Commutator challenge

[mrCage]

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:rolleyes:

Per

Correction:

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

 

[DavidWoner]

This one's pretty useful (highlight)

[R U R2, R U2 R2]

 

[trying-to-speedcube...]

Sexy move!

 

[Lucas Garron]

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

 

[LNZ]

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.

 

[Am1n-]

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

mvg

 

[mrCage]

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

 

[Am1n-]

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

 

[mrCage]

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

 

[trying-to-speedcube...]

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.

 

[darthyody]

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

 

[trying-to-speedcube...]

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

 

[Am1n-]

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

off-topic: whiiiii, my first personal facepalm


mvg

 

[trying-to-speedcube...]

My pleasure

Nice find btw

 

[Am1n-]

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

mvg

 

[Lucas Garron]

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

Another alg that achieves the same result (though I honestly can't say is faster to execute) is:

R2 D2 L2 U2 L2 D2

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.]

 

[Lucas Garron]

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

Another alg that achieves the same result (though I honestly can't say is faster to execute) is:

R2 D2 L2 U2 L2 D2

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.]

Nothing takes time from expanding your knowledge like doing your homework and applying to college.:fp

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

 

[siva.shanmukh]

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.

 

[Lucas Garron]

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.

 

[siva.shanmukh]

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.

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.