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Comment by kevindamm | Hacker News Reader

This was done in 2010 thanks to the analysis of the symmetries inherent in the underlying group theory, and about 35 CPU-years:

https://www.cube20.org/

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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.

A 3×3×3 already has multiple such solutions. The center squares can rotate over 90° without visible changes. See https://de.speedcube.com.au/blogs/speedcubing-solutions/how-...

That’s interesting. I originally thought these were isomorphic to rotating the entire cube but of course that doesn’t take into account cases where they rotate relative to each other

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.

That page also shows their results for quarter turns, there are 3 (or fewer) with 26 quarter turns being the shortest they can be solved in. 36 (or fewer) needing 25 quarter-turns.

https://www.cube20.org/qtm/

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