I am trying to implement this alghoritm for finding a new constraint:
In my case we take only 3 natural numbers i.e 1,2, 3.
The sets associated with those natural numbers are M1, M2 and M3. Instead of the Newton Method in II(2), I chose a solver provided by Matlab fmincon.
Here is my code that is not working!
function[s_new]= checking2(M1,M2,M3,x)
M1=linspace(0,1,10)';
M2=linspace(0,1,100)';
M3=linspace(0,1,1000)';
bool1=0;
eta = 10^-8;
pocz=[];
max=-100;
x = [0.1,0.1]'; % warunek poczÄ…tkowy
A = [];
b = [];
Aeq = [];
beq = [];
Set=[0,1];
g = #(x,s) 5*x(1).^2.*sin(pi.*sqrt(s))./(1+s.^2) - x(2);
g_new = #(s) -g(x,s);
for i=1:length(M1)
if g(x,M1(i,:))>eta
s_new=M1(i,:);
bool1=1;
end
end
if ~bool1
for i=1:length(M1)
if g(x,M1(i,:))>max
pocz=M1(i,:);
max=g(x,M1(i,:));
end
end
if max<-eta
bool1=1;
end
end
if ~bool1
s_maybe = fmincon(g_new,pocz,A,b,Aeq,beq,min(Set),max(Set));
if g(x,s_maybe)>eta
s_new=s_maybe;
bool1=1;
end
end
if ~bool1
for i=1:length(M2)
if g(x,M2(i,:))>eta
s_new=M2(i,:);
bool1=1;
end
end
end
if ~bool1
for i=1:length(M2)
if g(x,M2(i,:))>max
pocz=M2(i,:);
max=g(x,M2(i,:));
end
end
if max<-eta
bool1=1;
end
end
if ~bool1
s_maybe = fmincon(g_new,pocz,A,b,Aeq,beq,min(Set),max(Set));
if g(x,s_maybe)>eta
s_new=s_maybe;
bool1=1;
end
end
if ~bool1
for i=1:length(M3)
if g(x,M3(i,:))>eta
s_new=M3(i,:);
bool1=1;
end
end
end
if ~bool1
s_new = 1;
end
disp(s_new);
The problem is:
Undefined function or variable 's_new'.
Error in checking2 (line 70)
disp(s_new);
So basically everything might be wrong, but I suppose it is something with fmincon.
EDIT:
The purpose of the alghoritm is to find a minimum of an objective function f(x), satisfying all the constraints g(x,s)<=0 for all s in S, where S is an infinite set (some interval in our case).
What my alghoritm does, at first it takes some finite subset of S and calculates the minimum of f on this set, then I am trying to update S with some s_new. This alghoritm that I am trying to implement is exactly the procedure for creating s_new. Then if it works properly, I will add s_new to my subset and calculate the minimum on the new set, and so on until g(x,s)<=eta, where eta is a small number.
I rewrite the algorithm, read through the comments
clc
clear
lb = 0;
ub = 1;
% Given
l = 3;
M1=linspace(lb,ub,10)';
M2=linspace(lb,ub,100)';
M3=linspace(lb,ub,1000)';
% one boolean value for each Matrix
bool = zeros(1,3);
eta = 10^-8;
% Used as fmincon initial starting guess
pocz = nan;
% Used to store the new finding s that fits all the conditions
s_new = nan;
% Fixed x
x = [0.1,0]';
% fmincon linear constraints
A = [];
b = [];
Aeq = [];
beq = [];
% Main function
g = #(x,s) 5*x(1).^2*sin(pi*sqrt(s))/(1+s.^2) - x(2);
% Optimization concerns s only, don't include x as x is fixed
g_new = #(s) -g(x,s);
% Assuming the maximum is reached at the upper bound, used in(II)(2)
max_s = ub;
maxfun = g(x, max_s);
% Use a cell, for each iteration use a specific matrix M
M = {M1, M2, M3};
for j = 1: length(M)
% used in (II)(1)
check = 0;
step = 1;
% (I) step 1
for i = 1:length(M{j})
% Stopping criteria
if g(x, M{j}(i)) > eta
s_new = M{j}(i);
bool(j) = 1;
break;
else
% Function maximum value for next step (II)
if maxfun < g(x, M{j}(i))
maxfun = g(x, M{j}(i));
% To be used in fmincon as pocz
max_s = M{j}(i);
end
end
% To be used in (II)(1)
if maxfun < -eta
check = 1;
end
end
% End of (I)
% Put (II)(1) here step 2
if ~bool(j) && check
step = step + 1;
% Stopping criteria
if step >= l
disp('S_new not defined');
break;
end
% otherwise go to the next M
end
% (II)(2) step 3
if ~bool(j)
step = step + 1;
if maxfun >= -eta && maxfun <= eta
pocz = max_s;
bool(j) = 1;
end
end
%% EDIT: if bool(j) changed to if ~bool(j)
% (II)(2) Continue
if ~bool(j)
s_maybe = fmincon(g_new,pocz,A,b,Aeq,beq,lb,ub);
% End of (II)(2)
% (II)(2)-1 step 4
step = step + 1;
if g(x, s_maybe) > eta
s_new = s_maybe;
bool(j) = 1;
end
% End of (II)(2)-1
end
% Put (II)(2) here step 5
if ~bool(j)
step = step + 1;
% Stopping criteria
if step >= l
disp('S_new not defined');
break;
end
% otherwise go to the next M
end
end
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
Here is the code which is trying to solve a coupled PDEs using finite difference method,
clear;
Lmax = 1.0; % Maximum length
Wmax = 1.0; % Maximum wedth
Tmax = 2.; % Maximum time
% Parameters needed to solve the equation
K = 30; % Number of time steps
n = 3; % Number of space steps
m =30; % Number of space steps
M = 2;
N = 1;
Pr = 1;
Re = 1;
Gr = 5;
maxn=20; % The wave-front: intermediate point from which u=0
maxm = 20;
maxk = 20;
dt = Tmax/K;
dx = Lmax/n;
dy = Wmax/m;
%M = a*B1^2*l/(p*U)
b =1/(1+M*dt);
c =dt/(1+M*dt);
d = dt/((1+M*dt)*dy);
%Gr = gB*(T-T1)*l/U^2;
% Initial value of the function u (amplitude of the wave)
for i = 1:n
if i < maxn
u(i,1)=1.;
else
u(i,1)=0.;
end
x(i) =(i-1)*dx;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for j = 1:m
if j < maxm
v(j,1)=1.;
else
v(j,1)=0.;
end
y(j) =(j-1)*dy;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for k = 1:K
if k < maxk
T(k,1)=1.;
else
T(k,1)=0.;
end
z(k) =(k-1)*dt;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Value at the boundary
%for k=0:K
%end
% Implementation of the explicit method
for k=0:K % Time loop
for i=1:n % Space loop
for j=1:m
u(i,j,k+1) = b*u(i,j,k)+c*Gr*T(i,j,k+1)+d*[((u(i,j+1,k)-u(i,j,k))/dy)^(N-1)*((u(i,j+1,k)-u(i,j,k))/dy)]-d*[((u(i,j,k)-u(i,j-1,k))/dy)^(N-1)*((u(i,j,k)-u(i,j-1,k))/dy)]-d*[u(i,j,k)*((u(i,j,k)-u(i-1,j,k))/dx)+v(i,j,k)*((u(i,j+1,k)-u(i,j,k))/dy)];
v(i,j,k+1) = dy*[(u(i-1,j,k+1)-u(i,j,k+1))/dx]+v(i,j-1,k+1);
T(i,j,k+1) = T(i,j,k)+(dt/(Pr*Re))*{(T(i,j+1,k)-2*T(i,j,k)+T(i,j-1,k))/dy^2-Pr*Re{u(i,j,k)*((T(i,j,k)-T(i-1,j,k))/dx)+v(i,j,k)*((T(i,j+1,k)-T(i,j,k))/dy)}};
end
end
end
% Graphical representation of the wave at different selected times
plot(x,u(:,1),'-',x,u(:,10),'-',x,u(:,50),'-',x,u(:,100),'-')
title('graphs')
xlabel('X')
ylabel('Y')
But I am getting this error
Subscript indices must either be real positive integers or logicals.
I am trying to implement this
with boundary conditions
Can someone please help me out!
Thanks
To be quite honest, it looks like you started with something that's way over your head, just typed everything down in one go without thinking much, and now you are surprised that it doesn't work...
In the future, please break down problems like these into waaaay smaller chunks that you can individually plot, check, test, etc. Better yet, try simpler problems first (wave equation, heat equation, ...), gradually working your way up to this.
I say this so harshly, because there were quite a number of fairly basic things wrong with your code:
you've used braces ({}) and brackets ([]) exactly as they are written in the equation. In MATLAB, braces are a constructor for a special container object called a cell array, and brackets are used to construct arrays and matrices. To group things like in the equation, you always have to use parentheses (()).
You had quite a number of parentheses wrong, which became apparent when I re-grouped and broke up those huge unintelligible lines into multiple lines that humans can actually read with understanding
you forgot to take the absolute values in the 3rd and 4th terms of u
you looped over k = 0:K and j = 1:m and then happily index everything with k and j-1. MATLAB is 1-based, meaning, the first element of anything is element 1, and indexing with 0 is an error
you've initialized 3 vectors u, v and T, but then index those in the loop as if they are 3D arrays
Now, I've managed to come up with the following code, which runs OK and at least more or less agrees with the equations shown. But I think it still doesn't make much sense because I get only zeros out (except for the initial values).
But, with this feedback, you should be able to correct any problems left.
Lmax = 1.0; % Maximum length
Wmax = 1.0; % Maximum wedth
Tmax = 2.; % Maximum time
% Parameters needed to solve the equation
K = 30; % Number of time steps
n = 3; % Number of space steps
m = 30; % Number of space steps
M = 2;
N = 1;
Pr = 1;
Re = 1;
Gr = 5;
maxn = 20; % The wave-front: intermediate point from which u=0
maxm = 20;
maxk = 20;
dt = Tmax/K;
dx = Lmax/n;
dy = Wmax/m;
%M = a*B1^2*l/(p*U)
b = 1/(1+M*dt);
c = dt/(1+M*dt);
d = dt/((1+M*dt)*dy);
%Gr = gB*(T-T1)*l/U^2;
% Initial value of the function u (amplitude of the wave)
u = zeros(n,m,K+1);
x = zeros(n,1);
for i = 1:n
if i < maxn
u(i,1)=1.;
else
u(i,1)=0.;
end
x(i) =(i-1)*dx;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
v = zeros(n,m,K+1);
y = zeros(m,1);
for j = 1:m
if j < maxm
v(1,j,1)=1.;
else
v(1,j,1)=0.;
end
y(j) =(j-1)*dy;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
T = zeros(n,m,K+1);
z = zeros(K,1);
for k = 1:K
if k < maxk
T(1,1,k)=1.;
else
T(1,1,k)=0.;
end
z(k) =(k-1)*dt;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Value at the boundary
%for k=0:K
%end
% Implementation of the explicit method
for k = 2:K % Time loop
for i = 2:n % Space loop
for j = 2:m-1
u(i,j,k+1) = b*u(i,j,k) + ...
c*Gr*T(i,j,k+1) + ...
d*(abs(u(i,j+1,k) - u(i,j ,k))/dy)^(N-1)*((u(i,j+1,k) - u(i,j ,k))/dy) - ...
d*(abs(u(i,j ,k) - u(i,j-1,k))/dy)^(N-1)*((u(i,j ,k) - u(i,j-1,k))/dy) - ...
d*(u(i,j,k)*((u(i,j ,k) - u(i-1,j,k))/dx) +...
v(i,j,k)*((u(i,j+1,k) - u(i ,j,k))/dy));
v(i,j,k+1) = dy*(u(i-1,j,k+1)-u(i,j,k+1))/dx + ...
v(i,j-1,k+1);
T(i,j,k+1) = T(i,j,k) + dt/(Pr*Re) * (...
(T(i,j+1,k) - 2*T(i,j,k) + T(i,j-1,k))/dy^2 - Pr*Re*(...
u(i,j,k)*((T(i,j,k) - T(i-1,j,k))/dx) + v(i,j,k)*((T(i,j+1,k) - T(i,j,k))/dy))...
);
end
end
end
% Graphical representation of the wave at different selected times
figure, hold on
plot(x, u(:, 1), '-',...
x, u(:, 10), '-',...
x, u(:, 50), '-',...
x, u(:,100), '-')
title('graphs')
xlabel('X')
ylabel('Y')
I am solving Burger equation with Newton-Raphson method in Mathlab.
For the description of the problem see 1.
My problem is the following this code finds the solution upto time
$t=1$, but at this time a discontinuity develops and then the wave
moves forward (like a step function), but this code does not produce
correct solutions after time $t=1$.
Any suggestions or comments to improve the code.
Here is the Matlab code that I am using
function BurgerFSolve2
clc; clear;
% define a 1D mesh
a = -1; b = 3; Nx = 100;
x = linspace(a,b,Nx);
dx = (b-a)/Nx;
J = length(x);
% Iinitial condition
p_init = zeros(size(x));
p_init(x<=0) = 1;
p_init(x>0 & x<1)= 1-x(x>0 & x<1);
% storing results
P = zeros(length(p_init),3001);
P(:,1) = p_init;
% Boundary condition
pL = 1; pR = 0;
% solver
dt = 0.001;
t = 0;
T = zeros(1,3001);
c = dt/dx;
for i = 1:3000
t = t+dt;
T(i+1) = t;
options=optimset('Display','iter'); % Option to display output
p = fsolve(#(p) myfun1(p, pL, pR, c, J, P(:,i)), p_init, ...
options);
% Call solver
P(:,i+1) = p;
p_init = p;
figure(1);
plot(x, p, '-o');
title(['t= ' num2str(t) ' s']);
drawnow;
end
end
function F = myfun1(p, pL, pR, c, J, p_Old)
% Rewrite the equation in the form F(x) = 0
F(1) = p(1) + c*(p(1)^2 - p(1)*pL) - p_Old(1);
for i=2:J-1
F(i) = p(i) + c*(p(i)^2 - p(i-1)*p(i)) - p_Old(i);
end
F(J) = p(J) + c*(p(J)^2 - p(J-1)*p(J)) - p_Old(J);
end
Currently I'm doing a code on the Secant Method, and so far the code runs ok.
However, I still have to update my "count.funcCount" by counting the number of function calls that I used in the "secant" function. How should I modify my code ?
This is what I have so far for the code:
function [ root, fcneval, count ] = secant(f, x0, x1, my_options)
my_options = optimset('MaxIter', 50, 'TolFun', 1.0e-5);
count = struct('iterations',0,'funcCount',0,'message',{'empty'});
xp = x0; %# Initialize xp and xc to match with
xc = x1; %# x0 and x1, respectively
MaxIter = 100;
TolFun = 1.0e-5;
%# Secant Method Loop
for i = 2:MaxIter
fd = f(xc) - f(xp); %# Together, the ratio of d and fd yields
d = xc - xp; %# the slope of the secant line going through xc and xp
xn = ((xc * fd) - (f(xc) * d)) / fd; %# Secant Method step
if (abs(xc - xn) < TolFun) && (abs(f(xn)) < TolFun) %# Stopping condition
break;
elseif i > MaxIter %# If still can't find a root after maximum
%# 100 iterations, consider not converge
count.message = sprintf('Do not converge');
end
xp = xc; % Update variables xc and xp
xc = xn;
end
%# Get outputs:
root = xn;
fcneval = f(root);
count.iterations = i;
end %# end function
---------------------------------------
function [f] = fun(x)
f = cos(x) + x;
end
Please help me, thank you in advance
Although I did not understand your question but still I think you can use a counter. Initialize it with zero and increment it every time your function is called. You are using an inbuilt function, just make necessary changes in its source code( FYI :you can do this in matlab) then save it with a new name and then use it in your main code.
You could create a nested function which have access to the variables of its parent function (including the function f and the counter count.funcCount). This function would call the actual method which does the computation, and then increment the counter.
Here is a rather silly example to illustrate the concept:
function [output,count] = myAlgorithm(f, x0)
count = 0;
output = x0;
while rand()<0.99 %# simulate a long loop
output = output + fcn(x0);
end
%# nested function with closure
function y = fcn(x)
%# access `f` and `count` inside the parent function
y = f(x); %# call the function
count = count + 1; %# increment counter
end
end
Now you call it as:
[output,count] = myAlgorithm(#(x)cos(x)+x, 1)