I recently came across a problem where I had four circles (midpoints and radius) and had to calculate the area of the union of these circles.
Example image:
For two circles it's quite easy,
I can just calculate the fraction of the each circles area that is not within the triangles and then calculate the area of the triangles.
But is there a clever algorithm I can use when there is more than two circles?
Find all circle intersections on the outer perimeter (e.g. B,D,F,H on the following diagram). Connect them together with the centres of the corresponding circles to form a polygon. The area of the union of the circles is the area of the polygon + the area of the circle slices defined by consecutive intersection points and the circle center in between them. You'll need to also account for any holes.
I'm sure there is a clever algorithm, but here's a dumb one to save having to look for it;
put a bounding box around the circles;
generate random points within the bounding box;
figure out whether the random point is inside one of the circles;
compute the area by some simple addition and division (proportion_of_points_inside*area_of_bounding_box).
Sure it's dumb, but:
you can get as accurate an answer as you want, just generate more points;
it will work for any shapes for which you can calculate the inside/outside distinction;
it will parallelise beautifully so you can use all your cores.
Ants Aasma's answer gave the basic idea, but I wanted to make it a little more concrete. Take a look at the five circles below and the way they've been decomposed.
The blue dots are circle centers.
The red dots are circle boundary intersections.
The red dots with white interior are circle boundary intersections that are not contained in any other circles.
Identifying these 3 types of dots is easy. Now construct a graph data structure where the nodes are the blue dots and the red dots with white interior. For every circle, put an edge between the circle middle (blue dot) and each of its intersections (red dots with white interior) on its boundary.
This decomposes the circle union into a set of polygons (shaded blue) and circular pie pieces (shaded green) that are pairwise disjoint and cover the original union (that is, a partition). Since each piece here is something that's easy to compute the area of, you can compute the area of the union by summing the pieces' areas.
For a different solution from the previous one you could produce an estimation with an arbitrary precision using a quadtree.
This also works for any shape union if you can tell if a square is inside or outside or intersects the shape.
Each cell has one of the states : empty , full , partial
The algorithm consists in "drawing" the circles in the quadtree starting with a low resolution ( 4 cells for instance marked as empty). Each cell is either :
inside at least one circle, then mark the cell as full,
outside all circles, mark the cell as empty,
else mark the cell as partial.
When it's done, you can compute an estimation of the area : the full cells give the lower bound, the empty cells give the higher bound, the partial cells give the max area error.
If the error is too big for you, you refine the partial cells until you get the right precision.
I think this will be easier to implement than the geometric method which may require to handle a lot of special cases.
I love the approach to the case of 2 intersecting circles -- here's how i'd use a slight variation of the same approach for the more complex example.
It might give better insight into generalising the algorithm for larger numbers of semi-overlapping circles.
The difference here is that i start by linking the centres (so there's a vertice between the centre of the circles, rather than between the places where the circles intersect) I think this lets it generalise better.
(in practice, maybe the monte-carlo method is worthwhile)
(source: secretGeek.net)
If you want a discrete (as opposed to a continuous) answer, you could do something similar to a pixel painting algorithm.
Draw the circles on a grid, and then color each cell of the grid if it's mostly contained within a cirle (i.e., at least 50% of its area is inside one of the circles). Do this for the entire grid (where the grid spans all of the area covered by the circles), then count the number of colored cells in the grid.
Hmm, very interesting problem. My approach would probably be something along the lines of the following:
Work out a way of working out what the areas of intersection between an arbitrary number of circles is, i.e. if I have 3 circles, I need to be able to work out what the intersection between those circles is. The "Monte-Carlo" method would be a good way of approximating this (http://local.wasp.uwa.edu.au/~pbourke/geometry/circlearea/).
Eliminate any circles that are contained entirely in another larger circle (look at radius and the modulus of the distance between the centre of the two circles) I dont think is mandatory.
Choose 2 circles (call them A and B) and work out the total area using this formula:
(this is true for any shape, be it circle or otherwise)
area(A∪B) = area(A) + area(B) - area(A∩B)
Where A ∪ B means A union B and A ∩ B means A intersect B (you can work this out from the first step.
Now keep on adding circles and keep on working out the area added as a sum / subtraction of areas of circles and areas of intersections between circles. For example for 3 circles (call the extra circle C) we work out the area using this formula:
(This is the same as above where A has been replaced with A∪B)
area((A∪B)∪C) = area(A∪B) + area(C) - area((A∪B)∩C)
Where area(A∪B) we just worked out, and area((A∪B)∩C) can be found:
area((A∪B)nC) = area((A∩C)∪(B∩C)) = area(A∩C) + area(A∩B) - area((A∩C)∩(B∩C)) = area(A∩C) + area(A∩B) - area(A∩B∩C)
Where again you can find area(A∩B∩C) from above.
The tricky bit is the last step - the more circles get added the more complex it becomes. I believe there is an expansion for working out the area of an intersection with a finite union, or alternatively you may be able to recursively work it out.
Also with regard to using Monte-Carlo to approximate the area of itersection, I believe its possible to reduce the intersection of an arbitrary number of circles to the intersection of 4 of those circles, which can be calculated exactly (no idea how to do this however).
There is probably a better way of doing this btw - the complexity increases significantly (possibly exponentially, but I'm not sure) for each extra circle added.
There are efficient solutions to this problem using what are known as power diagrams. This is really heavy math though and not something that I would want to tackle offhand. For an "easy" solution, look up line-sweep algorithms. The basic principle here is that that you divide the figure up into strips, where calculating the area in each strip is relatively easy.
So, on the figure containing all of the circles with nothing rubbed out, draw a horizontal line at each position which is either the top of a circle, the bottom of a circle or the intersection of 2 circles. Notice that inside these strips, all of the areas you need to calculate look the same: a "trapezium" with two sides replaced by circular segments. So if you can work out how to calculate such a shape, you just do it for all the individual shapes and add them together. The complexity of this naive approach is O(N^3), where N is the number of circles in the figure. With some clever data structure use, you could improve this line-sweep method to O(N^2 * log(N)), but unless you really need to, it's probably not worth the trouble.
The pixel-painting approach (as suggested by #Loadmaster) is superior to the mathematical solution in a variety of ways:
Implementation is much simpler. The above problem can be solved in less than 100 lines of code, as this JSFiddle solution demonstrates (mostly because it’s conceptually much simpler, and has no edge cases or exceptions to deal with).
It adapts easily to more general problems. It works with any shape, regardless of morphology, as long as it’s renderable with 2D drawing libraries (i.e., “all of them!”) — circles, ellipses, splines, polygons, you name it. Heck, even bitmap images.
The complexity of the pixel-painting solution is ~O[n], as compared to ~O[n*n] for the mathematical solution. This means it will perform better as the number of shapes increases.
And speaking of performance, you’ll often get hardware acceleration for free, as most modern 2D libraries (like HTML5’s canvas, I believe) will offload rendering work to graphics accelerators.
The one downside to pixel-painting is the finite accuracy of the solution. But that is tunable by simply rendering to larger or smaller canvases as the situation demands. Note, too, that anti-aliasing in the 2D rendering code (often turned on by default) will yield better-than-pixel-level accuracy. So, for example, rendering a 100x100 figure into a canvas of the same dimensions should, I think, yield accuracy on the order of 1 / (100 x 100 x 255) = .000039% ... which is probably “good enough” for all but the most demanding problems.
<p>Area computation of arbitrary figures as done thru pixel-painting, in which a complex shape is drawn into an HTML5 canvas and the area determined by comparing the number of white pixels found in the resulting bitmap. See javascript source for details.</p>
<canvas id="canvas" width="80" height="100"></canvas>
<p>Area = <span id="result"></span></p>
// Get HTML canvas element (and context) to draw into
var canvas = document.getElementById('canvas');
var ctx = canvas.getContext('2d');
// Lil' circle drawing utility
function circle(x,y,r) {
ctx.beginPath();
ctx.arc(x, y, r, 0, Math.PI*2);
ctx.fill();
}
// Clear canvas (to black)
ctx.fillStyle = 'black';
ctx.fillRect(0, 0, canvas.width, canvas.height);
// Fill shape (in white)
ctx.fillStyle = 'white';
circle(40, 50, 40);
circle(40, 10, 10);
circle(25, 15, 12);
circle(35, 90, 10);
// Get bitmap data
var id = ctx.getImageData(0, 0, canvas.width, canvas.height);
var pixels = id.data; // Flat array of RGBA bytes
// Determine area by counting the white pixels
for (var i = 0, area = 0; i < pixels.length; i += 4) {
area += pixels[i]; // Red channel (same as green and blue channels)
}
// Normalize by the max white value of 255
area /= 255;
// Output result
document.getElementById('result').innerHTML = area.toFixed(2);
I have been working on a problem of simulating overlapping star fields, attempting to estimate the true star counts from the actual disk areas in dense fields, where the larger bright stars can mask fainter ones. I too had hoped to be able to do this by rigorous formal analysis, but was unable to find an algorithm for the task. I solved it by generating the star fields on a blue background as green disks, whose diameter was determined by a probability algorithm. A simple routine can pair them to see if there's an overlap (turning the star pair yellow); then a pixel count of the colours generates the observed area to compare to the theoretical area. This then generates a probability curve for the true counts. Brute force maybe, but it seems to work OK.
(source: 2from.com)
Here's an algorithm that should be easy to implement in practice, and could be adjusted to produce arbitrarily small error:
Approximate each circle by a regular polygon centered at the same point
Calculate the polygon which is the union of the approximated circles
Calculate the area of the merged polygon
Steps 2 and 3 can be carried out using standard, easy-to-find algorithms from computational geometry.
Obviously, the more sides you use for each approximating polygon, the closer to exact your answer would be. You could approximate using inscribed and circumscribed polygons to get bounds on the exact answer.
I found this link which may be useful. There does not seem to be a definitive answer though.
Google answers. Another reference for three circles is Haruki's theorem. There is a paper there as well.
Depending on what problem you are trying to solve it could be sufficient to get an upper and lower bound. An upper bound is easy, just the sum of all the circles. For a lower bound you can pick a single radius such that none of the circles overlap. To better that find the largest radius (up to the actual radius) for each circle so that it doesn't overlap. It should also be pretty trivial to remove any completely overlapped circles (All such circles satisfy |P_a - P_b| <= r_a) where P_a is the center of circle A, P_b is the center of circle B, and r_a is the radius of A) and this betters both the upper and lower bound. You could also get a better Upper bound if you use your pair formula on arbitrary pairs instead of just the sum of all the circles. There might be a good way to pick the "best" pairs (the pairs that result in the minimal total area.
Given an upper and lower bound you might be able to better tune a Monte-carlo approach, but nothing specific comes to mind. Another option (again depending on your application) is to rasterize the circles and count pixels. It is basically the Monte-carlo approach with a fixed distribution.
I've got a way to get an approximate answer if you know that all your circles are going to be within a particular region, i.e. each point in circle is inside a box whose dimensions you know. This assumption would be valid, for example, if all the circles are in an image of known size. If you can make this assumption, divide the region which contains your image into 'pixels'. For each pixel, compute whether it is inside at least one of the circles. If it is, increment a running total by one. Once you are done, you know how many pixels are inside at least one circle, and you also know the area of each pixel, so you can calculate the total area of all the overlapping circles.
By increasing the 'resolution' of your region (the number of pixels), you can improve your approximation.
Additionally, if the size of the region containing your circles is bounded, and you keep the resolution (number of pixels) constant, the algorithm runs in O(n) time (n is the number of circles). This is because for each pixel, you have to check whether it is inside each one of your n circles, and the total number of pixels is bounded.
This can be solved using Green's Theorem, with a complexity of n^2log(n).
If you're not familiar with the Green's Theorem and want to know more, here is the video and notes from Khan Academy. But for the sake of our problem, I think my description will be enough.
If I put L and M such that
then the RHS is simply the area of the Region R and can be obtained by solving the closed integral or LHS and this is exactly what we're going to do.
So Integrating along the path in the anticlockwise gives us the Area of the region and integrating along the clockwise gives us negative of the Area. So
AreaOfUnion = (Integration along red arcs in anticlockwise direction + Integration along blue arcs in clockwise direction)
But the cool trick is if for each circle if we integrate the arcs which are not inside any other circle we get our required area i.e. we get integration in an anticlockwise direction along all red arcs and integration along all blue arcs along the clockwise direction. JOB DONE!!!
Even the cases when a circle doesn't intersect with any other is taken
care of.
Here is the GitHub link to my C++ Code
I'm creating a circle dynamically based on two positions; the first is the centre, and second is the radius. I'm then determining how many of my objects will fit onto the perimeter of the circle by int fillCount = Mathf.RoundToInt(circumference / sizeOfObject);
So now I've got a circle, I know how many things will fit on it, I know the circumference, I know the radius, and I've cobbled together a function that will find the Vector3 of any given point on the circle.
What I can't figure out is how to evenly distribute each object. i.e, place an object every 'n' degrees. I don't think I need to worry about arc-reparameterization, or do I? I think the answer lies in finding the degree of placement by using the inverse cosine/sine, but I'm not sure and my maths isn't that good.
If anyone has any input or advice, I'd greatly appreciate it.
If you know how many objects are in the circle, you can divide 360deg by the number of objects to determine the angle on the circle each object will be placed. Since we know the point on a circle is located at x=r*cos(angle), y=r*sin(angle) you can add this Vector2 to the Vector2 center of your circle.
As an example, if you have 3 objects to place around the circle. Get your angle by dividing 360/3 = 120. Each object will be placed 120deg from each other.
Pick your starting point (we will pick 0 degrees),
Vector2(r*Mathf.Deg2Rad*cos(0), r*Mathf.Deg2Rad*sin(0)).
The next object will be at 120deg, which will be
Vector2(r*Mathf.Deg2Rad*cos(120), r*Mathf.Deg2Rad*sin(120)).
Similarly, the 3rd object will be at 240deg,
Vector2(r*Mathf.Deg2Rad*cos(240), r*Mathf.Deg2Rad*sin(240))
This seems to be a rather asked question - (hear me out first! :)
I've created a polygon with perlin noise, and it looks like this:
I need to generate a texture from this array of points. (I'm using Monogame/XNA, but I assume this question is somewhat agnostic).
Anyway, researching this problem tells me that many people use raycasting to determine how many times a line crosses over the polygon shape (If once, it's inside. twice or zero times, it's outside). This makes sense, but I wonder if there is a better way, given that I have all of the points.
Doing a small raycast for every pixel I want to fill in seems excessive - is this the only/best way?
If I have a small 500px square image I need to fill in, I'll need to do a raycast for 250,000 individual pixels, which seems like an awful lot.
If you want to do this for every pixel, you can use a sweeping line:
Start from the topmost coordinate and examine a horizontal ray from left to right. Calculate all intersections with the polygon and sort them by their x-coordinate. Then iterate all pixels on the line and remember if you are in or out. Whenever you encounter an intersection, switch to the other side. If some pixel is in, set the texture. If not, ignore it. Do this from top to bottom for every possible horizontal line.
The intersection calculation could be enhanced in several ways. E.g. by using an acceleration data structure like a grid, quadtree, etc. or by examining the intersecting or touching edges of the polygon before. Then, when you sweep the line, you will already know, which edges will cause an intersection.
I would like to know how to compute rotation components of a rectangle in space according to four given points in a projection plane.
Hard to depict in a single sentence, thus I explain my needs.
I have a 3D world viewed from a static camera (located in <0,0,0>).
I have a known rectangular shape (an picture, actually) That I want to place in that space.
I only can define points (up to four) in a spherical/rectangular referencial (camera looking at <0°,0°> (sph) or <0,0,1000> (rect)).
I considere the given polygon to be my rectangle shape rotated (rX,rY,rZ). 3 points are supposed to be enough, 4 points should be too constraintfull. I'm not sure for now.
I want to determine rX, rY and rZ, the rectangle rotation about its center.
--- My first attempt at solving this constrint problem was to fix the first point: given spherical coordinates, I "project" this point onto a camera-facing plane at z=1000. Quite easy, this give me a point.
Then, the second point is considered to be on the <0,0,0>- segment, which is about an infinity of solution ; but I fix this by knowing the width(w) and height(h) of my rectangle: I then get two solutions for my second point ; one is "in front" of the first point, and the other is "far away"... I now have a edge of my rectangle. Two, in fact.
And from there, I don't know what to do. If in the end I have my four points, I don't have a clue about how to calculate the rotation equivalency...
It's hard to be lost in Mathematics...
To get an idea of the goal of all this: I make photospheres and I want to "insert" in them images. For instance, I got on my photo a TV screen, and I want to place a picture in the screen. I know my screen size (or I can guess it), I know the size of the image I want to place in (actually, it has the same aspect ratio), and I know the four screen corner positions in my space (spherical or euclidian). My software allow my to place an image in the scene and to rotate it as I want. I can zoom it (to give the feeling of depth)... I then can do all this manually, but it is a long try-fail process and never exact. I would like then to be able to type in the screen corner positions, and get the final image place and rotation attributes in a click...
The question in pictures:
Images presenting steps of the problem
Note that on the page, I present actual images of my app. I mean I had to manually rotate and scale the picture to get it fits the screen but it is not a photoshop. The parameters found are:
Scale: 0.86362
rX = 18.9375
rY = -12.5875
rZ = -0.105881
center position: <-9.55, 18.76, 1000>
Note: Rotation is not enought to set the picture up: we need scale and translation. I assume the scale can be found once a first edge is fixed (first two points help determining two solutions as initial constraints, and because I then know edge length and picture width and height, I can deduce scale. But the software is kind and allow me to modify picture width and height: thus the constraint is just to be sure the four points are descripbing a rectangle in space, with is simple to check with vectors. Here, the problem seems to place the fourth point as a valid rectangle corner, and then deduce rotation from that rectangle. About translation, it is the center (diagonal cross) of the points once fixed.
I'm doing some image processing, and I need to find some information on line growing algorithms - not sure if I'm using the right terminology here, so please call me out on this is needs be.
Imagine my input image is simply a circle on a black background. I'd basically like extract the coordinates, so that I may draw this circle elsewhere based on the coordinates.
Note: I am already using edge detection image filters, but I thought it best to explain with a simple example.
Basically what I'm looking to do is detect lines in an image, and store the result in a data type where by I have say a class called Line, and various different Point objects (containing X/Y coordinates).
class Line
{
Point points[];
}
class Point
{
int X, Y;
}
And this is how I'd like to use it...
Line line;
for each pixel in image
{
if pixel should be added to line
{
add pixel coordinates to line;
}
}
I have no idea how to approach this as you can probably establish, so pointers to any subject matter would be greatly appreciated.
I'm not sure if I'm interpreting you right, but the standard way is to use a Hough transform. It's a two step process:
From the given image, determine whether each pixel is an edge pixel (this process creates a new "binary" image). A standard way to do this is Canny edge-detection.
Using the binary image of edge pixels, apply the Hough transform. The basic idea is: for each edge pixel, compute all lines through it, and then take the lines that went through the most edge pixels.
Edit: apparently you're looking for the boundary. Here's how you do that.
Recall that the Canny edge detector actually gives you a gradient also (not just the magnitude). So if you pick an edge pixel and follow along (or against) that vector, you'll find the next edge pixel. Keep going until you don't hit an edge pixel anymore, and there's your boundary.
What you are talking about is not an easy problem! I have found that this website is very helpful in image processing: http://homepages.inf.ed.ac.uk/rbf/HIPR2/wksheets.htm
One thing to try is the Hough Transform, which detects shapes in an image. Mind you, it's not easy to figure out.
For edge detection, the best is Canny edge detection, also a non-trivial task to implement.
Assuming the following is true:
Your image contains a single shape on a background
You can determine which pixels are background and which pixels are the shape
You only want to grab the boundary of the outside of the shape (this excludes donut-like shapes where you want to trace the inside circle)
You can use a contour tracing algorithm such as the Moore-neighbour algorithm.
Steps:
Find an initial boundary pixel. To do this, start from the bottom-left corner of the image, travel all the way up and if you reach the top, start over at the bottom moving right one pixel and repeat, until you find a shape pixel. Make sure you keep track of the location of the pixel that you were at before you found the shape pixel.
Find the next boundary pixel. Travel clockwise around the last visited boundary pixel, starting from the background pixel you last visited before finding the current boundary pixel.
Repeat step 2 until you revisit first boundary pixel. Once you visit the first boundary pixel a second time, you've traced the entire boundary of the shape and can stop.
You could take a look at http://processing.org/ the project was created to teach the fundamentals of computer programming within a visual context. There is the language, based on java, and an IDE to make 'sketches' in. It is a very good package to quickly work with visual objects and has good examples of things like edge detection that would be useful to you.
Just to echo the answers above you want to do edge detection and Hough transform.
Note that a Hough transform for a circle is slightly tricky (you are solving for 3 parameters, x,y,radius) you might want to just use a library like openCV