The New Method / Substep / Concept Idea Thread

[hyn]

so can i claim ownership of this method?

dont forget to patent it

 

[Athefre]

so can i claim ownership of this method?

You proposed but everyone said no. It may be heartbreaking, but it's time to find another.

It's a method not even a mother could love.

 

[Cuber987]

I have a new method (ortega variation) idea for 2x2. I'm calling it the BOP method, standing for bar, OBL, PBL.

So basically...

Example scramble: U2 R U F2 R' F2 R' F' R U'

Reconstruction

For the bar step:

You need to find the quickest way to attach two pieces correctly in the LFD and LBD spots. Here you can do that with y' z F'. This forms the bar.

For the OBL step:

There are many ways to do it, and I am open to suggestions, but how I currently do it is as follows:

You need to get the top and bottom sides to have only two colors. In this case they are white and yellow because of where we are holding the bar, which will not move. You could do this with an algorithm set, but that would be lots of algorithms, so currently, I orient two in a bar on the bottom right with the following moves: U2 R. Now the bottom is all oriented. You now identify the OLL like in Ortega. We have a T. Solve it with U R U R' U' R' F R F'. Now your OBL is done.

For the PBL step:

First you want to put all the yellow on the bottom no matter if it's a side or layer. Here we can do this with an R2.

Then there are four cases:

1. (the case we have) a t-perm. You should know this from Ortega.
2. a y-perm. You should know this from Ortega.
3. one bar on the bottom left (made in the beginning). My alg currently is R2 U R2 U R2 U' R2 U R2 U R2.
4. two bars on the left. My alg currently is F2 U' R2 U2 F2 U' F2

Do the t-perm. My alg is U2 R' F R F' R U2 R' U R U2 R'. The AUF is a U'.



I am open to suggestions for one bar and two bar PBLs and any suggestions in general.

Another example solve:

Scramble: U2 R' F' R2 F R U' R

z' // green bar solved
R U R2 U R U R' U' R' F R F' // OBL
U2 R2 U' R2 U F2 U' R2 U2 F2 U' F2 U // PBL and AUF

Reconstruction

 

[V Achyuthan]

I have a new method (ortega variation) idea for 2x2. I'm calling it the BOP method, standing for bar, OBL, PBL.

So basically...

Example scramble: U2 R U F2 R' F2 R' F' R U'

Reconstruction

For the bar step:

You need to find the quickest way to attach two pieces correctly in the LFD and LBD spots. Here you can do that with y' z F'. This forms the bar.

For the OBL step:

There are many ways to do it, and I am open to suggestions, but how I currently do it is as follows:

You need to get the top and bottom sides to have only two colors. In this case they are white and yellow because of where we are holding the bar, which will not move. You could do this with an algorithm set, but that would be lots of algorithms, so currently, I orient two in a bar on the bottom right with the following moves: U2 R. Now the bottom is all oriented. You now identify the OLL like in Ortega. We have a T. Solve it with U R U R' U' R' F R F'. Now your OBL is done.

For the PBL step:

First you want to put all the yellow on the bottom no matter if it's a side or layer. Here we can do this with an R2.

Then there are four cases:

1. (the case we have) a t-perm. You should know this from Ortega.
2. a y-perm. You should know this from Ortega.
3. one bar on the bottom left (made in the beginning). My alg currently is R2 U R2 U R2 U' R2 U R2 U R2.
4. two bars on the left. My alg currently is F2 U' R2 U2 F2 U' F2

Do the t-perm. My alg is U2 R' F R F' R U2 R' U R U2 R'. The AUF is a U'.



I am open to suggestions for one bar and two bar PBLs and any suggestions in general.

Another example solve:

Scramble: U2 R' F' R2 F R U' R

z' // green bar solved
R U R2 U R U R' U' R' F R F' // OBL
U2 R2 U' R2 U F2 U' R2 U2 F2 U' F2 U // PBL and AUF

Reconstruction

This is probably just worse ortega.

 

[Vo1d]

So... I just thought of something reallllly dumb...

Essentially, its just Roux and CFOP put together.

Step 1: F2B just like Roux

Step 2: Orient and Permute UB and UF edges

Step 3: M2

Step 4: perform CFOP OLL/PLL or CFCE CLL/ELL

Step 5: M2 to solve it


Example Solve:

Scramble: R2 B2 D' R2 B2 D' R2 U2 F2 R D' R2 F' D F2 D2 F' R

inspection: z2
first block: U2 R' L U L' U2 L2 U' L' U F2 U F'
second block: U B' R B R U2 R2 U' R U2 F' U' F U R U' R'
orient UB and UF edges: U2
M2: M2
OLL: U2 r U2 R' U' R U' r'
PLL: U' x R2 F R F' R U2 r' U r U2 x'
AUF: U
M2: M2

Ik i suck at roux plz dont bully me

 

[G DA CUBER]

I thought of a new CF skewb method.

1. Solve bottom corners.
2. Solve top corners while solving bottom Centre.
3.Last 5 Centres

If it's unoriginal I'm so sorry.
Ill delete it if it is

 

[ProStar]

This is probably just worse ortega.

It's closer to bad EG

 

[V Achyuthan]

So... I just thought of something reallllly dumb...

Essentially, its just Roux and CFOP put together.

Step 1: F2B just like Roux

Step 2: Orient and Permute UB and UF edges

Step 3: M2

Step 4: perform CFOP OLL/PLL or CFCE CLL/ELL

Step 5: M2 to solve it


Example Solve:

Scramble: R2 B2 D' R2 B2 D' R2 U2 F2 R D' R2 F' D F2 D2 F' R

inspection: z2
first block: U2 R' L U L' U2 L2 U' L' U F2 U F'
second block: U B' R B R U2 R2 U' R U2 F' U' F U R U' R'
orient UB and UF edges: U2
M2: M2
OLL: U2 r U2 R' U' R U' r'
PLL: U' x R2 F R F' R U2 r' U r U2 x'
AUF: U
M2: M2

Ik i suck at roux plz dont bully me

I don't think this is a good method. It just looks like a ZZ-Roux Variant but without full EO.

 

[OreKehStrah]

So... I just thought of something reallllly dumb...

Essentially, its just Roux and CFOP put together.

Step 1: F2B just like Roux

Step 2: Orient and Permute UB and UF edges

Step 3: M2

Step 4: perform CFOP OLL/PLL or CFCE CLL/ELL

Step 5: M2 to solve it


Example Solve:

Scramble: R2 B2 D' R2 B2 D' R2 U2 F2 R D' R2 F' D F2 D2 F' R

inspection: z2
first block: U2 R' L U L' U2 L2 U' L' U F2 U F'
second block: U B' R B R U2 R2 U' R U2 F' U' F U R U' R'
orient UB and UF edges: U2
M2: M2
OLL: U2 r U2 R' U' R U' r'
PLL: U' x R2 F R F' R U2 r' U r U2 x'
AUF: U
M2: M2

Ik i suck at roux plz dont bully me

This is just ZBRoux but worse

 

[V Achyuthan]

This is just ZBRoux but worse

AGREED

 

[Filipe Teixeira]

So... I just thought of something reallllly dumb...

Essentially, its just Roux and CFOP put together.

Step 1: F2B just like Roux

Step 2: Orient and Permute UB and UF edges

Step 3: M2

Step 4: perform CFOP OLL/PLL or CFCE CLL/ELL

Step 5: M2 to solve it


Example Solve:

Scramble: R2 B2 D' R2 B2 D' R2 U2 F2 R D' R2 F' D F2 D2 F' R

inspection: z2
first block: U2 R' L U L' U2 L2 U' L' U F2 U F'
second block: U B' R B R U2 R2 U' R U2 F' U' F U R U' R'
orient UB and UF edges: U2
M2: M2
OLL: U2 r U2 R' U' R U' r'
PLL: U' x R2 F R F' R U2 r' U r U2 x'
AUF: U
M2: M2

Ik i suck at roux plz dont bully me

is better to just solve DF and DB and finish like CFOP

 

[PapaSmurf]

Even better: just do Roux finish after F2B.

 

[Filipe Teixeira]

Even better: just do Roux finish after F2B.

or you could just use petrus

 

[Cuber987]

Okay, I have a pyraminx method that might be good.

So basically, you need to make a whole bar on the bottom left. You then orient edges which can then provide opportunities for solving the rest of it 2-gen.

Scramble: R U B' R' L B' R B' l b

Only do bolded moves to follow along with the reconstruction.

First step: Bar (average move count: 3-5)

I suggest solving the tips with the bar. This step is done intuitively. Here we can do it with the moves l' b' L U R' L y, and we have formed a bar on the bottom left.

Next step: EO (average move count: 6.6)

This step is the most confusing. We want to orient the rest of the edges so that they are solvable using only R and U moves. There are three cases; two need to flip, four need to flip, and EO skip. You may know two-flip as R' L R L' U L' U' L from many other methods. For four-flip, I use R U' R' L' U L R U' R' L' U L.

In this solve, I notice both the green edges are oriented, as the green-red can be solved with R' U R and the green-yellow with R' U R' U' R'. The yellow-blue can be solved with R, but both the red-blue and yellow-red cannot be solved two-gen. Therefore, those two need to be flipped. Do the alg: R' L R L' U L' U' L.

Next step: Two-gen (average move count: 10?)

I would solve one bottom layer edge first, with R' U R. Notice that a benefit of this method is that we can now force a last layer skip by preserving one top layer edge. We can do that with R U R'. We can then solve the rest with U' R U' R' U.

A benefit of this method is that it is intuitive with only two algorithms.

Any suggestions are welcome. I believe this method can be optimized for very fast times. The only disadvantage I see so far is recognition time for EO.

 

[Filipe Teixeira]

Okay, I have a pyraminx method that might be good.

So basically, you need to make a whole bar on the bottom left. You then orient edges which can then provide opportunities for solving the rest of it 2-gen.

Scramble: R U B' R' L B' R B' l b

Only do bolded moves to follow along with the reconstruction.

First step: Bar (average move count: 3-5)

I suggest solving the tips with the bar. This step is done intuitively. Here we can do it with the moves l' b' L U R' L y, and we have formed a bar on the bottom left.

Next step: EO (average move count: 6.6)

This step is the most confusing. We want to orient the rest of the edges so that they are solvable using only R and U moves. There are three cases; two need to flip, four need to flip, and EO skip. You may know two-flip as R' L R L' U L' U' L from many other methods. For four-flip, I use R U' R' L' U L R U' R' L' U L.

In this solve, I notice both the green edges are oriented, as the green-red can be solved with R' U R and the green-yellow with R' U R' U' R'. The yellow-blue can be solved with R, but both the red-blue and yellow-red cannot be solved two-gen. Therefore, those two need to be flipped. Do the alg: R' L R L' U L' U' L.

Next step: Two-gen (average move count: 10?)

I would solve one bottom layer edge first, with R' U R. Notice that a benefit of this method is that we can now force a last layer skip by preserving one top layer edge. We can do that with R U R'. We can then solve the rest with U' R U' R' U.

A benefit of this method is that it is intuitive with only two algorithms.

Any suggestions are welcome. I believe this method can be optimized for very fast times. The only disadvantage I see so far is recognition time for EO.

petrus for pyra?

 

[Cuber987]

petrus for pyra?

Basically

 

[G DA CUBER]

Petrus for Pyra already exists.

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