The New Method / Substep / Concept Idea Thread

[Z1hc]

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.

 

[ruffleduck]

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]

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.

 

[ruffleduck]

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]

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]

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]

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

 

[TipsterTrickster]

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]

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]

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]

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.

 

[V Achyuthan]

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]

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.

 

[ruffleduck]

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

 

[V Achyuthan]

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]

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]

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

OH K

 

[BenChristman1]

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

If all of the edges are already done, wouldn’t the only PLL cases be the A perms and E perm? (and skip, obviously)

 

[tsmosher]

what is NBRS?

Notation Based Reference System

 

[Quốc Hưng]

Hello, I'm was made this method one month ago, this a 4x4 method for the ZZ method.
Some think ZZ44 is better than 4Z4 but I'm not.
I think ZZ44 and 4Z4 are the same.

And I'm very sorry PaPa Smurf cubes because spam his page.


Link ZZ44: ZZ44 method inf0

How about you