# BCE methods

#### gogozerg

##### Member
Hi Kenneth,

That would be great, if you could achieve good times with such a method. Hang on!

When I considered this strategy, I gave up because of 2 things:
1) Building 4 pairs as the first step is a tough task. Fast recognition is a problem, and when an edge you need is located at DL or DR, it induces unpleasant moves (see what I mean?).
2) Learning all sequences needed for step 2 (6x8x3 minus identical cases and symmetries) is just too much for me. But people who can learn so many sequences could make the "last 4 corners" step more productive, as you propose (or make it prepare final edges orientiation).

Gilles.

#### Kenneth

##### Not Alot
Hi Kenneth,

That would be great, if you could achieve good times with such a method. Hang on!

When I considered this strategy, I gave up because of 2 things:
1) Building 4 pairs as the first step is a tough task. Fast recognition is a problem, and when an edge you need is located at DL or DR, it induces unpleasant moves (see what I mean?).
2) Learning all sequences needed for step 2 (6x8x3 minus identical cases and symmetries) is just too much for me. But people who can learn so many sequences could make the "last 4 corners" step more productive, as you propose (or make it prepare final edges orientiation).

Gilles.
Hi Gilles, I was actually about to mail you and ask for some comments, after all you are the master of this type of methods.

To build pairs when an edge is att DR just place the corner at URF or URB and do S', or if the edge is oriented the other way, place corner at ULF or ULB and do S2 = two turns! (do y and M for real) If the corner is oriented in U then it has to be turned so it is on the side instead. Exeption is if you got one or more sides free, then you can place the edge in the M-layer and do U to build the pair and simply R2/F2/L2 or B2 to place it in posistion. I can do all four pairs in about 15 STM using that style. Never place the second pair diagonally to the first, that way you block all sides = not good.

Jopp! recognition in the start is the worst problem, but that's the same for most methods, a lot of practice usally helps.

When all pairs are corretly oriented, then your MCLL works fine but here you can go even futher, creating a SMCLL.

Last edited:

#### Kenneth

##### Not Alot
Here is another one, also a grandmaster method I think:

1, Orient 4 pairs (as in the previous menthod), also oirent LL corners while placing the last pair (not so easy to learn but doable).

2, Compleate F2B (as 3, previous method)

3, Orient last six edges and compleate F2L

4, Use 3 x PLL to solve the rest

Just look at this diagonal pairs + N-PLL : Ra U2 Ra'

Edit: and this J : (R' U L') U2 (R U' L) : F-side pairs and N-PLL : (R d' L) U2 (L' d R') (same alg as the 0 + 1 PBL for Ortega step 3)

In step one you can of course simplfy a little by putting in the pair and then do corner OLL. Doing it that way this is pretty easy to learn, only the 3 x PLL that needs some work really, but 1/3 of them most of you know already.

Edt: just made up a stepping stone for the last step. You can in a first step solve pairs (2 + 1 solved cases) and corner permutation (2 + 1 solved = totaly 8 + 1 solved cases) = PF2L/CPLL and end in EPLL. For that method you don't need many algs at all. 7 x corner OLL, 9 x 6-Edge OLL, 8 x PF2L/CPLL, 4 x EPLL gives a total of 28 algs. Then when you can solve it like that (probably also using PLL if pairs are correct initially) just add algs to solve PF2L/PLL in one step to learn the full method, (maybe also learn to orient corners when putting in the last pair, it's intuitive, just like edge control but more cases, callt it "corner control").

Think this is the best of the methods so far for speedsolving, it's fairly easy recognition, not crazy many algs to learn and doing the pairs as I do them (see my reply to Gilles) is something you pick up rather fast, I'm alredy doing all four pairs with no serious breaks in maybe 1/3 of the solves after only 2-3 days of practice.

Anyone who likes to use a solver to find algs for permutation for this method (PF2L/PLL) is of course free to do so, I would welcome it because I won't do it myself, I stick to the previous method, at least for now.

Last edited:

#### David Pritts

##### Member
What does BCE mean?

#### Kenneth

##### Not Alot
If yo read the first post of this thread you will find it's explaination. It's in the second, short part

Anyone who logs in for the first time today, (see date of this post if you are unsure), also read the post two steps up please =)

Last edited:

#### gogozerg

##### Member

Of course, in step 3, you could choose arbitrarily the couple of edges to insert. DL+DR, DF+DB, UL+UR, etc.

I used to classify this method as a "4*1x1x3 method" (cube state after step 2).
The most simple method in this class is a pure corners-first technique:
1) 8 Corners.
2) + 4 edges.
By the way, I noticed that a slightly different approach was interesting regarding STM move count.
1) 8 corners.
2) Choose L and R side arbitrarily, and insert 2 opposite L-edges.
3) Insert 2 R-edges and solve L/R-centers.
But it's only good for STM FMC.

#### gogozerg

##### Member
4, Use 3 x PLL to solve the rest
Mmmmhhh... original!
Some sequences would be quite interesting indeed!

Last edited:

#### Kenneth

##### Not Alot
I used to classify this method as a "4*1x1x3 method" (cube state after step 2).
Four 1x1x3 = "columns first".

Of course, in step 3, you could choose arbitrarily the couple of edges to insert. DL+DR, DF+DB, UL+UR, etc.
One of many fine things about columns first.

#### Kenneth

##### Not Alot
Swapping pairs and doing CLL (the EG style) in one is not that hard to learn if you accept algs that are a little longer than the optimal ones.

Look at this CLL

R U R' U' f' U' F

Now the same case but also swap two diagonal pairs, it uses the same alg**

R U R' U' f' U' B' U2 B F

R U R' U' f' R' U2 R U' F

I only insert a few turns in the CLL to swap the pairs, otherwise the algs are almost** the same.

** The two second ones solves the mirror case for real, in this particular case the mirror is the case you get if you do N PLL on the original = the inverted case.

All CLL's has got a inverted case (case + N PLL) so you can always solve the CLL to N PLL state by using the inverted alg (the alg that solves the inverted case) and then end it by doing the similarity to Ra U2 Ra' if it's digonal pairs and R d' L U2 L' d R' for adjacent pairs, (often your CLL ends in U R' or likewise, try to solve the CLL's from the angle where you not have to do U R' (y) R U' to swap pairs, if you are in proper position non of those turns will be needed).

This is how you solve EG from using only 40 CLL's (not sure yet if all works but)

Last edited:

#### Kenneth

##### Not Alot
To build pairs when an edge is att DR just place the corner at URF or URB and do S', or if the edge is oriented the other way, place corner at ULF or ULB and do S2 = two turns! (do y and M for real) If the corner is oriented in U then it has to be turned so it is on the side instead. Exeption is if you got one or more sides free, then you can place the edge in the M-layer and do U to build the pair and simply R2/F2/L2 or B2 to place it in posistion. I can do all four pairs in about 15 STM using that style. Never place the second pair diagonally to the first, that way you block all sides = not good.
I just found you don't have to orient the first pairs you build, just put them thogether and leave them where they are, at least the first three is often easy to do like that, the last one is more often blocked by the others so after three I orient those or even start orientation of the first two while building the third, and then I do the last. First try this style I built the first two in 2 turns, had three oriented in 5 and all four in 11

This is also wery good for inspection, often you find one pair in the scramble, the second you can put togeter in 1-2 turns. So it's often easy to find also the third.

Last edited:

#### Kenneth

##### Not Alot
Just look at this diagonal pairs + N-PLL : Ra U2 Ra'

Edit: and this J : (R' U L') U2 (R U' L) : F-side pairs and N-PLL : (R d' L) U2 (L' d R') (same alg as the 0 + 1 PBL for Ortega step 3)
Adjacent pairs + J-PLL : R B U' F U2 F' B' R' F' U' F

Fast version: R (y) R U' L U2 Ra' (x) U' r' U L

Working on finding algs for the EG step, found this while doing that. EG algs will come up someday, I got like 80% of adjacent pairs done, digonal pairs are in many cases only a change in U/U' <-> U2 turns and faces but the same algs works. So it may take a week more before I got algs I lke for all cases.

OK, I could use a solver and get it done in a few hours, but I think that's cheating

#### Kenneth

##### Not Alot
Working on improving the last eight edges I solve after columns first (EG method).

Earlier I did it much like Roux solving F2B first and then the M slice and LL edges = orient all, place two and alg the last four, either M permut (EPML) or EPLL.

But I found it is much faster to simply place any edges/centres pairs into F2L until you got one or two left to do, then I orient the rest of the edges. Sometimes, if I got three unoriented in LL and one in FL I orient earlier because of the alg M' U/U' M/M'. Now I'm trying to learn all easy cases to orient the last ones and also place one into F2L in one go. If I got a case like that and normal for the rest I can do the edges in about 20 turns. If I fail to place one while orientating then it's about 25 (did 3 averages and got 24.4, 23.4 and 21.9).

While working on this I found this fun double alg:

Normaly you would do: M' U2 M U2 -- S' U2 S U2

But you can merge (logically OR) the algs to: M' S' U2 S M U2

Last edited: