Comment by kevindamm
4 days ago
This was done in 2010 thanks to the analysis of the symmetries inherent in the underlying group theory, and about 35 CPU-years:
4 days ago
This was done in 2010 thanks to the analysis of the symmetries inherent in the underlying group theory, and about 35 CPU-years:
For me, any position which requires the full maximum 20 moves to solve would qualify as the perfect shuffle.
I wonder how many positions qualify (perhaps only 1 discounting symmetry as there is 1 fully solved cube, again discounting symmetry). But that’s a rabbit hole that I am not going down.
But I can appreciate the choice made in the article.
> perhaps only 1 discounting symmetry as there is 1 fully solved cube
I don’t see how ”there is 1 fully solved cube” would even hint at “perhaps only 1”
Also, there isn’t only 1. https://www.cube20.org/: “Distance-20 positions are both rare and plentiful; they are rarer than one in a billion positions, yet there are probably more than one hundred million such positions. We do not yet know exactly how many there are”
A fun thing to think about is if you’re allowing generalizations to more than 3x3x3, there can be more than one fully-solved configuration. A 4x4x4 cube for example has 4 identical centre pieces on each face where the centre piece of a 3x3x3 cube is. I’m pretty sure these 4 pieces can be in any configuration as long as they are the correct colour and the cube is still completely solved. Likewise each edge piece in a 3x3x3 is replaced in a 4x4x4 with two “wings” which can be swapped without changing the fact that the cube is solved.
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There are about 490 million of them, the full list can be downloaded from that same cube20 site
https://www.cube20.org/distance20s
You can trim it down by counting in quarter moves.
That is, count a 180° move as two 90° moves. Which it is, though we usually don't think of it that way.
Assuming a random move length distribution, that would only leave a (2/3)^20 fraction of them, which is about 147000 positions.
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