computational-geometry

I've come to a conceptually difficult problem. In short, I need to find the vector path of two vector paths combined through different boolean operations. Such as the Union, the Difference, the Intersection, and Subtraction. It would be nice if I could do it similar to how Canvas does it's globalCompositeOperation. How in the world would I go about doing this? There is a JavaScript library that allows for boolean operations on SVG paths under the condition that the path is a polygon. Using a high enough sampling, the beziers can be polygonized to such a high quality that the visual result is

I have a need for a Python module/package that provides a mesh on which I can do computational science? I am not doing graphics, so I don't think the blender package is what I want. Does anyone know of a good package? The most useful packages out there are perhaps mshr , pygalmesh , dmsh , pygmsh , and MeshPy . I've created a repo showcasing a number of examples. In addition, there is optimesh for improving the quality of any mesh. (Disclaimer: I'm the author of pygmsh, pygalmesh, dmsh, and optimesh.) If you're trying to solve FE or CFD style equations on a mesh you can use MeshPy in 2 and 3

I have a convex polygon ABCDE... (it can have any number of points). I need to sort all its vertexes so none of the edges will intersect. example: A _____ B \ / \ / X / \ /___\ C D That polygon in ABCD order has intersecting edges. however in ABDC order: A _____ B | | | | | | | | |___| C D None of the edges intersect so ABDC is the expected output. How can I do this? andand Assuming your points are all on the convex hull of your polygon, you can use the following: Pick the two extreme points with the min and max X value, (call them X min and X max ) and draw the line between them. In the case

I am trying to write an efficient algorithm that counts the number of points inside a Sphere of Radius R and Dimension D. The sphere is always at the origin. Suppose we have a sphere of dimension 2 (circle) with radius 5. My strategy is to generate all possible points within the first quadrant, so for the above example we know that (1,2) is in the circle, so must all + / - combinations of that point which is simply dimension squared. So for each point found in a single quadrant of an n-dimensional sphere we add 2 ^ dimension to the total count. I'm not sure if there is a much more efficient

I'm coding renderer for road network, which based on the RoadXML format. Road curves from this format has four types: segment, circle arc, poly line, clotho arc. And I have problem with the last one. Clothoid is the same with Euler spiral and Cornu spiral. In the RoadXML clotho arc is given by three parameters: start curvature, end curvature, length. For arc triangulation I need a function like foo(t), which returns (x, y) coords for t = 0..length. I created similar methods for circle arc without problems, but I can't do it for clotho arc. Part of the problem is that I not totally understand

I am extracting image features from 10 classes with 1000 images each. Since there are 50 features that I can extract, I am thinking of finding the best feature combination to use here. Training, validation and test sets are divided as follows: Training set = 70% Validation set = 15% Test set = 15% I use forward feature selection on the validation set to find the best feature combination and finally use the test set to check the overall accuracy. Could someone please tell me whether I am doing it right? So kNN is an exception to general workflow for building/testing supervised machine learning

In 2D plane, I have a point and a line. How to get the mirror point along this line? Suppose the equation of the line is ax + by + c = 0 . Now imagine a line perpendicular to it, which can be represented by -bx + ay + d = 0 (product of slopes of two perpendicular lines is -1). Now the problem is to find d . Put the co-ordinate of the point on the second line, and you'll get the value of d easily. The second part is, to find a point on the second line which is equidistant as the first point from the first line. For that, you can find the intersection of the two lines. Calculate the differences