2-gen reduction algorithm count?

[Teoidus]

I could've sworn there was a statistics thread, but I can't seem to find it...

How many cases are needed to perform full reduction to 2-gen after 2x2x3 + 1 F2L Pair, leaving a 2-gen LSLL?

i.e. how many cases are needed to solve CP + EO + FD edge?

Here's what I have so far:

If we fix the FD edge at UF, there are 2^6/2 unique EO cases, and only 2 types of swaps are needed to reduce corners to 2-gen group. Yielding 64 algorithms?

 

[obelisk477]

I could've sworn there was a statistics thread, but I can't seem to find it...

How many cases are needed to perform full reduction to 2-gen after 2x2x3 + 1 F2L Pair, leaving a 2-gen LSLL?

i.e. how many cases are needed to solve CP + EO + FD edge?

Here's what I have so far:

If we fix the FD edge at UF, there are 2^6/2 unique EO cases, and only 2 types of swaps are needed to reduce corners to 2-gen group. Yielding 64 algorithms?

Do you mean fixing the FR edge? FD should be solved if you have a 2x2x3

 

[Teoidus]

Sorry, I should specify that the 2x2x3 is in BD (so all unsolved pieces are facing you)

 

[gyroninja]

8 eo * 6 cp = 48 algs

 

[Teoidus]

Where are you getting those figures?

 

[gyroninja]

Where are you getting those figures?

They are common values. For eo (2^4)/2. Cp is a little more confusing to explain pre last layer. Take a look on Jaap's website for a good understanding on cp. Another way to look at it is that you can insert the corner 2gen (preserve cp) and the last layer cp wise can only be 1 of 6 cases (solved, adj front, adj right, adj back, adj left, and diag)

 

[Teoidus]

You might've misread the OP--FD is not solved to begin with, so EO shouldn't be 2^4/2 (6 edges to work with, not 5).

I've looked at jaap's website but I can't quite understand it yet. I think I get where you're getting 1/6 but I keep seeing CP methods like Briggs or Noah's CP blocks where they only use 2 algorithms to do 2-gen corner permutation. Do you happen to know how this works?

 

[gyroninja]

You might've misread the OP--FD is not solved to begin with, so EO shouldn't be 2^4/2 (6 edges to work with, not 5).

I've looked at jaap's website but I can't quite understand it yet. I think I get where you're getting 1/6 but I keep seeing CP methods like Briggs or Noah's CP blocks where they only use 2 algorithms to do 2-gen corner permutation. Do you happen to know how this works?

Yeah I thought you meant FR since FD would give worse algs. The three algs are nothing, fix adj swap, fix diag swap.

 

[Teoidus]

Could you explain why only one algorithm is needed to fix adj swap, yet there are 4 adj swap possibilities for LL?

 

[gyroninja]

Rotational symmetry. You can solve a T perm with only one alg. You can auf T perm to be an adj swap on any side.

Edit: The figure I gave with the corrected eo value would be an upper bound.

 

[Cale S]

You can also get 1/6 for CP by fixing one corner, so there are then 3! = 6 permutations of the 3 remaining corners