cmowla said:
...note that this algorithm does not build on any of the parity algorithms that I found.
Okay, I lied, because the first piece is an algorithm I found before, but please be patient with me to explain it along with the second piece.
[1] With any 4x4x4 parity algorithm, it is a given to start out with an inner layer slice turn because OLL parity is referring to wing edges.
Rw'
[2] Four dedges are now unpaired. The closest thing we can do in comparison to just inverting Rw' with Rw is to replace one of these 4 unpaired dedges with a paired dedge from the outside first.
U R' U
Note that we do not do
U R' U', but we could also do
U' R' U', because besides replacing one of the 4 unpaired dedges from step 1, we want to have two 1x2 center blocks of the same color to be adjacent to each other in slice r (in this case with this alg.garron.us animation, these are white).
The reason we do this is because the second (middle) inner layer slice turn will become the extra quarter turn once the third (last) inner layer slice turn is done: with many 4x4x4 parity algorithms, when two 1x2 center blocks adjacent to each other in the same slice when the extra quarter turn is applied to that slice,
those two centers will be swapped with each other when the algorithm is complete (
Here's a simple example)
. This means that we can preserve one full center by just carefully following this fundamental rule.
[3] Now we are ready to do "the extra inner layer quarter turn". This move should pair back up one of the 4 originally unpaired dedges and unpair a new dedge. So we still have 4 dedges unpaired.
Rw
*It is very important that we don't pair back 2 dedges in this step because if we want to still keep 4 dedges unpaired because we can pair up 4 dedges with the third and final inner layer slice in this first main part of this 17 STM algorithm.
[4] For this step, all we are going to do is perform some outer layer turns to line up the 4 unpaired dedges so that with one more inner layer slice turn, all dedges will be paired up again.
Based on the route taken,
U F2 U is sufficient to do this. We can finish off with
Rw', since that's the obvious move from this point to pair back all of the dedges.
Note that U F2 U also positioned the two white 1x2 center blocks on opposite sides of the cube so that, with the final inner layer quarter turn slice, they shall be matched up.
[5] So now we have a 9/10 move (depending on how you count moves)
double parity algorithm which destroys the colors of 3 centers. Obviously the next step is to fix the centers. If you have ever used the 3 move sequence
Rw U2 Rw' to pair up centers (without worrying about dedge pairing), simply observe that those three moves "flip" the top front dedge and unpairs the top back and front bottom dedges (observe that the green-yellow wings are in the same slice and the white-blue wings are in the same slice.)
So, let's use this. The top center is complete, so let's keep it that way. If we observe the formation of the 3 discolored centers, if we want to use Rw U2 Rw' (and move equivalences of that) to fix up the centers, the only two choices we have in
our algorithm is to either do the move
F2 or
D2 first to line up the two green 1x2 blocks (we must solve the next center adjacent to the top center because the move sequence Rw U2 Rw' does not swap center pieces in opposite centers to be able to solve the centers should we instead pair up the bottom center). For a reason you will soon see, F2 is the proper choice for fewer moves.
So let's do F2 and one application of that 3 move sequence.
(Rw' U R' U Rw U F2 U Rw') F2 Rw F2 Rw'
[6] Now only two centers are discolored and two dedges are unpaired (which is actually great).
Although it may not be initially obvious, but the move
S2 will give us the same formation which we saw Rw U2 Rw' generate earlier. So we can now finish off the algorithm with that to get...
(Rw' U R' U Rw U F2 U Rw')(F2 Rw F2 Rw' S2 Rw' U2 Rw)
If we would have instead chose to do D2 (instead of the bold F2 move above, in step [5]), then the resulting path would be
(Rw' U R' U Rw U F2 U Rw') D2 Lw' F2 Lw S2 x', for which we would need to first do
S R2 S' to get the last two centers lined up to
finish off.
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