How to Solve the 3X3X3 Cube Without Algorithms
I always wanted to explain to someone why commutators work on a cube besides explaining the commutator itself. So, I am going to derive it with reason.
The First Layer
I believe many can figure out how to solve the first layer without complicated algorithms.
The Middle Layer
Our goal should be to form an algorithm that inserts an edge from D into E, while preserving the U and E layers.
(Now, this algorithm being developed is to move the edge in FD to FR).
We start out with the four moves
D F D' F'.
(Yes, the algorithm above is a commutator, but a very cube intuitive one...it's cube "common sense").
Although it moves the front-bottom edge to its correct location (the front-right edge slot), the top-front-right corner is messed up. So we need to find a way to preserve this top-front-right corner piece while still using our 4 moves D F D' F'.
If we add the preliminary moves L' U', this places the front-bottom-right corner (which is not solved anyway, and therefore, we can use it as our guinea pig) in the top-front-right corner slot (the location of the piece which we are aiming to preserve).
But why those two moves and not others? Answer: Those two moves do not move the front-bottom edge, which is the very edge we are depending on being in the FR slot single we are building off of the moves D F D' F'.
Now we have:
L' U' (preliminary to preserve the original top-right corner)
D F D' F' (original 4)
To finish off, we simply undo those two moves to have: U L.
L' U'
D F D' F'
U L
Notice that, when we are undoing the set-up moves L' U', if we add only U, there is a top corner in the front left bottom slot, and it should be in the back-bottom-left corner slot before the first preliminary move, L', is reversed.
L' U'
D F D' F'
U
So let's move that corner in the back-left corner slot by doing D'.
So we have:
L' U' (preliminary to preserve the original top-right corner)
D F D' F' (original 4)
U (this is the reverse of 2nd preliminary move)
D' (puts the top corner in the front-bottom-left in the back-bottom-left corner slot)
Now we can reverse the first preliminary move: we do move L, to get:
L' U' (preliminary to preserve the original top-right corner)
D F D' F' (original 4)
U (this is the reverse of 2nd preliminary move)
D' (puts the top corner in the front-bottom-left in the back-bottom-left corner slot)
L (this is the reverse of the 1st preliminary move).
Thus we have:
L' U'
D F D' F'
U D' L
Now, it is obvious that only one move is required to restore the cube significantly, D.
L' U'
D F D' F'
U D' L
D
This entire algorithm is obviously a commutator. An algorithm of the form X Y X-1 Y-1.
L' U' D F
D'
F' D' U L
D
Now, the first two layers are intuitively solved because we have an "algorithm" to put an edge from FD to FR. If the mirror is taken, we can easily put an edge from FD to FL...
The Last Layer
For the last layer, it would help if we take this commutator and write its slice turn equivalent:
L' E' y' F
D'
F' E y L
D
Then, rotating,
U' M' U L' U' M U L
Adding the set-up move U,
U
U' M' U L' U' M U L
U'
=
M' U L' U' M U L U'
We have an algorithm that affects only 3 edges in the same face (e.g. the last layer).
We can use this commutator for the corners if we convert the M slice to the R slice (since the other slice already used is L).
R U L' U' R' U L U'
From here, many know how to use conjugates to do just about everything there is.
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