#### blah

##### brah

I did my first 4x4x4 BLD yesterday, with pure r2 for edges, that means the VERY long algs for the r/l edges (at least I consider them long). So I consulted Mike Hughey, and he said he used commutators for r/l edges and r2 for R/L edges.

I wanted to be faster (actually I've only had 2 successful solves till now and about 10 DNFs ranging from 14 to 17 mins), so I decided to do the same. I sat down for half an hour with a cube, a pen and a piece of paper, and came up with all the commutators I needed for r/l edges

Actually I sorta Fridrichized them to minimize thinking during an actual solve, so there'll be 2 setup moves (c.f. the ideal 1) for almost all of them (but I guess 2 extra moves, the setup and the 'setdown', compensates for the extra thinking time wasted if I hadn't Fridrichized this).

I memorize in pairs (actually I memorize in six-letter groups, so I know I'm about to finish all the edges when I have ~4 groups, helps me keep track ), and I wanted to do commutators like Mike does, i.e. if

Notation: [A, B] = ABA'B'; [F] = F/F'/F2 (depending on the situation).

1. [l] to bring l-piece to UFl.

2. [R/L] to bring R/L-piece to U.

3a. If R/L-piece on f, commutator [D'[f]D, F2] or inverse.

3b. If R/L-piece on b, commutator [D

I wanted to be faster (actually I've only had 2 successful solves till now and about 10 DNFs ranging from 14 to 17 mins), so I decided to do the same. I sat down for half an hour with a cube, a pen and a piece of paper, and came up with all the commutators I needed for r/l edges

Actually I sorta Fridrichized them to minimize thinking during an actual solve, so there'll be 2 setup moves (c.f. the ideal 1) for almost all of them (but I guess 2 extra moves, the setup and the 'setdown', compensates for the extra thinking time wasted if I hadn't Fridrichized this).

I memorize in pairs (actually I memorize in six-letter groups, so I know I'm about to finish all the edges when I have ~4 groups, helps me keep track ), and I wanted to do commutators like Mike does, i.e. if

*any*one piece of the next pair is in r/l then I'll do a commutator. So here's my systematic approach:Notation: [A, B] = ABA'B'; [F] = F/F'/F2 (depending on the situation).

**Case A: one piece in l, the other in R/L.**1. [l] to bring l-piece to UFl.

2. [R/L] to bring R/L-piece to U.

3a. If R/L-piece on f, commutator [D'[f]D, F2] or inverse.

3b. If R/L-piece on b, commutator [D

**D', F2] or inverse.**

1. [x] to bring DFr (buffer) and r-piece to UFr and DFr.

2. [R/L] to bring R/L piece to U (the new U after the cube rotation).

3a. If R/L piece interchangeable with UFr, commutator [FdF',**Case B: one piece in r (not UBr), the other in R/L.**1. [x] to bring DFr (buffer) and r-piece to UFr and DFr.

2. [R/L] to bring R/L piece to U (the new U after the cube rotation).

3a. If R/L piece interchangeable with UFr, commutator [FdF',

__] or inverse.__

3b. If R/L piece not interchangeable with UFr, commutator [[D](R2/L2)[D], r'] or inverse.

1. [x] to bring DFr (buffer) and r-piece to UFr and DFr.

2. [l] to bring l-piece to UBl (the new UBl).

3. Commutator [FdF', U2] or inverse.

1. B2 to setup to Case C.

1. If 'adjacent' (hope you get what I mean), [l] to bring both to UFl and UBl, then it's Case C with a cube rotation, commutator [UbU', F2] or inverse.

2. If 'opposite', haven't came up with any solutions with 2 setup moves or less Suggestions, anyone?

I don't think I'm missing out any cases. Is this a good approach (to Fridrichize the commutators to minimize thinking)? Or are there better commutators for some of these cases? Thanks in advance.

P/S: I just tried to calculate the probability of at least one piece being in r/l, and it turns out to be more than half?! Is there something wrong with my calculation? It's simple: 1 - (16/24) * (15/23) = 0.565. Or from another perspective: 1 - (16C2)/(24C2) = 0.565.

Another question out of the blue, how do you (whoever's reading this) keep track of which edges have been memorized and which haven't? 'Cause if I have a 24-cycle then I have a very fast (sub-6) memo. If I have a 21- or 22-cycle I take all day finding out which are the 2 or 3 edges left to be cycled (and the worst part is finding that they're already permuted), and if it's REALLY bad the memo ends up being more than 10 mins Any techniques?3b. If R/L piece not interchangeable with UFr, commutator [[D](R2/L2)[D], r'] or inverse.

**Case C: one piece in r, the other in l.**(I almost died thinking how to setup these both-in-r/l cases, then I realized they were easier than Cases A and B -.-")1. [x] to bring DFr (buffer) and r-piece to UFr and DFr.

2. [l] to bring l-piece to UBl (the new UBl).

3. Commutator [FdF', U2] or inverse.

**Case D: both in r.**1. B2 to setup to Case C.

**Case E: both in l.**1. If 'adjacent' (hope you get what I mean), [l] to bring both to UFl and UBl, then it's Case C with a cube rotation, commutator [UbU', F2] or inverse.

2. If 'opposite', haven't came up with any solutions with 2 setup moves or less Suggestions, anyone?

I don't think I'm missing out any cases. Is this a good approach (to Fridrichize the commutators to minimize thinking)? Or are there better commutators for some of these cases? Thanks in advance.

P/S: I just tried to calculate the probability of at least one piece being in r/l, and it turns out to be more than half?! Is there something wrong with my calculation? It's simple: 1 - (16/24) * (15/23) = 0.565. Or from another perspective: 1 - (16C2)/(24C2) = 0.565.

Another question out of the blue, how do you (whoever's reading this) keep track of which edges have been memorized and which haven't? 'Cause if I have a 24-cycle then I have a very fast (sub-6) memo. If I have a 21- or 22-cycle I take all day finding out which are the 2 or 3 edges left to be cycled (and the worst part is finding that they're already permuted), and if it's REALLY bad the memo ends up being more than 10 mins Any techniques?

Last edited: Jun 20, 2008