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Speedcubing: Solving the Rubik's Cube > Speedcubing > 4x4x4 'K4' Method

Title: 4x4x4 'K4' Method

Kirjava - July 16, 2006 10:25 PM (GMT)
Here is my new method for the 4x4x4 cube:

I tryed to get an original method, as Per's centres step can only be done fast by Per.

I'd like to hear what people think. I prefer it to anything else I've used so far. I'm gonna do move count averages tomorrow.



PJK - July 16, 2006 11:51 PM (GMT)
Thom, thats looks very interesting. I'm going to have to try it out. Thanks for sharing.

Erik - July 17, 2006 10:20 PM (GMT)
I tried a few (4) solves with it with a best time of 2:30 or so... the last step is a bit difficult stil... I think it could be fast if you could speed it up a little and practise with it...

Erik - July 19, 2006 10:36 AM (GMT)
By the way, I can solve 5x5 with this too...

Me! - July 19, 2006 11:19 AM (GMT)
Looks very interesting, i'll try it some time.

DDRKirby(ISQ) - July 25, 2006 04:39 PM (GMT)
besides RUR'U'RwR'URU'r' and RwUR'U'Rw'URU'R'

the two algs

and its inverse,
are also useful for 3-edge commuters for the last layer (i use columns first)

(where Rw is a double layer turn, and Rww is a "triple" layer turn)

Kirjava - July 25, 2006 08:41 PM (GMT)
wow, you're right, they are really good. don't think much of the notation though :P

could you gimmie a little more info about the method? I don't know how columns first works...


Oh, I found some great 2-cycle algs today

lrD2l'r'UlrD2l'r'U' was the basis for most of them..

DDRKirby(ISQ) - July 25, 2006 08:57 PM (GMT)
The only major columns-first cuber that I know of is Akimoto--

his overview of the solution is here.

my method uses his as a starting point, then...changes some stuff. (first part and last part are different)

essentially, you first get the columns (akimoto does four white corners followed by insterting edge pairs...I instead match edge pairs, then do an F2L-like thing...), then you do white center pieces, followed by 3 pairs of white edges, then all other centers (using the free white edge pair), then complete the final white edge pair, then dooo last layer.

it's quite fun to do, actually. works on 5x5x5 kinda but it's...ugly. or maybe i just dont know enough to apply it there.

in any case, centers first -seems- like it's in general faster, looking at all the fastest times on however, you can't really make that good of a judgement because no one really uses columns first. except akimoto, and he got an average of 1:18.96, which is....pretty great. so i know the method has potential.

i dont know exactly how to get there though...there's not much documentation...watching akimoto's solving vids helped.

right now i'm struggling around the 2 min barrier. used to be able to break 2 mins fairly often but then i started experimenting with other things so i need to figure it all out.

for last layer (i use some of these for inserting the 4th edge pair too), i use 8 different algs, which are all based on RUR'U'rR'URU'r'...

1) RUR'U'rR'URU'r'
2) inverse of #1
3) R'U'RUr'RU'R'Ur ("reverse" of #1)
4) inverse of #3
5) RUR'U'RwwRw'URU'Rww'RwR' (this notation sucks but i dont really know how else to describe it accurately--its basically #1, but...moving a different slice)
6) inverse of #6
7) R'U'RURww'RwU'R'URwwRw'R ("reverse" of #5)
8) inverse of #7

so i use those, plus set-up moves if necessary.

in most cases I can pair up one pair of last layer edges while inserting the last white edge pair. then i usually do COLL->3x3x3 edge permute->1 or 2 of the above algs->fix parity if necessary.

Athefre - July 29, 2006 06:32 AM (GMT)
This is pretty similar to the way I solve the 5x5.

1. Solve left block
2. Make a line with all of the centers on the left and right, leaving the M center pieces unsolved
3. Solve right block.
4. Solve M except for the stuff in U
5. Solve U.

Kirjava - July 29, 2006 09:03 AM (GMT)
Hmm. Akimoto's method looks very interesting. I did a few practise solves with it before and I think it has potentional to get better. At the moment, with my method I'm averaging about 1:40.


Me! - October 2, 2006 01:33 AM (GMT)
think the same meathod can be applied to a 5x5?

PJK - October 2, 2006 02:33 AM (GMT)
Me: Erik mentioned above: "By the way, I can solve 5x5 with this too..."

My guess would be yes. I may try this method out. 5x5 is really slow for me now, and 4x4 average is at 2:20, so maybe I can get faster with this method.

PJK - April 15, 2007 02:38 AM (GMT)
Anyone else using this now? Try it on the 5x5. I know a couple people who use it on that.

Kirjava - April 15, 2007 02:44 AM (GMT)
I'm writing the new K4 guide for

wait until it comes out

then you can use it to it's full potential.

Erik - April 15, 2007 10:07 AM (GMT)
Have you also included these extra steps, i'd call them improvement:
1. while pairing up the last edge of your cross also solve a middle layer edge.
2. for solving the corners also solve one of the other edges like F2L
3. I have to show you this :P

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