From a purely mathematical point of view it's just as easy to think in 11 dimensions, as it is to think in three or four.
-- Stephen Hawking

Previously,

We haven't been moving for a long time now. I reset the result of applying Algorithm 2M yesterday, back to the status where the first cell was just finished.

Applying Algorithm 2M

Recall the net result of the algorithm,

• W-O3 was originally yellow-gray-green-turquoise, now gray-turquoise-green-yellow
• W-O4 was originally green-gray-turquoise-blue, now gray-blue-turquoise-green

Others (the whole other 14 cubies!) remain unchanged. As analysized before, the Z stickers are locked. The 3-cycle nature is thus the rotation of X+->W-->Y-->X+. As long as we can provide some setup sequence so that turquoise stickers are at X+O2 or X+O5, 2M can bring that to W-.

I will skip the gif animation today because the only core sequence is 2M. The setups, however, are quite different different, so I will provide the result each time we apply a 2M, which hopefully set one or two turquoise stickers to W-.

1. too many turquoise stickers in Z-

Let's deal with it! So the setup here will be counterclockwise Y-axis-twist to W-, bringing the three turquoise stickers to X+. This setup doesn't have to be canceled because a pure W- twist won't affect the cell we have done before at all.

The result is as follows,

This is just smooth.

2. one more pair

The turquoise stickers at W- form a weird shape. W-O0/O2/O3/O4, not very convenient huh? But theoretically no matter how complicated the setup sequence we use, we will be able to reverse it afterwards. We should consider how we can make the room of two adjacent non-turquoise stickers at W-O3/O4, and at the same time, we can also sort out the out-of-W- turquoise stickers, settling them at X+O2/O5.

The description above looks scary, but actually if we can make the second condition true, the first automatically holds. So let's focus on making a setup sequence. The turquoise stickers are now at X+O6, Y+O5, Z-O2 and Z+O6.

OK, I decide to make the setup counterclockwise Y-axis-twist to Y- so that Z-O2 goes to X+O5 joining X+O6; then, counterclockwise X-axis-twist to W- so that the two turquoise stickers are ready. For recovery later, only the first part is strictly required.

The result is as follows,

Actually, MC4D provides a feature called "macro", but I haven't learnt how to use it yet. 24 twists for me is still affordable.

3. final pair!

That looks promising, right? The rest two turquoise are at X-O3 and Y+O5. We can of course stick to the plan to move them together an settle them at X+O2/O5. This will require 5 twists. The other one is simpler and not too complicated to execute, which is simply a counterclockwise Z-axis-twist to Z-, which brings Y+O5 to X-O4 and perfectly fits a mirrored Algorithm 2M. In other words, 2M is Algorithm 2m follows Algorithm 2 at its core, but the mirrored version is the oposite.

The result is as follows,

Conclusion

Fun facts:

• The number of total twists from the beginning to finishing the first cell is 70.
• The core of Algorithm 2M requires 24.
• Now, finishing all turquoise at W-, the number of twists is 146.

Easier than expected, right? It is good to be optimistic, but as OLL finished, the future became full of unknown challenges again. Can we even find any useful PLL algorithms? Will they easily apply to the situation we are facing now? Let's move on!

小結

使用昨天開發的 2M 公式，將青松色全數歸位 W-。