Commutators for r/l edges on big cubes

blah

brah
I did my first 4x4x4 BLD yesterday, with pure r2 for edges, that means the VERY long algs for the r/l edges (at least I consider them long). So I consulted Mike Hughey, and he said he used commutators for r/l edges and r2 for R/L edges.

I wanted to be faster (actually I've only had 2 successful solves till now and about 10 DNFs ranging from 14 to 17 mins), so I decided to do the same. I sat down for half an hour with a cube, a pen and a piece of paper, and came up with all the commutators I needed for r/l edges

Actually I sorta Fridrichized them to minimize thinking during an actual solve, so there'll be 2 setup moves (c.f. the ideal 1) for almost all of them (but I guess 2 extra moves, the setup and the 'setdown', compensates for the extra thinking time wasted if I hadn't Fridrichized this).

I memorize in pairs (actually I memorize in six-letter groups, so I know I'm about to finish all the edges when I have ~4 groups, helps me keep track ), and I wanted to do commutators like Mike does, i.e. if any one piece of the next pair is in r/l then I'll do a commutator. So here's my systematic approach:

Notation: [A, B] = ABA'B'; [F] = F/F'/F2 (depending on the situation).

Case A: one piece in l, the other in R/L.
1. [l] to bring l-piece to UFl.
2. [R/L] to bring R/L-piece to U.
3a. If R/L-piece on f, commutator [D'[f]D, F2] or inverse.
3b. If R/L-piece on b, commutator [DD', F2] or inverse.

Case B: one piece in r (not UBr), the other in R/L.
1. [x] to bring DFr (buffer) and r-piece to UFr and DFr.
2. [R/L] to bring R/L piece to U (the new U after the cube rotation).
3a. If R/L piece interchangeable with UFr, commutator [FdF', ] or inverse.
3b. If R/L piece not interchangeable with UFr, commutator [[D](R2/L2)[D], r'] or inverse.

Case C: one piece in r, the other in l. (I almost died thinking how to setup these both-in-r/l cases, then I realized they were easier than Cases A and B -.-")
1. [x] to bring DFr (buffer) and r-piece to UFr and DFr.
2. [l] to bring l-piece to UBl (the new UBl).
3. Commutator [FdF', U2] or inverse.

Case D: both in r.
1. B2 to setup to Case C.

Case E: both in l.
1. If 'adjacent' (hope you get what I mean), [l] to bring both to UFl and UBl, then it's Case C with a cube rotation, commutator [UbU', F2] or inverse.
2. If 'opposite', haven't came up with any solutions with 2 setup moves or less Suggestions, anyone?

I don't think I'm missing out any cases. Is this a good approach (to Fridrichize the commutators to minimize thinking)? Or are there better commutators for some of these cases? Thanks in advance.

P/S: I just tried to calculate the probability of at least one piece being in r/l, and it turns out to be more than half?! Is there something wrong with my calculation? It's simple: 1 - (16/24) * (15/23) = 0.565. Or from another perspective: 1 - (16C2)/(24C2) = 0.565.

Another question out of the blue, how do you (whoever's reading this) keep track of which edges have been memorized and which haven't? 'Cause if I have a 24-cycle then I have a very fast (sub-6) memo. If I have a 21- or 22-cycle I take all day finding out which are the 2 or 3 edges left to be cycled (and the worst part is finding that they're already permuted), and if it's REALLY bad the memo ends up being more than 10 mins Any techniques?

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Mike Hughey

Super Moderator
Staff member
I don't think I'm missing out any cases. Is this a good approach (to Fridrichize the commutators to minimize thinking)? Or are there better commutators for some of these cases? Thanks in advance.
These work just fine. They're not all the same ones that I use, but they're generally just as good. For cases A and B, it's never necessary to do more than one setup move (at least I think so - I'm pretty sure I never use more than one setup move for those), and your rules will sometimes require two, but since it's systematic, I doubt that will affect your times much.

P/S: I just tried to calculate the probability of at least one piece being in r/l, and it turns out to be more than half?! Is there something wrong with my calculation? It's simple: 1 - (16/24) * (15/23) = 0.565. Or from another perspective: 1 - (16C2)/(24C2) = 0.565.
Is this taking into account that DFr and UBr don't count? Anyway, I will speak from experience that it is true that a fairly large percentage of the pairs include r/l - it feels like slightly less than half to me, but not a lot less. The nice thing about this method is that, as you have already seen by working this out, the commutators for these cases are actually pretty easy to work out compared to some of the other cases.

Another question out of the blue, how do you (whoever's reading this) keep track of which edges have been memorized and which haven't? 'Cause if I have a 24-cycle then I have a very fast (sub-6) memo. If I have a 21- or 22-cycle I take all day finding out which are the 2 or 3 edges left to be cycled (and the worst part is finding that they're already permuted), and if it's REALLY bad the memo ends up being more than 10 mins Any techniques?
The answer? Practice. I wish there were a better answer, but I really don't know of one. Let me know if you figure out a good solution. But I have found this to be becoming less and less of a problem for me the more I do them.

I have a spot in my "journey" that I know is the place for a single 24-cycle to end. Then I add one for each cycle I've had to add. So I'll know when I'm done, based on that. If I run out and don't immediately see the next cycle, I will then try to count all the pieces that are already in the correct position, to see if I'm done yet. That will give me a count of the number of pieces I still need to find. But sometimes it does still go bad for me, which is how I can sometimes still wind up with a 15 minute (or even more sometimes!) 4x4x4 BLD solve. I hate those! What's worst is when I accidentally skip or add a piece in my memorization; that can take minutes to figure out.

masterofthebass

What I usually do (if I bother to) is set up a [ ___, r] commutator. If the edge in the l slice, then you just bring it to the UB edge and do a U2 to set-up into the easy commutator. if they are both in the l slice then you just do the l slice omm. My ohly issue is when I have to shoot to DBr because it's annoying for the r slice comm. I just end up doing the long alg that I use for 3x3, as it's still pretty fast.

Member
I'm in trouble with your E cases too. For the second one :

Df' [U2,r'Dr] fD' But I don't like it :S

For the edges, i don"t have any tips to know if they are in a cycle or not, but for centers, just put a finger on each solved center (a finger could be on more than one center if they're adjacent).

Mike Hughey

Super Moderator
Staff member
I'm in trouble with your E cases too. For the second one :

Df' [U2,r'Dr] fD' But I don't like it :S

For the edges, i don"t have any tips to know if they are in a cycle or not, but for centers, just put a finger on each solved center (a finger could be on more than one center if they're adjacent).
Yeah, for the centers, Chris made that suggestion several years ago, and I do it. You always solve the centers in a particular order on a given face, and you put one finger on each center on the last piece you've solved on that face. I use 3 fingers from each hand - right hand is U, B, R, left hand is D, F, L. It works really well.

I've thought about this - you could actually do the same with edges, although it's much more complicated. My thought - if you've solved one piece on an edge, but not the other, put your finger on that one piece. If you've solved both pieces on an edge, put it between the two pieces on that edge (on the central edge piece on a 5x5x5, on the crack of a 4x4x4). If you've solved both pieces on two different edges, put your finger on the center piece that is diagonal to those two pieces. Etc. Put your finger on the center of the entire face if all the edges of that face are solved. It gets complicated, but I suspect with some work you could make a system like this work for all cases. It's already awkward enough holding the cube for centers, though - I can't imagine how bad this might get with the edges.

cmhardw

I've also thought about, for edges, a base 5 counting system with my feet in order to count edges.

It works like this:

1) Your left foot is the 5's place, and your right foot is the 1's place.

The digits are:
0: foot flat on floor
2: tilt foot to the right (your viewpoint)
3: tilt foot to the left and flex your toes
4: tilt foot to the right and flex your toes

so to count from 0 to 24 you count:
00, 01, 02, 03, 04, 10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 30, 31, 32, 33, 34, 40, 41, 42, 43, 44

It looks funny to do it, but it keeps track of the counting for you. This would be easier done when sitting and doing big cube BLD, standing and doing this is a bit uncomfortable and looks *really* um....... unique... Not to mention, when counting in the 40's I tend to have to be careful to avoid foot cramps ;-)

Right now I know how many more edges I need to memorize based on where my image line stops in my journey, and I have to account for the number of single letter images too. It's not exactly easy, but it isn't too bad. I have seriously considered several times learning to foot count, but I have never actually practiced it for competition use, only at home.

Chris

blah

brah
I saw (actually, skimmed through only) your (edit: referring to Mike, because Chris made a post while I was typing this ) centers tutorial somewhere on the forum earlier today/yesterday (forgot), and I think you mentioned Chris' suggestion there too, I tried it once, and didn't like it at all because my hand was all over the cube and it was just plain awkward. Good thing is, I don't have this damn-where-are-the-last-two-unswapped-pieces problem with centers, at all, for some unknown reason. I always seem to have a gut feeling which piece I last 'used' on each face, I guess it's because there are much fewer centers than edges. And I don't think it's ever been wrong every time I double check (though I don't double check often ).

My current 'solution' for the 'lost' edges is this: First, scan B, U, F, D faces (in that order, so I don't forget any face) for any solved edges (I look out for edge stickers with the same color as the face I'm scanning), this'll take about 2 seconds in total, of the few attempts I've had, most of the time, there's none or one. If the number of solved edges and my memorized cycle(s) adds up to ~24 (just a feeling, because I don't count the buffer and sometimes there are 2 or 3 cycles I've memorize so there'll be more than 24) then there are no 'lost' edges. If it's less than that, look for the lost edges by alphabetical order (this seems the only systematic way to do it), and every time I hit a letter I spend 3 seconds reciting my cycle to see if the letter's in there, sometimes when I'm sick of recalling, I just see where that piece has to go and if the letter pair is familiar than I've already memorized that piece. I really hope you understand what I mean, because it's hard to explain what I do... Is this a good approach?

By the way, my corners take 2 to 3 minutes for memorization, edges take 3 to 4 minutes and corners take ~30 seconds. What's a good target for each of these piece types (centers, edges and corners) if I want to sub-15 sometime soon? (I've had a hell lot of 14:xx and 15:xx DNFs, still have only had 2 successful solves out of more than 20 tries now, and it's getting frustrating Wonder how you get that 50% accuracy, mine's somewhere in the single-digit region )

Edit: @Chris, I don't have a problem figuring out how many edges left I have to solve (because I memorize in 6-letter groups, as mentioned in the first post of this thread), I have a problem locating them. So foot counting wouldn't really work would it? Though imo, it's such a good idea I'm definitely gonna use/modify it for something else, in the future

Edit again: @Chris again, can you take a look at Case E2 in the first post of this thread and help me come up with a nice commutator for it? Thanks.

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Mike Hughey

Super Moderator
Staff member
I think you mentioned Chris' suggestion there too, I tried it once, and didn't like it at all because my hand was all over the cube and it was just plain awkward.
If you thought it was bad for centers, you certainly don't want to try it for edges. I just gave my suggestion above a try for edges, and while it works, it was absolutely horrible trying to hold the cube while doing it - way worse than for centers. I can't imagine it would be worth it for edges after trying it, so I guess it's a bad idea.

I think your current approach for dealing with lost edges is great. It's not far from what I do. Lately, I've been able to just guess where they are pretty often, though, so generally I fall back to your approach if I don't spot a lost edge immediately.

By the way, my corners [I assume you meant centers here] take 2 to 3 minutes for memorization, edges take 3 to 4 minutes and corners take ~30 seconds. What's a good target for each of these piece types (centers, edges and corners) if I want to sub-15 sometime soon? (I've had a hell lot of 14:xx and 15:xx DNFs, still have only had 2 successful solves out of more than 20 tries now, and it's getting frustrating Wonder how you get that 50% accuracy, mine's somewhere in the single-digit region )
Those ratios seem very nice for sub-15. Just shave a few more seconds off and you'll be there. Like I said, memorization is usually almost exactly half of my time, so 7:30 for memorization = 15:00 total.

I've always been around 50% accuracy on 4x4x4 (except lately I've actually been more like 70%!!). The catch is, I've always been very slow compared to others with my experience level. When I had done 20 tries, I had 10 successes (I just checked my records), but my fastest solve was 27:05.94, and my fastest DNF was 23:51.74. (My first successful solve was 63:42.73, and my first attempt, which was a DNF, was over 88 minutes!) So you can see I was a LOT slower than you. Going that slow probably helped me to have fewer DNFs. Maybe it just comes from being more careful and not being as focused on super-fast speed.

cmhardw

Edit again: @Chris again, can you take a look at Case E2 in the first post of this thread and help me come up with a nice commutator for it? Thanks.
[l] to bring the pieces to DFl and UBl
then use U' l2 U l' D2 l U' l' D2 l' U

or if you prefer the other direction do the different [l] turn and do
U' l D2 l U l' D2 l U' l2 U

That alg is a BH alg, it's a commutator and is just a setup into a 9 move commutator (a commutator with a move cancellation).

In pure commutator notation (in all its glory) that alg is
[A : B , C] = A B C B' C' A'

[U' : [l : l U l' , D2] ] = U' l2 U l' D2 l U' l' D2 l' U
[U' : [l : D2 , l U l'] ] = U' l D2 l U l' D2 l U' l2 U

That's one of my favorite 11 move cases btw. A bit weird to see how it works at first, but if you take a look at it, it's really cool.

Chris

P.S. There are really only 2 cases here, both of which can just be solved with a BH alg or its inverse.

Case 1)
[U' : [l : l U l' , D2] ] = U' l2 U l' D2 l U' l' D2 l' U
or
[U' : [l : D2 , l U l'] ] = U' l D2 l U l' D2 l U' l2 U

Case 2)
[U L : B' u B , D2] = U L B' u B D2 B' u' B D2 L' U'
or
[U L : D2 , B' u B] = U L D2 B' u B D2 B' u' B L' U'

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blah

brah
I did it! I did it! I did it! I completed 3 out of 4 attempts, all sub-18! And the third one which was a DNF was only 2 edges off (though I still have no idea why). Thanks Mike, going slow and being careful really helps! I'll stick to this principle from now on, the speed will come with practice, hopefully The second to fourth ones are on Cubemania by the way, I did the first one on CCT.

@Chris: What does BH stand for and what is it? And wow, I've never been able to figure out how to do commutators involving 2 adjacent pieces. I'm too tired to figure out so many setup moves now though, I'll understand the whole thing intuitively by tomorrow, hopefully

By the way, how can you have 11 favorite cases when there are just about that many types? (I think, or are there more?) And it took me long enough to realize [A : B , C] actually means [ A : [B , C] ] instead of [ [A : B] , C ]

Member
Chris, you are an inspiration

dbeyer

Member
Blah:

BH = Beyer-Hardwick, our commutator method. Its basically a documentation process of every optimal algorithm for every possible cycle buffer bound. Like from the URb, or the UBl and every possible combination of 3 pieces with that buffer. In documenting these things we studied trends and characteristics. We noticed that there is sort of a relationship between the pieces in the cylce that determines the move count and the method of approach to solving the cycle.

Its just like the corner edge pairs in F2L, but 3 pieces, in freespace on the cube.

blah

brah
Thanks. Err... is this list of commutators available anywhere on the Internet? I've checked Chris' website but I couldn't find a link.

shelley

chang
Haha Chris, I see I'm not the only one who uses feet to keep track of things during BLD. My system's not nearly as complex as yours of course; I use r2 for edges and use my right foot to keep track of when my r slice is off. It's one less thing to keep track of in my head.

dbeyer

Member
Parts of it are online, but not available quite yet, because we simply want to publish it as a whole.

1/3 or so is done available and thats quite a milestone, but thats not even half the battle ... over one thousand cycles left. Like we said though, its really intuitive, and moreso a recognition of relationships in freespace, and methods of approach rather than "algorithms."

Later,
DB

blah

brah
There are that many?! But even 23C2 = 253 only (and that's including all isomorphic cases). And you could always rotate your buffer to a fixed position, e.g. UFr, so I don't see how there can be over one thousand cycles left...

Anyway, I've got a new question about commutators, but it's for centers. Almost all my DNFs are because 2 or 3 centers are off, and I absolutely hate that feeling of taking off the blindfold, thinking the cube is solved, only to do a little rotation and find 2 centers at B and D swapped

I'm not sure but I think it's because I do too many cube rotations during center commutators, I never have more than 2 setup moves, but there's usually at least 1 cube rotation for every 3-cycle. Thing is, I don't forget to rotate them back (that's why my edges and corners are still solved), I just keep making stupid mistakes thinking that one piece is another after a rotation, e.g. say ABCD are 4 different pieces on my U face, but after I do a complex cube rotation, say xyz, I tend to confuse which piece is which, anyone have the same problem? Or am I just supposed to stick to awkward turns (like b' or something like that) and keep cube rotations to a minimum? If I don't do cube rotations I find it very hard to visualize the commutators, is this something I've gotta start getting used to, seeing commutators from any direction?

By the way, are most people's DNFs the same as mine (center screw-ups)? I think of the 20+ DNFs I've had, only 1 or 2 are corner screw-ups, and probably 2 or 3 are edge screw-ups, and another 1 or 2 major DNFs where the cube looks nowhere near the solved state, other than these the rest are center swaps

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Mike Hughey

Super Moderator
Staff member
I think I just use a different approach, so it's hard to compare. I never rotate the cube for commutators. I'm slow enough of a "speed"cuber that it really doesn't make that much difference for me as to whether or not the move is awkward. And I've worked hard to get where I can see the commutator anywhere. I've worked for that so that I can always maximize solved centers on the 4x4x4 when orienting it, even if that means the top is completely solved and the bottom is completely unsolved. I'm not afraid of commutators to move pieces on the D and B faces, simply because I've gotten used to doing them.

But this has been a reasonably successful strategy for me because I'm not a fast cuber anyway, so having finger-friendly moves just doesn't mean that much to me. I doubt you should follow my lead on that.

Looking over the past 50 4x4x4 solves other than the ones for the weekly competitions that I've done, I have 1 where I was interrupted and had to stop, one where I did one move in a commutator the wrong direction (r instead of r'), 4 with only centers messed up, 3 with only edges messed up, 1 with only corners messed up (forgot parity), and 3 with centers, edges, and corners all messed up (apparently undid setup moves incorrectly - that's what usually causes that). So it's a fairly even distribution, really.

masterofthebass

Hmm.... I think i may try a couple of solves without rotating for centers. I always seem to struggle seeing the rotation, but I do see what I need when I don't rotate.

joey

Member
Whenever I've done 4x4 sighted, I don't really rotate. Maybe cos I've been using a buffer?

blah

brah
@DanCohen: Don't really get what you mean by 'struggle seeing the rotation', do you mean you struggle to follow where the center pieces go after a rotation (like me), or do you mean something else? And what do you mean by 'you see what you need when you don't rotate', if you mean what I think you mean, why do you need to rotate at all?

@joey: Buffer? You use a fixed buffer for centers? How?