OSPA(New 2x2 Method)

[Henrik]

It is hard to say how many cases there are, but I think those 33 are more than the half at least.
When I was playing around with this method yesterday, I got one of those L-cases in about 60-70% of the solves.

It doesn't mean that it is more than the half algs, but it might be.

If I understand your questions/ thoughts correct the the answer would be 36.

Each layer can end in 6 cases, 6*6 = 36

This is at least for the 3/4 layer cases. (where 3 out of 4 oriented is one color, and the last is opposite)

As promised: my work from April/May 2008:
View attachment Guidmond step 2.pdf

A few notes: as this was made in 2008, my aim of time was completely off it should be less than 2 sec.
Also I have not looked into mirrors/opposites/inverses, I have just listed the cases and how to recognize them.

 

[Jakube]

Also wanted to post some algs for the orientation step. This site will help you to see which situation it is. Cases in "[]" aren't changed beacuse they are pretty nice or beacuse I couldn't find any nice algs.

All algs aren't nice, but some of them really are!

What about the Guimond algs: Link
In 90% you can use them.
These Orientation algs are pretty short (3-6 moves), so one-looking should not be that hard.

 

[emolover]

Aww man! I thought I was on to something. Doesn't mean we can't find the algs for the other cases.

 

[Henrik]

Aww man! I thought I was on to something. Doesn't mean we can't find the algs for the other cases.

What do you mean? You thought you where on to something?
Why are you not on to something?

This is another step in the Orient opposite sides - permute area, and solving form.

The other is CLL/EG one side do all.

I think its good to explore new ideas in the 2x2 area, even though someone might have made the algs (but never published), does not mean it should not be fully developed.

I do think that 3/4 of a side is the easiest to recognize, when it comes to this method, I have not found a good way to do it with other cases, like diagonal-diagonal nor adj-diagonal. Someone might have a good idea for that.

 

[Ickenicke]

If I understand your questions/ thoughts correct the the answer would be 36.

Each layer can end in 6 cases, 6*6 = 36

Yeah, that was my thought from the beginning, but I can't find the last 3.

So if someone can find them it would be nice


edit: oh sorry. I saw the cases in the link.

 

[Ickenicke]

I tried to find out some cases when there are 2 of each colour at each side. I used yellow and white orientation.

I did like that: I seperated up the cases by watching if the 2 white should be next to each other when the cube was solved, or if they were diagonal. And also how the yellow was.

So there are 3 different groups:
1. Both white pieces and both yellow pieces at both layer will be next to each other when the cube is solved.
2. White or yellow pieces at both layer will be next to each other when the cube is solved. The other colours pieces is in both layer diagonal if the cube was solved.
3. Both white pieces and both yellow pieces at both layer will be diagonal to each other when the cube is solved.

So here is the beginning of my work:

F2R2U2 Double T-swap
R2U'RU'RF2R'UR' Solved 2x1 (1)
R'UR'U'RF2RFRU Solved 2x1 (2)
UF'UF'R2FU'F' Diagonal pattern

4 cases when both the D-layer white & yellow pieces should be next to each other at a solved cube. Of course will also the pieces at top be next to each other when solved. There would be 2 solved 2x1 block at the D-layer if you would do 2 T-perms, thatis why I named the cases as Double T-swap

U'R'FU'R2UF'R'U2 Double T-swap
R2U2F2U'R2UF2 Solved 2x1 (R)
F2UFU'FR2F'RFU2 Solved 2x1 (2)
R2U'R2U2F2U Diag pattern




4 cases when both the D-layer white & yellow pieces should be next to each other at a solved cube. But this time is the bottom pieces diagonal to each other from the beginning.

R2UR'FR'F2RU'R Double T-swap
F2UF2U2R2UR2U2 Solved 2x1 (R)
F2R'U'FU2F'UR'U Solved 2x1 (2)
F2U2R2U'R2U'R2 Diag pattern


4 cases when both the D-layer white & yellow pieces should be next to each other at a solved cube. Here are there one 2x1 block solved at the bottom from the beginning.
R2U2F2U'R2U2F2U'F2 Double T-swap
F2RUR'F2RF'RU' Solved 2x1 (1)
R2U2 Solved 2x1 (2)
F2U'R'UF'R2FR'F'U2 Diag pattern

4 cases when both the D-layer white & yellow pieces should be next to each other at a solved cube. Here are there two 2x1 block solved at the bottom from the beginning.

This is all group 1 cases, I started at group 2 but did not do that much. What do you think about this way to split up the rest of the cases?
Do you understand how I were doing? Have I being do something wrong?

Tips and criticism is welcome!

If someone haven't understand that, so would those 3 groups of cases complete the total pemutation step.

 

[emolover]

I like this way of splitting the cases up and can't wait for the cases to be fully finished. There won't be 36 cases for each but how many do you think there will be?

 

[Ickenicke]

I like this way of splitting the cases up and can't wait for the cases to be fully finished. There won't be 36 cases for each but how many do you think there will be?

I think it can be somewhere between 40-60 cases for all those 3 groups.

It also depends at the second group, because in many of those cases ( which not are having the diagonal pattern) can you just do R2 to get another case.

 

[Henrik]

I think it can be somewhere between 40-60 cases for all those 3 groups.

It also depends at the second group, because in many of those cases ( which not are having the diagonal pattern) can you just do R2 to get another case.

Just a quick guess on the cases with two-bars of 2 in the bottom and diagonal pattern on top would be 18 cases.
The diagonal pattern gives the opportunity to do a U2 and therefore cutting the cases by half. (from 36 to 18)
(I am not 100% sure, but ill find the last 6 cases I'm missing in my old sheet (Yes I also started that))

Here are the cases for what I call Guimond2-2
View attachment Guimond step 2-2.pdf
Btw: avg move count: 8 moves

I don't know about 2bars bottom 2bars top cases.

EDIT2:
Would 2bars - 2bars not be just 7 cases?
You can do a setup like U2R2 or U2L2 and get one of the known PBLs

If you already know how the "seperation" is going to end up, you might as well add U2R2 or what ever makes two full sides, to get a known PBL.
For 2bar-diag pattern you could learn the 18 cases. And if you are crazy you could also learn the 36 cases for 3/4 patterns.

Right now we are looking at a method with 61 cases for PBL. Maybe it should be Permute and Separate Both Layers (PSBL)
Also I just did some quick calculations, this approach looks like it can be done with an avg of 12-13 moves.

 

[Ickenicke]

@Henrik

Yeah, maybe it is a good idea to just do something like U2R2.

You haven't looked into those cases when there are the diagonal pattern at both sides?

 

[Henrik]

@Henrik

You haven't looked into those cases when there are the diagonal pattern at both sides?

Oh I had luckily forgotten about them.

Hmm I wonder how to recognize them. I will have to think about that.

EDIT:

Okay I have looked at the diag-diag patterns, thoughts about it and come up with 7 distinct cases with different algs. The rest of the cases can be oriented into one of these 7 cases.
I started out with 18 cases (after some thinking, and reducing the logical 36 cases by half), Then sitting with 7 2x2s and messing around with orientations and what not. I had to color-code the algs to keep track

Here is a messy draft: Algs in bold are the "original" or the one that others can be oriented to. Let me know if I missed anything. (I think I have)
View attachment Guimond step 2-3.pdf

 

[Henrik]

A cleaner setup of the 7 cases of diag-diag:
View attachment OSPA 7 cases diag-diag.pdf

 

[Zarxrax]

Ok, so what advantage does this have over 2-step guimond, if you combine the separation and permutation? Guimond has very short orientation algs, and there are only a few of them. So I'm guessing guimond would be both easier to learn and have shorter solutions (and easier to 1-look). Am I wrong?

 

[Ickenicke]

2-step Guimond? Isn't this method having shorter solutions?

Average would be something like 12-14 moves I think.

 

[Zarxrax]

Well, guimond orientation step averages about 4 moves, and only has 16 algs.
It looks like this method would probably average about 5 moves for orientation... plus there are like 50~ orientation algs.

 

[emolover]

For the first "side" of Guimond it's between 3-5 moves, for the orientation it's about 7 moves, for the separation it's 3-5 moves and the PBL 7-11. So that is between 20 and 28 for a full and normal solve. I too have been getting between 12 and 14 move solves but I quite often get 10 or lower.

 

[Jakube]

For the first "side" of Guimond it's between 3-5 moves.

That's not true. 90% of the time the first "side" (3 stickers of opposite colors on one side) is already made. 10% of the time you'll get a 1 or 2 move solution.

 

[Ickenicke]

Well, guimond orientation step averages about 4 moves, and only has 16 algs.
It looks like this method would probably average about 5 moves for orientation... plus there are like 50~ orientation algs.

You can use the Guimond orientation with this method too, but if you don't have 3/4 of a side solved you can still find solutions in about 4-5 moves. I really don't think you have to learn all, but you can learn some of the shortest as an extra thing which can help you.

 

[Henrik]

For the first "side" of Guimond it's between 3-5 moves, for the orientation it's about 7 moves, for the separation it's 3-5 moves and the PBL 7-11. So that is between 20 and 28 for a full and normal solve. I too have been getting between 12 and 14 move solves but I quite often get 10 or lower.

Emolover: I think you have messed up/ confused your understanding of Guimond and Ortega.

Ortega: do one side, orient opposite, PBL
Guimond: orient opposite sides, separate colors, PBL

SS combines the two: 3/4 of a side and then doing the separation and orientation in one.

 

[oll+phase+sync]

I just replaced the orient step in the first two exsample solves with some guimond moves.

Scramble: R2 F2 R F2 U' F U2 F2 R'
Presolve: x2 y'
Orient: U' y' R' U' R U R' U' R (8)
Guimond: U' y' R2 U' R (4)

__________________________________
Scramble: R' F2 U2 F' R2 U' R U2
Presolve: x2 y
Orient: R U' R' U R U' R' (7)
Guimond: D' R2 F2 U' R' (5)

As you see the Guimond thing needs less moves.
Few moves and double moves are a good thing, because the fewer moves you need, the easier it is to predict the final case.

al least it should be much easier to predict the final case than CLL. (not only because of the number of moves, but also because all corners are already orinted).


Regarding Guimond - you could call it 4 step:
0. Orient 3 opposit-colored stickers
1. Orient 8 opposit-colored stickers
2. Separate
3. PBL

But in fact it is more like a 2 Step methode because
0. this is less than on move on average (if you have to do a move, you have an option to force an fast orientation case most of the time. )
1. has few moves wich makes prediction of Separation case quite easy.


If you need algs ACUBE is decent (set all edges to @?)