- #1
edenstar
- 11
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I have been thinking about finding a way to define entropy on a rubik's cube. My idea is to use the number of cubies that are not in their solved position as the macrostate. This works well because there is exactly one way for all the cubies to be in the solved position so it has entropy of zero. On the other hand if the cubies are all in incorrect positions then this macrostate encompasses every state but the solved state so the entropy of that macrostate is maximum.
As an example just consider the corner cubies without considering their orientation:
If 6 cubies are in the correct position and 2 cubies are incorrect then there are 8x7 ways that this can be possible. so the entropy would be -k ln(56). In general the entropy would be -k ln(T/C)
where T is the total number of configurations and C is the number of those configurations which are correct.
Does this seem good to anybody? Does this give us any insight about the rubiks cube? Also do you think I should've posted this in a different section?
As an example just consider the corner cubies without considering their orientation:
If 6 cubies are in the correct position and 2 cubies are incorrect then there are 8x7 ways that this can be possible. so the entropy would be -k ln(56). In general the entropy would be -k ln(T/C)
where T is the total number of configurations and C is the number of those configurations which are correct.
Does this seem good to anybody? Does this give us any insight about the rubiks cube? Also do you think I should've posted this in a different section?