Chaos was the law of nature. Law was the dream of man.
-- Henry Adams

Yesterday we were just developing some sequence to be used, so the status remains. Total twists: 150.

Recall (the arrow stands for "should go to")

The status,

• W-O0 red-yellow-orange -> W-O7
• W-O1 blue-orange-red -> W-O6
• W-O2 green-blue-gray OK
• W-O3 gray-green-yellow OK
• W-O4 blue-green-red -> W-O1
• W-O5 blue-gray-orange OK
• W-O6 green-yellow-red -> W-O0
• W-O7 gray-yellow-orange -> W-O4

Algorithm 4XY,

X: A clockwise Y-axis-twist to Y+ followed by a counterclockwise X-axis-twist to X+
Y: A counterclockwise X-axis-twist to X+ followed by a counterclockwise Z-axis-twist to Z+
B: A clockwise Y-axis-twist to W- followed by a counterclockwise X-axis-twist to W- 

X B X' YYY B Y'Y'Y'

The effect of 4XY,

• The cubies on the rotation axis
  • W-O0/O5 stays but orientation X->Z->Y->X
• Other cubies that stay
  • W-O1 X->Y->Z->X
  • W-O4/O6 X->Z->Y->X
• 3-cycle movement
  • W-O2 goes to W-O7, orientation Y<->Z
  • W-O7 goes to W-O3, orientation X->Y->Z->X
  • W-O3 goes to W-O2, orientation X<->Y

Setup 1

As usual, we have decided the core algorithm being 4XY, so the tricky part remaining is the setup. The 5 non-settled cubies forms a 5-cycle, so we should run the sequence at least twice.

I would like to start with W-O0/O1/O6, because they should go to W-O7/O6/O0 respectively. We are not trying to find a perfect match, but to break the 5-cycle. It seems to me that choosing these three is a good choice, because we can settle W-O1/O6 at once.

Consider 4XY where the 3-cycle is W-O2/O7/O3. The challenge is how we do the setup. A viable way is to move O6 to O2, O0 to O7 (so that O6@O2->O0@O7), and O1 to O3 (so that O1@O3->O6@O2). This can be done in three twists:

1. counterclockwise Z-axis-twist to W-
2. clockwise Y-axis-twist to W-, settling O1@O3 and O6@O2
3. clockwise Y-axis-twist to Y+, settling O0@O7

The resuming steps should be performed in a reversed order. After all these are done, we have

• W-O0 yellow-red-green OK!
• W-O1 yellow-red-orange -> W-O7
• W-O2 blue-gray-green OK!
• W-O3 yellow-gray-green OK!
• W-O4 green-red-blue -> W-O1
• W-O5 gray-orange-blue OK
• W-O6 red-orange-blue OK
• W-O7 orange-gray-yellow -> W-O4

Setup 2

Its even trickier because we need a rotation of W-O1/O7/O4. Structurally speaking, it would be easier if O7/O4 pair can be set to O3/O2, because they are adjacent now. It turns out that it can be easily done in two twists,

1. clockwise X-axis-twist to W-
2. counterclockwise Y-axis-twist to W-

After the setup up, 4XY sequence, and resume steps, we have

• W-O0 red-green-yellow ... need X->Y->Z->X
• W-O1 blue-red-orange OK!
• W-O2 gray-green-blue ... need X->Y->Z->X
• W-O3 green-yellow-gray ... need X->Z->Y->X
• W-O4 orange-yellow-gray ... need X->Z->Y->X
• W-O5 orange-blue-gray ... need X->Z->Y->X
• W-O6 orange-blue-red ... need X->Z->Y->X
• W-O7 orange-yellow-red ... need X->Z->Y->X

Conclusion

I am not sure which of the following statements is correct: (1) In 2x2x2x2, not like 2x2x2, OLL followed by PLL is not a convenient sequence of solving the puzzle; the player still has to fix the in-place orientation issue after PLL; (2) We just have a bad PLL... Algorithm 4XY affects all cubies! There must be some thing that is better.

Hmm... in-place orientation sounds familar, isn't it? I believe we have seen it before?

小結

使用 4XY，確保所有 W- 方塊位置歸位，但仍有方向問題待解。