That means I can't solve any state with 2gen right.
That's correct. Even on the 3x3x3, where it may appear that some positions should be 2-gen friendly (solvable with only <R, U> moves), they are not. (Fun fact: For the <F, F', F2, B, B', B2, L, L', L2, R, R', R2> 3x3x3 move set, you are able to do a U or U' move, but a 4 edges must be flipped! See
my post. I posted such an algorithm at the end of that post, but see the 23 move algorithm at the beginning of
this later post.)
It makes sense if you think that you have to reduce the 3x3 to solve 2gen state. (Permuted corners, oriented edges)
Which makes me think that you understand what I just mentioned above pretty well.
What about big cubes? What are the conditions to be able to solve 2gen? And what can't I solve?
Well, it varies, depending on the "type of 2-gen" you have.
With
2-gen Type 1 from my video series (<U, r>, where r is just the inner layer slice), there isn't much of any restriction, at least on the 4x4x4. For example, Bruce wrote a LL wing 2-gen solver and found optimal solutions for all 8! wing LL cases with
this code (again, like me ever playing with Ksolve, I never ran this code). And clearly the bottom corners must be solved, and the top ones must be able to be solved by AUF. And, as long as it's not the supercube, there is no restriction for the X-centers. (But obviously, if you consider the 4x4x4 supercube, then all other centers (not in reach of <U, r> moves) need to be solved (supercube-wise) before attempting to complete the 4x4x4 supercube with <U, r> moves.)
However, as you can (will) see in my
2-gen Type 5 video, I mentioned (discovered) that with a position like this
r2 Uw Bw' Uw' r Uw Rw Bw r' Bw' Rw' Bw Uw' r, even though the X-center pieces affected appear to be in reach of
Uw, Rw> moves, it's only possible to "flip" that dedge in that moveset. The X centers cannot be touched . . . at least if you want to also exchange them simultaneously as the 2 wings . . . at least with my limited 3 hours experience worth of experimentation.
Edit:
However, as you can (will) see in my
2-gen Type 5 video, I mentioned (discovered) that with a position like this
r2 Uw Bw' Uw' r Uw Rw Bw r' Bw' Rw' Bw Uw' r, even though the X-center pieces affected appear to be in reach of
Uw, Rw> moves, it's only possible to "flip" that dedge in that moveset. The X centers cannot be touched . . . at least if you want to also exchange them simultaneously as the 2 wings . . . at least with my limited 3 hours experience worth of experimentation.
After watching the video on it, it's actually the opposite of what I said. It's possible to generate that position in the move set, but it could be impossible generate a
conjugate of it in this move set, such that the 4 center pieces are swapped with center pieces of the same color!
Resuming original post . . .
In fact, I don't know if it's even possible to flip 1 dedge and permute exactly 4 centers simultaneously (which is a very common occurrence in 4x4x4 single dedge flip algorithms). A (genuinely PURE) single dedge flip may be all you can do in that regard.
As you can see in
my post on math.stackechange regarding how to calculate exactly how many 3x3x3 positions are solvable with using only <U, R> moves, specifically:
Potential Sketch of Proof 1
Consider either the U face or R face. There are (4!)/4 = 6 possible 4-cycles and (4!)/(2^2*2!) = 3 possible 2 2-cycles in 4 objects. Using < R,U > moves alone, only two of the six possible 4-cycle permutations are reachable in < R,U > (one of the three distinct 4-cycles and its inverse) and only one of the three 2 2-cycles is reachable in < R,U >. Since we can write any of the 6 possible 4-cycles as a product of any of one of the other 4-cycles and a 3-cycle, clearly this shows that it's not possible to do a 3-cycle of corners in < R,U >. A similar argument holds true for the 2 2-cycle.
It could be (mostly likely be) that the move set
Uw, Rw> also restricts the type of cycles we can do with the centers. 3x3x3 movesets like <U2, F2, D2, B2, L2, R2> don't allow for all 43 quintillion 3x3x3 positions (see
Jaap's page for the exact amount of "only" 663,552 positions).
And so it may be for
2-gen types 2-4, but maybe these unexpected situations just happen with
Type 5. (Not sure.)
Basically, the key to answering your question (for a specific cube size and/or specific "type"/variation of 2-gen . . . or more generally, "subset"), the rule for permutations is, if you can do a 2-cycle of a certain piece type with a specific move set, then you can do ANY permutation of that piece type with that move set. (Because you can solve any permutation/unordered list of objects by a series of swaps.)
Therefore, for move sets which allow odd permutations (<U2, F2, D2, B2, L2, R2> is an example which does NOT allow/generate odd permutations), finding out if you can (with the limited number of possible moves you can use) do a swap (2-cycle) of pieces in EVERY orbit will be a good test on whether all positions, which
visually appear to be solvable in a big cube subset, can in fact be solved/generated with the available moves in the move set.
And for piece orbits in which a move set only allows even permutations for, see if you can do a 3-cycle in each of them. If so, then all (even) permutations all that piece orbit are able to be solved.
)
