The big cubes commutators?

[alkanova]

Hi,
I've been looking for help in commutators in the big cubes like the 5×5 ,4×4 etc.
but I couldn't find anyone presenting any explanation and help.can anyone help plz?(especially with the parities like the single edge flipped in 5×5 etc.)
thanks in advance

 

[Stefan]

Well, "single edge flipped in 5×5" is first of all impossible, and what you probably really mean is impossible with commutators.

And one nice thing about commutators and parity is that if you really understand these concepts, you can apply them to many puzzles. So commutator/parity tutorials for other puzzles might be helpful for you as well.

 

[alkanova]

what I meant was a single edge flipped left when paring the edges,srry if I wasn't clear,anyway,it was jst an example I don't face it but I face others .
The problem is that I've already read about commutators in 3×3 but when I go to 5×5 or 4×4(and obviously when I will go to 6×6 and 7×7) I get stuck not being able to grasp the concept ,so I was looking for some explanations but all that I could find is examples on 3×3.

Edit:and these things are only applicable once I have made the cube like a 3×3 .what I am looking for are things before making it like a 3*3 or when failing to make it exactly like a 3*3

 

[Stefan]

what I meant was a single edge flipped left when paring the edges

And that's impossible. What you probably mean is that the two wing edges around one middle edge need to be swapped, and that's impossible to fix only with commutators.

and these things are only applicable once I have made the cube like a 3×3.

Wrong. Very wrong.

 

[AvGalen]

You could try doing the last 2 centers on a 6x6x6 and 7x7x7. Without realising it you will find commutators for doing that yourself

 

[Stefan]

Nah, you can solve those centers without commutators (for example like I do, reduce to 5x5x5 centers). Plus you can find commutators without understanding the concept, so it doesn't necessarily help.

 

[kratos94]

Well, how would you do a commutator for permutation parity on a 4x4x4?

 

[Stefan]

Well, how would you do a commutator for permutation parity on a 4x4x4?

Who?

 

[AvGalen]

Nah, you can solve those centers without commutators (for example like I do, reduce to 5x5x5 centers). Plus you can find commutators without understanding the concept, so it doesn't necessarily help.

....but it probably would

Well, how would you do a commutator for permutation parity on a 4x4x4?

I simply wouldn't. I would learn/find one and perform it without caring if it was a commutator.

Not to confuse the topicstarter, but for edge-pairing I use this technique for "flipped edge", like the last two unpaired edges on 4x4x4
d (match 1)
R U' R' (store it in a safe place)
F' U2 F (restore it from the safe place without affecting any pieces I am interested in at that moment)
d' (match again)
Or this variation
d2 (R U' R') (F' U2 F) d2'
This isn't strictly a commutator because it doesn't follow the A B A' B' rule.
It isn't a conjugate either because the setup/undo-move influences the pieces that I am interested in solving.
Is there a name for this type of construction?

 

[tim]

Well, how would you do a commutator for permutation parity on a 4x4x4?

2 3-cycles with setup moves.

 

[Stefan]

Arnaud, that *is* a conjugate, both according to the definition of the term as well as to what you probably mean, i.e., setup-DoSomethingUsefulOnItsOwn-setdown. Note that your (R U' R') (F' U2 F) flips the FR dedge, or in other words, swaps the FRu and FRd edges. That's something useful on its own. And that's also the explanation I prefer, I like that better than your store+restore story.

 

[Stefan]

2 3-cycles with setup moves.

Meh, you only need to move two pieces, not 23.

 

[tim]

2 3-cycles with setup moves.

Meh, you only need to move two pieces, not 23.

For PLL parity? Don't you have to move 4 pieces?

 

[TMOY]

Here is a commutator which performs PLL parity: [L R u2 L' R' u2, U' r l U]
Of course it is definitely not the fastest way to do it, but this one also preserves centers on a supercube.

 

[deco122392]

2 3-cycles with setup moves.

Meh, you only need to move two pieces, not 23.

23? there was a space between the 2 & the 3. edit: just noticed his choice of casual and relaxed words like 'meh' and 'nah', is Stefan having a good day?

 

[Stefan]

For PLL parity? Don't you have to move 4 pieces?

Was thinking of swapping two corners. However, I just noticed you're right, as it was your answer for the question for doing it with commutators.

23? there was a space between the 2 & the 3.

I know, but my point is that it's still bad to read. You'll never see me writing like that, I would've written "Two 3-cycles".

 

[tim]

23? there was a space between the 2 & the 3.

I know, but my point is that it's still bad to read. You'll never see me writing like that, I would've written "Two 3-cycles".

I prefer "200 3-cycles" over "two hundred 3-cycles" .
I think it's ok to use words for stories like "Two dogs walked over the street." But in mathematically related topics i prefer to read the actual numbers.

 

[deco122392]

23? there was a space between the 2 & the 3.

I know, but my point is that it's still bad to read. You'll never see me writing like that, I would've written "Two 3-cycles".[/QUOTE]
oh.. well that does make scence. i apoligize. & with pll parity or at least with the r2 U2 r2 U2 r2 u2 U2 alg you also cycles the top four centers

 

[kratos94]

Well, how would you do a commutator for permutation parity on a 4x4x4?

Who?

Oh I'm sorry Stefan, I thought that most people would be able to understand the usage of the universal "you".

 

[AvGalen]

Arnaud, that *is* a conjugate, both according to the definition of the term as well as to what you probably mean, i.e., setup-DoSomethingUsefulOnItsOwn-setdown. Note that your (R U' R') (F' U2 F) flips the FR dedge, or in other words, swaps the FRu and FRd edges. That's something useful on its own. And that's also the explanation I prefer, I like that better than your store+restore story.

I always thought that a conjugate (like a commutator) wasn't allowed to change other pieces. That is why I mentioned "without affecting any pieces I am interested in at that moment".
So F (R U R' U') F' would be a conjugate as well.
Actually, inserting a pair like R (U) R' would be considered a conjugate?