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I am starting this thread because I hijacked another (sorry about that).

I came up with the idea of parking the UL and UR edge cubies in the DF and DB positions (either order), and solving one of eleven possible orientation patterns, so that all remaining edges are oriented, and UL and UR are still in DF and DB positions, ready to be slotted home with U (or U', or U2), M2, then U or U'.

The original algorithms that I found via Ron's sticker mode solver were:

M U' M U M' U2 M U' M U' M (11,12)

M U M' U2 M U M' (7,8)

M' U2 M' U2 M' U' M U2 M U2 M (11,15)

M' U M' U2 M' U M' U M U' M' (11,12)

M' U M U' M' U2 M (7,8)

M U' M U2 M U' M U' M' U M (11,12)

M U' M' U M U2 M' (7,8)

M' U' M' U M U' M (7,7)

M' U2 M' U2 M U' M' U' M2 (9,12)

M U M U M U M (7,7)

M' U M' U M' U M2 U' M U2 M' U' M (13,15)

Athefre was able to find several amazing shortcuts to several of the algorithms as above, and we learned about some blind spots that computer solvers have when looking for optimal solutions to the orientation step with the roux method.

I came up with the idea of parking the UL and UR edge cubies in the DF and DB positions (either order), and solving one of eleven possible orientation patterns, so that all remaining edges are oriented, and UL and UR are still in DF and DB positions, ready to be slotted home with U (or U', or U2), M2, then U or U'.

The original algorithms that I found via Ron's sticker mode solver were:

M U' M U M' U2 M U' M U' M (11,12)

M U M' U2 M U M' (7,8)

M' U2 M' U2 M' U' M U2 M U2 M (11,15)

M' U M' U2 M' U M' U M U' M' (11,12)

M' U M U' M' U2 M (7,8)

M U' M U2 M U' M U' M' U M (11,12)

M U' M' U M U2 M' (7,8)

M' U' M' U M U' M (7,7)

M' U2 M' U2 M U' M' U' M2 (9,12)

M U M U M U M (7,7)

M' U M' U M' U M2 U' M U2 M' U' M (13,15)

Athefre was able to find several amazing shortcuts to several of the algorithms as above, and we learned about some blind spots that computer solvers have when looking for optimal solutions to the orientation step with the roux method.