computational-geometry

How can I draw a perpendicular on a line segment from a given point? My line segment is defined as (x1, y1), (x2, y2), If I draw a perpendicular from a point (x3,y3) and it meets to line on point (x4,y4). I want to find out this (x4,y4). I solved the equations for you: k = ((y2-y1) * (x3-x1) - (x2-x1) * (y3-y1)) / ((y2-y1)^2 + (x2-x1)^2) x4 = x3 - k * (y2-y1) y4 = y3 + k * (x2-x1) Where ^2 means squared From wiki : In algebra, for any linear equation y=mx + b, the perpendiculars will all have a slope of (-1/m), the opposite reciprocal of the original slope. It is helpful to memorize the slogan

What is the simplest way to test if a point P is inside a convex hull formed by a set of points X? I'd like an algorithm that works in a high-dimensional space (say, up to 40 dimensions) that doesn't explicitly compute the convex hull itself. Any ideas? The problem can be solved by finding a feasible point of a Linear Program. If you're interested in the full details, as opposed to just plugging an LP into an existing solver, I'd recommend reading Chapter 11.4 in Boyd and Vandenberghe's excellent book on convex optimization . Set A = (X[1] X[2] ... X[n]) , that is, the first column is v1 , the

I have a lot of points on the surface of the sphere. How can I calculate the area/spot of the sphere that has the largest point density? I need this to be done very fast. If this was a square for example I guess I could create a grid and then let the points vote which part of the grid is the best. I have tried with transforming the points to spherical coordinates and then do a grid, both this did not work well since points around north pole are close on the sphere but distant after the transform. Thanks To add some other, alternative schemes to the mix: it's possible to define a number of

Given a set of points on a plane, find the shortest line segment formed by any two of these points. How can I do that? The trivial way is obviously to calculate each distance, but I need another algorithm to compare. http://en.wikipedia.org/wiki/Closest_pair_of_points The problem can be solved in O(n log n) time using the recursive divide and conquer approach, e.g., as follows: Sort points along the x-coordinate Split the set of points into two equal-sized subsets by a vertical line x = xmid Solve the problem recursively in the left and right subsets. This will give the left-side and right

The problem: N points are given on a 2-dimensional plane. What is the maximum number of points on the same straight line? The problem has O(N 2 ) solution: go through each point and find the number of points which have the same dx / dy with relation to the current point. Store dx / dy relations in a hash map for efficiency. Is there a better solution to this problem than O(N 2 )? jonderry There is likely no solution to this problem that is significantly better than O(n^2) in a standard model of computation. The problem of finding three collinear points reduces to the problem of finding the

If you have 2 points, (x1, y1) and (x2, y2), which represent two opposite corners of a rectangle, and 2 other points, (x3,y3) and (x4,y4), which represent 2 endpoints of a line segment, how can you check if the line segment intersects the rectangle? (The line segment is just the segment contained between the given endpoints. It is not an infinite length line defined by those two points.) templatetypedef One very simple option would be to use a standard algorithm for checking whether two line segments intersect to check whether the line segments intersects any of the four line segments that

Below are 2 rectangles . Given the coordinates of the rectangle vertices - (x1, y1)...(x8, y8), how can the area of the overlapping region (white in the figure below) be caclulated? Note that: Coordinates of points might be any Rectangles may or may not overlap Assume area is 0 when rectangles don't overlap, or they overlap at point or line. If one rectangle is inside the other, then calculate area of smaller rectangle. Since you stated that the rectangles may not be aligned, possible answers may be nothing, a point, a line segment, or a polygon with 3-8 sides. The usual way to do this 2d