OSPA(New 2x2 Method)

[Henrik]

@ oll+phase+sync: you might want to call "orient" for SS or SOAP instead.


Example 1:
Scramble: R2 F2 R F2 U' F U2 F2 R'
pre solve: z2
Orientation: U' R2 U' R' y' (4/4)
Separation /permutation : R2'y’R2U'R2'UR2 (6/10)
AUF: U' (1/11)

Example 2:
Scramble: R' F2 U2 F' R2 U' R U2
Presolve: x2 y
D' R2 F2 U' R' (5/5)
y' U'R2U'R2F2U'F2UR2 (9/14)
U' (1/15)

Example 2 - 2:
Scramble: R' F2 U2 F' R2 U' R U2
Presolve: x2 y
Orient: R U' R' U R U' R' (7/7)
U2 (1/8)
FU'R'F2R'F2R'U'R (9/16)

Cool

 

[oll+phase+sync]

@ oll+phase+sync: you might want to call "orient" for SS or SOAP instead.

Sorry I don't understand.

Another question: SS looks like a very solid method to produce 1look solutions to me. For SOAP ... I seen no reason why to use it ... am I missing something?

 

[emolover]

You have to solve 3 corners which can take quite a few moves. With OSPA you know what your alg does to all the pieces. If you know exactly where they all go immediately, it makes it easier to 1-look.

 

[Zarxrax]

Sorry I don't understand.

Another question: SS looks like a very solid method to produce 1look solutions to me. For SOAP ... I seen no reason why to use it ... am I missing something?

SOAP only has half as many algs to learn as SS.
Solutions are on overage 1-2 moves longer. That's the trade-off.
They both seem to complement each other well, and could be combined to form an even better method (SSOAP?)

 

[Zarxrax]

Today I wrote a solver that can generate algs for the orientation step of OSPA method, so I will try to post some algs soon.

 

[Zarxrax]

Here we go: http://www.amvhell.com/stuff/cubes/OSPA_ALGS.html

I didn't really test these algs or try to optimize them. So there is probably lots of room for improvement.
On average, you might save about 1 move whenever you get a bad guimond case, otherwise the guimond cases are usually better.

Maximum move count for OSPA first step is 6 moves, but I think this can ALWAYS be reduced to 5 by choosing a better starting case (you will often have more than 1 case to start from) (and if someone can find a case that needs 6 moves, please show me). So basically, look ahead through usually 5 moves or less during the inspection time to see the final step.
The separation+permutation step doesn't seem to have much worse recognition than CLL. This seems like a valid contender to EG method to me (though with just the guimond algs is ok).

 

[oll+phase+sync]

Most of th algs in Your list are not RU. In fact if most of them where RU, there might be an easy way to narrow down the options for Step 2 (seperation and permutation).

 

[Zarxrax]

Yea, Its possible to make them all RU, but I don't think its worth the trouble to learn them all. Probably better to stick with just the 16 alg guimond subset (which can also be RU).
I think the only way to reduce the number of cases for step 2 is if you had a bar of 1 color on the left side. Then it becomes something like SORTEGA method.

 

[cuBerBruce]

Maximum move count for OSPA first step is 6 moves, but I think this can ALWAYS be reduced to 5 by choosing a better starting case (you will often have more than 1 case to start from) (and if someone can find a case that needs 6 moves, please show me).

As I noted [post=152726]here[/post], there are 1152 configurations that do take 6 moves. For example, F' U F2 R2 U R' U' R' requires at least 6 moves to orient with respect to U/D stickers, L/R stickers, or F/B stickers. I note that this particular example, however, only requires a 5-move alg preceded by an alignment move.

I note that your page only generally shows one AUF case instead of all 4 (when applicable). So you seem to be assuming, in general, an alignment move plus an alg. Interestingly, in some cases, your alg starts with a U layer move, so your standard case isn't always even an optimal AUF case.

I also note that apparently you choose a face for the D layer such that it has at least two stickers of some pair of opposite colors. However, you seem to assume this choice will have at least a "bar" of these opposite color stickers. There are 5184 configurations that don't have a face with at least a bar, but rather diagonally opposite stickers only. I see nothing about this case. (Example: U R2 F R2 F2 R F2 U')

Finally I note there is some redundancy. For example, the first case in the "Asym[m]etrical Subset" category is really the same as a Guimond case, only with a different cube orientation. As you are allowing for alignment moves, you need to be very careful if you try to remove such redundancies, though.

 

[Zarxrax]

As I noted [post=152726]here[/post], there are 1152 configurations that do take 6 moves. For example, F' U F2 R2 U R' U' R' requires at least 6 moves to orient with respect to U/D stickers, L/R stickers, or F/B stickers. I note that this particular example, however, only requires a 5-move alg preceded by an alignment move.

Ah, well noted.

I note that your page only generally shows one AUF case instead of all 4 (when applicable). So you seem to be assuming, in general, an alignment move plus an alg. Interestingly, in some cases, your alg starts with a U layer move, so your standard case isn't always even an optimal AUF case.

I haven't really looked at the algs, I just pasted them in. Some of them start with a U, because otherwise I would have to draw a different diagram for the case.

I also note that apparently you choose a face for the D layer such that it has at least two stickers of some pair of opposite colors. However, you seem to assume this choice will have at least a "bar" of these opposite color stickers. There are 5184 configurations that don't have a face with at least a bar, but rather diagonally opposite stickers only. I see nothing about this case. (Example: U R2 F R2 F2 R F2 U')

This method as defined in the first post would have at least a bar. That's because it was said that SOAP algs could be used, and SOAP algs would require a bar. Basically all I did above was generate some SOAP algs optimized for opposite colors.

I was mainly just curious if it might end up with a lot of really short cases like guimond has. These do tend to be a bit longer though.

 

[cuBerBruce]

I also note that apparently you choose a face for the D layer such that it has at least two stickers of some pair of opposite colors. However, you seem to assume this choice will have at least a "bar" of these opposite color stickers. There are 5184 configurations that don't have a face with at least a bar, but rather diagonally opposite stickers only. I see nothing about this case. (Example: U R2 F R2 F2 R F2 U')

This method as defined in the first post would have at least a bar. That's because it was said that SOAP algs could be used, and SOAP algs would require a bar. Basically all I did above was generate some SOAP algs optimized for opposite colors.

Well, anyway, I confirm that you never need more than one move to make the bar (allowing opposite colors, of course). (SOAP's initial step includes guaranteeing you have a bar.)

 

[LongRong]

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' 对角线图案

4 种情况下，D 层的白色和黄色部分应该在已解决的立方体中彼此相邻。当然，当解决时，顶部的部分也会彼此相邻。如果你做 2 个 T-perms，那么在 D 层会有 2 个已解决的 2x1 块，这就是为什么我将这些情况命名为双 T-swap

U'R'FU'R2UF'R'U2 双 T 掉期
R2U2F2U'R2UF2 已解决 2x1 (R)
F2UFU'FR2F'RFU2 已解决 2x1 (2)
R2U'R2U2F2U 诊断模式




4 种情况下，D 层的白色和黄色部分应该在已解决的立方体中彼此相邻。但这一次是底部棋子从一开始就相互对角。

R2UR'FR'F2RU'R 双 T 掉期
F2UF2U2R2UR2U2 已解决 2x1 (R)
F2R'U'FU2F'UR'U 已解决 2x1 (2)
F2U2R2U'R2U'R2 诊断模式


4 种情况下，D 层的白色和黄色部分应该在已解决的立方体中彼此相邻。这是从一开始就在底部解决的一个 2x1 块。
R2U2F2U'R2U2F2U'F2 双T-swap
F2RUR'F2RF'RU' 已解决 2x1 (1)
R2U2 已解决 2x1 (2)
F2U'R'UF'R2FR'F'U2 诊断模式

4 种情况下，D 层的白色和黄色部分应该在已解决的立方体中彼此相邻。这是从一开始就在底部解决的两个 2x1 块。

这是所有第 1 组案例，我从第 2 组开始，但没有做那么多。您如何看待这种拆分其余案例的方式？
你明白我是怎么做的吗？我做错了什么吗？

欢迎指正和批评！

如果有人不明白这一点，那么这 3 组案例也会完成整个置换步骤。有

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.

i want to know more