magicsquares
Regular
- Joined
- Oct 31, 2007
- Messages
- 82
If there was a method that integrates the 4th F2L pair and OLL how many cases would there be? Or is it just not possible?
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(2*3) * 58 -- corner and edge in F2L
(2*3) * 216 + (2*3) * 216 -- one in LL, one in F2L
(8*3) * 216 -- both in LL
Could you fix it? I don't see what's wrong.@ johannes91 - Your calculation is a little out
It seems like there has been a lot of confusion about this lately. The ZB and VH methods both orient LL edges while solving the final F2L pair. The difference between the methods is that with VH, you first convert the pair into one of the basic three move insert cases, then solve the pair while orienting LL edges. With ZBF2L, you don't convert anything. You solve the pair as it is and orient LL edges. Essentially, VH is a simplified version of ZBF2L, since you first convert the pair into a three move insert and then use a ZBF2L alg. VHF2L has 32 cases, while ZBF2L has 306 cases. Also, VHLL typically is COLL + EPLL, which is 44 cases. ZBLL does everything in one step (orients corners, permutes edges, and permutes corners), and there are 494 algorithms in this step.@ masterofthebass - the cross on top is the Vanderbergh-Harris method I think. And the only reason to learn this is to avoid OLL if you find identification of oll cases hard. The zb method orientates the ll while doing the last pair.
I believe there is a 1/27 chance of that happening with ZBF2L.Ah sorry about that post. I tried a zb alg once and it orientated the whole ll O.o