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4x4: How do I fix this situation in w/ 1 alg?

pjk

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I get this situation quite often, where everything else is solved except the two shown:
center6.jpg

(sorry for the crummy pic)

Do I have do to the r2 U2 r2 U2 u2 r2 u2 and then solve it like a 3x3, or is there an alg that can fix it and have it solved with that 1 alg? Thanks
 

dougreed

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You can always do R d* B' [your normal alg for this parity] B d*' R'

where d* is rotating the bottom three layers in the direction of D, and [your normal alg [...]] is the r2 U2 r2 U2 u2 r2 u2 alg.

-Doug
 

cmhardw

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R2 D' x (Uu)2 (Rr)2 U2 r2 U2 (Rr)2 (Uu)2 x' D R2

That's what I use. I treat it as one alg and not a setup turn and my regular parity alg. It's pretty fast too, see what you think.

Chris
 

krnballerzzz

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Originally posted by PJK@May 2 2006, 02:02 AM
I get this situation quite often, where everything else is solved except the two shown:
center6.jpg

(sorry for the crummy pic)

Do I have do to the r2 U2 r2 U2 u2 r2 u2 and then solve it like a 3x3, or is there an alg that can fix it and have it solved with that 1 alg? Thanks
I do (R' U R U) r2 U2 r2 U2 u2 r2 u2 U' R' U R. This is single slices. u2 is only the inner layer of u (not including U).
 
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