I am working on images to detect and fit the largest possible circle in any of the free areas of an image containing distributed particles:
(able to detect the location of particle).
One direction is to define a circle touching any 3-point combination, checking if the circle is empty, then finding the largest circle among all empty circles. However, it leads to a huge number of combination i.e. C(n,3), where n is the total number of particles in the image.
I would appreciate if anyone can provide me any hint or alternate method that I can explore.
Lets do some maths my friend, as maths will always get to the end!
Wikipedia:
In mathematics, a Voronoi diagram is a partitioning of a plane into
regions based on distance to points in a specific subset of the plane.
For example:
rng(1)
x=rand(1,100)*5;
y=rand(1,100)*5;
voronoi(x,y);
The nice thing about this diagram is that if you notice, all the edges/vertices of those blue areas are all to equal distance to the points around them. Thus, if we know the location of the vertices, and compute the distances to the closest points, then we can choose the vertex with highest distance as our center of the circle.
Interestingly, the edges of a Voronoi regions are also defined as the circumcenters of the triangles generated by a Delaunay triangulation.
So if we compute the Delaunay triangulation of the area, and their circumcenters
dt=delaunayTriangulation([x;y].');
cc=circumcenter(dt); %voronoi edges
And compute the distances between the circumcenters and any of the points that define each triangle:
for ii=1:size(cc,1)
if cc(ii,1)>0 && cc(ii,1)<5 && cc(ii,2)>0 && cc(ii,2)<5
point=dt.Points(dt.ConnectivityList(ii,1),:); %the first one, or any other (they are the same distance)
distance(ii)=sqrt((cc(ii,1)-point(1)).^2+(cc(ii,2)-point(2)).^2);
end
end
Then we have the center (cc) and radius (distance) of all possible circles that have no point inside them. We just need the biggest one!
[r,ind]=max(distance); %Tada!
Now lets plot
hold on
ang=0:0.01:2*pi;
xp=r*cos(ang);
yp=r*sin(ang);
point=cc(ind,:);
voronoi(x,y)
triplot(dt,'color','r','linestyle',':')
plot(point(1)+xp,point(2)+yp,'k');
plot(point(1),point(2),'g.','markersize',20);
Notice how the center of the circle is on one vertex of the Voronoi diagram.
NOTE: this will find the center inside [0-5],[0-5]. you can easily modify it to change this constrain. You can also try to find the circle that fits on its entirety inside the interested area (as opposed to just the center). This would require a small addition in the end where the maximum is obtained.
I'd like to propose another solution based on a grid search with refinement. It's not as advanced as Ander's or as short as rahnema1's, but it should be very easy to follow and understand. Also, it runs quite fast.
The algorithm contains several stages:
We generate an evenly-spaced grid.
We find the minimal distances of points in the grid to all provided points.
We discard all points whose distances are below a certain percentile (e.g. 95th).
We choose the region which contains the largest distance (this should contain the correct center if my initial grid is fine enough).
We create a new meshgrid around the chosen region and find distances again (this part is clearly sub-optimal, because the distances are computed to all points, including far and irrelevant ones).
We iterate the refinement within the region, while keeping an eye on the variance of the top 5% of values -> if it drops below some preset threshold we break.
Several notes:
I have made the assumption that circles cannot go beyond the scattered points' extent (i.e. the bounding square of the scatter acts as an "invisible wall").
The appropriate percentile depends on how fine the initial grid is. This will also affect the amount of while iterations, and the optimal initial value for cnt.
function [xBest,yBest,R] = q42806059
rng(1)
x=rand(1,100)*5;
y=rand(1,100)*5;
%% Find the approximate region(s) where there exists a point farthest from all the rest:
xExtent = linspace(min(x),max(x),numel(x));
yExtent = linspace(min(y),max(y),numel(y)).';
% Create a grid:
[XX,YY] = meshgrid(xExtent,yExtent);
% Compute pairwise distance from grid points to free points:
D = reshape(min(pdist2([XX(:),YY(:)],[x(:),y(:)]),[],2),size(XX));
% Intermediate plot:
% figure(); plot(x,y,'.k'); hold on; contour(XX,YY,D); axis square; grid on;
% Remove irrelevant candidates:
D(D<prctile(D(:),95)) = NaN;
D(D > xExtent | D > yExtent | D > yExtent(end)-yExtent | D > xExtent(end)-xExtent) = NaN;
%% Keep only the region with the largest distance
L = bwlabel(~isnan(D));
[~,I] = max(table2array(regionprops('table',L,D,'MaxIntensity')));
D(L~=I) = NaN;
% surf(XX,YY,D,'EdgeColor','interp','FaceColor','interp');
%% Iterate until sufficient precision:
xExtent = xExtent(~isnan(min(D,[],1,'omitnan')));
yExtent = yExtent(~isnan(min(D,[],2,'omitnan')));
cnt = 1; % increase or decrease according to the nature of the problem
while true
% Same ideas as above, so no explanations:
xExtent = linspace(xExtent(1),xExtent(end),20);
yExtent = linspace(yExtent(1),yExtent(end),20).';
[XX,YY] = meshgrid(xExtent,yExtent);
D = reshape(min(pdist2([XX(:),YY(:)],[x(:),y(:)]),[],2),size(XX));
D(D<prctile(D(:),95)) = NaN;
I = find(D == max(D(:)));
xBest = XX(I);
yBest = YY(I);
if nanvar(D(:)) < 1E-10 || cnt == 10
R = D(I);
break
end
xExtent = (1+[-1 +1]*10^-cnt)*xBest;
yExtent = (1+[-1 +1]*10^-cnt)*yBest;
cnt = cnt+1;
end
% Finally:
% rectangle('Position',[xBest-R,yBest-R,2*R,2*R],'Curvature',[1 1],'EdgeColor','r');
The result I'm getting for Ander's example data is [x,y,r] = [0.7832, 2.0694, 0.7815] (which is the same). The execution time is about half of Ander's solution.
Here are the intermediate plots:
Contour of the largest (clear) distance from a point to the set of all provided points:
After considering distance from the boundary, keeping only the top 5% of distant points, and considering only the region which contains the largest distance (the piece of surface represents the kept values):
And finally:
You can use bwdist from Image Processing Toolbox to compute the distance transform of the image. This can be regarded as a method to create voronoi diagram that well explained in #AnderBiguri's answer.
img = imread('AbmxL.jpg');
%convert the image to a binary image
points = img(:,:,3)<200;
%compute the distance transform of the binary image
dist = bwdist(points);
%find the circle that has maximum radius
radius = max(dist(:));
%find position of the circle
[x y] = find(dist == radius);
imshow(dist,[]);
hold on
plot(y,x,'ro');
The fact that this problem can be solved using a "direct search" (as can be seen in another answer) means one can look at this as a global optimization problem. There exist various ways to solve such problems, each appropriate for certain scenarios. Out of my personal curiosity I have decided to solve this using a genetic algorithm.
Generally speaking, such an algorithm requires us to think of the solution as a set of "genes" subject to "evolution" under a certain "fitness function". As it happens, it's quite easy to identify the genes and the fitness function in this problem:
Genes: x , y, r.
Fitness function: technically, maximum area of circle, but this is equivalent to the maximum r (or minimum -r, since the algorithm requires a function to minimize).
Special constraint - if r is larger than the euclidean distance to the closest of the provided points (that is, the circle contains a point), the organism "dies".
Below is a basic implementation of such an algorithm ("basic" because it's completely unoptimized, and there is lot of room for optimizationno pun intended in this problem).
function [x,y,r] = q42806059b(cloudOfPoints)
% Problem setup
if nargin == 0
rng(1)
cloudOfPoints = rand(100,2)*5; % equivalent to Ander's initialization.
end
%{
figure(); plot(cloudOfPoints(:,1),cloudOfPoints(:,2),'.w'); hold on; axis square;
set(gca,'Color','k'); plot(0.7832,2.0694,'ro'); plot(0.7832,2.0694,'r*');
%}
nVariables = 3;
options = optimoptions(#ga,'UseVectorized',true,'CreationFcn',#gacreationuniform,...
'PopulationSize',1000);
S = max(cloudOfPoints,[],1); L = min(cloudOfPoints,[],1); % Find geometric bounds:
% In R2017a: use [S,L] = bounds(cloudOfPoints,1);
% Here we also define distance-from-boundary constraints.
g = ga(#(g)vectorized_fitness(g,cloudOfPoints,[L;S]), nVariables,...
[],[], [],[], [L 0],[S min(S-L)], [], options);
x = g(1); y = g(2); r = g(3);
%{
plot(x,y,'ro'); plot(x,y,'r*');
rectangle('Position',[x-r,y-r,2*r,2*r],'Curvature',[1 1],'EdgeColor','r');
%}
function f = vectorized_fitness(genes,pts,extent)
% genes = [x,y,r]
% extent = [Xmin Ymin; Xmax Ymax]
% f, the fitness, is the largest radius.
f = min(pdist2(genes(:,1:2), pts, 'euclidean'), [], 2);
% Instant death if circle contains a point:
f( f < genes(:,3) ) = Inf;
% Instant death if circle is too close to boundary:
f( any( genes(:,3) > genes(:,1:2) - extent(1,:) | ...
genes(:,3) > extent(2,:) - genes(:,1:2), 2) ) = Inf;
% Note: this condition may possibly be specified using the A,b inputs of ga().
f(isfinite(f)) = -genes(isfinite(f),3);
%DEBUG:
%{
scatter(genes(:,1),genes(:,2),10 ,[0, .447, .741] ,'o'); % All
z = ~isfinite(f); scatter(genes(z,1),genes(z,2),30,'r','x'); % Killed
z = isfinite(f); scatter(genes(z,1),genes(z,2),30,'g','h'); % Surviving
[~,I] = sort(f); scatter(genes(I(1:5),1),genes(I(1:5),2),30,'y','p'); % Elite
%}
And here's a "time-lapse" plot of 47 generations of a typical run:
(Where blue points are the current generation, red crosses are "insta-killed" organisms, green hexagrams are the "non-insta-killed" organisms, and the red circle marks the destination).
I'm not used to image processing, so it's just an Idea:
Implement something like a gaussian filter (blur) which transforms each particle (pixels) to a round gradiant with r=image_size (all of them overlapping). This way, you should get a picture where the most white pixels should be the best results. Unfortunately, the demonstration in gimp failed because the extreme blurring made the dots disappearing.
Alternatively, you could incrementelly extend all existing pixels by marking all neighbour pixels in an area (example: r=4), the pixels left would be the same result (those with the biggest distance to any pixel)
As described, I would like to find the largest empty rectangle in a given 2D point cloud. This information will be then used to get parallel lines shown as green lines in the images, which will then be used to get the angle the LiDAR has rotated.
I've tried convex hull, boundary, and alpha shape functions in matlab with this point cloud data. From the looks of it, I think i have to
(1) break the point cloud into two chunks (gift wrap algo?) and then
(2) apply boundary function; With coordinates extracts from boundary function,
(3) I'm thinking to run RANSAC line fitting.
However, RANSAC needs more points to judge the "fitness" of the line. Currently I'm exploring Hough Transform to see if that line detection would work for this case.
Thus the question here is,
(1) am I in the right track to find the green lines? Or is there a better way?
(2) How to get the angular value/gradient from the line when the line is vertical (tan(90) = inf)
The second image depicts the origin may be in the center of the cloud points or at an offset. One can assume there will be parallel lines with varying distance from the center
You can use kmeans clustering to find two clusters then find two points each from a cluster that distance between them is minimum
%a function for rotation of points
function out = rot(theta,data)
mat = [cos(theta), -sin(theta);
sin(theta), cos(theta)];
out = data * mat.';
end
a= [rand(505,1)*.4 , rand(505,1) ];
b= [rand(500,1)*.4 + .6 , rand(500,1)];
a = [a;b];
rt = rot(.4,a);
%kmeans clustering
km=kmeans(rt,2);
class1 = rt(km==1,:);
class2=rt(km==2,:);
%find min distance
[srch, d] = dsearchn(class1, class2);
[mn, I] = min(d);
plot(rt(:,1), rt(:,2),'.')
hold on
p1 = class1(srch(I),:);
p2 = class2(I,:);
plot(p1(1),p1(2),'*r')
plot(p2(1),p2(2),'*g')
xy = [class1(srch(I),:);class2(I,:)]
plot(xy(:,1),xy(:,2),'-g')
axis equal
I'm working in image segmentation, testing a lot of different segmentation algorithms, in order to do a comparitive study. At the moment i'm using Hough transform to find circles in the image. The images that i'm using have plenty objects, so when Í count the objects the result is hudge. I think the problem, is the overlaping circle. Do you know how can i maybe remove the overlaping circles to have a result more close to reality?
The code that i'm using is:
clear all, clc;
% Image Reading
I=imread('0001_c3.png');
figure(1), imshow(I);set(1,'Name','Original')
image used
% Gaussian Filter
W = fspecial('gaussian',[10,10]);
J = imfilter(I,W);
figure(2);imshow(J);set(2,'Name','Filtrada média');
X = rgb2gray(J);
figure(3);imshow(X);set(3,'Name','Grey');
% Finding Circular objects -- Houng Transform
[centers, radii, metric] = imfindcircles(X,[10 20], 'Sensitivity',0.92,'Edge',0.03); % [parasites][5 30]
centersStrong = centers(1:60,:); % number of objects
radiiStrong = radii(1:60);
metricStrong = metric(1:60);
viscircles(centersStrong, radiiStrong,'EdgeColor','r');
length(centers)% result=404!
You could simply loop over the circles and check if others are "close" to them. If so, you ignore them.
idx_mask = ones(size(radii));
min_dist = 1; % relative value. Tweak this if slight overlap is OK.
for i = 2:length(radii)
cur_cent = centers(i, :);
for j = 1:i-1
other_cent = centers(j,:);
x_dist = other_cent(1) - cur_cent(1);
y_dist = other_cent(2) - cur_cent(2);
if sqrt(x_dist^2+y_dist^2) < min_dist*(radii(i) + radii(j)) && idx_mask(j) == 1
idx_mask(i) = 0;
break
end
end
end
%%
idx_mask = logical(idx_mask);
centers_use = centers(idx_mask, :);
radii_use = radii(idx_mask, :);
metric_use = metric(idx_mask, :);
viscircles(centers_use, radii_use,'EdgeColor','b');
The picture shows all circles in red, and the filtered circles in blue.
The if clause checks two things:
- Are the centers of the circles closer than the sum of their radii?
- Is the other circle still on the list of considered circles?
If the answer to both questions is yes, then ignore the "current circle".
The way the loop is set up, it will keep circles that are higher up (have a lower row index). As is, the circles are already ordered by descending metric. In other words, as is this code will keep circles with a higher metric.
The code could optimized so that the loops run faster, but I don't think you'll have millions of circles in a single picture. I tried writing it in a way that it's easier to read for humans.
I have another problem today:
I have a binary matrix t, in which 1 represents a river channel, 0 represents flood plane and surrounding mountains:
t = Alog>10;
figure
imshow(t)
axis xy
For further calculations, I would like to expand the area of the riverchannel a few pixels in each direction. Generally speaking, I want to have a wider channel displayed in the image, to include a larger region in a later hydraulic model.
Here is my attempt, which does work in certain regions, but in areas where the river runs diagonal to the x-y axis, it does not widen the channel. There seems to be a flow in approaching this, which I cannot quite grasp.
[q,o] = find(t == 1);
qq = zeros(length(q),11);
oo = zeros(length(o),11);
% add +-5 pixel to result
for z=1:length(q)
qq(z,:) = q(z)-5:1:q(z)+5;
oo(z,:) = o(z)-5:1:o(z)+5;
end
% create column vectors
qq = qq(:);
oo = oo(:);
cords = [oo qq]; % [x y]
% remove duplicates
cords = unique(cords,'rows');
% get limits of image
[limy limx] = size(t);
% restrict to x-limits
cords = cords(cords(:,1)>=1,:);
cords = cords(cords(:,1)<=limx,:);
% restrict to y-limits
cords = cords(cords(:,2)>=1,:);
cords = cords(cords(:,2)<=limy,:);
% test image
l = zeros(size(img));
l(sub2ind(size(l), cords(:,2)',cords(:,1)')) = 1;
figure
imshow(l)
axis xy
This is the image I get:
It does widen the channel in some areas, but generally there seems to be a flaw with my approach. When I use the same approach on a diagonal line of pixels, it will not widen the line at all, because it will just create more pairs of [1 1; 2 2; 3 3; etc].
Is there a better approach to this or even something from the realm of image processing?
A blur filter with a set diameter should be working somewhat similar, but I could not find anything helpful...
PS: I wasn't allowed to add the images, although I already have 10 rep, so here are the direct links:
http://imageshack.us/a/img14/3122/channelthin.jpg
http://imageshack.us/a/img819/1787/channelthick.jpg
If you have the image processing toolbox, you should use the imdilate function. This performs the morphological dilation operation. Try the following code:
SE = strel('square',3);
channelThick = imdilate(channelThin,SE);
where SE is a 3x3 square structuring element used to dilate the image stored in channelThin. This will expand the regions in channelThin by one pixel in every direction. To expand more, use a larger structuring element, or multiple iterations.
You may apply morphological operations from image processing. Morphological dilation can be used in your example.
From the image processing toolbox, you can use bwmorth command BW2 = bwmorph(BW,'dilate') or imdilate command IM2 = imdilate(IM,SE).
Where IM is your image and SE is the structuring element. You can set SE = ones(3); to dilate the binary image by "one pixel" - but it can be changed depending on your application. Or you can dilate the image several times with the same structuring element if needed.