The inner box unpacks the outer box at the same time as the outer box
unpacks the inner.
-- The Ace of Heart

Previously,

In-place orientation fix?

Yeah, actually we have very similar algorithms alraedy! Algorithm 2, 2m and 2M fit but they fix Z stickers while we need some algorithm that fixes W stickers. Should we try to find such algorithms? Or, should we do some transformation to make those turquoise stickers go Z+/Z-, and then apply the 2-family algorithms?

Well, I believe the later is achievable. Recall the effect of Algorithm 2,

• W-O0/O7: X+->W-->Y+->X+
• W-O1/O3/O4/O6: X-/X+->Y+/Y-->W-->X-/X+
• W-O2/O5: unchanged

For now, only W-O1 is OK in both position and orientation, so it should go to O2 or O5 after the turquoise-to-Z setup is done. The setup can be done in the following manner: center Z+ (which should be all green but not yet), 180 X-axis-twist to W- (so that half of the turquoise goes to Z+, the done cubie, original W-O1 now at W-O2/Z+O5), and then 180 X-axis-twist to Z- (so that the other half of turquoise sticker face W-). As clip,

What is the goal for W- now? Well, If we perform the above setup sequence to a solved 2x2x2x2 puzzle, then the Z+ half of W- will be all green stickers; for similar reason, the Z- half will be all orange stickers. We can then foresee the outcome of Algorithm 2 that changes 6 cubies' orientation. It is lucky that (now) W-O0/O7 will automatically done after Algorithm 2, because their green and orange happen to be at X+; (now) W-O1/O3 are also good, because their green stickers are at Y. 4 out of 6 cubies will be settled at once, excellent!

Now, total number of twists: 216.

There are three left!

Theoretically we can apply Algorithm 2M twice to solve this. We are so close now. The first is to solve one of them and leave one that has the same orientation offset as the 16th cubie, so the second should be able to solve them.

pair 1

We still have to do some setup otherwise the W-O5/O6/O7 won't be transformed by Algorithm 2M. I choose to apply a quite different setup sequence here. I don't want to use the set-turquoise-to-Z setup because even if we use that, still refine twists are required to set them to the W-O3/O4.

Instead, I choose first to center Z-, the orange cell. Second, a 180 X-axis-twist to X+ (which brings W-O0 to W-O4). Finally, a counterclockwise Z-axis-twist to Z+ (which brings W-O2 to W-O3). Now, the Z cells are turquoise and pink, which are already settled. As follows,

Recall: the precise effect of Algorithm 2M is X+->Y-->W-->X+ for W-O3/O4, and they currently are

• W-O3: orange(X+)-gray(Y-)-blue(W-)
• W-O4: red-orange-yellow

Since W- = orange is the desired result, we can expect that after the sequence, W-O4 is settled. As expected, after resuming,

pair 2

Although the rest two cubies stick together, it doesn't mean that we should move them together. With the center color being orange, we should be able to set both out-of-center orange stickers together at X+O2/O5 so that we can apply Algorithm 2M. So the setup for two cubies must be seperated.

Once they are all there, applying Algorithm 2M for the last time should be the first 2x2x2x2 solve in my life. As the following clip,

Woo-Hoo~~!!!

Conclusion

How was that? How do you think about the PART II 2x2x2x2 journey? Was it exciting? At least it is not boring to me. Maybe we should have done PLL first and then OLL, since we need to do the final refinement like what we just did today.

Anyway, here I am now a real hypercuber. The main theme will finally start tomorrow! Yeah!!!

小結

生涯首次解決高維魔方：2x2x2x2。第二章完結。