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Model of a crowd on concert venue or how to distribute random points according to the 2D window distribution

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Y I on 28 Nov 2018
Edited: Bruno Luong on 26 Sep 2019
Could you please help me with the following issue: I need to "simulate" the crowd in a concert venue by the set of cylinders ( for example with height of 1.7 m and radius of 0.25 m with different concentration (number of cylinders per m^2) ).
In fact the task is to find the coordinates (x, y) of these cylinders.
In real case concentration of people decreases to the edges of the venue and to its back part. For example near the stage we have 2 pers/m^2 and it decreases to 0 (no people) to the edges of the venue and to its back part. Attached picture is an example what I need to do, but there is a constant concentration. пример.jpg
Using a 2D window, I modeled a change in concentration on the venue (I attach a picture and a mat file) concentration.jpg
contourf(x, y, Nu, 'ShowText','on')
shading interp
xlabel('X (Length), m')
ylabel('Y (Width), m')
title('People concentration variation through the venue')
Now, according to this distribution, I need to find the coordinates of the points (in fact coordinates of the cylinders.) ... and these points should be randomly placed within a given concentration, I mean not in line as an array ... that is, in fact, I need to simulate more or less real placement of crowd.
To understand better I need to do something like this пример1.JPG but taking into account than cylinders have a radius of 0.25 m.
I will be very grateful! Because I can not figure out how to do it.
Thank you! If it's not clear, don't hesitate to ask me, because I really hope for your help!

  1 Comment

Torsten on 28 Nov 2018
Discard the choice of a random point if the minimum distance to the points already generated is smaller than 0.5.

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Answers (1)

Bruno Luong
Bruno Luong on 28 Nov 2018
Edited: Bruno Luong on 26 Sep 2019
Torsten's rejection might bias the distribution.
One of the better approach is to repell the points when they are close to each other. The constraints are not always meet but it's very reasonable overall.
You can run Torsten's rejection afterward if needed. Otherwise consider the overlapping circles as mom carrying her baby :-).
L = 20; % <-- Choose length of square sides
x0 = L/2; y0 = L/2; % <-- Choose center of square
n = 500; % <-- Choose number of points
% Generate 2D epanechnikov-distribution
X = [x0,y0] + (sin(asin(2*rand(n,2)-1)/3))*L;
XYR = [x0,y0]+[[-1;1;1;-1;-1],[-1;-1;1;1;-1]]*L/2;
XB = interp1((0:4)'*L,XYR,linspace(0,4*L,200));
XB(end,:) = [];
nrepulsion = 50;
% Repulsion of seeds to avoid them to be too close to each other
n = size(X,1);
Xmin = [x0-L/2,y0-L/2];
Xmax = [x0+L/2,y0+L/2];
% Point on boundary
XR = x0+[-1,1,1,-1,-1]*L/2;
YR = y0+[-1,-1,1,1,-1]*L/2;
hold on
h = plot(X(:,1),X(:,2),'b.');
axis equal
dmin = 0.5;
d2min = dmin*dmin;
beta = 0.5;
for k = 1:nrepulsion
XALL = [X; XB];
DT = delaunayTriangulation(XALL);
T = DT.ConnectivityList;
containX = ismember(T,1:n);
b = any(containX,2);
TX = T(b,:);
[r,i0] = find(containX(b,:));
i = mod(i0+(-1:1),3)+1;
i = TX(r + (i-1)*size(TX,1));
T = accumarray([i(:,1);i(:,1)],[i(:,2);i(:,3)],[n 1],@(x) {x});
maxd2 = 0;
R = zeros(n,2);
move = false(n,1);
for i=1:n
Ti = T{i};
P = X(i,:) - XALL(Ti,:);
nP2 = sum(P.^2,2);
if any(nP2<2*d2min)
move(i) = true;
move(Ti(Ti<=n)) = true;
maxd2 = maxd2 + mean(nP2);
b = Ti > n;
nP2(b) = nP2(b)*5; % reduce repulsion from each point of the border
R(i,:) = sum(P./max((nP2-d2min),1e-3),1);
if ~any(move)
if k==1
v0 = (L*5e-3)/sqrt(maxd2/n);
R = R(move,:);
v = v0/sqrt(max(sum(R.^2,2)));
X(move,:) = X(move,:) + v*R;
% Project back if points falling outside the rectangle
X = min(max(X,Xmin),Xmax);
theta = linspace(0,2*pi,65);
xc = dmin/2*sin(theta);
yc = dmin/2*cos(theta);
% plot circles f diameter dmin around random points
for i=1:n


Bruno Luong
Bruno Luong on 3 Jun 2019
If I'm not mistaken, each statement of my code could be converted to 3D counter part. You migh adjust some hard code constant for the convergence.

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