The New Method / Substep / Concept Idea Thread

I might have a variation of LSFB from speed solving wiki:
(For this explanation make sure to have yellow as your bottom face.)
First step: make a 1x2x2 block at bd[R].
Second step: make another 1x2x2 block but at bd[L].
Now it should look like roux F2B right left but without the front 2 F2L pairs.

Third step: orient the L6C. They can be 2 gen if you choose to.

Fourth step: If you have your 1x2x2 blocks in the back with them in the first 2 layers then you can solve FL and FR but you dont really need to care about permuting them. Kind of like a square 1 middle slice.
Fifth step: Orient L6E while putting a yellow edge in BD.
Sixth step: this step is 2 gen and intuitive. Turn the cube so the two 1x2x2 blocks are at LD. Then you make a 1x1x3 yellow bar (doesent need to be a completed bar) and also solve the middle bar from step 4.

Seventh step: You can either permute everything in one algorithm (around 150?) or do 2 look way.
 
I just came up with this idea.
I realised it’s just SSC without EO but I’m posting it anyway(I took 10 minutes coming up with this don’t judge)
So basically,
First you make 3 quarters belt
Then you solve the remaining D layer corners (the last one separating edges and using WV)
Then EZD
 
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Feel free to let me know if I am dumb for inventing a method that already exists
So yesterday I came up with a method that goes like this (have no idea what to name it) :
EO
F2L (yes without cross)
solve top corners while preserving EO
Solve Midges in 1 look
Rotate and Solve S layer which is now the M layer 1 look

Here are some example solves
B2 U R2 D F2 D2 R2 F2 R2 B2 F2 U' F' U2 B' L D F' D2 L2 F U'
F R B D L U' F x2 // EO
D U' R D' // BR pair
R U2 R' // FR pair
U2 L D' L U2 L' S' U2 S D L' U2 L // BL and FL pair
R2 D' R U R' D R U R U' R' U' R // COLL
U2 R D2 M' D2 L F' M2 F R2 x // Midges
y' M' U2 M U2 // Rotate and solve S layer edges
53 STM

L2 R2 B F2 R2 U2 B F2 L2 F' U2 F' R U F2 R' D' L B' L2 U'
B' D R' U R F // EO
L2 // FL pair
R2 U L U L' // BL pair
D R' U2 R D' // FR pair
U' M' U2 r' U' R2 U2 R D R' U2 R D' R2 // BR pair cancelled into COLL
U R2 F2 U2 M2 U M2 U F2 R2 // Midges
y' M' F2 M F2 // Rotate and solve S layer edges
45 STM

This method averages around 47 moves. (All algs not generated yet).
Thank you
 
Alright, don't get angry with me for bringing up another method (fell free to call me dumb for inventing another method that already exists)
I came up with this in like 10 min.
Steps
Solve a 1x1x3 (should be connected to the centres) and place it in DL position
Belt
EO
6CO
6CP
Insert the cross edge that goes between the two corners on the right thus creating another 1x1x3. align both the 1x1x3s to their respective centres using u / u' / u2.
Solve LSE like Roux.

Here are some examples
B2 D F2 L2 D2 B2 R2 F2 U' B2 D' F' R B' L B2 F U2 R' U'
y' L2 U' L' // 1x1x3
R' U' R' u2 U R M R U R' M' // Belt
M2 U' M U' M // EO
U' R' U2 R U R2 U' R U2 R // 6CO
U' x U2 R' D' R U2 R' D R x' // 6CP
D2 // Roux F2B with EO and CMLL done
M' U2 M U M' U2 M2 U2 M' U // LSE
49 STM

U2 D' L D' B L2 F B2 U B' D2 R2 F2 L2 F2 D2 L2 B' U2 B' D2
x2 y L' B2 // 1x1x3
u' R2 u R2 u R // Belt
M' U' R2 U R' r' // EO
U D' R U R' U2 R U R' D // 6CO
U2 R2 D' R2 U R2 U' R2 U' D R2 U R2 // 6CP
M2 U' S' U2 S M2 u' // Roux F2B with EO and CMLL done
M U2 M U' M' U2 M' // LSE / L5E
51 STM

D' F' B D2 B U' B D2 L' B' L2 D2 F' L2 F L2 F' U2 L2 D2 B
F U' L' F2 y // 1x1x3
U' R u' U' R u' R2 r2 R U R' r2 // Belt
M R2 U' M r2 // EO
U' R' D' R U R' D R // 6CO
U' R U2 R' D' R' D R' D' R' U2 R' // 6CP
U2 M' U2 M D // Roux F2B with EO and CMLL done
U M' U2 M U x' U2 M U2 M' // LSE
55 STM

D' L2 R2 D F2 R2 D2 F2 L2 F2 R2 U F R B2 R2 D' B' F R' U
x2 F2 U F' // 1x1x3
u' U2 R' u2 R' U R' u' R U R' // Belt
U M2 U' M U' M // EO
R D' R U' R' D R' // CO
U F U F' R2 F U' F' U' R2 // 6CP
S' U2 S u2 // Roux F2B with EO and CMLL done
M U2 M' U' M2 U' M U2 M U' // LSE / L5E
51 STM

U L2 U' L2 R2 U' F2 L2 B2 D F2 U2 B' R' F2 L' U' B L2 U2 B' R'
y2 x D2 B U2 L // 1x1x3
u R' u2 R' U' R' u2 r2 B U2 B' r2 // Belt
M' U' M' U M U' M // EO
U' R2 U2 R' U R' U R U2 R' // 6CO
U2 F U F' R2 F U' F' U2 R2 // 6CP
S' U2 S u // Roux F2B with EO and CMLL done
M U2 M U' M' U2 M U2 M2 U2 // LSE
57 STM

U B2 F2 U2 B2 U' L2 U' R2 U2 L2 F' R U' F' D L' D B D2 F'
U B U' F2 // 1x1x3
r U' r' u U' R' u R u R U' R' // Belt
U' M2 U' M U' M // EO
R U' R' U R' U R' U' R // 6CO
R2 U' D' R2 U R2 U' R2 // 6CP
S' M' U M U2 S // Roux F2B with EO and CMLL done
U' M' U2 M U' // Somehow left with L3E
50 STM

F' L2 D2 L2 U2 F' R2 F' D2 F R2 B2 L U2 B' F R' D' F' D L
x' D2 L // 1x1x3
U' R2 U' R' u2 R2 U' R' // Belt
U R F R2 F' R // EO
R2 U2 R' U R' U2 R U R' // 6CO
U' R2 U R2 U R2 D R2 U' R2 U R2 // 6CP
D' // Roux F2B with EO and CMLL done
M' U2 M U M' U2 M2 U2 M' U' // LSE
48 STM

L' F2 D2 F2 U2 F2 R' B2 F2 R' F2 L' F' R U2 B' F' U' L' F U2
x' y U' L' // 1x1x3
u R u' R2 u' R U R' // Belt
U' F' L' F U F L F' D' // EO
U2 R U' R U2 R' U R' U R U R' // 6CO
U2 R2 U R2 U' D R2 U' R2 U R2 D' R2 // 6CP
S' M' U' M S u2 // Roux F2B with EO and CMLL done
M U2 M' U' M' U2 M U' // LSE
58 STM

U' L2 B2 D R2 B2 L2 F2 D2 U' R2 U2 F U' L D2 B2 D' B F' L
y2 U' B2 U2 F2 // 1x1x3
u R2 U2 R u2 U2 R2 M U R' M2 U' M // Belt + EO
R' D' R U' R' D R // 6CO
U2 D' R U' R' U2 D R2 U2 R2 D' R U R' // 6CP
D2 y M2 U2 M2 S' M' U M U2 S U' // LSE (But not Roux style)
49 STM

U2 F2 L2 D' B2 D F2 L2 U F2 R2 F2 L' B2 D' L U B2 D2 B' R2 F
x' y' R2 U L // 1x1x3
u' R2 u R2 u r' U r // Belt
U2 M' U' M // EO
U2 R2 U2 R' U R' U2 R U R' // 6CO
R2 U D R2 U' R2 U R2 // 6CP
y M2 U S' U2 S // Roux FB with EO and CMLL done
U' M2 U M2 U M' D2 M' D2 // LSE
47 STM

Average movecount = 51.5 STM
 
Last edited:
V Achyuthan said:
Alright, don't get angry with me for bringing up another method (fell free to call me dumb for inventing another method that already exists)
I came up with this in like 10 min.
Steps
Solve a 1x1x3 (should be connected to the centres) and place it in DL position
Belt
EO
6CO
6CP
Insert the cross edge that goes between the two corners on the right thus creating another 1x1x3. align both the 1x1x3s to their respective centres using u / u' / u2.
Solve LSE like Roux.

Here are some examples
B2 D F2 L2 D2 B2 R2 F2 U' B2 D' F' R B' L B2 F U2 R' U'
y' L2 U' L' // 1x1x3
R' U' R' u2 U R M R U R' M' // Belt
M2 U' M U' M // EO
U' R' U2 R U R2 U' R U2 R // 6CO
U' x U2 R' D' R U2 R' D R x' // 6CP
D2 // Roux F2B with EO and CMLL done
M' U2 M U M' U2 M2 U2 M' U // LSE
49 STM

U2 D' L D' B L2 F B2 U B' D2 R2 F2 L2 F2 D2 L2 B' U2 B' D2
x2 y L' B2 // 1x1x3
u' R2 u R2 u R // Belt
M' U' R2 U R' r' // EO
U D' R U R' U2 R U R' D // 6CO
U2 R2 D' R2 U R2 U' R2 U' D R2 U R2 // 6CP
M2 U' S' U2 S M2 u' // Roux F2B with EO and CMLL done
M U2 M U' M' U2 M' // LSE / L5E
51 STM

D' F' B D2 B U' B D2 L' B' L2 D2 F' L2 F L2 F' U2 L2 D2 B
F U' L' F2 y // 1x1x3
U' R u' U' R u' R2 r2 R U R' r2 // Belt
M R2 U' M r2 // EO
U' R' D' R U R' D R // 6CO
U' R U2 R' D' R' D R' D' R' U2 R' // 6CP
U2 M' U2 M D // Roux F2B with EO and CMLL done
U M' U2 M U x' U2 M U2 M' // LSE
55 STM

D' L2 R2 D F2 R2 D2 F2 L2 F2 R2 U F R B2 R2 D' B' F R' U
x2 F2 U F' // 1x1x3
u' U2 R' u2 R' U R' u' R U R' // Belt
U M2 U' M U' M // EO
R D' R U' R' D R' // CO
U F U F' R2 F U' F' U' R2 // 6CP
S' U2 S u2 // Roux F2B with EO and CMLL done
M U2 M' U' M2 U' M U2 M U' // LSE / L5E
51 STM

U L2 U' L2 R2 U' F2 L2 B2 D F2 U2 B' R' F2 L' U' B L2 U2 B' R'
y2 x D2 B U2 L // 1x1x3
u R' u2 R' U' R' u2 r2 B U2 B' r2 // Belt
M' U' M' U M U' M // EO
U' R2 U2 R' U R' U R U2 R' // 6CO
U2 F U F' R2 F U' F' U2 R2 // 6CP
S' U2 S u // Roux F2B with EO and CMLL done
M U2 M U' M' U2 M U2 M2 U2 // LSE
57 STM

U B2 F2 U2 B2 U' L2 U' R2 U2 L2 F' R U' F' D L' D B D2 F'
U B U' F2 // 1x1x3
r U' r' u U' R' u R u R U' R' // Belt
U' M2 U' M U' M // EO
R U' R' U R' U R' U' R // 6CO
R2 U' D' R2 U R2 U' R2 // 6CP
S' M' U M U2 S // Roux F2B with EO and CMLL done
U' M' U2 M U' // Somehow left with L3E
50 STM

F' L2 D2 L2 U2 F' R2 F' D2 F R2 B2 L U2 B' F R' D' F' D L
x' D2 L // 1x1x3
U' R2 U' R' u2 R2 U' R' // Belt
U R F R2 F' R // EO
R2 U2 R' U R' U2 R U R' // 6CO
U' R2 U R2 U R2 D R2 U' R2 U R2 // 6CP
D' // Roux F2B with EO and CMLL done
M' U2 M U M' U2 M2 U2 M' U' // LSE
48 STM

L' F2 D2 F2 U2 F2 R' B2 F2 R' F2 L' F' R U2 B' F' U' L' F U2
x' y U' L' // 1x1x3
u R u' R2 u' R U R' // Belt
U' F' L' F U F L F' D' // EO
U2 R U' R U2 R' U R' U R U R' // 6CO
U2 R2 U R2 U' D R2 U' R2 U R2 D' R2 // 6CP
S' M' U' M S u2 // Roux F2B with EO and CMLL done
M U2 M' U' M' U2 M U' // LSE
58 STM

U' L2 B2 D R2 B2 L2 F2 D2 U' R2 U2 F U' L D2 B2 D' B F' L
y2 U' B2 U2 F2 // 1x1x3
u R2 U2 R u2 U2 R2 M U R' M2 U' M // Belt + EO
R' D' R U' R' D R // 6CO
U2 D' R U' R' U2 D R2 U2 R2 D' R U R' // 6CP
D2 y M2 U2 M2 S' M' U M U2 S U' // LSE (But not Roux style)
49 STM

U2 F2 L2 D' B2 D F2 L2 U F2 R2 F2 L' B2 D' L U B2 D2 B' R2 F
x' y' R2 U L // 1x1x3
u' R2 u R2 u r' U r // Belt
U2 M' U' M // EO
U2 R2 U2 R' U R' U2 R U R' // 6CO
R2 U D R2 U' R2 U R2 // 6CP
y M2 U S' U2 S // Roux FB with EO and CMLL done
U' M2 U M2 U M' D2 M' D2 // LSE
47 STM

Average movecount = 51.5 STM
Click to expand...
I've had this idea many times with different finishes. The thing that makes this method hard to make viable would be 6CO/6CP algs which generally aren't the best. (A lot of R2 and such.) Not a terrible idea.

EDIT: For example, you could ignore edges and leave L7E for the end with this approach. Things like that.
 
tsmosher said:
I've had this idea many times with different finishes. The thing that makes this method hard to make viable would be 6CO/6CP algs which generally aren't the best. (A lot of R2 and such.) Not a terrible idea.
Click to expand...
You mean this method is not a terrible idea?
tsmosher said:
EDIT: For example, you could ignore edges and leave L7E for the end with this approach. Things like that.
Click to expand...
I will work on that. I will probably leave L7EP for the end.
 
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I would say this method is just a worse roux. Basically if you take out the pointless eo step in the middle and use dcal+cdrll to solve corners (better algs than 6CO+6CP) You have essentially just made up a different way to do SB.
 
PiKeeper said:
I would say this method is just a worse roux. Basically if you take out the pointless eo step in the middle and use dcal+cdrll to solve corners (better algs than 6CO+6CP) You have essentially just made up a different way to do SB.
Click to expand...
But half belt being ~5 moves or less and DCAL being ~9 moves, it is not a terrible SB (although it is partial). 64% algorithmic as well.

Yes, pro Roux would build entire SB in ~15 moves, according to Mr. Mansour. But more realistic for most people, I think, is ~17+ moves for SB. If SBLS is done algorithmically, I think that is 45-50% algorithmic in either case.

You're not wrong though. And Mehta could be said to simply be "a different way to do SB" as well...
 
tsmosher said:
But half belt being ~5 moves or less and DCAL being ~9 moves, it is not a terrible SB (although it is partial). 64% algorithmic as well.

Yes, pro Roux would build entire SB in ~15 moves, according to Mr. Mansour. But more realistic for most people, I think, is ~17+ moves for SB. If SBLS is done algorithmically, I think that is 45-50% algorithmic in either case.

You're not wrong though. And Mehta could be said to simply be "a different way to do SB" as well...
Click to expand...
Isn’t everything just a different method, copying a step, etc? Like you could say ZZ is just CFOP made harder and roux is just a weird Mehta, and it goes both ways. Yes I know I’m two months out of date but I doubt that someone has invented some genius method that solves the cube in 20 moves or less in 5 seconds. Basically what I’m trying to say is this is a useless post and I’m trying to get more posts for no effort isn’t every method an expansion of another one, or just a slightly more efficient way of doing something?

By the way I’m not hating on any of the methods I mentioned for God’s sake people get so hyped up over methods it’s kinda strange I use CFOP/ZZ and I know Mehta LEOR and Roux to some extent so yeah there’s your proof bye I’m out
 
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