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I was watching a video on Youtube of a youtuber named “Mathologer” (he’s a great guy, gi check it out!) about pigeonhole principle. It gave an example of how the pigeonhole principle could be applied to see that if you repeat a given algorithm to a solved Rubik’s cube you’ll eventually get back to a solved cube, and that the number of moves it takes must divide the number of configurations of the Rubik’s cube (it’s a huge number I won’t write here). This is basic algebra stuff about actions on a set or something like that, I am into analysis and I cannot remember if this result has a proper name but I surely have it written in some notes somewhere.

I was wondering though: is there an algorithm that before getting back to the solved cube takes every other configuration? In the video it was said that even the most complex algorithm must have a relatively small amount of moves (1260 at best), so I was wondering if even with the limitation of a quite simple and short algorithm like this (at least with respect to the number of configurations) we could get any state of the Rubik’s cube. Of course, after eventually proving its existence, I would be curious to know the shortest algorithm to do that, maybe with the moves written down.

tommy1996q
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  • Yeah, but the algorithm seems to be longer than 1260 moves. Maybe I read it wrong though, I’ll check it more carefully later – tommy1996q Jul 08 '21 at 00:03
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    1260 is the largest number of repetitions of any sequence of moves that will keep producing new configurations of the cube. So the length of a sequence that repeats to form a Devil's Algorithm must be a least $\frac{43252003274489856000}{1260}$ moves. – Troposphere Jul 08 '21 at 00:05
  • The reason why the number of moves it takes divides the number of configurations of the Rubik's cube is due to Lagrange's theorem; the structure of the Rubik's cube puzzle forms a group (the subgroup of $S_{48}$ generated by the 6 permutations that turn each face clockwise). – Prasun Biswas Jul 08 '21 at 00:12
  • Could you explain what exactly you mean by "before getting back to the solved cube takes every other configuration"? Because if a method takes all possible cube configurations before reaching the desired configuration, it is certainly not very useful as an algorithm to solve a puzzle because it is as best as working out by hand through trial-and-error with different move sequences each time from a starting position. – Prasun Biswas Jul 08 '21 at 00:13
  • @Prasun Biswa I mean that if you start, say, with the solved cube and start applying this algorithm over and over, before getting back to the solved cube you obtain every other possible configuration of the solved cube. It’s not about efficiency, it’s just that I am curious if it can be done and how – tommy1996q Jul 08 '21 at 08:17

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