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Hi Guys, I'm back again.....
So, I've worked out the missing solution...now my algs looks like that:
UF DB (U' M2) (R U' R' U) (M U') (R U r' U)
BU DB x (U' M2) (R U' R' U) (M U') (R U r' U) x '
UF ® BU x' (U' M2) (R U' R' U) (M U') (R U r' U) x
DB UF (U' r U' R' U) (M' U') (R U R') (M2 U)
DB BU x (U' r U' R' U) (M' U') (R U R') (M2 U) x'
BU UF x' (U' r U' R' U) (M' U') (R U R') (M2 U) x
BD UF (U M' U' M) U2 (M U M' U)
DB UB x (U M' U' M) U2 (M U M' U) x'
UB FU x' (U M' U' M) U2 (M U M' U) x
UF BD (U' M U' M') U2 (M' U M U')
UB DB x (U' M U' M') U2 (M' U M U') x'
FU ® UB x' (U' M U' M') U2 (M' U M U') x
UB UF (U M U M) U2 (M' U M' U')
DB FU x' (U M U M) U2 (M' U M' U') x
UF UB (U M U' M) U2 (M' U' M' U')
FU DB x (U M U' M) U2 (M' U' M' U') x'
BU FU (M' U2 M U2)
BD FU x (M' U2 M U2) x'
FU ® BU (U2 M' U2 M)
FU BD x (U2 M' U2 M) x'
UB BD U2 (D M' D M') U2 (M' D M' D)
BD UB (D' M D' M) U2 (M D' M D') U2
BD BU (M U2 M U2) M2
BU BD M2 (U2 M' U2 M')
And of course:
UB s U2 M2 (s) M (s)' M U2 Switch RU and LU
s UB U2 M' (s) M' (s)' M2 U2
UF s M2 (s) M (s)' M
s UF M' (s) M' (s)' M2
s FU (s) M2 (s)' M U2 M U2
DB s M2 (s) M' (s)' M'
s DB M (s) M (s)' M2
s BD (s) M2 (s)' U2 M' U2 M'
The two main algs for UF-> DB and Inverse seems quite long but are actually pretty fast...
I also had another Idea...that could be useful for somebody who whants to keep algs memorisation at the minimum:
For UF-> DB (and Inverse) you could also do "F2 then Alg of UF-> BD and thatn F2 again....
That means that the actual memorization of algs, would become of..about five? If you don'consider the really easy 4 moves one... of course.
I hope that can be useful for you...I don't know if that is gonna be faster than Maky's commutators way...I think so though.....
Have a grat Day!
Federico
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