One of the most important things to do before you start learning any speedcubing method is to know how to solve your Rubik's cube a beginner way. If you know this, you will understand that you cannot just get all of one color on one side, but you have to put the pieces in the right place as well. Knowing this will also help you to understand why many of the algorithms of a speedcubing method work. You also need to know the color scheme of your cube by heart. To get a fast time, you must be able to know where pieces go relative to each other without having to think about it.
The notation that will be used here is as follows:
R - Right face
L - Left face
U - Up face
D - Down face
F - Front face
B - Back face
M - Slice between R and L (Middle)
E - Slice between U and D (Equatorial)
S - Slice between F and B (Side)
x - whole cube in direction of R
y - whole cube in direction of U
z - whole cube in direction of F
No Suffix - quarter-turn clockwise
' - quarter-turn counter-clockwise
2 - half-turn
w or lowercase letter - double layer quarter-turn clockwise
w' or lowercase letter - double layer quarter-turn counter-clockwise
w2 or lowercase letter - double layer half-turn
Solving the cross is the first step in solving the Rubik's cube using CFOP. This involves solving the four edge pieces on one side. The cross can be solved on any side. The method that is preferred by most is solving on the bottom. The method explained here will solve the cross on the bottom. When solving the cross, you do not have to worry about matching the edges to the individual sides. You must only make sure that the pieces are correctly positioned in relation to all the other cross pieces. You can turn the bottom face to match the pieces after the cross is done. This is usually the simplest step and takes expert speedcubers 2-3 seconds.
Permutation of Last LayerEdit
The Permutation of the Last Layer, or PLL, is the final step in solving a cube using CFOP. This step involves moving the pieces of the top layer without reorienting them to move them into a solved state. This step requires the memorization of 21 algorithms, which is relatively few considering how many F2L and OLL required.