Reachable positions with <R,U>?

[qqwref]

Come on Stefan, I expected more out of someone as unbelievably arrogant as you. You should at least make a basic effort to understand what I'm saying before you start flaming my approach. Jaap's page clearly shows how a U and R turn will always transform one of the five permutations into another; for instance, if the corners are now in a V permutation and you do a U turn, they will be in a W permutation.

 

[Stefan]

Oh boy, this is really not my day... sorry. I think I did understand you, it's what I had tried myself before, but I had made a mistake and because of that thought it doesn't work and that U is a counterexample. However, I still believe the method isn't quite correct. You'd get 5*3!*2^3 possibilities, that's twice as many as the 120 actually possible ones. So I guess there's still some parity issue in there. Should've used that argument right away, but "U" was shorter.

Edit: Yeah, just swap the two corners of one pair.

 

[qqwref]

OK, I see what you mean. Hm, once we've found out that the corners fit into one of the permutations, how do we check if it's got the right parity (what's the 'solved' state to compare to)? Clearly just checking the parity of the corners won't work.

 

[meichenl]

It looks to me like qqwref's solution is almost there.

What we need to do, according to Jaap's page, is check that the corner permutation corresponds to a permutation of VWXYZ.

So, to work an example, let's use Jaap's proof to show that U2 R U2 is solvable by R and U. We start by saying that the solved position is V, and ask what the scrambled position is. It turns out to be Y.

Next, we say that the solved position is W, and find that with this assumption, the scrambled position is also W.

Going through and setting the solved position to each of the five pairings, we get

V -> Y
W -> W
X -> V
Y -> Z
Z -> X

this is the permutation

(VYZX)

Since it's a permutation of {V, W, X, Y, Z}, it is a solvable state.

We fail the test if at least one of the pairings does not get mapped to another pairing. For example, consider
the state reached by

U' F' U2 F' D2 F U2 F' D2 F2

This is a 4-cycle of corners, but it can't be solved by a 2-gen algorithm. If the solved state is called pairing V, then this corner permutation doesn't correspond to any of the other pairings, so it fails the condition qqwref set. This shows that some 4-cycles can be solved, but other cannot. I got that position by looking for a 4-cycle that obviously failed to map V into another pairing, and using CubeExplorer to get an algorithm for it.

Now consider switching all the pairs in V by the algorithm

F2 D' L2 U R2 D B2 R2 D' L2 F2 D' F2 D2

This clearly maps V to V, but it doesn't map W onto another pairing. This is why there was the overcount Stefan mentioned regarded qqwref's criterion. There are corner permutations that map some but not all pairings to other pairings.

It's now easy to see that swapping corners isn't allowed. That would leave four corners untouched, which means we'd have to be dealing with the identity permutation, but if we just swapped corners it's obviously not the identity!

We can also see why 3-cycles are disallowed. Suppose we could do an A-perm. Then by doing the A-perm, then a U, we'd have swapped a pair. If the three-cycle is on the R face, it's basically the same argument. If it's split between both faces, it's easy to see by playing with the cube a couple moments that it could be achieved by conjugating an A-perm with one or two allowed moves, so 3-cycles are not allowed.

 

[qqwref]

This clearly maps V to V, but it doesn't map W onto another pairing. This is why there was the overcount Stefan mentioned regarded qqwref's criterion. There are corner permutations that map some but not all pairings to other pairings.

Aha! That's it. So it looks like checking one of the pairings is not enough, and checking all five is sufficient. I wonder if we can make do by checking fewer than five. I think it might be enough to check 2, actually. Any permutation of the corners that cannot be 2gen'd must be some 2-cycle away from one that is, and we can use 2gen moves (permutations of the five pairings) to move the two pieces that need to be swapped into D. If we switched those corners all pairings would then go to another pairing, so it follows that in the position we have only one pairing (Y) goes to another pairing while the rest go to some pairing that isn't in our set of five. Hence in the original permutation of the corners (any one which isn't solvable with 2gen) only one of the five pairings can be mapped to another of those five pairings.

If that's correct, the method to do this would require you to know the groups of corners formed by two pairings (I suggest X and Y). You would have to check one, and if it didn't match then you would know the position wasn't solvable 2gen, but if it did you'd check the second one to make sure.

Try it out on this one: F2 R2 U F2 U2 L F2 L B2 D R2 D' L B2 L U2

Spoiler

Y works, but X doesn't.