RandomPoints &condition distance

Dear

I need to draw a figure that contains 20 randomized points in an area [-100 100]. If we have the distance between two points less than 6 meters .It is necessary to repeat the randamization of the points

Then each point must know its distance from the other points

close all;
clear all;
clc;
set(gca,'xtick',-100:20:100);
set(gca,'ytick',-100:20:100);

 axis([-100 100 -100 100]);
X = 200 * rand(1,20) - 100;
Y = 200 * rand(1,20) - 100;
plot(X, Y, 'b*');
grid on
hold on
datacursormode
%calculate the distance between a point and the other points
%exemple the distance between (A and B) & (A and C) &(A and D) & (A and E)...
%also (B and A) & (B and C) &(B and D) & (B and E).......
distance=[];
for i=1:20
dist =  sqrt((X(i)-X(:)).^2+(Y(i)-Y(:)).^2);  
distance=[distance dist];
end

thank you

[Moved from asnwer section:]

My question is to display an imege of 20 randomized points and if we have the distance between two points <6meters I repeat the randomization of these two points. And after all if the condition is fulfilled. Each point must know the distance with the other 19 points so I want to find a distance matrix in the workspace that contains 20 row and 20 column as indicated my code.

It is varied that in my code I find in the workspace the matrix distance that is to say each point know its distance with respect to the other points but the condition so that if the distance between two points is <6meters is not realized

I would like my question to be clear

Answer 1

[Jan]

Current version, [Last EDITED, 01.02.2017 16:52 UTC]

function [X, Y, D] = GetPointsRandom(nWant, XWidth, YWidth, R)

X       = zeros(nWant, 1);
Y       = zeros(nWant, 1);
dist_2  = R ^ 2;  % Squared once instead of SQRT each time
iLoop   = 1;      % Security break to yoid infinite loop
nValid  = 0;
while nValid < nWant && iLoop < 1e6
  newX = XWidth * (rand - 0.5);
  newY = YWidth * (rand - 0.5);
  if all(((X(1:nValid) - newX).^2 + (Y(1:nValid) - newY).^2) > dist_2)
    % Success: The new point does not touch existing points:
    nValid    = nValid + 1;  % Append this point
    X(nValid) = newX;
    Y(nValid) = newY;
  end
  iLoop = iLoop + 1;
end

% An error is safer than a warning:
if nValid < nWant
  error('Cannot find wanted number of points in %d iterations.', iLoop)
end

% [Edited start] "figure" inserted, 'Parent' used, 'XGrid' inserted:
FigH  = figure;
AxesH = axes('XTick', -100:20:100, 'YTick', -100:20:100, ...
     'XLim', [-100, 100], 'YLim', [-100, 100], ...
     'XGrid', 'on', 'YGrid', 'on', 'NextPlot', 'add', ...
     'Parent', FigH);   
plot(X, Y, 'b*', 'Parent', AxesH);
% [EDITED end]

if nargout > 2
  % D = pdist([X, Y]);   % Faster with statistics toolbox
  D = sqrt(bsxfun(@minus, X, X.') .^ 2 + bsxfun(@minus, Y, Y.') .^ 2);
end
end

Call this e.g. as:

[X, Y, D] = GetPointsRandom(1000, 1e6, 1e6, 6)

The list of coordinates is expanded, when a new random point does not touch any exitsing points.

If you do not have the Statistics Toolbox for pdist, use e.g. http://www.mathworks.com/matlabcentral/fileexchange/15145-distance-matrix.

The suggested method for the distance matrix computes the full matrix, although it is symmetric. For in the number of points is small (< 10000), this is not tragic.

This rejection method works reliably and fast up to a number of about 700 points for a side length of 200 and a diameter of 6. If it fails due to reaching the iteration limit, check if increasing the limit helps. If not, the area is nearly filled up by circles and there will not be a solution for your problem.

[cheulhee kwon]

hello sir. it is very helpful to my project.

But, i'm very poor at this matlab.

can you answer my question?

if you can't

how to measure two point distance

A(fixed) and B(random)

i wanna fix A point and

make random B point

and wanna find the distance

thank u for reading this

[Image Analyst]

Try

distance = hypot(A, B) 

or

distance = sqrt((ax-bx).^2+(ay-by).^2)

Answer 2

[John BG]

Marwen

please test the following

clear all;clc;
format bank
rng('Shuffle')

figure;ax=gca;ax.DataAspectRatio=[1 1 1]
ax.XLim=[-100 100];ax.YLim=[-100 100];
ax.XTick=[-100:20:100];ax.YTick=[-100:20:100];grid on;hold all;

R0=6;a=linspace(0,2*pi,20);xc=R0*cos(a);yc=R0*sin(a);

X=zeros(1,20);Y=zeros(1,20);
x_grid=[-100+R0:1:100-R0];y_grid=[-100+R0:1:100-R0]; % avoid circles hitting frame
[X_grid,Y_grid]=meshgrid(x_grid,y_grid);
X_grid0=X_grid;Y_grid0=Y_grid;

P=[X_grid(:)';Y_grid(:)'];

[sz1P sz2P]=size(P)

xc2_base=2*R0*cos(a);yc2_base=2*R0*sin(a); 
xc1_base=R0*cos(a);yc1_base=R0*sin(a);

for k=1:1:20

     [sz1P sz2P]=size(P);
     nP = randi(sz2P,1,1);
     X(k)=P(1,nP);Y(k)=P(2,nP);
     xc1=xc1_base+X(k);yc1=yc1_base+Y(k);
     in=inpolygon(P(1,:),P(2,:),xc1,yc1);

     in=in(:)';
     P(:,find(in>0))=[];

     plot(xc1,yc1,'b'); % R0 radius 
     plot(X(k),Y(k),'r*'); % centre circle

end

.

now regardless of the amount of the amount of times tested, the points are further than R0 each other, any pair, first run, no repetitions needed.

This removes the 'gamble' factor that Simon wants you to include in your testing.

.

L2=combinator(20,2);
Dmatrix=reshape(((X(L2(:,2))-X(L2(:,1))).^2+(Y(L2(:,2))-Y(L2(:,1))).^2).^.5,[20 20]);
D2=Dmatrix+100*eye(20);
D2(D2<6)

testing with the linear distance vector that I use, no need for the diagonal nulls

L=combinator(20,2,'c');
relD2=((X(L(:,2))-X(L(:,1))).^2+(Y(L(:,2))-Y(L(:,1))).^2).^.5
find(relD2<R0)

therefore

I consider this to be a valid answer that does not include faith in not hitting a toss of points too close.

Appreciating time and attention,

John BG

<mailto:[e-mail] [e-mail]>

Answer 3

[Image Analyst]

If you have list of the circle centers and radii, you can simply use the viscircles() function to plot the circles.

viscircles(centers, radii);

[Marwen Tarhouni]

@ Image Analyst

Error in GetPointsRandom (line 37)
centers=[X,Y];
radii=15;
viscircles(centers, radii);

[Walter Roberson]

centers=[X,Y];
radii = 15 * ones(size(centers,1), 1);
viscircles(centers, radii);

Answer 4

[Manik K]

Hi, @John BG

I would like to to this for multiple species ( particles of different sizes ) in box.

I have succesfully implementes this with one by one random points generation and selection condition. but it is very slow.

could you check weather it can be done with your combinatorics method

thanks

manik

[Image Analyst]

John BG hasn't been seen here in a long time. Not sure what you mean by different species, but you can pass a 'MarkerSize' property into plot(), otherwise you should start a new question and include your code and explain what you need that is different than your (hopefully very well commented) code.