← Back to context Comment by kevindamm 5 days ago There are about 490 million of them, the full list can be downloaded from that same cube20 sitehttps://www.cube20.org/distance20s 2 comments kevindamm Reply BurningFrog 5 days ago 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 5 days ago 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/
BurningFrog 5 days ago 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 5 days ago 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/
kevindamm 5 days ago 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/
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/