The New Method / Substep / Concept Idea Thread

[edward_9x]

No cuz you solve Corner permutation and edge orientation at same time in second step

Ah, like in RUP. Didn't pay attention to that. I forgot you solved EOCP before solving CO lol

Cool method

 

[Twisted101]

I think you need both corner orientation and also permutation to get EPLL, not just orientation, so that would require more algs (maybe it was a typo?)

But this sounds very interesting

No you do CP in second step

 

[Silky]

I have a new idea for a method:

1. 2x2x3 on left
2. CP+EO (from RUP)
3. F2L-1pair
4. perform an algorithm out of 486 algorithms to make and solve the last pair while doing corner orientation
5. EPLL

I recommend looking into CEOR methods first.

 

[Neuro]

I have a way to force either twisty TTLL or normal PLL with the same algorithm count and quality as forcing normal TTLL. The only drawback is recognition, but the quality of Twisty TTLL makes up for it.

Oh this is cool, thank you! Let me see if this can work with TOLS or not.

 

[THE_V01DL0RD]

I have a feeling this method exists but I think it's cool never the less.
1) Make a 2x2x3 Block

2) Intuititively solve 4 edges like this

3) Solve the last 4 edges to make a cross

4) Solve the last 6 corners to solve the cube

I have only basic knowledge of cubing terminology and how stuff works. I don't think this will be a speedsolving method or an FMC method. It's just a method to solve the cube. I might have worded things weirdly so I hope my visual aids help. Finally, if there are any problems or algorithms you came up with, just tell me.

 

[zzcuberman]

I have a feeling this method exists but I think it's cool never the less.
1) Make a 2x2x3 Block
View attachment 25135
2) Intuititively solve 4 edges like this
View attachment 25134
3) Solve the last 4 edges to make a cross
View attachment 25136
4) Solve the last 6 corners to solve the cube
View attachment 25137
I have only basic knowledge of cubing terminology and how stuff works. I don't think this will be a speedsolving method or an FMC method. It's just a method to solve the cube. I might have worded things weirdly so I hope my visual aids help. Finally, if there are any problems or algorithms you came up with, just tell me.

Just solving 5 corners is 614 algs

 

[THE_V01DL0RD]

Just solving 5 corners is 614 algs

Then you better get to studying

 

[zzcuberman]

Then you better get to studying

I've no intention of learning that

 

[THE_V01DL0RD]

I've no intention of learning that

How did you know that you need 614 Algorithms for 5 corners?

 

[Thom S.]

How did you know that you need 614 Algorithms for 5 corners?

Math
Also,
L5C first result

 

[zzcuberman]

How did you know that you need 614 Algorithms for 5 corners?

because I know the algorithm set thats L5C.

 

[Twisted101]

this is an extremely algorithmic direct solving method for 4x4 that I made


Solve D center

Make 3/4 of cross and a half center attached to each cross dedge

Solve the last cross dedge and half center while solving the U center

Solve F2L-1 pair

Intuitively solve 3rd layer dedges other than the one above the open slot

Perform L4HC to solve the last 4 half-centers in one of 64 algorithms

Perform L5C to solve the last 5 corners out of 614 algs

Perform OLTW to orient the last ten wings in one of 484 algorithms


Perform PLTW by identifying 2 L5EP+Parity cases and then executing PLW(permutation of left wings) then executing PRW(permutation of right wings). This is a 1-look 2-alg system like M-Cell (64 algorithms total for PLTW)

 

[edward_9x]

Just solving 5 corners is 614 algs

You could do 3-style commutators intuitively for a Heise-style finish. Or maybe splitting corners as 2 separate steps into CO+CP would yield less algs (?) but I'm too lazy to work it out. There are many more methods of approaching this but these are just off the top of my head.

 

[Silky]

Just solving 5 corners is 614 algs

Its actually larger than this, I believe, since you are solving with edges already solved.

 

[edward_9x]

Its actually larger than this, I believe, since you are solving with edges already solved.

It think that would be the same amount of algs, just with more moves to preserve the edges (like CMLL and COLL). If anything, you can't have only 2 corners swapped with all edges solved so it could even be less algs.

 

[Silky]

It think that would be the same amount of algs, just with more moves to preserve the edges (like CMLL and COLL). If anything, you can't have only 2 corners swapped with all edges solved so it could even be less algs.

Well consider that L4C>CLL

 

[edward_9x]

Well consider that L4C>CLL

I guess I agree, but for the quadrupled AUF cases I think an EPLL or something similar after the alg would suffice. So less algs but higher movecount

 

[zzcuberman]

You could do 3-style commutators intuitively for a Heise-style finish. Or maybe splitting corners as 2 separate steps into CO+CP would yield less algs (?) but I'm too lazy to work it out. There are many more methods of approaching this but these are just off the top of my head.

I know L3C I redid and am currently using LLE+1C and finishing with L3C

 

[Silky]

I know L3C I redid and am currently using LLE+1C and finishing with L3C

aka Synder. Have you posted the algs?

 

[zzcuberman]

aka Synder. Have you posted the algs?

Yeah snyder and no I haven't. It's 192 if you include dots if your wondering for the set