Not a solution, just a brainstorming idea: IIRC one common way to get approximate solutions to the TSP is to start with a random configuration, and then applying local operations (e.g. “swapping” two edges in the path) to try and get shorter and shorter paths. (Wikipedia link) I think something similar would be possible here: … Read more
Your best bet is to iterate over all the lines and find the minimum distance from a point to a line segment. To find the distance from a point to a line segment, you first find the distance from a point to a line by picking arbitrary points P1 and P2 on the line (it … Read more
Let’s label the points q = (x1, y1) and q + s = (x2, y2). Hence s = (x2 − x1, y2 − y1). Then the problem looks like this: Let r = (cos θ, sin θ). Then any point on the ray through p is representable as p + t r (for a scalar … Read more
I have a very similar piece of code. It got me very close results when compared to another implementation. I think the problem with yours is that you are using “distance” as linear distance in meters instead of angular distance in radians. /// <summary> /// Calculates the end-point from a given source at a given … Read more
The shape you have is one of so called “Goldberg polyhedra”, is also a geodesic polyhedra. The (rather elegant) algorithm to generate this (and many many more) can be succinctly encoded in something called a Conway Polyhedron Notation. The construction is easy to follow step by step, you can click the images below to get … Read more
To give you some more choices of libraries out there: Polyboolean. I never tried this one, but it looks promising: http://www.complex-a5.ru/polyboolean/index.html General Polygon Clipper. This one works very well in practice and does triangulation as well as clipping and holes holes: http://www.cs.man.ac.uk/~toby/alan/software/ My personal recommendation: Use the tesselation from the GLU (OpenGL Utility Library). The … Read more
Assuming that point1 and point2 are different, first you check whether the point lies on the line. For that you simply need a “cross-product” of vectors point1 -> currPoint and point1 -> point2. dxc = currPoint.x – point1.x; dyc = currPoint.y – point1.y; dxl = point2.x – point1.x; dyl = point2.y – point1.y; cross = … Read more
You can use FEniCS: from fenics import ( UnitSquareMesh, FunctionSpace, Expression, interpolate, assemble, sqrt, inner, grad, dx, TrialFunction, TestFunction, Function, solve, DirichletBC, DomainBoundary, MPI, XDMFFile, ) # Create mesh and define function space mesh = UnitSquareMesh(100, 100) V = FunctionSpace(mesh, “Lagrange”, 2) # initial guess (its boundary values specify the Dirichlet boundary conditions) # (larger … Read more
The problem described can be solved as follows. Let M = m_11 m_12 m_13 m_21 m_22 m_23 m_31 m_32 m_33 denote the desired rotation matrix. We require 1 0 0 * M + t = x_x x_y x_z 0 1 0 y_x y_y y_z 0 0 1 z_x z_y z_y where t denotes the translation; … Read more
Here is a sketch of an O(n^2 log n) solution. First, the preliminaries shared with other answers. When we have a line passing through some rectangles, we can translate it to any of the two sides until it passes through a corner of some rectangle. After that, we fix that corner as the center of … Read more