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Total circles area: Difference between revisions
Added notes to the second Haskell version
(→Python: 2D Van der Corput sequence: As mentioned on the talk page, this version does more ...) |
(Added notes to the second Haskell version) |
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This is how this solution works:
# Given a list of circles, give all of them the same winding. Say, imagine every circle turns clockwise.
# Find all the intersection points among all circles. Between each pair, there may be 0, 1 or 2 intersections. Ignore 0 and 1, we need only the 2 case.
# For each circle, sort its intersection points in clockwise order (keep in mind it's a cyclic list), and split it into arc segments between neighboring point pairs. Circles that don't cross other circles are treated as a single 360 degree arc.
# Imagine all circles are on a white piece of paper, and have their interiors inked black. We are only interested in the arcs separating black and white areas, so we get rid of the arcs that aren't so. These are the arcs that lie entirely within another circle. Because we've taken care of all intersections eariler, we only need to check if any point on an arc segment is in any other circle; if so, remove this arc. (The haskell code uses the middle point of an arc for this, but anything other than the two end points would do).
# The remaining arcs form one or more closed paths. Each path's area can be calculated, and the sum of them all is the area needed. Each path is done like the way mentioned on that stackexchange page cited somewhere. This works for holes, too. Suppose the circles form an area with a hole in it; you'd end up with two paths, where the outer one winds clockwise, and the inner one ccw. Use the same method to calculate the areas for both, and the outer one would have a positive area, the inner one negative. Just add them up.
There are a few concerns about this algorithm: firstly, it's fast only if there are few circles. Its complexity is maybe O(N^3) with N = number of circles, while normal scanline method is probably O(N * n) or less, with n = number of scanlines. Secondly, step 4 needs to be accurate; a small precision error there may cause an arc to remain or be removed by mistake, with disastrous consequences. Also, it's difficult to estimate the error in the final result. The scanline or Monte Carlo methods have errors mostly due to statistics, while this method's error is due to floating point precision loss, which is a very different can of worms.
=={{header|Perl 6}}==
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