Remove require calls #33
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mauriciopoppe
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Jul 27, 2024
… the origin is not in the polygon The PR #33 added detection for an edge case where all the points are part of a plane, the detection algorithm was: 1. Compute the initial tetrahedron containing 4 vertices. 2. Taking any 3 vertex from it create a plane. 3. Project all the points towards the plane using dot product, if the dot product is greater than some tolerance (not zero because of float point errors) then it means that a point is not part of plane. There is an error in (3), let's assume a simple input in 2d where all the points are not a polygon. - [2, 0], [2, 1], [2, -1] (they're points in x=2) - [1, 0] which is a point in x=1 An ascii diagram: ``` 1 x | | -0---x---x | | -1 x x=0 x=1 x=2 ``` If step (2) detected that all the points in x=2 were part of the plane where the normal is [1, 0] then by following (3) and projecting point [1, 0] to the plane (note that the point is not the normal) would result on it being behind the face and therefore it pass the check and consider the input to be a plane. The problem in (3) is that we should compare the projection of the point to test [1, 0] to the projection of any plane point to the plane - project [2, 0] to the plane with a normal [1, 0] is dot([2,0], [1,0]) = 2 - project [1, 0] to the plane with a normal [1, 0] is dot([1,0], [1,0]) = 1 - The diff is 1 and is greater than any tolerance (typically tolerance is a tiny value like 1e-9)
mauriciopoppe
added a commit
that referenced
this pull request
Jul 27, 2024
… the origin is not in the polygon The PR #33 added detection for an edge case where all the points are part of a plane, the detection algorithm was: 1. Compute the initial tetrahedron containing 4 vertices. 2. Taking any 3 vertex from it create a plane. 3. Project all the points towards the plane using dot product, if the dot product is greater than some tolerance (not zero because of float point errors) then it means that a point is not part of plane. There is an error in (3), let's assume a simple input in 2d where all the points are not a polygon. - [2, 0], [2, 1], [2, -1] (they're points in x=2) - [1, 0] which is a point in x=1 An ascii diagram: ``` 1 x | | -0---x---x | | -1 x x=0 x=1 x=2 ``` If step (2) detected that all the points in x=2 were part of the plane where the normal is [1, 0] then by following (3) and projecting point [1, 0] to the plane (note that the point is not the normal) would result on it being behind the face and therefore it pass the check and consider the input to be a plane. The problem in (3) is that we should compare the projection of the point to test [1, 0] to the projection of any plane point to the plane - project [2, 0] to the plane with a normal [1, 0] is dot([2,0], [1,0]) = 2 - project [1, 0] to the plane with a normal [1, 0] is dot([1,0], [1,0]) = 1 - The diff is 1 and is greater than any tolerance (typically tolerance is a tiny value like 1e-9)
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