RandomPoints &condition distance

7 次查看(过去 30 天)
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

采纳的回答

Jan
Jan 2017-1-30
编辑:Jan 2017-2-14
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.
  2 个评论
cheulhee kwon
cheulhee kwon 2019-3-7
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
Image Analyst 2019-3-7
Try
distance = hypot(A, B)
or
distance = sqrt((ax-bx).^2+(ay-by).^2)

请先登录,再进行评论。

更多回答(3 个)

John BG
John BG 2017-1-31
编辑:John Kelly 2017-1-31
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

Image Analyst
Image Analyst 2017-2-20
If you have list of the circle centers and radii, you can simply use the viscircles() function to plot the circles.
viscircles(centers, radii);
  2 个评论
Marwen Tarhouni
Marwen Tarhouni 2017-2-21
编辑:Walter Roberson 2017-2-22
@ Image Analyst
Error in GetPointsRandom (line 37)
centers=[X,Y];
radii=15;
viscircles(centers, radii);
Walter Roberson
Walter Roberson 2017-2-22
centers=[X,Y];
radii = 15 * ones(size(centers,1), 1);
viscircles(centers, radii);

请先登录,再进行评论。


Manik K
Manik K 2022-2-20
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
  1 个评论
Image Analyst
Image Analyst 2022-2-20
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.

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Marine and Underwater Vehicles 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by