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Official Blindfold Algorithm List

kinch2002

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kinch2002
How about this one: M2 U2 D R U' R' U R U' D' R2 U R D R U' D' R2 U'

It's the best I have found so far using CE

Also not bad: M2 x D' R2 U R' U R U R' U' R' D' R U R' D R U2 R' D
 
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Again, i am looking for an algorithm that switches 2 oblique center pieces on the 6x6. One on the U-face and the other one on the D-face, because i use U2 to solve big cubes blindfolded. Thanks already. Dennis
 
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Again, i am looking for an algorithm that switches 2 oblique center pieces on the 6x6. One on the U-face and the other one on the D-face, because i use U2 to solve big cubes blindfolded. Thanks already. Dennis
U 2r2 U' [3l2] U 2r2 U' [3l2] to cycle UB, DF and DR obliques. 2r2 refers to an r2 with the outer r-slice while the 3l2 is the same with the inner l-slice (excuse my noobish notation)
 
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U 2r2 U' [3l2] U 2r2 U' [3l2] to cycle UB, DF and DR obliques. 2r2 refers to an r2 with the outer r-slice while the 3l2 is the same with the inner l-slice (excuse my noobish notation)
Thanks Ollie, but i need an algorithm that switches only one center piece on the D-face and 2 on the U-face. I need an algorithm that does:

UF -> DB-> UB (for obliques of course)

Am i right? :D

For example on the 5x5x5 (M follows L):

M' d2 M - U2 - M' d2 M

I hope i made myself clear?

Greetings, Dennis
 
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Thanks Ollie, but i need an algorithm that switches only one center piece on the D-face and 2 on the U-face. I need an algorithm that does:

UF -> DB-> UB (for obliques of course)

Am i right? :D

For example on the 5x5x5 (M follows L):

M' d2 M - U2 - M' d2 M

I hope i made myself clear?

Greetings, Dennis
I do now :) The best I can find at the moment is UF->DB->DR, but it's a tricky case:

[3l2] D' 2r2 D [3l2] D' 2r2 D
 
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I do now :) The best I can find at the moment is UF->DB->DR, but it's a tricky case:

[3l2] D' 2r2 D [3l2] D' 2r2 D
I am totally confused right now. I do solve 4x4s and 5x5s blindfolded but when i do the algs for centers, it seems as if they only swap 2 center pieces, wich is not possible. So why am i so confused right now?! xD
I dont get it anymore: What happens here? I mean, wich pieces change the position?:

M' d2 M - U2 M' d2 M

And is there a way to do exactly the same: UF -> DB , but for obliques?

Greetings, Dennis
 
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I am totally confused right now. I do solve 4x4s and 5x5s blindfolded but when i do the algs for centers, it seems as if they only swap 2 center pieces, wich is not possible. So why am i so confused right now?! xD
I dont get it anymore: What happens here? I mean, wich pieces change the position?:

M' d2 M - U2 M' d2 M

And is there a way to do exactly the same: UF -> DB , but for obliques?

Greetings, Dennis
Haha, sorry, I forgot you wanted a U2 alg :/ Have you tried some of these?
 
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I am totally confused right now. I do solve 4x4s and 5x5s blindfolded but when i do the algs for centers, it seems as if they only swap 2 center pieces, wich is not possible. So why am i so confused right now?! xD
I dont get it anymore: What happens here? I mean, wich pieces change the position?:

M' d2 M - U2 M' d2 M

And is there a way to do exactly the same: UF -> DB , but for obliques?

Greetings, Dennis

I don't use U2, but as I see it, you use a comm (that cycles 3 pieces) that has a U2 in it, but you do not do the last U2 (just like in M2/R2/r2) so it seems like you are only switching two obliques because a U2 move causes double parity(this is why you need an even amount of cycles or you retain parity). Does this make sense or should I try to explain again?

And Ollie gave you the algs.

Good luck Dennis.
 
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I don't use U2, but as I see it, you use a comm (that cycles 3 pieces) that has a U2 in it, but you do not do the last U2 (just like in M2/R2/r2) so it seems like you are only switching two obliques because a U2 move causes double parity(this is why you need an even amount of cycles or you retain parity). Does this make sense or should I try to explain again?

And Ollie gave you the algs.

Good luck Dennis.
Oh man, now the penny has dropped. Of course! :fp
What i was talking about was like only half of the "algorithm".
Thanks man. I was getting crazy xD

Greetings, Dennis
 
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In Old-Pochmann for corners, there is the standard Y-perm for UBL-RFD:
R U' R' U' R U R' F' R U R' U' R' F R

Someone also mentioned a good algorithm for UBL-DFR:
R U R' F2 r F' r U r2 F2

Is there any good algorithm for UBL-FDR?
 

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