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BH Tutorial

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fanwuq
#25
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?
 

byu

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Thread starter #30
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.
 

byu

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Thread starter #33
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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#36
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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#37
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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#38
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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#40
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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