line-intersection

How line can be represented using Boost Geometry? I don't need finite segment, but I need infinite lines (maybe Segment or Linestring can be extended?) As I understand I can use boost::geometry::intersects , but I don't know how to define infinite line. If you want to test whether an infinite line A intersects a line segment B , this can be done using boost::geometry::strategy::side::side_by_triangle : template <typename Point> struct line { boost::geometry::model::segment<Point> segment; }; template <typename Point> bool intersects(line<Point> const& A, boost::geometry::model::segment<Point>

There are N line segments which are either Horizontal or vertical. Now I need to find out total number of Intersections and total number of Intersections per line segment. N can go upto 100000 . I tried checking every pair of lines. The answer is correct but I need to reduce it's time taken by it. Here's my code : using namespace std; typedef struct Point { long long int x; long long int y; } ; bool fun(Point p0, Point p1, Point p2, Point p3) { double s1_x, s1_y, s2_x, s2_y; s1_x = p1.x - p0.x; s1_y = p1.y - p0.y; s2_x = p3.x - p2.x; s2_y = p3.y - p2.y; double s, t; s = (-s1_y * (p0.x - p2.x)

I've seen many postings here on stackoverflow, which are discussing this topic. I took a solution from stackoverflow, but I couldn't find the posting. It was to say: If two lines are intersecting, then the crossproduct produces for the left and the right side two different results. One positive and one negative. Otherwise both have the same sign. So far it is okay. The used formula is, where AB is one line and CD another. dotproductleft = (B.x-A.x) (C.y-B.y)-(B.y-A.y) (C.x-B.x) dotproductright = (B.x-A.x) (D.y-B.y)-(B.y-A.y) (D.x-B.x) If I calculate this for the following given GPS coordinates

Given a list of line segments, the easiest way to find the intersection points is to loop through the line segment list, check whether they are intersecting and record the intersection point if they do. But the runtime of this method is O(n^2) , which is very inefficient. Is there any other algorithm that could speed up this process? The Bentley-Ottmann Algorithm may be what you are looking for. 来源： https://stackoverflow.com/questions/4125452/find-the-intersection-points-of-all-the-line-segments

I have two cubic bezier curve, curve 1:- 1st anchor-point(a1x,a1y), 1st control-point(c1x,c1y), 2nd control-point(c2x,c2y), 2nd anchor-point(a2x,a2y) curve 2:- 1st anchor-point(a3x,a3y), 1st control-point(c2x,c3y), 2nd control-point(c4x,c4y), 2nd anchor-point(a4x,a4y) Now I want to find the intersection points between these two bezier curve; How to do it? Any reference document with algorithm will help me; MBo There are two main methods to find a Bezier curve intersection: Recursive subdivision exploits the convex hull property of Bezier curves and usually checks the intersection of bounding

I've tried searching for a javascript function that will detect if two lines intersect each other. The function will take the x,y values of both the start end points for each line (we'll call them line A and line B). Is to return true if they intersect, otherwise false. Example of the function. I'm happy if the answer uses a vector object instead. Function isIntersect (lineAp1x, lineAp1y, lineAp2x, lineAp2y, lineBp1x, lineBp1y, lineBp2x, lineBp2y) { // JavaScript line intersecting test here. } Some background info: this code is for a game I'm trying to make in html5 canvas, and is part of my

How would I use numpy to calculate the intersection between two line segments? In the code I have segment1 = ((x1,y1),(x2,y2)) and segment2 = ((x1,y1),(x2,y2)). Note segment 1 does not equal segment2. So in my code I've also been calculating the slope and y-intercept, it would be nice if that could be avoided but I don't know of a way how. I've been using Cramer's rule with a function I wrote up in Python but I'd like to find a faster way of doing this. Hamish Grubijan Stolen directly from http://www.cs.mun.ca/~rod/2500/notes/numpy-arrays/numpy-arrays.html # # line segment intersection using