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That's not what we mean when we say "minimum", tim... it's more technically some kind of min-max problem, since in this (and similar God's Algorithm type computations) we're concerned with the algorithm which produces the minimum possible value for the maximum number of required cube rotations (or turns or whatever) over all allowed scrambled positions.
lol i wasn't aware that there were so many interpretations of my question!
Naturally, the cross is solved. I'm referring to F2L as the stage of Fridrich method, not the actual first two layers -- a common use of the terminology.
And naturally I am only concerned with y/y'/y2 rotations... I'm trying to apply this stuff to an actual solve... have you seen someone use x rotations to insert some F2L pairs?
The question where you are constrained to only <R,U> is interesting, but seems much harder to answer. And the answer probably does not yield any interesting insights, because people solve slots one by one, and dont really plan ahead to optimize number of rotations... for good reasons too! so never mind.
Thinking about this (and badmephisto's recent video) has changed the way I look at F2L during actual practice.
I used to spot an edge in the middle layer and not really notice whether or not it was flipped--I'd want to set up an easy case with its corner cubie which often required a cube rotation as I was ejecting it. To my dismay I would immediately need another cube rotation in order to execute the easy case. This was frustrating.
Now, if I see that it's flipped, I look for a way to pop it out without the rotation, even if its not my "preferred" setup with its corner, and voila! No more double-rotation!
For recognition, I've determined that if the edge's "outer" (not on the U face) stickers cannot match the L or R center pieces it means a cube rotation is required. This isn't too hard to recognize as an edge is either aligned with the red-orange axis or blue-green (for my color scheme).