Matlab 'Function Unset/Unused' - matlab

I am new to Matlab, so there is a lot I do not understand; I have three questions, as followed:
The biggest issue I have revolves around function B= upper_triang(A) where it states that it is 'a function unused'. Reading through the Matlab tooltip, and I am guessing that it has to do with a value that has not been defined? ANd B of course being unset, but I thought I defined B as a function, so why is it unset? I am assuming that B is U, but in the math part that is ahh, so I am not sure what it would be.
To give context to my code I sent the part of the homework code I have a question on.
%%%This is for part a for problem 1.
A=zeros(8,8);%%%For the size column and row size of matrix A.
for j=1:8
for k=1:8
if j==k
A(j,k)=-2;
end
if abs(j-k)==1
A(j,k)=1;
end
end
end
[L,U]=lu(A); %%%Proving L and U.
L*U; %%%Proving that L*U=A, thus it is correct.
%%%This is for part b for problem 1
function B= upper_triang(A)
%%%Input A--> Matrix for m*n for the size
%%%Output B--> Is the upper triangular matrix of the same size
[m,n]=size(A);
for k=1:m-1
for i=k+1:m
u=A(i,k)/A(k,k);
for j=(k):n
A(i,j)=A(i,j)-u*A(k,j);
end
end
end
B-A;
end

Function being unused
The problem here is that you've only created a function, but you've never used it. Everything after the function B = upper_triang(A) just tells Matlab what to do in case that function is ever called, but your never do.
I'm guessing that, after your LU decomposition but before the declaration of your function, you should add a line
B = upper_triang(A);
Output unset
There is a second problem which is that, while you declared that B should be the output of your function upper_triang(), it it not "created" anywhere in the function itself.
There should be a line somewhere with
B = something;
Now, my linear algebra is quite rusty but I think that what you're trying to do should rather be:
%%%This is for part b for problem 1
function B= upper_triang(A)
%%%Input A--> Matrix for m*n for the size
%%%Output B--> Is the upper triangular matrix of the same size
[m,n]=size(A);
B = A; % copy the input matrix in B, and work with it from now on
for k=1:m-1
for i=k+1:m
u=B(i,k)/B(k,k);
for j=(k):n
B(i,j)=B(i,j)-u*B(k,j);
end
end
end
B = B - A; % not really sure about this
end

Related

Fourier series graph generation from given coefficients an, bn in Matlab(Scilab)

I've calculated coefficients an, bn (100, T=2*pi) in c++ and checked that they are correct using few sources. Now i try to generate Fourier series graph for given example function in Scilab:
(x+2)*abs(cos(2*x*(x-pi/6)))
M=csvRead(filename, ";", [], 'double')
n=size(M,1)
for i = 1:n
A(i)=M(i)
B(i)=M(i + n)
end
function series=solution(x)
series=A(1)/2;
for i = 2:n
series=series+(A(i)*cos(i*x)+B(i)*sin(i*x));
end
endfunction
function series=solution2(x)
series=(x+2).*abs(cos(2.*x.*(x-%pi/6)));
endfunction
x = -%pi:%pi/100:%pi
plot2d(x, solution(x), 3)
x2 = -%pi:%pi/100:%pi
plot2d(x2, solution2(x2), 4)
Here is the result:
It clearly looks that tendency is ok but the beginning and the end of the period are wrong (reversed?). Do you see any issues in Scilab code? What could cause the problem - values in sin/cos in function solution(x)? Should i provide an, bn values and check for miscalculation there?
I don't know how did you calculated your A & B coefficients, but I assume that you used the usual notations to get the first line of the below formula:
Thus n starts from 1. Since Scilab starts vector indexing from 1, you correctly made your loop from 2, but forgot to compensate for this "offset".
Your function should be something like this:
function series=solution(x)
series=A(1)/2;
for i = 2:n
series=series+(A(i)*cos((i-1)*x)+B(i)*sin((i-1)*x));
end
endfunction
Since you didn't provided A & B, I can not check the result.
Additional note: Syntactically more correct if you explicitly define all input variables in a function, like this:
function series=solution(x,A,B)
This way you may be sure that your input is not changed somewhere else in the code.

Matlab: fmincon and inequality constraints (implementation issue)

I'm working on an optimization problem in Matlab, but unfortunately, I'm stuck.
I want to maximize \theta using the function fmincon, but this particular problem depends on $n$, and $n$ can get very large. There are $n-1$ inequality constraints, all defined with the relation:
For all i \neq j \leq n : \theta - (x_i - x_j)^2 - (y_i - y_j)^2 \leq 0.
So $c(x)$ is an (n-1)x1 - vector.
I'm looking for a way to implement this, so that I don't have to write a new matlab file for each different $n$. (and as you can imagine, when n gets large, that would be one heck of a job)
Any help would be dearly appreciated.
Cheers!
EDIT : I now have created an extra m.file, just for this constraint:
function constraint(n)
%this is a function which creates the constraints of the distance.
for i= 1: n
for j= 1:n
if j==i
continue;
end
(x(i)-x(j))^2 + (y(i)-y(j))^2;
end
end
But the problem now is that matlab goes over the elements one by one. For example: it doesn't calculate (x(1) - x(4))^2 + (y(1) - y(4))^2.
Any idea on how to solve this one?
Thanks again!
I don't see that your function wouldn't at some point calculate that value (when i = 1 and j = 4). The main issue seems to be that your function doesn't return anything, or take in x. According to this, a constraint function should return two things:
Nonlinear constraint functions must return both c and ceq, the
inequality and equality constraint functions, even if they do not both
exist. Return empty [] for a nonexistent constraint.
So first, we need to make sure our constraints are saved into an output vector, c, that c and an empty ceq are returned, and that our function takes both x and n. There might be prettier ways of doing it but
function [c, ceq] = constraint(x,n)
%this is a function which creates the constraints of the distance.
counter = 1;
for i= 1: n
for j= 1:n
if j==i
continue;
end
c(counter)=(x(i)-x(j))^2 + (y(i)-y(j))^2;
c = c+1;
end
end
ceq = [];
end
Next problem: this function takes two inputs, but nonlcon as an input to fmincon needs to take only one, x. We get around this by wrapping this function in an anonymous function (n needs to be predefined), so in the actual fmincon call you would set it to be something like #(x)constraint(x,n)

Gauss-Seidel method doesn't work for large sparse arrays?

Once again I have a problem with the Gauss-Seidel Method in Matlab. Here it is:
function [x] = ex1_3(A,b)
format long
sizeA=size(A,1);
x=zeros(sizeA,1);
%Just a check for the conditions of the Gauss-Seidel Method (if it has dominant diagonal)
for i=1:sizeA
sum=0;
for j=1:sizeA
if i~=j
sum=sum+abs(A(i,j));
end
end
if abs(A(i,i))<sum
fprintf('\nGauss-Seidel''s conditions not met!\n');
return
end
end
%Actual Gauss-Seidel Method
max_temp=10^(-6); %Pass first iteration
while max_temp>(0.5*10^(-6))
xprevious=x;
for i=1:sizeA
x(i,1)=b(i,1);
for j=1:sizeA
if i~=j
x(i,1)=x(i,1)-A(i,j)*x(j,1);
end
end
x(i,1)=x(i,1)/A(i,i);
end
x
%Calculating infinite norm of vector x-xprevious
temp=x-xprevious;
max_temp=temp(1,1);
for i=2:sizeA
if abs(temp(i,1))>max_temp
max_temp=abs(temp(i,1));
end
end
end
It actually works fine for a 100x100 matrix or smaller. However, my tutor wants it to work for 100000x100000 matrices. At first it was difficult to even create the matrix itself, but I managed to do it with a little help from here:
Matlab Help Center
Now, I call the ex1_3 function with A as a parameter, but it goes really slow. Actually it never ends. How can I make it work?
Here's my code for creating the specific matrix my tutor wanted:
The important part is just that it meets these conditions:
A(i; i) = 3, A(i - 1; i) = A(i; i + 1) = -1 n=100000
b=ones(100000,1);
b(1,1)=2;
b(100000,1)=2;
i=zeros(299998,1); %Matrix with the lines that we want to put nonzero elements
j=zeros(299998,1); %Matrix with the columns that we want to put nonzero elements
s=zeros(299998,1); %Matrix with the nonzero elements.
number=1;
previousNumberJ=0;
numberJ=0;
for k=1:299998 %Our index in i and j matrices
if mod((k-1),3)==0
s(k,1)=3;
else
s(k,1)=-1;
end
if k==1 || k==2
i(k,1)=1;
j(k,1)=k;
elseif k==299997 || k==299998
i(k,1)=100000;
j(k,1)=(k-200000)+2;
else
if mod(k,3)==0
number=number+1;
numberJ=previousNumberJ+1;
previousNumberJ=numberJ;
end
i(k,1)=number;
j(k,1)=numberJ;
numberJ=numberJ+1;
end
end
A=sparse(i,j,s); %Creating the sparse array
x=ex1_3(A,b);
the for loop works very slowly in Matlab, perhaps you may want to try the matrix form of the iteration:
function x=gseidel(A,b)
max_temp=10^(-6); %Pass first iteration
x=b;
Q=tril(A);
r=b-A*x;
for i=1:100
dx=Q\r;
x=x+1*dx;
r=b-A*x;
% convergence check
if all(abs(r)<max_temp) && all(abs(dx)<max_temp), return; end
end
For your A and b, it only takes 16 steps to converge.
tril extracts the lower triangular part of A, you can also obtain this Q when you build up the matrix. Since Q is already the triangular matrix, you can solve the equation Q*dx=r very easily if you are not allowed to use \ function.

need explanation about a matlab code snippet about moran process

I am new to Matlab. I was reading this code snippet, but in some parts (marked with asterisks) I don't understand what it means, so if anybody could help would be very much appreciated
function [A1nmb] = moran(initsize, popsize)
% MORAN generates a trajectory of a Moran type process
% which gives the number of genes of allelic type A1 in a population
% of haploid individuals that can exist in either type A1 or type A2.
% The population size is popsize and the initial number of type A1
% individuals os initsize.
% Inputs: initsize - initial number of A1 genes
% popsize - the total population size (preserved)
if (nargin==0)
initsize=10;
popsize=30;
end
A1nmb=zeros(1,popsize);
A1nmb(1)=initsize;
**lambda = inline('(x-1).*(1-(x-1)./N)', 'x', 'N');
mu = inline('(x-1).*(1-(x-1)./N)', 'x', 'N');**
x=initsize;
i=1;
while (x>1 & x<popsize+1)
if (lambda(x,popsize)/(lambda(x,popsize)+mu(x,popsize))>rand)
x=x+1;
A1nmb(i)=x;
else
x=x-1;
A1nmb(i)=x;
end;
i=i+1;
end;
nmbsteps=length(A1nmb);
***rate = lambda(A1nmb(1:nmbsteps-1),popsize) ...
+mu(A1nmb(1:nmbsteps-1),popsize);***
**jumptimes=cumsum(-log(rand(1,nmbsteps-1))./rate);**
jumptimes=[0 jumptimes];
stairs(jumptimes,A1nmb);
axis([0 jumptimes(nmbsteps) 0 popsize+1]);
The first line you marked
lambda = inline('(x-1).*(1-(x-1)./N)', 'x', 'N');
creates something called an inline function. It is equivalent to defining a mathematical function. Example:
y = inline('x^2')
would allow you to do
>> y(2)
4
This immediately explains the second line you marked.
rate = lambda(A1nmb(1:nmbsteps-1),popsize) ...
+mu(A1nmb(1:nmbsteps-1),popsize);
will compute the value of the function lambda(x,N) at x = A1nmb(1:nmbsteps-1) and N = popsize.
I will say immediately here that you should take a look at anonymous functions, a different format used to accomplish the same as inline. Only, anonymous functions are generally better supported, and usually a lot faster than inline functions.
Then, for the final line,
jumptimes = cumsum(-log(rand(1,nmbsteps-1))./rate);
is a nested command. rand will create a matrix containing pseudorandom numbers, log is the natural logarithm ("ln"), and cumsum creates a new matrix, where all the elements in the new matrix are the cumulative sum of the elements in the input matrix.
You will find the commands doc and help very useful. Try typing
doc cumsum
or
help inline
on the Matlab command prompt. Try that again with the commands forming the previous statement.
As a general word of advice: spend an insane lot of time reading through the documentation. Really, for each new command you encounter, read about it and play with it in a sandbox until you feel you understand it. Matlab only becomes powerful if you know all its commands, and there are a lot to get to know.
It defines an inline function object. For example this
lambda = inline('(x-1).*(1-(x-1)./N)', 'x', 'N')
defines lambda as a function with 2 variables. When you call lambda(A,n) Matlab simply expands the function you define in the first string. Thus lambda(A,n) using the variables you provide in the function call. lambda(A,n) would will evaluate to:
(A-1).*(1-(A-1)./n)
it just expands the function using the parameters you supply. Take a look at this link for more specific details http://www.mathworks.co.uk/help/techdoc/ref/inline.html
The cumsum function just returns the cumulative sum of a matrix along a particular dimension. Say we call cumsum on a vector X, then the value at element i in the result is equal to the sum of elements in X from index 1 to i. For example X = [1 2 1 3] we would get
AA = cumsum(X);
we would have
AA = [1 3 5 8]
See this link for more details and examples http://www.mathworks.co.uk/help/techdoc/ref/cumsum.html

nlfilter taking same values twice

I used nlfilter for a test function of mine as follows:
function funct
clear all;
clc;
I = rand(11,11);
ld = input('Enter the lag = ') % prompt for lag distance
A = nlfilter(I, [7 7], #dirvar);
% Subfunction
function [h] = dirvar(I)
c = (size(I)+1)/2
EW = I(c(1),c(2):end)
h = length(EW) - ld
end
end
The function works fine but it is expected that nlfilter progresses element by element, but in first two iterations the values of EW will be same 0.2089 0.4162 0.9398 0.1058. But then onwards for all iterations the next element is selected, for 3rd it is 0.4162 0.9398 0.1058 0.1920, for 4th it is 0.9398 0.1058 0.1920 0.5201 and so on. Why is it so?
This is nothing to worry about. It happens because nlfilter needs to evaluate your function to know what kind of output to create. So it uses feval once before starting to move across the image. The output from this feval call is what you see the first time.
From the nlfilter code:
% Find out what output type to make.
rows = 0:(nhood(1)-1);
cols = 0:(nhood(2)-1);
b = mkconstarray(class(feval(fun,aa(1+rows,1+cols),params{:})), 0, size(a));
% Apply fun to each neighborhood of a
f = waitbar(0,'Applying neighborhood operation...');
for i=1:ma,
for j=1:na,
x = aa(i+rows,j+cols);
b(i,j) = feval(fun,x,params{:});
end
waitbar(i/ma)
end
The 4th line call to eval is what you observe as the first output from EW, but it is not used to anything other than making the b matrix the right class. All the proper iterations happen in the for loop below. This means that the "duplicate" values you observe does not affect your final output matrix, and you need not worry.
I hope you know what the length function does? It does not give you the Euclidean length of a vector, but rather the largest dimension of a vector (so in your case, that should be 4). If you want the Euclidean length (or 2-norm), use the function norm instead. If your code does the right thing, you might want to use something like:
sz = size(I,2);
h = sz - (sz+1)/2 - ld;
In your example, this means that depending on the lag you provide, the output should be constant. Also note that you might want to put semicolons after each line in your subfunction and that using clear all as the first line of a function is useless since a function will always be executed in its own workspace (that will however clear persistent or global variables, but you don't use them in your code).