Zeroing LSLL (BT+): How to skip LL every time

ruffleduck

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Could you please point to the case that takes 3 commutators to solve? I fail to see that.
With 5 corners left:
A pure 5 cycle needs 2
A 5-twist (impossible with this method) needs 4
Everything else (3c2t / 2c2c1t) needs 3 (1 to solve any twisted corners and 2 to solve the resulting 5 cycle for example)
 
This could be used for megaminx right? Just would need mega OCELL algs and then the last step is intuitive and still requires a max of 2 commutators!
 
BT+ is a cool concept. Are you going to expand on the documentation? If yes, looking forward.
 
A

Actually now that I think about it.....it could work
Just spent a couple minutes toying around with this on the megaminx....and it looks like it could be pretty viable. Unfortunately, looks like 3 commutators are possible because stuff like y perms take 3 times to return the cube back to a solved state
 
Just spent a couple minutes toying around with this on the megaminx....and it looks like it could be pretty viable. Unfortunately, looks like 3 commutators are possible because stuff like y perms take 3 times to return the cube back to a solved state
After OCELL, you are left with 6 corners. Because of the zeroing pair and the fact that all but 2 corners will be oriented (1 being the corner for the LS which cannot be in the correct position), no twisted corners can occur. That means you are left with:
3c3c (2 3 cycles = 6 corners), 2 commutators needed
5c (1 5 cycle), 2 commutators needed
2c2c (a double 2 swap = 4 corners), 2 commutators needed
3c (1 3 cycle), 1 commutator needed.

Correct me if I'm wrong (because above is how L6C would work on 3x3), but it should be the same for megaminx
 
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