# My commutator BLD corner method



## cmhardw (Oct 12, 2007)

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.

--------------------
Advantages to this method:

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.

---------------------
Disadvantages to this method:

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


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## Pedro (Oct 12, 2007)

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


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## Pedro (Oct 12, 2007)

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


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## tim (Oct 12, 2007)

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 .


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## Mike Hughey (Oct 12, 2007)

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.


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## joey (Oct 12, 2007)

Mike Hughey said:


> 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.



Mike Hughey said:


> 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.


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## Lucas Garron (Oct 13, 2007)

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...


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## cmhardw (Oct 13, 2007)

Pedro said:


> 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
> ...




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


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## cmhardw (Oct 13, 2007)

Pedro said:


> 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


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## Pedro (Oct 13, 2007)

cmhardw said:


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



sure  I'll do it


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## Pedro (Oct 13, 2007)

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...


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## Pedro (Oct 13, 2007)

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...


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## cmhardw (Oct 13, 2007)

Pedro said:


> 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'
> 
> ...



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...
> 
> ...



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


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## cmhardw (Oct 13, 2007)

Pedro said:


> 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...
> 
> ...



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
> ...



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


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## Pedro (Oct 13, 2007)

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


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## dbeyer (Oct 13, 2007)

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 ...


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## Mike Hughey (Oct 13, 2007)

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.


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## hait2 (Oct 13, 2007)

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


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## tim (Oct 14, 2007)

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.


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## hait2 (Oct 14, 2007)

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


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## dbeyer (Oct 14, 2007)

Somebody, 
Please check my solution to Pedro's latest scramble
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

I didn't check myself, and probably won't ... I just looked at the stickers that he wanted cycled, and I visualized it. I know that I had a cycle wrong, correct stickers, wrong direction (but I quickly edited that after posting)

You know how it always looks good to you, but ... somebody else looks at it, and finds 100 mistakes (I hope there aren't that many, I mean there aren't even that many moves lol) (I hate English Term Papers!!)

Later,
Daniel Beyer


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## mrCage (Oct 15, 2007)

Hi 

How can you say that (R F' R' F)*3 U2 (R F' R' F)*3 U2
{ ((R F' R' F)*3 U2 )*2 }

is fast or has a low move count?? Replace (R F' R' F)*3 with R2 D' R2 D R2 and i may partially agree ;-)

-Per K


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## Stefan (Oct 15, 2007)

cmhardw said:


> 2 setup move cases:
> --------------------
> UBL->RBU->BRD
> 
> 1) R' F R2 B L B' R2 B L' B' F' R



D2 [ U L2 U', B' ] D2


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## Johannes91 (Oct 15, 2007)

U2 F' U B2 U' F U B2 U

a.k.a. COLL


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## Pedro (Oct 15, 2007)

StefanPochmann said:


> cmhardw said:
> 
> 
> > 2 setup move cases:
> ...



is it just me or that one moves edges too?


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## Stefan (Oct 15, 2007)

Don't know how that happened. I wish the applet I always use for checking supported commutator notation.

Corrected: D2 [ B', U F2 U' ] D2


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## UberStuber (Oct 16, 2007)

Why limit yourself by keeping the buffer at ULB, i.e., having to break into new cycles? You might end up with the wrong orientation, but I'd rather use an orienting alg than have to do two commuters, and I think it would ease some of the memorization troubles.

Thanks for sharing this.


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## cmhardw (Oct 16, 2007)

StefanPochmann said:


> Don't know how that happened. I wish the applet I always use for checking supported commutator notation.
> 
> Corrected: D2 [ B', U F2 U' ] D2



Hey Stefan,

Wow I like that alg! I hadn't thought to solve this case that way. I can see now how with a "change of viewpoint" that cycle pops out. I simply had not seen that alg to solve that case.

I think that was my only case that "required" 2 setup turns. So really I think every case is possible with either 1 setup turn or no setup turns at all. I don't know what the ratio of cases requiring no setup turns to cases requiring 1 setup turn is, but it's nice to know that no cases require 2 turns.

Cool, thanks Stefan! That alg rocks! Also I am very strongly considering switching to M2 for edges, but would you recommend it for the wings of a 5x5x5 cube for BLD? I don't see why it wouldn't work for the 5x5x5, but then again I have very limited experience with M2 right now.

Chris


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## cmhardw (Oct 16, 2007)

UberStuber said:


> Why limit yourself by keeping the buffer at ULB, i.e., having to break into new cycles?



Hi Uber,

I'm afraid I don't fully understand. In my actual solving I do use something called "pseudo-buffer" meaning that if my buffer piece is solved at the start of the scramble, rather than kick it out to start a new cycle, I use a different piece as my buffer. Also I'm a bit confused by this sentence:



> I'd rather use an orienting alg than have to do two commuters



the point of my method is to solve position and orientation at the same time using intuitive commutators rather than mystical algs. Also I am trying to do 2 steps at once rather than having to split up fixing the corners into 2 steps, orienting and then permuting. Do you mean that you would rather use orientation algs to solve the last 2 or 3 corners after orienting and permuting corners throughout the solve? Or do you mean you would rather orient all pieces first, and then following that permute all pieces?

Chris


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## joey (Oct 16, 2007)

cmhardw said:


> UberStuber said:
> 
> 
> > Why limit yourself by keeping the buffer at ULB, i.e., having to break into new cycles?
> ...


Coming from pochmann methods, I am used to a fixed buffer. I hope that I can start using all pieces as buffers, as it does reduce move count. I have sometimes used a different buffer, but that was on 2x2 BLD XD 

Also I'm a bit confused by this sentence:



> I'd rather use an orienting alg than have to do two commuters



the point of my method is to solve position and orientation at the same time using intuitive commutators rather than mystical algs. Also I am trying to do 2 steps at once rather than having to split up fixing the corners into 2 steps, orienting and then permuting. Do you mean that you would rather use orientation algs to solve the last 2 or 3 corners after orienting and permuting corners throughout the solve? Or do you mean you would rather orient all pieces first, and then following that permute all pieces?

Chris[/QUOTE]
I believe he means that when you use a fixed buffer, you have to do two extra algs when you start a new cycle.


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## Pedro (Oct 16, 2007)

I think what he means if that you can get 2 corners correctly placed but unoriented at the end...and he would rather use an alg to orient them than 2 commutators


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## Stefan (Oct 16, 2007)

cmhardw said:


> Wow I like that alg! I hadn't thought to solve this case that way.



Johannes' alg is also a conjugated commutator, btw, and shorter in QTM than mine:

U' [ U' F' U, B2 ] U



cmhardw said:


> I think that was my only case that "required" 2 setup turns.


I haven't checked all your algs, so I'm sorry if I missed this. How do you cycle UBL to BUR to LFU?



cmhardw said:


> Also I am very strongly considering switching to M2 for edges, but would you recommend it for the wings of a 5x5x5 cube for BLD?


Yes, I've already been using r2 for the 4x4 and 5x5 wings, and m2 for the 5x5 middle edges. Needless to say I like it a lot.


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## Pedro (Oct 16, 2007)

StefanPochmann said:


> I haven't checked all your algs, so I'm sorry if I missed this. How do you cycle UBL to BUR to LFU?



UBL -> BUR -> LFU

I tought about doing L as setup, so I have now FUL -> BUR -> LFD...but I won't be able to shoot LFD to one of the other 2 spots...

that's the kind of cycles that gives me more trouble...when I have, say, 2 on the U face and one on L/F/R/B, is kinda easy to do, as you can put a U sticker on both L and F (for example), but you can't put a sticker that is on B face in the L or R face with one move...

so, the only way I found was something like L for setup and Caltech moves...


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## UberStuber (Oct 16, 2007)

cmhardw said:


> UberStuber said:
> 
> 
> > Why limit yourself by keeping the buffer at ULB, i.e., having to break into new cycles?
> ...




Ah, I was under the impression that you used ULB as your ONLY buffer, and then when you finished a cycle, you break into a new cycle and keep that same buffer, as opposed to just switching to a new buffer. 



> Also I'm a bit confused by this sentence:
> 
> 
> > I'd rather use an orienting alg than have to do two commuters



I meant I'd rather use an orienting algorithm for the last 2 permuted but unoriented corners.

Sorry about my confusing wording.


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## cmhardw (Oct 17, 2007)

> I meant I'd rather use an orienting algorithm for the last 2 permuted but unoriented corners.
> 
> Sorry about my confusing wording.



Oh ok I get you now. Actually I also prefer to use orienting algs in that case. I am still trying to figure out the best option for if I get a case where I am left with a 2 cycle of corners, and 1 other correctly permuted but disoriented corner. As an example: UBL->LUB and the UFR corner needs to rotate counter clockwise to be solved. Sometimes I find that it is easier to memorize the cycle as UBL->ULF->UFR->FRU and sometimes I just memorize UBL<->ULF and that before I can execute that 2 swap I have to orient the UFL and UFR corners.

I don't know which is easier to memorize, but to me the solving is about equally as fast no matter which method I use.

Again I don't know how to memorize the cube in 14 seconds, but I have found that this kind of thing slows down my memorization on the corners *considerably*, which is why I hate that kind of case. But ironicly I can solve it very quickly, it's only the memo that bothers me about this case.

If anyone has any ideas on how they memorize this case quickly I would definitely be interested, I don't know of a consistently fast way to memorize cases like this. Also the similar case but for edges slows me down quite a bit during the memorization phase as well.

Chris


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## Lucas Garron (Jan 2, 2008)

Bringin' this thread back to mention a nice comm I use for 3x3x3:

[RU'R'U2RUR' , [D]]

RU'R'U2RUR' is its own inverse, so makes: RU'R'U2RUR' [D] RU'R'U2RUR' [D']
It reduce thinkings and time for annoying cases with oriented corners on different layer, and sometimes with comm solving/freestyle).

R2UR2U'R2 also works for RU'R'U2RUR' (qqwref's idea).
Stuff like RU'R'F'U2F is horrible and not self-inverting...

Also try RU'R'(Arne/H-perm)U2RUR' 

EDIT: (By the way, I never liked the "Caltech move"  )


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## LarsN (Jan 3, 2008)

Lucas Garron said:


> RU'R'U2RUR' is its own inverse, so makes: RU'R'U2RUR' [D] RU'R'U2RUR' [D']
> It reduce thinkings and time for annoying cases with oriented corners on different layer, and sometimes with comm solving/freestyle).



For those cases I use (R2 D R2 D' R2 U2)x2. Also inversed and rotated around the cube. I find that it reduces thinking and that it's fairly fast.


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