Well, it basically sound just like guimond, except no intuition the first step with algs, and what I predict to be 1260 algs for the final step (8! positions of cubies, divided by 8 orientations, making it 5040, divided by 4 AUFs, making it 1260). I could be wrong with the final number, but I know the answer is huge.
You're actually totally wrong with the final number. Even assuming you don't want to have to flip your cube and thus cases which are similar up to z2 are different cases, you get:
- with separation done (just PBL): 8 cases (not counting skip)
- with one wrong corner in each layer: that corner can be chosen arbitrarily, and you get 6 permutations of the other 3, hence 6*6=36 cases
- with two wrong corners in each layer, forming bars on both: choosing one of them arbitrarily, you have 3 possibilities for the other one and 2 permutations of the last two. That's 6 cases for each layer, and given that some cases may be identical up to a y2 you actually have only 21 different cases.
- with bar on top and diagonal on bottom: choose arbitrarily one cube of each layer to put on the top layer. You have 3 possible choices to fill each remaining slots, and with symmetries up to a U2 it yields to 6 different cases. For the bottom layer, you have only 2 possible choices (which differ by a Y-perm), so 12 cases.
- with diagonal on top and bar on bottom: similar as above
- with diagonal on both: again only 12 cases by the same reasoning.
That's 101 algs to learn, the number is quite large but not unmanageable. Recognition would be quite messy though.