4x4x4 reduction + cross idea

[Robert-Y]

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:

 

[StachuK1992]

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

 

[guitardude7241]

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

 

[fanwuq]

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

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

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

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.

 

[joey]

The start = the start of K4 - minus corners.

 

[Robert-Y]

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

 

[amostay2004]

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!

 

[Lofty]

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.

 

[masterofthebass]

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...

 

[Robert-Y]

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 )

 

[joey]

Kirjava, Thom Barlow, should definately get a mention!

 

[masterofthebass]

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.

 

[deadalnix]

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.

 

[King Koopa]

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?

 

[masterofthebass]

yes.

 

[Robert-Y]

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

 

[DavidWoner]

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.

 

[Robert-Y]

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.

 

[KConny]

Yawza!

 

[pjk]

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.