Rubik's Cube with Oriented Tiles - minimum moves and algorithms

Question

I know that there are mathematical descriptions of the Rubik's Cube based on finite group theory. And as far as I know those descriptions deal with just coloured tiles. Suppose however that in addition to the typical colourings the cube has letters (or markings similar to dice) applied to each tile and those letters are orientable. So that a correct solution to the cube not only requires you to match that correct coloured tiles to form a monochromatics face but also requires that every face have its letters in the same orientation.

How does orientation change our group theoretic description of the cube?

I am not certain if there is a better way to describe these cubes. It is a two years work anniversary gift.

Answer 1

With regards to the corners and edges, the solution of the cube, and its difficulty will be entirely unchanged by your addition of stickers since the cube has only one solution for each of those. Every piece fits in one and only one place, and in only one orientation, in the final solution of the Rubik's cube.

This is because there is only one "yellow and red" edge cube and its only correct position is between the red and yellow axes, and with the red on the side of the red axis. The same rule applies for all squares on the cube.

That is to say, provided you begin with a cube on which all the orientable stickers are correctly oriented, it will still be solvable in exactly the same way as before, and if you begin with a cube which is solved other than the stickers incorrectly oriented, you will never be able to solve it.

HOWEVER... That only speaks for the edges and corners, and leaves the exceptions, which are the centre squares. These are different in that they CAN each have 4 different orientations in the finished cube, at least in theory. These are always ignored in analysis of the solution to the cube since they are fixed in position in relation to each other and their choice of which of their 4 orientations they are in, is indistinguishable. Therefore I've not seen material on whether the cube can be solved with different centre rotations.

I just marked the yellow, red, white and orange centre-squares of a solved cube with green crayon on the edge nearest to the green face. Then I mixed up and solved the cube, to discover that the centre squares can indeed achieve different orientations from one solution to the next. Therefore if you marked them, there would be between $4^6$ and $2$ different solutions depending on the symmetry of the rotation group of the axes. I can't give you a rigorous answer as to how many more permutations this introduces but judging by the pattern on my cube after solving I can make a guess in the dark that it will multiply the currently achievable number of arrangements of the cube by $2^3=8$.

Bear in mind however, the centre square of your cube shown in the picture does not have a distinguishable orientation.

Answer 2

There are many ways you could describe such a Rubik's cube in group theory. One of the easiest ways is to divide each centre piece into four equal squares. Then clearly rotating the centre piece is simply a 4-cycle on those squares. The other pieces of the cube (edges and corners) are divided as usual into 2 and 3 faces respectively, which is sufficient all unique description of any state, because the position and orientation (flip or twirl) of each of those pieces already fully determines the state of each of its faces (unlike for the centre pieces).

Now as to the more interesting question of which states (of the centre pieces) are possible, it turns out that there are $\frac12 \cdot 4^6 = 2048$ states, namely all the states where the sum of the parities of the quarter turns on each centre piece is even. This is easy to prove. First note that each quarter turn moves exactly four corner pieces in a $4$-cycle (ignoring orientation), and thus by basic group theory you need an even number of quarter turns to perform any even permutation of the corners, which includes going from a solved state to a solved state (ignoring centre pieces). Thus the sum of the parities of the centre pieces' rotations (in quarter turns) must be even. Also, it is easy to check that we can achieve any such state by first performing the necessary quarter turns to set the centre pieces in the correct orientation (which performs an even permutation on the edges and an even permutation on the corners) and then use only commutators (as described here) to solve all the edges and corners. (If one is clever enough, one can alternatively use commutators to directly turn the centre pieces!)