How many cube rotations do you use?

[rubiknewbie]

1
0
1
1
0
using Breandan's Fridrich.
I use about 1-3 using my normal Fridrich f2l.

Ha! escher I was expecting you to reply and give some numbers like that. I looked through some of your F2Ls in another thread and they were mind boggling to me. I can never do it like that.

What is Breandan's Fridrich?

 

[yogonu]

3
4
3
3
4

Ok I guess

 

[Lt-UnReaL]

3
6
5
2
3

This isn't really going to reflect much on how efficiently we solve F2L since we're including the LL (but it looks like Escher isn't)...I got like 3 F perms. >_> Also, I counted any d or d' turns as rotations. With only F2L and not counting d or d' turns I get:
2
1
0
1
1

 

[Escher]

Ha! escher I was expecting you to reply and give some numbers like that. I looked through some of your F2Ls in another thread and they were mind boggling to me. I can never do it like that.

What is Breandan's Fridrich?

Thanks
'Breandans Fridrich' is the method he uses...
It involves solving cross with a corner/edge preserved in the BL slot, then optimally or near optimally solving the BL slot with a truckload of tricks.
During preinspection he plans this entire step, and looks ahead to the rest of the solve. During solving the step, he inspects EO of the next f2l slots.
Then, he'll solve either FL or BR slots, depending on EO.
To finish, he'll either rotate and solve what WAS FR and FL basically at the same time, or he wont have to rotate and he'll solve BR & FR.

During each of these steps, he knows lots of <R, U, F> and <R, U, u> tricks to avoid rotations, plus he solves lots of his pairs using crazy things like B and D moves to save moves...
The result is comfortably sub 30 f2l move count and practically no rotations.
Insane.