# The New Method / Substep / Concept Idea Thread

#### V Achyuthan

##### Member
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.
Ok

#### tsmosher

##### Member
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.
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...

#### Waffles

##### Member
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...
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 Yes I know I’m two months out of date but I doubt that someone has invented some genius method that solves the cube in 20 moves or less in 5 seconds.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

#### Z1hc

##### Member
Might be or not, I have a new method.

Step 1: EO

Step 2: make the first layer on the bottom. Make sure it’s solved with the centers.

Step 3: solve BLU and BRU corners using 1 algorithm. Or 2 if you are a beginner.

Step 4: Solve UL and UR edges in a single algorithm. Or 2 if you are a beginner. Make sure you don’t break EO after.

Step 5: solve 3 middle layer edges. Algorithms are necessary for fast solves. There are less algorithms due to there being EO. It doesent matter if LF or RF is unsolved, just make sure you don’t break EO after.

Step 6: L5P (Last 5 Pieces). There isn’t that many algorithms for this.

I am gonna guess this method has under 100 algorithms, correct me if I’m wrong.

#### zzoomer

##### Member
Might be or not, I have a new method.

Step 1: EO

Step 2: make the first layer on the bottom. Make sure it’s solved with the centers.

Step 3: solve BLU and BRU corners using 1 algorithm. Or 2 if you are a beginner.

Step 4: Solve UL and UR edges in a single algorithm. Or 2 if you are a beginner. Make sure you don’t break EO after.

Step 5: solve 3 middle layer edges. Algorithms are necessary for fast solves. There are less algorithms due to there being EO. It doesent matter if LF or RF is unsolved, just make sure you don’t break EO after.

Step 6: L5P (Last 5 Pieces). There isn’t that many algorithms for this.

I am gonna guess this method has under 100 algorithms, correct me if I’m wrong.
In step 3, why only solve BLU and BRU corners, why not solve all U corners?
oh right... that would make this method EO waterman

#### Z1hc

##### Member
In step 3, why only solve BLU and BRU corners, why not solve all U corners?
oh right... that would make this method EO waterman
I didn’t even think of it being EO waterman. I was thinking of just solving everything in one algorithm if I were to solve all the 4 corners. And it would also make the method Ofapel method with EO.

My current option for this method is kind of unique.

#### zzoomer

##### Member
My current option for this method is kind of unique.
But not promising... solving 2 corners and 3 edges in different layers would probably involve bad algorithms

#### V Achyuthan

##### Member
I thought of a beginner sort of method which goes like this
1st layer with block building
Solve 3 of E slice edges using commutators (not necessarily all 3 in one commutator)
Solve 2 LL edges
L3E
Or you could just combine the last 2 steps and call it L5E

This method averages 55 moves with advanced block building and maybe 65 with normal 1st layer (like doing cross and then inserting corners)

Let me know your thoughts on this
Here are some examples

L' F2 R' B' R B2 L' D R F2 U' B2 U R2 U F2 B2 R2 U2 F2
z2 // inspection
D U' R' D' L D' z' D R' F' U L' U' L R // 1st layer
U R' U L U' R U L' // CxLL
u' R' U' R E' R' U R E // 2 E layer edges using commutator
R U R' E R U' R' D // 3rd E layer edge
U' M' L' U' L U M U' L' U L // 2 LL edges
y U' R E2 R' U2 R E2 R' // L3E
58 STM

U R2 U' F2 R2 U' L2 R2 U B2 F2 U' L' U F' D' R2 B R B D' F2
y // inspection
D2 R2 L B D' B2 z' U L' U' L' U' L // 1st layer
U' R U2 R' F R' F' R U' R U R' E' R U2 R' E // Cxll cancelled into 2 E layer edges using commutator
L' U L E L' U' L E' // 3rd E layer edge
U M2 U' M U2 M' U' M2 // 2 LL edges
y B' r U R' U' M U R U' R' B // L3E
56 STM

D R B' D' F' L' B' R U' R2 U2 D2 R2 B' R2 B2 L2 F' U2 R2 U2
y' x // inspection
U2 r B' r' F2 U' r' F z' y' U2 R' F2 M F' r // 1st layer
U' R U2 R' F R' F' R U' R U' R' // CxLL
y' R' U' R E R' U R // 2 E layer edges using Commutator
y' R' U R E R' U' R // 3rd E layer edge
y' M F' L F M' F' L' F // 2 LL edges
U2 R' E2 R U2 R' E2 R U' // L3E
57 STM

F' R U2 D' F' D2 F' U F R2 B2 R2 U2 B2 L2 B D2 R2 B'
R2 B U' D B2 F z' U D2 R U' R' U R2 U R U R' // 1st layer
U R U2 R' F R' F' R U' R U' R' // CxLL
L U L' E L U' L' U' R' U R E' R' U' R // 2 E layer edges using commutator
y' E2 x' y2 M' U' M U2 M' U' M // L3E
52 STM

#### any name you wish

##### Member
I have made a new method which is really good for 5x5-7x7. I use this method for 6 and 7 but I use Yau on 5x5, but it's still pretty good.

Step 1: Centers
Step 2: Cross
Step 3: Last 8 Edges
Step 4: 3x3 Stage

#### AlgoCuber

##### Member
That's just... Reduction but you pair up cross first?

#### TipsterTrickster

##### Member
Yeah this is basically reduction, usually with reduction you want to pair up edges randomly, ie not doing all white then all yellow etc... Also with reduction you want to leave the entirety of 3x3 stage until after you have finished the reduction stage, this gives you a bit more freedom with edge pairing.

#### any name you wish

##### Member
Yeah this is basically reduction, usually with reduction you want to pair up edges randomly, ie not doing all white then all yellow etc... Also with reduction you want to leave the entirety of 3x3 stage until after you have finished the reduction stage, this gives you a bit more freedom with edge pairing.
Look ahead is way easier this way.

#### tsmosher

##### Member
Look ahead is way easier this way.
How is lookahead better than Yau (with 4 cross edges) though? It seems to me like you're just doing Yau with cross edges and L4C in a different order.

#### AlgoCuber

##### Member
Look ahead is way easier this way.
It's not worth the restriction in edge pairing. You can't use some edge pairing tricks that you would normally and it restricts what you can pair for the first 3/4 edges. The extra lookahead advantage for 6x6+ is negligible because you're taking away 3 seconds at most, but the disadvantages definitely add more time to your solve than it gains.

Last edited:

#### V Achyuthan

##### Member
Another method (lol)
Starting steps
1.FB (like Roux)
2. Belt
3. EO
4. CO
5. SB (like Roux)
(or)
4. CO
5. CP and insert DR edge
6. LSE
Next steps
1) Solve corners of LL and Finish with LSE
(or)
2) Solve DF and DB edges and finish with PLL
(or)
3) Place the DR and DL edges opposite while solving he DF and DB edges like Roux L4E. This reduces the number of PLLs to 7 (H perm, A perms, N perms, F perm and T perm)
(or)
4) If the corners are solved, just do LSE
Let me know you thoughts on this

Here are some examples
B' R F U2 D' B D L U D' F2 U2 R2 B R2 F2 L2 U2 R2 B R2 F
y x' // inspection
R2 U F L' B D' // FB (6)
R r U' R' U R // Belt
R2 U2 R2 S' U S // EO
R D R' U R D' R' // CO
M2 U' R2 U2 R2 // SB
U M' U2 M' U' M2 U2 M U2 M // LSE
40 STM

D' F2 D L2 F2 U2 B2 D2 R2 F2 R2 F2 R U2 B' U' F2 L2 B2 L' B
x // inspection
R2 D2 U2 L' D' B2 // FB
U r U R' u r U' r' // Belt
U M' U M // EO
U R U' D' R U' R' U2 D R U R2 U R2 U2 R2 u' // CO cancelled into SB
U M' U2 M' U2 M2 // DF and DB
R U R' U' R' F R2 U' R' U' R U R' F' U // PLL
56 STM

R2 F2 D2 L2 B' R2 D2 L2 D2 B' R2 F L' D2 F' D R' F2 D R U2 F2
z2 x // inspection
R2 U B2 L' F2 D // FB
R2 M U' R' // Belt
U' S' U' S // EO
R U R' U R U2 R U' M2 U' M2 U' R2 // CO cancelled into SB
U2 M' U2 M // DF and DB
U R' U R' U' y R' F' R2 U' R' U R' F R F U // PLL
47 STM

U' R2 B2 L2 D F2 R2 F2 U' F2 D L2 B' R' B' D2 R U' F L U'
y // inspection
R B U L B // FB
R' r U' r // Belt
U' R' F R2 F' D R' U R D' R' // EO cancelled into CO
R2 U2 R2 U' r2 U' R2 // SB
U' M' U2 M' // DF and DB
F R U' R' U' R U R' F' R U R' U' R' F R F' U' // PLL
49 STM

U2 B2 U F2 D B2 U' L2 B2 D' L2 F2 L B L2 R2 D' B' D U' R' U
y x2 // inspection
R D' B' D F // FB
U r U r' u r U' r' // Belt
R2 U2 M' U' M R2 // EO
U D' R U' R' U2 R U' R' D // CO
R2 M2 U2 M2 U R2 u' // SB
U x' M2 U2 M2 U2 x // DF and DB
U R2' F R F' R' U' F' U F R2 U R' U' R U' // PLL
57 STM

R2 U2 D' B' U2 F R' U2 L F2 D2 L2 B U2 F' R2 B L2 B2 U2
z2 y // inspection
L' D2 L U2 L' // FB
r U r' U2 r' U' r // Belt (EO skip)
U' F R2 U R2 U' F' // CO
U' r2 U2 M2 U R2 // SB
U2 M U2 M' U M' U2 M' U2 M2 U' // DF and DB
x R' U R' D2 R U' R' D2 R2 x' // PLL
45 STM

B D2 B' D2 F L2 F2 L2 R2 F L2 F R U B2 R' B' U L2 D2 U'
z2 // inspection
D2 L B2 R2 D2 F // FB
R2 U2 R // Belt
M2 S' U' S M2 // EO
U2 R2 U R' U R U2 R' U R' U2 // CO
R U' R' D' R U R2 D R U' R' D' R D // CP
S' U2 S // SB
U M' U2 M U M2 U' M' U2 M' U2 // LSE
53 STM

U2 B2 R2 F D2 U2 R2 B' L2 D2 B2 U2 L' B U B2 F2 L F U F
x2 // inspection
L R U L R2 D // FB
R u r U' r' // Belt
U2 S' U S // EO
L U2 L' U' L U' L' // CO
R2 U R2 U' D R2 U' R2 U R2 D' R2 // CP
U S' U2 S u' // SB
U2 x U2 M' U2 M // LSE
44 STM

F U2 F L2 B U2 B2 F' U2 R2 F R2 D U' L' D R2 F D L R'
y x' // inspection
D F B' U F' D2 // FB
U r R U' M2 R // Belt
M U' M' U' M U' M' // EO
U R' U' R U2 D R' U' R D' // CO
U2 R U R' B2 R U' R U R2 B2 // CP
S' U2 S // SB
M' U2 M U M2 U M U2 M' U2 M2 // LSE
54 STM

F2 R2 B2 D2 L' B2 R' D2 L B2 L' R2 F' U F' U2 F' U2 L' R2 F'
z2 x2 // inspection
U' R D B U' L2 B // FB
R U2 S' U' S U R2 U R' r2 // Belt + EO
y' U2 R D R' U R D' R' y // CO
U R2 U R2 U' R2 U R2 // SB
M' U2 M R U R' U' R' F R2 U' R' U' R U R' F' U' x2 M' U2 M U2 // Finish
54 STM

Average movecount = Around 49

#### PiKeeper

##### Member
Another method (lol)
Starting steps
1.FB (like Roux)
2. Belt
3. EO
4. CO
5. SB (like Roux)
(or)
4. CO
5. CP and insert DR edge
6. LSE
Next steps
1) Solve corners of LL and Finish with LSE
(or)
2) Solve DF and DB edges and finish with PLL
(or)
3) Place the DR and DL edges opposite while solving he DF and DB edges like Roux L4E. This reduces the number of PLLs to 7 (H perm, A perms, N perms, F perm and T perm)
(or)
4) If the corners are solved, just do LSE
Let me know you thoughts on this

Here are some examples
B' R F U2 D' B D L U D' F2 U2 R2 B R2 F2 L2 U2 R2 B R2 F
y x' // inspection
R2 U F L' B D' // FB (6)
R r U' R' U R // Belt
R2 U2 R2 S' U S // EO
R D R' U R D' R' // CO
M2 U' R2 U2 R2 // SB
U M' U2 M' U' M2 U2 M U2 M // LSE
40 STM

D' F2 D L2 F2 U2 B2 D2 R2 F2 R2 F2 R U2 B' U' F2 L2 B2 L' B
x // inspection
R2 D2 U2 L' D' B2 // FB
U r U R' u r U' r' // Belt
U M' U M // EO
U R U' D' R U' R' U2 D R U R2 U R2 U2 R2 u' // CO cancelled into SB
U M' U2 M' U2 M2 // DF and DB
R U R' U' R' F R2 U' R' U' R U R' F' U // PLL
56 STM

R2 F2 D2 L2 B' R2 D2 L2 D2 B' R2 F L' D2 F' D R' F2 D R U2 F2
z2 x // inspection
R2 U B2 L' F2 D // FB
R2 M U' R' // Belt
U' S' U' S // EO
R U R' U R U2 R U' M2 U' M2 U' R2 // CO cancelled into SB
U2 M' U2 M // DF and DB
U R' U R' U' y R' F' R2 U' R' U R' F R F U // PLL
47 STM

U' R2 B2 L2 D F2 R2 F2 U' F2 D L2 B' R' B' D2 R U' F L U'
y // inspection
R B U L B // FB
R' r U' r // Belt
U' R' F R2 F' D R' U R D' R' // EO cancelled into CO
R2 U2 R2 U' r2 U' R2 // SB
U' M' U2 M' // DF and DB
F R U' R' U' R U R' F' R U R' U' R' F R F' U' // PLL
49 STM

U2 B2 U F2 D B2 U' L2 B2 D' L2 F2 L B L2 R2 D' B' D U' R' U
y x2 // inspection
R D' B' D F // FB
U r U r' u r U' r' // Belt
R2 U2 M' U' M R2 // EO
U D' R U' R' U2 R U' R' D // CO
R2 M2 U2 M2 U R2 u' // SB
U x' M2 U2 M2 U2 x // DF and DB
U R2' F R F' R' U' F' U F R2 U R' U' R U' // PLL
57 STM

R2 U2 D' B' U2 F R' U2 L F2 D2 L2 B U2 F' R2 B L2 B2 U2
z2 y // inspection
L' D2 L U2 L' // FB
r U r' U2 r' U' r // Belt (EO skip)
U' F R2 U R2 U' F' // CO
U' r2 U2 M2 U R2 // SB
U2 M U2 M' U M' U2 M' U2 M2 U' // DF and DB
x R' U R' D2 R U' R' D2 R2 x' // PLL
45 STM

B D2 B' D2 F L2 F2 L2 R2 F L2 F R U B2 R' B' U L2 D2 U'
z2 // inspection
D2 L B2 R2 D2 F // FB
R2 U2 R // Belt
M2 S' U' S M2 // EO
U2 R2 U R' U R U2 R' U R' U2 // CO
R U' R' D' R U R2 D R U' R' D' R D // CP
S' U2 S // SB
U M' U2 M U M2 U' M' U2 M' U2 // LSE
53 STM

U2 B2 R2 F D2 U2 R2 B' L2 D2 B2 U2 L' B U B2 F2 L F U F
x2 // inspection
L R U L R2 D // FB
R u r U' r' // Belt
U2 S' U S // EO
L U2 L' U' L U' L' // CO
R2 U R2 U' D R2 U' R2 U R2 D' R2 // CP
U S' U2 S u' // SB
U2 x U2 M' U2 M // LSE
44 STM

F U2 F L2 B U2 B2 F' U2 R2 F R2 D U' L' D R2 F D L R'
y x' // inspection
D F B' U F' D2 // FB
U r R U' M2 R // Belt
M U' M' U' M U' M' // EO
U R' U' R U2 D R' U' R D' // CO
U2 R U R' B2 R U' R U R2 B2 // CP
S' U2 S // SB
M' U2 M U M2 U M U2 M' U2 M2 // LSE
54 STM

F2 R2 B2 D2 L' B2 R' D2 L B2 L' R2 F' U F' U2 F' U2 L' R2 F'
z2 x2 // inspection
U' R D B U' L2 B // FB
R U2 S' U' S U R2 U R' r2 // Belt + EO
y' U2 R D R' U R D' R' y // CO
U R2 U R2 U' R2 U R2 // SB
M' U2 M R U R' U' R' F R2 U' R' U' R U R' F' U' x2 M' U2 M U2 // Finish
54 STM

Average movecount = Around 49
This is literally roux-domino-duction.

#### zzoomer

##### Member
It would be nice if everyone here used NBRS to describe their method/substep ideas.

#### V Achyuthan

##### Member
This is literally roux-domino-duction.
and what is that?

It would be nice if everyone here used NBRS to describe their method/substep ideas.
what is NBRS?

This is literally roux-domino-duction.

There isn't anything like that on the speedsolving wiki.

#### PiKeeper

##### Member
and what is that?
A method proposed by Blobinati Cuber
It's
FB+EO
Belt
CO
SB
LSE
PLL but its only 7 cases

#### V Achyuthan

##### Member
A method proposed by Blobinati Cuber
It's
FB+EO
Belt
CO
SB
LSE
PLL but its only 7 cases
OH K