The New Method / Substep / Concept Idea Thread

[PapaSmurf]

Hello
in VH one obtains an oriented cross (as in ZZ)
I have a cross oriented AND PLACE following 2 shemas which limits the OLL to 5 instead of 57 and the PLL to 8 instead of 21 and especially in much much simpler

It's still not that great and again, has been thought of before. It's much easier and more beneficial to learn more algorithms and be fast than find a low alg method that pretends to be fast. That doesn't mean don't find new methods, but methods that aim to limit algs are generally slower (exception: Roux).

 

[pascal france]

It's still not that great and again, has been thought of before. It's much easier and more beneficial to learn more algorithms and be fast than find a low alg method that pretends to be fast. That doesn't mean don't find new methods, but methods that aim to limit algs are generally slower (exception: Roux).

My method is not to compete with CFOP Fridrich, ZZ or Roux, I do not have this claim, it is not adapted to the competition.
It's just a simple way to solve the Rubik in 30 seconds without difficulty (less than 20 seconds when you're good), in about fifty movements, and without learning many algorithms
I'm too old to learn a lot of formulas and I do not have the agility of fingers or reflexes anymore :confused: :?

 

[Angry_Mob]

I'm not sure if this is useful, but I came up with a new way of doing SB + CMLL in Roux.

Imagine you have a cube with solved blocks. When you do R2 U2 R2 U2 R2, the second block will have what is basically an equator flip on Square-1. In this method, you will solve SB into that state, then do an algorithm so solve CMLL and flip the equator.

Solving SB is super easy, you simply make each pair with only one color matching in each pair, and insert the pair based on where the corner goes.

There are a few ways to solve CMLL + equator flip. The easiest is to solve CMLL, but cancel into R2 U2 R2 U2 R2 at the end. This works really well, but I've almost finished generating unique algorithms for this step. Most of them are garbage, but some of them are pretty good. Examples: R' U2 R2 U' R2 U2 R2 U R2 U2 R, R U' R' U R U' R D R D' R D R2 D' and R U2 R' U2 F2 D R D' R' F2 R2 U' R'. Most of the time canceling into R2 U2 R2 U2 R2 works better though.

Obviously this wouldn't be used every solve, as it would add on average 4 moves to each solve, but I think it's useful if you already have a pseudo-pair solved, or a free pair.

Spoiler: Example solve

R2 B2 F2 D2 L2 U' R2 F D2 B L' B L2 R' D R' D2 B (FB solved)

U r' U' r U R U M U r U' R' //Psuedo-SB
R U2 R' U' R U' R U2 R2 U2 R2 //CMLL +equator flip

 

[Sue Doenim]

I'm not sure if this is useful, but I came up with a new way of doing SB + CMLL in Roux.

Imagine you have a cube with solved blocks. When you do R2 U2 R2 U2 R2, the second block will have what is basically an equator flip on Square-1. In this method, you will solve SB into that state, then do an algorithm so solve CMLL and flip the equator.

Solving SB is super easy, you simply make each pair with only one color matching in each pair, and insert the pair based on where the corner goes.

There are a few ways to solve CMLL + equator flip. The easiest is to solve CMLL, but cancel into R2 U2 R2 U2 R2 at the end. This works really well, but I've almost finished generating unique algorithms for this step. Most of them are garbage, but some of them are pretty good. Examples: R' U2 R2 U' R2 U2 R2 U R2 U2 R, R U' R' U R U' R D R D' R D R2 D' and R U2 R' U2 F2 D R D' R' F2 R2 U' R'. Most of the time canceling into R2 U2 R2 U2 R2 works better though.

Obviously this wouldn't be used every solve, as it would add on average 4 moves to each solve, but I think it's useful if you already have a pseudo-pair solved, or a free pair.

Spoiler: Example solve

R2 B2 F2 D2 L2 U' R2 F D2 B L' B L2 R' D R' D2 B (FB solved)

U r' U' r U R U M U r U' R' //Psuedo-SB
R U2 R' U' R U' R U2 R2 U2 R2 //CMLL +equator flip

That's a pretty interesting idea. As it is now, I don't think it's worth it, but if you take the idea bit further, you could have it so that you solve SB, but each edge can be in any of the three positions. Then, when it comes to CMLL, You do one of six algs to solve corners and SB at the same time. I might check that out some more.

 

[Angry_Mob]

That's a pretty interesting idea. As it is now, I don't think it's worth it, but if you take the idea bit further, you could have it so that you solve SB, but each edge can be in any of the three positions. Then, when it comes to CMLL, You do one of six algs to solve corners and SB at the same time. I might check that out some more.

I like the idea of solving SB in one of six ways, but it would make CMLL + SB have a ton of algorithms (43 x 5 = 216 (excluding CMLL)). Maybe since the 2G algs are by far the best one I've genned so far, doing CPFB would be useful? that would bring it down to 40 algs, which is super reasonable imo.

 

[Sue Doenim]

I like the idea of solving SB in one of six ways, but it would make CMLL + SB have a ton of algorithms (43 x 5 = 216 (excluding CMLL)). Maybe since the 2G algs are by far the best one I've genned so far, doing CPFB would be useful? that would bring it down to 40 algs, which is super reasonable imo.

The problem there is that CPFB is really difficult. I have my doubts as to whether or not anyone could get to the point where they could consistently plan it in inspection. At any rate, I don't really think ~250 algs is really that far-fetched. This would definitely be a pretty advanced technique, and that kind of alg count isn't far off from what an advanced 2x2 or CFOP solver would use. And that's not even taking into account the people that learn ZBLL. I think what really would make or break the idea is whether or not the algs are comparable to normal CMLL algs. I kinda get the feeling that they would be a bit worse, especially considering what you said about the opposite-swap algs you've genned so far.

 

[Angry_Mob]

The problem there is that CPFB is really difficult. I have my doubts as to whether or not anyone could get to the point where they could consistently plan it in inspection. At any rate, I don't really think ~250 algs is really that far-fetched. This would definitely be a pretty advanced technique, and that kind of alg count isn't far off from what an advanced 2x2 or CFOP solver would use. And that's not even taking into account the people that learn ZBLL. I think what really would make or break the idea is whether or not the algs are comparable to normal CMLL algs. I kinda get the feeling that they would be a bit worse, especially considering what you said about the opposite-swap algs you've genned so far.

Yeah, I was worried that most of the algs would be garbage. So far only about 7 of my R2 U2 R2 U2 R2 algs are any good. I'll try making algs for the whole set, but chances are they'll be much slower then CMLL (then again, I probably don't have the best judgement on what's good and I'm not spending a ton of time on each alg)

 

[PapaSmurf]

CPFB is definitely possible, it just brings 0 advantage to anything except maybe OH. And I also think that the concept will bring about some bad algs from overturning etc, and normal SB will be better. But prove me wrong.

 

[Sue Doenim]

CPFB is definitely possible, it just brings 0 advantage to anything except maybe OH. And I also think that the concept will bring about some bad algs from overturning etc, and normal SB will be better. But prove me wrong.

Yeah, I think that the fact that SB is solved makes it more likely that the CMLL algs will be better in a similar way to how having cross solved helps F2L to be done with a more neutral wrist position.

 

[efattah]

This is sort of what I have been saying for a long time, the Roux variant where the FR edge is solved in a disoriented state, then you use Waterman Set 3 to solve L6E (Set 3 solves UL+UR while flipping FR). The Set 3 subset is only 16 algorithms; unless you allow the BR edge to be solved disoriented where you end up at 32 algorithms, and if you allow DR edge to be solved disoriented it is 48 algorithms (these algs are already generated for a different purpose).

 

[White KB]

This is one of the most (if not the most) commonly proposed methods. It's just a worse version of Roux and CFOP. It's not worth doing at all.

I came up with that too, but someone else had come up with it as well.

 

[White KB]

2x2 method:
pair up 2 corners or find a pair (only the bottom color has to match)
use 1 (extremely efficient) alg to solve the rest of the first layer
then CLL and you're done
it would possibly aid people who don't want to learn EG-1 or EG-2, and the algs would be very short (2-4 moves, except for some special cases)
I'm calling it the BL (Bruce Layer) subset (my last name is Bruce, after all).
Let me know what you think, or if someone else has come up with it, or ways it can improve.*

*besides the name. **
**unless someone else has already come up with it. you can change it then.

 

[Sue Doenim]

2x2 method:
pair up 2 corners or find a pair (only the bottom color has to match)
use 1 (extremely efficient) alg to solve the rest of the first layer
then CLL and you're done
it would possibly aid people who don't want to learn EG-1 or EG-2, and the algs would be very short (2-4 moves, except for some special cases)
I'm calling it the BL (Bruce Layer) subset (my last name is Bruce, after all).
Let me know what you think, or if someone else has come up with it, or ways it can improve.*

*besides the name. **
**unless someone else has already come up with it. you can change it then.

I think that the case count would be too high. With your bar, there are 3 different CP cases (solved, opposite, and adjacent). Then there are 6*5 permutation cases and 3*3 orientation cases for the other two corners, leaving you with an upper bound of 810 cases. There would be some repeated cases, but not many, I wouldn't think.

 

[White KB]

I think that the case count would be too high. With your bar, there are 3 different CP cases (solved, opposite, and adjacent). Then there are 6*5 permutation cases and 3*3 orientation cases for the other two corners, leaving you with an upper bound of 810 cases. There would be some repeated cases, but not many, I wouldn't think.

Ok, that sounds cool. I wasn't sure how to calculate it, so you saved me a lot of work. Thanks!
And looking at the case count, it seems huge, but most of the cases would be almost intuitive. You would only have to put in real brain power for a small percentage of them, and realize that a lot of them would be intuitive. It could work, if one cared to put in the time to practice.

 

[PapaSmurf]

It also seems highly useless. EG would be so much better, and the algs are easy anyway. If you were dedicated, you could learn them in a week per set. (EG-1 in a week, EG-2 in a week etc.) So just do EG if you want to be fast. Or another method (2GR possibly).

 

[efattah]

White KB's proposal is not useless. In fact it echoes something that I have been wanting to do for some time. Making efficient faces (even for EG) is not always intuitive; I wish someone would generate a table of all the common cases and the fastest solution that is finger friendly. Rough guess is there are around 50 cases to make the 1st face. I have personally generated some of the most irritating ones to find the fastest solution; top 2x2 solvers would already know every single one of them, no algo generator would be needed, someone would just need to draw diagrams of the 50 cases and a top 2x2 solver could fill in the blanks. Kind of like that 2x2 CLL efficient 1st layer thread except for EG.

Even better would be the same table with a 1-look column that shows how the LL corners are affected and reorganized by the face solution.

 

[Zeke Mackay]

This is a message to everyone interested in making a method or who is about to post a method: Stop making RouxFOP

 

[SM cubing]

This is a message to everyone interested in making a method or who is about to post a method: Stop making RouxFOP

THIS IS VERY IMPORTANT!!!!!!

 

[WoowyBaby]

I'd like to propose a new method challenge:

Create the worst method you can think of! I'll be really interested to see what you guys make up!

But, I do need to set some basic rules so the methods aren't infinite or impossible:
- If you end up breaking progress of previous steps you must restore it in the same step that you break it.
- Maximum of 12 steps are allowed(sorry about your crazy ideas w/ 100 steps).

"Score" is judged by the average (stepwise optimal) movecount.
Good luck!!

 

[White KB]

I'd like to propose a new method challenge:

Create the worst method you can think of! I'll be really interested to see what you guys make up!

But, I do need to set some basic rules so the methods aren't infinite or impossible:
- If you end up breaking progress of previous steps you must restore it in the same step that you break it.
- No more than 12 steps are allowed(sorry about your crazy ideas w/ 100 steps).

"Score" is judged by the average (stepwise optimal) movecount.
Good luck!!

A very inefficient method for big cubes (not sure if you could accurately determine average move count, but it would be very inefficient):
LBL (based on a tutorial I actually saw once)
1: Solve the yellow center (pfft, who needs to be CN, even if you have an easy white face) one piece at a time, just using the slice turn Niklas every time.
2: Pair one yellow edge at a time, one piece at a time, where the unsolved edge is on the left, the piece you need is on the right, and x represents the number of layers you have to turn to match it: xU L' U L xU'.
3: Once you have done that, place the edges in their respective places (in odd-layered cubes, use centers to determine that, in even-numbered cubes, remember the color scheme).
4: form 2nd layer for centers (somehow)
5: insert proper edges into the second layer, breaking an edge or two if necessary (e.g. blue-orange being paired with blue-red)
6: form 3rd layer centers (somehow) but it's a little bit easier.
7: pair necessary pieces to their respective edges.
8: After this it depends on the N in NxNxN, but for 4x4, skip to step 9, and for 5x5 and up, pretty much keep repeating steps 6 and 7, except with the proper layer, until everything except the last layer edges and corners are done, and replace corners when necessary.
9: pair last layer edges by displacing the two front-middle edges, then pairing ALL the white edges, then replacing those F2L pairs.
10: for 5x5, 7x7, 9x9, 11x11, etc. just solve the last layer with beginner's method (orient edges, permute edges, permute corners, orient corners), then once you're done, skip to 'step "12"'
11: for even-numbered 'N's, pretty much do step 10, but for PLL parity do r2 U2 r2 Uw2 r2 u2, where a lowercase letter represents a slice turn, and for OLL parity (this is the least efficient algorithm I know) do r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2. then just solve with beginner's method for everything else.
"12" (not really a step). You're done! Yay...