Breaking in new cycle (M2)

[dvtpgm]

Hi guys,

I'm currently using M2 for edges, and pre-oriented R2 for corners. Few days ago, I finally got my first legitimate BLD solve at 7:21.90. However, currently I have a question about M2, and hope that you guys will help me out.

The main thing about my BLD attempt is that, I'm having trouble with breaking into new cycle (for edges). I understand that all you need to do is shoot the buffer piece into any unsolved position. For me, I take very long to determine where to shoot the buffer piece to, mainly because I must go through my memorization again to determine which edge is currently still unsolved. Is there a better way to deal with this?

Thanks for reading, and appreciate any help for a novice BLD solver.

 

[Ellis]

I usually quickly find an M slice edge (preferably the one that belongs to UB) to shoot to because 1) it's easier to figure out if you've already used it and 2) you can decide which sticker of the edge you can shoot to. In other words, you can pick the easier of the two cases. Even if all the M edges are already taken, I don't really have to go through my memo to see which free edge I can use, so I guess it just comes with a little practice.

 

[blah]

I think I'm the only one who doesn't shoot to UB to break into a new cycle.

I've mentioned my approach about a year ago in some other thread but I can't seem to find it now, so I'll just briefly describe it again. Might not be particularly useful for a beginner, but experienced cubers can give it a shot.

Don't shoot to random M-slice edges. Shoot to an edge that's supposed to belong to the M-slice (that hasn't already been used, duh). Shoot to its U/D sticker. It really doesn't take more than a second to spot an M-slice edge on some random position on the cube, literally, and it doesn't even take practice.

Why? Because I hate flipping more than 2 edges at the end. With my approach, I've always had 0 or 2 M-slice edges to be flipped at the end, never 4. Of course, we all know that the probability of this happening is nonzero. But I've been BLD-ing for more than a year and I've yet to come across such a case, so either I'm really lucky, or that nonzero probability is pretty damn low

Edit: You're Benson right? -.-

 

[Ellis]

Don't shoot to random M-slice edges. Shoot to an edge that's supposed to belong to the M-slice (that hasn't already been used, duh). Shoot to its U/D sticker. It really doesn't take more than a second to spot an M-slice edge on some random position on the cube, literally, and it doesn't even take practice.

I think that's essentially what I said. I even edited it from "preferably UB" to "Preferably the one that belongs to UB" to try and make the distinction more clear, although I guess it wasn't clear enough. I think it's pretty obvious that's the better way to do it, take the easy cases while you can. If you just did M2 to start the new cycle, you might end up having to shoot to BU later on and waste some moves.

 

[Stefan]

If you just did M2 to start the new cycle, you might end up having to shoot to BU later on and waste some moves.

Yeah but with a 50% chance you might end up having to shoot to UB later on and save some moves. Consider a two-cycle involving UB and the lengths of the steps:

Break into place UB:
(short+medium+short + short+medium+long) / 2 = short+medium+medium

Break into where piece UB is:
medium+short+medium

So on average, they're equivalent, and for single solve records you might want to break into place UB to have the chance for double short. Cause who do you want to be? The guy who always gets 40 seconds or the guy who averages 40 seconds by alternating between 38 and 42?

Why? Because I hate flipping more than 2 edges at the end.

But if you do break into the M places and get "unlucky" and more M edges end up flipped, that can be *good*. I certainly prefer flipping let's say DF+UL+UB+DB to flipping just DF+UL, cause the former only takes (M'U)4.

 

[Ellis]

Yeah but with a 50% chance you might end up having to shoot to UB later on and save some moves. Consider a two-cycle involving UB and the lengths of the steps:

Break into place UB:
(short+medium+short + short+medium+long) / 2 = short+medium+medium

Break into where piece UB is:
medium+short+medium

So on average, they're equivalent, and for single solve records you might want to break into place UB to have the chance for double short. Cause who do you want to be? The guy who always gets 40 seconds or the guy who averages 40 seconds by alternating between 38 and 42?

That's a good point. I guess it depends on how you actually solve BU though. But really, I want to be either of those guys... doesn't matter which. I may reconsider this now once I've changed how I solve BU. I'm still using a 16 move MU alg out of habit even though I know better ways to solve it. Once I've switched, I think doing it your way will actually save moves on average.

 

[MatsLuthman]

I had the same problem when switching from visual memorization to roman rooms. I thought this would be a major drawback with that method (and it actually is harder to remember which pieces you have already taken, with pure visual memo I simply remember them all) but it gets much easier by just practicing a lot, at least it did for me. You will probably be surprised when you see how much easier everything gets after practicing for just a couple of months.

 

[amostay2004]

I think I'm the only one who doesn't shoot to UB to break into a new cycle.

That makes 2 of us. My noobness makes me think it's complicated to shoot to M-slice edges..I usually shoot to RF or LF..if not it's LD or RD..or also RU (since it's the sexy move for setup)

 

[Sakarie]

Also using m2 with letter-memo, I always shoot to the lowest/"first in alphabet" letter unsolved. That means I often use A, or if it's taken C. and if it's taken too, E, and so on. That means sometimes a little more time, but to me more efficient ratio-wise.

 

[dvtpgm]

Also using m2 with letter-memo, I always shoot to the lowest/"first in alphabet" letter unsolved. That means I often use A, or if it's taken C. and if it's taken too, E, and so on. That means sometimes a little more time, but to me more efficient ratio-wise.

Sounds good. This idea seems identical to what most 3OP solvers do. But then though, to me, it may take time to get used to this. Why? If you realize A is not solved, it doesn't mean you shoot to it immediately, as you also have to check for M (same edge but different sticker). Just that you have to be careful with this. But yeah, sounds good. Will try it out.

 

[Sakarie]

Also using m2 with letter-memo, I always shoot to the lowest/"first in alphabet" letter unsolved. That means I often use A, or if it's taken C. and if it's taken too, E, and so on. That means sometimes a little more time, but to me more efficient ratio-wise.

Sounds good. This idea seems identical to what most 3OP solvers do. But then though, to me, it may take time to get used to this. Why? If you realize A is not solved, it doesn't mean you shoot to it immediately, as you also have to check for M (same edge but different sticker). Just that you have to be careful with this. But yeah, sounds good. Will try it out.

Yes, of course, you need to look for that piece/letter too. To me, it's B, which seems mor logical than M, but that's your choice!

 

[CT_Warrior]

When are you supposed to break into a new cycle?

After the buffer gets solved, or before the buffer gets solved?
I've been hearing some different answers.

 

[Ellis]

When are you supposed to break into a new cycle?

After the buffer gets solved, or before the buffer gets solved?
I've been hearing some different answers.

When you get to the buffer piece in your memo, the buffer will automatically get solved when you put the correct edge in whatever position the buffer was in... or at least the buffer will end up in the buffer position whether oriented or not. So whenever you start a new cycle, the buffer piece will start in the buffer position. The purpose of starting a new cycle is to load a piece that isn't the buffer into the buffer position so that you continue solving pieces from the buffer location (the buffer piece being the last piece you want in the buffer position).

And how many times can you say buffer?


Sorry I went off, that may have been too wordy without being direct enough. In short 1) Solve the piece that belongs where the buffer is, 2) The buffer is now in the buffer position, but may not actually be solved because of orientation- it doesn't matter either way 3) Start a new cycle to get the buffer piece out of the buffer position. Try it with your eyes open, it shouldn't be too hard to see what's going on.

 

[CT_Warrior]

Hmm, I see, so that's how you buffer do it.

Buffer Macky's buffer version of M2 avoids putting the buffer piece in the buffer area by breaking into a new cycle beforehand.

So either buffer way is fine I guess.
I like your buffer way better, so I'm just going to buffer it like that.

 

[Sakarie]

This isn't exactly on trrhe same topic, but I need help with m2, so I figured it best not to start a new thread.

"Scramble" the cube with U B2 R2 D' L2 F2 R2 U' L2 U' with the cube rotation you use while solving.

There's two 2-cycles, but since I'm not comfortable with 2-cycles blindfolded, but I always solve the M-slice pieces together with it's pair (if well-oriented, with variants of M' U2 M U2). I would originally solve it like this:

UR DL DB DL
=
R' U R U' M2 U R' U' R
U' L2 U M2 U' L2 U
U' L2 U M' D2 M D2 U' L2 U

The problem is of course how I do the re-setup move for DL, and the immideatly do the same moves back again. How would you solve this, without setting up for a M2 U2 M2 U2 or something like that?

How would you solve if the edges finished with
*M-slice off* RF FU FR ?

Very thankful for help!