# How To: Commutators for Big Cubes



## dbeyer (Mar 17, 2007)

Firstly, are you interested in solving the big cubes blindfolded, yet have trouble seeing the commutators?
http://tinyurl.com/pft5d
This is Chris' thread, it is very useful. Not comfy with using commutators for the wing edges still? Here is my solution to solve the wings, it is very easy, fast and is a stepping stone to using full blown commutators.

What I use is called a helper buffer method. It is almost like Pochmann's 2-cycle method for the 3x3x3, however you are cycling 3 edges back and forth.
Buffer->Target->Helper This is done w/ an ABA'B' commutator
Helper->Target->Buffer This is done w/ an BAB'A' commutator

There are three special cases that we will encounter:
Starting a new cycle
Solving the helper edge
Solving the OLL Parity

Let's establish the Helper, Buffer, and Targets (notice plural, makes setups easy, sorta like w/ 2-cycling the 3x3x3 blindfolded)

Buffer-- FRu
Helper-- UFr
Targets-- the other 6 edges on the u and r slices. 
FDr, DBr, BUr, LFu, BLu, RBu

The algs:
We use these two commutators
A: RU'R'; B: 
A: [r]; B: U'RU

let  = u, u2, or u' -- likewise [r] = r, r2, or r'

You alternate between ABA'B' commutators and BAB'A' commutators. 
AB commutators: RU'R'u2RUR'u2; RU'R'u'RUR'u; rU'RUr'U'R'U
BA commutators: uRU'R'u'RUR'; U'RUr2U'R'Ur2 ...
--so on and so forth

You do an AB commutator to solve the first wing, then you proceed to solve the piece after that, which was cycled to the Helper's location. Use the proper setup + a BA commutator to solve that piece, you alternate back and forth.

*Special Case 1:*
_Starting a new cycle:_ What happens, the Buffer is brought into the helper buffer area before the other 23 wings are solve. You must start a new cycle. Simply pick a new location and do the setup to solve that piece. The flow continues as normal, the algorithm used to start the new cycle should be the inverse of the last one used.

*Special Case 2:*
_Solving the Helper:_ This is what I call integration. You want to solve the helper and the piece paired with it.

Starting with 1, every other piece is the start of a new pair. 1,2 is a pair; 3,4 is a pair; 5,6 is a pair etc.

Odd numbers are solved w/ the AB commutator, Even Numbers are solved w/ the BA commutator. The rule of thumb the pair w/ the helper: The helper is integrated and the other wing is solved in one alg, using the commutator that would solve the target wing in a normal pair.

Solve this pair: The UFr and the LBd

LBd is the second wing in the pair, so a BA commutator would be used. I would solve that pair w/ (Ll)2 u'RU'R'uRUR' (Ll)2

Solve this pair: RDb and the UFr

RDb is the first wing in the pair, so an AB commutator would be used. I would solve that pair w/ (Dd)' rU'RUr'U'R'U (Dd)

Continuing with the solve after integration:
Ok there is nothing special about this integration thing (save for it's 1 alg to solve the pair rather than 2) The net result is you solved another pair, so you continue solving the next pair as you normally would. The first wing of the pair using an AB commutator; The second wing using a BA commutator. I did not realize this until recently, that there is no other significance to integrating the helper. 

*Special Case 3:*
_Solving the Orientation Parity:_ How do you determine if there is a parity? The answer, the last wing is unpaired, in other words, two solve the other 22 wings (everything but the helper and buffer) you'll require an odd number of algorithms. Starting a new cycle counts as an algorithm solving a piece counts as an algorithm, integration counts as two algorithms, because the net effect is solving two pieces. Using a memorization system that pairs locations, such as Letter Pairs or a Person Action system allows you to easily determine if there is a parity or not.

So, you've solved all of the other 22 wings. Now the Helper and Buffer are all that's left to be solved. We don't have an algorithm to directly swap the UFr and FRu. However, with a simple commutator, we can simply use a well known algorithm. 

r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2

Can you see the setup + commutator so that you can perform that algorithm with no cube rotations?
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(Ll)'(Bb)' RU'R'uRUR'u' (Bb)(Ll)

I would like to make a disclaimer. I've shown setup moves being done with double slice turns. I feel that double slice turns are safer for avoiding lockups. Since Commutators have no affect on other piece types, a double slice turn is fine.


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## dbeyer (Mar 17, 2007)

To Continue:
I took a look at big cubes. I was beginning to understand how to build the components of commutators. However ... the difficulty came with cycling in the correct direction. I broke it down to myself in a text file. I wrote all of my thoughts out, did some exploratory tests. Saved the file, and sent it to Chris. He checked the file, he approved of it.
Here it is:
http://dbeyer.110mb.com/centers_commutators.txt
This helped me better understand freestyling commutators across the cube.

There is the theory behind commutators and how to cycle them in the correct direction. 

I solve x-centers relatively. As I am cycling, each center is solved in the first unsolved location on that face. I use my own scheme.
I label the centers
12
34 
1 would include the locations Ful, Lub, Bur, Ruf, Dfl
2 would include the locations Fru, Lfu, Blu, Rbu, Drf
3 would include the locations Fld, Lbd, Brd, Rfd, Dlb
4 would include the locations Fdr, Ldf, Bdl, Rdb, Dbr

For the U face I solve them in this order
1B
23 
B is my buffer. I have a simple letter scheme as well, built upon this number scheme.

The basic method that I use involves the Buffer, I try to use the Cross section of the l and f slices as a big help to make easy commutators.

I look at the other two pieces involved in the cycle. There are three things I try if the pair isn't already interchangeable.
1: Seeing that one piece is in the cross section (that is it's on the l or f slice) I do a setup to make that piece and the buffer interchangeable on the U face. That would just take a f or l turn of course.
2: Seeing that one piece is is in the cross section, I do a setup making the other piece of the pair interchangeable on the u/d slices.
3: Seeing that one piece is interchangeable with the cross section, I do a setup to move the other piece of the pair into the cross section.

If the pair is already interchangeable, but not in the cross section, I would do a setup on the U face so that I can freestyle this basic commutator.

There are two basic commutator types:
When the pieces are interchangeable on the cross section
When the pieces are interchangeable on the U face

Int-Cross Cross section ... that leaves the Buffer as the action spot
Int-U Face ... that leaves the other piece (generally on the F/B, or L/R faces) as the action spot

Int-U Face commutators are very easy to see, in my opinion. 
I will show you the 4 builds where the action spot interacts with the Buffer's location first. It will be your job to explore and freestyle it across the cube.

A: r'r; B: 
A: r[d]r'; B: 
A: f[d]f'; B: 
A: f'f; B: 

Remember to use AB and BA commutator types!


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## dbeyer (Mar 19, 2007)

This is a short and simple post
Here is a list of the setups to make each piece interchangeable hence, one of the six commutators can be used. 

I will list the Location, the setup, and what move interchanges the target and the helper/buffer.

LFu:	--;	u'
BLu:	--;	u2
RBu:	--;	u
BUr:	--;	r'
DBr:	--;	r2
FDr:	--;	r
FLd:	L2;	u2
RFd:	d'L2;	u2
BRd:	B;	r'
LBd:	B;	r2
UBl:	B';	u
FUl:	l'B';	u
DFl:	D2;	r2
BDl:	D2;	r
ULf:	L';	u2
RUf:	f'L';	u2
DRf:	D;	r2
LDf:	D;	r
URb:	bL;	u'
LUb:	L;	u'
DLb:	L;	u2
RDb:	D';	r


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## AvGalen (Mar 20, 2007)

Thanks Daniel, these kinds of topics really lift the level of this forum!

Now all I have to do is finally attempt a blind 2x2x2 solve, then practise "a little" so I can do the 3x3x3 blind. "Tomorrow" I should be able to do the 4x4x4 and I will probably have a couple of hours left during the weekend to do my first 5x5x5 solves 

Seriously, how long did it take you to learn/develop big cubes blind?


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## dbeyer (Mar 20, 2007)

Off and on since G.wiz 2006, Chris is a real inspiration. I looked into it even sooner than that. 

I took it seriously October 19th, after a long day of competition, and a great afterparty, I go to my aunt's ... and pick up the cube. I played w/ commutator type moves. And I found URb and BUr as my buffer and helper. I then applied to rotation to the UFr and FRu.

I spent most of December learning a letter pair system. I wrote the guide for theory on free style late December or early January, that really helped me understand freestyle commutators for the centers! I put big cubes blindfolded on hold, then come February I picked them up again. 

I practiced for a weekend, intent on solving the 4x4x4 blindfolded. February 4th, my brother's birthday (Superbowl sunday too), I solved the 4x4x4 blindfolded for my first time. It wasn't a big deal, it was sloooow, and I knew satisfaction would only come from something fast on the 4x4x4. I competed in the last week of Ryan Heise's previous competition. I DNF the 4x4x4 solves, and I did this early in the week ... I was boreeeed ... so I picked up the 5x5x5 ... practiced and practiced the 5x5x5. My first successful solve was 46:15 on the 5x5x5, I was soooo stoked! I've now solved the 4x4x4 in as fast as 14:39. That was w/ an attrocious cube (I lubed it ... and I'm impressed!) My fastest 5x5x5 solve ... well that's confidential 

October 19th - March 20th; 5 months, no?
I started Full solves February 1st.
Since then even I've made a bunch of changes to my 4x4x4 method, wings method, centrals, and memorization techniques.

March ... I decided to publish my stuff here ... aren't you all lucky. Forget the yahoo forums ... the only activity there is spam. I trust that this will be pinned to the How-To Secotion  hehe


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## dbeyer (Mar 20, 2007)

*Corner Method*
_2-cycles:_ Well this is a very powerful method, problem comes when multiple pieces are disoriented, or more than additional one cycle needs to be intitated during the execution. I still use it and love it!

Stephan Pochmann describes using the Y perm Conjugated (setup) by F'

That cancels the initial F turn, and the ending F' turn. Leaving you with a very fast algorithm. I didn't like setting up for the Y perm as a beginner 3x3x3 bldist. I prefered setups to the DFL, I knew a conjugation of the J Perm (I didn't have an optimal PLL system then either)

During Summer 2006, I really disliked all the algorithms for Pochmann's method. I tried to develop something called simul block. 
Basically 12 algs to learn, all 2 Corner - 2 Edge cycles. You'd setup the Edge with an M/E move connecting the target to the buffer area (forming a 1x2x2 block) make a move to setup the Corner to the DFL, only 1 move setup required for this step too. The S slice was free ... so if any S slice edge became the target I just did an S' move ... after solving the F/B faces (2 1x3x3 blocks) I'd rotate the cube making the S slice the M slice, and cycle the edges. The method didn't work out. However, I took the new algs from it.

All 7 Corners can be setup to the DFL w/ 1 face move (Assuming the buffer is URB) I knew an alg to target each face of the DFL corner. 

After Rutgers 2006, I thought about learning 21 algs for corners ... because I liked 2-cycling the corners soo much. I noticed a pattern comming up on ACube ... I then tested it ...

I realized that my old alg could be interchanged w/ the Left Handed Y perm.

So ... the 4 Core Algs to my method are

DFL: L'U'LF2R'DR'D'R2F2 
LDF: L'ULUL'U'LFL'U'LULF'L'
FLD: y'RU'R'U'RUR'F'RUR'U'R'FRy
UFR: LU'R'UL'U2RU'R'U2R 
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F'L'ULUL'U'LFL'U'LULF'L'F
FRU'R'U'RUR'F'RUR'U'R'FRF'
F2L'U'LF2R'DR'D'R2
Recognize these?

Those are the Core algs as I said ... I've extended the method w/o learning anything else new really. Once I looked at ACube ... I was like oohh ...

*Deeper Conjugation:*
_Crazy Y Perms_
The method that I use for 3x3, 4x4, and 5x5 corners is really efficient now. I can target 10 locations with NO setups. I would consider these algs I'm about to list Deeper Conjugations: The Y perms are conjuated by an F move ... well I conjugate by [L]F and [R]F'. This leaves only the other 11 locations requiring 1 move to setup the (you can of course set them all to the DFL)
However of the remaining 11 possible targets, 8 of them can be targeted w/ a face move that is only 1 quarter turn.

The Tricky Locations: DBR, RUF, and BLU the setup is D2, F2, and L2 respectively, the other 8 locations can be setup w/ L/L',F/F',D/D'

Here are the extra six conjugations that I use along with the original two. You'll probably laugh at the simplicity of it. I'm just talking it up, because I think it's really great! Notice a pattern?

ULUL'U'LFL'U'LULF'L2
LULUL'U'LFL'U'LULF'L
L2ULUL'U'LFL'U'LULF'
L'ULUL'U'LFL'U'LULF'L' <-- Original

y'U'R'U'RUR'F'RUR'U'R'FR2y
y'RU'R'U'RUR'F'RUR'U'R'FRy <-- Original
y'R2U'R'U'RUR'F'RUR'U'R'Fy
y'R'U'R'U'RUR'F'RUR'U'R'FR'y

This leads to the next part of my solve. Handling the PLL parity. I just made an amazing breakthrough for the 4x4x4. But that's confidential.

The algorithm that Chris Hardwick gave me
(Ll)2 (Ff)2 U2 l2 U2 (Ff)2 (Ll)2

Chris orients and does three cycles, and he likes to solve the last 2 corners and two edges w/ a T perm and crazy setups if there is a parity. For the 5x5x5, I prefer to solve the Corners. Then: 
1.Restore the Wings/Centrals with LU'R'UL'U2RU'R'U2R (3x3x3 parity fix)
2.I will then proceed to solve the wings (fixing the OLL parity if necessary)
3.Solve the Centrals w/ Commutators, Using the UR as a buffer. The cube will be reduced down to the UR<->Last-Edge and URB<->UFR. Now this last algorhtim will swap the Wings on the UL and UR tredge. 

You must setup so that you don't disturb the 
URB and UFR corners, 
the U centers, 
the UR tredge
or the UL wings.

LoL! That's not as bad as it sounds. 

Dx edges set to the DL and do the setup S'D2S
yD edges set the the LD and do LEL'
E slice edges make sure the edge isn't on the L face, do an [e]L[E]L' or [e]L'[E]L setup

let the little e designate moving the FL, LF, BL, LB off of the L face/slice

The U layer is interesting. I visulized the setups wrong ... costing me a few 5x5x5 blindolded solves in the beginning ... so I'm going to GIVE them to you, rather than let you explore this part.


UF: FLE'L'F'
FU: FL'E2LF'
UB: B'L'ELB
BU: B'LE2L'B

so now ... that you've done the setup ... do this T perm (because I know this alg is safe for the 5 center faces)
RUR'U'R'FR2U'R'U'RUR'F'

Undo your setup
now the UL and UR wings are swapped, we know how to fix that. Rotate the Cube y or y', whichever you prefer and finish you're cube!
(Ll)2 (Ff)2 U2 l2 U2 (Ff)2 (Ll)2

I you want to start off basic: Use those setups, and 2-cycle all of the centrals.

If you are comfortable with Centrals Commutators: 
Those setups just aren't optimal ... which I just realized, that it is so even for the last edge. Explore your setups that you can use treating the professor as a 3x3x3. 

Consider this
Setup1:
Alg to swap the centrals
Setup2:
Parity Fix retoring the wings
Reset2:
Reset1:


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## AvGalen (Mar 26, 2007)

Hi Daniel,

I hope you had a good time last weekend. Due to a little bug in the "official results" database you are actually listed as the winner of the 3x3x3 bf competition. (http://www.worldcubeassociation.org/results/c.php?i=Chattahoochee2007) You should thank your parents for your names.

Seriously, what happened? You solved a 4x4x4 blind and Chris did a 4x4x4 and a 5x5x5, but neither of you did a succesfull 3x3x3? You are now ranked 4th in the world (http://www.worldcubeassociation.org...gionId=&years=&show=100+Persons&single=Single).

My plans for the weekend changed, so I still didn't read all of the above, but I think Erik has started to.


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## dbeyer (Mar 26, 2007)

Haha, Chris and I talked about it ... we don't really care about 3x3x3 blindfolded, I mean we are going for fast times, not accuracy. 

As for my blindfolded times:

3:15 DNF Off by a 3 cycle on the edges
3:05 DNF disoriented a pair of edges

4x4x4: Execution Error, off by a 3 cycle on the centers 14:45
4x4x4: Ranked me 4th in the world '_' ... a little hard time, forgot 2 letter pairs, but was able to skip and come back to them ... I should really add that to the guide!

5x5x5: Time 44:45ish I didn't undo a setup on the f slice, that was the only Error, there was a net turn of f2, I then continued to solve the Lfe and Rfe

Because of the net turn of f2, I permuted the f slice pieces, but not the four edges that were already there

An example, I had LP, DM ...

Put to roman rooms and images

This guy sits drunk at the barstool listening to a Linkin Park Concert
The Dungeon Master (DnD) rolls the dice and sends Jesus to the cross (the location is a porceline cross on the wall)

L was permuted, P was permuted (however D wasn't, because D came after a piece on the f slice)
M was permuted ...

the wings in the P and S locations were swapped, 

I think the letter pair using S was SW, SWAT team

So when I was permuting D, I actually placed the W wing into the D location.

Considering I drove 12 hours of the 15 hour long trip, and went straight into the school and started competing, w/ 1.5 hours sleep ... I did alright!


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## pjk (Mar 26, 2007)

Nice guide, I will put this into How-To soon. I want to start getting into bigger cubes BLD, but want to get better at 3x3 speedsolve and 3x3 BLD, along with 4x4 and 5x5 speedsolve. But within a year I will start 

Thanks for writing this up, it will certainly help me.


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## cmhardw (Mar 27, 2007)

> *Seriously, what happened? You solved a 4x4x4 blind and Chris did a 4x4x4 and a 5x5x5, but neither of you did a succesfull 3x3x3?*



Hey AvGalen, yeah Daniel and I talked about this a bunch in Georgia. I think the reason I still DNF the 3x3 under pressure is I don't have a memory system for it, only for the bigger cubes. Daniel and I talked a lot about potential memory systems for the 3x3x3, so we're working on it. Still that's my reason *cough* excuse *cough* ;-) for having DNF'd my 3x3x3 BLD solves.

Chris


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## dbeyer (Mar 27, 2007)

Meh, I was just tired *yawns* hehe, First place still


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## dbeyer (May 11, 2007)

So now I AM going to give several examples of how I would solve, and try and tell how you would solve as well.

Holding in the orientation that you consider solve: Please Scramble with these algorithms.
Scramble lower case letters as multi-slice moves.

1. D f U L' r' D' u F R' F2 U2 L2 r B2 D u U2 L2 F' r' U2 r2 R B' f2 u' B U B F L' r' R F u r2 R' f2 u' f' 


Anyway, ... you see that there is a rotation that solves 3 U centers and 5 other pieces? Well that is actually a very bad rotation for me ... I'd much rather see many D centers solved, in fact solving many U centers during optimal rotation makes the "forced cycle" appear frequently.

I'd take the 7 solved centers rotation of x'

Now the Ful, Fdr, Ufl, Lub, Ldf, Ruf, and Dbr are all solved.

Also just another reason to not take the rotation that solved 3 U centers, I'd look at the Urb and see the cycle Urb->Ulf->Dfl or whatever D center, anyway that is just a very nasty cycle!

I do also see how using the Ulf as a psuedo buffer could be very effective now (with the rotation solving the three U centers during rotations). But I'd take the rotation that solves 1 less center instead of memorizing something difficult or out of the usual.

I will give the letter pair + memo + execution. I don't really want to describe the locations in Location-Notation.

UL (which tells me Q->U->L) ...

UL - Unlucky - I'd see Aaron working on handling a belt full of packages, and it just never ends, how Unlucky!

AI - Artificial Intellgence - I'd see the data corrections being done by AI (like Brainiac)

RV - RV ... - I'd see an RV falling down the ladder crashing loudly into the floor below

JE - Jesus - Emergency: Jesus is Coming Jesus is coming!

KM - Kimono - I'd see a kimono being shipped in a large bag from QVC, and we shove it in the totes

WN - Window - I'd see a window being shot out by the scanner, I'd hear a loud beep (the scanner) then breaking of the glass

CF - Coffee - I'd see Chris Bringing some coffee upthe stairs

SO - Sofa - I'd see a huge sofa being sent down the belt (not sposed to be done at UPS) ... what dumbasses!

Hmm ... I don't even have to count how many algs I have to execute ... I just looked and I saw the last image was at the black belt (location 8, backwall behind me)

Ok ...
Execution of Centers.

UL: f2 [bub' U'; bu'b' U] f2
AI: [r'ur U; ru'r U']
RV: [r2 bL'b'; r2 bLb']
JE: F2 [f'Uf d; f'U'f d'] F2
KM: B [d lU'l'; d' lUl'] B'
WN: [b2 lB2l'; b2 lB2l']
CF: [rd2r' U'; rd2r' U]
SO: L' [fU2f' d; fU2f' d'] L

Corners:
D2 y'RU'R'U'RUR'F'RUR'U'R'FRy D2
L'ULUL'U'LFL'U'LULF'L'
LU'R'UL'U2RU'R'U2R (new cycle)
D L'ULUL'U'LFL'U'LULF'L' D'
LULUL'U'LFL'U'LULF'L
ULUL'U'LFL'U'LULF'L2
F2 L'ULUL'U'LFL'U'LULF'L' F2

There is a Parity. Now RU'R'U' Ll)2 Ff)2 U2 l2 U2 Ff)2 Ll)2 URUR' swaps the wings that were swapped. The UR and UB pairs.

With my new lettering scheme and using the bUR as the buffer. (I see commutators with clarity using this spot!) but the memo is still rough and choppy! 

I designed my lettering scheme 7 months ago, and now I'm changing it, it's kinda akward! 

PC - Peacock -- f' [R'uR U2; Ru'R U2] f
SN - Sonic -- l' [l'd'l U'; l'dl U] l
HX - Hexen -- [U f'u2f; U' f'u2f]
JL - Jack o Lantern -- r2 [b'd'b U'; b'db U] r2
EW - Ewww! -- L [Ld'L' U2; LdL' U2] L'
OB - Obelisk -- R [F'd'F D; F'dF D'] R'
KI - Killer Instinct -- [F'R'F r2; F'RF r2]

New Cycle
TM - Timer -- L [Bd'B' U'; BdB' U] L'
GU - G Unit -- [uL2u' R'; uL2u' R]

New Cylce
TV - Television -- [R'u'R U2; R'uR U2]
RF - Ref -- D' [B'lB R; B'l'B R'] D
V - Parity -- Rd2F *Parity Fix* F'd2R'

Well that looks all good! I hope that everybody enjoys this post. It's a new approach, I love what it's doing for my wing execution!!

Later,
Daniel Beyer


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## dbeyer (May 11, 2007)

Chris thanks a lot, you helped to mold my new Wing Execution methods, and to the rest of you sorry for the big delay of an example!


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## pjk (May 11, 2007)

Nice example, that is very handy. Thanks for the time you put into it. What do you avg for 4x4 BLD these days?


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## dbeyer (May 14, 2007)

Umm 15m ish ... I'm doing some adapting, learning trying new stuff. With my methodology (which Chris and I have been discussing a lot, he just set his new personal best of sub 6:30!)

I've been doing a lot of stuff non-cubing related (work/sleep) and writing very long emails to Chris rather than practicing =P

I can't wait to see a sub 30 solve on the 5x5 with my new methods, or a sub 10 solve on the 4x4!

Later,
Daniel Beyer


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## dbeyer (Jul 25, 2007)

To those of you who are already using commutators on the wings for big cubes. The centrals are just as easy, if you can view them as 1 piece that contains two orienations.

an example 

the FL central (that is the F sticker of the FL central) is the root of the dFL wing.

Whereas the LF (the L sticker of the FL) is the root of the uLF wing.

rUF -> dFL -> dRF

This cycle could be achieved by 

L'UL d' L'U'L d,
Now we'll apply this concept to the centrals

UF -> FL -> RF
L'UL E' L'U'L E (E relates to D)

It should become clear with a little bit of practice that a cycle such as

rUF -> dFL -> uFR requires a setup move

There is actually a nice cancelation.

A setup of L2 will make the dFL and uFR interchangeable on the u slice.

L2 [LUL' u2 LU'L' u2] L2 == L'UL' u2 LU'L' u2 L2

the same cycle can be done on centrals

UF -> FL -> FR
L'UL' E2 LU'L' E2 L2

Centrals have special cases, just like centers, and corners have. They are called blocked cycles.

Something such as 

UF -> DR -> DF 

On centrals a drop and slide with the action of 

x2 (let x be a random legal move for the second move of the part A)

Such as MD2M' or ER2E'.

In the cycle above.

D' M'U2M D M'U2M catches the D stickers of those pieces on the D face.

Yet

MDM' U MD'M' U' catches the L sticker of the DL rather than a D sticker.



Now for the Ts, it's rather simple.

You can do the drop and catches like

Ur -> Ub -> Lf

rewrite it
Lf -> Ur -> Ub

Lf is the lone T.
Drop and catch (A)
r'Er
Interchange (B)
U
A': r'E'r
B': U'
ABA'B' == r'Er U r'E'r U'

Some interesting cases

Fd -> Fu -> Ru

M'U2M u M'U2M u'

Yet M'UM u M'U'M u'
would actually cycle
Ul -> Fu -> Ru

Another interesting case

Df -> Ub -> Uf

U2 M'u2M U2 M'u2M

Some of my favorite cases are

are cases where you cycle a U center, a D center, and an F2L center.

Such as

Ur -> Dl -> ??

And T on the F, L, B or R face.

lets see...

Fu

so Ur -> Dl -> Fu
r'Fr S2 r'F'r S2

With that case interchanging the two pieces on the U and D by S2 or M2 you can always nail the commutator in 8 moves + a Simple U or D setup.

Don't be afraid to do BA commutators, that is to interchange first.

another interesting case
Ul and Dl are interchangeable by l2
You can cycle in any T on the L or R face

like

Ul -> Dl -> Lu

L [S'LS l2 S'L'S l2] L'

Work on the freestyling, and you'll have these concepts down in no time. There is a lot more than what I have to say, but I hope that you can learn it yourself. 

Later,
Daniel Beyer


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## Zeinest815 (Mar 25, 2008)

a question to big cubes BLDers.

the most difficult thing that I encounter when commutator-solving the wings is the case where in 2 of the 3 pieces are adjacent. how should I handle that?

in the case of the x-centers, the most difficult case is when pieces are on opposite faces (U/D especially). again how do you handle that?

also is it possible guys that the setups for the commutators need only a maximum of ONE move? i noticed this on dBeyer's sample 4x4x4 solve.

thanks for the reply.

(btw, I practice without the blindfold on. I just want to have a somewhat complete grasp of this big cube commutator thing)


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## Kenneth (Mar 25, 2008)

This is a cycle I use often. I show it as a general alg where sl (slice) marks the swap using either r2, l2, m2, M2 or any combination.

x' U2 sl U R U' sl U R' U x or mirror: x' U2 sl U' L' U sl U' L U' x

On a 4x4x4 this gives eight combinations for single edges and four if it's dedges. This if inverses are included in the group. but you do not need the algs for those, just do a y2 before and the inverse becomes the mirror instead.


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## Mike Hughey (Mar 25, 2008)

I think there are some cases that require two setup moves. At least, if there aren't, then I'm doing some of them inefficiently.

For wings, two adjacent are usually not a problem; just do a slice move to place them so the adjacent two are interchangeable. (And pick your slice move so that the third piece is in the right place to make it work; I think this works in all cases except the case where the third piece is on the opposite edge diagonally from the two adjacent pieces, in which case I need two setup moves.) But if you have, say, DFr, UFl, and UBr to switch, that's a hard case; I solve that particular one by doing B' R' to setup, then [D b' D', F2].

Also, as you say, opposite centers are ugly, especially when they're directly opposite. Again I use two setup moves. So if I had Ubl, Ubr, Dbr, I would do something like D' F'f' to setup, then [l u l', U'].

If Chris or Daniel can do those with just one setup move, I'd love to hear about it. Now before Chattahoochee so I can have a better chance of beating you would be nice.


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## Kenneth (Mar 25, 2008)

Mike: based on the alg I posted above: U R' U r2 U R U' r2 U R' U2 R U'

The r2's are the swaps.

Is that one ok?


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## Mike Hughey (Mar 25, 2008)

Sorry, Kenneth - for some reason that I don't fully understand, I completely failed to see your post. Yes, your algorithm works fine for this. And it's pretty easy to execute, with all the R's and U's. But it's actually one move more than mine. 

But that does look generally useful; I'm going to have to play with it a while and see if I can get where I really understand it. Thanks!


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## Kenneth (Mar 25, 2008)

U R' stores the Ubr at Fur, U moves the one that initialy was at Ulf to Ubr so a r2 can swap that one with Frd, then the piece at Ubr is chanced to the stored one using U R U'. Then the second swap follows (r2) and the rest is restoration, first the last exchange using U R' U' and then the storage is restored using U' R U' (the two following U' are of course merged to a U2).

Hope I got it right, have no cube here right now


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## Mike Hughey (Mar 25, 2008)

Instead of Fur, I think you meant FRd, but other than that, yeah, I think you got it right. Thanks - that explains it rather well.


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## mrCage (Mar 27, 2008)

Hi 

The way i prefer to explain (most common) commutators (for 3-cycles at least) is to show how it works on a "cutplane". The 2 different kinds of cutplanes available are outer and inner layers. Inner layers are of course not functioning for corner cycles. Cycling only 3 centers on a 3x3x3 cube is not possible.

The idea of almost all my commutators is to do the following:

1 - Replace a single cubie on a cutplane (A- commonly 3 or 4 turns)

2 - Rotate the cutplane (B - always a single turn!!)

3 - Return the cubie removed from the cutplane back into its new position on the cutplane (A' - inverse of A. To invert a sequence start from the back and move to the front. Invert every turn as you go along. The inverse of UR is NOT RU but R'U'.).

4 - Undo the cutplane rotation.

Easy example on corners (any size cube: 
1 - R' D R (A-part. Replaces one corner on the cutplane)
2 - U (B-part, the cutplane rotation)
3 - R' D' R (inverse of A. Returns the corner removed from cutplane back into a new position on the cutplane)
4 - U' (inverse of B-part)

This becomes R' D R U R' D' R U' (indeed a 3-cycle)

(Easy example on edges:
1 - r' D r (A-part)
2 - U (B-part, the cutplane rotation)
3 - r' D' r (inverse of A)
4 - U' (inverse of B-part)

This becomes r' D r U r' D' r U'

Note the intended similarity between the 2 examples.)

This whole thing becomes ABA'B' where the total length becomes length of A times two plus 2 (since length of B is 1!!)

Now for bigger cubes edges belong to different ORBITALS.
On a 4x4x4 cube UFr, URb, UBl and ULf all belong to the same ORBITAL as those positions are only a single turn of the cutplane apart. 
UFl, URf, UBr and ULb belong to the same (but different) orbital

What to do if you want to cycle 3 edges of different ORBITALS?
Simple answer: position the edges into same orbital first.

[UNFINISHED POST - lunch !!!!]

[edit - i will finish this post some other time, probably completely rewrite it by the look of it !!]

- Per


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## dbeyer (Apr 14, 2008)

Here is a special case that Chris and I have talked about for our commutator system.

Its actually rather interesting.
For wings, it's what I'd call an anti-slice commutator.

You have 3 pieces which all have no interchangeability. 

Such as two pieces on the same face-plane, but different orbitals, and then the 3rd is on a parallel slice-plane. Its on the same face-plane as one of the first two but again on a different orbital.

URb -> FUl -> LFu

I don't know how many of you remember Fridrich's notation where she used slice and anti slice notation. Well this is where I'm getting it from.

The Setup and the First move of the insertion are anti slice turns which I think make for nice commutators mostly. 

Now for that case. 

URb -> FUl -> LFu
Lets invert it.
URb -> LFu -> FUl
I can so you the SAB commutator rather than the SBA commutator. 

L' ] R'FR b' R'F'R b [ L

Another nice case:

These are called Ferris Wheels, just a nickname that Chris gave to them. It makes sense, I suppose.

These are cases where the S and B lead to a cancelation.

Basically: You'll see that two pieces are interchangeable. Such as the ULf and FLd. There is no 3 move insertion from either stationary point (where the cubies currently are) Yet you'll see somewhere on that face plane you can make an make a 3 move insertion.

So you move the first piece there, insert, move the second piece there, undo the insert, and then restore the main slice.

Like I said the ULf and the FLd, now the FUl.

I look and I see that doing l' moves the FUl to the UBl
I also see that doing U moves one of my main slice pieces to the UBl. So maybe I can do an insertion of Ul'U' and interchange?

No, it wouldn't work because the lone edge is distrubed by the turning of the U layer.

Now, I could move both pieces to say the DLb. 
So.

Setup: L
Insertion: DlD'
Interchange: L

L ] DlD' L Dl'D' L' [ L'
=
L DlD' L Dl'D' L2'
=
L DlD' L Dl'D' L2
Now if you did it the last way with that cancelation with a clockwise L2, you'd see the L slice making a complete 360 degrees rotation. Which is what a ferris wheel does, which I guess is why Chris named it like such.


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## PianoMan (Jan 12, 2009)

My brain is confuzzled!!!!


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## ConnorCuber (Jan 12, 2009)

PianoMan said:


> My brain is confuzzled!!!!



And you had to revive a thread that was last posted in april just to say that?

Good god.


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## Bomber (Jan 12, 2009)

ConnorCuber said:


> PianoMan said:
> 
> 
> > My brain is confuzzled!!!!
> ...



And you had to post that didn't you? Are you Admin? Or even a Moderator? No. 
Read this > http://www.speedsolving.com/forum/showthread.php?p=121233#post121233

I wasn't here 8 months ago so I obviously wasn't here to read it. I am going to read the thread and maybe post something *relevant*, I think it gives a lot more variety unlike the utter drivel you have just posted that is so repetitive on this forum.
Now I will read, and hopefully enjoy this topic.


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## joelwong (Jun 24, 2010)

For the edges commutators thing am i solving 1 or 2 edges at one time?


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