Full LMCF 3x3 method now available

[efattah]

The next E2L trick especially for beginners is to use the displacement method. This is an *extremely fast* intuitive method of edge solving which requires essentially no algorithms and has extraordinary lookahead, way easier/faster lookahead than probably any other solving method.

Hold white on top and orange on front, and scramble:
U M2 U' L' U M2 U' L' R U M2 U2 M' U L' R U M2 U' M U M' U2 M' U R' U M U' M'
We want to solve green-orange at DF, into UR. We see there is a blue-red piece there, we could solve it as a pair, but for this example we will only use displacement. We now want to displace ANY edge in the L face. So we do an L. Now we see we can displace the blue-yellow piece. We do U M' U'. We displaced the blue yellow to the BD position and we did that on purpose so we KNOW that is the next piece we can solve even though we can't see it any more. We do an r move. Now we want to solve blue-yellow at DF, into UL, and displace an edge piece on the R face. There aren't any available to displace so we just do U' M' U.
Now we want to solve the green-yellow piece at UF. We do an M move to put it at DF. Now we want to displace ANY edge piece on the L face. We do an L move. We see we can displace the blue white piece at UL. We do U M' U', and the blue-white piece is the next one we solve.

This wasn't a great example but it shows the method; you solve one edge while deliberately displacing another one and tracking it. Then you always solve that piece you displaced as the next one, and displace another one and so on. This way you have perfect lookahead as you always are tracking the next piece.

 

[Thermex]

I thought of an interesting variant of LMCF that would be more ergonomic and also a little more move efficient. It would start like this:

1. Solve a "hexagon" on the d-face (by hexagon I mean the block you make in hexagonal fransisco that consists of a full face minus one corner and one edge) ~9 moves
2. Put in the last corner of the d-face while silmultaniously orienting and permuting the u-layer corners (basically solve the last five edges in one algorithm. I'm currently working on a 2×2 algset that does this) ~10 moves
3. Solve either a triplet or pair of edges somewhere on the cube (I'm not sure what the most efficient thing to do here would be) 6-9 moves?
4. Solve the remaining edges using an L5E, L6E, or L7E set (depends on what you do in step 3) 12-15 moves?

Basically I'm sure about the first two steps, as with lots of practice you can probably one look the whole thing AND go into E2L with the 3 hardest edges to find already solved. In the last two steps, what's the most efficient way to solve DF, UF, UL, UB, UR, and the four e-layer edges with two algorithms? I was thinking you could solve three edges on the top layer and then solve the last 6 edges with one of the waterman L6E sets you made, but there might be faster and more efficient ways. If this works out we could work on this method together, as I'm very interested in method designing.

 

[bren077s]

What you have described here is basically a subset of the Waterman method. Waterman described part of his method that is identical to your method with one difference. He places the last corner on the face and then does CLL instead of doing last 5 corners. Your variant would save a few moves compared to that method at the cost of learning a lot more algs.

It is a fine variant of waterman in my opinion. I think that the issue with your method is that it gets rid of a few advantages of LMCF. First off, LMCF complete removes block building(your variant adds that back in). Secondly, LMCF allows for looking ahead to the entire corners during inspections. With your variant, you will still have to deal with recognizing your corner alg during the solve. I think you variant is viable and may save a move or two at the cost of good block building and a lot of corner algs.

 

[Thermex]

@brenn077s Yeah everything you said was also running through my head. Basically my thought process was that if you can one look your hexagon/first block during inspection, you can track your last d-layer corner and then try to recognize your L5C case as fast as possible (there are somewhere around 350 last 5 corner cases, including CLLs and TCLLs). Then you'll have 9 edges left, and you could do the rest of the solve exactly like waterman, but I was thinking somehow you could use E2L and L6E algs for the last two steps instead of the standard, kind of inefficient waterman way where you use the DF edge as a keyhole and solve 2 U-layer edges and DF, and then do an L6E case. I was thinking like maybe use an algorithm that solves 3 u-layer edges and then using one algorithm to solve the rest, or something like that. Basically I'm just asking: whats the most effiecient way to use two algorithms to finish off a waterman solve?

 

[bren077s]

That sounds interesting and could be a viable way to solve the cube pretty efficiently. I am not sure what the best way to solve 3 edges on the u face would be, but maybe efattah would have some good ideas. I mean he made the pdf full of algs after all. You are basically efficiently reducing the cube to corners along with a face-1 edge. Afterwards I feel that LMCF method of solving edges should work fine.

 

[Thermex]

@bren077s yeah basically this method is just using more algs and blockbuilding to have one triplet of edges already solved once you finish your corners, which makes lookahead easier and the movecount a bit lower. For the last two steps (L9E) I do think the most efficient way is to solve three u-layer edges and then use one of the L6E algs he has in his document, as L6E with UL and UR unsolved is more efficient than standard waterman L6E (two redges+m ring unsolved). I would just need to finish generating L5C cases (~4 months?) and get Efattah's help on the TEUL algs and this could be a seriously viable speedcubing method.

 

[bren077s]

That being said, the full version of LMCF has almost 800 algs. So this method would have 1000??? Cause it would have even more algs, correct? 800 algs is already more algs than full ZBLS and ZBLL. I feel like you are making a method that may be extremely alg heavy and time consuming to learn(yet still needing block building).

 

[efattah]

The best way and only currently existing way to quickly solve the last 9 edges is to use modified Waterman, where you have two options:
1. solve 2 redges at the same time in one algorithm, then finish with L7E (which is 2 algorithms, one to solve the L/R edges orient midges, then the ultra fast midge permutation)
2. solve 2 redges and the last ledge at the same time in a triplet algorithm, then finish with Waterman/LMCF L6E (this option can't always happen since a triplet set up doesn't occur every solve)

Waterman and its modernized variants have a lot going for it; slightly better ergonomics than LMCF, half the algorithms of LMCF, but LMCF beats it in two major ways; LMCF has much higher statistical chances of pre-solved edges and solving triplets, and LMCF puts the CLL/EG recognition in the inspection phase.

The problem with any block building method is you are forced to solve specific edges. In Waterman or even Roux, as an example, you are essentially forced to solve that first block. By constraining yourself and forcing yourself to solve such a limited set of pieces you are not putting luck on your side. The whole idea of LMCF was the 1-look corners (with algorithm selection in the inspection), and then by putting ZERO constraints on the next pieces you solve, you can allow the cube to partially solve itself. A true LMCF expert is color neutral and after solving the corners will choose the L/R sides based on which pieces on the cube are accidentally pre-solved, and then choose the next edge set to solve based on the first available natural triplet. This is how LMCF gets a lower move count than Waterman, because you get randomly solved edges on almost every solve. The drawback is higher algorithm count, and in some solves, less ergonomics on the edge solving phase.

 

[efattah]

I will comment that many times I have solved the corners and found FIVE edge pieces pre-solved by accident. This has often happened on the orange/red sides (which are not my choice for the L/R; I prefer blue/green). Nonetheless if I see such a lucky situation I still put orange/red on L/R and finish the solve anyway. The usual movecount for 5 pre-solved edges is 28-30.

As an example, a CFOP solver could solve the cross and find all four bottom corners pre-solved. But this would be useless as he would have to BREAK every single one of those corners to solve the F2L pairs. That is not putting luck on your side. If a piece is pre-solved you want to leave it in the pre-solved position.

Similarly a CFOP solver could solve the cross and find almost the entire upper layer solved. Again he would have to break it all up to finish his solve.

A roux solver could solve the first block and find all the U corners solved. He would have to break them to solve the second block.

 

[Thermex]

@efattah Ok, first the alg count. It definitely wouldn't be over a thousand. Evenn though L5C is ~350 algorithms, there would probably be around 160 TEUL or tiplet algs, depending on what you learn, and somewhere around 90-190 last _ edges algorithms depending on whether you learn waterman L6E, L7E, or L6E with UL and UR unsolved. The total probably wouldn't be more than full LMCF.

It's true, this proposed method flows much different and is more consistent and less lucky than LMCF. My whole idea was that I didn't really like that transition phase after the corners where you have ALL the edges unsolved and it can be really hard to lookahead and find a triplet. I'm sure there are people though (like you) who find this easier to do than I do, and yes, it does allow for a lot of lucky things to happen. By block building you're forcing the cube to have three edges solved, which in my opinion makes the solve feel a bit more like a CFOP/ROUX/ZZ solve where you do intuitive blockbuilding and then finish off the solve with 2-3 algoithms and you only have to look at one side of the cube. So to sum it up, it's just a slightly more efficient method (move count) with slightly better ergonomics (feels similar to the big 3 methods), but a lot less luck in it meaning in my opinion it's about the same as LMCF.

Moving on, these are the four ways I narrowed down for solving your last 9 edges (I don't care about the # of algorithms it has)

1. Solve two edges on the u-layer in one algorithm, then finish off the solve with Waterman L7E
2. Solve three edges on the u-layer in one algorithm, then finish off the solve with UL and UR unsolved L6E
3. Solve a regular triplet (DF, UL, and UR) in one algorithm, then finish off the solve with two-redges unsolved waterman L6E
4. Earlier in the solve blockbuild a FULL layer minus one corner, do step 2 normally, solve three edges on the top layer and finish off the solve with L5E and skip midge permutation

Either four of those would probably work in a solve, but a couple of questions about them:
1. I don't really understand how to solve the second triplet for option 2, would you use E2L triplet algs? If so, where would you place the DF edge before the algorithm, and when would normal triplet setup fail?
2. Would there be any way to generate TEUL algs for options 3 and 4? Do they already exist? What would be their average movecount?
3. For option 4, how bad is the recognition for L5E? Do algorithms for this already exist?
4. Of the above ways to finish off the solve of this method, which has the best recognition and the lowest movecount?
(Sorry for so many questions lol, hopefully you can answer them all)

 

[bren077s]

@Thermex I can totally understand your want for the method to feel more like one of the big three with better recognition. I currently am trying to get good at LMCF. I find that the first and last step are super easy for me to get decent times in. For me, decent times is about 10 seconds for Corners and about 10 seconds for L5E(please note that I normally average about 35 with ZZ, so those break outs are good for me). That being said, my recognition and execution of E2L is terrible. I average 40 seconds to do E2L currently(it is slowly getting better). My best E2L time is 20 seconds because it was lucky. Also, I can get better times(around 30 on average) with E2L if I just spam the first piece that I see. That being said, spamming pieces is very inefficient and I am trying to force myself to learn proper recognition and look ahead for E2L.

On a side note, Are you sure that your method is more efficient? Cause I am not completely sure that it would be. If step 3 and 4 of your method average towards the higher end of your estimate, then the move count would be the same as full LMCF.

If your method ends up having better recognition and some sort of beginner variant, so that people can use it without all of the algs. I think that it might be very useful, especially for most people who know a big three method and are terrible at E2L.

@efattah I am just wondering, how long did it take you to learn LMCF and get to major goals like sub 30, sub 20, and/or sub 15?

 

[Thermex]

@bren077s good idea. I just thought of a beginner method here:

1. Solve a full layer minus one edge piece
2. Use a 2-look CMLL (or 1-look if you already use 2×2 CLL or Roux)
3. Solve one pair of edges on the u-layer
4. Solve the last two edges on the u-layer
5. Rotate the cube so that the m-ring and FL are unsolved
6. Use one algorithm that solves FL and orients the midges
7. Permute the midges

Using 2-look CMLL, this variant would average in the low 60s and would probably use about the same amount of algorithms as beginner LMCF. This wouldn't be the most efficient method but still better than CFOP (lol). There are also easy ways to make an intermediate method that's close in move count to the advanced method(<50) but cuts off like 500 algs, you just solve a full layer-1 edge, CMLL, and the last two steps normally.

As for the efficiency, you're right. I would need the last two steps combined to be under 20 moves. I feel like there's DEFINITELY ways to do that, I just need Efattah's help on figuring that out (all my questions for that are in my last post). The first two steps can easily be done in 18-19 moves with L5C cases, so if the last two steps combine to be under 20 moves this would be a sub 40 move method (on average) which is slightly less than LMCF and the most efficient method I know.

 

[bren077s]

How exactly do you expect someone to do steps 3 and 4?

 

[Thermex]

@bren077s one of these four ways that I listed in a post a few hours ago:

1. Solve two edges on the u-layer in one algorithm, then finish off the solve with Waterman L7E
2. Solve three edges on the u-layer in one algorithm, then finish off the solve with UL and UR unsolved L6E
3. Solve a regular triplet (DF, UL, and UR) in one algorithm, then finish off the solve with two-redges unsolved waterman L6E
4. Earlier in the solve blockbuild a FULL layer minus one corner, do step 2 normally, solve three edges on the top layer and finish off the solve with L5E and skip midge permutation

I was wondering which of these four ways was the most efficeint and had the easiest recognition. You and efattah can both answer that if you can. Also in that post I asked these questions I hope can be answered:

1. I don't really understand how to solve the second triplet for option 2, would you use E2L triplet algs? If so, where would you place the DF edge before the algorithm, and when would normal triplet setup fail?
2. Would there be any way to generate TEUL algs for options 3 and 4? Do they already exist? What would be their average movecount?
3. For option 4, how bad is the recognition for L5E? Do algorithms for this already exist?

 

[efattah]

@efattah I am just wondering, how long did it take you to learn LMCF and get to major goals like sub 30, sub 20, and/or sub 15?

I started in early January 2016. I could solve the cube in 80 seconds using my ancient techniques from when I was young.
I developed LMCF at that point and started practicing it.
I got my first sub-30 Ao12 in 2 months (using Ortega and just the L5E DFL set)
I got my first sub-20 Ao12 in 5.5 months (using Ortega+EG1 and just the L5E DFL set)
I got my first sub-15 Ao12 in 13 months (using Ortega+EG1+CLL and 3 different L5E sets)

I got my first sub-20 single in 6 weeks.
I get my first sub-10 single in around 7 months.
I got my first sub-9 single in 13 months.

 

[bren077s]

@Thermex That makes sense. I don't know why I didn't realize that at first. I wish I had answers to your questions, but I only have guesses because I am pretty new to the method and just learning currently. I am trying not to simply say uneducated guesses.

I think I can answer question number 3 though. If you are trying to L5E plus midge permutation in one algorithm, I would bet that the recognition isn't terrible but will be bad enough that it will be much faster to to do L5E minus midge permutation and then a very short midge permutation algs. That honestly will probably be optimal or within a move or 2 of optimal most of the time.

 

[Thermex]

@Thermex I think I can answer question number 3 though. If you are trying to L5E plus midge permutation in one algorithm, I would bet that the recognition isn't terrible but will be bad enough that it will be much faster to to do L5E minus midge permutation and then a very short midge permutation algs. That honestly will probably be optimal or within a move or 2 of optimal most of the time

Makes sense. Sometimes I get too obsessed over movecount and forget that adding one or two moves to a step can actually make it faster. I actually found a post earlier in this thread where someone asked effatah for those L5E algs and he said the same thing as you.
This all means that we can now narrow down my list of possible L9E finishes to these 3 ways:

1. Solve two edges on the u-layer in one algorithm, then finish off the solve with Waterman L7E
2. Solve three edges on the u-layer in one algorithm, then finish off the solve with UL and UR unsolved L6E
3. Solve a regular triplet (DF, UL, and UR) in one algorithm, then finish off the solve with two-redges unsolved waterman L6E

And your response also leaves me with only three unanswered questions:

1. I don't really understand how to solve the second triplet for option 3, would you use E2L triplet algs? If so, where would you place the DF edge before the algorithm, and when would normal triplet setup fail?
2. Would there be any way to generate TEUL algs for option 2? Do they already exist? What would be their average movecount?
3. Of the above ways to finish off the solve of this method, which has the best recognition and the lowest movecount?

Hopefully you can answer these questions, efattah (Sorry for so many). Then this method can really get going!

 

[Neuro]

Hey so even though I'm not super good with the method I did get a 38 move solve with LMCF just now. Here's the link

 

[efattah]

Different solve on the same scramble with same number of moves:
Scramble: L F' U F2 R' D' B' D R U2 R' D2 R2 L U2 R B2 R2 B2
F R' f' R U' F R' F U' F2 R U R
U' M U' M' // red green edge
z' D M U M' U' D' // orange-blue and orange-green pair
x' M2 U M2 U' l' // red-blue and set up L5E
M U2 M U' M' U' // L5E rBDR set
r2 U2 M U2 // permute midges
38 STM

 

[efattah]

Cool untimed solve today:

Scramble: F U F2 D' R2 F2 D' F2 U' R2 B' L B' L U R2 D' U2
x' y2 F' R U' R F2 R' U R' // corners
z'
R2 M U' M2 U L' U M2 U' // E2L triplet
x L' U' M U R // E2L pair
R U' M' R' U' M U M R U // Waterman Set 1
U2 M2 U2 M2 // permute midges
Total 36 STM

https://alg.cubing.net/?setup=F_U_F...Waterman_Set_1 U2_M2_U2_M2_//_permute_midges