Create a square grid with some random points inside that follow the poisson distribution and use each of these points as a starting point for my function

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george korris 2023-1-13

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评论： george korris 2023-2-4

Hi everyone! Happy new year! I hope everyone is doing well!!

So I have a function that simulates dla (something like brownian motion). But I want to create a square grid with some random points inside of the grid that follow the poisson distribution and from each and everyone of those points call my function to simulate a small dla. Does anyone know how can I do this?

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Answer 1

William Rose 2023-1-13

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@george korris,

If you place a point at random in the unit square with rand(), then the the distribution of points inside small sub-squares will be approximately Poisson.

M=100; %number of simulations

N=200; %number of points per simulation

%Divide the unit square into a 10x10 grid of subsquares

counts=zeros(M,10,10); %number of points in each subsquare in each simulation

for i=1:M

pts=rand(N,2); %x,y coordinates of N points in the unit square

gridpts=floor(10*pts)+[1,1]; %convert point locaitons to indices in the subsquare grid

for j=1:N

counts(i,gridpts(j,1),gridpts(j,2))=counts(i,gridpts(j,1),gridpts(j,2))+1;

end

histogram(reshape(counts,1,M*100),'Normalization','pdf')

hold on

%% PLot expected Poisson distribution

lambda=N/100; %expected points per subsquare

k=0:10;

plot(k,lambda.^k.*exp(-lambda)./factorial(k),'-r')

legend('Observed','Expected')

This is the distribution of points in subsquares. The plot shows that the observed number of points per subsquare follows the expected Poisson distribution.

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William Rose 2023-1-14

在 MATLAB Online 中打开

@george korris,

I looked at your code and I asaved ot as file dla2D.m.

I added comments - see attached. For any function, it is useful to list and describe inputs and outputs in the comments, including variable type and allowed or expected values and size. I have attempted to do this. Include comments to describe how the function works.

I am not sure exactly how it is simulating diffusion linked aggregation.

The funciton initializes A to be NxN, all zeros except it is 1 in the middle element. What does A represent? It is an NxN array of 0s and 1s. It could be the occupancy of a particle of potential size NxN, which grows in a random and dendritic way over time. This interpretation is consistent with the fact that its inital value is all zeros except for a 1 in the middle. However, this interpretaiton is not consistet with setting A(x,y) to 1, in the absence of apther particles, which appears to happen. Perhaps A describes which points in space have particles in them. If this is the case, then there shoudl be more than 1 element of A that is 1 initially, and there is not.

The main part of the function is a loop over time steps: for j=1:max_iter.

d=2 is specified outside the for loop. This assigment is pointless, because d is overwritten with d=rmin+2 on every loop pass.

Since d=rmin+2 is the same on every loop pass, move this step outside the loop, before the loop.

On every loop pass, point A(x,y) is set to 1. x,y are integers representing discrete positions on a circle of radius d around the middle point (A(m,m)). The angle of the point chosen, relative to the middle, is selected randomly on each loop pass. Does this step represent particle movement, or particle growth, or something else? If it represents movement, then why don't you set A(m,m) to 0 when you set A(x,y) to 1?

Within each time step, a while loop executes. The while loop runs forever, unless a break occurs. What each pass around the while loop represent?

You said originally that you wanted particles to have a Posisson distribution. This implies that there may be 0, 1, or more than 1 particle at each locaiotn. I do not see the possibility in your code fro more than 1 particle per locaiotn. If there can only be 0 or 1 then it is not Poisson. Is that a problem?

How many particles are in the space? What are the bounds of the space? WHat varable(s) represent the particle positions at each time? Does each particle move by diiffusion with every time step?

When you write a funciton, write a main script that calls the function, as a way of testing it. I have written and attached a main script that calls dla2D(). The test script does not finish, because dla2D() always produces an error.

When I call dla2D() from the command line, I always get an error. The error is at line 36 or line 38 of the attached dla2D.m. The exact error message varies. See examples below.

>> clear; dla2D(1,.5)
Index in position 1 is invalid. Array indices must be positive integers or logical values.
Error in dla2D (line 36)
A(x,y)=1;
>> clear; dla2D(1,.5)
Index in position 1 is invalid. Array indices must be positive integers or logical values.
Error in dla2D (line 38)
sum=A(x+1,y)+ A(x-1,y)+ A(x,y+1)+A(x,y-1); % check if any neighbors
>> clear; dla2D(1,.5)
Index in position 1 exceeds array bounds. Index must not exceed 6.
Error in dla2D (line 38)
sum=A(x+1,y)+ A(x-1,y)+ A(x,y+1)+A(x,y-1); % check if any neighbors
>> clear; dla2D(1,.5)
Index in position 1 exceeds array bounds. Index must not exceed 5.
Error in dla2D (line 38)
sum=A(x+1,y)+ A(x-1,y)+ A(x,y+1)+A(x,y-1); % check if any neighbors

You use sum as a variable name. I would modify the variable name, because sum() is the name of a commonly used Matlab function.

William Rose 2023-1-18

@george korris,

Your function, dla2D.m, builds up an aggregate by adding points at random locations within a specified radius of the initial central point. Each new point sticks, with probability stickiness, iff it has existing neighbors to the N, E,W, or S. (Diagonal neighbors don't count.) Thus the final aggregate will be a connected structure. The aggregate is not formed by collision with other pre-existing particles. Rather, it forms all by itself. Do you want to keep this behavior? In other words, do you want each aggregate to assemble itself without regard to any other particles?

If the answer is yes, then we will assign an x,y location randomly to the center of each aggregate, and we will end up witha bunch of aggregates. The aggregates may overlap, if the density of aggregates is high enough.

If the answer is no, then I assume you want to revamp how the aggregates grow. An aggregate would grow iff it bumps into another particle. This will require a complete rewrite of the function, because every aggregate will have to know the position of every other particle and aggregate at each time step.

I wondered how the stickiness would affect the final mass. Here are images of an aggregate with stickiness=0.10 (left) and of an aggregate with stickiness=1.00 (right). The lower the stickiness, the more dense is the aggregate.

Here is an Excel plot of average mass of the final aggregate, as a function of stickiness. Each mass estimate is based on 1000 trials with the specified stickiness (stickiness = 0.1, 0.2, ..., 1.0, and 0.99). There are error bars (+- standard error of the mean) on each point, but the bars are too small to see.

William Rose 2023-2-3

Geroge, I see that you have posted this question separately elsewhere on this site, so I will think about it, and if I have an answer, I will post it there.

george korris 2023-2-4