Blindsolving Discussion and Help

[Ollie]

Where can I find an easy memo tutorial? Please name more than one.

Most tutorials will include something on how to actually memorize the pieces - if not then it's not a good tutorial. Do you mean how to get better at memo?

 

[rj]

Most tutorials will include something on how to actually memorize the pieces - if not then it's not a good tutorial. Do you mean how to get better at memo?

No, how to memo. I can't find one that I can understand.

 

[TheNextFeliks]

No, how to memo. I can't find one that I can understand.

Noah's video on cubing world is reaaaaaallllllllllllllllllllly easy to understand.

 

[rj]

Noah's video on cubing world is reaaaaaallllllllllllllllllllly easy to understand.

Thanks.

 

[BaconCuber]

I'm thinking about learning blind, but I don't know a good video tutorial. I've tried several times without success with Badmephisto's tutorial. If there are none better than that, I'll try with his again.

 

[TheNextFeliks]

I'm thinking about learning blind, but I don't know a good video tutorial. I've tried several times without success with Badmephisto's tutorial. If there are none better than that, I'll try with his again.

Literally two posts ago. Noah's tutorial on cubing world.

 

[BaconCuber]

Oh, derp.

 

[A Leman]

I have known some cases that are fast for 3style that I don''t use in solves because they are not based on commutators.

I want to make the switch to some of these MU cases, but I am worried about mistaking a U turn in execution or doing the wrong case by accident. I am talking about cycles like MUM'UMUM'U and M'UM'UMUM'UM2 which I would currently do as [y':MDM',U] and [x U2:M',ULU'] which are not as nice and need rotations. If someone could explain the thinking behind the MU cases so I could understand them in solves, then that would be great. Also, do other people have problems with cycles like this?

 

[Applecow]

i don't really like the MDM ones, but i have made up some other ones MU ones which are nice for these cases. i would have problems if i used them though.^^

 

[DrKorbin]

MUM'UMUM'U

So you confuse UF>UR>BD and UF>UL>BD and their backwards.
1) If your first target is UL or UR, then you do U or U', putting this target to a buffer (U for UR, U' for UL). Then you do M and M' repeatedly, turning upper face in one direction.
2) If your first target is BD, then you do M, "putting" this target to a buffer, your next move will be U or U', but the logic is opposite to a previous case: you move a buffer into your second target (U for UL, U' for UR).

 

[A Leman]

So you confuse UF>UR>BD and UF>UL>BD and their backwards.
1) If your first target is UL or UR, then you do U or U', putting this target to a buffer (U for UR, U' for UL). Then you do M and M' repeatedly, turning upper face in one direction.
2) If your first target is BD, then you do M, "putting" this target to a buffer, your next move will be U or U', but the logic is opposite to a previous case: you move a buffer into your second target (U for UL, U' for UR).

Thank you this helped a bit. I normally think of putting the buffer into a target or the last target into the buffer so I didn't think of sticking the first target into the buffer. I think that I am going to switch the cases now.

 

[TDM]

D2 L' D2 R2 F D' F' B' R2 L D2 F2 L2 U' B2 U R2 F2 U2 B2 R2
How I would solve the corners (using Boomerang):
x2 y' R' U R U2' R' U2' R' U' R U' R' U2 R U2' R U y x2
y' U' R U' R2' U' R U' R' U2' R' U2 R' U' R U' R2' y
D2 R2 U R2' U' R2 D2
D' R2 U R2' U' R2 D
M2' U2' M2' U' M2' U2' M2' U'
D R2 U R2' U' R2 D'
R2 U R2' U' R2
Odd number of cycles = parity
But when there's parity in Boomerang, UBR and UFL should be swapped. Which they aren't.
So is this parity or not?

 

[TheNextFeliks]

D2 L' D2 R2 F D' F' B' R2 L D2 F2 L2 U' B2 U R2 F2 U2 B2 R2
How I would solve the corners (using Boomerang):
x2 y' R' U R U2' R' U2' R' U' R U' R' U2 R U2' R U y x2
y' U' R U' R2' U' R U' R' U2' R' U2 R' U' R U' R2' y
D2 R2 U R2' U' R2 D2
D' R2 U R2' U' R2 D
M2' U2' M2' U' M2' U2' M2' U'
D R2 U R2' U' R2 D'
R2 U R2' U' R2
Odd number of cycles = parity
But when there's parity in Boomerang, UBR and UFL should be swapped. Which they aren't.
So is this parity or not?

Wtf. That's not boomerang. Boomerang uses triple sledgehammer not R2 U R2 U' R2. That messes up so much stuff.

 

[TDM]

Wtf. That's not boomerang. Boomerang uses triple sledgehammer not R2 U R2 U' R2. That messes up so much stuff.

No, that's Boomerang. I used the Messy trick; you sometimes don't have to use triple sledgehammer (see example solve 4).

 

[TheNextFeliks]

No, that's Boomerang. I used the Messy trick; you sometimes don't have to use triple sledgehammer (see example solve 4).

Oh. Sorry. Well like it says you have to do an even number of messies. I don't have a cube with me so I'll look at it later.

 

[hkpnkp]

i have never tried blindfold cubing. how is it actually done / what are the methods used ?

 

[TheNextFeliks]

i have never tried blindfold cubing. how is it actually done / what are the methods used ?

Ok. You memorize where each piece needs to go, then you put them there. Use M2/OP

 

[TDM]

Ok. You memorize where each piece needs to go, then you put them there. Use M2/OP

OP for edges may be easier as they already know the algs they need. You also don't have to remember where the M slice is. I know that the algs aren't difficult to learn for M2, but I found it easier when I first started learning BLD to know exactly what the algs I were using did.

 

[TheNextFeliks]

OP for edges may be easier as they already know the algs they need. You also don't have to remember where the M slice is. I know that the algs aren't difficult to learn for M2, but I found it easier when I first started learning BLD to know exactly what the algs I were using did.

Lol me too.

 

[Renslay]

D2 L' D2 R2 F D' F' B' R2 L D2 F2 L2 U' B2 U R2 F2 U2 B2 R2
How I would solve the corners (using Boomerang):
x2 y' R' U R U2' R' U2' R' U' R U' R' U2 R U2' R U y x2
y' U' R U' R2' U' R U' R' U2' R' U2 R' U' R U' R2' y
D2 R2 U R2' U' R2 D2
D' R2 U R2' U' R2 D
M2' U2' M2' U' M2' U2' M2' U'
D R2 U R2' U' R2 D'
R2 U R2' U' R2
Odd number of cycles = parity
But when there's parity in Boomerang, UBR and UFL should be swapped. Which they aren't.
So is this parity or not?

There isn't a parity.

Yes, you have to do odd number of cycles, but here is the trick: your UFL and UBR corner is actually swapped.
The side effect of the Boomerang is the UFL-UBR swap, so after odd number of swaps (odd number of targets), the UFL-UBR will be swapped.

You can see the same thing on Example 5.
http://renslay.byethost11.com/examples.html

Also this is mentioned in the "2 cycles" trick:

"Side effect only 2-cycle: Only one case remains... What if we have to swap UFL and UBR? Easy: do nothing! Just switch the parity case, so, if we solve the remaining corners with an even number of algorithms, then we have a parity (since UFL and UBR have to be swapped), and if we solve the remaining corners with an odd number of algorithms, then UFL and UBR is swapped, therefore solved: there is no parity. See Example 5."

Again: you have odd number of targets PLUS a single swap (UFL-UBR) = no parity.
"Solving the UFL-UBR swap" is the "do nothing" algorithm, because you have to do a double-swap: one is for solving UFL-UBR as targets, and one is for causing the side effect, as every swap does - but a double swap means nothing to do.

Rarely, but this can occur.

EDIT: a bit more explanation

example algorithms:
single swap UFR with DFR and cause side effect (swap UFL-UBR) = (F R' F' R)*3
single swap UFR with DFL and cause side effect (swap UFL-UBR) = D (F R' F' R)*3 D'
single swap DBL with DBR and cause side effect (swap UFL-UBR) = U2 F' R D R' F U2 F' R D' R' F (this is a shortcut: Adjacent downside 2-cycle)
...
single swap UFL with UBR and cause side effect (swap UFL-UBR) = do nothing

And every "single swap + side effect swap" is counted for determining the parity case, even the "do nothing" algorithm.
Therefore, in your case, it was not 5 algorithms (targets), but 6!