Commutator challenge

[Lucas]

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).

 

[Lucas Garron]

Please correct me if I am wrong anywhere. Thanks.

Let's see.

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.

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.)

 

[TMOY]

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.

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'.

 

[siva.shanmukh]

Please correct me if I am wrong anywhere. Thanks.

Let's see.

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.

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.)

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

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

 

[TMOY]

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).

 

[siva.shanmukh]

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).

Ooo! Okay.. I get it now.

 

[mrCage]

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).

Ooo! Okay.. I get it now.

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

 

[Cangurino]

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 ]

 

[hansho13]

3x3 commutator: R2 B2 R2 B2 R2 B2

very simple

 

[Tim Reynolds]

3x3 commutator: R2 B2 R2 B2 R2 B2

very simple

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

 

[Stefan]

3x3 commutator: R2 B2 R2 B2 R2 B2

very simple

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

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.

 

[Tim Reynolds]

3x3 commutator: R2 B2 R2 B2 R2 B2

very simple

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

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.

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.

 

[Johannes91]

3x3 commutator: R2 B2 R2 B2 R2 B2

very simple

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

Spoiler

[R2 B2 R2, z2 y]

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

 

[Stefan]

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

 

[Swordsman Kirby]

3x3 commutator: R2 B2 R2 B2 R2 B2

very simple

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

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.

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.

 

[siva.shanmukh]

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.

 

[Christopher Mowla]

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.

 

[TMOY]

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.

 

[siva.shanmukh]

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

 

[Christopher Mowla]

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?