# BH Tutorial

#### cmhardw

I like your videos Brian. To be honest, I visualize some of the cases differently than you do, and it was really neat to hear a very clear explanation of a different way to visualize some of the cases. All in all the videos I think did a very good job of explaining the different cases, they look great!

As for your comment on the Per Special vids, I think like anything they get easier with practice. Also, with Daniel's new speed optimized algs (the ones you posted) they are much more finger friendly than the old alg we were using. I only use the optimal algs now, because in my opinion once you can visualize them easily they are much faster than all the other available options. I did like how your video gave several ways to see them, especially since these cases are so rare that effectively you could use a fast sub-optimal alg and still get a really great solve time if it had a Per Special in it.

Chris

#### cmhardw

I'd be interested in hearing how you visualize these commutators Chris
It's just the little things. I guess I hadn't realized how, once I chose a way to practice each particular commutator type, how I stuck to using only that method.

For example, on Columns, I always setup into an A9 with no exceptions. Now I am trying to learn to see the cases that are easier done as setups into cyclic shifts. I knew theoretically that I could do this, but on a solve and while blindfolded I stuck with what I knew and never wavered.

Cyclic Shifts I view exactly like Daniel, as a cyclic shift of a commutator with a 4 move A. I think of it as
A = R F' R' F
B = U2

The only difference is that I view it as starting on the 3rd move, and thus you are "cyclic shifting" the alg. I like how you analyzed how it works as the two conjugates ABA' CB'C'. I knew that this is the structure the alg took after cyclic shifting it, but I like how you describe how each conjugate switches the two corners and also how it has a side effect on the edges which is fixed in the other conjugate. Cool stuff, I never would have though to think of it that way!

The only other example that I can think of where I visualize different is the 8 move case in your video (URB DFL DRF). I do that as:
A = F L F'
B = R2

So we use the same B part of the commutator but I insert to the other corner. I've seen Daniel do this too, but for some reason I've always stuck with inserting to the adjacent corner. I guess I never realized how I do only that now. I'm trying to go back through and learn to "see" the case the way you do.

Lastly, because I always view the cyclic shift as a cyclic shift I tend to execute it wrist turn style, since I am still somewhat thinking of which turns come next. I like your style of just blazing through it with finger tricks (something I do on the Per Specials). I think I will try to "see" those cases as two concatenated conjugates rather than as a cyclic shift so that I can also execute them as quickly as you do.

Just my thoughts after watching your videos. Overall I liked them a lot and thought they were very well done!

Chris

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#### trying-to-speedcube...

##### Member
Yay! After I understood the Per Specials, I think I can say I fully know BH corners, and to prove it, I will do a BH corners time attack this afternoon.

I think edges are easier for me, because I have used those a lot in my FMC insertions. I even think that I might have learned full BH for 3x3 earlier than you, Brian!

The only other example that I can think of where I visualize different is the 8 move case in your video (URB DFL DRF). I do that as:
A = F R F'
B = R2
That can't be right. In fact, it is so weird, that I can't even find out the correct solution.

#### cmhardw

Yay! After I understood the Per Specials, I think I can say I fully know BH corners, and to prove it, I will do a BH corners time attack this afternoon.

I think edges are easier for me, because I have used those a lot in my FMC insertions. I even think that I might have learned full BH for 3x3 earlier than you, Brian!
Congratulations! I think I can honestly say that Daniel and I are just ecstatic that others are starting to like the method, and to want to use it. Of course we love the method, but we're pretty biased haha. Glad that people are finding it helpful/useful for their cubing.

The only other example that I can think of where I visualize different is the 8 move case in your video (URB DFL DRF). I do that as:
A = F R F'
B = R2
That can't be right. In fact, it is so weird, that I can't even find out the correct solution.
Edited my original post, that was a typo. The alg should be:
A = F L F'
B = R2

#### trying-to-speedcube...

##### Member
I'm doing a time attack, and I split it up into 4 parts.

1. (URB UBL/ULF/URF/FUL/FRU xxx)
2. (URB FLD/FDR/RUF/RFD/RDB xxx)
3. (URB BLU/BRD/BLD/LUB/LFU xxx)
4. (URB LBD/LDF/DFL/DRF/DLB/DBR xxx)

I'm timing these seperately.

1. 23:06.61
2. 21:41.68
3.
4.

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#### byu

##### Member
Sorry Maarten, I finished "learning" BH optimally 2 days ago. However, my edge recognition is slow... Corners too. I guess I should update my sig

#### spdcbr

##### Member
Sorry Maarten, I finished "learning" BH optimally 2 days ago. However, my edge recognition is slow... Corners too. I guess I should update my sig
When will you do the notation for my page .

#### Steyler

##### Member
Wow... I can't memorize it all... by the way, this is the guy from swim practice you taught how to solve the rubik's cube ( I don't wanna say my real name). Anyway, this helped with my solving...
Thanks
______________________________________________________________
--Steyler

#### Musturd

##### Member
I don't entirely understand pure commutators.
Correct me if I'm wrong:
If B is first, interchange is what you think (you move the face so that the piece literally moves into it's spot)
If B is second, interchange move is opposite what you think (you move the face so that the piece moves away from it's spot)
If non-interchange has to go to the buffer -- B is first
If non-interchange has to go to other piece -- A is first

I'm not sure about the restrictions on the A moves either. Is it any combination of moves except the face that B is on?
And I don't understand how you find A if you don't know whether B is first or not.
For example, in EXERCISE 2 if you perform A before B the piece does not move to its correct spot, so how did you know to do B before looking for A?

Does anyone understand why I'm having trouble?
My questions seemed to have been overlooked...
Maybe now someone can help me?

#### dbeyer

##### Member
Musturd: You are slightly mistaken.

Pure commutators are optimal as such: ABA'B'
A is a 3 move insertion.
B is an interchanging

Two cubies are interchangeable on a plane or a slice.
URB and ULF are interchangeable on the U plane, by U2.
FLD and RFD are interchangeable on the D slice, by D/D'

When you have interchangeability on a plane, you have insertion points on the parallel slice.
You also have two orbits of insertion points on the slice. So you have the [U, and U']
U, is the inverse orbit of U'
U on the other hand is the U plane.

URB and ULF, R and L bring the cubies into their D,-orbit respectively
URB and ULF, B' and F' bring the cubies into their D'-orbit respectively

And cubie on the D slice, which would be the lone cubie, can be inserted to either the URB or the ULF location.

URB -> ULF -> FDR

FDR is the lone cubie, now we need an insertion point.
We see that by doing a one move setup we can make URB and FDR interchangeable. This move is B'. This means that FDR is on the D'-orbit

When you have interchangeability between the URB and the FDR after that B' setup, you will then interchange the two, and undo the B' -- do a B.
Now the FDR is inserted into the U layer from the RDB. RDB is the insertion point.

So
the part A of a commutator is a 3 moves.
Bring the a cubie from one layer to the parallel layer
Move the lone cubie to the insertion point
Bring the other cube back to the layer, undoing the first move.

A = B'DB in this example.
B = U2
FDR is the lone cubie, which goes to the URB.
B' (the move) sets the interchangeable cubies up to the insertion point (RDB)

URB needs to be brought to the insertion point first. It's already set for the insertion, so you insert first. The interchange. (ABA'B')

URB -> FDR -> ULF
A = B'DB
B = U2
FDR the lone cubie goes to the ULF
ULF needs to be brought to the insertion point first. It's not set above the insertion point, so interchange first, insert next. (BAB'A')

do you see?

So you have an insertion point. RDB is a good insertion point for the URB and ULF pair. No additional setups are required.

So determine where the lone piece goes.
Bring that piece to the insertion point. Insert, restore the layer, interchange, bring the other piece to the insertion point, undo the insertion, restore the layer, interchange, put the pieces back.

Its as simple as ABA'B'
ABA'B' is a little more simplified version, because there are only 8 moves in a pure commutator.

Lets say we chose a different insertion point such as the BDL.
And we were still cycling
URB -> ULF -> FDR

Bring the URB to the insertion point
U'L'
Insert
D2
Restore the Layer
L
Interchange
U2
Bring the other piece to the insertion point
L'
Undo the insertion
D2
Restore the Layer
L
Interchange
U2
Put the pieces back
U

U' L'D2L U2 L'D2L U'
That's way in depth that is an un-optimal execution of an 8 move case.
Just showing to see how you know the A and the B and when to do them.

Later,
DB

#### CubeLord

##### Banned
I kind of get the commutator stuff but why would you need it if the algorithms are already there?

#### puzzlemaster

##### Member
I kind of get the commutator stuff but why would you need it if the algorithms are already there?
Because it's easier to understand and execute commutators from intuition instead of memorizing 800 or so algorithms. hey sreeram

#### Musturd

##### Member
Musturd: You are slightly mistaken.

Pure commutators are optimal as such: ABA'B'
A is a 3 move insertion.
B is an interchanging

Two cubies are interchangeable on a plane or a slice.
URB and ULF are interchangeable on the U plane, by U2.
FLD and RFD are interchangeable on the D slice, by D/D'

When you have interchangeability on a plane, you have insertion points on the parallel slice.
You also have two orbits of insertion points on the slice. So you have the [U, and U']
U, is the inverse orbit of U'
U on the other hand is the U plane.

URB and ULF, R and L bring the cubies into their D,-orbit respectively
URB and ULF, B' and F' bring the cubies into their D'-orbit respectively

And cubie on the D slice, which would be the lone cubie, can be inserted to either the URB or the ULF location.

URB -> ULF -> FDR

FDR is the lone cubie, now we need an insertion point.
We see that by doing a one move setup we can make URB and FDR interchangeable. This move is B'. This means that FDR is on the D'-orbit

When you have interchangeability between the URB and the FDR after that B' setup, you will then interchange the two, and undo the B' -- do a B.
Now the FDR is inserted into the U layer from the RDB. RDB is the insertion point.

So
the part A of a commutator is a 3 moves.
Bring the a cubie from one layer to the parallel layer
Move the lone cubie to the insertion point
Bring the other cube back to the layer, undoing the first move.

A = B'DB in this example.
B = U2
FDR is the lone cubie, which goes to the URB.
B' (the move) sets the interchangeable cubies up to the insertion point (RDB)

URB needs to be brought to the insertion point first. It's already set for the insertion, so you insert first. The interchange. (ABA'B')

URB -> FDR -> ULF
A = B'DB
B = U2
FDR the lone cubie goes to the ULF
ULF needs to be brought to the insertion point first. It's not set above the insertion point, so interchange first, insert next. (BAB'A')

do you see?

So you have an insertion point. RDB is a good insertion point for the URB and ULF pair. No additional setups are required.

So determine where the lone piece goes.
Bring that piece to the insertion point. Insert, restore the layer, interchange, bring the other piece to the insertion point, undo the insertion, restore the layer, interchange, put the pieces back.

Its as simple as ABA'B'
ABA'B' is a little more simplified version, because there are only 8 moves in a pure commutator.

Lets say we chose a different insertion point such as the BDL.
And we were still cycling
URB -> ULF -> FDR

Bring the URB to the insertion point
U'L'
Insert
D2
Restore the Layer
L
Interchange
U2
Bring the other piece to the insertion point
L'
Undo the insertion
D2
Restore the Layer
L
Interchange
U2
Put the pieces back
U

U' L'D2L U2 L'D2L U'
That's way in depth that is an un-optimal execution of an 8 move case.
Just showing to see how you know the A and the B and when to do them.

Later,
DB
Thank you so much.
I'm sorry I haven't responded, but I'm in the middle of finals and if I don't get an 86.5% or better on my physics final or an 87% or better on my math final I DO NOT get an A minus for the year in either of those classes. Both the Math and Physics tests are on tomorrow .
I will work my way through your post Monday night or Tuesday.

#### dbeyer

##### Member
Musturd get that those As in the classes, trust me, I went through the same stuff on after I turned in my history paper. My final, I was frantic to do well. I somehow managed to get a very high A though ... It took me down to the last second to finish, and scramble to write down jibberish and incomplete thoughts on racism in the United States.

#### trying-to-speedcube...

##### Member
Maybe time for the edges? You said you had already learnt them
Also, I finally did a successful 2x2BLD with BH

#### trying-to-speedcube...

##### Member
I don't think it's too different to corners anyway. You have the same case names, the only thing that's really different is the wide setup into a pure commutator. And the 4-movers.

#### Musturd

##### Member
Musturd get that those As in the classes, trust me, I went through the same stuff on after I turned in my history paper. My final, I was frantic to do well. I somehow managed to get a very high A though ... It took me down to the last second to finish, and scramble to write down jibberish and incomplete thoughts on racism in the United States.
Nice

I finished my finals. Hopefully I did well . I do have SAT2s on Saturday, but I just took practice math and physics tests and got 790 and 760 respectively (this is without prestudying), so I think I am all set for today.

I understand your explanation of that particular commutator (or at least I think I do), but I still can't figure out any pures from byu's first post. Something isn't clicking in my mind.