Having posed the question about Two-Faced cubes, I could not resist
thinking about it some more and I have pretty well resolved it. If
you want to try your hand at Two-Faced Cubing, DON'T READ THIS !
Suppose we only allow twists of the Right and Back faces.
Edge Orientations: You cannot change the orientation of any edge
cubie. So there are 2^7 different ways to orient them.
Edge Permutations: Any permutation is possible. (Must be even if
corner permutation is.)
Corner Orientations: Just like the complete cube.
Corner Permutations: The twists generate a subgroup containing only
one-sixth of the 6! possible permutations of the 6 corners. I do not
know a simple way to see why this has to be so. (I have a messy
argument that is no more persuasive than enumerating them.)
Solving It -
Do whatever it takes to put the two right-front corners in their home
positions (without regard to orientation). Rotate the back face to
bring (say) the URB corner to correct position. Of the six possible
permutations of the remaining three corners, you must have the correct
one. (That is, assuming you started from a reachable configuration.
Otherwise, this is a way to find a unique representative of the
equivalence class of corner permuations.)
(RB')^3(R'B)^3 may be used to fix the orientations of the corners.
(RRBB)^3 may be used to permute the edges. (You can always get the
two edge cubie pairs you want to swap in vertical opposition.)
Counting Things -
There are 6!X3^6X12!/(6X3X2) reachable configurations.
There are 6X3X2X2^7 equivalence classes.
Anyone for Three-Faced Cubing?