The Yau method is a very effective way to solve the 4x4x4 cube. It is very practicable for Speedcubing. The Dutch cuber Mats Valk achieved a new world record (single time) by solving the 4x4x4 with this method on 17th December 2011 with a time of 30.02 seconds.

I will try to explain this method with a example solve.

Solve the first "center" (4 side pieces with the same colour). If you aren't colour neutral at solving the 3x3x3, you should begin with the same colour you would also begin at the 3x3x3, or with the colour of the opposite center (in my example white).

After that please solve the opposite center (in the example yellow).

1st center | 2nd center |

Begin with solving of those edges (consisting of 2 edge pieces with the same colours) which contain that colour, that is the one you are beginning with when solving the 3x3x3 cube (in the example solve it is the white colour). At first, you solve only 3 of the 4 edges. Bring each solved edge immediately to those position where it belongs to at the solved cube. You have to bring the 3rd edge also to the layer with the right center colour, but on the wrong of the remaining 2 positions (In the example the 3rd edge is the white/red one which we place opposite to the white/green one; on the solved cube there must be the white/blue edge on this position instead of the white/red one.) You will find out later the reason for that.

1st edge | 2nd edge | 3rd edge This edge is already solved by accident. |

Solve now the remaining 4 centers. Take care not to mix up the already solved pieces. If you hold the cube with the solved edges on the left side (and the not solved first layer edge on the top) then the only layers you should turn are the top and right layer and both middle layers next to the right layer.

Take care to place the centers at the right positions (where they have to be at the solved cube). If your cube has the same colour scheme as the example cube, then white must be opposite to yellow, orange opposite to red and green opposite to blue. Furthermore, when the white center is left and the red one is on the front layer, the blue center must be on top.

blue center | orange center | green and red center |

Hint:

To see the orange center which has been created on the bottom of the cube on the second applet, please rotate the cube a little bit.

A detailed instruction manual how to use the Java applets you'll find here.

Now solve the 4th edge of the first layer (in the example white/blue). If you would place this edge at the remaining position at the first layer, the centers wouldn't be solved and one side would be turned at 90°; therefore it wouldn't be possible to repair the centers with one turn. Since we have placed the white/red edge at the wrong position, we can now swap the white/red edge with the white/blue edge, turn the white layer and bring the white/blue edge to the right position. This makes it possible to repair the centers with only one more turn. Please also turn the white layer until the colours match with the centers.

4th edge |

Solve the remaining edges. If you arrange the edges correct, you can sometimes solve 3 (in rare cases even 4) edges at one time.

Solving edges 5 to 7:

In the example, left beside the red center there's a green/orange edge piece. We are searching now for a second unsolved edge which contains a green/orange edge piece and place it right beside the red center, with the green/orange edge piece on top. Below the green/orange edge piece there's a red/blue one. Therefore we place the second red/blue edge piece on top of the next edge. Below the red/blue edge piece there's a green/red one, therefore the second green/red edge piece must be the upper piece of the next edge. Now it is possible to solve 3 edges with one single turn. 4 already solved centers will be destroyed by this turn.

Solving edges 8 to 10:

We place 4 unsolved edges along the 4 destroyed centers in such a way that we can solve as much edges as possible when we repair the centers: The unsolved edge between the mixed centers ha a orange/yellow edge piece -> the edge to the right must have orange/yellow as the lower piece. Above there's a yellow/blue one -> the edge to the right must have yellow/blue as the lower piece. Above there's yellow/red -> yellow/red must be the lower piece of the 4th edge. The following turn will not only repair the centers but also solve 3 additional edges.

Solving of the remaining 2 edges:

The 3rd applet shows how the last edges can be solved: At first both unsolved edges will be placed side by side with both edges having the blue/orange edge piece as the upper and the yellow/green one as the lower edge piece. After that, the upper 2 layers will be turned 90° counterclockwise. With the next 7 turns, the right edge will be flipped (the upper piece goes down and the lower piece goes up). When you turn back the upper 2 layers (90° clockwise), all edges and centers will be solved.

5th to 7th edge | 8th to 10th edge | 11th and 12th edge |

Now you can solve the remaining cube just like a standard Rubik's Cube (3x3x3). As a matter of completeness, I will show you how I with solve the example cube with the Fridrich method. If you want to learn the Fridrich method, you will find plenty of websites explaining it, e.g. on solvethecube.110mb.com/

At "F2L" (first 2 layers) you have to connect a pair consisting of a corner of the first layer and the matching edge of the second layer and insert it in the right position.

1st F2L pair | 2nd F2L pair | 3rd F2L pair | 4th F2L pair |

AT "OLL" (orient last layer) all pieces of the last layer will be oriented, so that the colour of the last layer (in the example yellow) will be on top of each single piece. Don't care about the right position of each piece at this step.

OLL |

This is a special case of the 4x4x4 cube (see Complete solving the Cube (part 4) for another special case which can happen at "OLL") which isn't possible at the standard Rubik's Cube: The number of necessary permutations is odd (odd PLL parity). In this case you'll need the following algorithm:

PLL Parity |

At "PLL" (permute last layer) the pieces of the last layer will be placed at the right positions. Solving is now completed.

PLL |