# BH Tutorial



## byu (May 23, 2009)

This is an in-progress tutorial for BH.

BH Tutorial
by Brian Yu
Version 1.5

Consists of both written tutorial and video


*SECTION 1. CORNERS*

*PART 1 - INTRODUCTION*

This tutorial is meant to allow you to solve corners without memorizing 378 algorithms, one for each case. It will allow you to solve all of the cases, not only correctly, but also optimally.

In BH, algorithms take the form of a commutator. A commutator takes the form

ABA'B'

For example, if we had R U R' as A, and D as B, the commutator would be.

R U R' D R U' R' D'

Which is 8 moves. 8 moves is the standard 'pure' commutator length. There are many different kinds of commutators in BH. The ones that are needed for BH corners are:

Pure (8 moves)
A9 (9 moves)
Orthogonal (10 moves)
Cyclic Shift (10 moves)
Columns (11 moves)
Per Special (12 moves)

In this tutorial, I will go through each one of these commutators, and how to solve it. I HIGHLY recommend that you go through in order of this tutorial, and not skip around, because I will be introducing terminology as I go along.

PART 2 - PURE COMMUTATORS






The Pure Commutator is 8 moves, as you probably saw from the first section. I am going to use URB as buffer for this tutorial, but many other people also use UBL, and really you could potentially use any buffer.

The first term I am going to teach you is the term 'interchangeable'. If two stickers are interchangeable, that means that in one turn (R, D2, L', etc.) one can be moved to another. For example, UFR and URB are interchangeable with a U' move. However... FRU and URB are NOT interchangeable, because U' doesn't interchange them (FRU goes to RBU instead of URB).

So, in a Pure commutator, the first thing we are going to look for is two pieces that are interchangeable. Not only that, but those two pieces NEED TO BE on a different layer as the third one. For example, given this cycle:

(URB FRU LFU)

We see that LFU and FRU are interchangeable with U', and URB and FRU are interchangeable with R'. Which one is correct? Well, LFU and FRU are on the same layer as URB (U). However, URB and FRU are on different layers than LFU (R and L). So URB and FRU, with the interchange move R', is the correct interchange.

So, the interchange move will be the B of our commutator (ABA'B'). Using the same cycle as above, (URB FRU LFU), we are going to find out the A.

A needs to do the following: Insert the non-interchaging piece into one of the interchanging pieces without messing up the rest of the interchanging layer. Using (URB FRU LFU), we see that LFU can be inserted into FRU using F' L' F. Remember, that LFU must go to the FRU, exactly. It cannot go to RUF or UFR, otherwise the commutator has been messed up. You can see that with F' L' F, the LFU corner goes to FRU without the interchanging layer being messed up other than at FRU.

Now, we have the two parts, F' L' F and R'. But, we need to go back to the commutator.

(URB FRU LFU)

We look at the non-interchanging piece, LFU. Where does it need to go. It seems as if it just ends there, but it actually is a "cycle" so it loops back. LFU needs to go to URB. So, by doing F' L' F, we are actually inserting LFU into FRU, when it needs to go to URB.

So, we are actually doing an inverse pure commutator, instead of ABA'B', we are doing BAB'A'.

So, the first thing to do is to bring URB to FRU, so that we can insert it.

B = R'

Next, we have to insert the LFU to FRU, which right now holds the URB piece.

B = R'
A = F' L' F

Then, of course, we reverse the moves with B'A'.

B = R'
A = F' L' F
B' = R
A' = F' L F

So, the entire commutator that does the cycle (URB FRU LFU) is...

R' F' L' F R F' L F

Now, if the cycle had been done the other way around, as in, (URB LFU FRU), then we look at LFU, the non-interchanging one and say "where does this belong". It belongs to FRU. So this means we do F' L' F immediately.

So (URB LFU FRU) would be

F' L' F R' F' L F R

Which is the inverse of the first one I showed you. Commutators work just like that. I'll give you another example.

(URB LFU LUB)

The first thing that you should note, although it doesn't matter when you are really solving it, is that this case has more than one optimal solution.

There are two possible interchanges, URB and LUB, or LFU and LUB. I'm going to use URB and LUB, but the other one works too. So, for URB and LUB, the interchange move is B.

The A needs to send LFU to LUB without messing up the rest of the B layer. To do this, we do U' F' U.

Finally, we need to figure out whether we do ABA'B' or BAB'A'.

To do this, we look at the non-interchanging piece (LFU). It needs to go to LUB, which is what A does. So we do ABA'B'. If LFU had to go to URB, we need to do the B move first, therefore making a BAB'A'.

So the commutator is:

U' F' U B U' F U B'

Shortened down, here are the steps that you need to follow for a pure commutator:

1. Determine the interchanging pieces and the interchanging move (B).
2. Determine the A section, which inserts the non-interchanging piece.
3. Determine whether it is ABA'B' or BAB'A'
4. Execute

Follow these exercises to practice.

EXERCISE A. Earlier in this section, we worked on the commutator (URB LFU LUB) with the interchange move of B. Do this commutator again, but use L as the interchange.

EXERCISE B. Execute the commutator (URB FRU FUL)

EXERCISE C. Execute the commutator (URB FLD RFD)

*PART 3 - A9s*






The A9 is just an extension to a pure commutator. These, in my opinion, are fairly difficult to get optimal, since it really requires a lot of thinking.

An A9 can be used when you have two interchangeable pieces, but either there's a piece already in that layer, or there is no 3-move insertion (A) part.

The way it works is this. We have a setup to an 8-move commutator and then undo the setup. So, we then have:

SABA'B'S'

But, you're probably thinking, that would mean that it's 10 moves (1 for S, 1 for S', and then the 8 for the standard commutator). Isn't an A9 nine moves?

In fact, it is nine moves, because either S and A or B' and S' will cancel out one move (For example, R' and R2 become R). So, in an A9, you do this.

1. Find the possible setups
2. Find the one that cancels
3. Execute

So, let's take a really simple example, the A-Permutation that most speedcubers know:

R2 B2 R F R' B2 R F R

The A-Permutation is actually, believe it or not, an A9 commutator. Let's break this apart.

First, given an A-Permutation, there are several possible setups. L2 works, so does R2. Let's just take those. With L2, UFR and DFL are interchangeable with F2. Insertion would be R B R'. The commutator would then be.

L2 F2 R B R' F2 R B' R' L2

Although this works, it is NOT optimal, at 10 moves. Now look at if we do R2.

If we do R2, then UBL and DBR are interchangeable with B2. We can do an insertion with R F R' (inserting DRF to DBR).

So we now have.

R2 B2 R F R' B2 R F' R' R2

Now, look at what I mean about the cancellation. The last two moves, R' and R2, cancel out to become R.

So the entire commutator, at 9 moves is:

R2 B2 R F R' B2 R F' R

Congratulations. You've just derived the A-Permutation.

Now for a look at a different A9.

(URB FRU DRF)

URB and DRF are interchangeable, but there is no easy way to insert FRU. So, we're going to have to look for a setup. If we use the setup L, we'd have:

L F L F' R2 F L' F' R2 L'

This works, but once again, it's not optimal. We need to find a canceling setup.

After a little more looking, we see that U2 does the trick.

S = U2

Because then, we use a BAB'A' commutator, B being F2.

S = U2
B = F2

Then, we can use U B U' to insert.

S = U2
B = F2
A = U B U'

Now, the commutator is

U2 F2 U B U' F2 U B' U' U2

Of course, the last two moves cancel.

U2 F2 U B U' F2 U B' U

Now you have an A9.

Now, these examples both have cancellations at the end. I'm going to show you an example now with the cancellation in the beginning.

(URB BLU FRU)

So, the setup move in this case would be L2.

The insertion in this case is L B2 L'.

The interchanging move is F2.

So the algorithm we have is:

L2 L B2 L' F2 L B2 L' F2 L2

However, the first two moves cancel, and become:

L' B2 L' F2 L B2 L' F2 L2

A9s can be difficult to find cancellations for, but with practice, you'll find them faster.

EXERCISE A. Solve (URB FLD BDL)

EXERCISE B. Solve (URB DBR LUB)

EXERCISE C. Solve (URB RFD FRU)

*PART 4 - ORTHOGONALS*






So, now I have to introduce a little more terminology. The term I am going to use is AnI. AnI means Adjacent Non-Interchangeable. For example, RUF and URB are AnI, because the pieces are interchangeable, but not the stickers.

The next term I am going to use is an opposite. An opposite is a piece that cannot be moved to another piece in a quarter turn. For example, FRU and URB are not opposites, since a quarter turn will interchange the pieces themselves (not the stickers). However, ULF and URB are opposites. Every corner has 4 opposites.

URB, our buffer, has the following opposites: ULF, DLB, DRF, DFL. DFL

So now, on to the Orthogonal cases. To recognize an Orthogonal case, you will see URB (buffer) and two of its opposites, all of which are AnI. So, first, let's do some practice to determine an Orthogonal. In these tests, both parts must return yes in order for it to be an orthogonal.

(URB FRU UBL)
Are they opposites? - No, neither of the two pieces are opposite to URB.
Are they AnI? - No, an R' move can interchange URB and FRU.
Conclusion: Not an Orthogonal (This is a Pure)

(URB ULF DRF)
Are they opposites? - Yes, they are all opposites of URB.
Are they AnI? - No, they are in fact ALL interchangeable with one another, in other words, they are "mutually interchangeable"
Conclusion: Not an Orthogonal (This is a Per Special)

(URB FDR LFU)
Are they opposites? - Yes, all of these pieces are opposites.
Are they AnI? - Yes, all of them are not interchangeable.
Conclusion: This is an Orthogonal

(URB BLU DBR)
Are they opposites? - No, none of them are opposites of URB.
Are they AnI? - Yes, there are all AnI.
Conclusion: Not an Orthogonal (This is a Cyclic Shift)

So, now you should be able to recognize an Orthogonal. Once you can recognize one, they are VERY easy to execute. There is one setup move, then an 8-move commutator, then undo setup. Pretty much ANY QUARTER TURN will set up for an 8-move commutator from an orthogonal. The format is the same as the A9:

SABA'B'S'

Except, there is no cancellation.

For example, given the cycle (URB FDR LFU), which we tested earlier and we figured out IS an Orthogonal. Any quarter turn will give us an 8-mover. For example, let's randomly just choose U'.

We now see an interchange with F. An insertion can also be done with R' B2 R. So, we have.

U' R' B2 R F R' B2 R F' U

Which is 10 moves. Using the same cycle, let's randomly choose another quarter turn, L.

We now see an interchange with D'. A insertion can be done with F U2 F'. So, the commutator is:

L D' F U2 F' D F U2 F' L'

As you can see, any quarter turn will give you an 8 move commutator.

EXERCISE A. Find a minimum of 3 optimal solutions for (URB BDL LFU)

EXERCISE B. Find a minimum of 3 optimal solutions for (URB RFD FUL)

EXERCISE C. Find a minimum of 3 optimal solutions for (URB LBD FDR)

*PART 5 - CYCLIC SHIFTS*






Cyclic Shifts are, by far, the strangest commutator, my favorite type of commutator, and the most fingertrick friendly commutator in all of BH.

Cyclic Shifts take the form of two commutators in a row. We have:

ABA' CBC'

Notice that B is the same throughout.

The first step to cyclic shifts is to recognize them. The way to recognize a Cyclic Shift is if they are all AnI and all on the same layer (if you use URB as buffer, always the U, R, or B layer).

So, for example.

(URB LBD BLU) is a cyclic shift because they are all AnI to each other, and all on the same layer (the B layer).

Now, to execute. Given a simple Cyclic Shift, (URB FUL BLU), we are going to find the "middle" piece. In this case, the middle piece is BLU, since it is between URB and FUL.

The next step is to figure out where the middle piece needs to go, which is URB. 

Before we move on, I want you to try something.

Do F R' on a solved cube, and LOOK AT THE EDGES ON THE U LAYER. Memorize what colors they are, and where they are. Now, undo the moves, and do R' F, and LOOK AT THE EDGES ON THE U LAYER. They are the SAME edges. This is an important concept in a Cyclic Shift.

So, for the fist part ABA'.

We're going to bring URB to FRU, so that we can do a U2 to swap it with BLU. However, in order to preserve the edges, we have to do a F first (keep reading to figure out why). So we have F R' for the A. U2 for the B, and the R F' for the A'.

So we have:

F R' U2 R F'

as the first part of the cyclic shift. Now, for the second part, we need to do a U2 on the FUL which needs to go to FRU. So we do an F. But again, to preserve edges, we have to do an R' first. So the second part is.

R' F U2 F' R

If you apply both together, you will see that the corners swap and the edges are preserved.

F R' U2 R F' R' F U2 F' R

These can be fairly difficult to understand at first, but they will eventually become easier.

EXERCISE A. Execute (URB BDL DBR) optimally

EXERCISE B. Execute (URB BLU LBD) optimally

EXERCISE C. Execute (URB BLU FUL) optimally

*PART 6 - COLUMNS*






With Column cases (there are 12 of them), you have a freedom of choice between two options.

1. You do a 1 move setup, and then an A9, then undo the setup. (1+9+1 = 11)

2. You do a 1 move setup, and then a Cyclic Shift, then undo the setup, with a cancellation (1+10+1-1 = 11)

I prefer the cyclic shift, because it's faster for me, since finding a cancellation in an A9 is harder than finding a cancellation in a Column.

I will give one example

(URB LDF RUF)

So, let's do it with the 1 move then A9. U2 B2 allows a D2 interchange, so U2 will be the first part of the Column, the setup to the A9.

So U2 B2 D2, then the insertion is B U2 B'.

Now we have U2 B2 D2 B U2 B' D2 B U2 B' B2 U2.

Which then cancels to:

U2 B2 D2 B U2 B' D2 B U2 B U2

However, the same case can be done with a cyclic shift.

(URB LDF RUF)

L' sets up for a cyclic shift on the U-layer, so we'll go with that. The first part of the cyclic shift is L' B, so we have:

L' L' B U2 B' L B L' U2 L B' L

which cancels to:

L2 B U2 B' L B L' U2 L B' L

Note how both of these algorithms solve the SAME case optimally.

EXERCISE A. Solve (URB LDF RUF)	using both an A9 and a Cyclic Shift

EXERCISE B. Solve (URB DFL ULF) using both an A9 and a Cyclic Shift

EXERCISE C. Solve (URB DFL DBR) using both an A9 and a Cyclic Shift

*PART 7: PER SPECIALS*






Per Specials work like this.

As for recognition, all corners are mutually interchangeable, meaning any one corner is interchangeable with the other two.

Per Specials follow this format:

ABA'B'

A is 5 moves, which brings a corner (1) to another corner's spot (2) without messing up the rest of Corner 2's layer. Then, do a move to switch the other corner (3) and corner 2. Then undo A and B.

This can normally be done by hiding a 3x1x1 block for all but the corner you want to replace (corner 2), doing a move to bring it over to where corner 1 can be inserted, and then applying a move to replace that, followed by restoring the 3x1x1. You probably didn't understand that, so I'll give an example.

In (URB ULF DBL), we want to bring DBL to URB. To do this, we do R' U2 first, to bring URB to BLU, a quartern turn interchange from DBL (the interchange is L). Then do L, followed by a U2 to bring it back to the right side of the cube. Finally, we do an R turn to restore the rest of the U layer. So, the A part of the commutator is:

R' U2 L U2 R

Then, B becomes U2, since it swaps URB with ULF. The final commutator looks like this:

R' U2 L U2 R U2 R' U2 L' U2 R U2

EXAMPLE A. Execute (URB ULF DRF)

EXAMPLE B. Execute (URB DRF ULF)

EXAMPLE C. Execute (URB DRF DLB)

-------------------------------
Just a question, how would you rate my written tutorials (all of them in general, not just this one) on a scale from 1-10. I'm considering doing something (not telling yet) but I need your opinions first.


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## trying-to-speedcube... (May 23, 2009)

Your tutorials are great  A lot, and I really mean A LOT of people have learned from your video tutorials on youtube. Your written tutorials are very clear, even about something abstract like commutators.

P.S. The 8-movers are really clear to me now. I'm looking forward to the A9 part!

P.P.S. Maybe it's an idea that other people can submit to the tutorial as well, so you don't have to do everything


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## byu (May 23, 2009)

Thanks for the feedback, Maarten.

The A9 section is now up.

EDIT: Orthogonals section up.


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## fanwuq (May 23, 2009)

Very good!
The pure commutator part so far is very helpful!
When I do insertions, I usually can only look at the cycle from one perspective. This helps with looking at the commutator from any of the stickers.
Not sure what number rating I would give you, maybe 8 to 10, but it is a good guide!
Nice explanation of interchangeable.
I think Chris and Mike were right, it is kind of like learning intuitive F2L! Now to move on to A9...

For orthogonals, why aren't FRU and URB interchangeable? Ok, typo in the first paragraph.

Edit:
For A9, could you show an example where the cancellation is at the start?


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## trying-to-speedcube... (May 23, 2009)

Wow, thanks, the A9s are clear to me now, I've done the exercises, but it seems I still have trouble finding the cancelling setup. It's easy to find a setup that makes it into a pure commutator, but hard to make it cancel.

As for the number rating, I'd definitely give you a 10, just because I can't find any reason to subtract any points


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## byu (May 23, 2009)

Cyclic Shifts are up.

@fanwuq - Sure, I'll go ahead and do that now.

EDIT:

@fanwuq - Done. View near the bottom of the A9 section.

@fanwuq - Oh, you're right! I made a mistake about the Orthogonals. I meant to say RUF, does that make things clearer?


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## trying-to-speedcube... (May 23, 2009)

Wow. The cyclic shifts are pretty cool ^^. 



> Do F R' on a solved cube, and LOOK AT THE EDGES. Memorize what colors they are, and where they are. Now, undo the moves, and do R' F, and LOOK AT THE EDGES. They are the SAME edges. This is an important concept in a Cyclic Shift.



I don't understand that part. Of course they are the same edges, but not in the same positions. This is like saying that doing R' F R F' is an algorithm that doesn't affect edges.

Oh, and 



> ABA' CBC'



should be ABA' CB'C'.


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## byu (May 23, 2009)

trying-to-speedcube... said:


> Wow. The cyclic shifts are pretty cool ^^.
> 
> 
> 
> ...



No. Take a solved cube (white top and red front) and do F R'. We have

UR = White-Orange
UL = White-Green
UF = Green-Red
UR = Orange-Blue

Now take a solved cube (white top and red front) and do R' F. The edges are the same as above.



trying-to-speedcube... said:


> Oh, and
> 
> 
> 
> ...



It doesn't really matter, because B is always U2, R2, or B2, which the inverse is the same as the move itself.


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## trying-to-speedcube... (May 23, 2009)

Aha. I did a cyclic shift with U in the first and U' in the second part, but that turned out to be optimally solved in 8 moves...

So you only meant the 4 pieces on the U layer, because the rest is different. You should include that, because it's not really clear to a BH-n00b like me


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## byu (May 23, 2009)

Oh, yeah, I suppose that was a bit unclear. I'll edit that. Thanks for the suggestion. Columns should be out soon.

EDIT: Cyclic Shift Edge Preservation clarified, and Columns are up. Per Specials is going to take a while, hopefully by tonight.


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## stiwi griffin (May 23, 2009)

they are very good,but can you explain me how to identify the stikers following UBR and all that?


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## byu (May 23, 2009)

stiwi griffin said:


> they are very good,but can you explain me how to identify the stikers following UBR and all that?



It's the same as any other cycle blindfold method.


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## stiwi griffin (May 23, 2009)

edited,now is clear


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## trying-to-speedcube... (May 24, 2009)

How to recognize a Columns case?


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## cmhardw (May 24, 2009)

trying-to-speedcube... said:


> How to recognize a Columns case?



Two pieces are interchangeable by a double turn, and the third corner is AnI to one of those two. Remember that AnI means Adjacent non-Interchangeable.

Chris


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## trying-to-speedcube... (May 24, 2009)

Ah... Thanks. 

By the way, instead of seeing Cyclic Shifts as ABA'CBC' you could also see them as a simple SABA'B'S', where the A is 4 moves and the S is 2 moves, and the first 2 moves of A (or last 2 moves of A') cancel out with the setup.

Example: (URB FUL BLU) could be seen as: (F R') (U2, R F' R' F) (R F')

I think that it is easier to understand it like that, but nicer to understand it like you said, Brian


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## byu (May 25, 2009)

Video versions of Pure Commutators and A9s are now available. I'll get to Orthogonals, Cyclic Shifts, perhaps even time for Columns tomorrow.


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## dbeyer (May 25, 2009)

Oh Chris. Correction. We need to clarify something about columns cases. 
You stated 2 relationships. A commutator of 3 cubies, has 3 relationships.
A:B
A:C
B:C
we need to clarify all three. Leaving out one, to be assumed, you actually have given the cuber 3 choices. 

Taking your quote, allow me to name a column case, and then let me use your description to choose a wrong 3rd piece.

Two pieces are interchangeable by a double turn, and the third corner is AnI to one of those two. Remember that AnI means Adjacent non-Interchangeable.
URB -> ULF -> DFL is a case that is optimal in 11 moves (many possible solutions available)

Let's make a wrong assumption based on your incomplete description.
Mind you I'm not bashing Chris, but giving him constructive criticism so that you all can see the importance of relationships. Just like recognizing an F2L pair is only good if you know where to the F2L slot is!
So, URB, and ULF are interchangeable opposites.
URB:ULF - interchangeable opposites
URB and RUF are Adjacent non Interchangeable, picking a random piece that meets the criteria
URB:RUF - AnI
Lets just look at the arbitrary relationship between B and C.
ULF and RUF are Adjacent and interchangeable On the Slice
ULF:RUF - AoS 

Chris never clarified the relationship between the B:C.
In this case, you actually have a 9 move commutator.

B' R2 B'L2B R2 B'L2B2

By correctly describing all 3 relationships, you can pick any two interchangeable opposites and then find the 3rd piece.
A:B interchangeable opps
A:C AnI
B:C Polar Opps

URB and ULF are interchangeable opps, we know from here we can make a columns case still.
Now, we need to pick a 3rd piece, that is AnI to one, and the Polar Opp to the other.


ULF: URB - Opps
ULF: DFL - AnI
URB: DFL - Polar Opps

Lets try another one.

URB: ULF - Opps
URB: DBR - AnI
ULF: DBR - Polar Opps

Lets Try another one
URB: BLU - AnI
BLU: FLD - Opps
URB: FLD - Polar Opps


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## trying-to-speedcube... (May 25, 2009)

Brian - How far are you with the Per specials? Now you started with the videos, it almost seems like you're avoiding them...


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## byu (May 25, 2009)

Oh, wow, did I not finish Per Specials? I'll go do that now... sorry about anyone who was waiting for them.


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## Yes We Can! (May 25, 2009)

WOW!
It's nice to see someone, who does invest very much time to teach other cubers something! Well done!


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## byu (May 25, 2009)

Per Specials section up. I'm hoping to do the Orthogonals video today, maybe even Cyclic Shifts.


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## trying-to-speedcube... (May 25, 2009)

> Then, B becomes U2, since it swaps URB with ULF. The final commutator looks like this:
> 
> R' U2 L U2 R U2 R' U2 L' U2 R



No, it doesn't.

R' U2 L U2 R U2 R' U2 L' U2 R _*U2*_
Anyway, after a lot of F5'ing, I'm happy it's up now  I'm looking forward to the videos


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## byu (May 25, 2009)

Oh! Thanks for catching that Maarten. I'll fix it right away.

EDIT: Fixed.


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## fanwuq (May 25, 2009)

Consider
U R'F'R2FR U2 R'F'R2FR U
13 HTM, but only 16 QTM
Compared to the Per Special
R' U2 L U2 R U2 R' U2 L' U2 R U2
12 HTM, but 18 QTM
Which is better?


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## trying-to-speedcube... (May 25, 2009)

I was thinking that too. A setup into a Columns case, or even a 2 move setup into a Cyclic Shift, which is really nicely fingertrickable. 

I think you should choose what you want, whatever seems faster for you


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## rahulkadukar (May 25, 2009)

Nice videos can now do the 8 move commutators


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## byu (May 25, 2009)

Hm... the one move setup to columns is 13 HTM, but it's fast. I'll do some research, testing, etc. and figure some stuff out.


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## Musturd (May 25, 2009)

*Bookmarking page -- to be scrutinized after finals...


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## byu (May 26, 2009)

Orthogonal video has been added. Cyclic Shifts, Columns, and Per Specials are being uploaded at this moment, and I'll get this thread updated when they are.


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## byu (May 26, 2009)

All videos are now up for BH corners


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## masterofthebass (May 26, 2009)

Hmmm, your fix example in the orthogonals seems off. Unless pure comms aren't supposed to use half turns:

U' F' R B' R' F R B R' U
should be [R2, D' L D] instead, saving 2 moves.


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## byu (May 26, 2009)

Thanks for catching that Dan, I think that my Orthogonal alg doesn't even solve the case I meant for it to solve... very strange. I'll look into it right now.

EDIT: Fixed. Again, thanks for catching it.


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## trying-to-speedcube... (May 26, 2009)

I think there might be a lot of people that still don't understand the cyclic shift. Maybe it's a good idea to make a video that explains it as a 2-move setup into an 8 move commutator, with 4 moves cancel? (see post #16 to see what I mean)


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## byu (May 26, 2009)

I'll consider that once I get more feedback. Thanks for the suggestion.


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## dbeyer (May 26, 2009)

Hey guys. I would just like to comment that a Per Special (dubbed as such because Per has really crazy solutions, for these kind of things), is quite fast as a 12 move case. 

Transforming the Per Special case into a Columns case is fine and well. However, consider the chance of solving with an A9.

Note: I am not condoning this approach, I am merely stating the solutions, and the obvious transformations.

URB -> ULF -> DLB is a Per Special
URB -> ULF -> DFL is a Columns case

I can sub-3 R'U2LU2R U2 R'U2L'U2R U2.

The Columns case
Can be solved in numerous ways.
URB -> ULF -> DFL
R2DF2D'R DR'F2RD' R
B2D'L2DB' D'BL2B'D B'
L F2R'F' L2 FRF' L2 F' L'

However, here is a really nice case. 
R U2R'U' L2 URU' L2 U' R'

Method of approach for a columns case.
Moving the Lone Corner:
You can move the Lone corner 
in the example of URB -> ULF -> DFL
URB is the lone corner,
ULF and DFL are AnI
URB and DFL are polar opps
URB and ULF are interchangeable opps

but you can see that URB isn't next to any other cubie in the cycle so it's the lone corner.

When moving the lone corner, you only want to do a quarter turn setup.
Do not turn the layer, that the lone corner shares with it's interchangeable opposite.
URB -> ULF -> DFL -- Don't turn the U layer.

URB and ULF are interchangeable opposites
Your quarter turn setup with transform the case to where the interchangeable opposites are now either Polar Opposites, or AnI.
URB -> ULF -> DFL
R and B' transforms the relationship between the URB and ULF to twisted polar opposites.
R' and B transforms the relationship between the URB and ULF to Adjacent non-Interchangeable.

Transforming into a Twisted Polar Opposites relationship you will solve with an A9
such as R (U2R'U' L2 URU' L2 U') R'

Transforming the into an AnI relationship, you transformed the case into a cyclic shift. There is a cancellation with the Setup, and cyclic shift, giving you 11 moves.
R2DF2D'R DR'F2RD' R


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## dbeyer (May 26, 2009)

You can also move the AnI pair of cubies.
URB -> ULF -> DFL
You can't turn the U layer, or the D layer. You are transforming the case into a Per Special with any quarter turn setup, and a reflection of the Columns case with double turns.

When moving the AnI pair, you are looking for the cancellation. This is found by transforming the relationship between the interchangeable opposites into twisted polar opposites.

URB -> ULF -> DFL
F and L' bring the AnI pair to the U layer, both of which transform this case into a cyclic shift. From here, there is only one optimal solution, and there is no cancellation.
F BL'U2LB' L'BU2B'L F'

However by transforming the relationship between the Interchangeable opposites to Twisted Polar Opposites, we have chances for an 11 move case.

URB -> ULF -> DFL
L F2R'F' L2 FRF' L2 F' L'
The solution can be transformed quite nicely into
r U2R'U' L2 URU' L2 U' r'

There is also of course the solution if you use F' as the setup.


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## dbeyer (May 26, 2009)

There are 6 optimal solutions for each Columns case. Investigate your preferences, and I just showed you how to compose them, by each transformation.

Two setups to create a cyclic shift, moving the lone cubie
Two setups to create an A9, moving the lone cubie
Two setups to create an A9, moving the AnI pair


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## puzzlemaster (May 26, 2009)

I really appreciate this tutorial and the videos are very well made as well.. I finally understand the pure commutators.. on to A9's


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## Musturd (May 27, 2009)

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?


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## cmhardw (May 27, 2009)

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


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## byu (May 27, 2009)

I'd be interested in hearing how you visualize these commutators Chris


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## cmhardw (May 27, 2009)

byu said:


> 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... (May 27, 2009)

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.


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## cmhardw (May 27, 2009)

trying-to-speedcube... said:


> 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
> 
> ...



Edited my original post, that was a typo. The alg should be:
A = F L F'
B = R2


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## trying-to-speedcube... (May 27, 2009)

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 (May 27, 2009)

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


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## spdcbr (May 27, 2009)

byu said:


> 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 .


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## Steyler (May 29, 2009)

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


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## Musturd (May 29, 2009)

Musturd said:


> 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)
> ...



My questions seemed to have been overlooked... 
Maybe now someone can help me?


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## dbeyer (May 29, 2009)

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


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## CubeLord (May 31, 2009)

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


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## puzzlemaster (May 31, 2009)

CubeLord said:


> 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


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## Musturd (May 31, 2009)

dbeyer said:


> Musturd: You are slightly mistaken.
> 
> Pure commutators are optimal as such: ABA'B'
> A is a 3 move insertion.
> ...



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.


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## dbeyer (Jun 1, 2009)

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.


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## trying-to-speedcube... (Jun 2, 2009)

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


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## byu (Jun 2, 2009)

Can't take on edges now with school and exams, maybe afterwards


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## trying-to-speedcube... (Jun 2, 2009)

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.


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## Musturd (Jun 3, 2009)

dbeyer said:


> 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.


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## byu (Jun 3, 2009)

Follow the steps I explained I the tutorial. Where are you messing up?


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## Musturd (Jun 4, 2009)

byu said:


> Follow the steps I explained I the tutorial. Where are you messing up?



I'm going to go through B and C in this post:
EXERCISE B. Execute the commutator (URB FRU FUL)
B = R' or F'
I'm going to use R'
A = L' U L (puts corner in)
It puts it to the right spot, so ABA'B'
But that doesn't work...

EXERCISE C. Execute the commutator (URB FLD RFD)
B = D
A = F2 R' F2 (? this works but it seems like it would mess up everything)
B puts the corner to the right spot, so ABA'B'
and that also doesn't work....

(PS: I emailed my Physics teacher and he told me I got a 93! I nailed the A- for the year!!! That should be a ++ for engineering in college)


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## byu (Jun 4, 2009)

@Mustard

Good, you're getting the B section correct.
You're problems are:

in Exercise B:
Your A part has to be careful to not mess up any of the slice that B moves (in this case, the R slice). L' U L doesn't work because it moves away some of the existing R slice pieces. You should only move one. Instead, the proper setup move would be U L U', bringing FRU to FUL. Final commutator should be:

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

In exercise C:

Same problem is B, you're messing up too much of the D slice. The proper insertion should be L' U2 L. Final commutator is:

L' U2 L D' L' U2 L D


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## Musturd (Jun 5, 2009)

@byu
I think I understand what I am doing wrong.
Could you give me some more pure commutators to try out (or point me in the direction of where I could find some)?


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## byu (Jun 5, 2009)

Mustard-

Try these 3.

(URB FUL UBL)
(URB FLD UBL)
(URB FDR ULF)


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## Musturd (Jun 5, 2009)

byu said:


> Mustard-
> 
> Try these 3.
> 
> (URB FUL UBL)


B = L'
but that would ruin the other corner so
B = U
A = R' F' R (but that would ruin the other corner as well...)
so A = F R' F' (?)
B goes where it's supposed to so ABA'B'
and...
it didn't work still
I don't know what I'm doing wrong


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## byu (Jun 5, 2009)

Haha, you were right to start out with. Your B section can ruin other corners, A cannot. Assuming B is L', which it is, A can't mess up any other part of the L slice other than one corner.

I'm not going to tell you the full commutator, I want you to try again.


Oh, and this applies to anyone.

If you want personal live training on BH, contact me through AIM at rubikmaster100. If you don't have AIM, sign up at aim.com


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## Musturd (Jun 5, 2009)

byu said:


> Haha, you were right to start out with. Your B section can ruin other corners, A cannot. Assuming B is L', which it is, A can't mess up any other part of the L slice other than one corner.
> 
> I'm not going to tell you the full commutator, I want you to try again.
> 
> ...


B = L'
A = L' U' L
Messes too much up
A = no idea
I can't find anything that doesn't mess up the other corners...
U would work but it's not 3 moves...
(PS your not answering your AIM)


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## byu (Jun 5, 2009)

@Mustard-

Let me try to break this down even MORE. The A part of the commutator actually takes the form of

CDC'

Where C and D are each 1 move. D is always the opposite slice of B (B is L', so D would be R, R', or R2). C is one of the four sides that is not B or D.

Now, try and do it on your own.


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## Musturd (Jun 10, 2009)

byu said:


> @Mustard-
> 
> Let me try to break this down even MORE. The A part of the commutator actually takes the form of
> 
> ...



B = L'
A = F R' F'
didn't work
but I went back to your other commutators
B is opposite
so the whole thing: F R' F' L F R F' L'
Worked!!

At one point I think I understood why B or A went first, but now I don't understand it anymore...

(I'm going to try the next one)

(Sorry it took me a long time to respond, I am currently visiting a bunch of colleges)


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## Musturd (Jun 10, 2009)

byu said:


> (URB FLD UBL)



B = U
A = R D2 R'

R D2 R' U' R D2 R' U
doesnt work
U R D2 R' U' R D2 R'
works

So I can now choose A and B correctly, but I don't understand when to use which order.

This is progress 
BTW definately add your latest post to the first post, the CDC' truly enlightened me.


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## cmhardw (Jun 10, 2009)

Musturd said:


> So I can now choose A and B correctly, but I don't understand when to use which order.



Glad you're starting to like the method, and glad that it is starting to make more sense! It does seem like a lot at first, but trust me. Over time it becomes second nature to see which commutator you need for any given cycle.

As for choosing which direction, whether to use an ABA'B' commutator or a BAB'A' commutator, it all depends on the "lone piece" and which piece the A part of the commutator replaces in the B part slice turn.

Here is a cycle:
(URB RFD ULF)

Here I see that the interchangeability is on the U face. I can interchange URB and ULF with the turn U2. This means that I must insert the RFD corner into the U layer affecting either the ULF or the URB corner, but nothing else in the U layer.

I will do this by A = L D' L'

Now I need to see where the RFD corner needs to go to, and compare this to which spot the A part of my commutator places it. The RFD corner, the lone corner, needs to go to ULF. The A part of the commutator places the RFD corner at ULF. This means that where I need the corner to go from my cycle, and where the A part of my commutator places the RFD corner both line up at the ULF spot. This means I will do an ABA'B' commutator.

Now say I had the cycle (URB ULF RFD)

Here I see that the URB and ULF corners are interchangeable by a U2 turn. I also see that I can insert the lone corner, the RFD corner, into the U layer by A = L D' L'

Now I compare where I want the RFD corner to go with where the A part of my commutator sends it to. I want the RFD corner to go to URB, but the A part of my commutator places it at ULF. These do not line up, therefore I am going to interchange with the B part of my commutator first. I am going to do the turn U2 first, and thus do a BAB'A' commutator.

Whether you do ABA'B' or BAB'A' depends only on where you want to lone piece to go compared to where the A part of the commutator places it.

Hope that helps,
Chris


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## Musturd (Jun 10, 2009)

cmhardw said:


> Musturd said:
> 
> 
> > So I can now choose A and B correctly, but I don't understand when to use which order.
> ...




Thanks, I think I understand now. The A part can brings the piece to the layer it needs to go to. It doesn't just bring the piece to it's correct spot. Right?


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## cmhardw (Jun 10, 2009)

Musturd said:


> The A part can brings the piece to the layer it needs to go to. It doesn't just bring the piece to it's correct spot. Right?



Exactly 

Chris


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## stiwi griffin (Jul 4, 2009)

i have a question,how can you figure out the A move?

p.s.:sorry for the litle bump but i didn't want to make another full thread


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## stiwi griffin (Jul 6, 2009)

there is no one that can help me ?


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## grolich (Jul 13, 2009)

> However... FRU and URB are NOT interchangeable, because U' doesn't interchange them (FRU goes to RBU instead of URB).



Interesting. I've only been cubing for a week and a half, but it seems to me that if I get the terminology correct, and FRU is the sticker facing me on the F face on its upper right corner of that face, and URB is the sticker facing the top on the U face in its upper right corner,

an R move brings FRU to URB (and an R' does the opposite), so they do seem interchangeable to me.

Did I understand it wrong or was this a mistake in the original post?


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## Mike Hughey (Jul 13, 2009)

I think he was talking specifically about interchanging with U moves. They're not interchangeable with U, but as you point out, they are interchangeable with R.

Your understanding is correct.


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## grolich (Jul 18, 2009)

Mike Hughey said:


> I think he was talking specifically about interchanging with U moves. They're not interchangeable with U, but as you point out, they are interchangeable with R.
> 
> Your understanding is correct.



Thanks.
This makes a lot of sense.

The entire tutorial is great.


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## Sakarie (Nov 16, 2009)

I only wanted yo say that I have now looked over every possible cycle for corner, and tried to find an optimal solution myself, which worked almost always in the end. If I couldn't, I looked through and compared with the 15 pages I printed out. Right now, I slowly try to make my way through all the A9's and columns, which is those that I have a problem with. This tutorial have helped me a bit, especially with Per Special and Orthagonals. Hardwick's and Beyer's algorithms have helped me very much with everything! 

A question: I don't know if I want to start with BH edges now, since m2 still is very fast. But is there somewhere a slight guide to all those names? Sp9, B9, Wide A9 ? I have no idea!


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## rubiknewbie (Mar 1, 2010)

Thanks. I went through the tutorials and sort of understand. Now I want to try it out in actual corner solves. What do you do if the cycle is 2 or greater than 3? 

Is there a good place to learn cycle reduction and stuff?


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## Stini (Mar 1, 2010)

Let's say you have a cycle like (1 4 10 19 15), if you then apply a 3-cycle (1 4 10) then the original cycle reduces to (1 19 15). So in other words, the piece on place 1 goes to the correct place, piece on place 4 goes to the correct place and then piece on 10 goes back to place 1. One 3-cycle therefore solves two pieces at a time.

Now then if you have cycles like (3 10)(7 18), you can solve those by first making a cycle (3 10 7) and then (3 18 7). So (AB)(CD)=(ABC)(ADC).

If you only have a single 2-cycle of corners, that means that you will also get a 2-cycle of edges as well. You have to solve these 2-cycles simultaneously, for example by setuping to a PLL-case like T, J, R, V, Y, N that solve 2-cycles of edges and corners.


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## rubiknewbie (Mar 1, 2010)

Thanks I'll try it out.


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## joelwong (Jun 19, 2010)

Thank you very much for making such a fantastic tutorial like that! I am new to bh so I do not understand what the A does and should I use ABA'B' or BAB'A'. I do not really understand that.


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## Sune (Jun 22, 2010)

byu said:


> Now for a look at a different A9.
> 
> (URB FRU DRF)
> 
> ...



Isn't that a mistake? The given commutator cycles UBR-*FLU*-DFR.

Other than that, I'd give your tutorial 10/10, simply the best commutator tutorial I've ever read. Right now reading through orthogonals section. The "exercises" sections are extremely valuable, something that other tutorials usually miss.

... 1.5 years ago a friend of mine has gotten into cubing, the first method he found on the web was Petrus, and he stuck with it. He always kept telling me about how great Niklas alg is (RU'L'UR'U'LU), and I was always like, "Man, screw your Niklas+EPLL, watch me doing the same with Sune+PLL really fast!" I should have listened to him back then, commutators are actually fun and not that hard.


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## jms_gears1 (Jun 22, 2010)

byu said:


> The first term I am going to teach you is the term 'interchangeable'. If two stickers are interchangeable, that means that in one turn (R, D2, L', etc.) one can be moved to another. For example, UFR and URB are interchangeable with a U' move. However... *FRU and URB are NOT interchangeable, because U' doesn't interchange them (FRU goes to RBU instead of URB).*


Ok so ive read this a couple times, and ive been wodering is this information wrong?
FRU and URB are interchangeable with an R arent they?

F - > U
R - > R
U -> B


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## MrMoney (Jun 22, 2010)

jms_gears1 said:


> byu said:
> 
> 
> > The first term I am going to teach you is the term 'interchangeable'. If two stickers are interchangeable, that means that in one turn (R, D2, L', etc.) one can be moved to another. For example, UFR and URB are interchangeable with a U' move. However... *FRU and URB are NOT interchangeable, because U' doesn't interchange them (FRU goes to RBU instead of URB).*
> ...



Same here. Stopped reading after I could not understand that first example


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## amostay2004 (Jun 22, 2010)

Yea there're a couple of mistakes in the written tutorial iirc. But it's understandable overall..don't give up just continue reading


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## Sune (Jun 22, 2010)

jms_gears1 said:


> byu said:
> 
> 
> > The first term I am going to teach you is the term 'interchangeable'. If two stickers are interchangeable, that means that in one turn (R, D2, L', etc.) one can be moved to another. For example, UFR and URB are interchangeable with a U' move. However... *FRU and URB are NOT interchangeable, because U' doesn't interchange them (FRU goes to RBU instead of URB).*
> ...



Yes, but they are not interchangeable with U/U' move.
And if we talk about R/R', then UFR and URB are not interchangeable.


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## amostay2004 (Jun 22, 2010)

Sune said:


> jms_gears1 said:
> 
> 
> > byu said:
> ...



Yes but thinking in that approach is not good for learning commutators. The interchange move is not important compared to whether the 2 pieces are interchangeable or not. Thinking whether they are interchangeable with moving only a certain layer can make it quite confusing to beginners.


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## joelwong (Jun 24, 2010)

if 10 is the best, then i would rate it about 8-9. can you make an example solve? And can you tell how to set up the cases that I practice? Thanks a lot. Is there BH for edges?


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## amostay2004 (Jun 24, 2010)

joelwong said:


> if 10 is the best, then i would rate it about 8-9. can you make an example solve? And can you tell how to set up the cases that I practice? Thanks a lot. Is there BH for edges?



There's no need for example solves. You can have the whole list of algorithms for both BH corners and edges.

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


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## joelwong (Jun 27, 2010)

you surely don't expect me to remember all of them right??


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## aronpm (Jun 27, 2010)

The point is that you don't _need_ to remember algs for BH. You make up the commutators.


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## Mike Hughey (Jun 27, 2010)

aronpm said:


> The point is that you don't _need_ to remember algs for BH. You make up the commutators.



And the other point is that after you use them for a while, you'll start to realize that you have memorized them after all. You'll think your image (or whatever you use to memorize), and suddenly you'll find your fingers executing the algorithm. But you don't have to set out to remember algorithms - you start by making them up. The memorizing part just comes for free after practice.


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## aronpm (Jul 16, 2010)

Sorry for the bump but I went through basically all relevant threads on the forum about BH. I know the pure commutators, the orthogonals, the columns (although my solution isn't optimal, I like it; it reminds me of sq1), cyclic shifts, and per specials. The problem I have is with A9s, and finding the cancellation. 

I saw this post from dbeyer about the relationships between the three cubies and I was wondering what the relationships would be the rest of the cases.

I came up with this, but I'm really not sure about the pure or A9 ones:


> ==Pure==
> A:B - adjacent/opposite interchangable
> B:C - 3 move interchange
> A:C - not interchangable
> ...



Can someone who's good with commutators help me with this? Also, any advice on finding cancellations in A9s? I think once I understand them, I'll finally swap to BH corners, which should improve my execution times a lot.


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## cincyaviation (Jul 22, 2010)

Love this tutorial, worked my way through understanding pure and A9's, i'll get to the rest tomorrow but i have a feeling i may get confused.


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## AnnoyingApple (Nov 27, 2010)

Bump.

Will it be okay if I used letter pairs now, and proceed to visual later on? After all, I'll be tapping with letter pairing.


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## riffz (Nov 28, 2010)

AnnoyingApple said:


> Bump.
> 
> Will it be okay if I used letter pairs now, and proceed to visual later on? After all, I'll be tapping with letter pairing.


 
Yes, but your question is concerning memo, not solving using the BH method, so it doesn't really belong here.


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## AnnoyingApple (Nov 30, 2010)

riffz said:


> Yes, but your question is concerning memo, not solving using the BH method, so it doesn't really belong here.


 
I think it has some relevence, because I'm trying to learn the algorithms by heart using my lettering scheme like Mike Hughey suggested. So I was just wondering about if I could learn BH corners with letter pairing and then slowly neglecting letters and focusing on tapping (visual) instead, will it affect my progress.


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## XXGeneration (Dec 1, 2010)

Could someone please post solutions to the exercises?
I found a 10 move commutator for the very first one, but I know it should be 8.


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## aronpm (Dec 1, 2010)

XXGeneration said:


> Could someone please post solutions to the exercises?
> I found a 10 move commutator for the very first one, but I know it should be 8.


 
8 moves:
1) a) U'F'U B U'FU B'
b) L' URU' L UR'U'

2) R' ULU' R UL'U'
3) L'U2L D' L'U2L D

9moves:
1) F2 R' F' L2 F R F' L2 F'
2) U2 F' U B2 U' F U B2 U
3) U' R2 U' L' U R2 U' L U2

Orthogonals:
1) F R'D2R U R'D2R U' F'
2) F' RD2R' U' RD2R' U F
3) F' UL2U' R' UL2U' R F

Cyclic Shifts:
screw these, solve them by setting up to A9s, like columns.

Columns:
Protip: never set these up to cyclic shifts.
1) D' F2 D F U2 F' D' F U2 F D
2) R U L2 U R' U' L2 U R U2 R'
3) L' D2 L D R2 D' L' D R2 D L

Per Specials:
1) U2 L'U2RU2L U2 L'U2R'U2L
2) inverse of 1)
3) z2 U2 R'U2LU2R U2 R'U2L'U2R z2


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## XXGeneration (Dec 1, 2010)

For something like URB FRU FUL, for example, could you use both F and F' as an interchange move?


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## aronpm (Dec 1, 2010)

No. If you use F' there is no 3 move insertion to move UBR into FUL


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## XXGeneration (Dec 1, 2010)

How can you tell if there is a 3 move insertion or not?
My biggest problem right now is just finding the 3 move insertion.

Edit: also, why doesn't something like LB2L' work as an insertion?


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## Faz (Dec 1, 2010)

XXGeneration said:


> Edit: also, why doesn't something like LB2L' work as an insertion?


 
It does


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## aronpm (Dec 1, 2010)

LB2L' doesn't work for that case.

Unless you want to do F2 LB2L' F LB2L' F


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## XXGeneration (Dec 1, 2010)

It doesn't work when I try it. Why doesn't it work?


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## StachuK1992 (Dec 1, 2010)

For 'intuitive' BH edges similar to the BH corners here, is Chris' explanation on the other thread the best available currently?
I'd like to have some fun trying BH out.


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## rock1313 (Dec 12, 2010)

with the case URB FUR FLB I tried UF'U' F2 UFU' F2 but it didn't work why (I've also tried U2L'U2 R U2LU2 R' but that didn't work either)


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## aronpm (Dec 12, 2010)

rock1313 said:


> with the case URB FUR FLB I tried UF'U' F2 UFU' F2 but it didn't work why (I've also tried U2L'U2 R U2LU2 R' but that didn't work either)


 
I don't think that there's a FLB corner.


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## Zane_C (Dec 12, 2010)

You want the interchange to be the R face. Without rotations it can be done as B L' B' R B L B' R'


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## rock1313 (Dec 12, 2010)

oh sorry I think it's FBL but im not sure


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## rock1313 (Dec 12, 2010)

don't bother finally found a solution


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## aronpm (Dec 12, 2010)

rock1313 said:


> don't bother finally found a solution


 
If you tell us the ACTUAL cycle, we can tell you the solution.

FLB is NOT a corner because that FLB means "front, left, back" and you can't have a corner with opposite colours.


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## Zane_C (Dec 12, 2010)

Oh, I just looked back at it. For some reason I assumed FLB was LFU.


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## rock1313 (Dec 15, 2010)

sorry it's FLD isnt it


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## RyanReese09 (Dec 22, 2010)

Didn't feel like a new thread was appropriate. Plus it'll help others searching.




Also, how would I memo for BH when doing it in BLD?


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## aronpm (Dec 22, 2010)

Give me 14 minutes of my life back. That question was not worth a 14 minute video.

"How do you know whether to do the insertion or the interchange first?"

There. It didn't take me 14 minutes to ask that.

As to the question, I won't answer _solely_ because of the way you asked.


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## RyanReese09 (Dec 22, 2010)

aronpm said:


> Give me 14 minutes of my life back. That question was not worth a 14 minute video.
> 
> "How do you know whether to do the insertion or the interchange first?"
> 
> ...


 
Then thank you for the worthless post. My life has improved since I got to read that post.

And there were other questions there, aka questions on my logic of how to approach commutators.


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## MrMoney (Dec 22, 2010)

Look at the cycle A -> B -> C (A the buffer goes to B goes to C goes to A)

Find out where both the interchange and insertion will happen (is it at A, B or C?).

If it is for example B, you need to move the piece that goes to B (A) first. If the interchange and insertion will happen at C, interchange/insert B to C first and follow standard ABA´B´.

If this does not make sense then shout out!


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## RyanReese09 (Dec 22, 2010)

D' B2 D F D' B2 D F'
Do that backwards to set it up
A piece is URB
B is DLF
C is DFR

3 move setup is D' B2 D
Interchange is F'

Interchange/Insertion will happen at B. Thus I'll move A to B
D' B2 D
Then I know to move C to B using the interchange F

Undo D'B2D, then F'
Nice.


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## MrMoney (Dec 22, 2010)

URB - DLF - DFR is not solved by the cycle you wrote, Ryan 

It solves URB - DLF - RFD is solved with that algorithm. But you got the picture


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## RyanReese09 (Dec 22, 2010)

Bah ok. But yeah.
One more question. Do I pick any starting buffer position? Aka for this cycle, uh, UBR, DFR, FLU? Can I pick, for example, the UBR piece to be the buffer? Is the buffer piece the piece that when I do the interchangeable move, it will be in the "solve position?" If you guys understand. For example, doing the cycle I Just mentioned, UBR, when R2'd, is in the solved place, I normally would have that piece be the buffer piece (since it's solveable via interchangeability). Am I overthinking things?

And also, how would I do my memo for BH? I use lettering for corners and make it auditory, so would I just remember..M D X (ULB buffer normally) and make a commutator out of that?

R2 = Interchangeable 
D' L D = Setup

UBR = A
DFR= B
FLU = C
A has to go to B in this case, so R2, then the insertion to get C to B would be D' L D? Then undo the R2 and undo the D' L D

And one more thing, in your last post Mr Mooney (previous page) I am not quite sure where the insertion will happen for every case, for example..uh let me go find a random commutator I wouldn't know..
U' L U R2 U' L' U R2 . Do backwards. Ignoring what the commutator actually is, I'd approach this as..
y2 (just so I can see it)
Insertion as L2
A=FRD
B=UFL
C=DBL?
A goes to B, and it isn't the interchange so I'd do the insertion as U' R U
Interchange, L2
Undo the U' R U as U' R' U, then L2 to undo the interchange
YESS.
I think I finally have gotten this down.

Anyone care to give me a cycle to solve? Pure 8 move please


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## aronpm (Dec 22, 2010)

UBR LBU FUR


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## joey (Dec 22, 2010)

UBL RBU RFU.


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## RyanReese09 (Dec 23, 2010)

Interchange = B
Insertion = L F' L'
So insertion,then B, then L F L', then B' .

Alas! Didn't work.

Is it because I put the insertion piece at the wrong place? Aka after insertion, the piece in the UBL is not in the correct spot ? If that makes sense. Going off that logic, I do B, insertion, B' undo insertion. But I have 2 twisted corners now. 
I know I have to do the B interchange first..but I can't find a way to do the insertion <_<. I'd be delighted to have explained where my logic is wrong (please no answers to the commutator though)




Spoiler



I'm sorry for the long winded post. I'm going through my thought train as I'm doing this cycle



Edit-Saw Joeys post first, I think he deleted the original..? If so Joey, the cycle posted was..impossible? At least in 8 moves?


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## aronpm (Dec 23, 2010)

nubs can't do niklas


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## Toad (Dec 23, 2010)

lolryan


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## RyanReese09 (Dec 23, 2010)

aronpm said:


> nubs can't do niklas


 
I was eating dinner, and refreshed waiting for a possible answer from joey, since the original cycle he posted was, I believe, impossible. Once again, thank you for your fail post, and not being helpful. Gtfo now.

Edit-My above post was at Joeys original post, to clarify. I haven't even attempted Arons, though the answer has been revealed.


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## aronpm (Dec 23, 2010)

My cycle and Joey's cycle are the same


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## RyanReese09 (Dec 23, 2010)

aronpm said:


> My cycle and Joey's cycle are the same


 
His original post wasn't. His original post was before yours. His original had a headlights setup on the R face.


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## joey (Dec 23, 2010)

I've only posted once in this thread. There must have been an issue with the forum.


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## RyanReese09 (Dec 23, 2010)

joey said:


> I've only posted once in this thread. There must have been an issue with the forum.


 
If that's true then when setting it up (I slightly disassembled to make it easier visually) I must have made it into headlights. Meh. Anyone else not wanna be a smart*** and give me a real cycle?


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## aronpm (Dec 23, 2010)

How was that not a real cycle?


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## MrMoney (Dec 23, 2010)

My friend Ryan, there is something you need to understand about BH:

Every cycle has x number of mirrors and times two that many inverses. The 378 cycles can actually be broken down into about 30 TYPES. I call them "ideas" and once you understand the IDEA of how to solve the commutator you will instantly know what to look for. You can use anything as buffer when you understand the ideas. I plan on making a video to help others with this.

How to understand the interchange and insertion? Well, begin with 8 movers as they are the easiest. Focus on them.

Look at the cycle: URB -> UBL -> FDR (lets say URB is our buffer).
1. What can be interchanged? 
Well URB and UBL can be interchaned with U/U´
FDR can neither be interchanged with URB (R2 will change the pieces but NOT the stickers)
And FDR UBL are diagonal from each other so that is not possible.
Result: URB-UBL can be interchanged! The action will be at one of these.

2. What can be inserted?
We have found out that the interchange will happen at either URB or UBL, so these should not interchange (although it is possible, but not relevant in our cycle).
FDR can go to URB with B´DB! That is a possible insertion!
Lets not stop there, do we have any other options?
Well, FDR can also be inserted into UBL with L´D2L !

Let us choose the first one for the shortest amount of QTM.

URB -> UBL -> FDR
A = B´DB
B = U
A´= B´D´B
B´ = U´

B´DB U B´D´B U´

In an actual solve I would rotate the cube and perform this algorithm in less then 1.20seconds.

You have now learned an IDEA, think about this commutator and think of possible similarities with other commutators. Stick to URB -> UBL comms first and think about for example (FDR, RFD, DFR), (FLD LDF DFL). The idea I gave you will instantly learn you to solve these 6 commutators + inverses = 12 of the 378. Actually even more of them.

Does this make it clearer? We can have a videochat on skype if you want to SEE things happen and ask "on the go". Just let me know mate.

At work now, can´t write anymore.


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## RyanReese09 (Dec 23, 2010)

I understand the concept of it. I can find interchanges very easily, and normally setup moves aren't really much of a problem (many times they are though, since sometimes I do illegal moves). Many times my problem is doing the interchange first when I needed to do the setup first, or vice versa. This has it broken down
I swear somewhere in this thread, it had a list of rules to be followed when doing setup moves, aka something about the moves being in the opposite layer..did I imagine it? It was late last night when I "read it"
For R U' R' D R U R' D' I notice interchange as D'
*When I do interchange, if I can't do the 3 move setup then I do the setup first and then the interchange? What about when/if I can do either or first. Do I just do whatever I feel like doing?* (I choose between doing D or D' in this case, by just getting the 3rd cycle piece underneith the piece that belongs in the DFR peice, if that makes sense. So I did R U' R' which took out DFR, in the FR slot, so I'd do a D to get that LFD slot piece under the DFR slot, then undo everything.
I just need to know about the bolded above.


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## Kynit (Dec 23, 2010)

Let's take your example there: do R U' R' D R U R' D' on a solved cube. You should have 3 corners left to solve.

The moves to insert LUF into the bottom layer are R U' R.
The interchanging move is D.
However, since you want LUF to end up in DLF and not DFR, you have to do the interchange first.

Watch what happens when you do R U' R. The top piece goes into the bottom layer, but in the wrong slot. Undo that then try D R U' R. The bottom pieces are interchanged, then the top piece is inserted. Then you can finish the commutator with D' R U R.

Hopefully that made sense.

(For this example, there's a fairly simple 8 mover with the insertion before the switch. Can you see it?)


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## RyanReese09 (Dec 23, 2010)

L' U L D' L' U' L D? 
So if the insertion doesn't put it in the correct place it needs to be (the correct slot) then I know to do the interchange. Aka the setup move will take one piece, and put it into one slot in the cycle I ihave, if it's in the wrong place then I know to do interchange first? If that's a yes then I should be able to do any pure 8 move commutator, cycle please 

Also, above like joey said "UBL RBU RFU." Does this literally mean the UBL piece needs to go to the RBU (U peice goes to the R, B goes to the B, and L goes to the U?) and then RBU goes to RFU (same logic as above)?


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## uberCuber (Dec 23, 2010)

RyanReese09 said:


> Does this literally mean the UBL piece needs to go to the RBU (U peice goes to the R, B goes to the B, and L goes to the U?) and then RBU goes to RFU (same logic as above)?


 
When someone labels a corner with 3 letters, just look at the first one. The other two can be in any order, and it doesn't change the meaning.
UBL > RBU means that U sticker goes to R, and the other two just go wherever they go.
Whatever sticker is clockwise of the U sticker on the first piece goes to the position clockwise from the R sticker on the second piece. And whatever sticker is counterclockwise of the U sticker on the first piece goes to the position counterclockwise from the R sticker on the second piece.


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## Kynit (Dec 24, 2010)

RyanReese09 said:


> L' U L D' L' U' L D?


I think you mixed up your Us, but that's the right idea.



RyanReese09 said:


> So if the insertion doesn't put it in the correct place it needs to be (the correct slot) then I know to do the interchange. Aka the setup move will take one piece, and put it into one slot in the cycle I ihave, if it's in the wrong place then I know to do interchange first? If that's a yes then I should be able to do any pure 8 move commutator, cycle please


 
Sounds right to me. Also, you can do the 9-movers too; all of those commutators labeled A9 are, as far as I know, 8 move commutators with setup moves (with one of the setups cancelling).



RyanReese09 said:


> Also, above like joey said "UBL RBU RFU." Does this literally mean the UBL piece needs to go to the RBU (U peice goes to the R, B goes to the B, and L goes to the U?) and then RBU goes to RFU (same logic as above)?


...Yeah, but you don't need to worry about that. If you focus on one face of the piece, the entire thing will be solved.


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## Weston (Dec 24, 2010)

You're trying to learn commutators too mechanically.
Just learn how they work.
If you can't insert it in 3 moves, then use 4. Whats the big deal?


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## RyanReese09 (Dec 24, 2010)

Kynit said:


> I think you mixed up your Us, but that's the right idea.
> 
> 
> 
> ...


 
You sure I messed them up? Just setup the case right now and it worked. Ok I think I got it now..anyone care to test me out with a random cycle?

Weston, I'm only focusing on 8 move commutators, like I'll use an alg to setup the case and I'll try to solve that.
And to the mechanical approach comment. It's how I learn. I'm quite attrotious at intuitive crap.

Edit-It'll become intuitive in time. I just need to learn..the rules per say of commutators and then it'll click for me.


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## Cubenovice (Dec 24, 2010)

Some practice ideas:
- stickers (as in FMC), this way it is much easier to track the pieces in your cycle
- Hit a solved cube with a few Old Pochmann moves to set up three cycles (so you are not influenced by your set-up-moves)

I found byu's very first example very helpfull; he went straight to a BABA instead of ABAB getting the idea across right away.


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## RyanReese09 (Dec 24, 2010)

I tried the Old Pochmann idea. I did Yperm, then R Yperm R' and found that commutator . Onto A9s..


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## Cubenovice (Dec 24, 2010)

If you are a little more creative with your set up moves or even "double up" on some OP moves you can create random set ups that require all kinds of of commutators.


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## RyanReese09 (Jan 26, 2011)

How do you know by looking at the cycle if it is a 8 move commutator or, for example, an A9? I'm assuming you know if it's an A9 if the cycle has no 3 move insertion and you need to do a setup move.

For the smaller cases that take 11/12 moves (18 and 6 cases respectably) should I just learn the algs and drill it or still learn it intuitively?


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## amostay2004 (Jan 26, 2011)

The general rules for cases are explained in the tutorial (eg if all corners are opposite non-adjacent = orthogonal) and you'll learn to recognise them quickly through lots of practice.

I would suggest you learn cyclic shift as just one basic alg (F R' U2 R F' R' F U2 F' R). It's the same for all cyclic shift cases


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## RyanReese09 (Jan 26, 2011)

I didn't see the rules intially. THanks. I hadn't gotten that far down into the Orthogonals section

Take this rule.


> An A9 can be used when you have two interchangeable pieces, *but either there's a piece already in that layer*, or there is no 3-move insertion


I understand the last rule obviously, but this is..vague in the way he worded it.

I understand orthogonals now, the 3 pieces are not interchangeable, and need any quarter turn to setup a pure. And they are all opposites (from the buffer, the other two must need a turn such as R2, or L2, etc. No quarter turns, yes? Half turns.

I don't understand cyclic shifts at all, but would I just rotate, put the middle piece in the UBF place, do the alg, and undo rotations? I assume I'll understand the nature of the alg later.

For columns, do you recommend setting it up to A9? (</3 A9's) or setting it up to cyclic shifts (fingertrick friendly)? Your opinions on this?

Per specials are .. every corner is interchangeable..they look hard. I think I'll just learn the 6 algs and know which alg to apply where. Not the smartest move probably but whatever.


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## aronpm (Jan 26, 2011)

Screw intuition behind cyclic shifts and per specials. They are both 1 alg each. Just learn the alg.

Cyclic shift, move-optimal is crappy, here's a speed-optimal: x' U2 R' U2 R' D R U2 R' D' R2 U2, that's just setup to A9.
Per Special, this is pretty crappy but I haven't found a good speed-optimal alg: R' U2 L U2 R U2 R' U2 L' U2 R U2, just rotate to this position (or mirror) and do this or inverse.

For columns NEVER setup to a cyclic shift, they are the worst case. A9s are more fingertrick friendly than cyclic shifts. If your A9s are crap you're doing them wrong.

EDIT: cyclic shifts aren't the worst case, they are the second worst. Per specials are the worst.


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## RyanReese09 (Jan 26, 2011)

aronpm said:


> Screw intuition behind cyclic shifts and per specials. They are both 1 alg each. Just learn the alg.
> 
> Cyclic shift, move-optimal is crappy, here's a speed-optimal: x' U2 R' U2 R' D R U2 R' D' R2 U2, that's just setup to A9.


What do you mean it sets it up to A9? That alg solves cyclic shift yes? Setting it up by doing that alg backwards, it looks like it..



> Per Special, this is pretty crappy but I haven't found a good speed-optimal alg: R' U2 L U2 R U2 R' U2 L' U2 R U2, just rotate to this position (or mirror) and do this or inverse.


I'd rather move optimal but I suppose yours will do for now. Thank you.


> For columns NEVER setup to a cyclic shift, they are the worst case. A9s are more fingertrick friendly than cyclic shifts. If your A9s are crap you're doing them wrong.
> 
> EDIT: cyclic shifts aren't the worst case, they are the second worst. Per specials are the worst.


My A9's aren't bad. I haven't practiced them yet nor attempted them to be bad .


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## aronpm (Jan 26, 2011)

RyanReese09 said:


> What do you mean it sets it up to A9? That alg solves cyclic shift yes? Setting it up by doing that alg backwards, it looks like it..


Remove the U2 from the start and end and you've got an A9 (R'U2R' D RU2R' D' R2 which does UBR->DLF->BUL)



> I'd rather move optimal but I suppose yours will do for now. Thank you.


That per special is optimal, and is given in Byu's tutorial. Move-optimal isn't important.


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## RyanReese09 (Jan 26, 2011)

Got it. Understand the rules now, just have to memorize the rules and then I'll start attempting BH corners in solves :3. Thanks.


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## ilikecubing (Jan 28, 2011)

This commutator D R' D' R U R' D R U' D' 

UFR URB FLD

Is this an A9? From what I've heard A9s are pure 8 move commutators with setup moves and a cancellation at the end resulting in a 9 move algorithm.

But the above commutator is 10 moves,what is it if its not an A9 ?


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## Kynit (Jan 28, 2011)

A 10 move commutator like that is an orthogonal. Whether or not that's optimal, I don't really care to check, but orthogonals work that way.


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## aronpm (Jan 28, 2011)

ilikecubing said:


> This commutator D R' D' R U R' D R U' D'
> 
> UFR URB FLD
> 
> ...


That is an 8 move commutator, U' R D2 R' U R D2 R' (compact form is [U', R D2 R'])


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## cmhardw (Jan 28, 2011)

ilikecubing said:


> This commutator D R' D' R U R' D R U' D'
> 
> UFR URB FLD
> 
> ...


 
I would call your particular alg a "setup into a direct insert." And as Aron already showed, that situation is optimal at 8 moves.


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## rock1313 (Jan 31, 2011)

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?


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## aronpm (Jan 31, 2011)

D R2 D' L etc


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## JonnyWhoopes (Jan 31, 2011)

rock1313 said:


> With the commutator URB DFL FUL
> 
> it doesn't work with D R2 D' F D R2 D' F'
> 
> ...


 
D' B2 D F' D' B2 D F doesn't solve URB DFL FUL...


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## rock1313 (Jan 31, 2011)

I ment URB DFL LFU


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## JonnyWhoopes (Jan 31, 2011)

rock1313 said:


> I ment URB DFL LFU


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


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## rock1313 (Jan 31, 2011)

JonnyWhoopes said:


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


 
that solves URB FUL DFL


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## JonnyWhoopes (Jan 31, 2011)

rock1313 said:


> 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.


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## aronpm (Jan 31, 2011)

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.


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## rock1313 (Jan 31, 2011)

What does the D C D' mean.


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## JonnyWhoopes (Jan 31, 2011)

rock1313 said:


> What does the D C D' mean.


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


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## Joël (Jan 31, 2011)

rock1313 said:


> With the commutator URB DFL FUL
> 
> it doesn't work with D R2 D' F D R2 D' F'
> 
> ...


 
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.


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## JonnyWhoopes (Jan 31, 2011)

Joël said:


> 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?


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## Joël (Jan 31, 2011)

JonnyWhoopes said:


> 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.


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## Julian (Jan 31, 2011)

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?


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## aronpm (Jan 31, 2011)

Julian said:


> 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.


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## Julian (Jan 31, 2011)

aronpm said:


> 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.


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## cmhardw (Jan 31, 2011)

Julian said:


> 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.


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## RyanReese09 (Jan 31, 2011)

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.


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## rock1313 (Jan 31, 2011)

Joël said:


> 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.


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## RyanReese09 (Jan 31, 2011)

Joël said:


> 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..


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## MrMoney (Jan 31, 2011)

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


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## Kynit (Jan 31, 2011)

RyanReese09 said:


> 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.


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## Simboubou (Jan 31, 2011)

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.


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## RyanReese09 (Jan 31, 2011)

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?


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## cmhardw (Jan 31, 2011)

RyanReese09 said:


> 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.


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## RyanReese09 (Feb 1, 2011)

cmhardw said:


> 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'
> ...


 
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.


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## cmhardw (Feb 1, 2011)

RyanReese09 said:


> 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.



RyanReese09 said:


> 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.


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## RyanReese09 (Feb 1, 2011)

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?


cmhardw said:


> 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?


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## aronpm (Feb 1, 2011)

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.


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## cmhardw (Feb 1, 2011)

RyanReese09 said:


> 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.



RyanReese09 said:


> 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.


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## Cubenovice (Feb 1, 2011)

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.


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## Stini (Feb 1, 2011)

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.


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## pkvk9122 (Feb 8, 2011)

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


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## cmhardw (Feb 8, 2011)

pkvk9122 said:


> 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.



pkvk9122 said:


> 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
> 
> ...


 
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.


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## JonnyWhoopes (Feb 8, 2011)

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.


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## cmhardw (Feb 8, 2011)

JonnyWhoopes said:


> 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.



JonnyWhoopes said:


> 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 (*A*djacent and *n*on-*I*nterchangeable, 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.


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## JonnyWhoopes (Feb 8, 2011)

cmhardw said:


> Example:
> UBR -> LUB -> FDL


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


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## cmhardw (Feb 8, 2011)

JonnyWhoopes said:


> 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.


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## RyanReese09 (Feb 8, 2011)

JonnyWhoopes said:


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


 
Yes it is.

Edit-Ninjad


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## pkvk9122 (Feb 9, 2011)

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


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## JonnyWhoopes (Feb 13, 2011)

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].


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## Joël (Feb 13, 2011)

JonnyWhoopes said:


> 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.


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## cmhardw (Feb 13, 2011)

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.


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## DYGH.Tjen (Mar 18, 2011)

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


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## Marcell (Mar 18, 2011)

DYGH.Tjen said:


> Scramble: F' L U B' L2 U2 D2 F R' B' L2 F2 D2 L F' R2 U2 F2


(URB LDF RUF)	U2 B2 D2 B U2 B' D2 B U2 B U2 (11 HTM)	Columns
(URB DRF UBL)	R2 D' L2 D R2 D' L2 D (8 HTM)	Toss Up
(URB BRD LBD)	R B' L' B R' B' L B (8 HTM)	Direct Insert
You're left with a 2-cycle. You might want to orient the two remaining corners. Or you can just leave it there, then have UR and UL swapped at the end of the edges phase, and use F to set up to an T-perm to solve parity.
Now that is pure BH. I did nothing but copy the necessary algs from the corner list. In a real solve you might use speed-optimized algs.
For the first cycle you could also set up to a cyclic shift: L2 B U2 B' L B L' U2 L B' L or you could use (F2 RUR2U'R')2 or (F2 (RU'R'U)3)2.
For the third alg you can do y RUR' D' RU'R' D y' as well.


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## DYGH.Tjen (Mar 18, 2011)

Thanks Marcell.. You've been a great help! I do understand how to memo now, for BH corners. But is there a chance that you actually know how to do this WITHOUT using a fixed buffer? Is there such thing? No I guess, then forget it. Thanks a lot 

happy cubing, tjen

edit: then what is freestyle? And what is floating buffer? :O


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## Zane_C (Mar 19, 2011)

DYGH.Tjen said:


> Thanks Marcell.. You've been a great help! I do understand how to memo now, for BH corners. But is there a chance that you actually know how to do this WITHOUT using a fixed buffer? Is there such thing? No I guess, then forget it. Thanks a lot
> 
> happy cubing, tjen
> 
> edit: then what is freestyle? And what is floating buffer? :O


 
Freestyle can refer to BH using floating buffers.


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## Marcell (Mar 19, 2011)

Read this thread: http://www.speedsolving.com/forum/showthread.php?24310-Freestyle-method


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## riffz (Mar 20, 2011)

DYGH.Tjen said:


> Scramble: F' L U B' L2 U2 D2 F R' B' L2 F2 D2 L F' R2 U2 F2



I used ULB as my buffer so I hope this won't be too confusing to follow:

y
ULB -> RFU -> LDB: y' x [U' R2 U, L] (8 moves)
ULB -> LUF -> LFD: x U' [L2, U' R' U] U (A9)
ULB- > BRU -> DFR: L2 [D', L' U2 L] L2 (A9)

Now the last target is DRB and you can solve this in different ways depending on how you deal with parity.


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## Sakarie (Mar 20, 2011)

DYGH.Tjen said:


> 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
> ...


 
About when it's not in three-cycles, learn how to break in to new cycles.

Scramble: F' L U B' L2 U2 D2 F R' B' L2 F2 D2 L F' R2 U2 F2

With UBR as buffer:

UBR LFD RFU; L2 F2 L2 F' R2 F L2 F' R2 F' L2 - Columns
UBR DFR ULB; R2 D' L2 D R2 D' L2 D - A9
UBR BDR LDB; B U B' D' B U' B' D - 8-mover
Seting up for parity
UBR LUF ULB; B' U2 B' D' B U2 B' D B2 - A9

But normally you'd use some cuberotations for most of them.


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## evogler (Mar 30, 2011)

Here's a cheat sheet I came up with for BH corners: 


```
1+QI?
   1+QN or 3I?
     A9.
   Pure.
1+HI (0QI)?
   0Q?
      Pure.
   1QN?
      Columns. (1QN 1HI 1O)
   Per Special. (3HI)
0I?
   0O?
      0Q?
         Orthogonal. (3HN)
      Cyclic Shift. (2QN)
   A9.
```

Q = quarter turn apart
H = half turn apart
O = opposite corner (any orientation)
I = interchangeable (stickers)
N = not interchangeable (stickers)
1+ = 1 or more

The way you read it is: if the answer is yes, you go to the indented next line. If it's no, you go to the next line with the same indentation.


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## yoruichi (Apr 6, 2011)

wtf is that lol


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## evogler (Apr 26, 2011)

Ha, maybe it only makes sense to me. I was trying to reduce one aspect of BH (recognizing corner cases) to as few rules as possible.
I think there's a still a fair amount that could be done in terms of reducing the amount of information needed to present BH.


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## TheMachanga (Aug 6, 2011)

Bump.

I'm seriously trying to learn this now. Thank you for making such a great tutorial. I understand pure comms, and understand the concept of A9's (set up (that cancels), pure come, undo setup, right?). However I realized I only know 2 basic commutators, that show up in some of the BH algs that Chris has:

UFR - FDL - RDF (no with BH buffer, just an example). 
D (R U R') D' (R U' R') or in-other words, R U R' comms, I guess.

I also know the ones with a form of R2 U' L2 U R2 U' L2 U 

HOWEVER, when I choose a random A9 case, or ever a Pure C case, I get something I don't know. I mean, I can see how it's being solved in front of my eyes, but I don't really understand _why_. 

Ex. I understand D (R U R') D' (R U' R'). It's A B A' B' (wait...is it really? Shouldn't it end with R' U' R then? nevermind, my head hurts)
So I guess there are some commutators I haven't learned yet. What I'm asking is that you write it so I can see the A B A' B'. 

F L B2 L' F' L B2 L'. ---> URB FRD FLD 
This is F (L B2 L') F' (L B2 L'). (weird again, I thought it should be A B A' B', but this is A B A' B.........?)

After doing these moves slowly, I can kinda see what's going on. 
But this is very very scary. This case shows me that I'm not smart enough to recognize this, and apply this commutator to another case. So far, I have to evaluate every A9 case very very slowly, so I can make sure I understand what's going on. I have to break the algorithm apart, and really study it for about 5 minutes. 

Is there a list of Commutators that are not based on R U R' or R2 U' L2 U R2 U' L2 U (<--- btw I need someone to break this apart for me because I don't understand _why_ it does what it does, unlike the R U R' ones) that someone can give me?


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## aronpm (Aug 6, 2011)

No offense but reading your post, "I understand pure comms" just isn't true.



TheMachanga said:


> This is F (L B2 L') F' (L B2 L'). (weird again, I thought it should be A B A' B', but this is A B A' B.........?)


B is the same as B' for that. The inverse of LB2L' is LB2L'.


> R2 U' L2 U R2 U' L2 U (<--- btw I need someone to break this apart for me because I don't understand why it does what it does, unlike the R U R' ones)


R2 moves DFR to UBR. That's the interchange. U'L2U is the insertion, moving DLF to UBR without changing the R slice.


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## RyanReese09 (Aug 6, 2011)

TheMachanga said:


> Ex. I understand D (R U R') D' (R U' R'). It's A B A' B' (wait...is it really? Shouldn't it end with R' U' R then? nevermind, my head hurts)
> So I guess there are some commutators I haven't learned yet. What I'm asking is that you write it so I can see the A B A' B'.


Nope. It shouldn't end as R' U' R. Look at the first bracket. That is B.

For those 8 movers. It is always A B A' B'. R U R' is B.

Now we find out B'. Literally just do it backwards. (I don't do this, but perhaps you can see what I'm doing). Look at hte final R'. That inversed is R. So we have *R* so far. Opposite of U is U'. So we have R U'. And now opposite of the first R is R'. So we have * R U' R'*. That makes up B', which is the inverse of B.


> F L B2 L' F' L B2 L'. ---> URB FRD FLD
> This is F (L B2 L') F' (L B2 L'). (weird again, I thought it should be A B A' B', but this is A B A' B.........?)


In this case, B and B' are exactly the same. So it's technically A B A' B' but *it's also A B A' B. The inverse of B is B' and B. Make sense?



After doing these moves slowly, I can kinda see what's going on. 
But this is very very scary. This case shows me that I'm not smart enough to recognize this, and apply this commutator to another case. So far, I have to evaluate every A9 case very very slowly, so I can make sure I understand what's going on. I have to break the algorithm apart, and really study it for about 5 minutes. 


Is there a list of Commutators that are not based on R U R' or R2 U' L2 U R2 U' L2 U (<--- btw I need someone to break this apart for me because I don't understand why it does what it does, unlike the R U R' ones) that someone can give me?

Click to expand...

 
Do that commutator backwards. UBR - DFL to DFR. That is our cycle. First thing you do is find out what piece is going to be the piece where the stuff will happen. Notice that if you do R2 that does an interchange. So we have our interchange. Now we find out insertion. Since R2 affects DFR and UBR we know that we have to see if we can find a way to bring DFL to UBR. Since it's U/D stickers involved, we know that an L2 (something 2 moves) will be needed to get those stickers interchanged.

Side note-here is something I picked up. Take this comm. Same as above [R2, U' L2 U]. That A is the R2. Notice in B, that the 2nd move in the interchange (B) is L2. Note that the move L2 is the opposite face of the (A) interchange). So whenever you have an 8 move comm, and you know the interchange, you already have a glimmer of an idea how to do the insertion. It will be _ L(something) _

So back on topic. UBR to DFL to DFR. UBR is our point of excitement. So now we have R2, U' L2 U. The problem now becomes, which to do first? Think of it like this. If you do R2 first, will that bring the DFR piece to the UBR sticker (think stickers), and most importantly DOES THAT DFR PIECE NEED TO GO TO UBR? Does it actually *belong* there. Is that the solved position (aka where UBR belongs?). The answer to this is yes. So we do the R2. We know the insertion is U' L2 U. That will move the DFL to UBR. Now we undo everything. R2, and U' L2 U. Done.*


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## TheMachanga (Aug 6, 2011)

Thanks you guys. For [R2, U' L2 U], I guess I was looking at it the wrong angle. 



Is there a version of the list of the algs ordered differently? I want them ordered as PC's, A9's, ect. Would I have to do this manually?


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## RyanReese09 (Aug 6, 2011)

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


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## Cool Frog (Aug 6, 2011)

TheMachanga said:


> Thanks you guys. For [R2, U' L2 U], I guess I was looking at it the wrong angle.
> 
> 
> 
> Is there a version of the list of the algs ordered differently? I want them ordered as PC's, A9's, ect. Would I have to do this manually?


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


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## TheMachanga (Aug 6, 2011)

No no no, I've been using that already. I mean, those are ordered by cycles. I want them ordered by move counts. I assume I'll just have to do it on excel


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## Erdos (Aug 6, 2011)

Here's another question. I can easily recognize all >9-movers since they all have such strict properties/conditions. And I recognize A9s by memorizing the four non-isomorphic cases. But I'm having trouble recognizing pure comms as quickly as I can do for all other cases. Is it just based on practiced intuition that you can immediately determine if there is a 3-move insertion? It usually takes me a while to figure this out.


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## RyanReese09 (Aug 6, 2011)

You'll eventually get that knowingness that it's an 8mover.


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## Erdos (Aug 6, 2011)

^Thanks. I figured that'd be the case, but I was secretly hoping there was a way outside of practicing, as there is for all the other cases (sort of). Hmmm.. back to practicing random cases with my cube again..


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## FatBoyXPC (Aug 6, 2011)

David, in case you missed this thread:

http://www.speedsolving.com/forum/showthread.php?29367-How-to-learn-BH-efficiently


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## TheMachanga (Aug 6, 2011)

Erdos said:


> ^Thanks. I figured that'd be the case, but I was secretly hoping there was a way outside of practicing, as there is for all the other cases (sort of). Hmmm.. back to practicing random cases with my cube again..


 
That's what I've been doing. I guess it's like learning f2l for the first time (as mike said). At first, it actually is intuitive, but then after you do it a lot, you just use muscle memory for a case that comes up. So it's not really intuitive anymore.


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## Erdos (Aug 7, 2011)

After about 1.5 weeks of learning BH, I can recognize and solve everything easily now, except for A9 and Per Specials optimally. I figure that A9s are just going to take some more practice to find the cancellation easily. And I don't really mind not solving Per Specials optimally; I'm eventually going to switch to speed-optimal algs to solve for every Per Special anyway.


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## Stephen (Sep 6, 2011)

Thank you so much!!!! I finally get commutators!


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## RTh (Sep 21, 2011)

Have to say thanks, this tutorial was really helpful =]


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## lucarubik (Sep 21, 2011)

ooh byu's tutorial! really bad in my opinion but there's nothing else...


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## Cube Equation (Sep 30, 2011)

Thanks for these helpful tutorials. I have almost entirely learned the basic concepts behind BH corners through them. But here are a few questions for the experts.

1. I've found that certain A9 commutators cannot simply be identified with two interchangeable pieces and a third piece which is either in the same layer or not capable of being inserted with a three-move insertion. Would I be correct in assuming that the miscellaneous cycles which do not satisfy the criteria for other types of commutators belong to the A9 category?

2. I have been successful is finding the cancellation in all the A9s I have tried so far. But the time this takes is not very consistent and I am wondering if there is some general tips that the more experienced BH users can provide in finding the insertion.

3. As for BH edges, which I have decided to explore despite the lack of tutorials on the subject, would it be faster than TuRBo edges? The ELL algorithms seem to be a lot more fingertrick-friendly than some of the commutators containing S and E move.

4. On BH edges again, how do I recognise the A9 and B9 commutators and distinguish them from the 10 movers? It doesn't seem to be as simple as the corners.

Please answer these questions. Thank you.


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## aronpm (Sep 30, 2011)

Cube Equation said:


> Thanks for these helpful tutorials. I have almost entirely learned the basic concepts behind BH corners through them. But here are a few questions for the experts.
> 
> 1. I've found that certain A9 commutators cannot simply be identified with two interchangeable pieces and a third piece which is either in the same layer or not capable of being inserted with a three-move insertion. Would I be correct in assuming that the miscellaneous cycles which do not satisfy the criteria for other types of commutators belong to the A9 category?


3 of the 4 general types have an "adjacent non-interchangeable pair" (explanation of this term is in byu's tutorial). The fourth is an A perm (x R' U R' D2 R U' R' D2 R2).

One case has an interchange between two pieces that messes up the third piece (R' U2 R' D' R U2 R' D R2), and another case has an interchange that doesn't mess up the third piece (x R' U2 R' D2 R U2 R' D2 R2). The other case has no interchange (R' U2 R' D R U2 R' D' R2).



> 2. I have been successful is finding the cancellation in all the A9s I have tried so far. But the time this takes is not very consistent and I am wondering if there is some general tips that the more experienced BH users can provide in finding the insertion.


For all A9s (except A perm), you need to break the "adjacent non-interchangeable pair". You can only do this by turning two sides. When you turn that side with a quarter turn, it creates a possible interchange, and when you turn that side with a half-turn, it creates a possible interchange. You sound smart so I'll let you try to work from that 



> 3. As for BH edges, which I have decided to explore despite the lack of tutorials on the subject, would it be faster than TuRBo edges? The ELL algorithms seem to be a lot more fingertrick-friendly than some of the commutators containing S and E move.


You're supposed to rotate for the cases. Generally you want to either do the commutator so that slice moves are on the M slice, or come up with a new commutator (doesn't need to be optimal in my opinion) that is better. 



> 4. On BH edges again, how do I recognise the A9 and B9 commutators and distinguish them from the 10 movers? It doesn't seem to be as simple as the corners.


I can't really help with this sorry, I don't know "BH" edge commutators. I can make fast edge commutators but I might not always be able to find the optimal one, because I don't know the 'categories'.


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## Cube Equation (Oct 2, 2011)

> For all A9s (except A perm), you need to break the "adjacent non-interchangeable pair". You can only do this by turning two sides. When you turn that side with a quarter turn, it creates a possible interchange, and when you turn that side with a half-turn, it creates a possible interchange.



This seems to work for all the A9s I have attempted. And it also makes sense intuitively. I will continue to see if it holds for the other cases.

Edit: What would be the best way to deal with parity? Currently, I set the pieces up for a PLL algorithm.


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## aronpm (Oct 2, 2011)

Cube Equation said:


> This seems to work for all the A9s I have attempted. And it also makes sense intuitively. I will continue to see if it holds for the other cases.
> 
> Edit: What would be the best way to deal with parity? Currently, I set the pieces up for a PLL algorithm.



All of the A9 cases can be reduced, by rotating, inverting, or mirroring, to those 4 cases.

For parity I do 2 extra cycles. DF-x-UB and UBR-y-ULB, where x is my last edge and y is my last corner. Then parity is just y L2 T perm L2. This is inefficient though.


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## Cube Equation (Oct 2, 2011)

> All of the A9 cases can be reduced, by rotating, inverting, or mirroring, to those 4 cases.
> 
> For parity I do 2 extra cycles. DF-x-UB and UBR-y-ULB, where x is my last edge and y is my last corner. Then parity is just y L2 T perm L2. This is inefficient though.



Then A9s should be very easy. Thanks.

Is this the general approach for parity with a direct-solving 3-cycle method? Could you propose a more efficient method?


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## aronpm (Oct 2, 2011)

Cube Equation said:


> Then A9s should be very easy. Thanks.
> 
> Is this the general approach for parity with a direct-solving 3-cycle method? Could you propose a more efficient method?


 
A more efficient method would be setting up to a PLL, and even more efficient would be setting up to a ZBLL or LL alg. Setups are difficult for me because of the locations of my buffers.


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## Cube Equation (Oct 2, 2011)

aronpm said:


> A more efficient method would be setting up to a PLL, and even more efficient would be setting up to a ZBLL or LL alg. Setups are difficult for me because of the locations of my buffers.



Is this approach commonly used? I often have difficulties undoing the setup moves into a PLL and I don't plan to learn ZBLL anytime soon.


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## aronpm (Oct 2, 2011)

PLL is common, and so is some ZBLL. You don't need to know anywhere close to full ZBLL or LL, just some algorithms like RU'R' F'UF RUR2 FRF', rU'r U2 R'FR U2 r2F , F UR'U' RD'R2 UR'U' R2 D F', RU2R'U'R'FR2U'R'U'RUR'F'RU'R', U' R U R D R' U' R D' R' U2 R' U' R U' R', R U2 R' U2 R' F R U R U2 R' U' R U R' F'.

I wish I knew all of those (and more) but I don't yet.


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## isoq58 (Nov 1, 2011)

where can i learn BH edges from ?


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## RTh (Nov 1, 2011)

isoq58 said:


> where can i learn BH edges from ?


 
From here: http://www.speedsolving.com/forum/s...lanation-of-BH-Edge-Commutator-Types&p=310411


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## Nickmaovich (Dec 4, 2011)

Very, very, very nice tutorial.
Thanks you a lot!
I can just suggest to you writing setups for each of case. I mean, seeing only "URB FRU LFU" doesn't gives me a lot of info, and as i dunno BH yet, i want to see what i'm, doing. So while reading i was expecting smth like
*Lets give a look at URB FRU LFU (setup F' L' F R' F' L F R)*, and then explaining theory. btw, thanks you a lot!
My mark is *9.9* out of 10 (i see minor misclears for me in this tutorial). Theory is very nice!


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## RTh (Apr 26, 2012)

I'm bumping this thread because it has fallen into oblivion and I, for instance, found it incredibly useful in the past. May it be useful to many others.

Also, combined with this BH edges thread: http://www.speedsolving.com/forum/s...lanation-of-BH-Edge-Commutator-Types&p=310411 It's the perfect combo for BH. Someone should bump that one as well, if Chris is OK with it.
Maybe there's a new BH thread I haven't seen yet, but still these two should be active.

@Nickmaovich Since this is advanced stuff you are expected to have a good grasp of all these concepts, the BLD nomenclature for targets, buffer and so on is used from the beginning, and to learn BH you should at least know OP.


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## lkhphuc vietnam (Jul 14, 2012)

i have some question wanna ask you guy?
what if we have a cycle has more than 3 corners. should we separate the cycle?
what if we have a 2-cycle( i mean it has only 2 corners). if we set up with 2 edges, and then we must orient the wrong pieces(what we do in oll).
and what should we do if our buffer URB already correct.
tks in advance.
p/s: i haven't studied ClassicPochman so please don't tell me to learn it. i'm using m2/3op,please answer my question. tks



Sent from my SGH-I897 using Xparent Purple Tapatalk 2


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## JianhanC (Jul 14, 2012)

lkhphuc vietnam said:


> i have some question wanna ask you guy?
> what if we have a cycle has more than 3 corners. should we separate the cycle?
> what if we have a 2-cycle( i mean it has only 2 corners). if we set up with 2 edges, and then we must orient the wrong pieces(what we do in oll).
> and what should we do if our buffer URB already correct.
> ...



3OP corners is actually a subset of BH corners, except that it has a buffer piece. If you have a cycle that has more than 3 corners, you memorise the corners in pairs. For example, you have a cycle F > E > B > Q > T > J. You memorise FE, because the piece in the buffer position has to go to F, and the piece in F has to go to E. And the piece in E will go back to the buffer piece. Then the next cycle. The piece in the buffer now has to go to B, and the one in B has to go to Q, so after FE comes BQ, and so on. 

If you have a 2 cycle left, you should probably set it up into a PLL. Just remember to orient the corners and make sure they are in the same layer. If your buffer is already solved, 'store' it in another spot, lets say R. Then when you come across the piece that belongs in R you know you're done. I hope this made sense.


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## Zane_C (Jul 14, 2012)

lkhphuc vietnam said:


> what if we have a cycle has more than 3 corners. should we separate the cycle?


If you like, you can think of all the corners as being just the one big cycle. However, once you begin execution you need to treat the corners as separate 3-cycles. 

For example, your corner memorisation might look like this: UBR>RBD>FLD>DFR>UFR
However, a more accurate way of presenting the above cycle is to separate it into: UBR>RBD>FLD and UBR>DFR>UFR


> what if we have a 2-cycle( i mean it has only 2 corners).


There are 2 situations where this can occur:
_Parity:_ If there is only a 2-cycle left to solve the corners, you have parity. Parity with BH isn't that hard to fix, but I still wouldn't bother thinking about it until you understand the main concepts behind BH 3-cycles

_No parity:_ If you don't have parity, you'll be able to break into a new cycle by shooting to an unsolved piece and thus turning the '2-cycle' into a '3-cycle'. 


> if we set up with 2 edges, and then we must orient the wrong pieces(what we do in oll).


Sorry, I'm not sure what you're asking here.



> and what should we do if our buffer URB already correct.


Break into a new cycle. 

This is simply done by shooting UBR to any unsolved piece, then continuing like normal. An alternative to breaking into a new cycle is to use 'floating buffers', in that case you would switch to a different buffer (eg. UBL instead of UBR).


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## lkhphuc vietnam (Jul 14, 2012)

JianhanC said:


> 3OP corners is actually a subset of BH corners, except that it has a buffer piece. If you have a cycle that has more than 3 corners, you memorise the corners in pairs. For example, you have a cycle F > E > B > Q > T > J. You memorise FE, because the piece in the buffer position has to go to F, and the piece in F has to go to E. And the piece in E will go back to the buffer piece. Then the next cycle. The piece in the buffer now has to go to B, and the one in B has to go to Q, so after FE comes BQ, and so on.
> 
> If you have a 2 cycle left, you should probably set it up into a PLL. Just remember to orient the corners and make sure they are in the same layer. If your buffer is already solved, 'store' it in another spot, lets say R. Then when you come across the piece that belongs in R you know you're done. I hope this made sense.



thank a lot (zane+Jianhan) C. it made lots of sense. i can't find an optimal way to set up 2 corners into a Pll and orient it, i hope can u give me an example about that

Sent from my SGH-I897 using Xparent Purple Tapatalk 2


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## jonlin (Jul 14, 2012)

Say URB-> DRF-> ULB.

D' F2 (A-perm) F2 D


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## dbeyer (Jul 15, 2012)

The original intent of the method was to always end the same way.
I used a small algorithm set to solve parity. My set up was always 1 move.

Swapping both buffers w/ a random cubie leaves to many possibilities. So, I made sure to hold the Edge swap constant. For example, my buffers were UR and URB. I would always swap UR and UB for my edge cycle. The URB swaps randomly with XYZ. This works by adding one last cycle.

For example, UR ->XY -> UB for my last cycle. This leaves UR and UB swapped. Next, swapping URB w/ XYZ is easy. Use the algorithms I have provided 3 ... maybe 5 years ago. 

The cycle of edges is easier than figuring out a setup on the fly. Also, less time undoing the setup at the end of the solve. Bam, done. Rather than oh Geeze, how did I set this one up again?!


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## dbeyer (Jul 15, 2012)

jonlin said:


> Say URB-> DRF-> ULB.
> 
> D' F2 (A-perm) F2 D



R2 D'L2D R2 D'L2D
Or maybe BLB' R2 BL'B' R2


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## Jander Clerix (Nov 13, 2013)

How can you just swap 2 pieces with BH


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## TDM (Nov 13, 2013)

Jander Clerix said:


> How can you just swap 2 pieces with BH


You can't. Swapping just two pieces is impossible. If you have to swap two pieces of whichever type of piece you're executing first, you have parity.


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## sneze2r (Nov 13, 2013)

jonlin said:


> Say URB-> DRF-> ULB.
> 
> D' F2 (A-perm) F2 D



R' D R U R D' R' U' R D R'2 D' R


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## tseitsei (Nov 13, 2013)

sneze2r said:


> R' D R U R D' R' U' R D R'2 D' R



Why not R2 D' L2 D R2 D' L2 D ??


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## IQubic (Nov 24, 2013)

Can anyone find the cancellation in this A9 commutator I made?
the commutator is a 10 move A perm, not 9 like an A9 should be.
S=L2
A=R' D' R
B=U2
SABA'B'S
L2 R' D' R U2 R' D R U2 L2

Thanks
-IQubic


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## siva.shanmukh (Nov 24, 2013)

Try l2 instead of L2. This would work since it's a corner comm.

Instead of l2, you can write x2 R2. This will cancel an R. 

L2 R' D' R U2 R' D R U2 L2 becomes x2 R2 R' D' R U2 R' D R U2 R2 x2. And this on cancelling R becomes x2 R D' R U2 R' D R U2 R2 x2. Now if you want to remove the x2, you can transform it to R U' R D2 R' U R D2 R2 (replacing U with D and D with U)


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## adiwastu (Dec 25, 2014)

thanks! 8/10


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## Gorisanson (Mar 12, 2015)

byu said:


> To recognize an Orthogonal case, you will see URB (buffer) and two of its opposites, all of which are AnI.



I don't understand what this mean..
AnI = Adjacent and non-Interchangeable
I think "Adjacent" means one corner piece can be moved to other piece by quarter turn. So, I think two corner pieces can not be both "Adjacent" and "opposite." Isn't it?

So, I think we should say orthogonal is the case where three coners are mutually opposite and non-intergangeable.
Or... maybe I misunderstand the term "Adjacent".. What "Adjacent" mean? My mother tongue is not English...


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## cmhardw (Mar 12, 2015)

Gorisanson said:


> I don't understand what this mean..
> AnI = Adjacent and non-Interchangeable
> I think "Adjacent" means one corner piece can be moved to other piece by quarter turn. So, I think two corner pieces can not be both "Adjacent" and "opposite." Isn't it?
> 
> ...



Adjacent means two corners lie on the same edge of the cube. UFR and UFL are both on the edge of the cube at the intersections of the U and F faces, so they are adjacent.

Non-interchangeable means that if you focus on a particular sticker on one corner, and a particular sticker of another corner, then you can't move one sicker to the other's location in only 1 turn.

For example, when I say UFR I mean the U sticker on the corner at the intersection of the U, F, and R faces. When I say FUR I mean the F sticker on the corner at the intersection of the U, F, and R faces. When I say RUF I mean the R sticker on the corner at the intersection of the U, F, and R faces.

UFR and UFL are interchangeable. If I do the move U', then the UFL sticker moves to the UFR location. If I do the move U, then the UFR sticker moves to the UFL location.

UFL and FUR are non-interchangeable. I can move the UFL sticker to FUR if I do L' U2, but this is more than 1 turn. I can move the sticker at FUR to UFL if I do R U2, but this is more than 1 turn. Since interchangeability requires 1 turn, then these two stickers are non-interchangeable.

Notice the UFL and FUR are also adjacent corners since they are both on the UF edge of the cube (the intersection of the U and F faces). So this means that UFL and FUR are Adjacent and Non-Interchangeable, AnI.


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## Gorisanson (Mar 12, 2015)

Thank you so much! Now I understand very clearly


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## Berd (Mar 12, 2015)

This tutorial is amazing! I'm getting there...


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## IQubic (Apr 10, 2015)

Chris, or Daniel if you know, I have a question that I think only you'd be able to answer.
Why do you split the Pure Commutators (8 movers) into three categories, Drop and Catch, Toss up, and Direct Insert?
Also, how are the cases different from each other? I understand All of BYU's tutorial so you can use the various terms he introduced in his tutorial while answering me.

Thanks in advance,
IQubic


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## cmhardw (Apr 13, 2015)

IQubic said:


> Chris, or Daniel if you know, I have a question that I think only you'd be able to answer.
> Why do you split the Pure Commutators (8 movers) into three categories, Drop and Catch, Toss up, and Direct Insert?
> Also, how are the cases different from each other? I understand All of BYU's tutorial so you can use the various terms he introduced in his tutorial while answering me.
> 
> ...



Hi IQubic,

We named them based on how they appeared to us visually. Drop and Catch are cases where the two interchangeable pieces are in the U layer (or the "base" layer) and the lone piece is in the D layer (or the layer opposite the "base" layer). Toss up has the two interchangeable pieces on D, and the lone piece on U. Direct insert is a more general term for a toss up or a drop and catch on any face.

At one point Daniel and I thought about classifying the cases based on which pieces were affected by the first turn of the A or insertion part of the commutator. For example in the cycle UBR -> RFD -> UFR which can be solved as R' D' R U R' D R U', the first turn of the A is R' and this moves all three pieces that will eventually be cycled.

You also have cycles like UBR -> RFD -> UFL which can be solved as L D' L' U2 L D L' U2 and the first move of the A, L, moves one of the interchangeable corners. I think I remember that we thought that the pure theory of it, meaning which pieces are moved on the first turn of the A, is interesting but that during a blindsolve you will be thinking in a very visual way and that it would be better to think of the cases by how they looked, even if that meant that the "same" case from the theory standpoint had different names and was considered different things.


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## abunickabhi (Sep 30, 2018)

yay 5style which is inspired by 3style and BH

https://docs.google.com/document/d/1b3tT8Wv18WdzFyY7FujyptwzYeWFH8K-UXTddTvwUWw/edit?usp=sharing


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## Underwatercuber (Oct 2, 2018)

abunickabhi said:


> yay 5style which is inspired by 3style and BH
> 
> https://docs.google.com/document/d/1b3tT8Wv18WdzFyY7FujyptwzYeWFH8K-UXTddTvwUWw/edit?usp=sharing


1. Please don’t bump 3 year old post sharing something that’s not really that related to the subject
2. Your alg counts are wrong,
3. Not even humanly possible, MU 5style is a fun idea but that’s pretty much the max potential the method has
Math for alg counts:
Corners:
Say you have a buffer piece, that leaves 7 pieces, 21 stickers.
The number of pure cycles = 21x18x15x12 = 68040
Number of twists plus two different pieces = 14x18x15 = 3780
Number of non twist cycle breaks = 7x18x15x3x3 = 17010
Number of double twists = 21x4 = 84
That’s a total of 88914.
[thats a little less than what others generally expect, that’s because while cycle breaking, you can always choose a particular sticker of a piece to break to, that reduces the number of algorithms by a big number; and also since the place in the alg where you twist a piece doesn’t matter so if you have a single twist you better do it at the end]
Also, of course, you will need the 378+14 3-style algs (3style) and 21 parity case algs (pseudo-2style?)
If you prefer to learn pseudo 4style type parity as well, which you probably will given you chose 5style; that’s an additional 21x18x15+7x18x3+21x12 = 6300
This gives you a total of 95627 algs for full 5 style with parities for corners.

Edges:
Pure algs = 22x20x18x16 = 126720
Single flip = 22x20x9 = 3960
Double flip = 55
Cycle breaks = 11x20x18x2x3 = 23760
Net 5style = 154495
You must also learn 3 style, that’s 440+11 algs more
That’s 154946 algs for edges

Even if you don’t include flips or breaks and assume it is A -> B -> C -> D -> E where A-E are targets all are on their own unique pieces it still exceeds your algorithm count
edges: 12c5 * 2^4 * 4! = 304,128
corners: 8c5 * 3^4 * 4! = 108,864

Also full 5 style is not a human possibility I believe, I don’t know if you realize how huge those alg counts are and how long it would take for you to learn them. Assuming we want to only learn the edge algs without flips and breaks over the course of our life it’s going to take a while. Let’s calculate how many algs we need to learn a day assuming we start learning the day we are born and we are finish the day we die and that we live an average human life expectancy, 304,128/(365)(79)= ~10.547. So you are learning 10.547 algs per day every day of your life and that’s not even for edge cycles with flips and breaks and we haven’t even touched corners. That’s just not possible, to put that rate of learning algs into perspective you would be able to learn full 1LLL in (~3.9k algs) in just over a year.


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## abunickabhi (Oct 3, 2018)

Underwatercuber said:


> 1. Please don’t bump 3 year old post sharing something that’s not really that related to the subject
> 2. Your alg counts are wrong,
> 3. Not even humanly possible, MU 5style is a fun idea but that’s pretty much the max potential the method has
> Math for alg counts:
> ...


I agree with you, and also the alg count table was just considering pure 5-cycle cases, where the piece to a unique piece location everytime.

I know it seems an uphill task to master 5-style , but that is what I have been working on for the past 2 years , and I am a few percent through.
"So you are learning 10.547 algs per day every day of your life and that’s not even for edge cycles with flips and breaks and we haven’t even touched corners. That’s just not possible, to put that rate of learning algs into perspective you would be able to learn full 1LLL in (~3.9k algs) in just over a year." , Your statement considers a linear way of learning algs , which is not the case while learning cube algorithms, generally the learning process accelerates due to common idea between algs, symmetries and mirrors.
I am working on extracting these common idea for 5 style.

I agree this method can appear as unfeasible for any cuber who has just dabbled in blindfold solving , but in my case , I have put a lot of thought before deciding to take the plunge from 3-style to 5-style.

Let's consider a scenario , that I do 3-style for 5 years , and get really fast with it. There is high chance I will start losing motivation in it and quit blindsolving for a while.

5-style is just like a never ending (ZBLL^2 like method) , and it can keep motivation running for a BLDer to keep making new algs, along with the advanced technique algsets of parity algs, and floating buffer algs.


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## abunickabhi (Oct 3, 2018)

Underwatercuber said:


> Also full 5 style is not a human possibility I believe, I don’t know if you realize how huge those alg counts are and how long it would take for you to learn them. Assuming we want to only learn the edge algs without flips and breaks over the course of our life it’s going to take a while. Let’s calculate how many algs we need to learn a day assuming we start learning the day we are born and we are finish the day we die and that we live an average human life expectancy, 304,128/(365)(79)= ~10.547. So you are learning 10.547 algs per day every day of your life and that’s not even for edge cycles with flips and breaks and we haven’t even touched corners. That’s just not possible, to put that rate of learning algs into perspective you would be able to learn full 1LLL in (~3.9k algs) in just over a year.



Sorry for bumping up this post , but could not find another suitable thread to post this topic on.

If you give this method a quick thought (thought made upto a few days) , it would seem like an unrealistic pipedream when you realise the sheer number of algorithms in it , and that is the reason why I procrastinated for nearly 3 years before starting out on learning 5-cycles (from 2014 to 2016).

The doc had more content about just plain comparison , as I just wanted to address all kinds of scepticism that people might have when they hear about 5 cycles for the first time.

Few people are learning 1LLL and that itself is a massive undertaking, and by your linear scale of improvement , it should take ~1 year.

Your statement considers a linear way of learning algs , which is not the case while learning cube algorithms, generally the learning process accelerates due to common ideas between algs, symmetries and mirrors. So , I am banking on the acceleration in my learning to atleast memorise a size-able chunk of this algset. 

At first , this method might always feel like worth to give up on as there is no instant reward for quite a while, but ones you start to invest time and effort into it, the learning process can gain traction , and it would not seen like a mammoth task.

(Also I plan to do the 4-cycles , and the 5-cycles having flipped edge as target in them , after i pan out the Pure edge cycles = 22x20x18x16 = 126720 algs , pure corner cycles = 21x18x15x12 = 68040).

For flipped edges case, there is already cool resources out there, which have multiple algs for multiple cases of the flipped edge.


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## Underwatercuber (Oct 3, 2018)

abunickabhi said:


> Sorry for bumping up this post , but could not find another suitable thread to post this topic on.
> 
> If you give this method a quick thought (thought made upto a few days) , it would seem like an unrealistic pipedream when you realise the sheer number of algorithms in it , and that is the reason why I procrastinated for nearly 3 years before starting out on learning 5-cycles (from 2014 to 2016).
> 
> ...


 Even with mirrors and setups it’s still thousands of algs and not humanly possible to ever learn in your lifetime, also while going through the sheet I realized lots of the algs are pretty bad and you will have to generate thousands of algs. You can fail to learn it if you want but it would be far more benificial and humanly possible to learn floating and floating parity


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## abunickabhi (Oct 5, 2018)

Underwatercuber said:


> Even with mirrors and setups it’s still thousands of algs and not humanly possible to ever learn in your lifetime, also while going through the sheet I realized lots of the algs are pretty bad and you will have to generate thousands of algs. You can fail to learn it if you want but it would be far more benificial and humanly possible to learn floating and floating parity



Hmmm , I know the set seems humongous , and about half the algorithms pruned by me can be improved upon.

But I do not think it is humanly impossible.

The pattern recognition of our brain is even better than future supercomputers and quantum supercomputers that are going to come up.

If we focus enough , our pattern recognition will be able to digest a method like 5 cycle , in about 10 years at ease , and in 4 years in you really try hard. (It is just a rough estimate)

But , the way you are running your thought process of linearly extrapolating the time taken (10 algs per day or so) , it is not the best way to think about learning.

Again, my opinion will be of no value unless I prove it myself. (I am 2 years into this method , and almost got into the groove (15-20% confidence level))


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## Jezzular (Oct 5, 2018)

abunickabhi said:


> The pattern recognition of our brain is even better than future supercomputers and quantum supercomputers that are going to come up.



Thats wrong. Even today there are already lots of applications, where computers are better in pattern recognition than humans. It is estimated that in 10-20 years computers are better in almost every aspect of pattern recognition than humans.


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## Underwatercuber (Oct 5, 2018)

abunickabhi said:


> Again, my opinion will be of no value unless I prove it myself. (I am 2 years into this method , and almost got into the groove (15-20% confidence level))


Are you saying you have 15-20% of the algs down?


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## abunickabhi (Oct 9, 2018)

Underwatercuber said:


> Are you saying you have 15-20% of the algs down?


Yes 15-20% algs down (~17,500) , and my recall of the alg is improving at a slow pace , but it is still improving.


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## DGCubes (Oct 9, 2018)

abunickabhi said:


> Yes 15-20% algs down (~17,500) , and my recall of the alg is improving at a slow pace , but it is still improving.



That's extremely impressive. Do you learn the algs somewhat intuitively, or is it just straight memory? How much time do you/have you put into learning these algs?


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## Underwatercuber (Oct 9, 2018)

DGCubes said:


> That's extremely impressive. Do you learn the algs somewhat intuitively, or is it just straight memory? How much time do you/have you put into learning these algs?


As far as I’m aware most 5 style algs aren’t very intuitive so probably straight memory.

I would also like to know how much time you invest into this and also how you get the algs, do you gen them and have a sheet somewhere with them? How optimal are they?


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## mark49152 (Oct 9, 2018)

abunickabhi said:


> Yes 15-20% algs down (~17,500)


Edges or corners? Which sets, and how did you learn them?


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## abunickabhi (Oct 14, 2018)

DGCubes said:


> That's extremely impressive. Do you learn the algs somewhat intuitively, or is it just straight memory? How much time do you/have you put into learning these algs?



I have not measured the time I have put in for learning the algs, (but I started off 2 years back)

To memorise the algorithms , I use a story making process : https://drive.google.com/open?id=1SW4sAZ3RSENqmRNAah-CzCiPZm4nmj_T

If the alg cannot be brought to the form of [C:[A,B]] , then I just have to memorise them in blocks of 4 turns each.


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## abunickabhi (Oct 14, 2018)

mark49152 said:


> Edges or corners? Which sets, and how did you learn them?



For edges mainly , for corners , I am still skeptical on whether , the algs are any better than 2 3-style algs.

Cube explorer is basically weak in finding good 5 cycles for corners , and one of the reason can be due to bad orientation of one of the corner in the 4 corners that need to be cycled, the overall algorithm can be long and inefficient. To make 5 style work on corners , some addendum kind of work has to be done , where 3 style is extended out in some likely and unlikely cases , and categorised , and scaled to 5 cycles fully. 

So , my focus has been on learning edge algorithms till now.

(Corners are tough)


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## abunickabhi (Oct 14, 2018)

Underwatercuber said:


> As far as I’m aware most 5 style algs aren’t very intuitive so probably straight memory.
> 
> I would also like to know how much time you invest into this and also how you get the algs, do you gen them and have a sheet somewhere with them? How optimal are they?



I am getting the algs via cube explorer, there is no other viable and more fundamental tool I have found. (comparing to ksolve , korf solver)

They are optimal in <U,R,L,F,D> , I try to avoid B moves , and y rotations.

Sheet link : 5 style algs

Medium : https://medium.com/@abhijeetgokar/5...ger-tricky-3x3-blindfolded-method-582d05e8fcc


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## Underwatercuber (Oct 15, 2018)

abunickabhi said:


> I have not measured the time I have put in for learning the algs, (but I started off 2 years back)
> 
> To memorise the algorithms , I use a story making process : https://drive.google.com/open?id=1SW4sAZ3RSENqmRNAah-CzCiPZm4nmj_T
> 
> If the alg cannot be brought to the form of [C:[A,B]] , then I just have to memorise them in blocks of 4 turns each.





abunickabhi said:


> For edges mainly , for corners , I am still skeptical on whether , the algs are any better than 2 3-style algs.
> 
> Cube explorer is basically weak in finding good 5 cycles for corners , and one of the reason can be due to bad orientation of one of the corner in the 4 corners that need to be cycled, the overall algorithm can be long and inefficient. To make 5 style work on corners , some addendum kind of work has to be done , where 3 style is extended out in some likely and unlikely cases , and categorised , and scaled to 5 cycles fully.
> 
> ...


Just making sure I understand, so you have to recall a story everytime you want to use an alg? Do you ever get that mixed up with images/stories in bld/mbld?


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## abunickabhi (Oct 15, 2018)

Underwatercuber said:


> Just making sure I understand, so you have to recall a story everytime you want to use an alg? Do you ever get that mixed up with images/stories in bld/mbld?



Rather than a story , it is just a snippet or a mental note.
It does not get mixed up with MBLD images , as they are like mantras, and quite fixed into my head , as compared to the temporary sticking images of the MBLD/3BLD cube.


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## Gian21x (Oct 18, 2018)

Are the original videos of this tutorial still available? It seems I can't find It anymore


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## abunickabhi (Oct 22, 2018)

Gian21x said:


> Are the original videos of this tutorial still available? It seems I can't find It anymore



The original videos have been taken down (On youtube) a few years back. Luckily I have downloaded it a few years back and have it on my drive. PM for the link.


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## Gian21x (Oct 22, 2018)

abunickabhi said:


> The original videos have been taken down (On youtube) a few years back. Luckily I have downloaded it a few years back and have it on my drive. PM for the link.


Thnks mate, I don't really need It, because I'm using proto-3style, I was asking my self if the guy Who made this amazing tutorial was still cubing


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## Thom S. (Oct 22, 2018)

Gian21x said:


> I was asking my self if the guy Who made this amazing tutorial was still cubing



Brian Yu didn't login since 2011 and his last competition was in 2009 so I guess he's not as serious about cubing anymore


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## lucarubik (Oct 22, 2018)

i remember those videos... fun times


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## dbeyer (Dec 7, 2018)

Gian21x said:


> Thnks mate, I don't really need It, because I'm using proto-3style, I was asking my self if the guy Who made this amazing tutorial was still cubing



Not sure about the guy who curated this content. But I hear both the creators of the method itself are an actuary and engineer now.


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