User blog:Edwin Shade/An Algorithmic Analysis Of The 3x3x3 Rubik's Cube

In this blog post I will be showing ways to extract exponential-sized numbers from the combinatorics of various common twisty puzzles.

3x3x3 Functions
The first function I will introduce is the $$P(A)$$ function, where $$A$$ is an algorithm. The P stands for 'Peal' and functions similarly to how bell's peal the same tones in different combinations, (here the 'bell' is the puzzle, and the 'chimes' are the twists of the cube). $$P(A)$$ is calculated by multiplying together the cycle lengths of all algorithms of the form $$An$$, where $$n$$ is an additional move which is not a slice move or a double turn, but may nonetheless be a double turn.

For example, to calculate $$P(RUR'U')$$, we have to calculate the cycle length of any algorithm of the form $$RUR'U'n$$, where n is an additional move, of which there are 18 possibilities. The derivation of $$P(RUR'U')$$ is shown below, keep in mind that $$[A]$$ notates the cycle length of an algorithm $$A$$.

$$[RUR'U'U]=[RUR']=4$$

$$[RUR'U'U']=[RUR'U2]=12$$

$$[RUR'U'U2]=[RUR'U]=5$$

$$[RUR'U'D]=60$$

$$[RUR'U'D']=60$$

$$[RUR'U'D2]=18$$

$$[RUR'U'F]=6$$

$$[RUR'U'F']=18$$

$$[RUR'U'F2]=12$$

$$[RUR'U'B]=18$$

$$[RUR'U'B']=6$$

$$[RUR'U'B2]=12$$

$$[RUR'U'L]=60$$

$$[RUR'U'L']=60$$

$$[RUR'U'L2]=18$$

$$[RUR'U'R]=12$$

$$[RUR'U'R']=4$$

$$[RUR'U'R2]=5$$

$$4\cdot 12\cdot 5\cdot 60\cdot 60\cdot 18\cdot 6\cdot 18\cdot 12\cdot 18\cdot 6\cdot 12\cdot 60\cdot 60\cdot 18\cdot 12\cdot 4\cdot 5=$$

$$406,239,826,673,664,000,000$$

Therefore, $$P(RUR'U')=406,239,826,673,664,000,000$$.

Below is a list, (currently in progress), of all the values of the $$P(A)$$ function for algorithms equal to or less than five-moves in length. Not all algorithms have been included, but at least all the distinct ones have been. (Such ones as $$R'$$ and $$D$$ are really equivalent when viewed from the right perspective.

One-Move Algorithms
$$P(R)$$

$$[RU]=105$$ $$[RU']=63$$ $$[RU2]=30$$ $$[RD]=105$$ $$[RD']=63$$ $$[RD2]=30$$ $$[RF]=105$$ $$[RF']=63$$ $$[RF2]=30$$ $$[RB]=105$$ $$[RB']=63$$ $$[RB2]=30$$ $$[RL]=4$$ $$[RL']=4$$ $$[RL2]=4$$ $$[RR]=[R2]=2$$ $$[RR']=[]=0$$ $$[RR2]=[R']=4$$

Two-Move Algorithms
$$P(RU)$$

$$[RUU]=[RU2]=30$$ $$[RUU']=[R]=4$$ $$[RUU2]=[RU']=63$$ $$[RUD]=90$$ $$[RUD']=180$$ $$[RUD2]=168$$ $$[RUF]=80$$ $$[RUF']=60$$ $$[RUF2]=36$$ $$[RUB]=84$$ $$[RUB']=360$$ $$[RUB2]=9$$ $$[RUL]=$$ $$[RUL$$

4x4x4 Functions
The Peal function can of course be adapted to the 4x4x4.

5x5x5 Functions
So to the Rubik's revenge.

7x7x7 Functions
In can in fact be applied to a cube with any number of layers, but I will cease the examples at the level of the 7x7x7 cube due to the fact it is the highest order cube I have.

[IN PROGRESS]