Commutator "?": cycling edges&corners around the cube

[Lotsofsloths]

So how do you do this?
I know how-to cycle 3 corners/edges around the cube on the top layer without disturbing orientation. And I know set-up moves, but I want a much more clean way of doing that, does anyone(Chris Hardwick ) know how to do this?
Please explain how-to for both edges and corners.

Thanks in advance,

 

[fanwuq]

Can you clarify?
Isn't this what we usually do for BLD?

 

[joey]

You havn't really explained what you are looking for?

Do you mean corner AND edge cycles at the same time?

 

[Lotsofsloths]

@fanwuq:
Some people use commutators in BLD, but I don't.

@Joey:
I am asking 2 different "?"s,
1. how to cycle corners around the cube.
2. How to cycle edges around the cube.

 

[alexc]

Wait, you don't know how to use commutators?! It's a pretty important aspect of cubing that every cuber should know something about. I'll give you a tutorial:

Ok a commutator is simply an alg in the form ABA'B'. Typically part A is 3 moves and part B is one move. Take a look at this commutator:

A: R U R'
B: D2

(Note: You would first execute A, then B, then the inverse of A, and finally the inverse of B.)

You can see that this cycles three corners.

An edge example:

A: F E F'
B: U2

You can see that this one cycles three edges.

Now, you may be asking, "This is all very good, but how do I use a commutator to do something useful in solving the cube?" I will explain below:

Say you need to solve the cube by cycling RDF UFR and UBL. Here's how you would solve it with a comm. The most important thing in commutators is to have two pieces in the cycle you need the solve to be interchangeable. That means by doing a single turn you can make one piece occupy the spot of another one. Look at the example, can you find the two interchangeable pieces and what move you would use to interchange them?

I see that currently you're online, so I'll let you guess then tell you the answer in the next post.

 

[fanwuq]

The U and A perms I use are commutators. I could to the commutator to orient Corners and edges if I want. So I do use commutators in BLD.

You use M2 right? I don't really know much about it, but I know there are conjugates. ABA'. Setup, M2, reverse setup.

 

[Lotsofsloths]

Wait, you don't know how to use commutators?! It's a pretty important aspect of cubing that every cuber should know something about. I'll give you a tutorial:

Ok a commutator is simply an alg in the form ABA'B'. Typically part A is 3 moves and part B is one move. Take a look at this commutator:

A: R U R'
B: D2

(Note: You would first execute A, then B, then the inverse of A, and finally the inverse of B.)

You can see that this cycles three corners.

An edge example:

A: F E F'
B: U2

You can see that this one cycles three edges.

Now, you may be asking, "This is all very good, but how do I use a commutator to do something useful in solving the cube?" I will explain below:

Say you need to solve the cube by cycling RDF UFR and UBL. Here's how you would solve it with a comm. The most important thing in commutators is to have two pieces in the cycle you need the solve to be interchangeable. That means by doing a single turn you can make one piece occupy the spot of another one. Look at the example, can you find the two interchangeable pieces and what move you would use to interchange them?

I see that currently you're online, so I'll let you guess then tell you the answer in the next post.

LOL.
Sorry, I do know commutators just not a direct solution to that one you said in the last paragraph.
SO to solve that though I would do this:
(B' D B)(R' D2 R U2 R' D2 R U' R' D2 R U' R' D2 R)(B' D' B)

 

[Jason Baum]

For that last commutator mentioned in Alex's post (RDF-UFR-UBL), there are a couple of options. RDF and UFR are interchangeable by doing R, but UFR and UBL are also interchangeable by doing U2. That is probably the easier option here. So now, we need to get RDF (the "lone edge") to UFR without disturbing any other part of the interchangeable layer, U. To do that, simply do D' R' D R (part A). Now, do U2 to interchange (part B). Then undo parts A and B: R' D' R D U2. So the whole commutator becomes D' R' D R U2 R' D' R D U2.

 

[Lotsofsloths]

Wow!
That is...simple!
I love commutators..they are so brain working?
Yea they make you think!
Thanks Jason!

 

[Lucas Garron]

For that last commutator mentioned in Alex's post (RDF-UFR-UBL), there are a couple of options. RDF and UFR are interchangeable by doing R, but UFR and UBL are also interchangeable by doing U2. That is probably the easier option here. So now, we need to get RDF (the "lone edge") to UFR without disturbing any other part of the interchangeable layer, U. To do that, simply do D' R' D R (part A). Now, do U2 to interchange (part B). Then undo parts A and B: R' D' R D U2. So the whole commutator becomes D' R' D R U2 R' D' R D U2.

Not FDR. ;-)
But you'd probably find [ R'D'R , U2 ].

EDIT: Probably shoulda mentioned [ U2 : B'D2B ].