# Commutator challenge



## mrCage (Aug 11, 2009)

Hi 

I challenge everyone to come up with a (useful) commutator involving only exactly 2 faces (or layers) of the cube. The good ones i know use 3 faces or more

Per

Correction:

I actally know this one usng 2 layers: m D2 m' D2, in commmutator notation [m,D2]


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## DavidWoner (Aug 11, 2009)

This one's pretty useful (highlight)

[R U R2, R U2 R2]


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## trying-to-speedcube... (Aug 11, 2009)

Sexy move!


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## Lucas Garron (Aug 11, 2009)

Something like [(R2 U2)3, U] ?


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## LNZ (Aug 11, 2009)

Will this do?

To place the last few cubies on the last two centres of a 7x7x7 cube. Get as close as possible the normal way. And use these commutators to finish the job.

5x5 block of a 7x7 cube (both the same relative positions on front face and top face)

A A A B B
X C D X X
X X X X X
X X X X X
X X X X X

"A" move: A(up) U r U' 
A(down) U r' U'

"B" move: A(up) U' l' U 
A(down) U' l U

"C" move: A(up) U (Slice 5)(up) U'
A(down) U (Slice 5)(down) U'

"D" move: A(up) U (Slice 5)(up) U'
A(down) U (Slice 5)(down) U' 

The positions marked "X" can be rotated on both the top and front face to get into the stated positions for the above commutators.

Very similar commutators for the 6x6 and 5x5 cube too, but just with slighty different slices to move.


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## Am1n- (Aug 11, 2009)

I'd go for [U2, R2]
It's a row 3-cycle (or something) 

mvg


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## mrCage (Aug 12, 2009)

Vault312 said:


> This one's pretty useful (highlight)
> 
> [R U R2, R U2 R2]


 
Cool way to write a sune!! I had to write it out fully in order to realise.

Per


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## Am1n- (Aug 13, 2009)

I've been thinking about this, you can use (correct me if I'm wrong) [X,Y] as commutator, where X and Y are 2 random sequences using only 2 layers. You than have a commutator. Afterwards, you only have to specify what parts of the cube are affected by this. 

mvg


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## mrCage (Aug 13, 2009)

Am1n- said:


> I've been thinking about this, you can use (correct me if I'm wrong) [X,Y] as commutator, where X and Y are 2 random sequences using only 2 layers. You than have a commutator. Afterwards, you only have to specify what parts of the cube are affected by this.
> 
> mvg


 
I dont get that last part. Specify what? Hmmm ...

Per


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## trying-to-speedcube... (Aug 13, 2009)

Am1n- said:


> I've been thinking about this, *you can use (correct me if I'm wrong) [X,Y] as commutator, where X and Y are 2 random sequences using only 2 layers.* You than have a commutator. Afterwards, you only have to specify what parts of the cube are affected by this.
> 
> mvg


:fp

That's indeed pretty much the definition which Per used in the challenge. But the challenge isn't to make a commutator with 2 sides, but making a *useful* commutator with 2 sides. 

The problem is that it's not so easy to find something with 2 sides that affects little part of the cube. That's the whole challenge.


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## darthyody (Aug 13, 2009)

(R' F R F')*3 for corner switching.


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## trying-to-speedcube... (Aug 13, 2009)

That's not exactly a single commutator, is it? (R' F R F' is, but by itself it's not particularly useful)


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## Am1n- (Aug 13, 2009)

found 1: its an edge 3-cycle
[(U2 R U' R' U') , R2 ]

off-topic: whiiiii, my first personal facepalm 


mvg


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## trying-to-speedcube... (Aug 13, 2009)

My pleasure 

Nice find btw


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## Am1n- (Aug 15, 2009)

And there are ofcourse the edge flips and variations;
[(M U)4, U]
[(M U)4, U2]
[(M2 U2 M2), U]
enz...

mvg


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## Lucas Garron (Nov 6, 2009)

cmowla said:


> Lucas Garron said:
> 
> 
> > Something like [(R2 U2)3, U] ?
> ...


1) It doesn't achieve the same result. (Did you watch the link?)
2) That's not what this thread is about.

[3) It's almost certainly slower to execute than (R2U2)3 in practice.]


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## Lucas Garron (Nov 6, 2009)

cmowla said:


> Lucas Garron said:
> 
> 
> > cmowla said:
> ...


Certainly (see the NCLB thread). That reminds me, I've wanted to change that signature since a while (thanks!).


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## siva.shanmukh (Nov 12, 2009)

Just wondering if R2 U R U R' U' R' U' R' U R' can be broken down as [X Y]. Tried for half a minute and didn't get to anything.


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## Lucas Garron (Nov 12, 2009)

siva.shanmukh said:


> Just wondering if R2 U R U R' U' R' U' R' U R' can be broken down as [X Y]. Tried for half a minute and didn't get to anything.


Half a minute is certainly not enough.
If you mean a commutator [X, Y], then the answer is: Yes, with a conjugation. But you probably wouldn't be happy with it, unless there's a neat comm I don't know about.

And of course , if you just want a 2-gen comm with the same net effect, that's easy.


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## siva.shanmukh (Nov 12, 2009)

Lucas Garron said:


> siva.shanmukh said:
> 
> 
> > Just wondering if R2 U R U R' U' R' U' R' U R' can be broken down as [X Y]. Tried for half a minute and didn't get to anything.
> ...



See, here is what I got.

Removing the setup R2
R2 [U R U R' U' R' U' R' U R'] R'2

Breaking it into half. (The first half = X Y and the second half = X' Y')
U R U R' U' | R' U' R' U R'

Beyond this I am not able to break it into X Y (used brute force).

Thats why I had to stop after half a minute. I am sure even if tried longer I wouldn't go anywhere.

Please correct me if I am wrong anywhere. Thanks.


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## Lucas (Nov 12, 2009)

Mmm...

A pattern [M, E]
Edge perm [M, U2] but this doesn't behave exactly as most useful commutators do.

It is really difficult, every 8 move commutator that I use for bld requires 3 faces (one is the "main", and another two to insert).


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## Lucas Garron (Nov 12, 2009)

siva.shanmukh said:


> Please correct me if I am wrong anywhere. Thanks.


Let's see.



siva.shanmukh said:


> Removing the setup R2
> R2 [U R U R' U' R' U' R' U R'] R'2


You can't really do that. Now you're asking for a commutator for a different permutation, which won't give you a solution [X, Y] to the original alg.



siva.shanmukh said:


> Breaking it into half. (The first half = X Y and the second half = X' Y')
> U R U R' U' | R' U' R' U R'


That's a bad assumption. Consider [R2 U2, U2 F2] = R2 U2 U2 F2 U2 R2 F2 U2 = R2 F2 U2 R2 F2 U2.
Splitting it in half as R2 F2 U2 | R2 F2 U2 will not give you the parts X Y and X' Y'.

Neither of your two steps are necessarily bad, but you need to know why you're allowing yourself to do them, and understand what you might be ignoring.

(Not trying to be mean, just straightforward and (de-?)constructive.)


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## TMOY (Nov 12, 2009)

Lucas Garron said:


> siva.shanmukh said:
> 
> 
> > Just wondering if R2 U R U R' U' R' U' R' U R' can be broken down as [X Y]. Tried for half a minute and didn't get to anything.
> ...


Half a minute is more than enough to see that it is impossible, at least with Rs and Us only. There is an odd number of quarter-turns of each face in the algorithm and any commutator of the form [X,Y] requires an even number of each. 
If you allow cube rotations, then you can get R U' [U R U R U, z' y2] U R'.


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## siva.shanmukh (Nov 12, 2009)

Lucas Garron said:


> siva.shanmukh said:
> 
> 
> > Please correct me if I am wrong anywhere. Thanks.
> ...



I got it. I was too hasty to put it that way. Thanks for correcting me.



TMOY said:


> There is an odd number of quarter-turns of each face in the algorithm


Its even. 10. If it is odd, you also wouldn't have been able to come up with that conjugate.

Just wondering: Can every 3 cycling algorithm be written in the form of a conjugate?

This time I will give this a thought before replying


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## TMOY (Nov 13, 2009)

siva.shanmukh said:


> Its even. 10. If it is odd, you also wouldn't have been able to come up with that conjugate.


I count 7 Rs and 5 Us. Both 7 and 5 are odd 
It means that you can't find a pure RU commutator (or even a pure RUDLBF commutator), but you still can find one using other types of moves (in my example, cube rotations, but you could also consider slice moves or double layer moves).


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## siva.shanmukh (Nov 13, 2009)

TMOY said:


> siva.shanmukh said:
> 
> 
> > Its even. 10. If it is odd, you also wouldn't have been able to come up with that conjugate.
> ...



Ooo! Okay.. I get it now.


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## mrCage (Nov 15, 2009)

siva.shanmukh said:


> TMOY said:
> 
> 
> > siva.shanmukh said:
> ...


 
Well, one could obtain partial cancellations and get an odd number of either U or R turns !!


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## Cangurino (Dec 3, 2009)

Here's a couple of commutators I've come up with. Whether or not they're useful (or new for that matter) you have to decide for yourselves.


[ R2, U2 R U R' U R U2 ]
[ R', U2 R2 U2 R2 U2 ]
[ (R U2)2 R, (U2 R')3 U2 ]


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## hansho13 (Dec 24, 2009)

3x3 commutator: R2 B2 R2 B2 R2 B2 

very simple


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## Tim Reynolds (Dec 24, 2009)

hansho13 said:


> 3x3 commutator: R2 B2 R2 B2 R2 B2
> 
> very simple



Is this a commutator? I don't see immediately that it is.


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## Stefan (Dec 24, 2009)

Tim Reynolds said:


> hansho13 said:
> 
> 
> > 3x3 commutator: R2 B2 R2 B2 R2 B2
> ...



That's cause you immediately see that it isn't, right? First half and second half don't even begin with the same layer. Too easy.


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## Tim Reynolds (Dec 24, 2009)

StefanPochmann said:


> Tim Reynolds said:
> 
> 
> > hansho13 said:
> ...



I wouldn't jump to the assumption that there's no "interesting" way of writing it as a commutator. For instance, sune=[R U R2, R U R'] is somewhat surprising. There could be something like [R2 B, B R2 B']...that doesn't work, but comes somewhat close. I guess the fact that it's all half turns makes it unlikely that anything pretty will work out...x y x' y' must have x starting with R2 and y starting with B2, which really doesn't seem like it will magically work out.

But yeah. You can't just write R2 B2 R2 B2 R2 B2 as a commutator without doing something clever. I was giving hansho the chance to find something pretty, but I'm now becoming convinced there isn't anything pretty to be found.


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## Johannes91 (Dec 24, 2009)

Tim Reynolds said:


> hansho13 said:
> 
> 
> > 3x3 commutator: R2 B2 R2 B2 R2 B2
> ...





Spoiler



[R2 B2 R2, z2 y]


But not the kind of commutator this thread is (was) about.


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## Stefan (Dec 24, 2009)

I hate being wrong (shouldn't stay up until 6am where I don't have the motivation to think much).


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## Swordsman Kirby (Dec 26, 2009)

StefanPochmann said:


> Tim Reynolds said:
> 
> 
> > hansho13 said:
> ...



I guess there are commutators that generate the same position as that.


Spoiler



[U R2 U2 R' U, U R2 U2 R2 U2 R2 U]


"very simple"

Actually 2n-2m cycles are pretty hard for the corners-solved positions. I'm not sure whether they're solvable by one commutator, though I'm fairly certain that any combination of odd cycles is doable.


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## siva.shanmukh (May 17, 2011)

TMOY said:


> Half a minute is more than enough to see that it is impossible, at least with Rs and Us only. There is an odd number of quarter-turns of each face in the algorithm and any commutator of the form [X,Y] requires an even number of each.
> If you allow cube rotations, then you can get R U' [U R U R U, z' y2] U R'.


 
I know it has been a long time since this was said. I am just wondering, how you arrived at it. Is there an algorithm to find if a given move can be broken down into a commutator/conjugate.


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## Christopher Mowla (May 17, 2011)

Several 2-gen algorithms I have encountered in the past could not be expressed as a commutator. This does not mean that non can be, just the parity algorithms' base structures I tried to make into the traditional form.

_(For those who are questioning my intelligence, I know that a U-Perm is NOT a parity algorithm, but I am making a comparison with what I am already familiar with._)

TMOY's approach to introduce cube rotations is quite clever. I noticed that 2-gen algorithms can be modified in weird ways and still get positive results. For example, you can invert all the moves in a 2-gen alg and get a reflection.  This very fact could explain why TMOY could use cube rotations to express it as a commutator (maybe, I haven't really thought this through).

Anyway, not that I am going to speak for TMOY on how he found that breakdown, but here are some I made of the same algorithm without introducing cube rotations.

R2 U R U R' U' R' U' R' U R'
=
[R, R U R U] U R'
=
[R: [R U R, U]] U R'
=
[R2: [U R U, R']] U R'
=
R2 U [R, U] R2 [R, U'] R'

For me, it just takes experimentation. For example, if you have R2, break it into R R or R' R' and rearrange the moves until you can get a commutator form (not always possible without adding in extra moves which cancel to get the original alg).

Before starting with 2-gen, I would recommend analyzing non-2-gen algorithms.


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## TMOY (May 18, 2011)

My approach was actually very simple:
- R2 U R U R' U' R' U' R' U R' is a conjugate of R U R U R U' R' U' R' U' (an alg which is useful to know for BLD and very easy to remember);
- you can swap the U and R faces with a cube rotation and then U' R' U' R' U' becomes R' U' R' U' R'.
So all I had to do was to find a relevant cube rotation.


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## siva.shanmukh (May 29, 2011)

TMOY : I still am baffled at your idea of using a cube rotation. I would never have had tried that way.

cmowla : I have a lot to ask you. Will ask you in the thread where you posted links for the OLL parity derivation videos.

Coming back to my query.
To summarize, its more based of intuition and observation rather than a procedure. 

We already know that any 3 cycle move can be solved using a conjugate. I was wondering if the converse is true.
I was wondering if one can come up with an algorithm which can convert any given algorithm to solve a 3 cycle into a conjugate form.

Let us keep the question to that and later extend it probably to converting any even numbered quater turn move(non-parity) into a conjugate form


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## Christopher Mowla (May 30, 2011)

siva.shanmukh said:


> We already know that any 3 cycle move can be solved using a conjugate.


I am not sure what you mean here. Are you saying that we already know how to create algorithms to solve 3-cycles which are of the form:
A B A' , where B is not a 3-cycle algorithm?

And since that's true (not that I agree that it is or that you are saying that it is), is there a process to write any algorithm that solves 3-cycles into that same form?


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## irontwig (May 30, 2011)

cmowla said:


> I am not sure what you mean here. Are you saying that we already know how to create algorithms to solve 3-cycles which are of the form:
> A B A' , where B is not a 3-cycle algorithm?



That can't be true since if (A B A')^3=I then B^3=I since the three A'A cancel out.


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## Rpotts (May 30, 2011)

(A B A')*3 = A B*3 A'


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## Christopher Mowla (May 30, 2011)

irontwig said:


> That can't be true since if (A B A')^3=I then B^3=I since the three A'A cancel out.


Yeah, I know but I was just trying to see what he meant. By my interpretation of the wording, this is what appears to be what he meant. So I guess we'll see soon what he meant.


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## siva.shanmukh (Nov 25, 2011)

cmowla said:


> I am not sure what you mean here. Are you saying that we already know how to create algorithms to solve 3-cycles which are of the form:
> A B A' , where B is not a 3-cycle algorithm?
> 
> And since that's true (not that I agree that it is or that you are saying that it is), is there a process to write any algorithm that solves 3-cycles into that same form?


 
Oh.. What I meant is basically that we can solve any 3-cycle algorithm using a commutator. Even if we didn't want to go through the exercise of figuring out A and B, we can do some setup moves, do a known commutator and then undo the setup moves.

To be even more clear, we have found a conjugate version for the U perm as R U' [U R U R U, z' y2] U R'. My question is "Is it possible to write any even move sequence in this form C A B A' B' C' (Assuming this form as conjugate)


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