cubers

Huh, I've always seen people recommend learning PLL before OLL. Hope that's going well for you, anyway. I'd recommend you learn algs that work well for you rather than whatever your source suggests (check some algs out over on algdb.net).

Also, you may want to check out the Speedsolving Forums. I've used them and here to supplement my lack of visiting the cubing subreddit.

I guess I’m a speed solver, just a slow one. I’m nearly 40 and just wanted something to occupy my mind rather than Reddit (social media in general) and it’s certainly done that. Currently I’m working through learning OLL, then it’s PLL which is apparently easier.

In my opinion, if you time yourself, you’re a speedsolver. I’m older than you (43), but I know that I won’t reach sub-15 since I can train "properly". I’ll be happy to be sub-20 !

Learning 78 algs (even though I knew more than half of the PLL) was definitively a deal-breaker for me. Even with CMLL I struggle to reco the pattern and then reco the alg associated… I know that I lack practice, but I’m glad other methods exist  ! (I looked at Petrus and Heise before Roux).

I’m a CFOP man but Roux interests me, I just can’t seem to wrap my head around the last algs yet.

The LSE stage? At first, you can go without algs. Here how I process it:

After CMLL, the last step is divided into 3 parts:

• 4a: edge orientation (EO)
• 4b: finish L and R sides (if done with EO: EOLR, but don’t learn that yet. I still don’t know it :þ)
• 4c: permut M edges

4a: If you take the time to analyse EO, you’ll see it’s really simple. Let’s say that top/bottom is yellow and white:

• The goal is to have only yellow/white on top and yellow/white on bottom.
• in order to achieve that, you "reduce" all your cases to the "arrow case".

If you don’t know yet, arrow case is 3 non-white/yellow on top, and 1 non-white/yellow on bottom. Once you have an arrow, you do M'U*M*, with *=' or *=nothing (or MU*M* depending on the arrow facing you or not)

Except for the 6 flipped-edges, there are 3 cases that are not solved/arrow cases:

• 2 adjacent flipped on top, 2 on bottom : M2 and you have the arrow example
• 4 flipped with 0 adjacent (4 on top, or 2 on top opposite, and 2 on bottom) : the goal is to swap UF and DF (or UB and DB): M' U2 M can achieve that (in fact, I think that M* U2 M* are ok) example (4 on top), ex2(2 on top, 2 on bottom)
• only 2 flipped : M* U* M* will do the trick (they can’t be on the same face). I put the one on the M slice on D with M or M', then I put the one on r/L on top of the other one with u/U' and I bring the first one on top again with M/M' :

ex1 both on top

ex2, symmetrical to ex1

ex3 both on top again

ex4 1top, 1 bottom

ex5 1 top, 1 bottom, diag case

ex6 1 top, 1 bottom, same face

ex7 1 top, 1 bottom, same face2

ex8 2 bottom

You can see that in a lot of cases, the last M can be M'. But I like to keep it simple ;)

For the 6 flipped edges, either you learn an alg, or you M*U*M* to reduce to only 2 flipped: example

See also the flow chart in https://rouxl.es/lse.html

4b: in this step, your U and D faces should have yellow/white stickers only.

The goal is to finish the left and right sides (that is, put UL and UR). In order to do that, the easiest way, is to put UL/UL both in D, adjust the U face, do M2, and put U correctly.

• if they already are on the D face, AUF, M2 AUF
• if they are on U, opposite, AUF to put them on the M slice, M2 to put them on D (so we are back to first case), AUF,M2,AUF

In all other case, we first bring them diagonally of each other on the M slice.

• if one on U, one on D, AUF
• if they are on U, but adjacent, M2 to bring one on U, one on D. Then, AUF.

And then M* U2 M* to bring them both on D:

ex1

ex2

ex3

And then you finish with AUF,M2,AUF

Sometimes, you can merge the two steps:

ex4

That can happen when one of UL/UR are between L/R corners. I’ll let you experiment with it.

4c: L and R are solved now. You just have to permut M edges now.

There is really only 3 cases: bars, dots, and cycles.

• for bars, it’s just U2 M2 U2 (with M* setup)
• for dots, E2 M E2 M', or (U2 M' U2 M2 )2, or (U2 M2 U2 M')2, or any other alg
• for 3 cycles, on the M slice, you will have 4 "blocks": 1x2, 1x2, 2x2 and 1x1:

ex 2x2 is yellow-blue, 1x2 are green-white, and yellow-green, and then 1x1 is white-blue.

You’ll have to do U2 with the 1x1 without breaking the 2x2.

In the example, if you do U2, you’ll break the blue line of the 2x2, so you first have to do M

Doing so, you will have a 1x3 line (here white line) and a 2x3 block (here with yellow line).

Then you bring the 1x3 on top, U2 ( and sometimes M*)

​

Hope it helps