L5EOP

[fei193]

http://www.speedsolving.com/wiki/index.php/L5EOP I don't know how it work. Please help me. , i need some scramble of this method

 

[Kenneth]

It is easy, just use a 'Roux' scramble and solve the centres and the BD edge (0-5 moves).

qq has got MU-scrambles in the megascrambler : http://mzrg.com/miniSites/scrambler...cramble=roux&num=5&len=18&subbutton=Scramble!

Example : U' M U2 M2 U2 M2 U M U2 M2 U' M' U M U' M' U' M'

Centres and BD : U2 M2
L5EOP : U' M' U' M U2 M' U M ... leaves U-PLL and AUF.

 

[fei193]

thanks for your help. So use L5EOP and PLL, i thounght it can finish the cube^^

 

[Kenneth]

Not PLL, edge permutation = EPLL, the corners you are supposed to solve on an earlier stage (CMLL if it is Roux).

It is possible to do L5EOP in one look, but you need 960 algs

 

[Kirjava]

I don't see the advantage over Gilles' method of L6E.

 

[fei193]

I really don't like fixing bad edges step in Roux method(3 bad edges in U and D face ). So Im finding some ways not to do or restriction it

 

[Kenneth]

I don't see the advantage over Gilles' method of L6E.

The advantage I see is that you don't need to look at anything but the top layer and on occasions the FD edge.

But really, I know like 5 methods to solve L6E, including the original and to me all are about the same in speed and number of moves (besides just end F2L and use ELL, that is a bit faster, but I hate some cases so I don't use it).

If you use the original some cases are really easy (10-11 moves NL) but the worst case is still 18 moves optimally, if you know the optimal alg, else it will be more. You will spread in the range from lucky 5 moves or so to 20+ no matter the method.

It is more a matter of taste than anything else...

 

[Kirjava]

The advantage I see is that you don't need to look at anything but the top layer and on occasions the FD edge.

That's the same as with Gilles' normal approach.

 

[nigtv]

I just recently started using l5e, as I was having a rough time with just gilles l6e, which I understood as orient edges, then place UR and UL, then permute M-slice. Seemed like most of the time, it was easier just to solve DF and DB with centers, then solve the last 4 edges in one alg, but still felt a little weird to me (again, no idea why).

Not that anyone is going to find this helpful, i'm not even close to being a master with roux, or anything, but I figured i'd type this out, at the very least, for my own reference later. Maybe someone will find it helpful, or at least it'll start some conversation.

Right now, I'm using the order of the 1st block, then the 2nd block, then orienting corners in one alg, then permuting them in one alg, then orienting the last 6 edges while placing DB with centers using one alg (changing the last 2 or 3 turns to their mirror/inverse usually seems functional for this, but I haven't really worked it all out into black and white yet), then permuting the last 5 edges in one alg.

For the orienting of edges, I use the algs on gilles site, and change the last few turns.

For the l5e permuting, I use some algs I found on the speedcubing wiki. Learning them was pretty easy (most are short), but recognition killed me when 0 edges were solved. I worked out some basic recognition scheme, maybe someone can look at it and tell me if it's stupid, or if you can think of something better, or give me some feedback. I searched and didn't really find anything (yea, I searched first, I'll say it again), but still sorry if I am rehashing.

(With 0 solved U edges, things get tough for me to spot quickly, so my system is probably very sloppy compared to something someone else has worked out, if they have even worked it out, for all I know this recognition could come intuitively to people...)

(Also, I promise the process here isn't as complicated as my confusing explanation makes it seem)

With all edges oriented, centers solved and DB placed;

1. If DF is solved, just use EPLL (I have five cases, one solved)

If DF isn't solved, look at the number of solved U edges;
With 2 solved U edges, check if they are opposite or adjacent
-If opposite, there's only one case
-If adjacent, there's 2 cases, mirrors of eachother

With 1 solved U edge, check if the 2 unsolved U edges are opposite or adjacent
-If opposite, there's only one case
-If adjacent, there's only one case (I think...)

With 0 solved U edges, look at the edge opposite the DF edge.
-If this edge belongs in the spot DF is in, there's only two cases, mirrors of eachother
-If this edge doesn't belong in the spot DF is in, check which U edge is in DF, then
which spot in the U layer it needs to go
*If its spot in the U layer is adjacent to the DF piece, there's only 2 cases,
mirrors of eachother
*If its spot in the U layer is opposite the DF piece, there's only 2 cases, mirrors
of eachother

As you can probably tell, it's needlessly complicated, it'd be nice to have at least some system other than doing it enough times that it just *pops*