LCBM (Long-Comeau Belt Method) Guide and Algorithms

Erzz

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Introduction​

New method! LCBM is a belt method, concieved by Nick Long, aka OMGitsSolved. He contacted me about generating some algorithms to permute both layers and orientate both layers. Not in the same step, of course. Now you're probably thinking "wat, that's like 3768 algorithms!" Well, it would be, but Nick cleverly came up with ways to reduce this number to just 161. Let's get into the method.

Step 1) Edge Orientation​
The first step is to orientate all the edges. This is the same as ZZ except you don't make a line. Well, you could make a line, but it would have to be somewhere on the E layer, rather than D. Check out some ZZ guides to see how to do this.

Step 2) Make a Belt​
Solve all the edges on E. You can make the belt on M or S, but it has to end up on E, and you can't rotate or else EO will get messed up. So making it on E is probably the best. This is completely intuitive, similar to the cross.
(You actually can make it on M without messing up the EO.)

Step 3) Fix Corner Orientation Parity (if any)​
If you have an OLL case on U that you don't recognize, it's because a corner is twisted that shouldn't normally be twisted. See this post from qqwref on how to deal with parity.
qqwref said:
The idea is this: R2 U' R2 basically replaces DRF with ULF. So what you want to do is replace one corner on D with a differently-oriented one, fixing the corner parity. If you want to rotate a bottom corner clockwise, for instance, you could replace solved with clockwise, or clockwise with counterclockwise, or counterclockwise with solved.

Another option (if you're willing to have the belt be either solved or an R2 off, which might be worthwhile for corner separation) is to simply do an R2, moving two corners to the bottom layer. I am not sure this can always create a solvable parity case, but it seems to work most of the time, so it's probably still useful (and hey, one move is better than 3 or 7, eh?). I guess, yeah, if you're going to separate later, it wouldn't hurt to leave the belt up to an R2 off - no point wasting moves keeping it that way when you're just going to intuitively mess it up later.

Step 4) E Preserve Orient Both Layers​
Orientate both U and D layers while preserving the E layer. The U and D layers will get mixed up in this step. Algorithms are below. Since we did EO, there are only 56 (7 are just OLL) cases, rather than 3249.

Step 5) Separate and Corners of D Layer​
In this step, you solve the corners of the D layer while bringing all pieces to their correct layer. Solving the corners is done intuitively. If you can't figure it out, try R2 (some U move) R2 and see what that does. You can swap the UF and DF edges by using M' U2 M.
If you have just one corner out of place, put that corner in UBR, and put the location it needs to go in DFR, then do
R2 U' R2 U' R2 U R2

OR See this post.

Step 6) E Preserve Permute Both Layers​
Permute both layers. If you don't recognize the PLLs, do M2 U2 M2 to fix parity. (you can also control what PLL you get after fixing parity. I recommend avoiding V perms) Algorithms below. Since we solved the D layer corners, there are only 105 (21 are just PLL) cases, rather than 441.



Here are the algorithms, arranged by which case is on D.
All the cases are the same as in the wiki, except Gs, which have headlights on L. D layer cases should be recognized as if they were on the U layer with an x2 (not z2)
Average move count for EPPBL seems to be around 17 or 18.
Average move count for EPOCLL seems to be around 9 or 10.

EPPBL
H
H: M2 U2 M2 y M2 U2 M2
Ua: D M2 U M' U M2 D' M2 U M U M2
Ub: D M2 U2 M2 D' M2 U M' U2 M U' M2
Z: M2 U' L2 (U D') M2 (U' D) l2 U M2
Aa: L R2 D R' U2 R D' L' R2 D L R U2 L' R' D
Ab: R2 (U' D) R2' (U' D) L2 R2 U' l2 U L2' U' l2 U R2 U2
E: r2' U' R2 U l2 U' L2 U2 L2 R2 (U D') R2 U' l2 U' l2 U L2'
F: R2 (U' D) r2' U' R2 D r2' R2 (U D') l2 U' l2 U R2 U'
Ga: U L2 (U' D) R2 (U D') r2 R2 U' r2 U r2 U' r2 U r2 D'
Gb: R2 U' l2 U l2 U' l2 U l2 L2 (U' D) L2 (U D') l2
Gc: R2 U l2 U' l2 U R2 U' L2 U D' L2 U l2 L2 U l2 L2
Gd: R2 U l2 U' l2 U l2 U' l2 L2 (U D') L2 (U' D') l2
Ja: R2 U' L2 (U D') R2 U' R2 U R2 D' L2 U2 L2 R2 U' L2 R2
Jb: L2 R2 U' M2 U2 L2 U' l2 U R2 U' R2 (U' D) R2 U' L2
Na: l2 U l2' U' r2 U' r2' U2 l2' U' l2 U' r2 U R2' U L2 l2'
Nb: M2 U R2 U r2 U' l2 U' l2 U2 r2 U' r2 U' l2 U l2
Ra: R2 (U' D) L2 U r2 U' L2 l2 (U' D) r2 U l2 U' R2 U
Rb: R2 (U' D) R2 U' l2 U l2 L2 (U' D) r2 U' l2 U R2 U'
T: R2 U' R2 U' r2 U2 r2 U' D' R2 U' r2 U2 r2 U R2 D
V: U' r2 U' l2 U l2 U R2 U l2 U' L2 U' R2 (U' D) L2 U' l2 U L2 (looong)
Y: R2 U' R2 U L2 U l2 U2 r2 U R2 U L2 U' L2 U'

Ua
H: U M2 D M' D M2 U' M2 D M D M2
Ua: M' (U' D) M2 D2 M' U' M' (U' D) M U'
Ub: L2 (U' D) R2 U M2 U' L2 (U' D) L2
Z: M2 D M' U2 M' (U D') M2 U M2
Aa: U R2 (U' D) r2 R2 U R2 U' r2 (U D') R2 U' l2 U R2 U'
Ab: U' l2 U' l2 U R2 U' r2 U' L2 U r2 U2 L2 U' l2 U'
E: U' L2 U D L2 U l2 U L2 R2 U L2 U' L2 U' D' L2 U'
F: U R2 U' l2 U l2 U' R2 U r2 R2 U' r2 R2 U R2 (U D') R2
Ga: U' L2 U D' L2 U l2 U' R2 U r2 U' L2 U'
Gb: R2 U' L2 U2 l2 U L2 U r2 U' R2 U' D' R2 U2 L2
Gc: R2 U' r2 U r2 R2 (U' D) R2 U' R2 U r2 (U D') r2
Gd: R2 U' l2 U L2 U' r2 U R2 U D' R2 U2
Ja: L2 U r2 U' r2 U l2 U' l2 (U' D) L2 U'
Jb: U' R2 (U' D) r2 U' r2 U l2 U' l2 U l2
Na: L2 U2 L2 U' l2 U'r2 U L2 U L2 U2 D L2 U R2 U'
Nb: U L2 U l2 U' r2 U L2 U r2 U' l2 U' L2 U2 r2 U2
Ra: U L2 (U' D) L2 R2 U R2 U' L2 U r2 U' r2 (U D') L2 U2
Rb: R2 (U' D) r2 U' r2 U L2 U' R2 U L2 R2 (U D') R2 U'
T: U2 L2 U l2 U R2 U' r2 U' L2 U' L2 U R2 U R2
V: R2 U2 D R2 U L2 U' R2 (U D') L2 U l2 U' l2 U R2 U'
Y: L2 U l2 U L2 U2 r2 U' R2 U l2 U' L2 D l2 U'

Ub
H: R2 (U' D) M2 D M' U2 M' U' M2 D2 L2
Ua: L2 (U' D) r2 U M2 U' l2 (U' D) L2
Ub: M' (U' D) M2 U2 M' D M' (U' D) M D
Z: M2 D' M' U2 M' U' D M2 U' M2
Aa: U' r2 U r2 U' L2 U l2 U' r2 U2 r2 U' r2 U' R2 D'
Ab: (U' D) l2 U' l2 U R2 (U' D) l2 U R2 U' L2 l2 (U D') l2
E: U R2 U' D' R2 U' r2 U' R2 L2 U' R2 U R2 U D R2 U
F: U' L2 U r2 U' r2 U L2 U' l2 L2 U l2 L2 U' L2 (U' D) L2
Ga: R2 U l2 U' L2 U l2 U' D' l2 L2 U l2 U' D l2
Gb: R2 U' L2 U2 l2 U L2 U r2 U' R2 U' D' R2 U2 r2
Gc: R2 U' r2 U r2 R2 (U' D) R2 U' R2 U r2 (U D') r2
Gd: R2 U2 L2 U l2 U L2 U' r2 U R2 U D R2 U2 r2 U2
Ja: U L2 (U D') l2 U l2 U' r2 U r2 U' r2
Jb: R2 U' l2 U l2 U' r2 U r2 (U D') R2 U
Na: U R2 U' L2 U2 D' L2 U' L2 U' L2 D R2 U L2 U2 L2
Nb: U' R2 U2 R2 U r2 U R2 D U2 r2 U' r2 U l2 U' L2
Ra: U2 L2 U D' l2 U l2 U' R2 U L2 U' L2 R2 U' D L2 U'
Rb: U R2 U' D L2 R2 U' R2 U L2 U' r2 U r2 (U D') R2
T: R2 U' r2 U' L2 U l2 U R2 U R2 U' L2 U' L2 U2
V: U l2 U2 D L2 U r2 U' R2 (U' D) L2 U R2 U' r2 U R2
Y: R2 U' r2 U' R2 U2 L2 U L2 U' R2 U l2 U' L2 U'

Z
H: M2 D' R2 U D' M2 U' D L2 U M2
Ua: M2 U M' D2 M' U' D M2 D M2
Ub: M2 D' M2 U D' M D2 M U' M2
Z: M2 D' M2 U D M2 U' M2
Aa: U D l2 U' l2 U L2 (U D') L2 U l2 U' l2
Ab: U' l2 U l2 U' L2 (U' D) L2 U' l2 U R2 U'
E: R2 (U' D) l2 U2 L2 U' l2 U' D' L2 U' R2 U2 R2 (U D') R2
F: U L2 U L2 U' l2 U' r2 U L2 U' R2 U r2 U l2 U' l2 L2
Ga: D U2 L2 U' r2 U r2 U' R2 U D L2 l2 U' L2 (U' D) L2
Gb: r2 U L2 U l2 U L2 U2 l2 U' r2 U r2 U D' r2 U
Gc: D R2 U l2 U' l2 U L2 U D l2 L2 U L2 (U D') L2 U
Gd: D' r2 U' L2 U' l2 U' L2 U2 l2 U r2 U' r2 (U' D) r2
Ja: R2 U L2 U' D r2 U l2 U2 L2 U' r2 U' r2 U' r2 U2
Jb: r2 U r2 U D r2 U' r2 U' r2 U' l2 U' R2 U r2 U'
Na: l2 U' l2 U' L2 U' L2 U2 L2 U' l2 U D l2 U' l2 U L2 D' l2
Nb: R2 U l2 U' R2 U' L2 R2 U2 R2 U l2 U' D l2 U l2 U' L2 U'
Ra: D r2 U' R2 U' r2 U r2 U l2 U' l2 U r2 U' R2 U
Rb: r2 U R2 U r2 U' r2 U' l2 U l2 U' r2 U R2 U' D'
T: U R2 (U D') L2 U' L2 R2 U D r2 U' l2 U l2 U' l2
V: l2 U L2 U2 L2 l2 U L2 U' l2 U D l2 U' l2 U r2 U' R2 U
Y: l2 U' l2 U' D' L2 U l2 U' l2 U' D' R2 U' L2 U2 R2 U
See also: This pretty table I made

EPOCLL
Sune
Sune: R' U' R' U R U2 D R' D' R
antiSune: R' U' R U2 R (U' D) R2 U R'
H/Double Sune: U' R (U' D) r2 R2 U' r2 U R'
U/Headlights: R' U2 R' U D R2 U2 R' U' R
T/Chameleon: L D' L' U' r2 U2 L' U' L'
L/Bowtie: R D r2 R' U R U2 r2 R'
Pi/Wheel: L U D L' U D2 L' U2 D L'

antiSune
Sune: L U L' U2 L' (U D') L2 U' L
antiSune: L' U L U' L U' D2 L U L'
H/Double Sune: U' L' (U D') l2 L2 U l2 U' L
U/Headlights: U R U' R2' D2 R U2 R U' R'
T/Chameleon: R' (U2 D2) R' D R2 U2 D R
L/Bowtie: L U R U' L U2 L2 U' R'
Pi/Wheel: R U2 L2 U R U D r2' R U R

H / Double Sune
Sune: D R (U D') L2 R2 U' r2 U R'
antiSune: D' L' (U' D) L2 R2 U l2 U' L
H/Double Sune: R2 (U' D) L (U D') L2 l2 (U D') L
U/Headlights: R D R' U D2 l2 U2 R' U' R'
T/Chameleon: L D L' U' D2 r2 U2 L' U' L'
L/Bowtie: L U' L2 R2 U' L' l2 D U2 L
Pi/Wheel: R' U' L2 U R2 (U' D) r2 R' (U' D) R'

U / Headlights
Sune: R U' R2 D L' R2 U R' U' L
antiSune: L' U R U' L l2 U l2 U R'
H/Double Sune: R U R' U2 D R2 U2 R' U' R'
U/Headlights: U2 L' U L' U' r2 U' L U L'
T/Chameleon: R (U' D) R U' R' U2 D' R'
L/Bowtie: D' L' U' L' U2 L U' L
Pi/Wheel: L' U R U2 R' U L R U R'

T / Chameleon
Sune: L U' L' D' L2 U2 L' U' L'
antiSune: R' (U2 D2) R' U R2 U D2 R
H/Double Sune: L U L' U2 D' L2 U2 L' U' L'
U/Headlights: L2 U2 L' U' L U2 L' U' L'
T/Chameleon: R (U2 D2) R' (U2 D2) R'
L/Bowtie: U R U D2 R' U' D2 R'
Pi/Wheel: L' U' L l2 U' l2 U' L' U2 L

L / Bowtie
Sune: R U L2 U2 L' U R' U' L'
antiSune: x2 L U R U' L U2 L2 U' R' (or do all the Us as Ds, and don't rotate)
H/Double Sune: L U L2 U' L U' D2 L U2 L
U/Headlights: U' L' D' L' D2 L D' L
T/Chameleon: D R U2 D R' U2 D' R'
L/Bowtie: L U' D' L U D L'
Pi/Wheel: L' U' L2 U' L' U2 D' L2 U' L'

Pi / Wheel
Sune: L U D L' U2 D L' U D2 L'
antiSune: D R U L2 U D R U2 r2 U' R'
H/Double Sune: R' U L2 U' R2 (U D') L2 R (U' D) R'
U/Headlights: R' U' r2 D2 U' R' U r2 U' R
T/Chameleon: L' U' L2 D2 L U2 D' L U' L
L/Bowtie: D L' U' D' L D R2 U L R2
Pi/Wheel: L U D L' U2 D2 L U D L'
See also: This pretty table I made


Example Solves
B' U' B R' L2 D B U2 F' L2 R2 U' R' U B2 R L U2 F B' L2 F' L D R

EO: U2 D2 R y [3/3] (STM)
Belt: D' U2 R' L2 U L [6/9]
Corner parity: L' U2 L' D2 L U2 L [7/16]
EPOBL: x2 L' U2 L2 U L2 U L2 U2 L' [9/25]
Seperation+D Layer Corners: D' L2 U L2 U' R2 U2 R2 y' M' U2 M U' M U2 M' [15/40]
EPPBL: D2 r2 U' l2 U l2 U R2 U l2 U' L2 U' R2 (U' D) L2 U' l2 U L2 (D) [22/62]
(that's the longest EPPBL case)

L' B2 D' L F' R' D U2 B' F2 L' F2 B D L' B2 F2 U' F B2 R' B'

EO: y U D' R2 F [4/4] (3/3 STM)
Belt: U D R D L2 D L' [7/11] (7/10 STM)
Corner Parity: none
EPOBL: U2 x2 R' U L2 U' R2 (U D') L2 R (U' D) R' [13/24] (11/21 STM)
Separation+D Layer Corners: U2 R2 U' R2 L2 U' L2 d' M' U2 M [13/37] (11/32 STM)
PLL Parity: U' M2 U2 M2 [6/43] (4/36 STM)
EPPLL: U' D' l2 U' l2 U' D' L2 U l2 U' l2 U' D' R2 U' L2 U2 R2 U [20/63] (20/56 STM)

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

EO: D' R' U R F (5/5)
Belt: D' R' D R D L2 D2 L' (8/13)
Corner Parity: U R2 U R2 (4/17)
EPOCLL: x2 U L U' L' D' L2 U2 L' U' L' (10/27)
Seperation+ D Corners: R2 D2 U R2 U' R2 U2 R2 S U2 S' U M' U2 M (15/42)
EPPBL Parity: M2 U2 M2 (3/45)
EPPBL: D' U' L2 U D L2 U l2 U L2 R2 U L2 U' L2 U' D' L2 U' D' (20/65)

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

EO: L' D' F R' B R B' (7/7)
Belt: L F2 B2 (3/10)
Corner Parity: U L2 U2 L2 (4/14)
EPOCLL: U x2 U' L U L' U2 L' (U D') L2 U' L (11/25)
Separation: F2 U F2 D' R2 U' R2 M' U2 M2 U2 M' (d U) M' U2 M (17/42)
PLL Parity: D2 M2 U2 M2 (4/46)
EPPLL: U2 r2 U L2 U l2 U L2 U2 l2 U' r2 U r2 (U D') r2 U D (18/64)

more later.

Have fun.

(Also, this is my 500th post on Speedsolving!)

See also:
A different belt method.
A method that could become a belt method. (Guimond style separation of all layers)
 
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The EPOCLL algs are generally pretty nice, actually, but EPPBL looks just awful. So many double turns... It seems like the quarter-turn movecount would be very high.

Also, I don't think your orientation parity algs will always solve the case. Flip URF counterclockwise and DLF clockwise. Now neither of your parity algs will do anything. I have a feeling, though, that you can always navigate to a solvable OLL case using something of the form U^a D^b R2 U^c R2. In this case it would be something like U R2 U' R2, and now you have 2 flipped corners on D and 0 on U.
 
Looks interesting, i will have to try this method sometime,

although once you have solved the belt, you could orient the corners and solve it like a domino...?
note: ive not tried this and i dont own a domino so i might be completely off the mark...

EDIT: thought about the above, it could potentially mix up the belt pieces... but this is fixed with z or z' and a roux cycle
 
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The methods of reducing the cases are pretty good. I've heard lots of people mention PBL on 3x3, but didn't think anyone would come up with a working system (and I never imagined anyone would try OBL :) )
So well done for that. It's an impressive bit of work coming up with the whole system.

However..... the complexity of some of the stages and high alg count mean I can't really see anyone adopting it. It certainly doesn't jump out as a method that could be very fast (though I'd be happy to be proven wrong)

One question, I had both layer orientated and they were all sorted except for one corner in each layer... what is the best way of fixing that?
 
Nice idea and you have clearly put some effort into this.

But I do not think this method is for me, way to alg-heavy (and with looong algs at that).

I will try your step 3 "parity alg" in my Human Thistlethwaite solves though.
Takes the "thinking" out of turning "new OLL's" into standard OLL's.

Butthis reminds me to start working again on my HTA alg list for the "new" OLL's.
Should be about 10 algs using a Guimond style approach.
Alg restrictions of keeping EO and keeping belt edges in the belt (but may permute in it).
 
I done think you could solve it like a domino, there would probably be some sort of parity (probably just a pll, its wortha shot if you want to try it

Naw, the domino's corner's can't be rotated, the 3x3's can.
 
The EPOCLL algs are generally pretty nice, actually, but EPPBL looks just awful. So many double turns... It seems like the quarter-turn movecount would be very high.

Also, I don't think your orientation parity algs will always solve the case. Flip URF counterclockwise and DLF clockwise. Now neither of your parity algs will do anything. I have a feeling, though, that you can always navigate to a solvable OLL case using something of the form U^a D^b R2 U^c R2. In this case it would be something like U R2 U' R2, and now you have 2 flipped corners on D and 0 on U.

He told me to generate the EPPBL as <R2,L2,U,D>. The QTM counts are like 25 or 30. (Snyder metric is most flattering for these algs) That move group lets you just do r2 or l2 to change Ds to Us, and some of the algs turned out quite nicely. (though some are bad, they should still be sub 2 able with practice)
We had a lot of trouble with the corner parity. I'm not sure how you would recognize which moves to do for your way. If you figure something out, let us know.


The EO isn't for reducing movetypes, it's for reducing OLL cases.

However..... the complexity of some of the stages and high alg count mean I can't really see anyone adopting it. It certainly doesn't jump out as a method that could be very fast (though I'd be happy to be proven wrong)

One question, I had both layer orientated and they were all sorted except for one corner in each layer... what is the best way of fixing that?

Corner parity and recognizing cases on both sides are really the only complex steps.
You can do some PLL/commutator stuff to swap just the corners if you don't want to mess up your edges. If you don't care about edges, put the corner which is in U to the UBR position, and the spot it needs to go in the DFR position, and do R2 U' R2 U' R2 U R2.

But I do not think this method is for me, way to alg-heavy (and with looong algs at that).

I will try your step 3 "parity alg" in my Human Thistlethwaite solves though.
Takes the "thinking" out of turning "new OLL's" into standard OLL's.

The orientation algs are only like 9 or 10 moves.
The parity alg messes up a lot of stuff, not sure what Thistlethwaite solves before parity. See qqwref's post though.
 
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The EO isn't for reducing movetypes, it's for reducing OLL cases.

How does that matter? The fact still remain that your first 4 steps does reduce the cube to said generators. The only difference is that when doing reduction E might be unsolved, but takes at most two moves. Not that I think that any method based on the same idea(s) has much of a (speed) potential.
 
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One question, I had both layer orientated and they were all sorted except for one corner in each layer... what is the best way of fixing that?

How about placing one corner in UBR and one in DLF and do L2 A(a) L2 ?

Also, I came up with a somewhat different belt method a few months ago. Not sure if anything in there might be useful or if the 2 methods can be combined in any way to make them more efficient but might be worth a look.
 
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How does that matter? The fact still remain that your first 4 steps does reduce the cube to said generators. The only difference is that when doing reduction E might be unsolved, but takes at most two moves. Not that I think that any method based on the same idea(s) has much of a (speed) potential.

I'll have to go look up Kociemba to understand what you're doing in that solve. It looks hard to recognize though.
 
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