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I'm looking for the easiest/fastest (computationally) way to determine whether a shape, more specifically a GeneralPath object, contains any given line segment (of type Line2D.Double).

GeneralPath has methods to determine whether a point or a rectangle is contained, but not a line (that I can find). The line can be at a slant, so I can't just model a really thin rectangle. One point in the line will definitely be inside the shape, but I need to check the rest of the line segment. So potentially, I could also check to see if the line intersects any of the boundary edges, but I'm not sure how that would look just given the shape and the line.

scaevity
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2 Answers2

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Does your GeneralPath contain straight line segments or quadratic or bezier segments? This matters a lot. The algorithm for the straight line is the most straightforward:

Iterate over all points in the path. If two consecutive points are on opposite sides of the line, you have a potential crossing. Then you need to check if the line segment end point is inside or outside the shape relative to the potential crossing, which you get by solving for the intersection of two points (the line and the line formed by the two consecutive points) and seeing if the result is contained in your line segment.

Unfortunately, the curved paths can have two consecutive points with a parenthesis shape between them ")", which your line can pass through while still keeping the iterable points on the same side. If you can get the format with two end points and a single (double) control point(s) that form a bounding triangle (quadrilateral), you can get easy simple solutions (since the curve is guaranteed to fit inside the triangle/quad formed by the three/four points, as long as the line doesn't intersect the triangle/quad, you are good). Unfortunately, this has an ugly part as well--if the line does intersect the triangle/quad, you aren't guaranteed anything and have to check closer. As a double unfortunate, I don't know a technique other than normalizing the coordinate system and solving for the zeroes. And that's something I'd look up in a book that I can't seem to find (or wait until another nice SO poster comes along).

... actually, since the curvature properties of the shapes are invariant under rotation, for the closer inspection part you can just rotate the curve points (whether 3 or 4) to be axis aligned. Then do your skinny rectangle trick. That's probably not the cleanest, but it's the most obvious trick.

Come to think of it, why not do the rotation of all the points in the first place? The whole intersection problem is rotation invariant. That would save a lot of code. Just axis-align the line, apply the transformation to the shape, and do your cheeky rectangle trick.

ccoakley
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  • The path is made up of just straight line segments, so it's not overly complicated. I'll try what you suggested with iterating, thanks! – scaevity Jun 14 '11 at 04:39
  • The two points on the same side of the line part can be googled, BTW. The cross-product technique is the easiest to turn into code. – ccoakley Jun 14 '11 at 04:44
  • Oh, and if you want a quick proof of *why* you need to iterate: consider an algorithm that skips a single point. For the two points on opposite sides of the skipped point, assume they are on the same side of the line segment. Place the skipped point so that the path crosses the line segment. Alternatively, assume the two points are on opposite sides of the line. Move the skipped point out so that the path bends around the line. So no algorithm can skip a point and be correct with just the path information. – ccoakley Jun 14 '11 at 04:50
  • ok, so I think I understand exactly what I need to do... but I'm not sure how to iterate over every point in a path (or a line segment). any ideas/suggestions? – scaevity Jun 14 '11 at 06:37
  • Well, there's an SO question for that: http://stackoverflow.com/questions/5803111/obtain-ordered-vertices-of-generalpath – ccoakley Jun 14 '11 at 06:41
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Unless I am missing something, why can't you check if the the path contains x1,y1 and x2,y2 and AND the two as follows:

generalPath.contains(line.getX1(),line.getY1()) &&
generalPath.contains(line.getX2(),line.getY2())
Paul W
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Ram
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  • almost, except I don't just want the endpoints to be inside the shape, but every single point between the ends of the line segment (which isn't always the same thing with complicated shapes). – scaevity Jun 14 '11 at 06:07