Deterministic algorithm to determine whether a point is in a polygon #17
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seangransee
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I made an additional change to go back to the way you were originally checking whether a point was on a boundary. When a point is not on a boundary, it uses the deterministic algorithm I found as a replacement for All tests in |
| vertices.each_with_index do |vertex, i| | ||
| previous_vertex = vertices[i - 1] || vertex.last | ||
| if ((vertex.y > point.y) != (previous_vertex.y > point.y)) && | ||
| (point.x < (previous_vertex.x - vertex.x) * (point.y - vertex.y) / (previous_vertex.y - vertex.y) + vertex.x) |
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Align the operands of a condition in an if statement spanning multiple lines.
Line is too long. [121/80]
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| vertices.each_with_index do |vertex, i| | ||
| previous_vertex = vertices[i - 1] || vertex.last | ||
| if ((vertex.y > point.y) != (previous_vertex.y > point.y)) && |
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Use next to skip iteration.
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Thanks @seangransee! Will look into it tomorrow 👍 |
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Status update: still trying to figure out the idea of the algo author. From the description, I conclude that it is the same algorithm as the current one in this gem, but instead of a random ray they choose a horizontal positively-oriented ray, the rest is the same (they just calculate whether the number of intersections with edges is odd or even). If I understood everything correctly, why would it work well with edge-cases when the semi-infinite ray contains a vertex or a whole edge? @seangransee, can you please help me to understand? I'd really appreciate assistance as it would save me a lot of time digging/drawing/testing/thinking. |
I haven't gotten a chance to spend much time with this, but I've dropped in the algorithm I'm using in my application to replace
Polygon.contains?.There's one failing test, which is the check for a point being on the edge of a convex polygon. As stated in the section Point on a (Boundary) Edge from the algorithm's source page, this algorithm isn't able to predict when a point is exactly on the edge of a polygon. Perhaps a hybrid of this new algorithm with your current algorithm will do the trick.
Feel free to mess around with this and push commits to my branch.