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BH Tutorial

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.
 
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.
 
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'

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.
 
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
 
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

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

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

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

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

==A9==
A:B - 1 move setup to adj/opp interchangable
B:C - same layer
A:C - 1 move setup to adj/opp interchangable

==Orthogonal==
A:B - opposite non-interchangeable
B:C - opposite non-interchangeable
A:C - opposite non-interchangeable

==Columns==
A:B - adjacent non-interchangeable
B:C - opposite interchangeable
A:C - polar opposite

==Cyclic Shift==
A:B - adjacent non-interchangeable
B:C - adjacent non-interchangeable
A:C - adjacent non-interchangeable

==Per Special==
A:B - opposite interchangeable
B:C - opposite interchangeable
A:C - opposite interchangeable

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.
 
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.

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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