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I have a polygon of which I know:

  • Area
  • $x_{\max}$, $x_{\min}$
  • $y_{\max}$, $y_{\min}$

and I would like to establish to what extend the polygon can be considered a circle.

From what I found, for calculating the circularity or the compactness I need to know the perimeter, which I don't have. So far, my idea is to calculate the roundness:

    4 * Area / ( Pi * Major axis)^2

Is there anything else I could calculate?

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  • $\begingroup$ What do you mean by "X,Y max and min values"? By the way, I think your title don't reflect your question very well... $\endgroup$
    – Censi LI
    May 8, 2015 at 13:16
  • $\begingroup$ @CensiLI The maximum and minimum values of the polygon in both the $x$ and $y$ axis. Any suggestion for the title? $\endgroup$
    – Daniel
    May 8, 2015 at 14:10
  • $\begingroup$ Oh, I don't how to express "to what extend the polygon can be considered a circle" in shorter way, but I think "circularity" doesn't convey your intended meaning . Maybe you could try a straightforward, long title? $\endgroup$
    – Censi LI
    May 8, 2015 at 15:19
  • $\begingroup$ By the way given only such conditions, I think I can construct polygons which are as far away from circle as possible... (recall Koch snowflake) $\endgroup$
    – Censi LI
    May 8, 2015 at 15:23
  • $\begingroup$ @CensiLI a Koch snowflake could be acceptable by my standards. The method is to be applied on a computer vision program, which will often run with noisy images (I'm filtering the image) and missing parts of the circle. $\endgroup$
    – Daniel
    May 9, 2015 at 15:48

1 Answer 1

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How about comparing $x_{\max}-x_{\min}$ with $y_{\max}-y_{\min}$?

We can say the followings :

  • If it is a circle, then $x_{\max}-x_{\min}=y_{\max}-y_{\min}$.
  • If $x_{\max}-x_{\min}\not=y_{\max}-y_{\min}$, then it is not a circle.

Also, how about comparing $\pi$ with $\frac{\text{area}}{\left(\frac{x_{\max}-x_{\min}}{2}\right)^2}$?

We can say the followings :

  • If it is a circle, then $\pi=\frac{\text{area}}{\left(\frac{x_{\max}-x_{\min}}{2}\right)^2}$.

  • If $\pi\not =\frac{\text{area}}{\left(\frac{x_{\max}-x_{\min}}{2}\right)^2}$, then it is not a circle.

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  • $\begingroup$ Area +1. As for the $\Delta x = \Delta y$, I could also compare the midpoints of $x$ and $y$, which should but coincide, right? The problem with these approaches is that the solution is binary, 1 or 0, and I would like to "measure" somehow the similarity with a circle, e.g. 0.9 is more similar than 0.8 $\endgroup$
    – Daniel
    May 8, 2015 at 14:08
  • $\begingroup$ @kender6: Sorry, but I don't understand what 'Area +1' means. And yes, if it is a circle, then the midpoints of $x$ and $y$ coincide. $\endgroup$
    – mathlove
    May 8, 2015 at 16:57
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    $\begingroup$ I wanted to say that I like the idea of comparing PI with the area. $\endgroup$
    – Daniel
    May 9, 2015 at 15:42

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