BH Tutorial

[RyanReese09]

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.

 

[MrMoney]

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

 

[RyanReese09]

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

 

[aronpm]

UBR LBU FUR

 

[joey]

UBL RBU RFU.

 

[RyanReese09]

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?

 

[aronpm]

nubs can't do niklas

 

[Toad]

lolryan

 

[RyanReese09]

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.

 

[aronpm]

My cycle and Joey's cycle are the same

 

[RyanReese09]

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.

 

[joey]

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

 

[RyanReese09]

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?

 

[aronpm]

How was that not a real cycle?

 

[MrMoney]

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.

 

[RyanReese09]

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.

 

[Kynit]

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?)

 

[RyanReese09]

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)?

 

[uberCuber]

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.

 

[Kynit]

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

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

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).

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.