I have created this code from scratch. I want to make a plot and/or histogram of my "Observed" and "State" (these are 2 matrices). Some problem occurs at the 200th iteration, where my State matrix just becomes all 0's, there is no data being input into the State matrix. Can anyone troubleshoot the code? My possible states are {1,2,3}.
UPDATE:
When I adjust my n value, it adjusts how much of length T it will fill. So, n=5, only runs for 1/5 of T and n=1, run for entire length of T. I need an nxT matrix at the end (5X1000). The problem lies in the way I setup my for loops.
I still cannot solve the error though.
%Initialize A,pi,T
N = 3; # of states
%A is transition prob matrix
A = [.99,.005,.005;.005,.990,.005;.005,.005,.990];
%pi is initial state vector
pi = [1/3,1/3,1/3];
%T is # of observations per simulation
T = 1000;
%n is # of simulations
n = 5;
%Allocate space for the state matrix
State = zeros(n,T);
Observe = zeros(n,T);
%Create dummy emission matrix, must be row stochastic
B = ones(n,T)./T;
%loop over # of simulations
for i=1:1:n
x = rand(1);
if x <= (1/3)
State(i,1) = 1;
elseif x > (1/3) && x <= (2/3)
State(i,1) = 2;
else
State(i,1) = 3;
end
if State(i,1) == 1
b = -1;
elseif State(i,1) == 2
b = 0;
else
b = 1;
end
Observe(i,1)= normrnd(b,1);
for k=2:1:T
%Possible state 1,2,3
State(k) = randsample(N, 1, true, A(State(k-1),:));
if State == 1
c = -1;
elseif State == 2
c = 0;
else
c = 1;
end
Observe(i,k)= normrnd(c,1);
end
end
State(k) = randsample(N, 1, true, A(State(k-1),:));
This line is missing index (i) in position 1 inside State(k-1). It should be:
State(i,k) = randsample(N, 1, true, A(State(i,k-1),:));
Related
I am modelling flow through a tube in tube heat exchanger using MATLAB using the nodal port method. I need to populate a matrix with differential equations in order to solve for the enthalpies at each point. I have divided my pipe into multiple sections, each with 3 nodes. 1 node for the inner fluid, 1 node for the outer fluid, and 1 node for the pipe. However, due to the fact that there is counter flow I need to populate the matrix in a way that the outer fluid counts in reverse to the inner fluid. For example, if I have 9 nodes, my column vector will be [Eq 1 Eq 2 Eq 9 Eq 4 Eq 5 Eq6 Eq 7 Eq 8 Eq 3]. My code seems to work, but it doesn't enter anything for the 3rd position of the vector. Thank you in advance for the help.
NXP = 5; %Number of Divisions
HX = zeros(NXP,1);
cntr=2; %To see which number equation is being input
for j=1:NXP
if j==1
HX(1+(j-1)*3,1) = 125; %Boundary Condition
else
HX(1+(j-1)*3,1) = cntr;
cntr = cntr+1;
end
HX(2+(j-1)*3,1) = cntr;
cntr = cntr+1;
if j==NXP
HX(3+(j-1)*3,1) = 40; %Boundary Condition
else
HX(3*NXP-3*(j-1),1) = cntr;
cntr = cntr+1;
end
end
'''
I changed a bit your attempt, most notably moving the boundary conditions outside of the loop, but it now returns what I guess you were looking for:
NXP = 5;
HX = zeros(3*NXP, 1);
for j = 1:NXP
idx = (j - 1)*3;
HX(idx + 1, 1) = idx + 1;
HX(idx + 2, 1) = idx + 2;
HX(length(HX) - idx, 1) = idx + 3;
end
% Boundary conditions
HX(1) = 125;
HX(end) = 40;
EDIT:
Even better, just initialize the array as
HX = (1:3*NXP)';
and then just change the order for your 3rd nodes
HX(3:3:length(HX)) = flipud(HX(3:3:length(HX)));
and finally set your boundary conditions
HX(1) = 125;
HX(end) = 40;
I'm currently working on a code that makes use of a 2D cellular automata as an epidemic simulator in MATLAB. The main basic rule I'm trying to implement is that if any neighbour within a Moore Neighbourhood with a 1-cell radius is infected, the cell will become infected. But I can't seem to get a good code working for it.
Basically what I'm trying to do is say with for a cell with a one cell radius Moore neighbourhood, if any values in this neighbourhood = 2, then the initial cell will become 2.
I've tried using the forest fire code on the rosetta code as a basis for my code behaviour but it doesnt work very well. The rules don't really work that well when applying it to mine. I've tried using the mod function and a series of if loops to attach. I'll put in some code of each to give context.
This example doesn't really function well as an epidemic simulator to be honest.
matlab
clear; clc;
n = 200;
N = n/2;
E = 0.001; % Creating an arbitrary number for population exposed to
the disease but not infected
p = 1 + (rand(n,n)<E);
%p = ceil(rand(n,n)*2.12) - 1;
% ratio0 = sum(p(:)==0)/n^2;
% ratio1 = sum(p(:)==1)/n^2;
% ratio2 = sum(p(:)==2)/n^2;
% ratio3 = sum(p(:)==3)/n^2;
S = ones(3); S(2,2) = 0;
ff = 0.00000000002;
p(N,N) = 3;
%% Running the simulation for a set number of loops
colormap([1,1,1;1,0,1;1,0,0]); %Setting colourmap to Green, red and
grey
count = 0;
while(count<365) % Running the simulation with limited number of runs
count = count + 1;
image(p); pause(0.1); % Creating an image of the model
P = (p==1); % Adding empty cells to new array
P = P + (p==2).*((filter2(S,p==3)>0) + (rand(n,n)<ff) + 2); % Setting
2 as a tree, ignites based on proximity of trees and random
chance ff
P = P + (p==3); % Setting 3 as a burning tree, that becomes 1,
p = P;
end
second idea. this basically returns nothing
matlab
clear;clf;clc;
n = 200;
pos = mod((1:n),n) + 1; neg = mod((1:n)-2,n) + 1;
p = (ceil(rand(n,n)*1.0005));
for t = 1:365
if p(neg,neg) ==2
p(:,:) = 2;
end
if p(:,neg)==2
p(:,:) = 2;
end
if p(pos,neg)==2
p(:,:) = 2;
end
if p(neg,:)==2
p(:,:) = 2;
end
if p(pos,:)==2
p(:,:) = 2;
end
if p(neg,pos)==2
p(:,:) = 2;
end
if p(:,pos)==2
p(:,:) = 2;
end
if p(pos,pos)== 2
p(:,:) = 2;
end
image(p)
colormap([1,1,1;1,0,1])
end
third I tried using logic gates to see if that would work. I don't know if commas would work instead.
matlab
clear;clf;clc;
n = 200;
pos = mod((1:n),n) + 1; neg = mod((1:n)-2,n) + 1;
p = (ceil(rand(n,n)*1.0005));
%P = p(neg,neg) + p(:,neg) + p(pos,neg) + p(neg,:) + p(:,:) + p(pos,:)
+ p(neg,pos) + p(:,pos) + p(pos,pos)
for t=1:365
if p(neg,neg)|| p(:,neg) || p(pos,neg) || p(neg,:) || p(pos,:) ||
p(neg,pos) || p(:,pos) || p(pos,pos) == 2
p(:,:) = 2;
end
image(p)
colormap([1,1,1;1,0,1])
end
I expected the matrix to just gradually become more magenta but nothing happens in the second one. I get this error for the third.
"Operands to the || and && operators must be convertible to logical scalar values."
I just have no idea what to do!
Cells do not heal
I assume that
Infected is 2, non-infected is 1;
An infected cell remains infected;
A non-infected cell becomes infected if any neighbour is.
A simple way to achieve this is using 2-D convolution:
n = 200;
p = (ceil(rand(n,n)*1.0005));
neighbourhood = [1 1 1; 1 1 1; 1 1 1]; % Moore plus own cell
for t = 1:356
p = (conv2(p-1, neighbourhood, 'same')>0) + 1; % update
image(p), axis equal, axis tight, colormap([.4 .4 .5; .8 0 0]), pause(.1) % plot
end
Cells heal after a specified time
To model this, it is better to use 0 for a non-infected cell and a positive integer for an infected cell, which indicated how long it has been infected.
A cell heals after it has been infected for a specified number of iterations (but can immediately become infeced again...)
The code uses convolution, as the previous one, but now already infected cells need to be dealt with separately from newly infected cells, and so a true Moore neighbourhood is used.
n = 200;
p = (ceil(rand(n,n)*1.0005))-1; % 0: non-infected. 1: just infected
T = 20; % time to heal
neighbourhood = [1 1 1; 1 0 1; 1 1 1]; % Moore
for t = 1:356
already_infected = p>0; % logical index
p(already_infected) = p(already_infected)+1; % increase time count for infected
newly_infected = conv2(p>0, neighbourhood, 'same')>0; % logical index
p(newly_infected & ~already_infected) = 1; % these just became infected
newly_healed = p==T; % logical index
p(newly_healed) = 0; % these are just healed
image(p>0), axis equal, axis tight, colormap([.4 .4 .5; .8 0 0]), pause(.1) % plot
% infected / non-infected state
end
I am sending a matrix to my function modifikuj, where I want to replace the elements of the matrix with:
1 if element is a prime number
0 if element is a composite number
0.5 if element is 1
I dont understand why it is not working. I just started with MATLAB, and I created this function:
function B = modifikuj(A)
[n,m] = size(A);
for i = 1:n
for j = 1:m
prost=1;
if (A(i,j) == 1)
A(i,j) = 0.5;
else
for k = 2:(A(i,j))
if(mod(A(i,j),k) == 0)
prost=0;
end
end
if(prost==1)
A(i,j)=1;
else
A(i,j)=0;
end
end
end
end
With
A = [1,2;3,4];
D = modifikuj(A);
D should be:
D=[0.5, 1; 1 0];
In MATLAB you'll find you can often avoid loops, and there's plenty of built in functions to ease your path. Unless this is a coding exercise where you have to use a prescribed method, I'd do the following one-liner to get your desired result:
D = isprime( A ) + 0.5*( A == 1 );
This relies on two simple tests:
isprime( A ) % 1 if prime, 0 if not prime
A == 1 % 1 if == 1, 0 otherwise
Multiplying the 2nd test by 0.5 gives your desired condition for when the value is 1, since it will also return 0 for the isprime test.
You are not returning anything from the function. The return value is supposed to be 'B' according to your code but this is not set. Change it to A.
You are looping k until A(i,j) which is always divisible by itself, loop to A(i,j)-1
With the code below I get [0.5,1;1,0].
function A = modifikuj(A)
[n,m] = size(A);
for i = 1:n
for j = 1:m
prost=1;
if (A(i,j) == 1)
A(i,j) = 0.5;
else
for k = 2:(A(i,j)-1)
if(mod(A(i,j),k) == 0)
prost=0;
end
end
if(prost==1)
A(i,j)=1;
else
A(i,j)=0;
end
end
end
end
In addition to #EuanSmith's answer. You can also use the in built matlab function in order to determine if a number is prime or not.
The following code will give you the desired output:
A = [1,2;3,4];
A(A==1) = 0.5; %replace 1 number with 0.5
A(isprime(A)) = 1; %replace prime number with 1
A(~ismember(A,[0.5,1])) = 0; %replace composite number with 0
I've made the assumption that the matrice contains only integer.
If you only want to learn, you can also preserve the for loop with some improvement since the function mod can take more than 1 divisor as input:
function A = modifikuj(A)
[n,m] = size(A);
for i = 1:n
for j = 1:m
k = A(i,j);
if (k == 1)
A(i,j) = 0.5;
else
if all(mod(k,2:k-1)) %check each modulo at the same time.
A(i,j)=1;
else
A(i,j)=0;
end
end
end
end
And you can still improve the prime detection:
2 is the only even number to test.
number bigger than A(i,j)/2 are useless
so instead of all(mod(k,2:k-1)) you can use all(mod(k,[2,3:2:k/2]))
Note also that the function isprime is a way more efficient primality test since it use the probabilistic Miller-Rabin algorithme.
I am programming a simple perceptron in matlab but it is not converging and I can't figure out why.
The goal is to binary classify 2D points.
%P1 Generate a dataset of two-dimensional points, and choose a random line in
%the plane as your target function f, where one side of the line maps to +1 and
%the other side to -1. Let the inputs xn 2 R2 be random points in the plane,
%and evaluate the target function f on each xn to get the corresponding output
%yn = f(xn).
clear all;
clc
clear
n = 20;
inputSize = 2; %number of inputs
dataset = generateRandom2DPointsDataset(n)';
[f , m , b] = targetFunction();
signs = classify(dataset,m,b);
weights=ones(1,2)*0.1;
threshold = 0;
fprintf('weights before:%d,%d\n',weights);
mistakes = 1;
numIterations = 0;
figure;
plotpv(dataset',(signs+1)/2);%mapping signs from -1:1 to 0:1 in order to use plotpv
hold on;
line(f(1,:),f(2,:));
pause(1)
while true
mistakes = 0;
for i = 1:n
if dataset(i,:)*weights' > threshold
result = 1;
else
result = -1;
end
error = signs(i) - result;
if error ~= 0
mistakes = mistakes + 1;
for j = 1:inputSize
weights(j) = weights(j) + error*dataset(i,j);
end
end
numIterations = numIterations + 1
end
if mistakes == 0
break
end
end
fprintf('weights after:%d,%d\n',weights);
random points and signs are fine since plotpv is working well
The code is based on that http://es.mathworks.com/matlabcentral/fileexchange/32949-a-perceptron-learns-to-perform-a-binary-nand-function?focused=5200056&tab=function.
When I pause the infinite loop, this is the status of my vairables:
I am not able to see why it is not converging.
Additional code( it is fine, just to avoid answers asking for that )
function [f,m,b] = targetFunction()
f = rand(2,2);
f(1,1) = 0;
f(1,2) = 1;
m = (f(2,2) - f(2,1));
b = f(2,1);
end
function dataset = generateRandom2DPointsDataset(n)
dataset = rand(2,n);
end
function values = classify(dataset,m,b)
for i=1:size(dataset,1)
y = m*dataset(i,1) + b;
if dataset(i,2) >= y, values(i) = 1;
else values(i) = -1;
end
end
end
I can't seem to find a fix to my infinite loop. I have coded a Jacobi solver to solve a system of linear equations.
Here is my code:
function [x, i] = Jacobi(A, b, x0, TOL)
[m n] = size(A);
i = 0;
x = [0;0;0];
while (true)
i =1;
for r=1:m
sum = 0;
for c=1:n
if r~=c
sum = sum + A(r,c)*x(c);
else
x(r) = (-sum + b(r))/A(r,c);
end
x(r) = (-sum + b(r))/A(r,c);
xxx end xxx
end
if abs(norm(x) - norm(x0)) < TOL;
break
end
x0 = x;
i = i + 1;
end
When I terminate the code it ends at the line with xxx
The reason why your code isn't working is due to the logic of your if statements inside your for loops. Specifically, you need to accumulate all values for a particular row that don't belong to the diagonal of that row first. Once that's done, you then perform the division. You also need to make sure that you're dividing by the diagonal coefficient of A for that row you're concentrating on, which corresponds to the component of x you're trying to solve for. You also need to remove the i=1 statement at the beginning of your loop. You're resetting i each time.
In other words:
function [x, i] = Jacobi(A, b, x0, TOL)
[m n] = size(A);
i = 0;
x = [0;0;0];
while (true)
for r=1:m
sum = 0;
for c=1:n
if r==c %// NEW
continue;
end
sum = sum + A(r,c)*x(c); %// NEW
end
x(r) = (-sum + b(r))/A(r,r); %// CHANGE
end
if abs(norm(x) - norm(x0)) < TOL;
break
end
x0 = x;
i = i + 1;
end
Example use:
A = [6 1 1; 1 5 3; 0 2 4]
b = [1 2 3].';
[x,i] = Jacobi(A, b, [0;0;0], 1e-10)
x =
0.048780487792648
-0.085365853612062
0.792682926806031
i =
20
This means it took 20 iterations to achieve a solution with tolerance 1e-10. Compare this with MATLAB's built-in inverse:
x2 = A \ b
x2 =
0.048780487804878
-0.085365853658537
0.792682926829268
As you can see, I specified a tolerance of 1e-10, which means we are guaranteed to have 10 decimal places of accuracy. We can certainly see 10 decimal places of accuracy between what Jacobi gives us with what MATLAB gives us built-in.