I need to generate a fixed number of non-overlapping circles located randomly. I can display circles, in this case 20, located randomly with this piece of code,
for i =1:20
x=0 + (5+5)*rand(1)
y=0 + (5+5)*rand(1)
r=0.5
circle3(x,y,r)
hold on
end
however circles overlap and I would like to avoid this. This was achieved by previous users with Mathematica https://mathematica.stackexchange.com/questions/69649/generate-nonoverlapping-random-circles , but I am using MATLAB and I would like to stick to it.
For reproducibility, this is the function, circle3, I am using to draw the circles
function h = circle3(x,y,r)
d = r*2;
px = x-r;
py = y-r;
h = rectangle('Position',[px py d d],'Curvature',[1,1]);
daspect([1,1,1])
Thank you.
you can save a list of all the previously drawn circles. After
randomizing a new circle check that it doesn't intersects the previously drawn circles.
code example:
nCircles = 20;
circles = zeros(nCircles ,2);
r = 0.5;
for i=1:nCircles
%Flag which holds true whenever a new circle was found
newCircleFound = false;
%loop iteration which runs until finding a circle which doesnt intersect with previous ones
while ~newCircleFound
x = 0 + (5+5)*rand(1);
y = 0 + (5+5)*rand(1);
%calculates distances from previous drawn circles
prevCirclesY = circles(1:i-1,1);
prevCirclesX = circles(1:i-1,2);
distFromPrevCircles = ((prevCirclesX-x).^2+(prevCirclesY-y).^2).^0.5;
%if the distance is not to small - adds the new circle to the list
if i==1 || sum(distFromPrevCircles<=2*r)==0
newCircleFound = true;
circles(i,:) = [y x];
circle3(x,y,r)
end
end
hold on
end
*notice that if the amount of circles is too big relatively to the range in which the x and y coordinates are drawn from, the loop may run infinitely.
in order to avoid it - define this range accordingly (it can be defined as a function of nCircles).
If you're happy with brute-forcing, consider this solution:
N = 60; % number of circles
r = 0.5; % radius
newpt = #() rand([1,2]) * 10; % function to generate a new candidate point
xy = newpt(); % matrix to store XY coordinates
fails = 0; % to avoid looping forever
while size(xy,1) < N
% generate new point and test distance
pt = newpt();
if all(pdist2(xy, pt) > 2*r)
xy = [xy; pt]; % add it
fails = 0; % reset failure counter
else
% increase failure counter,
fails = fails + 1;
% give up if exceeded some threshold
if fails > 1000
error('this is taking too long...');
end
end
end
% plot
plot(xy(:,1), xy(:,2), 'x'), hold on
for i=1:size(xy,1)
circle3(xy(i,1), xy(i,2), r);
end
hold off
Slightly amended code #drorco to make sure exact number of circles I want are drawn
nCircles = 20;
circles = zeros(nCircles ,2);
r = 0.5;
c=0;
for i=1:nCircles
%Flag which holds true whenever a new circle was found
newCircleFound = false;
%loop iteration which runs until finding a circle which doesnt intersect with previous ones
while ~newCircleFound & c<=nCircles
x = 0 + (5+5)*rand(1);
y = 0 + (5+5)*rand(1);
%calculates distances from previous drawn circles
prevCirclesY = circles(1:i-1,1);
prevCirclesX = circles(1:i-1,2);
distFromPrevCircles = ((prevCirclesX-x).^2+(prevCirclesY-y).^2).^0.5;
%if the distance is not to small - adds the new circle to the list
if i==1 || sum(distFromPrevCircles<=2*r)==0
newCircleFound = true;
c=c+1
circles(i,:) = [y x];
circle3(x,y,r)
end
end
hold on
end
Although this is an old post, and because I faced the same problem before I would like to share my solution, which uses anonymous functions: https://github.com/davidnsousa/mcsd/blob/master/mcsd/cells.m . This code allows to create 1, 2 or 3-D cell environments from user-defined cell radii distributions. The purpose was to create a complex environment for monte-carlo simulations of diffusion in biological tissues: https://www.mathworks.com/matlabcentral/fileexchange/67903-davidnsousa-mcsd
A simpler but less flexible version of this code would be the simple case of a 2-D environment. The following creates a space distribution of N randomly positioned and non-overlapping circles with radius R and with minimum distance D from other cells. All packed in a square region of length S.
function C = cells(N, R, D, S)
C = #(x, y, r) 0;
for n=1:N
o = randi(S-R,1,2);
while C(o(1),o(2),2 * R + D) ~= 0
o = randi(S-R,1,2);
end
f = #(x, y) sqrt ((x - o(1)) ^ 2 + (y - o(2)) ^ 2);
c = #(x, y, r) f(x, y) .* (f(x, y) < r);
C = #(x, y, r) + C(x, y, r) + c(x, y, r);
end
C = #(x, y) + C(x, y, R);
end
where the return C is the combined anonymous functions of all circles. Although it is a brute force solution it is fast and elegant, I believe.
Related
I have some Matlab code lines for drawing Double Moon Classification:
function data=dm(r,w,ts,d)
clear all; close all;
if nargin<4, w=6;end
if nargin<3, r=10;end
if nargin<2, d=-4;end
if nargin < 1, ts=1000; end
ts1=10*ts;
done=0; tmp1=[];
while ~done,
tmp=[2*(r+w/2)*(rand(ts1,1)-0.5) (r+w/2)*rand(ts1,1)];
tmp(:,3)=sqrt(tmp(:,1).*tmp(:,1)+tmp(:,2).*tmp(:,2));
idx=find([tmp(:,3)>r-w/2] & [tmp(:,3)<r+w/2]);
tmp1=[tmp1;tmp(idx,1:2)];
if length(idx)>= ts,
done=1;
end
end
data=[tmp1(1:ts,:) zeros(ts,1);
[tmp1(1:ts,1)+r -tmp1(1:ts,2)-d ones(ts,1)]];
plot(data(1:ts,1),data(1:ts,2),'.r',data(ts+1:end,1),data(ts+1:end,2),'.b');
title(['Perceptron with the double-moon set at distance d = ' num2str(d)]),
axis([-r-w/2 2*r+w/2 -r-w/2-d r+w/2])
save dm r w ts d data;
Results:
My question is how to put a line into the Double Moon Classification so that can separate both classification in Matlab code?
I have made a simulation of your problem and wrote a pragmatic solution based on the steps
generates a grid of points fine enough(step size as small as
possible)
find which of these points are in the space between both points
groups
find which points have an equal distance to both groups
plot points
The main part (without my data generation code and the plot I did) of the code is
PBlues = data(1:ts,:);
PReds = data(ts+1:end,:);
Xs = linspace( min(data(:,1)), max(data(:,1)), 100);
Ys = linspace( min(data(:,2)), max(data(:,2)), 100);
[Xs,Ys] = meshgrid( Xs, Ys);
%% compute point relative distance to each point set (ble and red)
bCouldbeInbetween = false(size(Xs));
minDistsb = zeros(size(Xs));
minDistsr = zeros(size(Xs));
for p=1:numel(Xs)
distb = sqrt( (Xs(p)- PBlues(:,1)).^2+(Ys(p)- PBlues(:,2)).^2);
distr = sqrt( (Xs(p)- PReds(:,1)).^2+(Ys(p)- PReds(:,2)).^2);
minDistsb(p) = min(distb);
minDistsr(p) = min(distr);
i = find(distb == minDistsb(p));
j = find(distr == minDistsr(p));
bCouldbeInbetween(p) = (sign(PBlues(i,1)-Xs(p)) ~= sign(PReds(j,1)-Xs(p))) || ...
(sign(PBlues(i,2)-Ys(p)) ~= sign(PReds(j,2)-Ys(p)));
if bCouldbeInbetween(p)
end
end
% point distance difference to each point set
DistDiffs = abs(minDistsb - minDistsr);
% point with equal distance using proportional decision strategy
bCandidates = DistDiffs./ max(minDistsb,minDistsr) < 0.05;
medianLine = [Xs(bCandidates),Ys(bCandidates)];
[~,I] = sort(medianLine(:,1));
medianLine(:,1) = medianLine(I,1);
medianLine(:,2) = medianLine(I,2);
I hope this helps
I have just made a submission which can be downloaded here: https://de.mathworks.com/matlabcentral/fileexchange/66618-linebetweentwopointsgroups-exchange--
This a program that I have, I already asked before for how to find the intersection on my image with a circle, and somebody has an answer it (thank you) and I have another problem...
a = imread('001_4.bmp');
I2 = imcrop(a,[90 5 93 180]);
[i,j]=size(I2);
x_hist=sum(I2,1);
y_hist=(sum(I2,2))';
x=1:j ; y=1:i;
centx=sum(x.*x_hist)/sum(x_hist);
centy=sum(y.*y_hist)/sum(y_hist);
BW = edge(I2,'Canny',0.5);
bw2 = imcomplement(BW);
circle = int32([centx,centy,40]);%<<----------
shapeInserter = vision.ShapeInserter('Fill',false);
release(shapeInserter);
set(shapeInserter,'Shape','Circles');
% construct binary image of circle only
bwCircle = step(shapeInserter,true(size(bw2)),circle);
% find the indexes of the intersection between the binary image and the circle
[i, j] = find ((bw2 | bwCircle) == 0);
figure
imshow(bw2 & bwCircle) % plot the combination of both images
hold on
plot(j, i, 'r*') % plot the intersection points
K = step(shapeInserter,bw2,circle);
[n,m]=size(i);
d=0;
k=1;
while (k < n)
d = d + sqrt((i(k+1)-i(k)).^2 + (j(k+1)-j(k)).^2);
k = k+1;
end
Q: How can I calculate all the existing intersection values(red *) in a clockwise direction?
Not sure, but I think that your teacher meant something different from the answer given in the previous question, because
CW direction can be a hint, not a additional problem
I've seen no clue that the circle should be drawn; for small radius circles are blocky and simple bw_circle & bw_edges may not work. For details please see "Steve on image processing" blog, "Intersecting curves that don't intersect"[1]
May be your teacher is rather old and wants Pascal-stile answer.
If my assumptions are correct the code below should be right
img = zeros(7,7); %just sample image, edges == 1
img(:,4) = 1; img(sub2ind(size(img),1:7,1:7)) = 1;
ccx = 4; % Please notice: x is for columns, y is for rows
ccy = 3;
rad = 2;
theta = [(pi/2):-0.01:(-3*pi/2)];
% while plotting it will appear CCW, for visual CW and de-facto CCW use
% 0:0.01:2*pi
cx = rad * cos(theta) + ccx; % gives slightly different data as for
cy = rad * sin(theta) + ccy; % (x-xc)^2 + (y-yc)^2 == rad^2 [2]
ccoord=[cy;cx]'; % Once again: x is for columns, y is for rows
[ccoord, rem_idx, ~] =unique(round(ccoord),'rows');
cx = ccoord(:,2);
cy = ccoord(:,1);
circ = zeros(size(img)); circ(sub2ind(size(img),cy,cx))=1;
cross_sum = 0;
figure, imshow(img | circ,'initialmagnification',5000)
hold on,
h = [];
for un_ang = 1:length(cx),
tmp_val= img(cy(un_ang),cx(un_ang));
if tmp_val == 1 %the point belongs to edge of interest
cross_sum = cross_sum + tmp_val;
h = plot(cx(un_ang),cy(un_ang),'ro');
pause,
set(h,'marker','x')
end
end
hold off
[1] https://blogs.mathworks.com/steve/2016/04/12/intersecting-curves-that-dont-intersect/
[2] http://matlab.wikia.com/wiki/FAQ#How_do_I_create_a_circle.3F
I have written a code to simulate the motion of circular particles in a 2d box. Whenever they move out of the box, I put them inside the box and near the wall. I want to add the diameter (2R) of particles in the code, which means when the distance between the center of two circles become less than 2R, they separate along the line connecting their centers so that the distance between the centers of the circles becomes equal to 2R.
Could anyone suggest a code to perevent the overlapping of particles?
This is my code in which overlap is not considered:
clear all
close all
l = 224; nn = 800; %number of particles
time = 1000; dd = 1;
x= l*rand(1,nn);
y= l*rand(1,nn);
for t = 1:time;
x= x + rand(1,nn)-0.5* ones(1,nn);
y=y+rand(1,nn)-0.5* ones (1,nn);
index = (x < 0); x(index) = abs(normrnd(0,1,1,nnz(index)));
index = (y < 0); y(index) = abs(normrnd(0,1,1,nnz(index)));
index = (x > l); x(index) = l-abs(normrnd(0,1,1,nnz(index)));
index = (y > l); y(index) = l-abs(normrnd(0,1,1,nnz(index)));
end
Here is some commented code which does what you want. Notably:
psize is some defined particle size for interaction.
point-to-point distances found using pdist2.
points that are too close are moved away from each other by some amount (dp times their current distances, if dp=1/2 then their x and y distances double) until there are no clashes.
See comments for details.
clear; close all;
l = 224; nn = 800; % number of particles
time = 100;
x = l*rand(1,nn); y = l*rand(1,nn);
psize = 2; % Particle size for interaction
dp = 0.1;
figure; hold on; axis([0 l 0 l]);
for t = 1:time;
% Random movement
movement = 2*rand(2,nn)-1;
x = x + movement(1,:);
y = y + movement(2,:);
index = (x < 0); x(index) = abs(normrnd(0,1,1,nnz(index)));
index = (y < 0); y(index) = abs(normrnd(0,1,1,nnz(index)));
index = (x > l); x(index) = l-abs(normrnd(0,1,1,nnz(index)));
index = (y > l); y(index) = l-abs(normrnd(0,1,1,nnz(index)));
% Particle interaction. Loop until there are no clashes. For
% robustness, some max iteration counter should be added!
numclash = 1;
while numclash > 0
dists = pdist2([x;y]', [x;y]'); % Distances between all particles
dists(dists < psize) = NaN; % Those too close are assigned NaN
tooclose = isnan(tril(dists,-1)); % All NaNs identified by logical
[clash1,clash2] = find(tooclose); % Get particles which are clashing
numclash = numel(clash1); % Get number of clashes
% All points where there was a clash, move away from each other
x(clash1) = x(clash1) + (x(clash1)-x(clash2))*dp;
x(clash2) = x(clash2) - (x(clash1)-x(clash2))*dp;
y(clash1) = y(clash1) + (y(clash1)-y(clash2))*dp;
y(clash2) = y(clash2) - (y(clash1)-y(clash2))*dp;
end
% Plot to visualise results. Colour fade from dark to bright green over time
scatter(x,y,'.','markeredgecolor',[0.1,t/time,0.4]);
drawnow;
end
hold off
Result:
Edit:
For a clearer diagram, you could initialise some colour matrix C = rand(nn,3); and plot using
scatter(x,y,[],C*(t/time),'.'); % the (t/time) factor makes it fade from dark to light
This would give each particle a different colour, which also fade from dark to light, rather than just fading from dark to light as before. The result would be something like this:
I need to evaluate a function (say)
Fxy = 2*x.^2 +3 *y.^2;
on a ternary grid x-range (0 - 1), y-range (0-1) and 1-x-y (0 - 1).
I am unable to construct the ternary grid on which I need to evaluate the above function. Also, once evaluated I need to plot the function in a ternary contour plot. Ideally, I need the axes to go counter clockwise in the sense (x -> y--> (1-x-y)).
I have tried the function
function tg = triangle_grid ( n, t )
ng = ( ( n + 1 ) * ( n + 2 ) ) / 2;
tg = zeros ( 2, ng );
p = 0;
for i = 0 : n
for j = 0 : n - i
k = n - i - j;
p = p + 1;
tg(1:2,p) = ( i * t(1:2,1) + j * t(1:2,2) + k * t(1:2,3) ) / n;
end
end
return
end
for the number of sub intervals between the triangle edge coordinates
n = 10 (say)
and for the edge coordinates of an equilateral triangle
t = tcoord = [0.0, 0.5, 1.0;
0.0, 1.0*sqrt(3)/2, 0.0];
This generated a triangular grid with the x-axis from 0-1 but the other two are not from 0-1.
I need something like this:
... with the axes range 0-1 (0-100 would also do).
In addition, I need to know the coordinate points for all intersections within the triangular grid. Once I have this I can proceed to evaluate the function in this grid.
My final aim is to get something like this. This is a better representation of what I need to achieve (as compared to the previous plot which I have now removed)
Note that the two ternary plots have iso-value contours which are different in in magnitude. In my case the difference is an order of magnitude, two very different Fxy's.
If I can plot the two ternary plots on top of each other then and evaluate the compositions at the intersection of two iso-value contours on the ternary plane. The compositions should be as read from the ternary plot and not the rectangular grid on which triangle is defined.
Currently there are issues (as highlighted in the comments section, will update this once the problem is closer to solution).
I am the author of ternplot. As you have correctly surmised, ternpcolor does not do what you want, as it is built to grid data automatically. In retrospect, this was not a particularly wise decision, I've made a note to change the design. In the mean time this code should do what you want:
EDIT: I've changed the code to find the intersection of two curves rather than just one.
N = 10;
x = linspace(0, 1, N);
y = x;
% The grid intersections on your diagram are actually rectangularly arranged,
% so meshgrid will build the intersections for us
[xx, yy] = meshgrid(x, y);
zz = 1 - (xx + yy);
% now that we've got the intersections, we can evaluate the function
f1 = #(x, y) 2*x.^2 + 3*y.^2 + 0.1;
Fxy1 = f1(xx, yy);
Fxy1(xx + yy > 1) = nan;
f2 = #(x, y) 3*x.^2 + 2*y.^2;
Fxy2 = f2(xx, yy);
Fxy2(xx + yy > 1) = nan;
f3 = #(x, y) (3*x.^2 + 2*y.^2) * 1000; % different order of magnitude
Fxy3 = f3(xx, yy);
Fxy3(xx + yy > 1) = nan;
subplot(1, 2, 1)
% This constructs the ternary axes
ternaxes(5);
% These are the coordinates of the compositions mapped to plot coordinates
[xg, yg] = terncoords(xx, yy);
% simpletri constructs the correct triangles
tri = simpletri(N);
hold on
% and now we can plot
trisurf(tri, xg, yg, Fxy1);
trisurf(tri, xg, yg, Fxy2);
hold off
view([137.5, 30]);
subplot(1, 2, 2);
ternaxes(5)
% Here we plot the line of intersection of the two functions
contour(xg, yg, Fxy1 - Fxy2, [0 0], 'r')
axis equal
EDIT 2: If you want to find the point of intersection between two contours, you are effectively solving two simultaneous equations. This bit of extra code will solve that for you (notice I've used some anonymous functions in the code above now, as well):
f1level = 1;
f3level = 1000;
intersection = fsolve(#(v) [f1(v(1), v(2)) - f1level; f3(v(1), v(2)) - f3level], [0.5, 0.4]);
% if you don't have the optimization toolbox, this command works almost as well
intersection = fminsearch(#(v) sum([f1(v(1), v(2)) - f1level; f3(v(1), v(2)) - f3level].^2), [0.5, 0.4]);
ternaxes(5)
hold on
contour(xg, yg, Fxy1, [f1level f1level]);
contour(xg, yg, Fxy3, [f3level f3level]);
ternplot(intersection(1), intersection(2), 1 - sum(intersection), 'r.');
hold off
I have played a bit with the file exchange submission https://www.mathworks.com/matlabcentral/fileexchange/2299-alchemyst-ternplot.
if you just do this:
[x,y]=meshgrid(0:0.1:1);
Fxy = 2*x.^2 +3 *y.^2;
ternpcolor(x(:),y(:),Fxy(:))
You get:
The thirds axis is created exactly as you say (1-x-y) inside the ternpcolor function. There are lots of things to "tune" here but I hope it is enough to get you started.
Here is a solution using R and my package ggtern. I have also included the points within proximity underneath, for the purpose of comparison.
library(ggtern)
Fxy = function(x,y){ 2*x^2 + 3*y^2 }
x = y = seq(0,1,length.out = 100)
df = expand.grid(x=x,y=y);
df$z = 1 - df$x - df$y
df = subset(df,z >= 0)
df$value = Fxy(df$x,df$y)
#The Intended Breaks
breaks = pretty(df$value,n=10)
#Create subset of the data, within close proximity to the breaks
df.sub = ldply(breaks,function(b,proximity = 0.02){
s = b - abs(proximity)/2; f = b + abs(proximity)/2
subset(df,value >= s & value <= f)
})
#Plot the ternary diagram
ggtern(df,aes(x,y,z)) +
theme_bw() +
geom_point(data=df.sub,alpha=0.5,color='red',shape=21) +
geom_interpolate_tern(aes(value = value,color=..level..), size = 1, n = 200,
breaks = c(breaks,max(df$value) - 0.01,min(df$value) + 0.01),
base = 'identity',
formula = value ~ poly(x,y,degree=2)) +
labs(title = "Contour Plot on Modelled Surface", x = "Left",y="Top",z="Right")
Which produces the following:
I am working on rotating image manually in Matlab. Each time I run my code with a different image the previous images which are rotated are shown in the Figure. I couldn't figure it out. Any help would be appreciable.
The code is here:
[screenshot]
im1 = imread('gradient.jpg');
[h, w, p] = size(im1);
theta = pi/12;
hh = round( h*cos(theta) + w*abs(sin(theta))); %Round to nearest integer
ww = round( w*cos(theta) + h*abs(sin(theta))); %Round to nearest integer
R = [cos(theta) -sin(theta); sin(theta) cos(theta)];
T = [w/2; h/2];
RT = [inv(R) T; 0 0 1];
for z = 1:p
for x = 1:ww
for y = 1:hh
% Using matrix multiplication
i = zeros(3,1);
i = RT*[x-ww/2; y-hh/2; 1];
%% Nearest Neighbour
i = round(i);
if i(1)>0 && i(2)>0 && i(1)<=w && i(2)<=h
im2(y,x,z) = im1(i(2),i(1),z);
end
end
end
end
x=1:ww;
y=1:hh;
[X, Y] = meshgrid(x,y); % Generate X and Y arrays for 3-D plots
orig_pos = [X(:)' ; Y(:)' ; ones(1,numel(X))]; % Number of elements in array or subscripted array expression
orig_pos_2 = [X(:)'-(ww/2) ; Y(:)'-(hh/2) ; ones(1,numel(X))];
new_pos = round(RT*orig_pos_2); % Round to nearest neighbour
% Check if new positions fall from map:
valid_pos = new_pos(1,:)>=1 & new_pos(1,:)<=w & new_pos(2,:)>=1 & new_pos(2,:)<=h;
orig_pos = orig_pos(:,valid_pos);
new_pos = new_pos(:,valid_pos);
siz = size(im1);
siz2 = size(im2);
% Expand the 2D indices to include the third dimension.
ind_orig_pos = sub2ind(siz2,orig_pos(2*ones(p,1),:),orig_pos(ones(p,1),:), (1:p)'*ones(1,length(orig_pos)));
ind_new_pos = sub2ind(siz, new_pos(2*ones(p,1),:), new_pos(ones(p,1),:), (1:p)'*ones(1,length(new_pos)));
im2(ind_orig_pos) = im1(ind_new_pos);
imshow(im2);
There is a problem with the initialization of im2, or rather, the lack of it. im2 is created in the section shown below:
if i(1)>0 && i(2)>0 && i(1)<=w && i(2)<=h
im2(y,x,z) = im1(i(2),i(1),z);
end
If im2 exists before this code is run and its width or height is larger than the image you are generating the new image will only overwrite the top left corner of your existing im2. Try initializing im2 by adding adding
im2 = zeros(hh, ww, p);
before
for z = 1:p
for x = 1:ww
for y = 1:hh
...
As a bonus it might make your code a little faster since Matlab won't have to resize im2 as it grows in the loop.