Blindsolving Discussion and Help

[Username]

I just learnt OP and haven't got a success with it yet, but I have got edges success. People tell me to switch to M2 for edges. But I cannot understand M2 for the sake of god. I tried watching many video tutorials. I do not want to blindly memorize the setup moves and algorithms. How do I understand M2?

The setup moves should be intuitive, just like in OP. You just set the piece up to UB without disturbing the M-slice, do M2 and undo the setup. The algorithms are easy. There are 5 algs, (2 of them are inverses) so they won't be hard to remember. Check out Noah's or Zane's tutorials for more info

 

[Cubenovice]

What are some good ways for memorizing large amounts of twisted corners for multi?

I also memo them in letter pairs just like regular pieces.
But to keep memo short I only memo the current location of the U / D sticker. (it is obvious where they need to go)
So two twisted corners are coded into one word.

I just learnt OP and haven't got a success with it yet, but I have got edges success. People tell me to switch to M2 for edges. But I cannot understand M2 for the sake of god. I tried watching many video tutorials. I do not want to blindly memorize the setup moves and algorithms. How do I understand M2?

Set up moves are no more difficult than those for OP. They’ll become intuitive in a short while.
M2 set ups are basically: “pick up” the target piece with your UB sticker (as opposed to “bring” the target piece to UL in Old Pochmann)
Algorithms: unless you are skilled in edge commutators: just learn the algs.

You have probably learned many algs for sighted solving so what’s the deal in learning 4 extra ones specifically for M2?

 

[cxinlee]

I just learnt OP and haven't got a success with it yet, but I have got edges success. People tell me to switch to M2 for edges. But I cannot understand M2 for the sake of god. I tried watching many video tutorials. I do not want to blindly memorize the setup moves and algorithms. How do I understand M2?

Noah is making a tutorial on Cubing World. M2 is explained at part 3. I'm like you and I can only do edges, but watching his video actually helped me understand M2.

 

[mande]

I just learnt OP and haven't got a success with it yet, but I have got edges success. People tell me to switch to M2 for edges. But I cannot understand M2 for the sake of god. I tried watching many video tutorials. I do not want to blindly memorize the setup moves and algorithms. How do I understand M2?

The non M slice cases should be pretty easy to understand if you know and understand OP. Just memorize the M slice algs for now. I'll try to come down to the meet this weekend and teach you (if I don't make it, I'm sure Shubhayan will explain it to you).

 

[kunparekh18]

Re: One answer BLD question thread blind

The setup moves should be intuitive, just like in OP. You just set the piece up to UB without disturbing the M-slice, do M2 and undo the setup. The algorithms are easy. There are 5 algs, (2 of them are inverses) so they won't be hard to remember. Check out Noah's or Zane's tutorials for more info

Have seen Noah's but didn't quite get it, will have to rewatch it. Haven't seen Zane's yet.

Set up moves are no more difficult than those for OP. They’ll become intuitive in a short while.
M2 set ups are basically: “pick up” the target piece with your UB sticker (as opposed to “bring” the target piece to UL in Old Pochmann)
Algorithms: unless you are skilled in edge commutators: just learn the algs.

You have probably learned many algs for sighted solving so what’s the deal in learning 4 extra ones specifically for M2?

Hmm, thanks.

Noah is making a tutorial on Cubing World. M2 is explained at part 3. I'm like you and I can only do edges, but watching his video actually helped me understand M2.

Will have to rewatch it.

The non M slice cases should be pretty easy to understand if you know and understand OP. Just memorize the M slice algs for now. I'll try to come down to the meet this weekend and teach you (if I don't make it, I'm sure Shubhayan will explain it to you).

Thanks .. will try understanding it before Sunday, pls try to come.



Sent from my A75 using Tapatalk 2

 

[Crowned xerxes]

I have a question about 3x3 parity, I have yet to have parity on any of my attempts and I have made about 9 attempts. Is there something wrong with this? I always fine if I remember 11 edges that there is always a flipped edge or solved one so parity isn't needed to do or something, same for corners.

 

[JasonK]

I have a question about 3x3 parity, I have yet to have parity on any of my attempts and I have made about 9 attempts. Is there something wrong with this? I always fine if I remember 11 edges that there is always a flipped edge or solved one so parity isn't needed to do or something, same for corners.

The chance of your first nine solves being non-parity is 1/512. Unlikely, but not impossible.

I'm not sure what your second sentence means at all... could you reword it?

 

[Crowned xerxes]

I think when parity happens is when you remember 11 edges right? Well when I have to remember 11 edges then there is always a flipped or solved edge meaning I don't have to do parity.

 

[Renslay]

I think when parity happens is when you remember 11 edges right? Well when I have to remember 11 edges then there is always a flipped or solved edge meaning I don't have to do parity.

Uhm, sorry, that doesn't make any sense.

If you have to remember 11 edges, then you do 11 algorithms, which is odd, therefore, you have parity. If you solve 11 edges, then the last edge (originally which has to go to the buffer) is solved automatically. How can it be you have flipped edge, if you solved all 12 edges? Unless you mixed something up.

Besides that, the number of originally flipped edge has nothing to to with parity. Parity is only depends on how many times you ha to use an edge-insertion algorithm, which cause the side effect (i.e., swap two corners in OldPochmann). If you have flipped edges in the beginning, then you either solve it with two insertion algorithms (first bring it to the buffer, second put it back correctly), which does not change your parity - or you flip them after the solving of the remaining edges, using edge-flipping algorithms - but you can't count those algorithms as insertion algorithms (since it does not swap those two corners mentioned before!), therefore, your parity does not change.

If you have flipped edges at the end of the solve, but you didn't have them in the beginning, then obviously you did some insertions in a wrong way. But that is nothing to do with parity. If you have even number of insertion algorithms, then you don't have parity. If you have odd number of insertion algorithms, then you do have parity. Regardless of the number of flipped edges.

 

[JasonK]

I think when parity happens is when you remember 11 edges right? Well when I have to remember 11 edges then there is always a flipped or solved edge meaning I don't have to do parity.

Parity just means you have an odd number of edge/corner targets, not necessarily 11. Doesn't matter if there's a flipped or solve edge, if the number of targets is odd, you have parity.

 

[Crowned xerxes]

Uhm, sorry, that doesn't make any sense.

If you have to remember 11 edges, then you do 11 algorithms, which is odd, therefore, you have parity. If you solve 11 edges, then the last edge (originally which has to go to the buffer) is solved automatically. How can it be you have flipped edge, if you solved all 12 edges? Unless you mixed something up.

Besides that, the number of originally flipped edge has nothing to to with parity. Parity is only depends on how many times you ha to use an edge-insertion algorithm, which cause the side effect (i.e., swap two corners in OldPochmann). If you have flipped edges in the beginning, then you either solve it with two insertion algorithms (first bring it to the buffer, second put it back correctly), which does not change your parity - or you flip them after the solving of the remaining edges, using edge-flipping algorithms - but you can't count those algorithms as insertion algorithms (since it does not swap those two corners mentioned before!), therefore, your parity does not change.

If you have flipped edges at the end of the solve, but you didn't have them in the beginning, then obviously you did some insertions in a wrong way. But that is nothing to do with parity. If you have even number of insertion algorithms, then you don't have parity. If you have odd number of insertion algorithms, then you do have parity. Regardless of the number of flipped edges.

But I have had 3 successes so far and On all three of them there was a flipped edge from the start and I memorized 11 edges then just flipped the edge after I did all of the other edges and I never did a parity alg and I still solved it.

 

[Renslay]

But I have had 3 successes so far and On all three of them there was a flipped edge from the start and I memorized 11 edges then just flipped the edge after I did all of the other edges and I never did a parity alg and I still solved it.

I still don't understand. The cube has 12 edges. The slot has never to be memorized. Therefore, you have 11 edges to solve. Do you count the flipped edge here? How do you solve flipped edges? You know that the parity depends on only the number of insertion algorithms, not the actual number of unsolved edges? (i.e., breaking into new cycles)

Can you give us an example solve, where this is the case what you're talking about?

 

[A Leman]

I still don't understand. The cube has 12 edges. The slot has never to be memorized. Therefore, you have 11 edges to solve. Do you count the flipped edge here? How do you solve flipped edges? You know that the parity depends on only the number of insertion algorithms, not the actual number of unsolved edges? (i.e., breaking into new cycles)

Can you give us an example solve, where this is the case what you're talking about?

I am going to jump into this conversation because there may be a misunderstanding that is coming from how people memorize pieces. Some tutorials (especially the 3OP stuff) suggest you memorize the buffer first and then the targets.
memo would look like (A Bk VP Nq) instead of (Bk Vp Nq). The first memo shows the actual number of pieces that would be solved and there is not parity. If you use letter pairs you will know you have parity when you finish with a single letter. If you are trying to find out if you shot to all of your targets, realize that the initial target in a cycle break does not solve any pieces. I hope this helps.

 

[Crowned xerxes]

How many pieces do you need to remember for a 9x9 bld? For example you have to remember about 20 pieces for the 3x3, so how many for 9x9?

 

[KongShou]

which is the method with the least memo? i dont care about speed, i just want to be able to solve a 3x3 blindfolded. so any method with high move count is fine.

Edit: maybe that should be reworded as which is the easiest method.

 

[JianhanC]

M2 OP, BH/comms if you have a good understanding of how they work

 

[kunparekh18]

which is the method with the least memo? i dont care about speed, i just want to be able to solve a 3x3 blindfolded. so any method with high move count is fine.

Edit: maybe that should be reworded as which is the easiest method.

how can there be a method with least memo as long as you have to memorize the same number of pieces

 

[Friso]

where did you learn to solve blindfolded? a youtube tutorial? can you please send me a link where i can learn it. (i already understand the basics)

 

[JianhanC]

http://www.youtube.com/watch?v=ncpkVvIEcKc

 

[Friso]

thanks