# "Cross" Meta-Method for 4x4x4

#### qqwref

##### Member
More methods than your cube has room for.

The Theory

The idea behind this meta-method (and I call it that because it encompasses so many normal methods, as you'll see in a bit) is to divide the solve into a small number of steps, each of which can be done relatively efficiently/quicky and in many different ways. There are in fact only two steps, which are as follows:
- 1: Solve the centers and all 8 of the edge pieces on D (or L, if you prefer).
- 2: Solve the rest of the puzzle.

So, why are these two steps good? The key is that the first step is very simple to do, while still leaving a lot of freedom to the solver. In this step, we build a structure similar to the cross from Fridrich; there is a lot of freedom here, because until the entire step is completed you have the freedom to do most outer layer turns and a few Rw U Rw'-type triggers. It may seem constraining, but in practice it is quite easy to set things up without destroying what you have. There are many ways to do this, from blockbuilding to a more reduction-oriented approach.

After we have done the first step, what we are left with is set up quite well for speedsolving: not only have we paired up the centers (the pieces that typically require either a lot of freedom or a lot of moves), but we have also fixed the bottom edges so we can guarantee never to have to look there. We are now free to do U, Uw, and D moves, as almost all moves here will keep the orientation of the center blocks constant, and we can of course use any of the common F2L triggers on the outer layers. Even though there aren't a lot of layers that we can freely turn without breaking up the solved pieces, almost everything we can do has a useful effect on the cube and is also very suitable for speedcubing.

Step 1: The Cross
There are a lot of ways to do the cross step. Here are some of the ways I've tried, although they are by no means all of the ways to do this step:

K4 Style
The actual K4 method sets up some corners as well, so this is a variation on it. The first step is to set up two opposite centers, one of which should be your cross color. Next, using the M layer, you pair up 3 of the cross edges, placing them correctly on the left face (if you're right-handed). Finally, you finish the centers without disturbing the cross edges, and then pair up the last cross edge and insert it.

Reduction
First, simply solve all centers like in reduction. Then pair up the four cross edges (you can do this by pairing two edges at a time, or with a more 'freestyle' approach where you freely do double layer turns on one slice) and place them.

Columns
Start by doing only the D center, and place it on the bottom. Then set up the rest of the pieces by doing eight columns - each column has two center pieces and one edge piece. You can do the columns in pretty much any order, but it is important to realize that you can insert a column with Ru'R' type moves, not just rU2r'.

M-Slice
First, solve two centers and place them on R/L, but make sure they are not your cross color centers. Next, pair up (using the M slice) and solve the DL and DR cross edges. Finally, use the M slice to build the remaining three centers and two edge pairs in whatever order seems most convenient, making sure not to move the DL/DR edges. Notice that there's still a lot of freedom since B, U, and F are free, so you can do any kind of rUr' type triggers, and you shouldn't have to use too many slice moves.

Step 2: The Rest
Again, there are many ways to finish off the method; here are the ones I've played around with:

First, place the corners of the first layer (note that it is a bit more efficient to do this during Step 1 if you can). Next, fill in the second and third layer by inserting the 8 edge pieces with commutators. Finally, solve the last layer with CLL and then ELL.

F2L-Style K4
Place the corners of the first layer, using F2L techniques to also place one edge piece (on average) per slot. Next, fill in the rest of the second and third layer with commutators, although you should only need about 4 commutators on average instead of 8. Finally, use CLL and ELL to finish the last layer.

Reduction (Yau)
First, pair up the edges; 3-2-2 pairing (start with a u, place three edges and do a u', then do 2-pairing for the rest) is suggested. Next, solve the F2L and LL with whatever variant of Fridrich you are most comfortable with.

Commutators
Solve the corners and then finish the rest of the edges two at a time with commutators.

Commutator Fridrich
Place the first layer's corners along with as many edges as you can, using F2L techniques. Then use OLL and PLL to try to place as many last layer edge pieces as you can. Finally, solve the remaining edges two at a time with commutators.

The Method(s) In Practice
The three well-known 4x4 methods built around this are K4 (with two major variants) and Yau. The standard version of K4 uses the "K4 style" cross and the "traditional K4" rest, although it has the slight optimization of doing the four first-layer corners during the cross step. There is also a variation of K4, sometimes used by Dan Cohen, which is a "K4 style" cross followed by an "F2L-style K4" rest. Finally, the Yau method uses a "K4 style" cross with a "reduction" rest.

Sub-minute averages have been achieved with all three of these methods, and Dan Cohen has even broken the 50-second barrier using Yau. Yau was also the method he used to achieve the current WR single of 36.46 seconds. It is clear that this meta-method has some serious potential; all that remains is to practice a lot

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#### Lucas Garron

##### Moderator
Staff member
Have you timed averages?

I just decided to try centers > place 8 cross wings > K4/F2L-style pairing > 3x3x3, and I kinda liked it.

Do you think anything other than cross could be good?

EDIT: Wow, I just got a really nice solve on my second timed try. Woulda been mid-50s if I hadn't made mistakes on LL. I really like only having to search one side of the cube fore edges.

EDIT 2: 1:30.98, 1:03.07, 1:22.20, 1:43.20, 1:14.71, 58.57, 1:03.37, 57.48, 1:32.60, 1:04.03, 1:18.83, 1:08.74

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#### rubixfreak

##### Member
I like doing K4 - Fridrich hybrid, but somehow the LL is too ineffiecient. as im too lazy to work out CLL im doing OLL - PLL instead and it takes up to whopping 6 algorythms to solve it (3 for OLL and 3 for PLL).
qqwref do you know any good system for that which doesnt requiere CLL?

#### fanwuq

##### Member
Just another variation:

Reduction to solve centers up to last 2 adj centers

Unsolved centers at U and B.
Set up 2 cross edges to pair at UB and UF. Slice to pair up; <L, U, R> to place next 1 cross edge at UB or UF and trigger in its complement; slice back. Solve the 2 paired edges; solve centers.
2 pair to pair up 2 more pairs in which one is a cross dedge. Solve the 3rd cross dedge. 3-2-2-2 to finish reduction.

Example:
http://alg.garron.us/?alg=D-_f-_d_ ..._r2_R2_F_L2_B-_R_r_L_u-_L2_u-_F_L-&cube=4x4x4

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#### qqwref

##### Member
I like doing K4 - Fridrich hybrid, but somehow the LL is too ineffiecient. as im too lazy to work out CLL im doing OLL - PLL instead and it takes up to whopping 6 algorythms to solve it (3 for OLL and 3 for PLL).
qqwref do you know any good system for that which doesnt requiere CLL?
Uh, how about doing OLL and PLL in one step each? >_> I don't know any kind of CLL.

fanwuq: that's an interesting idea. If the last two centers are free you can pair up edges with Rw U/U' Rw'. You could actually pair up all edges like this, as there are 5 free slots, and then insert them later. I'll try a few solves.

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#### Crystl

##### Member
maybe that's a better way to solve the 4x4x4, I should try it

#### Meep

##### Member
What I said earlier today they got pushed away by the argument was that you could also make corner-edge-edge pairs which isn't very hard to do =P

#### Robert-Y

##### Member
What I said earlier today they got pushed away by the argument was that you could also make corner-edge-edge pairs which isn't very hard to do =P
What methods are there of doing this?

#### Meep

##### Member
What I said earlier today they got pushed away by the argument was that you could also make corner-edge-edge pairs which isn't very hard to do =P
What methods are there of doing this?
You can pair the corner to one edge, then pair the last one right before you slot it in. It's hard to explain with words but I had a video of a solve doing it. =P I fail at commutators so the LL really killed the solve

I also built the 'cross' with center-center-edge pairs

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#### Robert-Y

##### Member
Well you just pair up 4 edges, 3 on the top (or left or right) layer, and one on the middle layers. Then you should pair up the other 4 by cycling 3 edges.

You could attempt to deliberately solve one 1 edge, so you're left with a 3 unsolved edges (which might be nicer than having 2 edges left).

However, the move count could be quite high because you have to maintain the cross on the bottom (or left or right) layer.

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#### Am1n-

##### Member
So, on topic. The document (on page 1) posted by miniGOINGS was about a ZZ reduction.
I've worked out an example solve for this. My strategy is as followed:
- centers
- line- dedges + as many as possible random dedges (most of the time 3 extra)
- Line + EO of the already made dedges (place these in B and D slots)
- chain solving dedges (on the F-slots): depending on the orientation do u/u' or d/d'
- parity fix if necessary

example:
scramble (same as AvG's scramble for his edge-pairing method for 5x5x5 ==> centers are already done):
r U r' U r U2 r'
F' U
l' U2 l U l' U l
L2 R2 F2 B2

r U r' U r U2 r'
F' U
l' U2 l U l' U l
L R' F2 D' B2

r U r' U r U2 r'
F' U
l' U2 l U l' U l

ZZ- reduction: (I do orange on bottom, white on front)
(U2 F) u' (F' U F)(L' U L F U2 F')(B' U B) u
//make line dedges = some other dedges
L' F' L' z' y'
// Line
R2 L' F U F' L2
// EO-Line
(U' L' U L) u (L' U' L) u'
R
(U L' U' L) d' (L' U2 L) d
(U F U' F') u (L' U' L) u'
//rest of dedges while orienting them.

(U2 F) u' (F' U F)(L' U L F U2 F')(B' U B) u L' F' L' z' y' R2 L' F U F' L2 (U' L' U L) u (L' U' L) u' R (U L' U' L) d' (L' U2 L) d (U F U' F') u (L' U' L) u'

- The movecount is quite high if compared to other dedge- pairing methods
- Pairing is in E-slice (I do M-slice, so I'm not fast with it)
+ you have a ZZ solve after that (minus PLL parity, that should be a pain in the ass if you use the phasing ZZ- thingy)

Off topic: sorry for starting that discusion yesterday :'(

mvg

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#### qqwref

##### Member
Bump. I just realized that the Hoya method fits into this scheme too, which means the idea is definitely still relevant. In fact, you could make some new methods (e.g. a Hoya+K4 hybrid) by simply combining a Hoya start with another Step 2 method.

The Hoya Step 1 variant (as I understand it) is as follows: solve two opposite non-bottom centers and put them on L/R; solve the bottom center and put it on B; solve the center that should be on U and do an x rotation to leave the U and F centers unsolved and the bottom center on D; pair up and solve all four cross pieces on D using the freedom of the two unsolved centers; do an F move, finish the last two centers, and do an F'.

#### siva.shanmukh

##### Member
I do this.

1. Solve opp centers.
2. Solve three dedge pairs of one of the solved centers and put them around the corresponding center
3. Solve the remaining centers
4. Solve the fourth dedge pair of the same colour as the first three pairs.
5. Key hole, the eight dedges and four courners. The last triad may be solved as an f2l pair and an f2l commutator.
6. COLL and dedge commutators for LL

The sixth step is like in K4. The first four are from yau. Keyhole is the key change in my method.

#### ottozing

##### Platinum Member
I'm thinking that using Yau and Hoya could be good for someone who is either fixed cross or dual cross. This way, if white/yellow F2C was easy (assuming you use whte/yellow/both), you could use Yau. Whereas if blue, green, red, or orange was good for F2C, you could use Hoya. Any thoughts?

#### Kirjava

##### Colourful
I do this.

1. Solve opp centers.
2. Solve three dedge pairs of one of the solved centers and put them around the corresponding center
3. Solve the remaining centers
4. Solve the fourth dedge pair of the same colour as the first three pairs.
5. Key hole, the eight dedges and four courners. The last triad may be solved as an f2l pair and an f2l commutator.
6. COLL and dedge commutators for LL

The sixth step is like in K4. The first four are from yau. Keyhole is the key change in my method.