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BH Tutorial

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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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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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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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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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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...
 

cmhardw

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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...
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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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
 

cmhardw

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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
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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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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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.
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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