Blindsolving Discussion and Help

what are the M2 algs I need to know? (when i need to shoot to a place in the m slice) I can use a 2 edge flip and learn less algs.
where can I learn R2? or is there a more popular (faster) method?

Do really good bld solvers do something to enhance there general memorizing abilities?
 
the only m2 algs you need to know for the m slice are

U2 M' U2 M'

M U2 M U2

M2

this means that you may end up with some flipped M slice pieces but as you said you can use pure edge flips (such as a 2flip) to fix edges at the end.

You can learn R2 on stefan pochmann's site. In terms of "faster methods" there really isn't a definitive answer. Methods don't have speeds, Cornelius Dieckmann can solve the corners using Old pochmann and use M2 edges and get 40 second solves. Most people would agree that using commutators is one of the fastest and most efficient ways to solve corners BLD. The BH method is the most popular complete method for BLD solving with comms, though many just use freestyle (using comms/PLLs/whatever) with or without a fixed buffer.

The thing that really good bld solvers do to enhance their general memorizing abilities is

practice
 
rubiksarlen said:
Hey everyone,

can you guys take some time out to do me a favor? just post the memo for M2 edges of this scramble:

R D B2 U' F2 B' U' B U' L2 U B' U F2 U F' B' D' L F R2 U2 R' D R2

and if you still have time, help post the memo for this scramble too :):

B' R' L' U2 L R D' B' R' L2 U2 F2 R F2 D2 U' F2 B2 D2 L2 R D2 R D' F

remember, just for M2 edges. i need your help cos right now i don't know if i''ve really mastered this method or whether i'm just making mistakes. i've about more than 20 attempts with M2 and still not even one single success: :(
Click to expand...

I'm scrambling with the WCA standard and holding Y/U and O/F while solving

1. UR BR FR LF LU BU DL DR UF LB BD FU -> CNGR QvuM Wagon Volks DSPEeee
2. Buffer Filled - UB FU DL LU FR BD LB RB RF LU DL UB -> A E WeQ (pronounced Wek) GP (moto GP) KeSRI VAaaa
 
MaeLSTRoM said:
How fast is it possible to get using Old Pochmann for edges+corners?
Click to expand...

Very theoretical, but let us look at it this way:

An "average" BLD solve contains 7 corner-targets and 11 edge-targets.

Lets say you are Breandon-class and perform your T´s, J´s, R´s and Y´perms at an average 1sec per algorithm. This means you can shoot 11 edge-targets + Rperm (Parity) + 7corner-targets in 19 seconds. If you take into account the setup-moves and undo´s, this may result in 30 second solves. But trust me, your hands will get tired of spamming these algorithms. I believe in a 30sec OP solve.
 
Hey, I wasnt sure if this is the right place to be asking this, but:

How many unique corner orientations are there? I mean full cube. I figured that unique would mean that it could be solved in the form x2(Opt.) U(x) D(y) Alg

Thanks :D Alternatively, if there is something you can just tell me to edge me on to do the math, feel free :P Also, if you have on that doesnt take into account x2, that is good too.
 
Can someone post the r2 memo for this 4x4 scramble please? i'm having a hard time learning this method and I usually misunderstand the edge orientation

U2 r F R' u' B' D B D L U u D2 F B2 D' F' u' f D' F r f u2 r' f D2 U f' F2 B' U u2 D2 R2 r2 D2 u2 F' f
 
y235 said:
what is Cyclic shifts?
Click to expand...

Cyclic Shifts are BH commutators that are length 10, optimally. They are in the form A= 2 (_ _, usually R' F or F R' when rotated) move setup B = 1 move swap (_2, usually U2 when rotated) . _r means the direct reversal of it (doing it backwards without switching primes etc) A B A' Ar B Ar'
So A basic one would be R' F U2 F' R F R' U2 R F'

They are used when the 3 corners are in the same layer, but nothing is interchangable. If you were wondering how to use them, just let me know and I could continue an explanation.
 
minime12358 said:
Cyclic Shifts are BH commutators that are length 10, optimally.
Click to expand...

Yes, we use cyclic shifts for this case in BH. More generally a cyclic shift is starting an algorithm somewhere other than the first term, executing it all the way to the end, and then wrapping back around to the beginning.

The BH "cyclic shift" case is: F R' U2 R F' R' F U2 F' R

It's really based on this commutator: F' R F R' U2 R F' R' F U2

This commutator is a non-optimal way to perform the cycle: UBL -> RDF -> URF

However, it just so happens that you create an optimal length alg for the case UBL -> BRU -> RFU if you start on the 3rd move and perform a cyclic shift.

F' R F R' U2 R F' R' F U2

Start on the highlighted F and execute all the way until the end. After you reach the last U2, then perform the underlined F' R to finish off the cyclic shift.

What you're really doing is taking the original algorithm and performing the setup sequence (R' F) before beginning the algorithm.

(R' F) (F' R F R' U2 R F' R' F U2) (F' R)

Notice the cancellations, and this creates the case that Daniel and I have named the "cyclic shift" case. The reason we gave it that name is because it is the only case where we mentally performed a cyclic shift on a "common" commutator to create an optimal length algorithm for a particular case.
 
cmhardw said:
Yes, we use cyclic shifts for this case in BH. More generally a cyclic shift is starting an algorithm somewhere other than the first term, executing it all the way to the end, and then wrapping back around to the beginning.

The BH "cyclic shift" case is: F R' U2 R F' R' F U2 F' R

It's really based on this commutator: F' R F R' U2 R F' R' F U2

This commutator is a non-optimal way to perform the cycle: UBL -> RDF -> URF

However, it just so happens that you create an optimal length alg for the case UBL -> BRU -> RFU if you start on the 3rd move and perform a cyclic shift.

F' R F R' U2 R F' R' F U2

Start on the highlighted F and execute all the way until the end. After you reach the last U2, then perform the underlined F' R to finish off the cyclic shift.

What you're really doing is taking the original algorithm and performing the setup sequence (R' F) before beginning the algorithm.

(R' F) (F' R F R' U2 R F' R' F U2) (F' R)

Notice the cancellations, and this creates the case that Daniel and I have named the "cyclic shift" case. The reason we gave it that name is because it is the only case where we mentally performed a cyclic shift on a "common" commutator to create an optimal length algorithm for a particular case.
Click to expand...

Ahh yes, that clarifies a lot to me too, thanks ^^

And you dont happen to know the number of corner orientations do you? I was figuring a raw number _might_ be something like:

8 cases for no parity = 64

6 (?) cases for the next parity = 36

the same thing over = 36

= 136. Then exclude x2 = 68 ?
 
minime12358 said:
And you dont happen to know the number of corner orientations do you? I was figuring a raw number _might_ be something like:

8 cases for no parity = 64

6 (?) cases for the next parity = 36

the same thing over = 36

= 136. Then exclude x2 = 68 ?
Click to expand...

I don't know this number. I'm thinking that the calculation would be quite a bit more difficult to do than a very simple lines like this. There will be many isomorphic cases for each true possible case. Richard Carr studied the corner orientations in quite a lot of depth a couple years ago. If anyone would know the answer to this fairly definitively, then he would. I would try to send him an e-mail, he's a member of this forum.
 
How would you guys solve these following cases with freestyle/BH with a DF buffer?
(DF-FU-BU) (DF-BU-FU) (DF-BU-BD) (DF-BD-BU) (DF-FU-BD) (DF-BD-FU)
 
U' M U M U2 M' U M' U
U M U M U2 M' U M' U'
x' U M' U M' U2 M U M U' x
x' U' M' U M' U2 M U M U x
xy' U M' U R2 U' M U R2 U2 yx'
xy' U2 R2 U' M' U R2 U' M U' yx'
 
F' E F U2 this idea for the first 4
F2 D2 F E F' D2 F E' F this for the last 2
I would make z to fingertrick better
you can also transform into a half slice doing something as M U M and finding a cancell this would be 9 2gen moves i think...
U M U' M U2 M' U' M' U'
 
U' M U M U2 M' U M' U
U M U M U2 M' U M' U'
x' U M' U M' U2 M U M U' x
x' U' M' U M' U2 M U M U x
xy' U M' U R2 U' M U R2 U2 yx'
xy' U2 R2 U' M' U R2 U' M U' yx'
Click to expand...
Do these work on big cubes too(M' becomes r)?
If not, what would you do for these cases on wings for 4x4 and 5x5?
 
Last edited:
No

For big cubes I would do those for edges, but for wings they are different:
xy [U' r U, R2] y'x'
x' [l D2 l', U2] x
Same alg but with r instead of M'
 
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