A question I strangely could not find on the internet. Given a complicated curve C (i.e. a curve that you can't fit with polynomials) defined by N points and centered around x0=0.5,0 (blue curve in figure), how can I rescale the curve so that the center is the same and the new curve is located at a constant distance d from the curve C (e.g. green curve in figure)?
So far the only way I could find is using the MATLAB function bwdist (https://fr.mathworks.com/help/images/ref/bwdist.html) which computes the Euclidean distance map of a binary image (see code below). However, I'm constrained by the size of my matrix i.e. a curve of 1e5 points is fine but a matrix of size (1e5,1e5) is big for bwdist...so the results using a coarse matrix is an ugly step-wise function. The code is
%%% profile
x = linspace(0,1,1e5);
y = -(x-0.5).^2/0.5^2 + 1 - 0.5*(exp(-(x-0.5).^2/2/0.2^2) - exp(-(-0.5).^2/2/0.2^2));
%%% define mask on a region that encompasses the curve
N=512;
mask = ones(N,N);
xm = linspace(0.9*min(x),1.1*max(x),N);
ym = linspace(0.9*min(y),1.1*max(y),N);
[Xm,Ym] = meshgrid(xm,ym);
%%% project curve on mask (i.e. put 0 below curve)
% get point of mask closer to each point of y
DT = delaunayTriangulation(Xm(:),Ym(:));
vi = nearestNeighbor(DT,x',y');
[iv,jv] = ind2sub(size(mask),vi);
% put 1 to indices of mask that are below projected curve
for p=1:length(iv)
mask(1:iv(p)-1,jv(p)) = 0;
end
%%% get euclidean distance
Ed = bwdist(logical(mask));
Ed = double(Ed);
%%% get contours of Ed at given values (i.e. distances)
cont = contour(Ed,linspace(0,1,50));
% cont has the various curves at given distances from original curve y
I add that I first tried moving a point of curve C for a distance d using the normal of the tangent but since the curve is non-linear, this direction is actually not necessarily the one giving the appropriate point. So at some distance, the curve becomes discontinuous because using the tangent does not give the point at a given distance from the curve, only from the considered point on curve C.
The code is
% profil
x = linspace(0,1,1e5);
y = -(x-0.5).^2/0.5^2 + 1 - 0.5*(exp(-(x-0.5).^2/2/0.2^2) - exp(-(-0.5).^2/2/0.2^2));
% create lines at Dist from original line
Dist = linspace(0,2e-1,6);
Dist = Dist(2:end);
Cdist(1).x = x;
Cdist(1).y = y;
Cdist(1).v = 0;
step = 10; % every step points compute normal to point and move points
points = [1:1:length(y)];
for d=1:length(Dist)
xd = x;
yd = y;
for p=1:length(points)
if points(p)==1
tang = [-(y(2)-y(1)) (x(2)-x(1))];
tang = tang/norm(tang);
xd(1) = xd(1) - Dist(d)*tang(1);
yd(1) = yd(1) - Dist(d)*tang(2);
elseif points(p)==length(y)
tang = [-(y(end)-y(end-1)) (x(end)-x(end-1))];
tang = tang/norm(tang);
xd(end) = xd(end) - Dist(d)*tang(1);
yd(end) = yd(end) - Dist(d)*tang(2);
else
tang = [-(y(p+1)-y(p-1)) (x(p+1)-x(p-1))];
tang = tang/norm(tang);
xd(p) = xd(p) - Dist(d)*tang(1);
yd(p) = yd(p) - Dist(d)*tang(2);
end
end
yd(yd<0)=NaN;
Cdist(d+1).x = xd;
Cdist(d+1).y = yd;
Cdist(d+1).v = Dist(d);
end
% plot
cmap=lines(10);
hold on
for c=1:length(Cdist)
plot(Cdist(c).x,Cdist(c).y,'linewidth',2,'color',cmap(c,:))
end
axis tight
axis equal
axis tight
Any idea ?
What you want to do is not possible.
Scaling a curve with respect to a center point while remaining equal distance to the original curve means that all the points on this curve are moving along its normal direction towards the center of scaling, and will eventually, reduce to a point.
Imagine drawing the normal direction of each point on this curve, and extend them to infinity. All these lines should pass through a same point, which is the center of scaling. Unfortunately, this is not the case for your curve.
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)
I'm working on an application to determine from an image the degree of alignment of a fiber network. I've read several papers on this issue and they basically do this:
Find the 2D discrete Fourier transform (DFT = F(u,v)) of the image (gray, range 0-255)
Find the Fourier Spectrum (FS = abs(F(u,v))) and the Power Spectrum (PS = FS^2)
Convert spectrum to polar coordinates and divide it into 1º intervals.
Calculate number-averaged line intensities (FI) for each interval (theta), that is, the average of all the intensities (pixels) forming "theta" degrees with respect to the horizontal axis.
Transform FI(theta) to cartesian coordinates
Cxy(theta) = [FI*cos(theta), FI*sin(theta)]
Find eigenvalues (lambda1 and lambda2) of the matrix Cxy'*Cxy
Find alignment index as alpha = 1 - lamda2/lambda1
I've implemented this in MATLAB (code below), but I'm not sure whether it is ok since point 3 and 4 are not really clear for me (I'm getting similar results to those of the papers, but not in all cases). For instance, in point 3, "spectrum" is referring to FS or to PS?. And in point 4, how should this average be done? are all the pixels considered? (even though there are more pixels in the diagonal).
rgb = imread('network.tif');%513x513 pixels
im = rgb2gray(rgb);
im = imrotate(im,-90);%since FFT space is rotated 90º
FT = fft2(im) ;
FS = abs(FT); %Fourier spectrum
PS = FS.^2; % Power spectrum
FS = fftshift(FS);
PS = fftshift(PS);
xoffset = (513-1)/2;
yoffset = (513-1)/2;
% Avoid low frequency points
x1 = 5;
y1 = 0;
% Maximum high frequency pixels
x2 = 255;
y2 = 0;
for theta = 0:pi/180:pi
% Transposed rotation matrix
Rt = [cos(theta) sin(theta);
-sin(theta) cos(theta)];
% Find radial lines necessary for improfile
xy1_rot = Rt * [x1; y1] + [xoffset; yoffset];
xy2_rot = Rt * [x2; y2] + [xoffset; yoffset];
plot([xy1_rot(1) xy2_rot(1)], ...
[xy1_rot(2) xy2_rot(2)], ...
'linestyle','none', ...
'marker','o', ...
'color','k');
prof = improfile(F,[xy1_rot(1) xy2_rot(1)],[xy1_rot(2) xy2_rot(2)]);
i = i + 1;
FI(i) = sum(prof(:))/length(prof);
Cxy(i,:) = [FI(i)*cos(theta), FI(i)*sin(theta)];
end
C = Cxy'*Cxy;
[V,D] = eig(C)
lambda2 = D(1,1);
lambda1 = D(2,2);
alpha = 1 - lambda2/lambda1
Figure: A) original image, B) plot of log(P+1), C) polar plot of FI.
My main concern is that when I choose an artificial image perfectly aligned (attached figure), I get alpha = 0.91, and it should be exactly 1.
Any help will be greatly appreciated.
PD: those black dots in the middle plot are just the points used by improfile.
I believe that there are a couple sources of potential error here that are leading to you not getting a perfect alpha value.
Discrete Fourier Transform
You have discrete imaging data which forces you to take a discrete Fourier transform which inevitably (depending on the resolution of the input data) have some accuracy issues.
Binning vs. Sampling Along a Line
The way that you have done the binning is that you literally drew a line (rotated by a particular angle) and sampled the image along that line using improfile. Using improfile performs interpolation of your data along that line introducing yet another potential source of error. The default is nearest neighbor interpolation which in the example shown below can cause multiple "profiles" to all pick up the same points.
This was with a rotation of 1-degree off-vertical when technically you'd want those peaks to only appear for a perfectly vertical line. It is clear to see how this sort of interpolation of the Fourier spectrum can lead to a spread around the "correct" answer.
Data Undersampling
Similar to Nyquist sampling in the Fourier domain, sampling in the spatial domain has some requirements as well.
Imagine for a second that you wanted to use 45-degree bin widths instead of the 1-degree. Your approach would still sample along a thin line and use that sample to represent 45-degrees worth or data. Clearly, this is a gross under-sampling of the data and you can imagine that the result wouldn't be very accurate.
It becomes more and more of an issue the further you get from the center of the image since the data in this "bin" is really pie wedge shaped and you're approximating it with a line.
A Potential Solution
A different approach to binning would be to determine the polar coordinates (r, theta) for all pixel centers in the image. Then to bin the theta components into 1-degree bins. Then sum all of the values that fall into that bin.
This has several advantages:
It removes the undersampling that we talked about and draws samples from the entire "pie wedge" regardless of the sampling angle.
It ensures that each pixel belongs to one and only one angular bin
I have implemented this alternate approach in the code below with some false horizontal line data and am able to achieve an alpha value of 0.988 which I'd say is pretty good given the discrete nature of the data.
% Draw a bunch of horizontal lines
data = zeros(101);
data([5:5:end],:) = 1;
fourier = fftshift(fft2(data));
FS = abs(fourier);
PS = FS.^2;
center = fliplr(size(FS)) / 2;
[xx,yy] = meshgrid(1:size(FS,2), 1:size(FS, 1));
coords = [xx(:), yy(:)];
% De-mean coordinates to center at the middle of the image
coords = bsxfun(#minus, coords, center);
[theta, R] = cart2pol(coords(:,1), coords(:,2));
% Convert to degrees and round them to the nearest degree
degrees = mod(round(rad2deg(theta)), 360);
degreeRange = 0:359;
% Band pass to ignore high and low frequency components;
lowfreq = 5;
highfreq = size(FS,1)/2;
% Now average everything with the same degrees (sum over PS and average by the number of pixels)
for k = degreeRange
ps_integral(k+1) = mean(PS(degrees == k & R > lowfreq & R < highfreq));
fs_integral(k+1) = mean(FS(degrees == k & R > lowfreq & R < highfreq));
end
thetas = deg2rad(degreeRange);
Cxy = [ps_integral.*cos(thetas);
ps_integral.*sin(thetas)]';
C = Cxy' * Cxy;
[V,D] = eig(C);
lambda2 = D(1,1);
lambda1 = D(2,2);
alpha = 1 - lambda2/lambda1;
I have a binary image of a human. In MATLAB, boundary points and the center of the image are also defined, and they are two column matrices. Now I want to draw lines from the center to the boundary points so that I can obtain all points of intersection between these lines and the boundary of the image. How can I do that? Here is the code I have so far:
The code that is written just to get the one intersection point if anyone can help please
clear all
close all
clc
BW = im2bw(imread('C:\fyc-90_1-100.png'));
BW = imfill(BW,'holes');
[Bw m n]=preprocess(BW);
[bord sk pr_sk]=border_skeleton(BW);
boundry=bord;
L = bwlabel(BW);
s = regionprops(L, 'centroid');
centroids = cat(1, s.Centroid);
Step #1 - Generating your line
The first thing you need to do is figure out how to draw your line. To make this simple, let's assume that the centre of the human body is stored as an array of cen = [x1 y1] as you have said. Now, supposing you click anywhere in your image, you get another point linePt = [x2 y2]. Let's assume that both the x and y co-ordinates are the horizontal and vertical components respectively. We can find the slope and intercept of this line, then create points between these two points parameterized by the slope and intercept to generate your line points. One thing I will point out is that if we draw a slope with a vertical line, by definition the slope would be infinity. As such, we need to place in a check to see if we have this situation. If we do, we assume that all of the x points are the same, while y varies. Once you have your slope and intercept, simply create points in between the line. You'll have to choose how many points you want along this line yourself as I have no idea about the resolution of your image, nor how big you want the line to be. We will then store this into a variable called linePoints where the first column consists of x values and the second column consists of y values. In other words:
In other words, do this:
%// Define number of points
numPoints = 1000;
%// Recall the equation of the line: y = mx + b, m = (y2-y1)/(x2-x1)
if abs(cen(1) - linePt(1)) < 0.00001 %// If x points are close
yPts = linspace(cen(2), linePt(2), numPoints); %// y points are the ones that vary
xPts = cen(1)*ones(numPoints, 1); %//Make x points the same to make vertical line
else %// Normal case
slp = (cen(2) - linePt(2)) / cen(1) - linePt(1)); %// Solve for slope (m)
icept = cen(2) - slp*cen(1); %// Solve for intercept (b)
xPts = linspace(cen(1), linePt(1), numPoints); %// Vary the x points
yPts = slp*xPts + icept; %// Solve for the y points
end
linePoints = [xPts(:) yPts(:)]; %// Create point matrix
Step #2 - Finding points of intersection
Supposing you have a 2D array of points [x y] where x denotes the horizontal co-ordinates and y denotes the vertical co-ordinates of your line. We can simply find the distance between all of these points in your boundary with all of your points on the line. Should any of the points be under a certain threshold (like 0.0001 for example), then this indicates an intersection. Note that due to the crux of floating point data, we can't check to see if the distance is 0 due to the step size in between each discrete point in your data.
I'm also going to assume border_skeleton returns points of the same format. This method works without specifying what the centroid is. As such, I don't need to use the centroids in the method I'm proposing. Also, I'm going to assume that your line points are stored in a matrix called linePoints that is of the same type that I just talked about.
In other words, do this:
numBoundaryPoints = size(boundry, 1); %// boundary is misspelled in your code BTW
ptsIntersect = []; %// Store points of intersection here
for idx = 1 : numBoundaryPoints %// For each boundary point...
%//Obtain the i'th boundary point
pt = boundry(:,idx);
%//Get distances - This computes the Euclidean distance
%//between the i'th boundary point and all points along your line
dists = sqrt(sum(bsxfun(#minus, linePoints, pt).^2, 2));
%//Figure out which points intersect and store
ptsIntersect = [ptsIntersect; linePoints(dists < 0.0001, :)];
end
In the end, ptsIntersect will store all of the points along the boundary that intersect with this line. Take note that I have made a lot of assumptions here because you haven't (or seem reluctant to) give any more details than what you've specified in your comments.
Good luck.
I'm trying to generate a cloud of 2D points (uniformly) distributed within a triangle. So far, I've achieved the following:
The code I've used is this:
N = 1000;
X = -10:0.1:10;
for i=1:N
j = ceil(rand() * length(X));
x_i = X(j);
y_i = (10 - abs(x_i)) * rand;
E(:, i) = [x_i y_i];
end
However, the points are not uniformly distributed, as clearly seen in the left and right corners. How can I improve that result? I've been trying to search for the different shapes too, with no luck.
You should first ask yourself what would make the points within a triangle distributed uniformly.
To make a long story short, given all three vertices of the triangle, you need to transform two uniformly distributed random values like so:
N = 1000; % # Number of points
V = [-10, 0; 0, 10; 10, 0]; % # Triangle vertices, pairs of (x, y)
t = sqrt(rand(N, 1));
s = rand(N, 1);
P = (1 - t) * V(1, :) + bsxfun(#times, ((1 - s) * V(2, :) + s * V(3, :)), t);
This will produce a set of points which are uniformly distributed inside the specified triangle:
scatter(P(:, 1), P(:, 2), '.')
Note that this solution does not involve repeated conditional manipulation of random numbers, so it cannot potentially fall into an endless loop.
For further reading, have a look at this article.
That concentration of points would be expected from the way you are building the points. Your points are equally distributed along the X axis. At the extremes of the triangle there is approximately the same amount of points present at the center of the triangle, but they are distributed along a much smaller region.
The first and best approach I can think of: brute force. Distribute the points equally around a bigger region, and then delete the ones that are outside the region you are interested in.
N = 1000;
points = zeros(N,2);
n = 0;
while (n < N)
n = n + 1;
x_i = 20*rand-10; % generate a number between -10 and 10
y_i = 10*rand; % generate a number between 0 and 10
if (y_i > 10 - abs(x_i)) % if the points are outside the triangle
n = n - 1; % decrease the counter to try to generate one more point
else % if the point is inside the triangle
points(n,:) = [x_i y_i]; % add it to a list of points
end
end
% plot the points generated
plot(points(:,1), points(:,2), '.');
title ('1000 points randomly distributed inside a triangle');
The result of the code I've posted:
one important disclaimer: Randomly distributed does not mean "uniformly" distributed! If you generate data randomly from an Uniform Distribution, that does not mean that it will be "evenly distributed" along the triangle. You will see, in fact, some clusters of points.
You can imagine that the triangle is split vertically into two halves, and move one half so that together with the other it makes a rectangle. Now you sample uniformly in the rectangle, which is easy, and then move the half triangle back.
Also, it's easier to work with unit lengths (the rectangle becomes a square) and then stretch the triangle to the desired dimensions.
x = [-10 10]; % //triangle base
y = [0 10]; % //triangle height
N = 1000; %// number of points
points = rand(N,2); %// sample uniformly in unit square
ind = points(:,2)>points(:,1); %// points to be unfolded
points(ind,:) = [2-points(ind,2) points(ind,1)]; %// unfold them
points(:,1) = x(1) + (x(2)-x(1))/2*points(:,1); %// stretch x as needed
points(:,2) = y(1) + (y(2)-y(1))*points(:,2); %// stretch y as needed
plot(points(:,1),points(:,2),'.')
We can generalize this case. If you want to sample points from some (n - 1)-dimensional simplex in Euclidean space UNIFORMLY (not necessarily a triangle - it can be any convex polytope), just sample a vector from a symmetric n-dimensional Dirichlet distribution with parameter 1 - these are the convex (or barycentric) coordinates relative to the vertices of the polytope.