[Unofficial] Rex Cube Blindfolded

Nice. How'd you do it?

Always cool to see people BLDing weird puzzles. What algs did you use?

tl;dr - commutators for everything.

Details:

Notation:
1 - Turn URF corner clockwise
2 - URB clockwise
3 - ULB
4 - ULF
5 - DRF
6 - DRB
7 - DLB
8 - DLF

1', 2', 3' - turn it counterclockwise

17 - turn a belt between 1 and 7 corners clockwise if you look at 1 corner (17 = 1' 7 x y)
28, 35, 46 - similarly.
17' - turn it counterclockwise.

0) Orientation.
You can choose any orientation in which central centers form an even permutation. You may want to choose orientation in which more "petals" (X-centers) are solved.

1) Central centers.
U -> F -> R: [1 35' 4 35 1', 2']
U -> F -> D: Do a setup to previous case, or [1 46 2 46' 1', 5]

2) "Petals" (X-centers):
Note that all X-centers are divided into two orbits, i.e. Ubr can't go to Ubl. Try to figure out orbits by yourself (cuz I'm too lazy to write them out).
Basically you need to know algs for Ubr -> Fdr -> *. All other cases can be done as setups to these; or Ubr -> A -> B can be done as Ubr -> A -> Fdr; Ubr -> Fdr -> B.
There can be a "parity" when you got a 2-cycle Ubr -> A. You solve it like x-centers in 4x4 bld: make a 3-cycle, Ubr -> A -> Ufl.
Ubr -> Fdr -> Ufl: setup to Luf.
Ubr -> Fdr -> Rub: [3' 6 3, 46']
Ubr -> Fdr -> Rdf: [46, 8 6' 8']
Ubr -> Fdr -> Bur: [1 6' 1', 46'] or [46, 8 6 8']
Ubr -> Fdr -> Bdl: [46, 1' 6' 1] or [3, 1' 35 1]
Ubr -> Fdr -> Luf: [1 28 1', 8] or [3', 1' 35 1]
Ubr -> Fdr -> Lbd: setup to Dlb
Ubr -> Fdr -> Dfr: [46, 1' 6 1] or [3' 6' 3, 46']
Ubr -> Fdr -> Dlb: [1 6 1', 46'] or [1 28 1', 8']

3) Edges.
Note that you need to memorize only permutation of edges, without orientation. An edge cannot be flipped on it's place.
Since every single move has an order 3, there is no edge parity.
Basically you need to learn only alg for UF -> RU -> DF and its rotations, all other cases can be done as setups.
UF -> RU -> DF: [46, 17 4' 17']
 
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