riffz
Veteran
An MU alg for UB-BD-FU (Optimal, if possible)
I tried to generate some MU algs but acube is failing me.
Anyway, <MU> kind of sucks for this case. I would do:
y [U; M',UR2U'] --> y U M' U R2 U' M U R2 U2
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An MU alg for UB-BD-FU (Optimal, if possible)
A little pedantic, but if you're using my extended notation, then [U: M', UR2U'] (= [U; M', R2]) = [U M' U', U R2 U'] = U M U' U R2 U' U M U' U R2 U' --> identityy [U: M',UR2U'] --> y U M' U R2 U' M U R2 U2
A little pedantic, but if you're using my extended notation, then [U: M', UR2U'] (= [U; M', R2]) = [U M' U', U R2 U'] = U M U' U R2 U' U M U' U R2 U' --> identity
(The reason is that it's really common to conjugate an inner part of the commutator; conjugating an entire alg is larger scope, for which the semicolon is a stronger separation: [U; M' ,UR2U'] )
IS there any alg for the Dot case(2 look OLL) quicker than F(RUR'U')F' f(RUR'U')f'? (F' and f' kill me)
IS there any alg for the Dot case(2 look OLL) quicker than F(RUR'U')F' f(RUR'U')f'? (F' and f' kill me)
M U R U R' U' M' R' F R F'
IS there any alg for the Dot case(2 look OLL) quicker than F(RUR'U')F' f(RUR'U')f'? (F' and f' kill me)
IS there any alg for the Dot case(2 look OLL) quicker than F(RUR'U')F' f(RUR'U')f'? (F' and f' kill me)
Wrong case man.
You can avoid the dot cases. When you are on your last pair and you don't see any oriented edges, just use the insert R' F R F' and then you will get a better OLL.
lrn2read
I did read. I was just saying it was better to use R' F R F' and then you can do whatever you want to do OLL.
:fp he wasn't talking to you. He quoted Ryan's post.
I did read.
It's just that people say I don't read well.
OLL skip algorithm for this case