BH Tutorial

[cmhardw]

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

This is actually a B9. What A9 refers to is that the setup move cancels with the A part, the insertion, of the standard 8 move commutator. B9 means that the setup move cancels with the B part, the interchange, of the commutator. Your alg has the undoing of the F' setup move canceling with the B part of the commutator at the end, so a B9 alg.

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.

It's possible that you may have multiple options for either an A9 alg or a B9 alg. In this case, though, remember that for corners specifically no B9 alg is optimal. Every corner B9 case can also be done in 8 moves. As to your question in a more general case, then if you have multiple options that each do a 9 turn alg, then both options are viable B9's.

Take the two ways to do: UBL -> UBR -> BLD
1) U2 R' D2 R U' R' D2 R U'
and
2) U' F' D F U' F' D' F U2

Both of these algs are B9's, and both of them solve the corner case. My point before was that neither of these is optimal. Optimal for this case is:
B D B' U' B D' B' U

And before people get upset about how crappy that alg is, of course you would rotate the cube to a more comfortable angle first.

To be honest, in an actual speedsolve, I do use the 1st B9 alg I listed for this case. I consider it to be faster than the optimal alg, personally.

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?

It is, but again it's for B9's. If you apply this same style logic to BH edges, then you can come up with some useful B9 algs (which can be optimal by the way). A B9 is never necessary for corners, but they can make good, fast options for certain cases.

 

[RyanReese09]

This is actually a B9. What A9 refers to is that the setup move cancels with the A part, the insertion, of the standard 8 move commutator. B9 means that the setup move cancels with the B part, the interchange, of the commutator. Your alg has the undoing of the F' setup move canceling with the B part of the commutator at the end, so a B9 alg.

GRR. Ok, so my alg where I do the interchange first, then insertion is a B9, which is sorta like an A9, but whatever

It's possible that you may have multiple options for either an A9 alg or a B9 alg. In this case, though, remember that for corners specifically no B9 alg is optimal. Every corner B9 case can also be done in 8 moves. As to your question in a more general case, then if you have multiple options that each do a 9 turn alg, then both options are viable B9's.

Take the two ways to do: UBL -> UBR -> BLD
1) U2 R' D2 R U' R' D2 R U'
and
2) U' F' D F U' F' D' F U2

Both of these algs are B9's, and both of them solve the corner case. My point before was that neither of these is optimal. Optimal for this case is:
B D B' U' B D' B' U

And before people get upset about how crappy that alg is, of course you would rotate the cube to a more comfortable angle first.

To be honest, in an actual speedsolve, I do use the 1st B9 alg I listed for this case. I consider it to be faster than the optimal alg, personally.



It is, but again it's for B9's. If you apply this same style logic to BH edges, then you can come up with some useful B9 algs (which can be optimal by the way). A B9 is never necessary for corners, but they can make good, fast options for certain cases.

Ah, so basically all that thinking I did today was put to shame as I wasn't even contemplating A9s, but rather B9s. Never heard of them before, good to know. I never tried solving that case in 8 moves, I'll go try it out.
The case is URB to DRF to BDL..hmm .. [U L U', R2]
Wee..
So basically to convert my logic to A9, I just have to do something like..take this
F2 L' B2 L F2 L' B2 L

If I just wanted to make that into an A9 I'd have to take the beginning of the insertion (aka L slice in this case) so a setup on the L slice, do the alg, and then it'll cancel. So the first move of the insertion will contain the slice of which I'll need my setup? So since my logic was proven correct by Chris, er, well, holdup.

I use L' and L in this setup, so..for A9, can my setup be anything BUT the final move? Look above at what is bolded. It's the final move of the insertion. L. If I had L' as my setup then I would have no need for a final move when I finished the commutator. So my logic is, I can't have my setup move be the final move of the insertion. Aka for this A9, L is an invalid setup move, so only L' and L2 is a valid setup move (and the only two setup moves for an A9 in this case since the setup moev has to be on the same slice as the beginning of the insert.

 

[cmhardw]

So basically to convert my logic to A9, I just have to do something like..take this
F2 L' B2 L F2 L' B2 L

If I just wanted to make that into an A9 I'd have to take the beginning of the insertion (aka L slice in this case) so a setup on the L slice, do the alg, and then it'll cancel. So the first move of the insertion will contain the slice of which I'll need my setup? So since my logic was proven correct by Chris, er, well, holdup.

Yes, exactly. Every A9 is done this way.

I use L' and L in this setup, so..for A9, can my setup be anything BUT the final move? Look above at what is bolded. It's the final move of the insertion. L. If I had L' as my setup then I would have no need for a final move when I finished the commutator. So my logic is, I can't have my setup move be the final move of the insertion. Aka for this A9, L is an invalid setup move, so only L' and L2 is a valid setup move (and the only two setup moves for an A9 in this case since the setup moev has to be on the same slice as the beginning of the insert.

Yes, it sounds like you definitely have the idea. Also, notice that if you do make your setup turn the same as the last turn of the commutator, you just make a different 8 move commutator.

L F2 L' B2 L F2 L' B2 L L'
L F2 L' B2 L F2 L' B2
L F2 L' B2 L F2 L' B2

And notice that the new commutator has B2 as the interchange move and L F2 L' as the insertion.

 

[RyanReese09]

Alright so when it's not an orthoganol/cyclic etc and it's either an 8 or 9 move commutator, I know it's an A9 if there is no 3 move insertion that gets the pieces in the correct orientation yes?

Yes, it sounds like you definitely have the idea. Also, notice that if you do make your setup turn the same as the last turn of the commutator, you just make a different 8 move commutator.

That was my idea I had, if I made my setup the same turn as the last move of the commutator then it'd just be an 8 mover, which would not be an A9

So really, there are only two possible setup moves for A9 if you go off what slice the setup move can be on.

My final question is, how can I always know what setup move to do? Aka in that example I had F' and F2 as a possible setup. Is there perhaps a rule of thumb?

 

[aronpm]

Something that I wrote a while ago that might be useful to someone:

Finding a cancellation for these 9-move commutators is tricky. When you look at the cycle you'll see that there is (probably; the A perm is an exception) a Non-interchangeable Adjacent pair (AnI pair). Basically an AnI pair means that the two stickers cannot be interchanged with a 3-move insertion. The basic strategy for finding the setup is to break the AnI pair. There are only two possible faces to turn to do this, so there are only two options: one moves the third piece away while breaking the AnI pair and the other doesn't.

Also:

One additional thing to note is that when the AnI pair is broken with a half-turn, the part of the AnI that was moved must become interchangeable with the third piece. Because there are two options for breaking the AnI pair, you may have to try both possibilities. It's best to try the option that doesn't affect the third piece because this is more common. Also, the insertion, A, is in the form CDC', where D is parallel to the B interchange move. Once you know which half turns makes the piece interchangeable with the third piece, and you know what layer to turn to make that interchange, you can easily work out the insertion and solve the 9-move commutator.

 

[cmhardw]

Alright so when it's not an orthoganol/cyclic etc and it's either an 8 or 9 move commutator, I know it's an A9 if there is no 3 move insertion that gets the pieces in the correct orientation yes?

Thinking about it, yeah I think that works. If you can narrow out that it's not a 10, 11, or 12 mover, and no workable 3 move insertion exists, then yes it's an A9. This is not necessarily the case for edges, because edges have both A9's and B9's.

So really, there are only two possible setup moves for A9 if you go off what slice the setup move can be on.

My final question is, how can I always know what setup move to do? Aka in that example I had F' and F2 as a possible setup. Is there perhaps a rule of thumb?

Actually, it doesn't always work that both setup moves are options for an A9.

Using your line of reasoning let's start with the 8 mover case: ULB -> URF -> LFD
U2 R' D R U2 R' D' R

Now, since the alg ends with R, I don't want to create an A9 by doing a R setup. This would cancel and create another 8 move commutator as you put before. My options for what would cancel with the last R are to start with a setup move with either R' or R2.

Let's look at each case. Setup with R' :
R' U2 R' D R U2 R' D' R R
R' U2 R' D R U2 R' D' R2

This is definitely a case that is optimal as an A9. Now let's look at the other case.

Setup with R2 :
R2 U2 R' D R U2 R' D' R R2
R2 U2 R' D R U2 R' D' R'

Now, although this alg is an A9 alg, this case is not an A9 case. The case that this alg solves is actually optimal at 8 moves. The following alg is equivalent to the one we just discovered:

U' F U B2 U' F' U B2

This would seem to suggest that for A9 cases, only one of your setup move options would work. I'm not sure if this is actually the case, but I will try to look into it.

------------

I would suggest not to think of A9's this way. Let's take the case of: R' U2 R' D R U2 R' D' R2
This solves the case: UBL -> BRU -> LFD

What I would notice here, and the way Daniel and I think of A9's, is the following. I'm searching for a double turn that will move exactly one piece, and it will be interchangeable with both of the other pieces in the cycle through this double turn. By this I mean if I do the quarter turns in the proper direction, then after each quarter turn the corner I am moving will be interchangeable with one of the other corners.

Ok, so now onto the example. For the case ULB -> BRU -> LFD notice that the BRU corner is not interchangeable with either of the other two corners. If I do the turn (R')2 I will I can make the BRU corner interchangeable with each of the other corners.

After the first R' I notice that BRU ends up at URF. This makes it interchangeable with UBL via a U2. As I do the second R' the original BRU corner ends up at FRD. This makes it interchangeable with LFD via a D layer turn. So that means that R2 is the move that I need to use after the cancellation. Now I need to search for the setup move, at the proper point in the commutator, that will cancel to an R2 move. The setup move that does this is R'.

Now, again, this sounds much more complicated than it actually is. What I really do is do the proper setup turn such that I can send the corner that is supposed to go to BRU to it's new spot.

Again an example. The piece that is supposed to go to BRU is UBL. So that means I start with R', since after doing R' I can interchange the corner that was at BRU with UBL (the one that is supposed to go to BRU). This makes the alg:
R' U2 R' D R U2 R' D' R2

If I had the cycle UBL -> LFD -> BRU then I would see that BRU is the corner that can be made interchangeable with both of the other corners. The corner that is supposed to go to BRU is LFD. This means that I need to start with R2 so that I can interchange LFD with BRU on the D layer. The rest of the commutator just falls into place. The alg you would get for this case is:
R2 D R U2 R' D' R U2 R

The part I have bolded in each of the two cases mentioned here is the only part where I actually do any thinking during the A9. My earlier statement that "the rest of the commutator just falls into place" means that you just follow the structure of the normal 8 move commutator for the rest. You only need to think for the first two moves when doing an A9, the rest of the commutator is already built as long as you follow the usual rules based on doing an 8 move commutator as if you had only done the single setup turn (not the double turn).

I see people post things like "Don't waste your time searching for an A9" or "It takes way too long to spot the proper setup turn." It's really not that hard. Just do the setup turn that allows you to interchange the piece that is supposed to go to the corner you're moving, then do that interchange turn. It's that easy. A9's are some of my favorite cases to be honest, because they're so easy to see once you know what to look for.

Hope this helps.

 

[Cubenovice]

Great stuff guys! This thread keeps getting better and better.
Wish I had more time so I could look into all this...

I learned commutators from this thread and Joels page but currently only use them for FMC and some full intuitive solving.
In the future I might try some real BH stuff but first get my BLD accuracy and times down with OP.

 

[Stini]

Okay, here's one approach to A9s. There are only four fundamentally different A9s (that is, non-isomorphic cases) and here's a list of these:

A-perm: x R2 [R U R', D2] R2
Variation of A-perm x R2 [R U2 R', D2] R2 (notice the similarity with the case above)
Headlights: R2 [D, R' U2 R] R2
Variation of headlights: R2 [D', R' U2 R] R2 (notice the similarity with the case above)

That's it! You can also consider headlights to be a variation of A-perm if you want, but since headlights should be familiar to everyone, there's really no need for it. I think this is a pragmatic way to learn A9s, just learn to recognize those cases and learn the inverses and mirrors and you are good to go. This way you should also be able to execute the commutators fairly quickly without much practice. Of course I also encourage everyone to think for themselves and try solving the cases from different angles intuitively and so on.

 

[pkvk9122]

Hi guyz,

I have attempted to learn BH commutators several times and I have failed all the way. So i decided to ask someone here who might help me . Alright so i have two problems.
The first problem I have is to recognize in the pure commutators and a9's whether it is ABA'B', BAB'A'.

The second problems is with the a9's. When ever I am up against an a9 I never get it right. I have to go and look it up on this website for help.

http://www.speedcubing.com/chris/bhcorners.html

However this doesnt have a tutorial

Please help me or I'll die in the world of competitions =( Oh and btw i've been previously blindfolding with M2/Old pochmann.

Best Regards

Pkvk9122

 

[cmhardw]

Hi guyz,

I have attempted to learn BH commutators several times and I have failed all the way. So i decided to ask someone here who might help me . Alright so i have two problems.
The first problem I have is to recognize in the pure commutators and a9's whether it is ABA'B', BAB'A'.

Hi pkvk9122,

As to your first question, just try doing lots of commutators to get used to how they are cycling the pieces. For BH the A part of the commutator is always the insertion. Typically this is 3 turns, but for the Per Specials and Cyclic shifts it is more. The B part of the commutator is always exactly one turn, and this is always called the interchange move.

Let's look at a commutator written in ABA'B' form.

A B A' B'
(R' D' R) (U) (R' D R) (U')
Notice that the A part, the R' D' R, takes the corner at RDF and inserts it to URF. I am conveying the sticker cycle information here is well. By writing the corner names as I did I am also conveying that the R sticker of RDF went to the U sticker of URF. This is the only change to the U layer at all. The corner that was at URF got replaced by the one that started at RDF.

Now the interchange move comes next, the U turn. The U layer was the one that was not affected anywhere except to replace the URF corner. Now that we have replaced this corner, we need to move another corner to URF such that we can change another corner in that same spot. The URF location is sometimes referred to as the "action spot" as this is where all the action, the actual cycling, is happening.

After the U turn is done we need to do the A' part. This is basically the same insertion we did before, only backwards. So R' D R. This also changes the corner that was at URF (it replaces it with the one that was at LFD.

The very last turn is the B' or the undoing of the interchange move. We had done U as our interchange move before, so now go in the other direction with U' and you're done.

Now to tell whether you have an ABA'B' or a BAB'A' commutator you can either look at the insertion group, or you can find the interchange move.

B A B' A'
(U) (R' D' R) (U') (R' D R)

Notice that the concepts are the same as before, but this time we start with the B part of the commutator, we interchange on the U layer first.

Now obviously a commutator will not already be marked out with parentheses when you either have it in your head or find it on a website. Let's take a different alg as an example. In fact, let's do a pretty typical alg that cycles corners (with side effects to their orientations) from a lot of beginner methods.

L' U R U' L U R' U'

When looking at an alg I personally try to always focus on the interchange face, as this face is the one with that "action spot" where all the actual cycling is taking place. In my mind I would see the alg above like this:

L' U R U' L U R' U'

Notice that a turn on the L face only happens twice in the whole algorithm. Now, the same can be said for the R face as well, but the problem with the R face is that it is lodged always between two U face turns. This means that the R turn is part of the insertion (insertions are always conjugates, or sequences of the form M N M'). So we know that R is not the interchanging face, but L is. A turn on the L face comes first in this alg, so this alg is a BAB'A' commutator, since the B part of the commutator is what we do first.

The second problems is with the a9's. When ever I am up against an a9 I never get it right. I have to go and look it up on this website for help.

http://www.speedcubing.com/chris/bhcorners.html

However this doesnt have a tutorial

Please help me or I'll die in the world of competitions =( Oh and btw i've been previously blindfolding with M2/Old pochmann.

Best Regards

Pkvk9122

As for A9's, they can probably be saved for last to be perfectly honest. I would suggest to spend more time on the 8 move commutators, until you become more comfortable with them. After that cover all of the 10, 11, and 12 move cases. For the time being you can just do the A9's as if they were 10 move cases. Basically just do a setup turn that makes the situation solvable with a regular 8 move commutator, and afterward undo your setup turn. You may find that sometimes the setup turn you use is on the same slice as the first move of the interchange part of the commutator, the A part. This is what an A9 really is. Don't worry about mastering them right away, learn the 10 moves and higher cases first. After that come back to the A9's, and you'll be more comfortable with how the different cases relate to each other and can tackle the A9's with a fresh perspective.

I know there is no tutorial on the main alg site, and that alg site is really only a reference anyway. We only listed one alg for each case, when often there is more than 1 option. This tutorial thread is really great for corners, and I did a short write-up on BH edges on this forum as well. Search for "BH" in the wiki on this site to find it.

Hang in there, BH is a lot of fun! Hope this helps.

 

[JonnyWhoopes]

Wow, that's beautiful Chris. A9s were the one thing really holding my back. Any cancelations I found were always on accident!

My question: Is is there any simple/quick way to recognize columns? Often I just get stuck during execution, sitting there trying to figure out what to do with them. I eventually end up turning a side and recognizing a cyclic shift or an A9, but I never recognize the columns before hand.

 

[cmhardw]

Wow, that's beautiful Chris. A9s were the one thing really holding my back. Any cancelations I found were always on accident!

To be honest that's how Daniel and I discovered the A9's in the first place. I see no reason why this can't be a good way to learn those cases. They were also the last cases that Daniel and I optimized. The more I hear others say it, the more I think it really is good advice. Just do 10 movers for the A9's and slowly start to optimize them to the actual A9 algs one by one.

My question: Is is there any simple/quick way to recognize columns? Often I just get stuck during execution, sitting there trying to figure out what to do with them. I eventually end up turning a side and recognizing a cyclic shift or an A9, but I never recognize the columns before hand.

Two corners are AnI (Adjacent and non-Interchangeable, pronounced like the name "Annie") and one of those corners is interchangeable with another corner diagonally across a face (and is not on the same face as the 3rd corner).

Example:
UBR -> BLU -> FDL

UBR and BLU are AnI. BLU is interchangeable with FDL via an L2 turn. FDL is diagonally opposite UBR (They're not on the same face for any option of faces). That makes this a columns case.

 

[JonnyWhoopes]

Example:
UBR -> LUB -> FDL

Am I reading this wrong? Isn't that solveable via [R' F2 R, B']?

 

[cmhardw]

Am I reading this wrong? Isn't that solveable via [R' F2 R, B']?

Gah, wrote the message too quickly. I meant: UBR -> BUL -> FLD

I edited my first post to show this.

 

[RyanReese09]

Am I reading this wrong? Isn't that solveable via [R' F2 R, B']?

Yes it is.

Edit-Ninjad

 

[pkvk9122]

Chris,

Thanks sooooo much!
I finally understand the ABA'B' and BAB'A' now!

I'm now trying to just go with the other commutators and not the a9's yet. Cant thank you enough!

Thanks again,

Best regards

Pkvk9122

 

[JonnyWhoopes]

Ok Chris (or others who can explain), I've got a cycle that I'd like to be walked through.

Please solve, and explain how you found, the cycle URB->FUR->DFR. It's quite obviously an Aperm on the R-layer, but I can't solve it from this angle unless I use D' U' L2 U R U' L2 U R' D.

Also, just out of curiosity, is this notation correct? D':[U' L2 U, R] as well as D':[[U':L2],R].

 

[Joël]

Ok Chris (or others who can explain), I've got a cycle that I'd like to be walked through.

Please solve, and explain how you found, the cycle URB->FUR->DFR. It's quite obviously an Aperm on the R-layer, but I can't solve it from this angle unless I use D' U' L2 U R U' L2 U R' D.

I don't use BH, but I would probably be doing the same thing if I was. If that appeals to your intuition, I don't see any reason not to use it... If you really want to do it with 9 moves, you can see how you can modify this alg a little bit and change it to something that would be 9 moves. Instead of D' U', you could start with a D2 or U2 and use the same basic idea from there, since after U2 the relative position of the corners is similar to the position after D' U' (but with a cube rotation).

Doing this, the whole alg would be something like U2 F2 U B U' F2 U B' U

or in other words: U' (U' F2 U B U' F2 U B') U,

so a commutator with P = U'F2U, Q = B, and U' as setup. This way, the setup and the first move of the commutator combine in just 1 U2 move.

 

[cmhardw]

Hey Jonny,

Yes like Joël I would recommend to basically look for a setup into a commutator that cancels to 9 turns. You can do a 10 move alg from that angle if it makes more sense to you too. I'm finding that sometimes rather than doing an awkward 9 move alg that I know is optimal, during a real solve I will do a much more easily executed 10 mover instead.

As to the UBR -> UFR -> DFR cycle I'll take the alg off the BH site:
U2 F2 U B U' F2 U B' U

Now, obviously I would not execute it this way. I would more realistically do this starting from the DFR corner instead with the following:
y x L F' L B2 L' F L B2 L2 x' y'

This makes the execution exactly the same as for the normal A perm basically.

Now, as to the recognition I would viewpoint shift that case. I know Chester and others have recommended that I stop doing this, and I am slowly learning to recognize more and more cases without it, but for me it just makes perfect sense on a case like this.

When I see UBR -> FUR -> DFR I would notice that for both UBR and FUR that if I rotate one sticker counterclockwise, then both stickers I cycle to end up on the R face. I don't care about the effect to the DFR sticker as of right now, but I will rotate it the same way.

Ok so rotating every sticker cycled to by one counterclockwise gives me: RUB -> RFU -> RFD. Oh, that's an obvious A perm. Nice, just rotate the cube and solve it.

That is literally the thought process I would use if I saw that cycle during a real BLD solve.

Hope this helps.

 

[DYGH.Tjen]

One thing I wanna know is that for BH corners (ok I already printed out the BH corners list and thoroughly went through it with cube in hand, so I know probably everything except A9 and Columns with setup-A9-undosetup. Stupid A9 lol.) It was the BH corners tutorial by Brian Yu, and what I wanna know is for this kind of solves, how do you memo? As previously, with M2/OP corners, I had letters for every single sticker, and just memo'ed them and shot into a new cycle everytime I reached buffer. But for 3-cycles, the scramble wont always be in 3-cycles right. So could you kindly explain, how to memo for BH? (in detail?)

And oh, to further clarify, could you give me an example solve with BH corners? Only the corners. Write the alg, and specify the type of corner commutator used. Lol.
Scramble: F' L U B' L2 U2 D2 F R' B' L2 F2 D2 L F' R2 U2 F2

Take your time. Thanks,
tjen happycubing