scatter randomized points with circles
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Hello,
How we can draw circles of radius R on randomized points in a area100*100? Knowing that its points are located in the following way:
The minimum distance between all points> 6 meters.
do you have an idea ?
Thank you
1 个评论
Jan
2017-2-14
What is the difference to your question https://www.mathworks.com/matlabcentral/answers/322431-randompoints-condition-distance? Do you have the coordinates of the points already and only want to draw the circles?
采纳的回答
John BG
2017-2-20
Hi Marwen
Here are 2 functions that solve the question, not randomly guessing points from the entire area, but excluding gradually the area sized by already allocated points:
function [X,Y,Nmax,Dmatrix,delay]=scatter_points7(void)
% given H and W size of 2D rectangle scatter_points7.m does the following:
% 1. calculate Nmax, the maximum amount of circles radius R0 that orderly fit in rectangle HxW
% 2. scatter Ap random points with spatial resolution dx =- 0.1 and dy=0.1
% and spaced at least distance R0
% if requested amount Ap>Nmax rand2Dpoints breaks because there
% is no space to fit in so many random points complying with min
% distance R0
% 3. calculate distance matrix Dmatrix among all points
% 4. return the coordinates of all generated points in X and Y, along with Dmatrix and Nmax
% 5. plot points
% 6. plot safety circles to visually verify
%
% the amount of generated points is numel(X) and cannot be larger than Nmax
% because of the request for the points to be randomly generated.
%
% in this initial version, only manual input through message box.
%
% call examples:
% 1.
% [X,Y,Dmatrix,Nmax]=scatter_points7
%
% 2.
% [X,Y,Nmax]=scatter_points7
% 2 examples how to verify distance values and check distances meet requirement > R0
% 1.
% L=combinator(Ap,2,'c');
% relD2=((X(L(:,2))-X(L(:,1))).^2+(Y(L(:,2))-Y(L(:,1))).^2).^.5
% find(relD2<R0)
%
% 2.
% L2=combinator(Ap,2);
% Dmatrix=reshape(((X(L2(:,2))-X(L2(:,1))).^2+(Y(L2(:,2))-Y(L2(:,1))).^2).^.5,[Ap Ap]);
% Dmatrix([1:21:end])=NaN ;
% Dmatrix(Dmatrix<6)
%
% February 10th 2017
% version: 1.0
% author: John Bofarull Guix, any feedback to build next version is welcome at
% jgb2012@sky.com or through the Mathworks website
% this script was inspired by Mr Marwen Tarhoumi marwentarhoumi@rocketmail.com
% and Matt Fig's popkenai@yahoo.com mighty function combinator.m available
% from Mathworks File exchange
clear all;clc;close all
format bank
rng('Shuffle')
prompt = {'rectangle width W: ','rectange height H: ','amount points to scatter: ','safety distance Radius: '};
dlg_title = 'Input';
num_lines = 1;
defaultans = {'100','100','20','3'};
input_answer = inputdlg(prompt,dlg_title,num_lines,defaultans);
Diam0=2*str2double(input_answer{4});
W=str2double(input_answer{1})+Diam0;
H=str2double(input_answer{2})+Diam0;
Ap=str2double(input_answer{3});
Ap=floor(Ap);
Nmax=calc_amount_circles(W-Diam0,H-Diam0,Diam0/2);
fprintf('\nRectangle %f x %f has\n max capacity: %i circles radius %f\n',W-Diam0,H-Diam0,Nmax,Diam0/2);
if Nmax<Ap
fprintf('\nCannot fit in more than %i circles.\n',Nmax);
X=0;Y=0;Nmax=0;Dmatrix=0;
return;
end % error message in case requested amount points above Nmax
As=21; % amount sides polygon approximating circles
a=linspace(0,2*pi,As); % angle for circles
dx=1;dy=1;refine=0;
X=zeros(1,Ap);Y=zeros(1,Ap);
if W*H>=900^2
dx=1;dy=1;
refine=1;
else
dx=.1;dy=.1;
refine=0;
end
x_grid=[(-W+Diam0)/2:dx:(W-Diam0)/2];y_grid=[(-H+Diam0)/2:dy:(H-Diam0)/2]; % avoid circles hitting frame
[X_grid,Y_grid]=meshgrid(x_grid,y_grid);
P=[X_grid(:)';Y_grid(:)'];[sz1P,sz2P]=size(P);
if refine==0
xc2_base=Diam0/2*cos(a);yc2_base=Diam0/2*sin(a);
elseif refine==1
xc2_base=(Diam0/2+1)*cos(a);yc2_base=(Diam0/2+1)*sin(a);
end
figure;ax=gca;ax.DataAspectRatio=[1 1 1];
ax.XLim=[-W/2 W/2];ax.YLim=[-H/2 H/2];
ax.XTick=[(-W+Diam0)/2:10:(W-Diam0)/2];ax.YTick=[(-H+Diam0)/2:10:(H-Diam0)/2];grid on;hold all;
perimeter=[(-W+Diam0)/2-.25 (W-Diam0)/2+.25 (W-Diam0)/2+.25 (-W+Diam0)/2-.25 (-W+Diam0)/2-.25;
(-H+Diam0)/2-.25 (-H+Diam0)/2-.25 (H-Diam0)/2+.25 (H-Diam0)/2+.25 (-H+Diam0)/2-.25];
plot(perimeter(1,:),perimeter(2,:),'Color',[.3 .3 .3]);
tic;
for k=1:1:Ap
[sz1P,sz2P]=size(P);
if sz2P>0
nP = randi(sz2P,1,1);
else
break
end
if refine==0
X(k)=P(1,nP);Y(k)=P(2,nP);
elseif refine==1
dec_xP=randi([0 499],1,1);dec_xP=dec_xP/1e3; % worst case added jitter bringing 2 points on crash course
dec_yP=randi([0 499],1,1);dec_yP=dec_yP/1e3;
X(k)=P(1,nP)+dec_xP;Y(k)=P(2,nP);Y(k)=P(2,nP)+dec_yP;
end
xc2=xc2_base+X(k);yc2=yc2_base+Y(k);
in=inpolygon(P(1,:),P(2,:),xc2,yc2);
P(:,find(in>0))=[]; % exclude area already busy
figure(1);plot(X(k),Y(k),'r*'); % centre circles
figure(1);plot(xc2,yc2,'Color',[0.8 0.8 1]); % circles radius Diam0/2
end
L2=combinator(Ap,2); % test 2
Dmatrix=reshape(((X(L2(:,2))-X(L2(:,1))).^2+(Y(L2(:,2))-Y(L2(:,1))).^2).^.5,[Ap Ap]);
Dmatrix([1:Ap+1:end])=NaN ;
hold off
delay=toc;
end
function Anxny=calc_amount_circles(H_,L_,D_)
% calculates 1. amount of circles in hex pattern that fit within 2D rectangle L (columns) x H (tall, lines or rows)
% used graph from http://www.engineeringtoolbox.com/circles-within-rectangle-d_1905.html to calibrate
R_=D_/2;
if L_<(2*R_)
Nx=0;
else Nx=floor(L_/(2*R_));
end;
if H_<(2*R_)
Ny=0;
else
s=1;
while H_/(2*R_+s*R_*3^.5)>1;
s=s+1;
end
Ny=s;
end;
if rem(L_,2*R_)>=R_
min1_evenlines=0;
else min1_evenlines=1;
end;
Anxny=Nx*Ny-floor(Ny/2)*min1_evenlines;
end
%
% function handling_input_errors
% % input checks
% narginchk(5,5);nargoutchk(4,4);
%
% err_message={'error input type W';'error input type H';'error input type R0';'error input type R0';'error input saturate'};
%
% if(~isreal(H) || ~isscalar(W) || W<=0 )
% error(err_message{1});
% end
% if(~isreal(H) || ~isscalar(H) || H<=0 )
% error(err_message{2});
% end
% if(~isreal(R0) || ~isscalar(R0) || R0<=0 )
% error(err_message{3});
% end
%
% if(~isreal(Ap) || ~isscalar(Ap) || Ap<=0 )
% error(err_message{3});
% end
% if(~isreal(saturate) || ~isscalar(saturate) || saturate<0 || saturate>1)
% error(err_message{3});
% end
% end
and the saturation:
function [X,Y,Nmax,Dmatrix]=scatter_points_saturate(W0,H0,R0)
% scatter_points_saturate.m
% given H and W size of 2D rectangle scatter_points7.m does the following:
% 1. calculate Nmax, the maximum amount of circles radius R0 that orderly fit in rectangle HxW
% 2. scatter as many random points as possible with spatial resolution dx =- 0.1 and dy=0.1
% and spaced at least distance R0
% 3. calculate distance matrix Dmatrix among all points
% 4. return the coordinates of all generated points in X and Y, along with Dmatrix and Nmax
% 5. plot points
% 6. plot safety circles to visually verify
%
% the amount of generated points is numel(X) and cannot be larger than Nmax
%
%
% call examples:
% 1.
% [X,Y,Dmatrix,Nmax]=scatter_points_saturate(100,100,3)
%
% 2.
% [X,Y,Nmax]=scatter_points7
% 2 examples how to verify distance values and check distances meet requirement > R0
% 1.
% L=combinator(Ap,2,'c');
% relD2=((X(L(:,2))-X(L(:,1))).^2+(Y(L(:,2))-Y(L(:,1))).^2).^.5
% find(relD2<R0)
%
% 2.
% L2=combinator(Ap,2);
% Dmatrix=reshape(((X(L2(:,2))-X(L2(:,1))).^2+(Y(L2(:,2))-Y(L2(:,1))).^2).^.5,[Ap Ap]);
% Dmatrix([1:21:end])=NaN ;
% Dmatrix(Dmatrix<6)
%
% February 10th 2017
% version: 1.0
% author: John Bofarull Guix, any feedback to build next version is welcome at
% jgb2012@sky.com or through the Mathworks website
% this script was inspired by Mr Marwen Tarhoumi marwentarhoumi@rocketmail.com
% and Matt Fig's popkenai@yahoo.com mighty function combinator.m available
% from Mathworks File exchange
% clear all;clc;close all
% format bank
% rng('Shuffle')
% R0=35
Diam0=2*R0; % distance requirement among adjacent points not to be closer than
% W0=50
% H0=70
W=W0+Diam0; % rectangle width, columns
H=H0+Diam0; % rectangle tall, row
Nmax=calc_amount_circles(W0,H0,Diam0/2)
Ap=20
As=21; a=linspace(0,2*pi,As);xc=Diam0*cos(a);yc=Diam0*sin(a);
dx=1;dy=1;
X=zeros(1,Ap);Y=zeros(1,Ap);
if W*H>=1e6
dx=1;dy=1;
else dx=.1;dy=.1;
end
X=zeros(1,Ap);Y=zeros(1,Ap);
x_grid=[-W0/2:dx:W0/2];y_grid=[-H0/2:dy:H0/2]; % 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=Diam0/2*cos(a);yc2_base=Diam0/2*sin(a);
% xc1_base=Diam0*cos(a);yc1_base=Diam0*sin(a);
figure;ax=gca;ax.DataAspectRatio=[1 1 1]
ax.XLim=[-W/2 W/2];ax.YLim=[-H/2 H/2];
ax.XTick=[(-W+Diam0)/2:10:(W-Diam0)/2];ax.YTick=[(-H+Diam0)/2:10:(H-Diam0)/2];grid on;hold all;
perimeter=[(-W+Diam0)/2-.25 (W-Diam0)/2+.25 (W-Diam0)/2+.25 (-W+Diam0)/2-.25 (-W+Diam0)/2-.25;
(-H+Diam0)/2-.25 (-H+Diam0)/2-.25 (H-Diam0)/2+.25 (H-Diam0)/2+.25 (-H+Diam0)/2-.25];
plot(perimeter(1,:),perimeter(2,:),'Color',[.3 .3 .3]);
k=1
while sz2P>Diam0*.2
k=k+1;
[sz1P sz2P]=size(P);
if sz2P>0
nP = randi(sz2P,1,1);
else
break
end
X(k)=P(1,nP);Y(k)=P(2,nP);
% xc1=xc1_base+X(k);yc1=yc1_base+Y(k);
xc2=xc2_base+X(k);yc2=yc2_base+Y(k);
in=inpolygon(P(1,:),P(2,:),xc2,yc2);
in=in(:)';P(:,find(in>0))=[]; % exclude area already busy
figure(1);plot(X(k),Y(k),'g*'); % centre circles
figure(1);plot(xc2,yc2,'Color',[0.7 .9 0.7]); % circles radius Diam0/2
end
L2=combinator(Ap,2); % test 2
Dmatrix=reshape(((X(L2(:,2))-X(L2(:,1))).^2+(Y(L2(:,2))-Y(L2(:,1))).^2).^.5,[Ap Ap]);
Dmatrix([1:Ap+1:end])=NaN ;
end
function Anxny=calc_amount_circles(H_,L_,D_)
% calculates 1. amount of circles in hex pattern that fit within 2D rectangle L (columns) x H (tall, lines or rows)
% used graph from http://www.engineeringtoolbox.com/circles-within-rectangle-d_1905.html to calibrate
R_=D_/2;
if L_<(2*R_)
Nx=0;
else Nx=floor(L_/(2*R_));
end;
if H_<(2*R_) Ny=0;
else
s=1;
while H_/(2*R_+s*R_*3^.5)>1;
s=s+1;
end
Ny=s;
end;
if rem(L_,2*R_)>=R_
min1_evenlines=0;
else min1_evenlines=1;
end;
Anxny=Nx*Ny-floor(Ny/2)*min1_evenlines;
end
function handling_input_errors
narginchk(5,5);nargoutchk(4,4);
err_message={'error input type W';'error input type H';'error input type R0';'error input type R0';'error input saturate'}
if(~isreal(H) || ~isscalar(W) || W<=0 ) error(err_message{1}); end
if(~isreal(H) || ~isscalar(H) || H<=0 ) error(err_message{2}); end
if(~isreal(R0) || ~isscalar(R0) || R0<=0 ) error(err_message{3}); end
Ap=floor(Ap); if(~isreal(Ap) || ~isscalar(Ap) || Ap<=0 ) error(err_message{3}); end
if(~isreal(saturate) || ~isscalar(saturate) || saturate<0 || saturate>1) error(err_message{3}); end
end
the function combinator is available from here
attached scatter_points7.m and scatter_points_saturate.m
Appreciating time and attention
John Bofarull Guix
更多回答（2 个）
You can obtain the coordinates from the solution posted here: https://www.mathworks.com/matlabcentral/answers/142322-find-n-random-points-with-a-minimum-distance-r-inside-a-2d-rectangular-box#answer_254413. Then to draw the circles:
nWant = 20;
Dist = 6;
[X, Y] = GetPointsRandom(nWant, 100, 100, Dist);
alpha = linspace(0, 2*pi, 32).';
% cx = X.' + 0.5 * Dist * cos(alpha); % Matlab >= 2016b
% cy = Y.' + 0.5 * Dist * sin(alpha); % Matlab >= 2016b
cx = bsxfun(@plus, X.', 0.5 * Dist * cos(alpha));
cy = bsxfun(@plus, Y.', 0.5 * Dist * sin(alpha));
figure;
AxesH = axes('NextPlot', 'add');
plot(cx, cy, 'b', 'Parent', AxesH);
axis equal
set(AxesH, 'XLim', [0, 100], 'YLim', [0, 100]);
7 个评论
Marwen Tarhouni
2017-2-16
编辑：Marwen Tarhouni
2017-2-16
@Jan Simon With execution Of this I get nWant = 20;
Dist = 6;
[X, Y] = GetPointsRandom(nWant, 100, 100, Dist);
alpha = linspace(0, 2*pi, 15).';
cx = bsxfun(@plus, X.', 0.5 * Dist * cos(alpha));
cy = bsxfun(@plus, Y.', 0.5 * Dist * sin(alpha)); axes('XLim', [0, 100], 'YLim', [0, 100], 'NextPlot', 'add');
plot(cx, cy, 'b');
axis equal
@Marwen: Are you sure, that you use the code of GetPointsRandom from Answers: 142322? Here the complete code again with the bsxfun version for < R2016b:
function myTest
for nWant = [20, 200]
Dist = 6;
[X, Y] = GetPointsRandom(nWant, 100, 100, Dist);
alpha = linspace(0, 2*pi, 15).'; % 15 instead of 32, as you like
cx = bsxfun(@plus, X.', 0.5 * Dist * cos(alpha));
cy = bsxfun(@plus, Y.', 0.5 * Dist * sin(alpha));
figure; % Modified for nice XLimits in R2016b:
AxesH = axes('NextPlot', 'add');
plot(cx, cy, 'b', 'Parent', AxesH);
axis equal
set(AxesH, 'XLim', [0, 100], 'YLim', [0, 100]);
end
end
function [X, Y] = GetPointsRandom(nWant, XWidth, YWidth, MinDist)
X = zeros(nWant, 1);
Y = zeros(nWant, 1);
dist_2 = MinDist ^ 2; % Squared once instead of SQRT each time
iLoop = 1; % Security break to avoid infinite loop
nValid = 0;
while nValid < nWant && iLoop < 1e6
newX = XWidth * rand;
newY = YWidth * rand;
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
% Throw an error, if the area is filled too densely:
if nValid < nWant
error('Cannot find wanted number of points in %d iterations.', iLoop)
end
end
I added a figure, because in your screenshot two axes are drawn on eachother. But this could not lead to the strange overlapping circles. The result on R2016b:
This differs substantially from you screenshot and I cannot explain why. Does your code differ from this one?
You see that 200 is near to the possible limit and then all methods to find such a dense random packing can fail. In the worst case an algorithm sets the points to a regular triangular grid with the double distance minus eps, such that the area is almost empty, but this very unlikely.
Please note, that this code allows a distance greater equal 6. If only greater 6 is wanted replace ">= dist_2" with "> dist_2" .
Marwen Tarhouni
2017-2-21
编辑：Marwen Tarhouni
2017-2-21
I completely agree with Marwen: the centre points have to be clearly indicated because each point is a mobile phone, the R circles tell each range (a fair approximation of how far each mobile phone may reach individually, am I right?), and the overlapping circles are visually helpful to verify no two points are closer than the required minimal distance.
Marwen merçi beacou und Viel Spaß
John BG
@Marwen: The center point of the circles are not displayed in my figures, because this detail has not been mentioned in this question. If you want to draw the circles with a radius of 15, use:
alpha = linspace(0, 2*pi, 32).';
R = 15;
cx = bsxfun(@plus, X.', R * cos(alpha));
cy = bsxfun(@plus, Y.', R * sin(alpha));
Add this to get the center points also:
plot(X, Y, '*');
You mention "'XLim', [-100, 100]" and "the area 100 * 100". The first is an area of 200*200.
"The maximum points that respect the 6 meter condition" is not uniquely defined, because this depends on the random locations. If you want the maximum number, and not a specified number, this might be a different problem. So I'm starting to get confused about what you need.
"Cell phones"? This detail has not been mentioned before and for the solution it does not matter.
John BG
2017-2-23
the centre points are not 'a detail' but the utmost important point, it's the first thing you should plot when considering visual presentation of anything attempting to answer this and similar questions.
John BG
Jan
2017-2-24
@John: The question was:
How we can draw circles of radius R on randomized points in
a area 100*100?
Then the center is "a detail [which] has not been mentioned".
KSSV
2017-2-14
0 个投票
2 个评论
Marwen Tarhouni
2017-2-14
KSSV
2017-2-14
That link....creates pints.....around those points circles can be drawn. Isn't it?
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