rectangles

I have a rectangle that has a rotation already applied to it. I want to get the the unrotated dimensions (the x, y, width, height). Here is the dimensions of the element currently: Bounds at a 90 rotation: { height 30 width 0 x 25 y 10 } Here are the dimensions after the rotation is set to none: Bounds at rotation 0 { height 0 width 30 x 10 y 25 } In the past, I was able to set the rotation to 0 and then read the updated bounds . However, there is a bug in one of the functions I was using, so now I have to do it manually. Is there a simple formula to get the bounds at rotation 0 using the info

I am looking for an algorithm as follows: Given a set of possibly overlapping rectangles (All of which are "not rotated", can be uniformly represented as (left,top,right,bottom) tuplets, etc...), it returns a minimal set of (non-rotated) non-overlapping rectangles, that occupy the same area. It seems simple enough at first glance, but prooves to be tricky (at least to be done efficiently). Are there some known methods for this/ideas/pointers? Methods for not necessarily minimal, but heuristicly small, sets, are interesting as well, so are methods that produce any valid output set at all.

This is a language-agnostic question. Given a rectangle's dimensions with l,t,w,h (left, top, width, height) and a point x,y , how do I find the nearest point on the perimeter of the rectangle to that point? I have tried to resolve it in Lua, but any other language would do. So far this is my best effort: local function nearest(x, a, b) if a <= x and x <= b then return x elseif math.abs(a - x) < math.abs(b - x) then return a else return b end end local function getNearestPointInPerimeter(l,t,w,h, x,y) return nearest(x, l, l+w), nearest(y, t, t+h) end This works for a point outside of the

I have code that receives a specific area already defined before on server side and creates a hole on Form in client side. Instead of this, i want to get a screen capture of this same area but without appears my Form in final result, like a normal desktop capture, but in this case will be only captured, this small area. So, how i can adapt this my code below for this? procedure TForm1.CS1Read(Sender: TObject; Socket: TCustomWinSocket); var X1, X2, Y1, Y2: Integer; List: TStrings; FormRegion, HoleRegion: HRGN; StrCommand: String; begin StrCommand := Socket.ReceiveText; if Pos('§', StrCommand) >

Let's say you have a set of ranges: 0 - 100: 'a' 0 - 75: 'b' 95 - 150: 'c' 120 - 130: 'd' Obviously, these ranges overlap at certain points. How would you dissect these ranges to produce a list of non-overlapping ranges, while retaining information associated with their original range (in this case, the letter after the range)? For example, the results of the above after running the algorithm would be: 0 - 75: 'a', 'b' 76 - 94: 'a' 95 - 100: 'a', 'c' 101 - 119: 'c' 120 - 130: 'c', 'd' 131 - 150: 'c' I had the same question when writing a program to mix (partly overlapping) audio samples. What

I'm designing a collision detection game tutorial for young adults, so I want this to be as simple as possible to make it easier to explain. The requirements are very simple. The world is 2D and contains only rectangles (of arbitrary sizes). BSP and even quadtrees seems like it would be overkill (again, the emphasis is on simplicity) but I would like something more efficient than brute forcing through all n(n-1)/2 possible collisions. 2D, rectangles only, and simple. Can anyone point to an algorithm I can look up? Is a quadtree algorithm what I'm looking for? EDIT: Also, the rectangles will

For example we have two rectangles and they overlap. I want to get the exact range of the union of them. What is a good way to compute this? These are the two overlapping rectangles. Suppose the cords of vertices are all known: How can I compute the cords of the vertices of their union polygon? And what if I have more than two rectangles? There exists a Line Sweep Algorithm to calculate area of union of n rectangles. Refer the link for details of the algorithm. As said in article, there exist a boolean array implementation in O(N^2) time. Using the right data structure (balanced binary search