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Kenneth's big cube method explained

Kenneth

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Kenneth's big cubes method

This guide is not for beginners who are looking for a method to solve a big cube, it is written mostly for advanced cubers who are looking for something that is maybe faster than the usual reduction method, I'm pretty sure it is if one is willing to paractice it enough.

I myself can't prove it doe because I'm not fast enough, mostly because I suck in look ahead and finger trixing. My 4x4 turn rate is about 1/sec, if you are twice that fast and does this method, then you are in the top rankings =)

People who tried Thom Barlow's K4 will find this wery familair, this method is definitly in the same group.

Introduction:

The most common way to solve a big cube is to use reduction, that is to block up certain pieces (centres and edges) so it functions as a 3x3x3 in the end solve. The worst problem using reduction is that it is turn intensive.

My method do not use reduction, here everything is put in place directly and that saves turns. The problem with direct solving is that it is slice intensive (seems to go for all direct methods, not just mine) and slices are slower than the face and double face turns that are mostly used in reduction.

As always, there are pros and cons...

But something can be done about that, see the post about speed optimizations below.

Method outline

There are three mayor steps, F2B, M-slice and LL, these are divided into smaller steps:

F2B builds two opposite centres, R and L (I never use my U and D colours for this but any of the other two pairs). After that the RD and LD deges (tredges and so on depending on the size of the cube) are paired and placed. A variation is to place edges while building the centres, it saves some turns but has slower recognition. Next pairs are built and placed to compleate the first two blocks.

At this point you can use MCLL to solve the first LL step because from here the cornes are not affected by the rest of the steps and it saves a few turns on average (see post about speed optimizations/MCLL4 lover in this tread).

The M-slice part compleates the first layers (F3L on 4x4x4, F4L on 5x5x5 and so on). Easiest is to build the D centre first and then add B centre, use the empty F-side as keyhole to pair the BD dedge and then place it. But you can also solve centres slice by slice, it's sometimes easier. Then the M-part ends by solving the F-side, at 4x4 I usaly block up two centres and one edge to a tripple that is placed, on 5x and bigger the same is not that easy to do so I usaly bring as many centres as possible while doing the edges and then I solve the centres that was left out, sometimes in blocks using double slices.

The Last layer is pure brute force, first the four corners are solved using CLL and then there are three ELL steps, the first solves the RU dedge, the second the LU dedge and the third solves both BU and FU, paritys and all in one alg. For 5x5x5 you have to add another ELL step for the mid edges, best is to use 3x3x3 ELL as the first of the ELL steps.

For 5x5x5 LL you can use OLL/PLL to solve corners and mid edges and then use the theree ELL steps for the dedges.

There are many cases in the ELL but not many algs are used, once understood it's pretty easy to learn. The part that is hardest to learn for this method is CLL if you do not know it before.
 
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Kenneth

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Example solve #1

Example solution #1:

This example is a 4x4x4 solution and I think it is the only size that I need to use, there are only small differnces between 4x and 5x and so on using my method.

Note that I do not use cube rotations in the first two parts here. That is to make the texts cleaner to read, naturally the real life solving of the first layers includes a lot of orientations.

Scramble:

B2 f F2 u2 B D2 f2 u B2 f2 F2 r U2 L' D2 r2 f R2 U' B R' F u2 L u2 U' f' R' F D' u U' r2 D' u U2 L2 r' R u2

First two blocks:

R centre: (r2 R2) f
L centre: u' F' u l L2 u' F2 u
RD dedge: F' r' F R'
LD dedge: B U' r' U' L2
Pair 1: F B' l L' U L
Pair 2: U2 l2 U' F' U2 F B' U' B
Pair 3: U l2 U M L' B' L
Pair 4: r2 U2 r L F L'

This is pretty strait forward intuitive solving, first two opposite centres, add the two corresponding D layer dedges and then build and place pairs (triplets really but let's say it's a corner dedge/tredge... pair so it fits all sizes)

M-part:

B centre: M' U M' U M2
D centre: U' l' U r' U2 r' U
BD dedge: M' r2 U M' U2 r U' l' U l U' M r' U2 M'
F-side : U l' U' l U' r U' r' U2 r U2 r'

At first two centres are built, D and B (or F it does not matter). then the BD dedge is paired and placed (here one of the pieces was stuck, it is possible to fix it using an alg but that's advanced, here I'm using only intuition. So I had to get it out before I could do the pairing, that's why that part got pretty long in this example) Last the F side is solved by building pairs of centres + an edge that are placed.

The M-part is the most special part of my method, it can be done in so many ways, I will try to cover the best parts (found in a post lower in the tread).

Last layer:

CLL: B' R B R' U2 R' U2 R ... Sune case
AUF: U
ELL 1: (y' x') l' U L' U' l r U L U' r' (x) ... dbl 3-cycle, solves UL dedge.
ELL 2: (y2 x') R U l' U L' U' l r U L U' r' U' R' (x) ... same as last but with setup, solves UF dedge.
ELL 3: (y' x2) r2 U2 r' U2 l U2 l' U2 (x) U2 l' U2 r' (x) ... the shortest parity, solves UF and UB dedges.

Not so much to say, it is 100% brute force, no intuitive parts at all. To learn this you have to learn algs, CLL you can find elsewhere, the algs for my ELL steps will be added to this thread later (now done).

------------
With reservation for errors, I have checked the solution and will check it once again when I have got the time.
 
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Kenneth

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M-part examples

M-part, example 1:

I constructed a 12 turn "scramble" for the example: 2x(2x(r' U) l U)

Solution:

D centre: r2 U2 l2 U l2
F centre, first half: l r U2 r'
F centre, second half: r' U' l2 U' l r
Pair FD dedge: l U' r' U r U l'
Place FD dedge: M' U2 M

Last side, B in this case so we start with a cube orientation to move it to the F side instead. It's not needed but it gives easier look ahead.

F side first half: (y2) r U l' U M
F side second half: M U2 M' U' r U2 r' U' M U2 M' ... tricky case but a nice and intuitive solution for it.

As you understand you do not have to restore the slices to their correct positions all the time as I do here, (there are some l2 - l' and such above). As long as the blocks of pieces are in their proper slice and not in the U layer they will be protected from destruction.

----------

M-part, example 2:

A little more advanced, the scramble is the same as the previous: 2x(2x(r' U) l U)

Solution:

D and F centres + FD dedge: M U2 M U2 l2 U' l' U' r' U' l2 U2 l U M' r U M

Last side: (y2) U M' U2 M U' r U2 r' U2 r U2 r'

----------

M-part, example 3:

Cool turns this time, same scramble again: 2x(2x(r' U) l U)

Solution:

All in one: U' l' U l' U l U r U' M' U2 r U' l U2 l' U r U' r' U' r U2 r' U2 r

----------

If you practice these examples a couple of times you will soon get used to do it, it's pretty easy intuitive solving and the number of uniqe cases are not that many.

----------

For the last side it is sometimes easier to place all pices but one or two centres that are solved one by one on 4x4 and in some cases on 5x5 in blocks using double layer slice turns.

Here is an F side example for 5x5, the scramble is the inverse of the solution, the situation I had after solving the first steps of a full scramble.

Scramble: m' U' r U m U' r' U2 r U' l' U r' U' l U m' U2 m U m l' U r U2 M U'

Solution:

r slice: U M' U2 r'
l slice: U' l
m slice: m U' m' U2 m
Missing centre 1: U' l' U rU' l U r'
Missing centre 2: U2 r U m' U' r' U m

I my next try I had this situation:

Scramble: l' r U' m' U M U' r U r' U' r U r' l' U2 l U2 l U l' U

Solution:

l slice: U' l' U' l U2 l' U2 l
r slice: r U' r' U r U' r'
m slice + two missing centres in one go: U M' U' m U r' l ... that's a nice one =)

----------

Some useful algs:

swap U and F centres: M' U l U2 l' r' U2 r U' M ... only for even cubes, use double, triple... layer r/l slices for the larger ones. The same alg is also useful for uneven cubes but it does something else there, try it.

Solve upper F centre of r slice: r U' l' U r' U' l <-- inverse to setup, mirror for l slice
Solve lower F centre of r slice: r U l' U' r' U l <-- inverse to setup, mirror for l slice

If one edge is missing you can use these:

l D2 r U2 r' D2 r U2 M ... commutator, inverse to setup.
3x(r U2 r' U2) ... don't mind the last U2, inverse to setup.

And mirrors for l side.

With reservation for errors...
 
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Kenneth

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Content of this post is now a copy of this post at the Y-group It may need some edits (one I already did =) and some algs I have changed since I first wrote this, I do the edits later...

ELL algorithms:

ELL 1:

Format for case description:

B-side

..8..7..
5.....6
4.....3
..1..2..

F-side

case number: (nr of moves) (cube orientations) alg

The numbering in the case descrition tells which two edges that are
being solved, nr 1 and 2. These edges initially are palced at
positions 1-8. Even edges (really only nr 2) are oriented at even
places and the other way around at odd places. Same thing goes for
the odd edge (nr 1).

The cases:

..8..7..
5......6
4......3
..1..2..

1-1 + 2-2 : (0) Solved
1-1 + 3-2 : (8) (y' x') r U' R U r' U' R' U
1-1 + 4-2 : (8) (y' x') U' R U r U' R' U r'
1-1 + 5-2 : (8) (y x') r U L' U' r' U L U'
1-1 + 6-2 : (8) (y x') U L' U' r U L U' r'
1-1 + 7-2 : (8) (y2 x') l' U L' U' l U L U'
1-1 + 8-2 : (9) (x') U2 l2 U L U' l2 U L' U

(49 moves / seven cases = 7.0 average)
-----------

..8..7..
5......6
4......3
..1..2..

2-1 + 1-2 : (12) (x') M' U' R U M U2 M2 U R' U' M2 U2
2-1 + 3-2 : (10) r' M D2 l' U l D2 l' U' r2
2-1 + 4-2 : (10) l2 U' r' D2 r U r' D2 l' M'
2-1 + 5-2 : (10) r' M D2 l' U' l D2 l' U r2
2-1 + 6-2 : (10) l2 U r D2 r' U' r' D2 l' M'
2-1 + 7-2 : (10) r U2 l D2 l' U2 l D2 l' r'
2-1 + 8-2 : (10) r' l' D2 r U2 r' D2 r U2 l

(72 = 10.30)
-----------

..8..7..
5......6
4......3
..1..2..

3-1 + 1-2 : (10) r2 U l D2 l' U' l D2 r M'
3-1 + 2-2 : (8) (y x') U L' U' l' U L U' l
3-1 + 4-2 : (14) U2 r U' r' U' r' U r2 U r' U' r' U' r
3-1 + 5-2 : (10) (y x') r U L' U' r' l' U L U' l
3-1 + 6-2 : (8) (y' x') M' U' R U M U' R' U
3-1 + 7-2 : (10) (x') l' U' R U l r U' R' U r'
3-1 + 8-2 : (9) (y) Ra' u2 R U' R' u2 R U L

(69 = 9.86)
-----------

..8..7..
5......6
4......3
..1..2..

4-1 + 1-2 : (10) M l D2 r U' r' D2 r U l2
4-1 + 2-2 : (8) (y x') l' U L' U' l U L U'
4-1 + 3-2 : (11) (y x') U2 l2 U' L' U M2 U' L U r2 U2
4-1 + 5-2 : (8) (y x') M' U L' U' M U L U'
4-1 + 6-2 : (10) (y x') l' U L' U' l r U L U' r'
4-1 + 7-2 : (10) (y2 x') l' U' R U l r U' R' U r'
4-1 + 8-2 : (11) (x') U2 l2 U' L' U M2 U' L U r2 U2

(68 = 9.71)
-----------

..8..7..
5......6
4......3
..1..2..

5-1 + 1-2 : (10) r2 U' l D2 l' U l D2 r M'
5-1 + 2-2 : (8) (x') l' U L' U' l U L U'
5-1 + 3-2 : (10) (y' x') r U' R U r' l' U' R' U l
5-1 + 4-2 : (8) (y x') M' U L' U' M U L U'
5-1 + 6-2 : (14) (y2) l' U l U l U' l2 U' l U l U l' U2
5-1 + 7-2 : (10) (x') l' U L' U' l r U L U' r'
5-1 + 8-2 : (14) (y' x') L' U' r U' R U r' l' U' R' U l U L

(74 = 10.57)
-----------

..8..7..
5......6
4......3
..1..2..

6-1 + 1-2 : (10) l M D2 r U r' D2 r U' l2
6-1 + 2-2 : (8) (y' x') l' U' R U l U' R' U
6-1 + 3-2 : (8) (y' x') M' U' R U M U' R' U
6-1 + 4-2 : (10) (y' x') l' U' R U l r U' R' U r'
6-1 + 5-2 : (11) (y' x') U2 r2 U R U' M2 U R' U' l2 U2
6-1 + 7-2 : (10) (y2 x') l' U L' U' l r U L U' r'
6-1 + 8-2 : (11) (x) U2 l2 U' R' U M2 U' R U r2 U2

(68 = 9.71)
-----------

..8..7..
5......6
4......3
..1..2..

7-1 + 1-2 : (10) r l D2 l' U2 l D2 l' U2 r'
7-1 + 2-2 : (9) (x') U2 r2 U' R' U r2 U' R U'
7-1 + 3-2 : (11) (x) U2 r2 U' R' U M2 U' R U l2 U2
7-1 + 4-2 : (14) (y' x') L' U' l' U' R U l r U' R' U r' U L
7-1 + 5-2 : (11) (x) U2 r2 U L U' M2 U L' U' l2 U2
7-1 + 6-2 : (14) (y x') R U l' U L' U' l r U L U' r' U' R'
7-1 + 8-2 : (6) r2 U2 r2 Uu2 r2 u2

(74 = 10.57)
-----------

..8..7..
5......6
4......3
..1..2..

8-1 + 1-2 : (10) l' U2 r' D2 r U2 r' D2 r l
8-1 + 2-2 : (8) (y2 x') r U' R U r' U' R' U
8-1 + 3-2 : (10) (y2 x') r U L' U' r' l' U L U' l
8-1 + 4-2 : (10) (x') r U L' U' r' l' U L U' l
8-1 + 5-2 : (10) (y2 x') r U' R U r' l' U' R' U l
8-1 + 6-2 : (10) (x') r U' R U r' l' U' R' U l
8-1 + 7-2 : (8) (y2 x') M' U' R U M U' R' U

(68 = 9.71)
-----------

Total moves / cases ratio: 524 / 56 = 9.68

--------------------------------------------
--------------------------------------------

ELL 2:

Cases here are the same as for ELL 1 but after a y2 cube orientation
minus all cases that has got an 7 or an 8 in the number.

.Solved.
5......6
4......3
..1..2..

1-1 + 2-2 : (0) Solved
1-1 + 3-2 : (8) (y' x') r U' R U r' U' R' U
1-1 + 4-2 : (8) (y' x') U' R U r U' R' U r'
1-1 + 5-2 : (8) (y x') r U L' U' r' U L U'
1-1 + 6-2 : (8) (y x') U L' U' r U L U' r'

(32 = 6.4)
-----------

.Solved.
5......6
4......3
..1..2..

2-1 + 1-2 : (12) (x') M' U' R U M U2 M2 U R' U' M2 U2
2-1 + 3-2 : (10) r' M D2 l' U l D2 l' U' r2
2-1 + 4-2 : (10) l2 U' r' D2 r U r' D2 l' M'
2-1 + 5-2 : (10) r' M D2 l' U' l D2 l' U r2
2-1 + 6-2 : (10) l2 U r D2 r' U' r' D2 l' M'

(52 = 10.4)
-----------

.Solved.
5......6
4......3
..1..2..

3-1 + 1-2 : (10) r2 U l D2 l' U' l D2 r M'
3-1 + 2-2 : (8) (y x') U L' U' l' U L U' l
3-1 + 4-2 : (14) U2 r U' r' U' r' U r2 U r' U' r' U' r
3-1 + 5-2 : (10) (y x') r U L' U' r' l' U L U' l
3-1 + 6-2 : (8) (y' x') M' U' R U M U' R' U

(50 = 10.0)
-----------

.Solved.
5......6
4......3
..1..2..

4-1 + 1-2 : (10) M l D2 r U' r' D2 r U l2
4-1 + 2-2 : (8) (y x') l' U L' U' l U L U'
4-1 + 3-2 : (11) (y x') U2 l2 U' L' U M2 U' L U r2 U2
4-1 + 5-2 : (8) (y x') M' U L' U' M U L U'
4-1 + 6-2 : (10) (y x') l' U L' U' l r U L U' r'

(47 = 9.4)
-----------

.Solved.
5......6
4......3
..1..2..

5-1 + 1-2 : (10) r2 U' l D2 l' U l D2 r M'
5-1 + 2-2 : (8) (x') l' U L' U' l U L U'
5-1 + 3-2 : (10) (y' x') r U' R U r' l' U' R' U l
5-1 + 4-2 : (8) (y x') M' U L' U' M U L U'
5-1 + 6-2 : (14) (y2) l' U l U l U' l2 U' l U l U l' U2

(50 = 10.0)
-----------

.Solved.
5......6
4......3
..1..2..

6-1 + 1-2 : (10) l M D2 r U r' D2 r U' l2
6-1 + 2-2 : (8) (y' x') l' U' R U l U' R' U
6-1 + 3-2 : (8) (y' x') M' U' R U M U' R' U
6-1 + 4-2 : (10) (y' x') l' U' R U l r U' R' U r'
6-1 + 5-2 : (11) (y' x') U2 r2 U R U' M2 U R' U' l2 U2

(47 = 9.4)
-----------

Total moves / cases ratio: 278 / 30 = 9.27

--------------------------------------------
--------------------------------------------

ELL 3:

Here the cases has got a little diffrent description:

The solved case:

43
12

12 = UF-dedge, 43 = UB-dedge.
Even edges are oriented at even places and so on...

(1:24) Number of these cases in the whole group.

"alg" (number of turns)

-----
PLL-parity:

21
34

(1:24)

r2 U2 r2 (U2+u2) r2 u2 (6)

-----
Orientation only:

34
21

(1:24)

M' U M' U M' U2 M U M U M U2 (12)

-----
Orientation + PLL-parity:

12
43

(1:24)

x' U (l+r'+R') U x' (PLL-parity) x U' (l'+r+R) U' x (12)

Looks weird but it is three set up turns + cube orientations at
first, then the PLL-parity + undo set up. (l+r'+R') is a three layer
turn =)
-----

OLL-parity:

43
21

(2:24)

F2 l2 F2 U2 l' U2 l U2 l2 F2 r' F2 r U2 l' (15)

-----
O + P (both paritys):

34
12

(2:24)

r U2 r' E2 F2 l' F2 l F2 r F2 r' D2 l' (14)

-----
These are four cases of three cycles solved using a commutator. The
same alg is also used in step ELL 1, for example case "7-1 + 1-2".

41
23

(2:24)

r l D2 l' U2 l D2 l' U2 r' (10)

23
41

(2:24)

l' r' D2 r U2 r' D2 r U2 l (10)

42
31

(2:24)

r U2 l D2 l' U2 l D2 l' r' (10)

13
24

(2:24)

l' U2 r' D2 r U2 r' D2 r l (10)

-----
Orient mixed:

14
23

(1:24)

r' U2 r2 U2 r U2 r' U2 r U2 r2 U2 r' (13)

Mirror:

32
41

(1:24)

l U2 l2 U2 l' U2 l U2 l' U2 l2 U2 l (13)

-----
Swap two diagonaly:

41
32

(1:24)

r S2 U2 r' U2 r U2 l' U2 l F2 r' B2 l' (14)

Mirror:

23
14

(1:24)

l' S2 U2 l U2 l' U2 r U2 r' F2 l B2 r (14)

----
Swap two opposite:

42
13

(2:24)


r2 D2 r' D2 l D2 l' D2 B2 l' B2 r' (12)

-----
Same as above + PLL-parity:

31
24

(2:24)

F2 l2 F2 l F2 l' F2 r U2 l U2 r' U2 l U2 l' (16)

-----

Total moves / cases ratio: 278 / 24 = 11,58

--------------------------------------------
--------------------------------------------

ELL-1: 524 / 56 = 9.68
ELL-2: 278 / 30 = 9.27
ELL-3: 278 / 24 = 11,58

All three ELL steps: 30,53

Whole LL:

CLL = 9,18
ELL = 30,53
= 0.75

Total: 40,46

For real it is better, you can work on ELL1 while solving CLL, you can choose 1 of 4 possible ELL1's, you can work on ELL2 while solving ELL1 and work on ELL3 while doing ELL2. My true move count for LL is about 35-37.
 
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If you are new to the method you will probably do around 100 for first layers, after the amount of practice I have it's some 85-90 but I'm sure it is possible to go lower by using more advanced moves. LL is, if you use my algs about 40 turns.

Sum 120-140 depending of how advanced the first layers are solved. That includes paritys.

For 5x5, if you are skilled you can probably average sub 200 but I don't, the few times I have tried an average I have got 200-210 something mostly because I do not practice 5x5 that much. My best 5x5 FMC using this method is 172, for 4x4 it's 89 but it was pretty lucky.
 
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speed optimizations:

The worst problem using my method are all the slice turns. But in many cases it is possible to use double layer turns instead of slices, aspecially for the R and L sides. When doing the first steps (two centres and two dedges) double layer turns can be used most of the time. When building the pairs some of the turns can be double, to place a pair rather do R'r' U2 Rr than R' U R. Then, when F2B is done you can use my special CMLL4 = speed optimized CMLL algs for 4x4x4.

I will list those here and later I will also edit my ELL algs above in the same way (I'm learning to do this right now so it's not compleatly readymade yet)

CMLL4

Here I will list all algs needed to do CMLL on big cubes, these algs preserves only F2B, not the M slice so these are used after F2B is compleated, before the M-part (some of the algs preserves frst layers).

For all algs, do the inverse to setup (easiest way to see the case when there are no images).

Keep in mind that F2B always must be at the R and L sides. Because of that it is sometimes faster to not do like this, for some cases rather use a normal CLL to skip AUF and cube orientations before the alg.

Notation:

To save space and improve readability I will use a special notation for double layer R and L turns here.

> = R r
>' = R' r'
>2 = R2 r2
< = L l
<' = L' l'
<2 = L2 l2

It's easy and intuitive, left side angle is the left side of the cube. For 6x6 it's three layers and so on, for 5x5x5 use either 2 or 3 layers = the turns are optitional, you can do only the face or multi layer turns for the > and < turns, pick your choise =)

Sune:

> U R' U > U2 >' .... normal Sune
> U2 >' U' R U' >' .... anti Sune

(x) U R' U' < U2 > U2 >' ... FRFR Sune
> U2 >' U2 <' U R U' (x') ... anti FRFR Sune

(x) U' L U >' U2 <' U2 < ... mirror FRFR Sune
>' U2 < U2 > U' L' U (x') ... mirror anti FRFR Sune

>' U L U' R U <' ... Niklas
< U' R' U L' U' > ... mirror Nicklas

> U R' U <' U R U' < U2 >' ... inverse Sune
<' U' L U' > U' L' U >' U2 < ... mirror inverse Sune

(x') U L F2 L U' L'U F2 U' > ... worst Sune, Gilles CMLL alg is 12 turns and not easier than this :p
(x') U' R F2 R' U R U' F2 U <' ... mirror of worst Sune

T cases:

<' U2 >' D2 > U2 >' D2 >2 (x') ... or mirror. (The same alg is used in ELL but then using only r/l instead of R r/L l)

>' U' R U < U' R' U (x) ... a bit like Niklas
< U L' U' >' U L U' (x) ... mirror

>' U > U2 <' R' U R U' < ... or miror or inverse or mirror-inverse

> U2 >' U' R U' >2 U2 > U R' U > ... anti Sune twice (also mirror works)

>' U R2 D L' B2 L D' R2 U' > ... X Y -X style Y is the B2 turn

U cases:

(F f) R U R' U' (F' f') ... the shortes LL alg but double F turns

< U' R' U L' U2 F' U F > ... Niklas + FRURUF merged

<' >' D2 > U2 >' D2 > U2 < ... inverse of the first of the T cases

>2 D >' U2 > D' >' U2 >' ... inverse makes a L case
<2 D' < U2 <' D < U2 < ... mirror of previous

> U R' U > U2 >2 U' R U' >' U2 > ... Sune twice

L cases:

(x') U L' U' > U L U' <' ... there is also the 7 turn used for 2x2 but I think this is faster, inverse is used for one of the T cases.
(x') U' R U <' U' r' U > ... mirror of previous

>' U2 < U' R' F2 L2 D' <' ... orient x before the F and also D so they becomes U-turns (it's so messy to notate but if you do the alg you will get it)

> U2 > D >' U2 > D' >2 ... nice and fast, I like it =)
<' U2 <' D' > U2 <' D <2 ... mirror

> U' R' (L' U2 L U L' U <) > U2 >' ... only orientation, parentesis is a mirror anti Sune. can be done as 11 turns because of some anti slice turn possibilities.

pi cases:

> U2 >2 U' r2 U' >2 U2 > ... the usual pi OLL

> U2 >' U' R U >' U2 <' U R U' (x') ... same as one of the sunes but with repetition.
<' U2 > U L' U' < U2 > U' L' U (x') ... mirror

> U R' U > U' (y) R U' R' F' ... probably you use it for OLL.

> U L U' R U' <' U' L U' <' ... Niklas + anti Sune (also mirror works), inverse does a H case

(x) U R' U' < U2 > U L' U R' U' < ... FRFR + Niklas, inverse does a H case

H cases:

> U R' U > U' R' U > U2 >' ... double Sune

(F f) 3x(R U R' U') (F' f') ... triple FRURUF with doble layer F's, bloody worst CLL case I think, there are no good algs to solve it.

> U R' U > U L' U R' U' < ... Sune + Niklas, inverse does a pi case

>' U L U' R U' <' U2 >' U L U' (x) ... Niklas + FRFR, inverse does a pi case.

with reservation for errors

The algs here are also good for the Roux 3x3x3 method (CMLL) but then some of the turns must be only the face and not double layer turns, if you use the algs you will find out where and how =)
 
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Well, that was the CLL's, next I will present my basic algs for the ELL. It is supposed to function as a stepping stone to full ELL. If you learn the algs then you can improvise the ELL using them. Later you can swith to full brute force ELL (it saves time but not always turns).

Notation:

These are genral algs so when I write [m], [m]' or [m]2 it is any of the x axis slices (r and l on 4x4, r, l and m on 5x5x5 and so on). All turns the same direction as the R-face. If there is a r, l, m (no bracets) etc in the alg then it is the specific slice given you shall turn (direction is then the normal = l and m turns opposite to r).

Greatest alg in the world:

(x') U' R U [m] U' R' U [m]' (x) ... gives two algs on 4x4, 3 on 5x5 and so on.
(x') U L' U' [m] U L U' [m] (x) ... mirror

(x') [m] U' R U [m]' U' R' U (x) ... inverse of previous
(x') [m] U L' U' [m]' U L U' (x) ... mirror

This first alg gives in total 8 three cycles on a 4x4x4. But you can also use it to solve two in one. Using M instead of r or l is pretty obvious, it does two parallell cycles. But if the cycles goes in diffrent directions you can do this:

(x') r U' R U r' l' U' R' U l (x) .. double three cycle
(x') l' U' R U l r U' R' U r' (x) ... the inverse

There are also mirrors so in total this alg gives 12 combinations. Because the first two ELL steps solves only two edges and these are three cycles the same alg can be used for several cases, I'm not sure how many of the ELL's I use this one for but it's many, no wonder it's the "greatest alg in the world" (x' R U L' U' R' U L U' x does a corner cycle, mirror and inverses makes 4, I use it in the CMLL =)

----------
Homework: How many combinations will it give on a 7x7x7? Note that up to 5 cycles can be done in one go. (post your answer here =)
----------

This also explains why my ELL is easy to learn, wery few algs are used, the worst part is recognition (to learn, not ot master, once learned it's really fast because you only need to find a few pices at the time).

T.B.C..
 
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#10
Well, that was the CLL's, next I will present my basic algs for the ELL. It is supposed to function as a stepping stone to full ELL. If you learn the algs then you can improvise the ELL using them. Later you can swith to full brute force ELL (it saves time but not always turns).

Notation:

These are genral algs so when I write [m], [m]' or [m]2 it is any of the x axis slices (r and l on 4x4, r, l and m on 5x5x5 and so on). All turns the same direction as the R-face. If there is a r, l, m (no bracets) etc in the alg then it is the specific slice given you shall turn (direction is then the normal = l and m turns opposite to r).

Greatest alg in the world:

(x') U' R U [m] U' R' U [m]' (x) ... gives two algs on 4x4, 3 on 5x5 and so on.
(x') U L' U' [m] U L U' [m] (x) ... mirror

(x') [m] U' R U [m]' U' R' U (x) ... inverse of previous
(x') [m] U L' U' [m]' U L U' (x) ... mirror

This first alg gives in total 8 three cycles on a 4x4x4. But you can also use it to solve two in one. Using M instead of r or l is pretty obvious, it does two parallell cycles. But if the cycles goes in diffrent directions you can do this:

(x') r U' R U r' l' U' R' U l (x) .. double three cycle
(x') l' U' R U l r U' R' U r' (x) ... the inverse

There are also mirrors so in total this alg gives 12 combinations. Because the first two ELL steps solves only two edges and these are three cycles the same alg can be used for several cases, I'm not sure how many of the ELL's I use this one for but it's many, no wonder it's the "greatest alg in the world" (x' R U L' U' R' U L U' x does a corner cycle, mirror and inverses makes 4, I use it in the MCLL =)

----------
Homework: How many combinations will it give on a 7x7x7? Note that up to 5 cycles can be done in one go. (post your answer here =)
----------

This also explains why my ELL is easy to learn, wery few algs are used, the worst part is recognition (to learn, not ot master, once learned it's really fast because you only need to find a few pices at the time).

T.B.C..
Hi :)

What do you mean by m on 4x4x4 ? Do you mean m = (l r') ??
And what would m be on larger even cube?

In my own private notation m does not exist on even size cubes and is a single (middlemost) layer on odd sized cubes.

M is s block turn meaning all layers on LR axis except L and R themselves. Likewise for E and S of course. I know this is different in other notations - but i insist that my definition is the most logical one ;-)

Hence your m is M in my notation :D

-Per
 
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Kubbuth
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Hi :)

What do you mean by m on 4x4x4 ? Do you mean m = (l r') ??
And what would m be on larger even cube?

In my own private notation m does not exist on even size cubes and is a single (middlemost) layer on odd sized cubes.

M is s block turn meaning all layers on LR axis except L and R themselves. Likewise for E and S of course. I know this is different in other notations - but i insist that my definition is the most logical one ;-)

Hence your m is M in my notation :D

-Per
Hi Per, yes it is [m] = l or r on 4x4x4 (you can use any or both) while m (not [m]) does not exist on even cubes. for 5x5x5 [m] is l, m, r, l+m, m+r, l+r, 0r M (l+m+r) and so on.
 
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