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I Dimension using box counting technique

  1. Apr 14, 2017 #1
    On an exam we just took, we were asked to find the dimension of a set using the box counting technique. So choose an epsilon, and cover your object in boxes of side length epsilon, and count the minimum number of boxes required to cover the object. Then use a smaller epsilon and and count the new minimum number of boxes and find the dimension using $$D=\frac{log(N_{i+1}/N_i)}{log(\epsilon_i/\epsilon_{i+1})}$$.

    The problem given was essentially the cantor set, except instead of removing the middle third of each layer, you removed the middle fifth. So if the first segment went from ##[0, 1]## then the second segment was ##[0, \frac{2}{5}]\cup[\frac{3}{5}, 1]## and the third was ##[0, \frac{4}{25}]\cup[\frac{6}{25}, \frac{2}{5}]\cup[\frac{3}{5}, \frac{19}{25}]\cup[\frac{21}{25}, 1]## and so on.

    Here's where the issue comes in. This should work no matter the epsilon sequence you choose, but we are getting 2 separate answers.

    Using ##\epsilon=\frac{1}{5}## on the third segment requires ##\frac{4*\frac{4}{25}}{\frac{1}{5}} \approx 4## boxes. and ##\epsilon=\frac{1}{25}## on the same segment requires ##\frac{4*\frac{4}{25}}{\frac{1}{25}} = 16## boxes. This would then give ##D = \frac{log(\frac{16}{4})}{log(\frac{1/25}{1/5})} = \frac{log(4)}{log(5)}## which is what most people found, including the professor.

    However, using ##\epsilon=\frac{2}{5}## on the third segment requires ##\frac{4*\frac{4}{25}}{\frac{2}{5}} \approx 2## boxes. and ##\epsilon=\frac{4}{25}## on the same segment requires ##\frac{4*\frac{4}{25}}{\frac{4}{25}} = 4## boxes. This would then give ##D = \frac{log(\frac{4}{2})}{log(\frac{4/25}{2/5})} = \frac{log(2)}{log(\frac{5}{2})}## which is what I thought the answer should have been.

    Since there is a discrepancy here, there is also a formula using fractals given in the book (however the problem specifically said to use the box counting method) which gives the dimension to be $$D = \frac{log(N)}{log(\epsilon)}$$ but here ##N## is the number of self similar copies and ##\epsilon## is the length of the original relative to the copies (so essentially the factor to scale from the copies to the original). Using THIS method on the third segment, gives 4 copies, each of length ##\frac{4}{25}## relative to the original. So ##N=4## and ##\epsilon=\frac{25}{4}##. Plugging into the formula gives ##D=\frac{log(4)}{log(\frac{25}{4})} = \frac{2*log(2)}{2*log(\frac{5}{2})} = \frac{log(2)}{log(\frac{5}{2})}##. Which matches what I believe the answer should have been.

    Ultimately, my question here is why is there this discrepancy. How can using 2 different sequences of ##\epsilon## and ##N(\epsilon)## yield 2 completely different dimensions for the same set? Is there a reason that a particular ##\epsilon##-sequence does not work? I read that having an epsilon smaller than your "particle" or "cell" doesn't contribute new information so it essentially would not work, but using the same ##\epsilon##-sequences on the second segment rather than the third yields the same results.
     
  2. jcsd
  3. Apr 14, 2017 #2

    mfb

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    Staff: Mentor

    With ##\epsilon_i = 0.4^i##, you cannot cover [1,0] with an integer number of boxes, and the boxes you choose are not arranged in a regular pattern. I'm quite sure you cannot do that.
     
  4. Apr 14, 2017 #3
    Using ##0.4^i## is what matches the dimension formula though, so I would assume that it actually works. But as far as I understand from the method, it doesn't have to be an integer number of boxes, you just have to round up if you get a decimal because that is the minimum number of boxes required. For example, the third segment with ##\epsilon=0.2## doesn't use an integer number of boxes either. The book in my senior level math class, is written for non math majors....so it doesn't exactly explain the fine details of how things work. So it could just be coincidence then that ##\epsilon_i=0.4^i## actually worked.
     
  5. Apr 15, 2017 #4

    mfb

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    You always cover the "final segment". You don't care about what you call "first segment", "second segment" and so on. You cover the final set, and you count all boxes you have to include to cover it.

    In the case of ##\epsilon_i = 0.4^i##, your first step is ##\epsilon=1##, and you use the box from 0 to 1 to cover all of it. Your second step are 0.4 boxes. You need the [0, 0.4] box, the [0.4, 0.8] box and the [0.8,1.2] box. Three boxes. But that number ultimately doesn't matter, because you are looking at a limit process.
    With 0.16 boxes, you can leave at least one gap in the middle. With 0.064 boxes, you can leave even more gaps, so you slowly save boxes (leading to a dimension smaller than 1).
     
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