I use the Old Pochmann method: placing one piece at a time from a buffer position using "pair swap" PLLs (PLLs that swap 2 edges and 2 corners each). Basically, this entails learning 6 algorithms to shoot the buffer piece to its appropriate slot (I might decide to do a video on this). This is not that hard, and there are plenty of good YouTube videos on it, though, again, I still might make on.
There are 6 algs to know:
T permutation
Ja permutation (swap blocks to the left (1), to the back (2), and to the right (3))
Jb permutation
Y permutation
Ra permutation (corner swap to the right)
My memorization time is down to about 10 minutes on a good attempt and at about 6 if I write down piece positions. The problem is, I always remember the last few corners wrong. So how do I memorize? Every edge has a number of 1 to 12 in this order: UF, UR, UB, UL, DF, DR, DB, DL, FR, FL, BR, BL (notice though that you will never send to position 2- it's the buffer). I then mark whether the edge is oriented or not, marked with an 'o' or 'u', respectively. Basically, an edge is oriented if it can be set up with the group (basically, it isn't out of your way to solve it with a T perm). Then I get a string of number-letter pairs. So one edge sequence might be: 12u 9u 4o 10u 11u 8u 5u 7o 6u 3o 6u. This edge sequence is bad for memo for 2 reasons: (1) there is an odd number of edges to place so there will be parity and (2) the two 6 spot edge placements means I broke into a cycle. blah...
I memorize the corners in much the same was: counterclockwise around the top starting with UFR and the same around the bottom starting with DFR, numbered from 1 to 8 (so 3 never gets used). I use the letter 'h' to stand for a top/bottom sticker on the top of the buffer piece. 's' stands for a side-colored sticker (in my case that is always red/orange) and 'd' stands for the front/back-colored sticker (blue/green for me). So a corner position might be: 8s 1d 2d 4h 6d 5h 4d 7h 7s. Again, an odd number of corners to be moved means parity, and the appearance of more than 1 4 and 7 indicates that cycles have been broken.
My solution to the above cube configuration? It's complicated.
d L' [T] L d2 L' [T] L d [T] d' L [T] L' d2 L [T] L' d' D M' [J(b)] M D' M' [J(b)] M' [J(b)] M2 D' M' [J(b)] M D [J(a)1] D' M' [J(b)] M D [R(a)] F [J(a)3] F' R [J(a)2] R2 [Y] R [J(a)3] R' [J(a)2] R' [J(a)2] R2 F R [J(a)2] R' F' D2 R2 [J(a)2] R2 D R' [J(a)2] R D
