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BH Tutorial

DYGH.Tjen said:
Scramble: F' L U B' L2 U2 D2 F R' B' L2 F2 D2 L F' R2 U2 F2
Click to expand...
(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.
 
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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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
Click to expand...

Freestyle can refer to BH using floating buffers.
 
DYGH.Tjen said:
Scramble: F' L U B' L2 U2 D2 F R' B' L2 F2 D2 L F' R2 U2 F2
Click to expand...

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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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

Take your time. Thanks,
tjen (: happycubing
Click to expand...

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

Code: 
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.
 
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.
 
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?
 
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.........?)
Click to expand...
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)
Click to expand...
R2 moves DFR to UBR. That's the interchange. U'L2U is the insertion, moving DLF to UBR without changing the R slice.
 
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'.
Click to expand...
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.........?)
Click to expand...
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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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.
 
^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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