BH Tutorial

[JonnyWhoopes]

I ment URB DFL LFU

Kay. I prefer [F R' F', L'].

 

[rock1313]

Kay. I prefer [F R' F', L'].

that solves URB FUL DFL

 

[JonnyWhoopes]

that solves URB FUL DFL

Whoops. [L', F R' F']

Nice catch. Gotta love typos.

::EDIT:: Dangit, I'm still solving your old cycle anyway. Ignore me. Lulz.

 

[aronpm]

The insertion is in the form D C D' and the interchange is in the form A. A has to be parallel to C, they can't change the same pieces.

 

[rock1313]

What does the D C D' mean.

 

[JonnyWhoopes]

What does the D C D' mean.

It stands for the form that your three move insertion should take.

 

[Joël]

With the commutator URB DFL FUL

it doesn't work with D R2 D' F D R2 D' F'

but it works with D' B2 D F' D' B2 D F

why?

I think I see what you are trying to do here. First of all, the first alg does have the structure of a commutator, so way to go! However, this commutator does not have the structure of a corner three cycle. The reason is that F 'touches' many pieces that are also moved by the D R2 D' part. For a corner three cycle, if you take P = D R2 D', you want Q to turn the face opposite of the middle move in P, in this case Q would be L, L', L2. If you take Q = L, it will cycle the same pieces, but with different orientations: D R2 D' L D R2 D' L' cycles UBR -> LFD -> LUF.

In order to cycle URB DLF LUF (not FLU), you have to notice that the the DFL piece is solved by doing F. You want F to be your last move, but first you have to move that piece out of the way without effecting the F layer. You only want to replace that piece with the URB corner. P = D' B2 D does exactly that, the only piece changed in the F layer is the DLF corner. For Q you take the inverse of what the last move is going to be, in this case Q = F'. Then finish the commutator by undoing P and Q: D B2 D' F.

 

[JonnyWhoopes]

In order to cycle URB DLF LUF (not FLU, that's not possible)

Don't the cycles I posted solve those?

::EDIT:: To clarify, I meant don't they solve URB DLF FLU?

 

[Joël]

Don't the cycles I posted solve those?

::EDIT:: To clarify, I meant don't they solve URB DLF FLU?

Oh yeah, whoops. My mistake.

 

[Julian]

I'm partway through the OP, and was just wondering, during a blindsolve, for an A9-type case, is it necessary to find the setup the cancels? Maybe I'm missing something, but it seems that you have to think for a while to find the right setup, just to save one move?

 

[aronpm]

I'm partway through the OP, and was just wondering, during a blindsolve, for an A9-type case, is it necessary to find the setup the cancels? Maybe I'm missing something, but it seems that you have to think for a while to find the right setup, just to save one move?

It's not necessary, you'll just be doing 1 more move. When you are more experienced with A9s and start to recognize the patterns between them you'll see how the 9mover works and you can do that pretty much instantly instead.

 

[Julian]

It's not necessary, you'll just be doing 1 more move. When you are more experienced with A9s and start to recognize the patterns between them you'll see how the 9mover works and you can do that pretty much instantly instead.

Thanks, thought so.

 

[cmhardw]

I'm partway through the OP, and was just wondering, during a blindsolve, for an A9-type case, is it necessary to find the setup the cancels? Maybe I'm missing something, but it seems that you have to think for a while to find the right setup, just to save one move?

You do at first, but eventually the (setup + first part of the A) move makes intuitive sense as one turn. Now I don't think of it as Setup THEN the first part of the A. I just know that a double turn is required at exactly this point in the commutator. Yes it's more work at first, but you learn to instantly see the setup move that causes the cancellation with enough practice.

 

[RyanReese09]

MrMoney and I had a skype session together and my understanding of BH is pretty firm now thank to him. Not to put him on the spot, but I'd highly recommend talking to him and seeing if he would be up for the idea of doing a skype session with you.

Just for those who are strugging and would like some "IRL" BH examples.

 

[rock1313]

I think I see what you are trying to do here. First of all, the first alg does have the structure of a commutator, so way to go! However, this commutator does not have the structure of a corner three cycle. The reason is that F 'touches' many pieces that are also moved by the D R2 D' part. For a corner three cycle, if you take P = D R2 D', you want Q to turn the face opposite of the middle move in P, in this case Q would be L, L', L2. If you take Q = L, it will cycle the same pieces, but with different orientations: D R2 D' L D R2 D' L' cycles UBR -> LFD -> LUF.

In order to cycle URB DLF LUF (not FLU), you have to notice that the the DFL piece is solved by doing F. You want F to be your last move, but first you have to move that piece out of the way without effecting the F layer. You only want to replace that piece with the URB corner. P = D' B2 D does exactly that, the only piece changed in the F layer is the DLF corner. For Q you take the inverse of what the last move is going to be, in this case Q = F'. Then finish the commutator by undoing P and Q: D B2 D' F.

So what your saying is that with the one move insertion, it can move any edges that got affected by the 3 move insertion.

 

[RyanReese09]

I think I see what you are trying to do here. First of all, the first alg does have the structure of a commutator, so way to go! However, this commutator does not have the structure of a corner three cycle. The reason is that F 'touches' many pieces that are also moved by the D R2 D' part. For a corner three cycle, if you take P = D R2 D', you want Q to turn the face opposite of the middle move in P, in this case Q would be L, L', L2. If you take Q = L, it will cycle the same pieces, but with different orientations: D R2 D' L D R2 D' L' cycles UBR -> LFD -> LUF.

In order to cycle URB DLF LUF (not FLU), you have to notice that the the DFL piece is solved by doing F. You want F to be your last move, but first you have to move that piece out of the way without effecting the F layer. You only want to replace that piece with the URB corner. P = D' B2 D does exactly that, the only piece changed in the F layer is the DLF corner. For Q you take the inverse of what the last move is going to be, in this case Q = F'. Then finish the commutator by undoing P and Q: D B2 D' F.

Just saw this.

This is, I believe, the only thing that I recognized that was holding me back from finding 3 move insertions (interchanges=no problem). Did not know about this. Perfect . Once I get A9 cancellations down I'm good to attempt BH in solves..

 

[MrMoney]

I am up for guiding anyone through pures, A9 and other types whenever. Just contact me here and we can discuss the time. I am glad you understood something of the session, Ryan

 

[Kynit]

Once I get A9 cancellations down I'm good to attempt BH in solves..

You shouldn't even need to get too good at the cancellations - a regular setup move should suffice. One extra move in your solve is going to be better than a 3 second pause to figure out the next comm.

 

[Simboubou]

Well, your main concern is speed. So if you lose even half a second just to figure out how you can have a cancelation move, it's too long. Not to mention one move saved does not implies a faster execution.

 

[RyanReese09]

I just realized something at work today trying to think about A9's and practicing them. Take this A9 I just randomly made. Off the top of my head I dunno what cycle it solves, but ..yeah.
F' L' B2 L F2 L' B2 L F2 F

Assuming you do ABA'B' as in the example above (insert then interchange) the cancellation move has to be a move on the face that of the interchange, and in this case, the interchange is F2, so that means the setup move (assuming A9 case) has to be F, F' or F2. And then say we did F' as the setup, then I would know that F' + F2 = F so at the end of the cycle I'd just do an F, so far so good?

Could I potentially pick any of the F/F'/F2 for the setup? I don't believe I could do F2 because that would be redundancy I think due to the fact that it is my setup, so assuming the setup move can't be the same as the interchange, I'd have F or F' as my potential setup. 2 choices, can A9's be done with either? Or do I have to find the setup that brings it into the same plane as the beginning of A, Aka F' brings the piece to the L face thus able to start the insertion.

I'm sorta rambling on now, but say I know the setup is L' B2 L, And the interchange is F2, the setup has to be F or F', so I pick the one that will bring the "spot of action" to the L plane.

Is my logic correct in its' entirety here?