The New Method / Substep / Concept Idea Thread

[Michael Womack]

TCLL is a kind of CLL that orients some of the bottom corners at the same time as orienting and permuting the top corners. What I came up with is not a kind of CLL. It is a recognition system for CLL.

Well then what you're saying now is a method that would be used for Learning CLL if i'm correct.

 

[Athefre]

I thought about learning CLL using hyperorientation which also allows for non-matching CLL

This is a very common misconception. Hyperorientations is no better for non-matching blocks than the standard CLL recognition. At least you see the reason why.

The only developed recognition that works for non-matching blocks without issue is linked in my signature.

Could you provide images for your idea?

 

[elrog]

I followed some links from your signature and got to this: https://sites.google.com/site/athefre/



From what I can tell, this is like hyperorientation, but it accounts for a R2 rotation. To do a non-matching block (or a pair of corners) that is off by a 90 degree turn makes you end up with 3 stickers that are the color that your left side pieces should be on top and the same situation for the right. This means that you should recognize the left and right stickers before recognizing the orientation pattern. This is very doable and is by far the best method for non-matching CLL recognition, though I think it would be hard to get used to.



I would like to point out that if one were to do TCLL or something of the sort, you could not use just the top layer to recognize the case using my recognition system. You would need to know how the bottom is oriented. This isn’t really a big deal though.



I was having trouble with the spoilers when I tried to add the images to my first post on this topic, so I will remake them and add them here later.

 

[Athefre]

This means that you should recognize the left and right stickers before recognizing the orientation pattern. This is very doable and is by far the best method for non-matching CLL recognition.

You almost have it. The first step is to recognize the left and right side stickers as you said. But after that, you look in a set place for matching stickers, not for the U color orientation. Just as easy, and sometimes even easier than the popular CLL recognition.

 

[elrog]

I was thinking about how RoFL actually makes shorter solutions that standard CLL even though less is solved (this was actually a long time ago before RoFL, but I dismissed the idea thinking it would make the PLLs longer). I was wandering how effectively this concept could be applied to PLL. When doing OLL, rather than solving the orientation of the last layer, you could turn it into some specific OLL case that produces shorter PLLs. You would want to pick a case that has some symmetry such as the H COLL. You would essentially double the number of PLLs (though its not a big deal because there's only 21 of them), but you would have shorter OLLs and probably PLLs aswell. You may even be able to use the same OLL algorithm for more than one OLL case, which is a plus. The OLL recognition would be the exact same, just different algorithms. The PLL recognition would be a little different, but it would be easier to recognize it from the front/right/top view because you actually have more information showing.

 

[UberStuber]

I think it'd be about the same or worse. If you orient everything to an H case, you end up with 40 algs averaging 13 (PLL is 11.8). I think the symmetry actually makes H cases a little harder to recognize edge permutation.

 

[elrog]

Did you create algorithms and find their average? If so, couldn't you do that for a lot of different OLLs to find the best one. The H case was just an example btw.

 

[Suratha]

A (New?) Petrus method variant:Rotten 2x2x2 block
Step-1:Solve the 2x2x2 block;the corner can be twisted-so fewer moves than normal 2x2x2.
Step-2:Finish 2x2x3.
Step-3:Get all the wrong edges to the top layer(0-3 moves).If six edges are wrong,get four on top.We can solve the twisted corner and edge orientation using a single alg and finish the F2L.If there are no wrong edges,finish the F2L and do CLS or CPLS(fewer cases because the corner is in place).
Step-4LL or 2-gll.
Questions:
i)Is this exists anywhere else?
ii)Will it be faster than the normal Petrus method?

 

[elrog]

As far as I know, this has not been proposed before. It seems like it could be pretty good. I think it could be faster than petrus for speedsolving, but I'm not sure about it being better for FMC. I have a couple questions through. I am a fan of making 2x2x3 blocks by starting with a 1x2x3. You could do this, but start of with a 1x2x3 with a corner misoriented.

How did you come up with the O-3 moves to move the edges to the top layer? I'm assuming you didn't count the AUF before the algorithm, but you still can have situations that require more than 3 moves.

Also, how did you come up with one algorithm? Would you not need one for having 2 misoriented adges, one for 4, and one for 6? You would also need their inverses for the corner being flipped the other way.

 

[Suratha]

Sorry.That was not one alg.I need to know the number of cases.That 0-3 moves,I just took my cube and did some moves to get all wrong edges on top and I got all 0-3 moves.So I put 0-3;Sorry if that is wrong.

 

[elrog]

Well, getting the misoriented edges to the top never takes more than 2 moves (will be 1 move most of the time) with 2 misoriented edges, but it can take up to 6 moves with 4 misoriented edges (this won't happen often). With 6 misoriented edges, no moves are necessary before the algorithm. You also won't need to AUF if you have 4 edges wrong.

You would need 6 algorithms to flip the corner and orient the edges if you plan on doing CLS or CPLS when edges are oriented.

It would be handy to be able to take advantage of using a misoriented corner, but don't do it if it is easier to just use one that is oriented. If you don't use a misoriented corner, just continue with regular Petrus.

 

[TDM]

As far as I know, this has not been proposed before.

It has. By the same person, in the same thread, less than a month ago.

I can see no advantage to using this method instead of normal Petrus. How does the movecount compare to Petrus? If we knew that, that'd be more helpful.

 

[notfeliks]

So this is just raw thought and I've barely acted on it at all, but suppose we finish our F2L and we end up with all four corners oriented and edges in any orientation and permutation. Could we, in ZBLL style, use one alg to solve the rest of the LL?

Of course even if this is plausible, it would be difficult to recognize cases and also much less useful than ZBLL as it is probably a fair bit less common to finish your F2L with CO than it is with EO.

anyway I have neither the means nor knowledge to make anything come of this on my own so I will offer it up for criticism from the community.

Not sure how many algorithms it would have but I'm guessing around 250, including mirrors.

If this has already been proposed by someone earlier then I apologize.

 

[Suratha]

I have come up with algorithms and cases that are

needed for the variant that I posted earlier.These algs

are for 2x2x3 on the left and twisted corner in the

left and can be mirrored to 2x2x3 on the right and

twisted corner in the right.
Keep the 2x2x3 in left(R & U sides are unsolved)or

right(L & U sides are unsolved) but keep the twisted

corner in the front.
Get the wrong edges to the top layer with as few

moves as possible.
If 6 edges are wrong,then put the oriented edge in

the BR or LR depending on where you've kept your

2x2x3.(Left-BR ,Right-BL)
Twisted Corner-Bottom color facing front:
2 wrong edges:
i) Opposite: F U2 R2 U' R2 U R U2 F' (Front & Back)
ii)Adjacent: R2 F U2 F' R' F R2 U2 F' (Front & Right)
4 wrong edges:
i) R2 F U2 R' F R2 F' U2 F'
6 wrong edges:
i) F U2 R' F R2 F U' F2 U' F' (BR edge is oriented)
Twisted corner-Bottom color facing left:
2 wrong edges:
i) Opposite: F2 U R F' R2 F' U' R F2
ii) Adjacent: R F U2 R2 F' U F U F'
4 wrong edges:
i) R F U2 F R2 F' R U2 F'
6 wrong edges:
i) F R U' F2 U' F' R2 F' U2 F' (BR edge is oriented)

Example solve:
1) Scramble: B2 D2 U2 L U2 B2 R' D2 U2 B2 R2 B

U2 R D' B L2 D B2 D2 L' R'
Normal Petrus:
2x2x2: R U' F2 B2 L F2 (6 moves)
2x2x3: U2 B' D L' U' L' D' (7 moves)
Orient all the edges: F U2 F2 L F (5 moves)
Remaining F2L: U2 L' U L' U' L2 U L2 (8 moves)
OLL: U' R U2 R' U' R U R' U' R U' R' (12 moves)
PLL: U R U R' U' R' F R2 U' R' U' R U R' F' U (16

moves)
Total: 54 moves.
Rotten 2x2x2: x2 y U2 R2 U' R' (4 moves)
2x2x3: y U2 R U R2 U2 B' R2 (7 moves)
Bring bad edges to the top: y' U' L' (2 moves)
Twist the corner and orient all edges: L2 F' U2 F L

F' L2 U2 F (9 moves)
Remaining F2L: U L2 U2 L2 U' L' (6 moves)
OLL: L U2 L' U' L U' L' (7 moves)
PLL: R U R' F' R U R' U' R' F R2 U' R' (14 moves)
Total : 49 moves
This variation of Petrus will be useful if you have a twisted corner with fewer moves than oriented corner.I don't know exactly the movecount difference between these two.

 

[Suratha]

So this is just raw thought and I've barely acted on it at all, but suppose we finish our F2L and we end up with all four corners oriented and edges in any orientation and permutation. Could we, in ZBLL style, use one alg to solve the rest of the LL?

Of course even if this is plausible, it would be difficult to recognize cases and also much less useful than ZBLL as it is probably a fair bit less common to finish your F2L with CO than it is with EO.

anyway I have neither the means nor knowledge to make anything come of this on my own so I will offer it up for criticism from the community.

Not sure how many algorithms it would have but I'm guessing around 250, including mirrors.

If this has already been proposed by someone earlier then I apologize.

ZBLL has 7 sets excluding PLL and this would have 7 sets(I am not sure) but the number of algorithms will be as high as ZBLL.

 

[TDM]

ZBLL has 7 sets excluding PLL and this would have 7 sets(I am not sure) but the number of algorithms will be as high as ZBLL.

You sure? I thought it'd be 72 cases for adj edge flip and opp edge flip, 22 for no edges oriented and 21 for all oriented (total = 187).

 

[Suratha]

You sure? I thought it'd be 72 cases for adj edge flip and opp edge flip, 22 for no edges oriented and 21 for all oriented (total = 187).

I am sure that the case count will be as high as ZBLL.Because you would have 4 sets in adj edge flip case because for the same corner case you will have the front and right edges oriented,right and back edges oriented,back and left edges oriented and front and left edges oriented and similarly for opposite edges case you will have Front and back edges oriented and left and right edges oriented.

 

[TDM]

I am sure that the case count will be as high as ZBLL.Because you would have 4 sets in adj edge flip case because for the same corner case you will have the front and right edges oriented,right and back edges oriented,back and left edges oriented and front and left edges oriented and similarly for opposite edges case you will have Front and back edges oriented and left and right edges oriented.

Nope. The 72 for each case covers every possible AUF for each PLL; by rotating how you view the cube you're only repeating cases. For no edges oriented, because it can be rotated and the orientation of the LL stickers will always look the same, you only have one case for each PLL, plus one for solved PLL which is why there are 22 cases.

 

[Suratha]

Nope. The 72 for each case covers every possible AUF for each PLL; by rotating how you view the cube you're only repeating cases. For no edges oriented, because it can be rotated and the orientation of the LL stickers will always look the same, you only have one case for each PLL, plus one for solved PLL which is why there are 22 cases.

I already posted this idea earlier and some have disagreed with me and I don't understand your explanation. http://www.speedsolving.com/forum/s...-New-Method-Substep-Concept-Idea-Thread/page9

 

[notfeliks]

Sample case and alg, supplied by cube explorer:

The case is a Ga perm, headlights on B and the UF and UR edges flipped. This case is selected randomly.

Headlights on B:

L F' L F2 R' F R F2 L2

Very nice alg, move optimal, 3-gen and definitely fingertrick-able. Also if we do an x rotation and perform:

L D' L D2 R' D R D2 L2

we have an alg strangely reminiscent of A-perms. I'm quite liking this.

Second sample, also randomly selected:

Rb perm with swapped corners on R and UF/UR edges flipped.

F R' F' U R U2 R U2 R' U R' U2 R U

Apart from the sledgehammer sort of trigger at the start, we have a 2-gen algorithm, and very similar to a typically alg used for the normal Rb perm.

I'm starting to really like this concept.