RUF Last Layer

[IRNjuggle28]

So, I was wondering what the fewest number of F moves needed for an otherwise RU last layer would be.

Explanation: So, F2L is solved. Every last layer position can be solved with a minimum of n F moves plus R and U. What is God's number for fewest F moves needed to solve any given last layer case given as many R and U moves as needed?

 

[GlowingSausage]

2 - risky guess

 

[supercavitation]

You'd need to (worst case) orient 4 edges, and either straight or diagonal swap the corners, all with F turns. I'd guess more than 2.

Then again, a God's Algorithm doesn't need to be a nice, friendly algorithm, so maybe it could be 2?

 

[IRNjuggle28]

There's an easier way to figure this out: which CPEOLL algorithm is worst when optimized to be as close to <RU> as possible?

 

[GlowingSausage]

There's an easier way to figure this out: which CPEOLL algorithm is worst when optimized to be as close to <RU> as possible?

:tu nice approach indeed :tu

edit: every oll case can be solved using only R, U & r moves - so... the RUF pll with the most F moves solves this problem :tu :tu
edit: it's 2, I just tried

 

[IRNjuggle28]

:tu nice approach indeed :tu

edit: every oll case can be solved using only R, U & r moves - so... the RUF pll with the most F moves solves this problem :tu :tu
edit: it's 2, I just tried

Lowercase r moves don't count as <RU>. Doing r U is basically the same thing as doing L F x.

 

[mDiPalma]

One F move can only affect the orientation of one LL edge, given that we start the analysis from a fully solved F2L.

You can have a maximum of four misoriented LL edges.

You do the math.

 

[GlowingSausage]

You do the math.

r U R' U R U2 r2 U' R U' R' U2 r ---> my "math" wasn't wrong cuz I assumed that "r" moves did count

You do the reading.

 

[IRNjuggle28]

One F move can only affect the orientation of one LL edge, given that we start the analysis from a fully solved F2L.

You can have a maximum of four misoriented LL edges.

You do the math.

You can use R and U moves before the F to make a single F move affect more than one LL edge.

F moves are also needed for CP. There are a bunch of things that explanation didn't take into account.

 

[qqwref]

I do already know that you can solve every ZBLL with only R and U moves and at most two F's (or two L's, or two B's, or two D's). It may also be true that every LL case with at most two edges misoriented can be solved with two F's. However, as several people have mentioned, you can't solve a four-edge-misoriented case with only two F's.