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Blindfold solving edges question

TheChaiCuber

Regular
Joined
Sep 30, 2015
Messages
224
Location
San Francisco
WCA
2016CHAI03
Hi all,

I've just started reading the forum tutorials on how to blindfold solve and I'm practicing the edge portion of the old pochmann method. Right now I understand how the solving is supposed to work but i've come at a crossroads while regular solving using the method.

The cube is held with red as the F face and white as the U face. the cubie in UR is white (U) and green (R). To swap it into it's correct position, all I have to do is do a T-perm. However, the piece that I'm going to swap that cubie with is the white (U) blue (L) piece!. So if I perform the algorithm, my UR piece will be solved as well as the UL piece. If my UR piece is solved, doesn't that mean I'm stuck? in a blindfold situation where I can't see the cube, I won't be able to just change sides of the cube to continue solving I think. How am I supposed to fix this problem? thanks guys.
 
When UR is solved, thats a cycle break. Shoot the UR edge to any random (unsolved) edge and continue. You will end up back at that random edge at the end/earlier, so remember which one it is.
 
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I'm encountering the parity multiple times when i swap with a random unsolved edge piece with the solved or flipped UR piece; is this normal? also, I'm remembering which edge pieces I'm swapping the UR piece with but It doesn't have any significance later on, am I doing anything wrong?
 
Don't you mean cycle break?

yeah oops im dumb

I'm encountering the parity multiple times when i swap with a random unsolved edge piece with the solved or flipped UR piece; is this normal? also, I'm remembering which edge pieces I'm swapping the UR piece with but It doesn't have any significance later on, am I doing anything wrong?

its called cycle break (my bad!)

when you shoot to a random unsolved edge, you keep solving edges until your target is the random edge again. The edge will be solved, and since you left the buffer there, it is solved too.

I think most scrambles are solvable by a maximum of 2 cycle breaks (not counting flipped edges)
 
yeah oops im dumb



its called cycle break (my bad!)

when you shoot to a random unsolved edge, you keep solving edges until your target is the random edge again. The edge will be solved, and since you left the buffer there, it is solved too.

I think most scrambles are solvable by a maximum of 2 cycle breaks (not counting flipped edges)

Uhm, sorry if I seem stupid but I don't quite understand the concept of cycle breaking and the language in how you described it. So when you say "shoot to a random unsolved edge", you mean swapping an unsolved edge with the solved/flipped UR? If so then the next edge to be solved will be the edge I swapped the UR piece with. Therefore what do you mean by "you keep solving edges until your target is the random edge again"? Are you saying the random edge is the originally solved UR? or the piece I swapped UR with? This is very confusing, thanks man.
 
Uhm, sorry if I seem stupid but I don't quite understand the concept of cycle breaking and the language in how you described it. So when you say "shoot to a random unsolved edge", you mean swapping an unsolved edge with the solved/flipped UR? If so then the next edge to be solved will be the edge I swapped the UR piece with. Therefore what do you mean by "you keep solving edges until your target is the random edge again"? Are you saying the random edge is the originally solved UR? or the piece I swapped UR with? This is very confusing, thanks man.

its totally normal to be a little confused with blindfolded concepts, dont worry about it.

yes, I mean you swap solved/flipped UR (called buffer) edge with an unsolved edge. example, swap UR with UL.

The buffer is now at UL, and whatever was at UL is at UR. Now you solve that edge, lets say it goes to UF. UF has to go to UL, then because you know UL is where you left the buffer, it will now be solved. thats what I mean by "keep solving edges until buffer is solved again"
 
its totally normal to be a little confused with blindfolded concepts, dont worry about it.

yes, I mean you swap solved/flipped UR (called buffer) edge with an unsolved edge. example, swap UR with UL.

The buffer is now at UL, and whatever was at UL is at UR. Now you solve that edge, lets say it goes to UF. UF has to go to UL, then because you know UL is where you left the buffer, it will now be solved. thats what I mean by "keep solving edges until buffer is solved again"

OMG! I think I get what you mean now. This is really cool. Hopefully I can try memorizing the edges soon, then I can work on the corners! thank you!
 
awesome!
Once you get edges, you will get corners very quickly since its the exact same concept as edges.

I've run into one more problem. Doing solves while looking has become very easy to understand and do but there's been some solves where there are two cubies left, one of them is the buffer and they are both solved AND flipped! by switching up the two pieces and putting them back I can fix this problem, but is this something i can prepare for during a blind solve? am I still doing something wrong that could cause this issue? Basically I'm asking if theres any way to avoid this or a way to fix it while blindfolded.
 
I've run into one more problem. Doing solves while looking has become very easy to understand and do but there's been some solves where there are two cubies left, one of them is the buffer and they are both solved AND flipped! by switching up the two pieces and putting them back I can fix this problem, but is this something i can prepare for during a blind solve? am I still doing something wrong that could cause this issue? Basically I'm asking if theres any way to avoid this or a way to fix it while blindfolded.

flipped edges: at the end of memo, every edge should be solved. if one edge is flipped in its place, it will not be solved (obviously). you cant have one flipped edge, so the other one is the buffer. you just remember which edge is flipped and flip it and the buffer at the end of edge execution.
 
flipped edges: at the end of memo, every edge should be solved. if one edge is flipped in its place, it will not be solved (obviously). you cant have one flipped edge, so the other one is the buffer. you just remember which edge is flipped and flip it and the buffer at the end of edge execution.

I have edges completely figured out, all that's left is the memorization. However, looking at the tutorial the corners are very hard to understand IMO. It says the UBL piece ( yellow, blue, red) is supposed to be at DFR (yellow, red, blue). I don't understand this. Shouldn't UBL be in DRF instead of DFR? Here's the link to the tutorial I found in these forums - https://www.speedsolving.com/forum/showthread.php?10595-How-to-Solve-a-Rubik-s-Cube-Blindfolded

And because I don't know what orientations the corners should be in I have no idea how to intuitively determine the set up moves for the corners (I figured that out on the edges). I know that memorizing set up moves isn't as good as intuitively learning them since there's already so much to memorize anyway. I'd really like to know what the rules are for the corner execution in blindfold solving. Thanks in advance!
 
I have edges completely figured out, all that's left is the memorization. However, looking at the tutorial the corners are very hard to understand IMO. It says the UBL piece ( yellow, blue, red) is supposed to be at DFR (yellow, red, blue). I don't understand this. Shouldn't UBL be in DRF instead of DFR? Here's the link to the tutorial I found in these forums - https://www.speedsolving.com/forum/showthread.php?10595-How-to-Solve-a-Rubik-s-Cube-Blindfolded

And because I don't know what orientations the corners should be in I have no idea how to intuitively determine the set up moves for the corners (I figured that out on the edges). I know that memorizing set up moves isn't as good as intuitively learning them since there's already so much to memorize anyway. I'd really like to know what the rules are for the corner execution in blindfold solving. Thanks in advance!

this should help
 
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