computational-geometry

I have a question very similar to this: How to know if a line intersects a plane in C#? I am searching for a method (in C#) that tells if a line is intersecting an arbitrary polygon. I think the algorithm by Chris Marasti-Georg was very helpful, but missing the most important method, i.e. line to line intersection. Does anyone know of a line intersection method to complete Chris Marasti-Georg's code or have anything similar? Is there a built-in code for this in C#? This method is for use with the Bing Maps algorithm enhanced with a forbidden area feature. The resulting path must not pass

I have bought a nice little book about computational geometry. While reading it here and there, I often stumbled over the use of this special kind of binary search tree. These trees are balanced and should store the data only in the leaf nodes, whereas inner nodes should only store values to guide the search down to the leaves. The following image shows an example of this trees (where the leaves are rectangles and the inner nodes are circles). I have two questions: What is the advantage of not storing data in the inner nodes? For the purpose of learning, I would like to implement such a tree.

Good afternoon. My situation: In two-dimensional space . Input : a set of rectangles ( overlapping rectangles too). Rectangles coordinates are integer type . There are not any constraints on rectangle-size and rectangle-location (only extent of integer). No rectangles have width=0 or height=0. I need to find : all rectangles that contain entered point (with integer coordinates). Questions: What is the efficient structure to keep rectangles? What alghorithm is efficient in this case? And what algorithm is efficient only for adding rectangles without removing? Thanks :-). R-Tree is the best data

I am trying to determine the (x,y,z) coordinates of a point p. What I have are the distances to 4 different points m1, m2, m3, m4 with known coordinates. In detail: what I have is the coordinates of 4 points (m1,m2,m3,m4) and they are not in the same plane: m1: (x1,y1,z1), m2: (x2,y2,z2), m3: (x3,y3,z3), m4: (x4,y4,z4) and the Euclidean distances form m1->p, m2->p, m3->p and m4->p which are D1 = sqrt( (x-x1)^2 + (y-y1)^2 + (z-z1)^2); D2 = sqrt( (x-x2)^2 + (y-y2)^2 + (z-z2)^2); D3 = sqrt( (x-x3)^2 + (y-y3)^2 + (z-z3)^2); D4 = sqrt( (x-x4)^2 + (y-y4)^2 + (z-z4)^2); I am looking for (x,y,z). I

first question to StackOverflow, please be gentle. I am trying to find the equation (and then the algorithm for) the center point of three different points on a 2D cartesian plane, given a certain magnitude or "signal strength". These signal strengths are all on a scale relative to each other, but shouldn't necessary be conflated with "radius" of a circle. Wikipedia entry on trilateration: http://en.wikipedia.org/wiki/Trilateration I've also checked out this thread, but it's a little different than what I need Trilateration using 3 latitude and longitude points, and 3 distances A general

My requirements: A user should be able to draw something by hand. Then after he takes off his pen (or finger) an algorithm smooths and transforms it into some basic shapes. To get started I want to transform a drawing into a rectangle which resembles the original as much as possible. (Naturally this won't work if the user intentionally draws something else.) Right now I'm calculating an average x and y position, and I'm distinguishing between horizontal and vertical lines. But it's not yet a rectangle but some kind of orthogonal lines. I wondered if there is some well-known algorithm for that,

I'm searching for an algorithm for path simplification and smoothing for 2D trajectories. So I have a ordered list of 2D points. These points should be simplified, e.g. with the Ramer–Douglas–Peucker algorithm. But the output must be smooth, so the resulting path should be constructed from Bezier curves or splines. Is there any modification of the Ramer–Douglas–Peucker algorithm which could handle this? I found a path simplification algorithm in the paper.js library, which does exactly what I'm searching for: http://paperjs.org/examples/path-simplification/ But I was not able to understand the