S = B D2 R U B D' R D2 F U D' B2 D' L2 D' F2 R2 B2 D2 R2
M1 = B2
S M1 has two blocks: DB&DBL (now in RU&RUB), LB&LBU (now in UL&ULB).
For the first one, X = z y', so after executing S M1 X, DB&DBL are "solved". As a result, DB&DBL are also "solved" if we execute X' M1' S'.
In fact, DB&DBL...
You are right. I didn't explain clearly.
Maybe we can see the cube from a different point of view. That is, the center cubies can change positions (by rotation), just like edges and corners.
Then "a block is solved" means the block is in the same position as it was in the initial state. For...
My answer is regarding unsolved blocks too. I meaned to answer you question using what you have known in #1 :)
What I said is that we can extend the cube group by adding rotations x,y,z. Then for a unsolved block A of S M1, we can rotate the whole cube (the rotation is X) so that we put the...
S M1 has a block A, then there exists a sequence X (exactly, X is some rotation of the cube), and block A is among the solved cubies of S M1 X.
As you posted, block A is among the solved cubies of X' M1' S', too. Since X' is a rotation, M1' S' has a block B.
I've read Inside the Rubik's cube and beyond(by Christoph Bandelow) and Handbook of cubik math(by A.H.Frey & D.Singmaster). These might help.
I tried and solved, turning only L-U-R.
At first, I only thought of the alg I used: R' U L' U2 R U' R' U2 R2. Then I realized that L' R, then F turns...
Void cube is the cube without the centre caps ^_^
Turning a middle layer for 90° results in a center-4-cycle(invisible in this situation) and an edge-4-cycle.
4-cycle is an odd permutation. That's why parity can occur.
Scramble 1: F' U' F U R U R' U (1 to go with the help of hint)
Scramble 2: F' U' F U R U R' U' (done with the help of hint)
Scramble 3: F' U' F U R U R' U2 (done)
Scramble 4: F' U' F U R U R' (done)
After seeing the hints, I still can't find this one.
As to the Dutch Open 2009 one, I just tried and got a start like this: U' L' R2 U' R2 U' R2 F2 D' R F' R2 D F2 D' F.
But I'm not good at the last layer at all. After this start, I only find
NISS: R2 B R B2 D' B D' R D2 F D F' D2, leaves only two twisted corners……
Aha, the scramble below also has kind of symmetry, that no 2 adjacent(even adjacent in diagonal sense) cubies are of the same color.
L' R2 B2 D2 B2 R' B L' D' U F' L' F2 D' R2 B' U2 R F2 U'
So can it be a hard scramble?