[stoneman24]

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.

[Someone]

> 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”

[seanhunter]

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.

[Someone]

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-...

[seanhunter]

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

[kevindamm]

There are about 490 million of them, the full list can be downloaded from that same cube20 site

https://www.cube20.org/distance20s

[BurningFrog]

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.

[kevindamm]

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/