My commutator BLD corner method

cmhardw

Hi everyone,

At the World Championships I had several people ask me about my corner method when they saw me BLD cubing. I use commutator freestyle cycling to solve the position and orientation of 2 pieces at a time. Unlike Turbo though I try to ban setup turns as much as I possibly can. Of course this is not completely possible, but my worst cases require 2 setup turns only, and almost all cases that require a setup turn can be done in 1 setup turn. Often I don't need any setup turns.

Ok here are the algorithms I use, and what I call them. If you notice these are pretty much the exact same algorithms as I use in big cube BLD. Also you'll notice that the names *are* the exact same names that Daniel Beyer and I use for the similar cycle on the bigger cubes BLD for centers.

I try to treat the corners of the 3x3x3 like the X-centers on the 5x5x5 cube. This doesn't work exactly the same way, but it's close enough for most of the cases.

My buffer is UBL (the U sticker of the UBL corner) and do cycles all over the cube, from all angles. Just like Pochmann and Turbo you need to sometimes break into other cycles, etc..

Here are some of the algs I use, and their names. In the notation I am using, a turn done in brackets means that you can replace it with any other turn of the same layer. The second turn in brackets *must* be the inverse turn of the first turn in brackets. You'll see what I mean after the first example.

-----------------------
Easy no-setup-moves cases

"Toss-up" : R U2 R' [D] R U2 R' [D']
again the brackets mean that I can also do: R U2 R' D' R U2 R' D; or R U2 R' D2 R U2 R' D2. These algs are all 3 "toss-up" cases.

"Drop-and-catch" : R' D R U R' D' R U'
You can also use R' D2 R U R' D2 R U'; R' D R U2 R' D' R U2; R' D2 R U2 R' D2 R U2; etc.

"Direct-Insert" : R' D' R R' D R [U']

"Caltech-move" (I don't know who deserves full credit for this move, so I call it Caltech-move because I learned it from the Caltech crew) : (R F' R' F)*3 U2 (R F' R' F)*3 U2

"Pure 3-cycle" (this is the pure form of the standard Fridrich 3-corner cycle) :
B2 R F R' B2 R F' R'
This move should be able to be done from all angles and on all sides. As an example I would do F2 L' B' L F2 L' B L; L2 F' R' F L2 F' R F; etc.

"Ferris-Wheel" : D' R' U R D' R' U' R D2

-----------------------
medium 1-setup-move-cases

The case name is the name of the case from the above group that you can setup into.

"Toss-up" cases:
1) F' R U2 R' D2 R U2 R' D2 F

"Drop-and-catch" cases:
1) R R D' R' U R D R' U' R' combines to R2 D' R' U R D R' U' R'
This can also be considered a "toss-up" case if you view the first setup move as R2. So again you have some freedom here to setup into different cases if you prefer one over the other.
2) F' L' D L U L' D' L U' F
3) R' U2 F D' F' U2 F D F' R
4) etc..

Ferris wheel cases:
1) F' D' R' U R D' R' U' R D2 F

Pure 3-cycle cases:
1) U' B2 R F R' B2 R F' R' U

Direct-Insert Cases:
1) F2 L' D' L U L' D L U' F2

hard 1 move setup cases:
-------------------------

Change of viewpoint case 1:
UBL->BUR->RDB

At first this cycle looks very difficult to use setup moves. But if you change your view point to be BLU->URB->DBR suddenly this is an easy Caltech move case.

1) L U2 (R' B R B')*3 U2 (R' B R B')*3 L'

Change of viewpoint case 2:
UBL->BUR->RUF

Again this looks like a hard case where you need 2 setup moves. But actually you just need to change your viewpoint to be BLU->URB->UFR in which case it's just 1 setup turn and a 3 cycle.

1) L' B2 R' F' R B2 R' F R L

2 setup move cases:
--------------------
UBL->RBU->BRD

1) R' F R2 B L B' R2 B L' B' F' R

----------------
EXAMPLE SOLVE
----------------

I've already provided a couple example solves of this approach, I think one on this forum and one on the blindfold solving forum. But for clarity here is one more. Scramble the cube in the orientation where you would normally blindfold solve.

F B' L2 U F2 R2 F2 L' F2 D2 L2 F2 U2 R' U' R2 B2 R U2 B2 L' D2 U2 B L

Again my buffer is UBL

1) UBL->FUL->DFL
1 setup move into a "Drop-and-catch"
F2 U2 L' D2 L U2 L' D2 L F2

2) UBL->LBD->DBR
1 setup move into a "Drop-and-catch"
R2 F D2 F' U2 F D2 F' U2 R2

3) Here we have to break into another cycle. I would do
UBL->URB->RFD "Drop-and-Catch"
B D2 B' U' B D2 B' U

4) UBL->UFR->RBU
1 setup move into a "drop-and-catch"
F F L2 F' R2 F L2 F' R2 F' which combines to F2 L2 F' R2 F L2 F' R2 F'
This is a drop and catch on the R face. Again you need to be able to do these on other faces as well as the U face.

Corners are solved.

--------------------

Solving is incredibly quick. Racing Stefan Pochmann and Joey Gouly with sighted BLD method solves at RWC2007 I got a 15 second solve, and lots of ones around 20-25 seconds. I think a super-fast solve might be about 10-15 seconds and a super slow solve might take 30-40 seconds to give an idea of a range for this method for solving corners.

Move count is very low, so you're less likely to mess up assuming you can correctly "see" which cycle you have, which like big cube BLD just takes practice. For the solve above it took 37 moves to solve corners counting move cancelations (which I actually do perform).

Because all algs are commutators you can use this method on big cubes for BLD, which because of the low turn count is extraordinarily fast. This is actually why I started with this method, but eventually it got to be fast enough that I was overtaking my 3x3x3 BLD times using it rather than orient first and permute.

---------------------

Sometimes I find it hard to memorize very quickly, especially if you have a lot of corners that are correctly permuted, but twisted. Also having a lot of 2 cycles can slow me down too. I memorize with images so perhaps this is just a fault of memorizing with images rather than visually. I can memorize the entire cube with images using about 40-45 seconds at my very fastest and 1:10-1:20 at my very slowest to give an idea of the range.

When just getting started it is sometimes hard to tell which corner cycle you have, but the more you practice the easier this gets. Also if you practice big cube BLD you get pretty much the exact same cycles on 5x5 BLD for X-centers, at least for the most part. It is not a perfect match, but X-centers on 5x5 is excellent practice for 3x3 corners.

-------------

Ok that's about it. I'm not saying this method is better than other methods that are out there, I just think it is a different way to approach it. And because I know people don't like methods unless they are proven in competition I used this method for corners, and a similar one for edges, at RWC2007 and came in 5th place for 3x3 BLD. My fastest competition solve using this method, and the similar one for edges, was 1:49.84. If I could only improve my memorization of this method I think I could get faster. My personal record with this method at home is 1:19.xx

If you have any questions about particular cycles or setup moves, just post here and I'll tell you what I do.

Also I noticed that Stefan using M2 was always faster than my method with commutator 3 cycles for edges. I'm going to try to practice my commutators and see if I can improve my speed. If not I will most likely switch to M2 for edges, or maybe Turbo. But for corners I like my method ;-)

Chris

Pedro

Member
hey Chris

it was really nice meeting you there

and thanks for this tutorial

I tried solving with this sometimes, but like you said, some bad cases were specially bad for me

but I think I'll give it more shots

Pedro

Member
One thing I didn't understand is where you say:

"Drop-and-catch" : R' D R U R' D' R U'
You can also use R' D2 R U R' D2 R U'; R' D R U2 R' D' R U2; R' D2 R
U2 R' D2 R U2; etc.

"Direct-Insert" : R' D' R R' D R [U']"

I don't see why these 2 are different...looks the same "type"...for me, at least : )

Pedro

tim

Member
wow, thanks Chris. I also wanted to ask you some questions about commutators, but i saw many people asking you the same question, so i decided to just listen to you. Unfortunately i didn't understand everything. It's really cool, that you decided to write such a tutorial .

Mike Hughey

Super Moderator
Staff member
Very cool, Chris; thanks for this!

Are those "sighted BLD" times of 15-25 seconds including a pre-memorization of the cube, or are you able to work out the commutators and solve them in 15 to 25 seconds? And is this time just for the corners, or edges as well?

Either way, I'm curious how fast you can do a 2x2x2 BLD using this method. I'd think it might be really fast for that. (Especially if you can choose a good cube orientation when you start.) The typical solve would just be 2 or 3 commutators in most cases, right? Might you consider entering this week's competition for 2x2x2 BLD using your corners method, just to show us how well it works? After all, it should only take you a couple of minutes, including scramble time.

I'm definitely going to try to at least learn how to do this. Somehow the commutators are still very much harder for me to see on the 3x3x3 than they are on the 5x5x5 X-centers, but your explanation makes them make sense pretty well, so I'm sure that mainly it will take practice.

joey

Member
Very cool, Chris; thanks for this!

Are those "sighted BLD" times of 15-25 seconds including a pre-memorization of the cube, or are you able to work out the commutators and solve them in 15 to 25 seconds?
We had about 2-4s for preinspection, just to get a gist of the cube before we started racing, but not actual memorising. So I think potentially, you could do it faster blindfolded, because you know the next cycle that is going to come up. I am definitely starting to look into this method.

And is this time just for the corners, or edges as well?
The time was just for corners.

At one point, Chris did the corners, then passed the cube to stefan and he would do the edges. They were solving it in 40-45s that way.

For me, I use the J perm and the Y perm to solve corners, and it takes me 30s or so. Since I use 160~ moves that way. As chris showed in this example, these comms takes ~40. Thats such a massive improvement, I really want to switch!

By the way, I am Joey Gouly, who was racing them.

Lucas Garron

Member
I was (well, still am) just chatting with dbeyer on #rubik, and I get these sorts of commutators better.
I was using r r' [D] r r' [D'] almost exclusively on 4x4x4 centers, but I think soon I'm going to go very commutatory. Since I'm shopping around for 3x3x3 BLD edge ideas, I'll definitely consider this for corners...

cmhardw

One thing I didn't understand is where you say:

"Drop-and-catch" : R' D R U R' D' R U'
You can also use R' D2 R U R' D2 R U'; R' D R U2 R' D' R U2; R' D2 R
U2 R' D2 R U2; etc.

"Direct-Insert" : R' D' R R' D R [U']"

I don't see why these 2 are different...looks the same "type"...for me, at least : )

Pedro

Hey Pedro, yes it was very nice to meet you in person as well! Brazil isn't too far from the states, I'd love to one day make it down south for a competition, though I'm really broke right now.

Also about what you said, I agree the "direct-insert" case really is technically a "drop and catch" but for some reason I do mentally see them differently. There is a similar situation for the "toss-up" case for the wings of a bigger cube. I guess it would be better to in theory just call the direct insert cases a "drop and catch", but for some reason I do mentally perceive them to be different cases. Even for the similar type of "toss-up" case for wings I also view to be different, even though technically it is not.

Just chalk that up to the weird way in which I think, not sure why I view it as different but I do. If you prefer though don't consider the direct insert as a different case, because technically it isn't, I do agree on that.

Chris

cmhardw

I tried solving with this sometimes, but like you said, some bad cases were specially bad for me
Hey Pedro,

One last thing, if you do get a hard cycle post it here and I'll post back with how I would view the case and what alg I would use to solve it.

Chris

Pedro

Member
I tried solving with this sometimes, but like you said, some bad cases were specially bad for me
Hey Pedro,

One last thing, if you do get a hard cycle post it here and I'll post back with how I would view the case and what alg I would use to solve it.

Chris
sure I'll do it

Pedro

Member
F B U2 B U2 F' B U B2 D2 F B2 R' U R2 U D R F L' R2 D U' L2 F

is a good example

the first cycle is Ulf (I think I'll start from there, as I'm used to start from ULF)
Ufl -> Bdr -> Fur

I tought about doind D2, to make the first and third sticker interchangeable...but that way I wouldn't be able to shoot Fur to any of the 2 positions (now Ufl and Fdl)

then I tought about doing R or R' and using (R' F R F')*2 (or in other faces)

but...I saw L'

that makes Ufl -> Bul, and so interchangeable with Fur

but that way I wouldn't be able to shoot Bdr

so I guess the only option is R/R' and that moves...

next cycle would be Ufl -> Ubl -> Dfr

I think I'd do R2 and a normal cycle on U...

next is Ufl -> Dbl -> Rub

that one I'd do as (F R' F') L2 (F R F') L2

and I'm left with Ufl -> Fdl for parity fix...

Pedro

Member
L2 U B' F' R2 L2 F' R2 L' D U2 R' B D L F' R2 F' D' B F2 U L' B L2

Ufl -> Ubl -> Ldb

D' (L D2 L') U' (L D2 L') U D

Ufl -> Bur -> Rdb

that's a tough one...R' or B would give me 2 stickers on U face, but the other one in a hard position...

so I tought about F, to have Ruf -> Bur...but I wouldn't be able to shoot Rdb...

F' would give me Ldf and Rdb interchangeable...but again I couldn't shoot..

so I guess that's an impossible case

haha...not really
maybe R2 D...no

damn...I can't do that one

so, the only solution I saw was R' B and a normal 3 cycle on U

next I'd have Ufl -> Fur -> Rdf

I think I'd do...hmm...don't know that's a tough one too

maybe L'...no...

cmhardw

F B U2 B U2 F' B U B2 D2 F B2 R' U R2 U D R F L' R2 D U' L2 F

is a good example
Hey Pedro, Actually this solve looks good. If you're curious here is how I would do the same solve, but I mean the way you are doing it is fine I think. I would recommend to use the commutators that you can "see" the easiest. I mean better would probably be to always be able to see the optimal one, but as long as you can see any one you can still get a fast time.

the first cycle is Ulf (I think I'll start from there, as I'm used to start from ULF)
Ufl -> Bdr -> Fur

I tought about doind D2, to make the first and third sticker interchangeable...but that way I wouldn't be able to shoot Fur to any of the 2 positions (now Ufl and Fdl)
Actually that does work, and you can shoot to either position. Here is how. Start with D2. Then you can shoot Fru if you do the setup turn R as well. It's not as efficient because you need 2 setup moves, but sometimes that happens. Plus if it's easy to see for you with 2 setup moves then do 2 setup moves. Better to be fast than to be super efficient but slightly slower I say.

then I tought about doing R or R' and using (R' F R F')*2 (or in other faces)

but...I saw L'

that makes Ufl -> Bul, and so interchangeable with Fur

but that way I wouldn't be able to shoot Bdr
If you do L' that makes the cycle Blu -> Brd -> Fru which is actually a Caltech move. I would do L' F' (L' U L U')*3 B2 (L' U L U')*3 B2 F L

Another thing to notice though, this is already a Caltech move, just on the R face. Do: R2 (F U' F' U)*3 R2 (F U' F' U)*3

next cycle would be Ufl -> Ubl -> Dfr

I think I'd do R2 and a normal cycle on U...

next is Ufl -> Dbl -> Rub

that one I'd do as (F R' F') L2 (F R F') L2

and I'm left with Ufl -> Fdl for parity fix...
Looks good. I would say as long as you find a commutator that solves your cycle just pick that one and go quickly. Better to be fast than super super optimal. I would recommend to do sighted solves without the time though to try to be able to always spot the optimal commutator though.

Chris

cmhardw

L2 U B' F' R2 L2 F' R2 L' D U2 R' B D L F' R2 F' D' B F2 U L' B L2

Ufl -> Ubl -> Ldb

D' (L D2 L') U' (L D2 L') U D
This looks good, but slightly faster I think is U L' D' L U' L' D L but again that is just my opinion. I'm pretty sure Daniel would recommend to use the move you are using, so it just comes down to personal preference.

Ufl -> Bur -> Rdb

that's a tough one...R' or B would give me 2 stickers on U face, but the other one in a hard position...

so I tought about F, to have Ruf -> Bur...but I wouldn't be able to shoot Rdb...

F' would give me Ldf and Rdb interchangeable...but again I couldn't shoot..

so I guess that's an impossible case
This one is tough. I think the fastest move to execute would be R' B' making it a "pure 3 cycle" where you could do F' L' F R2 F' L F R2 then undo with B R. But if you are trying for fewest setup turns I would do F' to make it a Caltech move. Remember you can change your viewpoint as to how to view the cycle. Rotate counter clockwise one sticker on each corner before you do the setup move. This gives you the cycle Lfu->Urb->Dbr which is easier to see as a caltech move where you need 1 setup turn.

F' (R B' R' B)*3 D2 (R B' R' B)*3 D2 F

haha...not really
maybe R2 D...no

damn...I can't do that one

so, the only solution I saw was R' B and a normal 3 cycle on U
That works too, that's how I used to do these cases, but it's a bit faster to execute if you try for R' B' and do a "pure 3-cycle" instead. Just my two cents.

next I'd have Ufl -> Fur -> Rdf

I think I'd do...hmm...don't know that's a tough one too

maybe L'...no...
Actually I agree with L'. Remember to change your viewpoint on harder cycles. Rotate one sticker clockwise on each corner. This makes it Ful->Ufr->Drf which is much easier to see as a Caltech move. I'd do L' U2 (R F' R' F)*3 U2 (R F' R' F)*3 L

Hope that helps. I agree that this solve is a much harder solve than the first one. But with practice you learn to see those "change of viewpoint" cases. I don't even change my viewpoint on them anymore, I just know that when I that type of hard looking cycle that it is just a Caltech move, I am only seeing it from some of the other stickers rather than the regular U and D ones.

Again I hope this helps. Feel free to post more solves if you want. Again I don't claim to always do the most optimal solve, I'll just post what I personally would do for whatever scramble you guys post. Sometimes I do like to do a sub-optimal commutator in terms of move counts if it gives me something fast like a pure 3 cycle or things like that.

Chris

Pedro

Member
L2 U2 B R F U L' F B2 D2 R' F' U' L U2 F L U' L2 B D' F2 R' B2 U2

another one...
Ufl -> Rdf -> Dfl

I'd do D2 R2 (F D2 F') U' (F D2 F') U R2 D2

Ufl -> Lub -> Dbl

I can't do this with a Caltech move...it will mess up UBL...

so...

B2 (U2 (F' D2 F) U2 (F' D2 F) B2

Ufl -> Bdr -> Lfu...so, breaking to Ufr

U (L D2 L') U' (L D2 L')

Ufl -> Ubr -> Ruf

hmm....that one is hard...

I'd orient the 2 on the right and cycle

dbeyer

Member
U] L2 B'R2B L2 B'R2B [U'
B DF'D' B' DFD'
U LD2L' U' LD2L'
F'] U2 LDL' U2 LD'L' [F

I am just going by how you notated the stickers. I hope I read it all correctly. As I didn't have a cube in front of me ...

Last edited:

Mike Hughey

Super Moderator
Staff member
Chris, thanks for taking the time to try some competition solves. The 2x2x2 times definitely look pretty reasonable. I'm going to try to learn how to do this with the corners, and as soon as I start to see them well, I'll start doing my 2x2x2 BLD competition solves with this method. It looks like fun!

It could be a little while, though - it still seems so much easier to do these on the 5x5x5 centers than on the 3x3x3.

hait2

Member
i think i will learn this, it looks very interesting
maybe m2 for edges as well, i keep hearing good things about it

i don't get a lot of things about m2 though, none of the sites i've found (pochmann's/erik's etc.) explain it well enough>_<. what happens if you have like 3-4 different cycles, how do u break into a new cycle, how do you pick a new buffer if yours is already 'taken' etc, there's so many cases that are not covered (here's a specific question: how would you solve edges that are basically a bunch of 2 cycles? like 3-4+ of them and the FD is already in its place oriented incorrectly. i'd guess you have to pick a different buffer, but you can't do that for every cycle, or am i missing something? =x)

maybe ill just stick with my version of the 3cycle for now and use this for corners

tim

Member
hait2: imagine you would have a 3-cycle on the left layer (FL -> BL -> DL) and the rest of your cube is solved (buffer = FD = solved). so your targets would be: FL (breaking in), BL, DL and FL again. Play around with your cube and i think you'll get it.

Last edited:

hait2

Member
hey, thanks a lot tim =)
i just tried your ideas and got the hang of it in about 30seconds. i guess it's my fault for not really trying to learn (resistance to change and all). seems like a piece of cake now

anyway m2 is going to be lots of fun, i love the double m trigger, and the fact that all the algs are purely intuitive <3