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Entropy of a Rubik's Cube

  1. Jun 28, 2015 #1
    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?
     
  2. jcsd
  3. Jun 28, 2015 #2

    anorlunda

    Staff: Mentor

    It sounds like you have the max, the min, and the log relationships correct. You are talking about the information view of entropy, not the thermodynamic view.

    But I wonder if the number of faces is the correct measure as compared to the number of moves to solve it.

    Consider starting with a solved cube, then rotating the top level 90 degrees. One move, but 45 faces out of position.
     
  4. Jun 28, 2015 #3
    two things: I should've said this but by cubies I mean one of the smaller cubes which contains three tiles of different colors (if its a corner) or two different colors (if its an edge)

    Second, you're completely right. It is unclear whether this has any relation to the number of moves required to solve the cube. Because an algorithm which solves the cube in a minimum number of moves can only be found with the help of a computer an entropy based on those numbers doesn't give us very much insight into the cube.

    On the other hand basing entropy on the number of moves required using a specific rubiks cube solving strategy runs the risk of defining entropy for that strategy and not the rubiks cube in general.
     
  5. Jun 28, 2015 #4

    anorlunda

    Staff: Mentor

    I guess you could go either way. But given the definiton of your choice, can you use it to make some useful predictions?
     
  6. Jun 28, 2015 #5

    Vanadium 50

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    The Wikipedia article on Rubik's Cube enumerates the permutations: 43,252,003,274,289,856,000. That means the entropy is 45.2k. (Subtracting off the 24 correct solutions is negligible)
     
  7. Jun 30, 2015 #6
    Vanadium, at least a lot of those configurations would have some pieces in the correct position so the actually number for maximum entropy would be less.
     
  8. Jul 1, 2015 #7

    Vanadium 50

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    Why do you think the entropy would be less?
     
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