7•11•13•17•19•23=10,708,457,760, which is the maximum cycle length of an algorithm **any** puzzle cube can have. This means the maximum cycle length of an algorithm on a 1,000 x 1,000 x 1,000 cube is the exact same as on a 10100 x 10100 x 10100 cube.

The "sufficiently large cube" in the last paragraph would have to have at least five center/edge groups, one of which must be an edge group (for 16eCo). The 6 x 6 x 6 is thus the first cube allowing this 10 billion long cycle length. There are 1,936 possible cycle lengths for cubes of tier 6 and up. The demonstration of this is left as an exercise to the reader.

The "Devil's Algorithm" is the shortest algorithm which when applied repeatedly will bring the cube to a solved state. Since the cube has 43,252,003,274,489,856,000 states but a maximum cycle length of 1,260, the Devil's Algorithm would have to be at least 34,326,986,725,785,600 moves long. At 1,000 moves per second assuming a machine that was fast enough, it would take a few centuries over 1.088 million years to complete one algorithm execution, and over a billion years to solve the cube like this. 7 and a half Devil's algorithms is equal to about one Deep Thought, execution time-wise.

## Determining Isomorphism

To determine if two algorithms are isomorphic, a variety of methods can be used, described like so:

### Transfoisomorphism

With 24 rotations, three possible ways of mirroring or not mirroring (making 8 choices), and reversal and parity (adding 4 choices), there are 768 isomorphic algorithms for any given cube algorithm (including that algorithm). Unless a program were written which could take two algorithms written in standard form and search the 768 algorithm space for the other algorithm, using all possible transformations is highly impractical. A benefit to this method however, is that if programmed, it could ensure no equivalence is overlooked or that two fundamentally identical algorithms aren't listed twice.

## The List

See List of 3x3x3 algorithms for the list.