LCBM (Long-Comeau Belt Method) Guide and Algorithms

[uberCuber]

00:35:24 <+Kirjava> TFM, Salvia and Belt fall under a method category that I like to call "bad ideas"

Off topic, curious, what do you think of Tripod?

 

[Kirjava]

Sorry, but if you are proposing this as anything other than fun you are deluded about what makes a good speedsolving method.

EDIT:

Off topic, curious, what do you think of Tripod?

Trying to solve the tripod efficiently is fun. And hard.

 

[OMGitsSolved]

B2 D B' R' D' F' D' U2 B2 F2 R2 F D' B2 U' F' U2 F' R' U' L U L' U' D'
Quarter turn metric
EO: D F B U B' (5,5)
Belt: R B2 R2 D' L2 D' L (7,12)
EPOCBL: U2 x2 R U' R2' D2 R U2 R U' R' (10,22)
1x2x3 Block: x2 U R2 D' R2 D' U' R2 U2 R2 U' R2 (11,33)
1x1x3 Block: D' R2 U2 R2 U' R2 D (7,40)
PLL: U R' U R U' R2 F' U' F U R x U R' U' R2 (15,55) stupid e perms.

qq, is this what you were talking about? Would this be better?
I'm liking both of them but the first version does have more algs.
EDIT:

This is how it would've gone done with the original version...
Quarter Turn Metric
EO: D F B U B' (5,5)
Belt: R B2 R2 D' L2 D' L (7,12)
EPOCBL: U2 x2 R U' R2' D2 R U2 R U' R' (10,22)
Seperation+D Layer Corners: x2 D' U' D' U R2 M' E2 M' (8,30)
EPPBL Parity: M2 U2 M2 D' (4,34)
EPPBL(U PLL Subgroup): U' R U R' F' R U R' U' R' F R2 U' R' U (15,49)

Hm?

 

[cuBerBruce]

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.

The above is not really correct.

Assuming you orient using the <U,D,L2,R2,F2,B2> EO scheme, then the belt must be on E or M (without making any cube rotations). If the belt is made on M, the cube can be rotated with a z or z' so the belt is on E in the rotated cube. Making the belt on S, and rotating the cube to make the belt on E, would generally mess up EO.

(Of course, making the belt on M instead of E affects what colors you would then use for CO.)

If you did happen to make an EOLine, with the line on D as usual for ZZ, then that line could be used as a start for a belt on M. A z/z' can then be used to make the line (and belt-to-be) on E. Of course, if you try to build an EOLine, you could simply try to build the line on L or R (directly in the E layer to avoid needing a z/z' rotation).

 

[qqwref]

B2 D B' R' D' F' D' U2 B2 F2 R2 F D' B2 U' F' U2 F' R' U' L U L' U' D'
Quarter turn metric
EO: D F B U B' (5,5)
Belt: R B2 R2 D' L2 D' L (7,12)
EPOCBL: U2 x2 R U' R2' D2 R U2 R U' R' (10,22)
1x2x3 Block: x2 U R2 D' R2 D' U' R2 U2 R2 U' R2 (11,33)
1x1x3 Block: D' R2 U2 R2 U' R2 D (7,40)
PLL: U R' U R U' R2 F' U' F U R x U R' U' R2 (15,55) stupid e perms.

qq, is this what you were talking about? Would this be better?

Yeah, that's the general idea (although you're not counting in quarter-turn metric (if you were it'd be 74 moves, I think) and that's an F perm, not E).

The first example solve in the first post (the one that's listed as 62 moves) is 92 sqtm. That's the big problem with the PLL+EPLL idea, it's just a huge number of quarter turns and thus takes a lot of time to perform, plus you do have to adjust two layers.

 

[OMGitsSolved]

Yeah, that's the general idea (although you're not counting in quarter-turn metric (if you were it'd be 74 moves, I think) and that's an F perm not an E.

Oh, silly me I used to mix those up when I learned PLL and it came back to haunt me. And I'm nub at turn metrics.

 

[Erzz]

Oh, silly me I used to mix those up when I learned PLL and it came back to haunt me. And I'm nub at turn metrics.

QTM = every 90 degree turn is 1, slices are 2 per 90
HTM = every turn of a face is 1 (90 or 180), slices are 2
STM = every turn of a face or slice is 1 (90 or 180)
Snyder Metric = every turn of opposite faces is 1 (90 or 180) ie U D F B R U is 4. U D are opposite, F B are opposite. No one uses this though

 

[uberCuber]

SQTM = every 90 degree turn of a face or slice is 1

 

[qqwref]

I think Snyder Metric and Axial Turn Metric (ATM) are the same thing, although I could be wrong. A long time ago I used AQTM (axial quarter turn metric) to describe how the Ryan Heise sim functioned.

 

[OMGitsSolved]

Updated version

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.
Spoiler:
Quote Originally Posted by qqwref
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) 1x2x3 Block On Bottom

This is done using only R2, U, and D moves. This can be completed by making a 1x2x2 block on bottom then adding a 1x1x2 block. The belt can be a R2 off at the end.

Step 6) Expand 1x2x3 to 1x3x3

This is done with the same turn mask (R2, U,and D). Simply move the 1x2x3 block to the left of the D layer and the make the 1x1x3 by connecting the corners to the remaining edge. While completing the block make sure you have the belt completely solved. Then align the D layer.

Step 7) PLL
Last step of the Fridrich method and many other methods.
http://www.speedsolving.com/wiki/index.php/Special:AlgDB?mode=view&view=default&puzzle=3&group=PLL

This does decrease the alg count and it may be more efficient.
Leaving the algs with just EPOCBL and PLL. Thanks, qqwref!

 

[Erzz]

Example solve of the new way.
B' R' L2 B R B' D F2 U' F D' R D2 F' R' F L2 U2 R' D L2 U B' D' R
STM
EO: D B D' F' D F (6/6)
Belt: L2 D' U R' (4/10)
Corner Parity: R2 U2 R2 (3/13)
EPOCBL: U2 x2 U' L' (U D') l2 L2 U l2 U' L (10/23)
D layer: U' L2 U2 L2 U' M' U2 M2 U2 M' y M' U2 M (13/36)
PLL: U' L' U L' U' L' U' L' U L U L2 (12/48)

Maybe more later.

 

[OMGitsSolved]

Example solve of the new way.
B' R' L2 B R B' D F2 U' F D' R D2 F' R' F L2 U2 R' D L2 U B' D' R
STM
EO: D B D' F' D F (6/6)
Belt: L2 D' U R' (4/10)
Corner Parity: R2 U2 R2 (3/13)
EPOCBL: U2 x2 U' L' (U D') l2 L2 U l2 U' L (10/23)
D layer: U' L2 U2 L2 U' M' U2 M2 U2 M' y M' U2 M (13/36)
PLL: U' L' U L' U' L' U' L' U L U L2 (12/48)

Maybe more later.

Not exactly the new way but close.

 

[Erzz]

Not exactly the new way but close.

Combining the blockbuilding steps.

 

[OMGitsSolved]

Decided to do at least 5 example solves tonight with qq's idea. I like it a lot.

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

EO: y L B2 L B'
Belt: U D' L' R2 D R'
Corner Parity: R2 (qq's idea; since the next after orientation it doesn't need to be alligned)
EPOCBL: x2 U' L D' L' U' r2 U2 L' U' L'
1x2x3: D R2 D' R2 L2 D' L2 R2 U' R2 B2 U2 B2 R2 (horrible...)
1x1x3: D' U R2 U' R2 U' R2 U R2 D
PLL: R' U2 R U2 R' F R U R' U' R' F' R2 U2

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

EO: y L' D2 B
Belt: L' D' R L D R'
Corner Parity: L2 y2
EPOCBL: U' L' U L' U' r2 U' L U L'
1x2x3: D2 U' R2 D' R2 D2 U R2 U2 R2 U R2
1x1x3: D U R2 U' R2 U' R2 U' R2 U' R2 U2 R2
PLL: y2 R U R' F' R U R' U' R' F R2 U' R' U'

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

EO: U L2 B
Belt: D L F2 R2 U R
EPOCBL: U2 x2 U R U R' U2 D R2 U2 R' U' R'
1x2x3: U' R2 D2 R2 D' U F2 U' F2 R2
1x1x3: D U R2 U2 R2 U' R2 U R2 U' R2 U' R2 U' R2
PLL: R2 u' R U' R U R' u R2 y R U' R' U'

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

EO: L' F B L B
Belt: D U L' R U' R
Corner Parity: R2
EPOCBL: x2 U2 R (U' D) R U' R' U2 D' R'
1x2x3: D L2 D L2 D' U' R2
1x1x3: D' U' R2 U R2 D2
PLL: R2 U R U R' U' R' U' R' U R'

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

EO: y R' B F U' F
Belt: R D L' D' L'
Corner Parity: L2 U' L2
EPOCBL: x2 U R' (U2 D2) R' U R2 U D2 R
1x2x3: U R2 D' R2 D' R2 D2 U2 R2 U R2
1x1x3: D U' R2 U' R2 U' R2 U' R2 U2 R2 D'
PLL: U' R U R' F' R U R' U' R' F R2 U' R' U2

 

[Godmil]

ooh, that new way is actually a lot of fun. I'll never learn the single orientation stage, but I can see myself doing more solves with this method.

 

[OMGitsSolved]

lol It's fun doing R2 U R2 U multiple times in a rapid motion? It is to me lol.

 

[gbcuber]

Decided to do at least 5 example solves tonight with qq's idea. I like it a lot.

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

EO: y L B2 L B'
Belt: U D' L' R2 D R'
Corner Parity: R2 (qq's idea; since the next after orientation it doesn't need to be alligned)
EPOCBL: x2 U' L D' L' U' r2 U2 L' U' L'
1x2x3: D R2 D' R2 L2 D' L2 R2 U' R2 B2 U2 B2 R2 (horrible...)
1x1x3: D' U R2 U' R2 U' R2 U R2 D
PLL: R' U2 R U2 R' F R U R' U' R' F' R2 U2

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

EO: y L' D2 B
Belt: L' D' R L D R'
Corner Parity: L2 y2
EPOCBL: U' L' U L' U' r2 U' L U L'
1x2x3: D2 U' R2 D' R2 D2 U R2 U2 R2 U R2
1x1x3: D U R2 U' R2 U' R2 U' R2 U' R2 U2 R2
PLL: y2 R U R' F' R U R' U' R' F R2 U' R' U'

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

EO: U L2 B
Belt: D L F2 R2 U R
EPOCBL: U2 x2 U R U R' U2 D R2 U2 R' U' R'
1x2x3: U' R2 D2 R2 D' U F2 U' F2 R2
1x1x3: D U R2 U2 R2 U' R2 U R2 U' R2 U' R2 U' R2
PLL: R2 u' R U' R U R' u R2 y R U' R' U'

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

EO: L' F B L B
Belt: D U L' R U' R
Corner Parity: R2
EPOCBL: x2 U2 R (U' D) R U' R' U2 D' R'
1x2x3: D L2 D L2 D' U' R2
1x1x3: D' U' R2 U R2 D2
PLL: R2 U R U R' U' R' U' R' U R'

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

EO: y R' B F U' F
Belt: R D L' D' L'
Corner Parity: L2 U' L2
EPOCBL: x2 U R' (U2 D2) R' U R2 U D2 R
1x2x3: U R2 D' R2 D' R2 D2 U2 R2 U R2
1x1x3: D U' R2 U' R2 U' R2 U' R2 U2 R2 D'
PLL: U' R U R' F' R U R' U' R' F R2 U' R' U2

Make a video plz

 

[OMGitsSolved]

a video of some solves using "the new way"
filmed and uploaded on june 24th, before it was even mentioned... i had planned to develop this myself but looks like a few people have beaten me to it
Note: i dont know eoline

eo, belt, co, bottom layer, fix belt (parity), pll

i originally tried the first method, and then wondered if instead... if all the pieces could be oriented and the puzzle be solved like a domino... or a sq1

You can avoid Belt parity... that's what your supposed to do. But I am happy people are trying it.
And really? You were going to develop this yourself before we posted it? I doubt it.

Make a video plz

Of what?
The examples? Okay. George those weren't my examples right? Because I spent awhile doing those.

 

[OMGitsSolved]

With that said, im not trying to "steal" your method, im simply stating that we have both come up with the same idea...

No, please don't play this game with me. You called it intuitive in your video which it really isn't. My original idea was to place D layer corners then PBL Then qq gave me the D layer blockbuild idea. I really don't think you came up with the same idea. I checked your website, too. I would stop.

 

[OMGitsSolved]

I called it "intuitive" because it can be

You cant see the page from my website because its locked for my eyes only, there are 4 method variants on there so i wont make it public willy nilly -.-

I will give you this however:

Spoiler

24/6/2011 - LCBM was released today and my first impressions were possitive, however i think i have come up with a way of solving it with less algorithms and possibly faster...

Eo,belt, corner orient... solve as a domino... z then roux cycle

i have yet to coutn the algorithms but im sure this could potentially be another method, i will have to try a few solves and then count the algs before deciding wether its worth continueing this as another side project of mine

Algorithm Count:
eo = intuitive
belt = intuitive
co = 7? + paity case (done on u and d)

solve as a domino = 44 - http://sites.google.com/site/rectangularcuboids/3x3x2-domino/pll

roux cycle = intuitive... think there is only 3

Afterthought: did a few solves using this last night and after a certain comment was posted on "said thread" afterorientation it is exactly the same as a SQ1, and therefor, this is entirely possible... oe, belt, orient corners, solve the rest R2UD gen (my favourite 3 gen) by first placing bottom edges, then bottom corners, and then PLL ... either R2UD gen or plain old CFOP PLL

Afterthought #2: Solve the corners with Ortega? permuting at least, using R2uD/R2UD gen algorithms... ill have to try this

I never said it was "Exactly" the same, but it is similar

That's not the first version or second version. Stop trying. I really don't want to argue with some who has like 10 spelling mistakes in his spoiler.