Well, I just learned the Ortega method, as I previously solved it just like corners on a 3x3x3 (LBL method). Because the Ortega method doesn't require the bottom face to have correct permutation, and because half the algorithms are faster than the T and Y perms I use at the moment, I assumed I would see an immediate improvement in my times, but in fact it's made my average about 4 seconds slower. Wondered if I was going about it the right way, here's my method.

0) During inspection work out the moves to solve bottom face (the corners don't have to be correct relative to each other) and see whether it will have 0, 1 or 2 correct 'bars' (i.e. 2 stickers the same colour next to each other on the D face) when complete.

1) Solve bottom face.

2a) If it's 0 bars, leave it as it is then do OLL, then step 3ai/ii/iii

2b) If it's 1 bar, put the bar at the back, then do OLL, then step 3bi/ii/iii

2c) If it's 2 bars, leave it as it is then do OLL, then step 3ci/ii/iii

3ai)If the top has 0 bars, perform R2 F2 R2 and AUF

3aii)If the top has 1 bar, put it at the front and perform (R U' R) (F2) (R' U R') then AUF.

3aiii)If the top has 2 bars, perform x2, then Y perm, then AUF.

3bi) If the top has 1 bar, put it at the back and perform (R2 U R2) (Y U2) (R2 U R2) F2 then AUF.

3bii) If the top has 0 bars, perform x2 (R U' R) (F2) (R' U R') then AUF.

3biii) If the top has 2 bars, perform x2 y' and then T perm

3bi) If the top has 0 bars, perform Y perm, then AUF.

3bii) If the top has 1 bar, put it on the left and perform T perm, then AUF.

3biii) If the top has 2 bars, the cube is complete!

As well as a question this threat can sort of also act as a guide for anyone else wondering about Ortega, as all the methods I've found on the internet seemed a little hard to understand.

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