# 4x4x4 reduction + cross idea



## Robert-Y (Sep 2, 2009)

Copied from description:

Erik Akkersdijk introduced the idea of solving 3 cross pairs after the first two centres, then finishing the centres without destroying the 3 cross pairs. Someone (possibly Syuhei Omura I'm not sure) discovered a way of pairing 3 edges after the centres by doing d, replace 3 edges, then do d'. Chris Hardwick invented an edge pairing method called "2 pair chain solving".

What I have done is combined all three to create this method of solving the centres + edges + cross.

17/10/09 EDIT: Here's a video tutorial with annotations:


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## StachuK1992 (Sep 2, 2009)

this...is k4. At least to some extent.


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## guitardude7241 (Sep 3, 2009)

This is k4. There are many variants, but it's k4.


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## fanwuq (Sep 3, 2009)

Stachuk1992 said:


> this...is k4. At least to some extent.





guitardude7241 said:


> This is k4. There are many variants, but it's k4.




http://www.speedsolving.com/forum/showthread.php?t=14875
Sure it is...  

Rob, 
It's an interesting idea. Not for me though, I dislike all inner slice turns. It is much like k4 for the first step, but afterwards, I like how you do the 3-2-2-2. 
I was thinking about maybe while setting up certain pieces for regular reduction, if you could see it, try to place a few cross pieces. I think that feels more natural. I just got 1:16 that way. I solved 2 cross pieces while setting up last 2 edges to pair. I think this is similar to how Erik Johnson does reduction-Petrus.

Edit:
Just got 1:58 your way.


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## joey (Sep 3, 2009)

The start = the start of K4 - minus corners.


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## Robert-Y (Oct 14, 2009)

Since Dan Cohen has apparently broken the 4x4x4 WR using my method* I will make a tutorial video (with annotations if necessary) explaining in more detail about this idea, including how to tackle all of the problems involved with the edge pairing part. (Or perhaps maybe Dan should do it since he is the fastest with this method as far as I know...)

Also, I realised just a few minutes ago that this method works well for left and right cross solvers if you keep the cross pieces on the left (or right) after you've finished the centres.

*I don't really feel like this should be called a method because it's composed of other people's ideas but I did think of combining all of the ideas together...
oh well... no-one really cares I guess


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## amostay2004 (Oct 14, 2009)

Yes, please do. I still fail to recognize how this would be better than the regular reduction due to the restriction in preserving the cross edges. Thanks!


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## Lofty (Oct 15, 2009)

I don't even own a 4x4 and don't really have time to add another event into my practicing but watching your videos kinda makes me want to go out and order a 4x4 and learn your method. 
I would be very interested in at least watching your video and trying to use it when I do get my hands on a 4x4.


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## masterofthebass (Oct 15, 2009)

I think I'll try and make a video about this. I doubt I will be able to cover all the issues that come up though...


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## Robert-Y (Oct 15, 2009)

Problems and solutions:

1. If the last cross edge is already solved, solve the cross and continue solving the edges however you like (you can even use nakajima's edge pairing method).

2. If after you've done d/d' and realise that one of the dedges you want to replace (with one on the E layer) is on the D layer, simply do one of the following moves (depending on the situation):

a. x L' U' L U x'
b. x R U R' U' x'

or

c. F' R' F R
d. F L F' L

3. If after you've done d/d' and realise that one of the dedges you want to replace (with one on the E layer) is on the E layer but in the wrong position, simply take it out to place it on the U layer, then put it back into the right position.

4. If after you've solved the cross and placed two other solved dedges in the BL and BR slots, BUT the FL and FR dedges are already solved, simply place two unsolved dedges from the U layer into the FL and FR slots.

These are all of the problems I can think of for now... (well the first and fourth ones aren't really problems, they're actually good things )


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## joey (Oct 16, 2009)

Kirjava, Thom Barlow, should definately get a mention!


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## masterofthebass (Oct 16, 2009)

after watching this video fully for the first time, I do a slight varation of this. After centers, I actually just pair hte last cross edge in a 2-pair style and then place it. I then do 3-2-2-2 from there, to solve the rest of the edges. I sometimes do some extra to pair more edges, but only if the edges are right in front of me.


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## deadalnix (Oct 16, 2009)

I was working on a similar system, but I seen it's already done.

I will wait for a proper tutorial before going deeper in my reflexions.


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## King Koopa (Oct 16, 2009)

masterofthebass said:


> after watching this video fully for the first time, I do a slight varation of this. After centers, I actually just pair hte last cross edge in a 2-pair style and then place it. I then do 3-2-2-2 from there, to solve the rest of the edges. I sometimes do some extra to pair more edges, but only if the edges are right in front of me.



Isn't this what you got WR with?


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## masterofthebass (Oct 16, 2009)

yes.


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## Robert-Y (Oct 16, 2009)

Hmm, maybe we should call this Yau-Cohen or Cohen-Yau?


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## DavidWoner (Oct 16, 2009)

YAUHEN

Or maybe Yau and Yau-C (Cohen variation)

But yeah, I solve the fourth cross edge immediately after centers as well. Edge pairing becomes ridiculously easy when you can ignore the L face entirely.


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## Robert-Y (Oct 16, 2009)

Yeah Yau and Yau-C sounds good to me...

Hmm I wonder if anyone actually interpreted the edge pairing part correctly...

My idea was to

1. Pair up 3 dedges including the last cross edge, 

2. Place the last cross edge in the last cross slot, and leave the other 2 paired up dedges in the BL and BR slots.

3. Use 2-pair chain solving to solve the rest. This is the best step as you never have to look at the D and B faces for any unpaired dedges, and you can continue with the F2L without much hesitation.


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## KConny (Oct 16, 2009)

Yawza!


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## pjk (Oct 16, 2009)

Did you know that Richard Meyer has been using a method very similar to this for over a year? The only difference is that he solves all the remaining edges after centers using 2-at-a time, and instead of solving the last "cross" edge and placing it, he places the 2 corners to complete a 1x3x4, (since he solves Roux, he already has the first block built). It is essentially the same method, only sets up for Roux style instead of Fridrich.


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## PatrickJameson (Oct 16, 2009)

I came up with this http://www.speedsolving.com/forum/showthread.php?t=4443 a while ago. At that time(and still now :/) I figured solving the cross pieces would be a waste of time.

I'm just sticking with normal reduction for 4x4.


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## blah (Oct 16, 2009)

I had a similar idea more than a year ago: http://www.speedsolving.com/forum/showthread.php?t=6855

Scroll down to post #17.


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## rachmaninovian (Oct 17, 2009)

I guess the only part that is unique to Yau is...the 3-2-2-2 part? no one has exploited the 3 pairing part yet =p


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## mazei (Oct 17, 2009)

You sure? I remember one Chinese guy who held the WR did 3-2-2-2-3. I use it now.


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## DavidWoner (Oct 17, 2009)

mazei said:


> You sure? I remember one Chinese guy who held the WR did 3-2-2-2-3. I use it now.



He means Yau is the only k4-redux hybrid to use 3-2-2.

Some thoughts:
I am terrible at the first 3 cross edges
restricted centers can get annoying
3-2-2 is incredibly fast when you can ignore the entire left (or bottom) side of the cube.
I've taken to doing 2-4-3 for the last 9 edges. I 2-pair the final cross edge and place it, set up and make one pair with r/r' (often this does not even require setup, you just have to make sure your initial slice is in the right direction), set up 3 pairs and slice back, then do the remaining 2 or 3 edges in one comm.

Not having to make the cross after edges pwns.
Forcing x-crosses pwns even more.


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## Robert-Y (Oct 17, 2009)

What move counts are you (to everyone) getting for reduction + cross after finishing the first two centres?

For normal reduction + normal cross, I'm getting about 55-60, and with my method, I'm getting around 65-70.

I'm going to try Yau-C now...


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## mazei (Oct 17, 2009)

I doubt I'll be doing this method as the restriction on centers a real turn off point for me. I can't stand not being able to do centers freely.

As for the 1st 3 cross pieces, I kind of fail to see how it is significantly better than just doing redux then cross(yes I know that you finish the whole cross during pairing). I kind of have a problem with 3-2-2-2-3 now due to the time I take to form the first 3.

Don't get me wrong, I'm sure this method is a beast in the right hands(Dan) but I think its not for me.


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## Robert-Y (Oct 17, 2009)

I've finished creating a video tutorial, you can just go to my first post of this thread to watch it.

However I haven't covered any issues in the video...


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## Jake Gouldon (Oct 18, 2009)

You destroyed an already paired yellow-green edge in the video, I would have preserved it.

I find that doing 2 pairing is better, at least for me. What I do is do the first 4 (including the final cross edge), and place the cross edge. Then I place 2 other solved pairs in the back right and back left slots. Then the rest of the edge pairing is easier since I don't have to do any cube rotations.


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## amostay2004 (Oct 18, 2009)

Thanks for the tutorial! It made me understand this method better and I'm definitely gonna practise this cos normal reduction is just so boring 

Looking forward to the next video!


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## DavidWoner (Oct 18, 2009)

67, 71, 68, 68, 60 -> 68 for yau

73, 71, 67, 70, 68 -> 70 for redux

uh... wut?


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## miniGOINGS (Oct 18, 2009)

pjk said:


> Did you know that Richard Meyer has been using a method very similar to this for over a year? The only difference is that he solves all the remaining edges after centers using 2-at-a time, and instead of solving the last "cross" edge and placing it, he places the 2 corners to complete a 1x3x4, (since he solves Roux, he already has the first block built). It is essentially the same method, only sets up for Roux style instead of Fridrich.



Explain please !!


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## mazei (Oct 18, 2009)

miniGOINGS said:


> pjk said:
> 
> 
> > Did you know that Richard Meyer has been using a method very similar to this for over a year? The only difference is that he solves all the remaining edges after centers using 2-at-a time, and instead of solving the last "cross" edge and placing it, he places the 2 corners to complete a 1x3x4, (since he solves Roux, he already has the first block built). It is essentially the same method, only sets up for Roux style instead of Fridrich.
> ...



Stadler method, look it up.


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## rachmaninovian (Oct 18, 2009)

mazei said:


> miniGOINGS said:
> 
> 
> > pjk said:
> ...



false. Meyer's RouxRedux is nothing like stadler method. as explained by Patrick, it is basically solving a 1x3x4 block on the left after the first 2 centres, finish the rest of the centres, followed by avg pairing to reduce the cube into a 3x3. from here, solve roux style, and this method method provides an advantage for roux because there is a free 1x3x4 block after the reduction.
Stadler method is basically 1x3x4 block on L, 1x3x4 block on R, CMLL, centres, pair up edges with *comms* then roux finish. different methods.


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## mazei (Oct 18, 2009)

rachmaninovian said:


> mazei said:
> 
> 
> > miniGOINGS said:
> ...



Woops. Well in my head it went like this - Richard Meyer = Roux = Roux on big cubes = Stadler method. Didn't read the whole post and this kinda shows how much I don't know.


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## Robert-Y (Oct 18, 2009)

I have an idea of how to get faster on the last 4 centres:

Solve the left half of 3 centres, now do an L move so that each cross piece is next to a half centre. Now you can solve the rest of what's left of the centres using r and U moves! 

(I had this idea whilst chatting to Ibrahim on facebook )


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## mazei (Oct 18, 2009)

That's really neat.


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## miniGOINGS (Oct 18, 2009)

rachmaninovian said:


> false. Meyer's RouxRedux is nothing like stadler method. as explained by Patrick, it is basically solving a 1x3x4 block on the left after the first 2 centres, finish the rest of the centres, followed by avg pairing to reduce the cube into a 3x3. from here, solve roux style, and this method method provides an advantage for roux because there is a free 1x3x4 block after the reduction.
> Stadler method is basically 1x3x4 block on L, 1x3x4 block on R, CMLL, centres, pair up edges with *comms* then roux finish. different methods.



So it's 

1: L&R centers
2: Left 1x3x4 block
3: M-slice centers
4: Edge pairing
5: Right 1x3x4 block
6: Corners of the last layer
7: Last six dedges

are there any tutorials or advanced tips that you know of?


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## Lofty (Oct 21, 2009)

I finally got my hands on a 4x4 today! I must say that I really like this method! I'm no good with it, about 2:30, but I've only done 3 or 4 solves and was only at about 1:30 with regular 2 at a time redux.


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## esquimalt1 (Oct 21, 2009)

Robert-Y said:


> I have an idea of how to get faster on the last 4 centres:
> 
> Solve the left half of 3 centres, now do an L move so that each cross piece is next to a half centre. Now you can solve the rest of what's left of the centres using r and U moves!
> 
> (I had this idea whilst chatting to Ibrahim on facebook )



Haha, yes!


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## pjk (Oct 21, 2009)

miniGOINGS said:


> rachmaninovian said:
> 
> 
> > false. Meyer's RouxRedux is nothing like stadler method. as explained by Patrick, it is basically solving a 1x3x4 block on the left after the first 2 centres, finish the rest of the centres, followed by avg pairing to reduce the cube into a 3x3. from here, solve roux style, and this method method provides an advantage for roux because there is a free 1x3x4 block after the reduction.
> ...


It is practically the same thing as the method Yau is explaining in this post. His video is pretty straight forward. The only difference is you insert 2 corners to complete the 1x3x4 block instead of inserting an edge piece for the cross.


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## ~Phoenix Death~ (Nov 1, 2009)

Robert-Y said:


> Copied from description:
> 
> Erik Akkersdijk introduced the idea of solving 3 cross pairs after the first two centres, then finishing the centres without destroying the 3 cross pairs. Someone (possibly Syuhei Omura I'm not sure) discovered a way of pairing 3 edges after the centres by doing d, replace 3 edges, then do d'. Chris Hardwick invented an edge pairing method called "2 pair chain solving".
> 
> ...



He made a separate video for annotations? Doesn't he know you can just turn 'em off?
But to my point:
The idea of solving one center, then the opposite center, then to the cross is a nice idea. I may try this when I get my ES 4x4x4.


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## Jake Gouldon (Nov 1, 2009)

I wonder how fast a 5x5, 6x6, or 7x7 version of this could get.


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## amostay2004 (Nov 1, 2009)

Jake Gouldon said:


> I wonder how fast a 5x5, 6x6, or 7x7 version of this could get.



Much, much slower than normal reduction. As you move to bigger cubes the centres just get more complicated and the move restrictions when finishing off the centres after pairing 3 cross edges will just kill you.


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## qqwref (Nov 1, 2009)

Yeah, it's really not very good for bigcubes. The biggest problem is that edge pairing during F2L becomes extremely inefficient if not impossible on 5x5 up... you can't do 3-edge pairing every time, and AvG is pretty icky if you have to do an RUxR' trigger to set up each pair. If you add that to the fact that centers (which are 1/3 of the solve on 5x5 and even more on bigger cubes) are significantly slowed down by having to keep 3 cross pieces on L, it looks like this method would end up quite a bit slower than reduction on 5x5+.


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## Robert-Y (Nov 1, 2009)

qqwref said:


> Yeah, it's really not very good for bigcubes. The biggest problem is that edge pairing during F2L becomes extremely inefficient if not impossible on 5x5 up... you can't do 3-edge pairing every time, and AvG is pretty icky if you have to do an RUxR' trigger to set up each pair. If you add that to the fact that centers (which are 1/3 of the solve on 5x5 and even more on bigger cubes) are significantly slowed down by having to keep 3 cross pieces on L, it looks like this method would end up quite a bit slower than reduction on 5x5+.



+1

It seems that reduction (centres, then edges) is probably the fastest method for solving the 5x5x5 and up.

I have come up with a way of reducing + solving 2 opposite cross edges for the 5x5x5 (but I'm pretty sure it's bad because of the high move count).

1. Solve any centre, solve the opposite centre

2. Solve two cross pieces which are opposite to each other (e.g. if you are using the western colour scheme, and your cross colour is white, you would solve either the red and orange cross pieces or the blue and green cross pieces)

3. Finish the rest of the centres (without disturbing the already solved cross pieces and other centres)

4. Use Chris Hardwick's chain solving idea to solve the rest of the tredges. (Sometimes you can solve 2 wings at once if you have a tredge in which only the midge doesn't match the wings)

Again, I think this isn't a good idea for now.


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## Lt-UnReaL (Nov 2, 2009)

Move counts after centers + 3 cross edges (up to 3x3):

Yau - 38, 43, 42, 37, 40, (33), 34, 34, 48, 37, 43, (52) -> 39.6
Yau-C - 38, 37, 46, (35), 37, 42, (46), 38, 40, 36, 42, 36 -> 39.2

I simulated them as much to a speedsolve as I could, I wasn't trying to find weird ways to save moves. Yau-C is somehow more consistent. I guess just choose whatever seems better for look ahead...


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## ~Phoenix Death~ (Nov 7, 2009)

What does he mean by the E Layer?


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## Robert-Y (Apr 15, 2010)

Bump...

I just read about Blah's method idea in his thread:



blah said:


> blah blah blah
> 
> 2. Solve 3 cross dedges, one of them in the _wrong order_, this becomes surprisingly easy to do after a while, because you have more freedom. If you don't understand what I just said, I'll give an example: say you have white center on D, then you'd solve, for example, red-white on F, green-white on R, and orange-white on L. Notice that red and green are in the correct order, but orange is intentionally placed at the wrong position, that's what I meant by _one of them in the wrong order_. Of course, any other combination will work. Use whatever method you want to solve them, direct solving, or pairing then solving, I haven't found the most efficient method yet.
> 
> ...



This idea is actually kinda useful for when your first 3 cross edges pairing goes slightly wrong.

I might be colour neutral for the 3x3x3 but to be honest, I think that's because I'm fully aware of which colours are opposites and which colours aren't opposites of each other. For this reason, I sometimes place a cross edge in the position opposite to where it's supposed to be whilst I'm solving the 3 cross edges.

So uh... yeah, if it goes wrong, you could just use Blah's idea:

1. ADF(adjust D face),
2. Replace the incorrect cross edge with the newly paired cross edge
3. ADF again
4. Place (which was) the incorrect edge in it's correct position.

Now you just continue with 3-2-2 edge pairing...

This idea makes it easier to do colour neutral solves, thanks Blah!



Now for something random about the name of this method for anyone who cares:

I think since what I've done is that I've used other people's ideas and combined them altogether, maybe we could form a name by taking the surnames of the people involved in the creation of this "method". Maybe an abbreviation?...

Akkersdijk
Barlow (or maybe his username: Kirjava)
Cohen
Hardwick
Lian (or maybe his username: Blah)
Omura
Yau

I'm not sure whether to include Erik (Akkersdijk) or Thom (Barlow) or both because I got the idea of pairing and placing 3 cross edges from one of Erik's video, but it seems that this idea sorta originally came from K4 (created by Thom).

I don't mind just sticking to "Yau" though 


(I have a good feeling that many don't understand or just care about what I've typed )


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## masterofthebass (Apr 15, 2010)

ya... I just switched to blah's way of doing the last cross edge after 1 solve. SO much smoother.


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## Weston (Apr 15, 2010)

Ive been doing this ever since I started using yau.


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## Robert-Y (Apr 15, 2010)

I just reread my post and now I'm thinking "that's so obvious" :confused:

oh well


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## cbracketdash (Dec 30, 2022)

This thread aged well.


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