I needed 6 decimal places of accuracy, so first way of solution will not work for my case. However, your second strategy seems promising. Working on it. Thanks :D ~l0nwlf
On Thu, Feb 4, 2010 at 5:49 PM, Bearophile <bearophileh...@lycos.com> wrote: > Shashwat Anand: > > > Given 'n' circles and the co-ordinates of their center, and the radius > of > > > all being equal i.e. 'one', How can I take out the intersection of > their > > > area. > > I can see two possible solutions, both approximate. In both solutions > you first look if there are a pair of circles that don't intersect, in > this case the intersect area is zero. You also remove all circles > fully contained in another circle. > > The first strategy is easy to do, but it probably leads to a lower > precision. Then you can sample randomly many times the rectangular > area that surely contains all circles. You count how many of those > random points are inside all circles. This gives you an approximate > area of their intersection. Increasing the points numbers slowly > increases the result precision. > > The second strategy is more complex. You convert your circles into > polygons with a fixed number of vertices (like 50 or 100 or 1000 or > more. This number is constant even if the circles don't have all the > same radius). So you can turn this circle into a simple mesh of > triangles (all vertices are on the circumference). Then you "subtract" > the second polygonalized circle, this can create a hole and split > triangles in pieces, and so on with successive circles. At the end you > can compute the total area of the triangles left. This is doable, but > you need time to do implement this. The advantage is that the > numerical precision of the result is probably higher. If you implement > this second solution you can implement the first one too, and use it > as a test to avoid bugs. Visualizing the triangles with Pygame or > MatPlotLib can be useful to look for bugs. > > Bye, > bearophile > -- > http://mail.python.org/mailman/listinfo/python-list >
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