Slowcubers-Blind for Megaminx edges
Only one alg for all your problems! And even better: it can be used for Gigaminx-midges, too!
The basic concept is to solve one edge at the time and bring it to the front first with D+ or D++ and so on. The first edge is solved with <alg>, the second with <alg'>. It is therefore very useful to memorize in pairs.
Shengshou color-sheme: white on top, upper half green in front, to the right red-blue-yellow-purple, lower half starting with brightyellow (under green and red) and to the right pink-brightgreen-orange-brightblue, grey bottom.
Memo is layer by layer counterclockwise (looked from above) starting top-front with 1 (white-green) and counterclockwise to 5 (white-purple). Next "layer" starts with 6 (purple-green), then 11 (brightblue-green), 16 (brightyellow-green), 21 (brightblue-brightyellow) and 26 (grey-brightyellow).
11 and 16 are in the same „horizontal layer“, but they need different setups move, so I take them as different „memo-layers“. Look at the basic concept: 16 (brightyellow-green) and 17 (pink-red) are the „same“, because I can bring the 17 with a D- to the front (at the place of the 16) - after that, they have the same setup.
Flipped edges: I add a 30, if the wing has to be flipped. But when is it flipped? This rule you can make by yourself. I have 1, 6, 11, 16 as not flipped, if green is in front (be careful: the green-red is number 7 and belongs to the red side!), 21 with brightblue on the left and 26 with the grey side down. The highest number is the 60: blue-grey.
Another, maybe better explanation: „the first color of a color-pairs (like I called them in the memo above) is the side, where you have to look at while memorizing“.
(Sorry, I have huge problems with the orientation and solve 3x3x3 bld with orientation first. I haven’t found the trick yet and so I don’t know, whether someone can understand my explanations - just find it out by yourself!)
The alg (found it in this forum):
L' R F2' L R' U' L' R F2 L R' U
and its inverse:
U' L' R F2' L R' U L' R F2 L R'
The buffer is 5 (white-purple), the target 16 (brightyellow-green) and the helper 1 (white-green).
If you have the pieces-pair brightblue-brightyellow,blue-red (21,38), it goes like this:
setup21 alg setup21' setup38 alg' setup38'
Exception: if a top-edge is involved, you will take this as the helper and will only need one alg and it is used as a normal commutator (see below).
Here are my setups:
6: U2' F2' U2
36: D+ U2' F U2
11: D+ DL'
41: DR DL
16: F5

46: D' DL2 DR
21: D+ DR2'
51: DR
26: D- DL' DR
56: DL2
All the others need only some D+ before and some D- afterwards.
Example for a top-edge: pair (3,53):
D-- (bring 53 to front, to 51)
DR (setup for 51)
U' (bring buffer to 1)
L' R F2' L R' (first half of alg)
U2' (bring the 3 to 1 - it is now in position 4!)
L' R F2 L R' (second half of the alg)
U2' (bring 5 to 5 again)
DR' (setup' for 51)
D++ (bring 53 back to 53)
Example for a top-edge: pair (51,3):
DR (setup for 51)
U2 (bring 3 to 1)
L' R F2' L R' (first half of alg)
U2 (bring the buffer 5 to 1 - it is now in position 3!)
L' R F2 L R' (second half of the alg)
U (bring 5 to 5 again)
DR' (setup' for 51)
Ok, if both edges belong to the top (e.g. 2,3), you can give up this attempt

… or bring the 2 to 16 with U F2 U', then solve it like a 16,3 and do a U F2 U' to bring the 2 back.
A top-edge flipped?
Setup the other edge of the pair, then flip the top-edge without affecting the buffer 5 and the target 16. The easiest way for the 33 will be BR' R' and you will solve it then like a 2 (and don't forget to bring it back afterwards).
If you have a edge in place but flipped, e.g. 51, just solve 51,21 at the end.
Now look again at the alg: there is a U at the end and alg' begins with an U' (that is clear, because it is the inverse). You will only need this U and U', if one or both edges of the pair have setups that affect the top-face (only 6 and 36). Otherwise they cancel each other.
(Another idea for flipped top-edges: solve them without flipping and flip them at the end in pairs. The alg would be e.g.:
L F R' F' L' U2' R U R U' R2' U2 R
from
https://sites.google.com/site/permuteram.../old-p-bld
But this doesn’t work for the Gigaminx. When I made that I hadn’t thought of the easy way I described above. I solved it at the end by bringing one edge to 16, executed the alg, flipped it, executed alg' and brought it back to the top).