What's your favorite commutator?

[cubernya]

I'm still a fan of [R U R2, R U R'].

I saw this thread and came to post Sune

Funny, I thought the exact same thing

 

[antoineccantin]

M' U M' U M' U M'

wtf is this

 

[CubeRoots]

U R U' R'

 

[Lucas Garron]

Funny, I thought the exact same thing

Yeah, I've seen some cool commutators, but there's something classic about that Sune decomposition.

 

[TheNextFeliks]

[R U R', D]

 

[antoineccantin]

U R U' R'

[U, R]

 

[Cubo largo]

2(M' U M U)

 

[TDM]

2(M' U M U)

Not a commutator, but I tried [M' U M,U] and used it to find this:
[F2:[M' U M,U]2]
which does the same as:
[(M' U)4,y2]

 

[Cubo largo]

Why not? FD>UL>UB

 

[mDiPalma]

[y, x z x2]

 

[TDM]

Why not? FD>UL>UB

A commutator is not the same as a 3-cycle.

 

[Cubo largo]

So what is this? So U2 M' U2 M' is not a commutator? Let me know please I ate to do mistakes in commutators...

 

[Christopher Mowla]

So what is this? So U2 M' U2 M' is not a commutator? Let me know please I ate to do mistakes in commutators...

U2 M' U2 M' isn't a commutator, and (M' U M U)2 is not, neither is M' U M U.

 

[Cubo largo]

I'm really sorry. What an error! So I think my preferite is [E2, R U2 R']

 

[TDM]

So what is this? So U2 M' U2 M' is not a commutator? Let me know please I ate to do mistakes in commutators...

U2 M' U2 M' isn't, but U2 M' U2 M and U2 M U2 M' are. They can be written as [U2,M'] and [U2,M].

 

[Christopher Mowla]

I'm really sorry. What an error! So I think my preferite is [E2, R U2 R']

There's no need to ask for an apology.

When I first started cubing, and I learned M' U2 M U2, I too thought that it might be possible that all 3-cycles are commutators. All 3-cycles of corners, at least (I'm not sure about middle edges, because I haven't proved it yet) can be reached with a single commutator, but as far as actual maneuvers are concerned, as TDM pointed out, this is not the case.

EDIT:
For big cubes, I like mine [U' Rw U2 r2 U2 Rw' U, R'].

 

[Ollie]

M' E2 M, U

 

[elrog]

I like these 3 the most. The 1st and 3rd combined do the same as the second.
[R, U' M' U]
[R, U' l' U]
[R, U' L' U]

I also like ones like these for FMC though they are hard to find. I am also aware that M moves count as two for FMC though I think they shouldn't.
[U' L' U2 L2 U', R']
[U' L' U2 L U', R2]
[U' M U, M']

 

[TDM]

[U' Rw U2 r2 U2 Rw' U, R'].

[[U' Rw U2:r2],R']

Can the 4x4 PLL parity alg be written as a commutator?

 

[Christopher Mowla]

Can the 4x4 PLL parity alg be written as a commutator?

If you are specifically talking about the algorithm: r2 U2 r2 Uw2 r2 u2, no. It can be expanded to (r2 U2 r2 U2 r2)(r2 u2 r2 u2), but you probably already know that. However, you could solve that PLL parity case with a single commutator if you wanted to.

You could even solve this or even this with a single commutator using my commutator method, but the commutators created using my method are not going to be something you would use in practice I suspect. In the future, I might be able to make my method create custom commutators and find a way to shorten them, but for now, my method is solely being used to complete the proof of Per's conjecture.

Maybe someone could start a "Request a commutator" thread for which they would ask only for commutators to solve certain positions or to create certain patterns, even if those commutators aren't practical to use in practice.

As I mentioned in the "My Ultimate Commutator Challenge" thread, I can solve any even cube even permutation configuration with a single commutator, and, if I can complete proving middle edges (I've been too busy to spend any time on cubing lately), then the same holds true for odd cubes as well. You could ask for any odd cube size even permutation, and I probably could create a single commutator solution for any position you ask, however (I just didn't technically solve all possible positions yet, but I have the method to solve positions as I need to).