# Improvements for M2/R2

Discussion in 'Blindsolving Discussion' started by Stefan, Oct 23, 2007.

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For FU I use this one for a long time. But I use it only if FU is the last edge to solve or if another M edge follows. In other cases, I have a simple technique :

1/ Do M2
2/ Do the set-up for the next piece
3/ Do M
4/ Do the reverse of the set-up
5/ Do M

And you solve the two piece in one time

The same scheme can be applied for BD :

1/ Do M2
2/ Do the set up for the next piece
3/ Do M'
4/ Do reverse set-up
5/ Do M'

And for UB :
1/ Do U2M2
2/ Do the set-up for the next piece. Theyre is trap here : UR and UL slot are swapped !
3/ Do M
4/ Do reverse set-up
5/ Do MU2

2. ### InusagiMember

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Oct 6, 2007
I liked thee third one. Anyway, what do you do if the next piece is in the M slice? What is the set-up moves for that one?

3. ### SimboubouMember

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I also found this with Acube :

UF2U'MUF2U'M' and UF2U'M'UF2U'M

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If the third piece is in the M slice, I use the alg given by stefan few post ago (is it correct ?).

Another tip to avoid x regrip during set-ups (for the case they need) :
1/ do M'
2/ Place the piece in the UF slot using only RU or LU.
3/ do M

Use this one only for the x cases.

5. ### timMember

Nov 22, 2006
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The FU alg is wrong. M and M' have to be switched:
FU: D M' U R2 U' M U R2 U' D' M2

It took me more than 5 minutes to figure that out, but at least i understand the alg now .

Last edited: Jul 17, 2008
6. ### timMember

Nov 22, 2006
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These are my R2 algs for BRU and RUB:

RUB: F U' L2 U F' R2 F U' L2 U F'
BRU: x2 F U' L2 U F' R2 F U' L2 U F' x2

7. ### blahbrah

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I've read through the entire thread and I've found similar ideas to this but not the exact same idea, so I'm posting it anyway:

Edit: I use DF as my buffer so do the necessary changes in your head if you're using FD (or any other buffer for that matter).

Whenever you hit an M-slice sticker, just do a 3-cycle. Here's how: skip whatever M-slice sticker you hit, and move on to the next non-M-slice sticker, set that up to UB as usual and do one of the following algorithms (they're actually just commutators with setups and move cancellations and stuff, but "considering" them algorithms makes the method more braindead)

UF: U2 M' U2 M
FU: y' U' M U' R2 U M' U' R2 U2 y (I've tried many, and though this one has a cube rotation, it's the most finger-friendly one for me, tell me if you've got a better one)
DB: M' u2 M' u2
BD: U2 F2 U M' U' F2 U M U

(I'm pretty sure almost everyone is doing the UF and DB one already.)

There are 2 ways to do this. The first is the "normal" way to do it, i.e. whenever you hit an M-slice sticker, just do one of the algorithms, if the M-slice is "bad", use the "other" algorithm (UF/DB, FU/BD), nothing special. The second one goes like this: memorize in pairs, you'll need a bit of thinking ahead, and if an M-slice sticker falls on the "first part" of your pair, do one of the algorithms as usual, if an M-slice stickers falls on the "second part" of your pair, then do the inverse algorithms, so you never ever need to do "the other algorithm".

So to deal with cases when the M slice is "bad", you can "learn more algorithms (the inverses)" or remember to do the "opposite" stickers, neither is better than the other imo, you be the judge.

It's better than some of the previously suggested ideas because it eliminates the need to come up with a commutator on the fly, making it more braindead. You're just doing your usual stuff, while skipping M-slice stickers in the process, and doing slightly different "algorithms" rather than the usual M2. Any comments on this? Thanks.

Edit: Oh, and if you've got 2 M-slice stickers in a row, I'm pretty sure there are algs somewhere else in this thread, I forgot to add them, I'll add them later once I've compiled all the cases (less than 10?).

Last edited: Jul 30, 2008
8. ### blahbrah

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Okay I've found algorithms for all the cases with 2 M-slice edges in a row to shoot to. 24 of them in total, including inverses. And I've decided to just use 4-move 3-cycle algorithms for these cases, regardless of orientation, and flip them later, because most of the algorithms are too similar and can cause confusion very easily, plus most of them are not even finger-friendly.

I've decided that it's simply not worth learning 24 different algorithms to handle all these cases (at least the edge-flipper algorithms (omega snaps?) are more finger friendly).

So my latest M2 modification involves using 3-cycles every time I encounter an M-slice edge. Since it takes as many moves to solve 1 piece as it does for 2 pieces, why not solve 2 pieces with a 4-move 3-cycle?

9. ### Mike HugheySuper ModeratorStaff Member

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I have the same problem with your new approach which I had with Stefan's original approach. I struggle too much with dealing with the pieces that need to flip. I don't know why, but I just can't seem to handle it with enough reliability. Somehow I just need to solve them and forget about them. Having to flip them later just ruins it for me.

I wish I could get over it, because it seems like it makes a lot of sense. And your approach especially. I'm tempted to try your method for a bit, but I still think I'll have the same problem I had with Stefan's original approach - the flipped edges will confuse me too much and slow me down and reduce my accuracy. Maybe this sort of method only works for really smart people.

I do like your M2 approach conceptually - it seems very nice.

10. ### blahbrah

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You're wrong. Either it works for all people, or all people are smart. Either way, you're wrong

Here's how I keep track of the bad M-slice edges: my feet. Actually, foot Because I originally intended to use the other foot to track the good/bad-ness of the M-slice, but umm, I didn't use this in the end because firstly, I looked retarded (or rather, my feet did), and secondly, there was too much "action" going on under the table that it became more of a distraction than a memory aid.

So with my left foot, this is what I do: when the UF piece is bad, I raise my toes and keep the rest of my foot on the ground; when the DB piece is bad, I raise my heel, so my foot gets into a tiptoe position; when the UB piece is bad, I shake my entire leg, violently. Since shaking my leg and raising certain parts of it are two mutually exclusive actions, I only have one "combo" to worry about, that is, if DB and UF are bad, for this I raise my entire foot so it's "floating" in the air. Over time, you'll learn to translate these actions directly into algorithms, rather than positions of bad pieces.

If you think this is too complex p) create your own system with your feet, it's worth a try, and it's good, at least for me

Edit: And don't worry, I have the exact same problem as you do, I just insist on this flip-everything-at-the-end approach because I believe it's gonna be beneficial in the long run. Ever so often, after I hit the spacebar and before I take off the blindfold I go "Oh crap!" because I realized my leg was still shaking, but I'm still practicing and I'm trying to make it a habit to remember to flip them at the end of every solve, so yeah.

Last edited: Aug 1, 2008
11. ### Mike HugheySuper ModeratorStaff Member

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I think I would definitely need to modify the system, since I always stand when doing 3x3x3 BLD. Also, I currently use my stance to tell me whether or not there are other edges that were already in the proper place but flipped (before I started doing this my accuracy was WAY too bad because I would forget to flip edges that were in the proper place but flipped) - wide stance is no flipping needed, narrow stance is flipping needed.

But I guess it should be possible to come up with some sort of variation that would account for all of this.

My current method of using commutators to solve anything with a "bad edge" seems simple in comparison at the moment to me. But I think it might be giving it a try, since I'm sure I slow down on the commutators, typically.

12. ### blahbrah

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Just came up with another solution for 2 M-slice stickers in a row! Though it does require a lotta thinkahead. Alternatively (in fact, this is the most logical way to do it since no one thinks ahead that far), make a mental note while memorizing the stickers.

Solution: Think ahead far enough to see 2 M-slice stickers coming, and separate them on purpose. E.g. Say this is your cycle, A B C M N D E, where M and N are M-slice edges, you shoot to A, then shoot to B, then 3-cycle buffer-C-M, then 3-cycle buffer-N-D, then shoot to E. The downside of this is when you get a cycle like this: A B M N C D, you've gotta cycle to the "opposite" stickers of M and N.

As for cases when there are 3 in a row, well, how unlucky can you be? That solve is doomed to suck anyway, so you might as well solve 'em one by one Or you can do 4-move 3-cycles for two of 'em and shoot to the third one, this solution seems more logical.

Last edited: Aug 1, 2008
13. ### blahbrah

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Oh I just thought of something that I've been doing for quite a while but haven't seen anybody post about it here yet, so I'm double posting, yeah.

To avoid "ugly" M-slice cases, when I break into a new cycle, I don't go for easy cases (e.g. UB), I go for "good" M-slice stickers, i.e. if I haven't encountered some of the M-slice edges in my first cycle, I locate them and start my new cycle from there. Would you rather break into a new cycle with an easy case (e.g. UB), just to find out later that you've got an ugly case? Or would you rather spend 2 more seconds looking for unsolved M-slice edges and break into your new cycle with a few more setup moves? I'd pick the latter. With this approach I only encounter ugly cases about once in every 3 solves, really (or is this because I haven't solved much yet? ).

Edit: I realized something. The first suggestion (break into a new cycle from an easy spot) can give you faster singles. The second suggestion (break into a new cycle from a good M-slice sticker) gives you faster averages. So for competition, the first approach is better actually Analogy: for 4x4x4 edge pairing, 6-pair is definitely gonna give you faster singles in the event that you don't have awkward cases, because it takes less moves as a whole; on the other hand, 2-pair is gonna give you faster averages because you don't encounter awkward cases at all, even though it takes slightly more moves than 6-pair. Is this a valid analogy?

Last edited: Aug 1, 2008
14. ### timMember

Nov 22, 2006
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R2 algorithm for BDR i use (found with ACube):

y R U' R D2 R' U R D2 R2 F2 y'

I very often lost grip on Stefan's alg ((R U R' D r2 U') (R U r2' U') (D' R)). I even dropped the cube twice in a row while executing his alg.

15. ### blahbrah

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Doesn't that alg just look A perm-ish?

16. ### joeyMember

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It is an Aperm

17. ### blahbrah

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Yeah, I realized too Didn't have a cube with me when I first saw the post and made that comment, but I just tried it, so yeah.

18. ### StefanMember

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At a recent party I did a demo solve under tough circumstances and when I got to the BDR (or FUR?) target, I was afraid I'd screw up the alg so I chose to do the "normal" PLL A-perm alg, too. So yeah, those are alternatives. When success is less important, I do use the other R2 algs, though.

19. ### KennethNot Alot

I used it for ELL since the 80's but starting from doing x'

Note that it is flexible, you can change the R/R' with R2's and the same for the M/M' -> M2 giving 4 combinations, factor that with the mirror and inverses and it is a total of 16 algs.

Edit: I just noted that Stefans alg is not the same as min, It is U' R U M' U' R' U M but they both work.

This also gives 24 diffrent combos (dunno if all are uniqe, have not checked)

Last edited: Aug 12, 2008

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