--- In blindfoldsolving-rubiks-cube@yahoogroups.com, "cmhardw"
<foozman17@h...> wrote:
>
> Alright here's my average for edges:
>
> 2:11.49, 2:08.62, 2:22.31, (1:36.52), 2:41.81, 2:25.22, 2:14.40,
> 2:25.49, 2:15.28, (3:02.38), 2:59.92, 1:56.67 = 2:22.12

Wow, that's really good! Quite inconsistent, but I guess that will go
once you treat odd cycles better.

> I'd say I average 1 setup move overall. Sometimes I can see a
> commutator with the way they're setup, and equally as often I need 2
> setup moves. Most of the time I do 1 setup move. So I'd say I
> average right at 1 setup move overall.

Wow, I have to try it again then. I think I did and it took me at
least 2 most of the time. Question: do you have a certain edge place
that is always part of every 3-cycle you do? Or do you do completely
free 3-cycles?

Ok, now here's my method: First scramble the cube and then solve
centers and corners speedstyle. Now we can begin.

First please also solve the URf edge any way you like. It's not
necessary to do this at the start but it will make the explanation
much easier.

Do you know my 3x3 blindsolving method and how I solve the edges? I
have a fixed buffer position from which I solve the edges one by one.
That fixed buffer position is UR.

For my 4x4 edges, I use the same idea: Solve the edges one by one,
shooting from a certain buffer. But this time, the buffer is not fixed
at one position: It alternates between UBr and URf. At the start
(after you solved the URf edge) the buffer is UBr. So look at the edge
in that position and bring its goal place (i.e. the edge currently in
that place) to FUr. Now do this algorithm:
R U R' U' (Rr) R' U R U' (Rr)'
This brings the edge from UBr to FUr so now undo your setup moves and
the edge will be solved. Now the buffer is at URf. So look at the edge
in that position and bring its goal place (i.e. the edge currently in
that place) to FUr. Now do the inverse of the above algorithm, i.e.:
(Rr) U R' U' (Rr)' R U R U' R'
This brings the edge from URf to FUr so now undo your setup moves and
the edge will be solved. Now the buffer is at UBr again. So repeat
from now on.

Is that clear enough to understand it? I pretend to use 2-edge-swaps
by alternating between two buffer positions. So I think of it as
solving with swaps, solving one edge at a time, exactly like my 3x3
method.

Cheers!
Stefan