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antech101

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Anti-Sune Twice?

(This is for CFOP)

During OLL has anyone ever had to do an Anti-Sune twice? For some reason whenever i do the alg on something that looks quite a lot like the alg, it just makes an actual anti-sune making it 3 look oll...
 

TDM

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(This is for CFOP)

During OLL has anyone ever had to do an Anti-Sune twice? For some reason whenever i do the alg on something that looks quite a lot like the alg, it just makes an actual anti-sune making it 3 look oll...

You're getting a Sune case. To solve this, hold the oriented corner in UFL and do sune (R U R' U R U2 R'). They look similar because they both only have one oriented corner, but are different cases.
 
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mafergut

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I'm not getting a sune (Im sure of it) its both not a sune nore anti-sune, its weird...

With one corner oriented and the other 3 not oriented it can only be either a sune or an antisune so your must have seen it wrong or executed the alg in the wrong orientation of the last layer or something like that.
 

TDM

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Algs for Corner 2-swaps (only corners) on 4x4+?

So PLL parity but edges and 2 corners are solved like t-perm U PLL parity

Thanks

SSE

I use:

Opposite: F R U' R' U' R U R' F' y [PLL parity] y' R U R' U' R' F R F'. Just a Y perm with PLL parity done in the middle.
Adjacent: R U R' F' y [PLL parity] y' R U R' U' R' F R2 U' R'. J perm with PLL parity done near the start.

With both of these algs you can swap the y and y' or even use AUFs instead.
 

mafergut

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I just do PLL parity for the edge in between the corners, if adjacent, and then T-perm or RUR'U' PLL parity URU'R' and then Y-perm, if opposite, because I hate N-perms in big cubes :) Which I'm sure is not optimal by far. I will try what TDM suggested for opposite as it saves me the adjacent PLL parity setup/undo, which would be nice. For the other I don't really see that it's better than just doing what I do now.
 

Sam N

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sq1optim gives this six twist solution: 6,0/3,0/-3,0/-4,-1/-2,1/-3,0/ [6|14] vs. the setups 7.

I'm wondering if he is referring to the OBL case, and not the optimal case to solve the puzzle. If not, then I guess this is a side note for the OBL case lol.

setup: (1,0)/(3,0)/(3,0)/(-1,-1)/(-2,1)/(-3,0)/(3,0)/

Here is a nice solution for the OBL case. I'm not sure if it's optimal, but it's the best I could come up on my own. I haven't had much time to learn OBL cases, I just know the super nice ones.

NOTE: The setup algorithm has the top layer misaligned by (1,0), so I am keeping it that way. When you use this algorithm, make sure the top is misaligned by (1.0) before you do it.

(-4,-1) / (4,1) / (-3,0) / (2,-1) /

BTW, if you want to look at the full spreadsheet for all the OBL case, here is a nice one that I believe was developed by Micheal Young.


http://www.its.caltech.edu/~skuroyam/obl_data_numstates_1110.pdf
 
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