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[video] Permutation of last layer without algorithms

circular

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Hello, here is a video showing how to permute the last layer using only moves that you can understand. So no algorithm to learn, just pure logic. If you have any comment, or suggestion on how to make it even better, feel free to propose any improvement.

The general explanation for 3-cycle is at 4:25. The first example on rotating corners is just a way of applying this principle to corners.


EDIT: More explanation on how to create algorithms using commutators :
 
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circular

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Well by algorithm that I mean no "sequence of move that you don't understand what is going on you just apply them in some case and you got the result".

What I am doing here, I think you can understand everything that is going on. Basically it's about using commutators. But I show the result so you don't have to figure it out.
 

cubernya

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Well by algorithm that I mean no "sequence of move that you don't understand what is going on you just apply them in some case and you got the result".

What I am doing here, I think you can understand everything that is going on. Basically it's about using commutators. But I show the result so you don't have to figure it out.

I understand everything that goes on in my PLL algs (more so than your explanation in this video anyway)
 

circular

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Well if it's not useful to you in particular, that's ok for me. But if you're so smart, you should have understood what I am showing here...
 

cubernya

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Well if it's not useful to you in particular, that's ok for me. But if you're so smart, you should have understood what I am showing here...

I understand commutators very well, but the way you explained it made no sense at all. If somebody that doesn't know commutators understood that, I need to figure out how :p
 

Rpotts

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Your A perm and H perm are longer, slightly more complicated versions of already very intuitive commutators/conjugates. The E perm was cool though.

Also, your second U perm and J perm /are/ the optimal algs for the case, you're just explaining them, which is good. I might suggest starting with pure block commutators to give viewers who don't know about them a better grasp of them before moving on to 'A9' equivalents.
 
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circular

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Your A perm and H perm are longer, slightly more complicated versions of already very intuitive commutators/conjugates.
Which sequence would you propose instead ?

I find the H perm in this video very intuitive, and it's 8 moves long, whereas the optimal one is 7 moves. So I do not agree.

The E perm was cool though.
Thanks. I noted that in fact it can be optimised by avoiding the U2 and rotating the whole algorithm instead. So then it becomes optimal (14 moves).

Also, your second U perm and J perm /are/ the optimal algs for the case
The U perm is optimal, but not the J perm, which is 11 moves whereas the optimal is 10 moves (F2 D F D' F L2 B' U B L2)

you're just explaining them, which is good.
Thanks.

I might suggest starting with pure block commutators to give viewers who don't know about them a better grasp of them before moving on to 'A9' equivalents.
Well I wanted to avoid complex vocabulary so that someone who just don't know anything would understand what I am talking about.

I don't understand what you mean by A9.
 

Rpotts

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Which sequence would you propose instead ?
H perm - M2 U M2 U2 M2 U M2 which can be notated in commutator notation as [M2 U: [M2, U2]]

A perm - R2 D2 R U R' D2 R U' R which can be notated as [R': [R' D2 R, U]]

I find the H perm in this video very intuitive, and it's 8 moves long, whereas the optimal one is 7 moves. So I do not agree.

The H perm in the video is intuitive, but so is the standard alg for the case. Might as well teach how to understand the better one, shorter SQTM one.

The U perm is optimal, but not the J perm, which is 11 moves whereas the optimal is 10 moves (F2 D F D' F L2 B' U B L2)

That J perm is optimal, 10 moves is optimal +1 AUF.

EDIT: My mistake, the alg you posted is 10 moves from the common 'solved' AUF, didn't know that.

I don't understand what you mean by A9.

A9 is a type of commutator that is a pure, 8-mover that is conjugated by one move, which cancels into the commutator at either the beginning or end. A perm is an example of an A9.
 
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Nico1

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I understand commutators very well, but the way you explained it made no sense at all. If somebody that doesn't know commutators understood that, I need to figure out how :p

I don't understand commutators (need to find a good video on that...) and I didn't understand anything in this video except that he was performing pre-memorized algorithms.

@circular, maybe you could edit your videos to teach how to make an algorithm yourself to solve pll instead of explaining the algorithms that you have memorized (that would help me a lot as there aren't many videos on this!)
 

circular

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H perm - M2 U M2 U2 M2 U M2 which can be notated in commutator notation as [M2 U: [M2, U2]]

A perm - R2 D2 R U R' D2 R U' R which can be notated as [R': [R' D2 R, U]]
Euh, how would you explain them?

A9 is a type of commutator that is a pure, 8-mover that is conjugated by one move, which cancels into the commutator at either the beginning or end. A perm is an example of an A9.
Ok, thanks.

I don't understand commutators (need to find a good video on that...) and I didn't understand anything in this video except that he was performing pre-memorized algorithms.

@circular, maybe you could edit your videos to teach how to make an algorithm yourself to solve pll instead of explaining the algorithms that you have memorized (that would help me a lot as there aren't many videos on this!)
Didn't you see in the video? I explain the 3-cycle at 4:24.

I didn't watch the video yet, but here is an intuitive 2-edge and 2-corner swapper I made without using any methods, commutators, conjugates, etc.

B L U' F U L' B' U R U R' U2

Just conjugate with U2 to get a J-Perm.
How would you explain it?
 

Christopher Mowla

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A perm - R2 D2 R U R' D2 R U' R which can be notated as [R': [R' D2 R, U]]
Euh, how would you explain them?

Didn't you see in the video? I explain the 3-cycle at 4:24.
You didn't really explain that the affected corners and edges in the bottom would be undone when you repeat the swaps an even number of times, though that is how your 3-cycle algorithm for swapping corners was: L' R' D2 L R U' R' U L' D2 L U' R U. I'm not criticizing you at all: your video was absolutely beautiful, but I felt it was only "telling one side of the story." But the key thing you mentioned was "the buffer."

In the A-perm Rpotts showed, look at the cube after R' D2 R is executed: the whole purpose was to separate a corner in UFR for which the turn U replaces that corner with another in the U-Layer. Then you undo R' D2 R with R' D2 R, and then undo U with U'. So now you have 3 corners swapped.

R' D2 R
U
R' D2 R
U'


They are not all in the same layer, but we can make it so by adding a setup move, just as you did for your 3-cycle of 3 edges to be placed in the U-Layer. Same idea.

I didn't watch the video yet, but here is an intuitive 2-edge and 2-corner swapper I made without using any methods, commutators, conjugates, etc.

B L U' F U L' B' U R U R' U2

Just conjugate with U2 to get a J-Perm.
How would you explain it?
See here. Take note that this algorithm does consist of a commutator and two conjugates, but that's not how I made it (as I have already explained) = [B L U': F] [U: [R, U] ].

Also, I don't know why you didn't use your idea to at least do a 3-cycle of corners and edges at the same time:

M2 U M2 U' M2
L R d2 L' U' L d2 L' U R'
M2 U M2 U' M2


Again, great job on the video. I too love simplicity, and you expressed it very well by using the same type of commutator, but you didn't even mention why you must place edges and corners facing a certain way before you exchange. It happens to work out, but you still didn't explain why you get PLLs from the moves you chose. But keep up the good work.
 
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circular

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In the A-perm Rpotts showed, look at the cube after R' D2 R is executed: the whole purpose was to separate a corner in UFR for which the turn U replaces that corner with another in the U-Layer. Then you undo R' D2 R with R' D2 R, and then undo U with U'. So now you have 3 corners swapped.

R' D2 R
U
R' D2 R
U'


They are not all in the same layer, but we can make it so by adding a setup move, just as you did for your 3-cycle of 3 edges to be placed in the U-Layer. Same idea.
Ok I got it. The setup move is merged with the commutator, and it moves the two corners instead of moving the buffer corner.

I still do not understand your move. But I kinda understand the principle you've used.

Also, I don't know why you didn't use your idea to at least do a 3-cycle of corners and edges at the same time:

M2 U M2 U' M2
L R d2 L' U' L d2 L' U R'
M2 U M2 U' M2
That's interesting. I didn't think about it. Is it really shorter than combining with 3-cycle of edges ?
 
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