Blindsolving Discussion and Help

[mati rubik]

Copied from an earlier posst in the "official BLD alg list" thread:

Old Pochmann corner alternative corner target to reduce set up moves: Shoot to DRF instead of RFD
My favorite from the Cube Explorer results:

F2 U' F2 D R2 D' R2 U R2 U' R2
=
y' R2 U' R2 D y R2 D' R2 U R2 U' R2 as I execute it
=
y' R2 U' R2 u R2 D' R2 U R2 U' R2

Thanks, I love it

now I just need one for the most anoying set-up, FDR

 

[Kynit]

F' R cancels into a Jperm; is that good enough?

EDIT: or did you mean for targets that set up to FDR?

 

[amostay2004]

Why would you wanna find algs for different targets? All setups can be done in 2 or less moves, and I think the Y perm with cancellation is much faster than the algs you found. Also you can always use J perms too

 

[mati rubik]

F' R cancels into a Jperm; is that good enough?

EDIT: or did you mean for targets that set up to FDR?

yes

@Amos: yes, but I hate some set-up, and it's always good to know more algs

 

[RyanReese09]

If you're trying to setup FDR to be solved then just do R, Y perm R'. Unless I'm misunderstanding. No setups are annoying at all except RUB, which isn't even that bad.

 

[Kynit]

If you're trying to setup FDR to be solved then just do R, Y perm R'. Unless I'm misunderstanding. No setups are annoying at all except RUB, which isn't even that bad.

RUB: R2 F' Yperm F R2
Both Fs are canceled, and the R2s become R'/R.

 

[RyanReese09]

RUB: R2 F' Yperm F R2
Both Fs are canceled, and the R2s become R'/R.

Thanks, I was previously doing either R' F' U2 (J perm no AUF) U' F R

Or R' D' R Yperm undo

I like your way better.

 

[RyanReese09]

I know that column recognition is one interchange, and all corners twisted a different way. Though how would you recognize this blindfolded? Even if I look at the cycle URB LDF RUF (like placing my fingers on the stickers) I can't tell it's a column, I can tell there is the interchange but how do most people tell if the corners are twisted?

BYU says I can just setup into Cyclic shift (which I use x' U2 R' U2 R' D R U2 R' D' R2 U2) for, and the inverse for the opposite direction of the cyclic shift (aka that alg solves counter clockwise, inverse clockwise)
How would I solve URB LDF RUF via setup into cyclic shift?

 

[cmhardw]

I know that column recognition is one interchange, and all corners twisted a different way. Though how would you recognize this blindfolded? Even if I look at the cycle URB LDF RUF (like placing my fingers on the stickers) I can't tell it's a column, I can tell there is the interchange but how do most people tell if the corners are twisted?

Columns are when you have two corners that are interchangeable across a face. The 3rd corner is AnI (Adjacent non-Interchangeable) to one of the first two, and diagonally opposite the other corner across the entire cube. I don't really think about the twists of the corners per se. I look for the two interchangeable corners, then make sure that the 3rd is AnI to one of them, and diagonally across the cube from the other. Basically you form a right triangle which includes the longest diagonal of the cube. In that triangle you have 2 interchangeable corners, and one AnI corner.

BYU says I can just setup into Cyclic shift (which I use x' U2 R' U2 R' D R U2 R' D' R2 U2) for, and the inverse for the opposite direction of the cyclic shift (aka that alg solves counter clockwise, inverse clockwise)
How would I solve URB LDF RUF via setup into cyclic shift?

Pure alg is: D2 B' R2 B D' B' D R2 D' B D'
I would probably execute it as: z' x' R2 U' F2 U R' U' R F2 R' U R' x z

Having said that, for the case URB -> LDF -> RUF I would actually do:
x L U2 L' U' R2 U L U' R2 U' L' x'

 

[RyanReese09]

Ok so that case is a column because the interchange on the one face can be F2, and the Ani is the URB piece with the RUF?

I just have to think that the 3rd piece (the first two being the two pieces that are interchangable) is Ani with one of them, and a polar opposite with the other (does it have to be opposite or polar opposite?

How would I go about finding the correct alg for solving each case?

 

[cmhardw]

Ok so that case is a column because the interchange on the one face can be F2, and the Ani is the URB piece with the RUF?

Yep, that's a good way to notice these from different angles/parts of the cube.

I just have to think that the 3rd piece (the first two being the two pieces that are interchangable) is Ani with one of them, and a polar opposite with the other (does it have to be opposite or polar opposite?

Yes, for it to be a columns case that 3rd corner must be polar opposite to one of the corners. Again, use the right triangle visualization if that helps. That's how I view all columns cases: a right triangle of corners where two are interchangeable and one is AnI (the triangle implies that one corner is polar opposite to the another corner).

How would I go about finding the correct alg for solving each case?

I don't solve these by doing a setup into a cyclic shift, but Daniel does. I use a setup into an A9. Either way, the first move still moves the same corner.

We call this case columns because the two AnI corners visually form a "column". The UBR and DBR stickers form a very obvious 3x1x1 column, visually, on the cube. When solving a columns case, the first move will move the 3rd corner, the one not part of the column.

Solve by setting up into a cyclic shift:
Move the 3rd corner 1 quarter turn such that it is adjacent to the AnI column.

Example: UBR -> DBR -> UFL
Setup move is F, as this moves the UFL corner adjacent to the column. This is now the cycle UBR -> DBR -> RFU, which is a cyclic shift case on the R face that will cancel 1 turn with the setup move.

Notice that you could also have done L' as a setup turn at the start.

Solve by setting up into an A9:
Move the 3rd corner 1 quarter turn such that it is still opposite to the column across a face.

Example: UBR -> DBR -> UFL

The setup move is L, as this brings the UFL corner (the one not in the column) down to the DFL spot. DFL is still opposite to the column across the D face. This creates the cycle UBR -> DBR -> FDL. This cycle is an A9 that will cancel 1 turn with the setup move.

Notice that you could also have done F' as a setup turn at the start.

 

[Sakarie]

A tip for Columns and Cycle Shifts that I learned from Riffz is to learn how to handle R U R2 U' R' to insert pieces. That could make
UBR -> DBR -> UFL: (U2, l U R2 U' l')
URB -> LDF -> RUF: (F2, R U R2 U' R')

 

[cmhardw]

A tip for Columns and Cycle Shifts that I learned from Riffz is to learn how to handle R U R2 U' R' to insert pieces. That could make
UBR -> DBR -> UFL: (U2, l U R2 U' l')
URB -> LDF -> RUF: (F2, R U R2 U' R')

Wow, I really like this approach! It's very reminiscent to me to how U' L2 U L2 U inserts UFL to DBL for the Per Special: UBR -> UFL -> DBL

For those who don't solve Per Specials the same way, for UBR -> UFL -> DBL I would do:
U' L2 U L2 U B2 U' L2 U' L2 U B2

Sakarie/Riffz, I like this approach to columns - it's very easy to visualize and understand what's happening. :tu

 

[JonnyWhoopes]

Dear Lord I love this thread so much.

 

[RyanReese09]

For UBR -> DBR -> UFL cycle I'd just do D' and do a per special to be honest, do you have a better example Chris? Look at this cycle URB LDF RUF

How would you go as to finding the alg for that case? I solve A9's via setup, 8 mover, undo setup. 10 movers. (I can't recognize A9 either for some reason...)

 

[amostay2004]

Look at this cycle URB LDF RUF

How would you go as to finding the alg for that case? I solve A9's via setup, 8 mover, undo setup. 10 movers. (I can't recognize A9 either for some reason...)

[R U R2 U' R', F2]

 

[RyanReese09]

What's the theory behind it though? How did you create that commutator? I'd rather not just learn the alg for each case. Sorry if my question was unclear.

 

[aronpm]

RUR2U'R' inserts UBR into RFU. Pretty easy to understand past that if you know 8 movers.

 

[RyanReese09]

So it's 5 moves to insert it always..any general rules to know for the insertions that are different then the rules for 8 movers? I assume the 3rd corner that's being used for the interchange can't be effected. And is the 5 move insertion always something like RU style where only 2gen moves are used? Or should I find the 5 move insertion using any combination of moves to just get the insertion in the correct twist?

BYU didn't cover columns well at all in my opinion.

 

[cmhardw]

For UBR -> DBR -> UFL cycle I'd just do D' and do a per special to be honest, do you have a better example Chris? Look at this cycle URB LDF RUF

How would you go as to finding the alg for that case? I solve A9's via setup, 8 mover, undo setup. 10 movers. (I can't recognize A9 either for some reason...)

Hey Ryan,

Before I explain how to do UBR -> LDF -> RUF you first have to understand how the more "standard" rotation of this case works.

Let's first do UBR -> DBR -> UFL

You'll see in a moment that this is the exact same case that you just asked me about, only at a different angle.

First I would see that the pieces form a right triangle. One leg of the triangle is formed by the line from UBR to DBR. The other leg is from UBR to ULF. This leaves the hypotenuse to be from ULF to DBR, the polar diagonal through the cube.

Once I see a triangle that has the hypotenuse as the polar diagonal I care only about two things:
1) Find the AnI corners (the "column")
2) Move the other corner so that it is still diagonally across from the column across a face

1) The AnI corners are UBR and DBR.
2) I have to move the other corner, ULF, such that it is still across from the column across a face. The move to do this is L. Now, instead of UFL being across from the URB piece across the diagonal on the U face, now the corner is at FLD and is across from the DBR corner (the "bottom" of the column) across the D face.

The rest of the commutator is a 9 mover.
UBR -> DBR -> FLD

Setup turn is U. This makes the cycle: UFR -> DBR -> FLD which is done via:
U L' U' R2 U L U' R2

Combine this with the setup move U, which will cause a 1 turn cancellation, and you get the 9 mover:
U2 L' U' R2 U L U' R2 U'

Now, combine this 9 mover with the initial L move setup to get the full 11 mover column alg which is:
L U2 L' U' R2 U L U' R2 U' L'

This alg is also very finger trick friendly for me.

-------

Now onto your case: UBR -> LDF -> RFU

Find the triangle. Legs are from UBR to RFU and from RFU to LFD. Hypotenuse is from UBR to LFD. This is an awkward angle, so rotate the cube with x.

This makes the cycle: BRD -> LFU -> RBU
Or, if you prefer to viewpoint shift as I do rotate every sticker once clockwise to get:

DBR -> UFL -> UBR or UBR -> DBR -> UFL. Now it's the exact same cycle as above.

Trust me in that it sounds SO much harder than it is. Columns are so easy to recognize once you know to keep the column intact during the first setup turn. All you do is move the "other corner" either from the "top" of the column to the "bottom" or vice versa. In your example we had to move the other corner from the "top" of the column" to the "bottom" of the column with our setup turn L. As long as you're comfortable with that particular 9 mover case that follows, then the only real thought is either: A) The other corner is at the top of the column, ok move it to the bottom, or B) The other corner is at the bottom of the column, ok move it to the top.