hello everybody i am a student master second year i have a problem i have no knowledge about Matlab anyone can help me explain this code and thanks in advance

function kansa1_MQ_1D% 
clear ALL;
% Local MPS method; Bi-Harmonic equation
% ni-----# of interior points
% nb-----# of boundary points
% c------shape parameter
% scale with maximum radius
%generate evenly distribute grid points on a unit square
tic
ns=3;  nn=30; c=50;
n=nn+1 ; nb=2; ni=n-nb; bb=1;
x1=0:bb/nn:bb;

intnode=zeros(ni,1);
j=0;k=0; %generate the interior and boundary nodes evenly
for i=1:n
    if(x1(i)==0 || x1(i)==bb)
       j=j+1;
       bdpt(j,1)=x1(i);
    else 
        k=k+1;
        intnode(k,1)=x1(i);
    end
end
x=[intnode(:,1); bdpt(:,1)]';
B=localmpsmatrix(x,ni,nb,n,ns,c);%produce sparse matrix
spy(B)
% Exemple 1
% f(1:ni)=exp(x(1:ni)); % for excat solution exp(x)
% f(ni+1:n)=exp(x(ni+1:n));% for excat solution exp(x)

%Exemple 2
% f(1:ni)=- sin(x(1:ni)); % for excat solution sin(x)
% f(ni+1:n)=sin(x(ni+1:n));% for excat solution sin(x)
% 
%Exemple 3
% f(1:ni)=- sin(x(1:ni))+exp(x(1:ni)); % for excat solution sin(x)
% f(ni+1:n)=sin(x(ni+1:n))+exp(x(ni+1:n));% for excat solution sin(x)

%Exemple 4
%f(1:ni)=3.*x(1:ni).^2-x(1:ni); % for excat solution x(x-1)
%f(ni+1:n)=0;% for excat solution x(x-1)

f(1:ni)=(x(1:ni).^2/2+x(1:ni)).*exp(x(1:ni))-x(1:ni).^2/2.*sin(x(1:ni))+x(1:ni).*cos(x(1:ni)); % for excat solution x(x-1)
f(ni+1:n)=sin(x(ni+1:n))+exp(x(ni+1:n));% for excat solution x(x-1)

alpha=B\f';
%u=exp(x)'; %exact solution
%u=sin(x)'; %exact solution
%u=(sin(x)+exp(x))'; %exact solution
%u=(x.*(x-1))';
u=(exp(x(1:n))+sin(x(1:n)))';
error1=max(abs((alpha(1:n)-u(1:n)))); %absolute maximu error
rms1=norm((alpha(1:n)-u(1:n)),2)/sqrt(n); % RMS error
toc
fprintf('n = %6d, nb = %4d, ns= %2d, c =%4d, \n',n,nb,ns,c);
fprintf('max error1: %e\nRMS1 error: %e\n', error1,rms1);
%hold on
%semilogy(param,error1,'b'),xlabel 'shape parameter' , ylabel 'error'
return
%%Produce the local sparse matrix
function localmpsmatrix=localmpsmatrix(x,ni,nb,n,ns,c)
t=x'; i1=0;
tree=kdtree_build(t); %search
[id, D]=kdtree_k_nearest_neighbors(tree,t(1,:),ns); 
ctrs=x(id(1:ns))'; %center of locale domain
DM= DistanceMatrix(ctrs, ctrs); %Distance matrix  
cc=D(1)*c;  
c2=cc*cc;
% r2 = D.*D; 
% mq=sqrt(DM+c2);
% A=mq;
% V1=sqrt(r2);
% V=c2./(r2 + c2).^(3/2);
% 
% coef=A\V; 

B1=zeros(1,ns*ni+nb); B2=B1; B3=B1;
for i=1:ni   
   % [id]=kdtree_k_nearest_neighbors(tree,t(i,:),ns); 
   [id, D]=kdtree_k_nearest_neighbors(tree,t(i,:),ns); 
ctrs=x(id(1:ns))'; %center of locale domain
DM= DistanceMatrix(ctrs, ctrs); %Distance matrix  
% cc=D(1)*c;  
% c2=cc*cc;
r2 = D.*D; 
mq=sqrt(DM+c2);
A=mq;
%%for solving -div(K*grad(u))=f
V1=(t(i,:)-ctrs(:))./(r2 + c2).^(1/2);
V2=c2./(r2 + c2).^(3/2);
V=t(i,:).^2*V2/2.+t(i,:).*V1;
coef=A\V; 
    for k=1:ns %sparse matrix storage for interior points
       i1=i1+1;
       B1(i1)=i; B2(i1)=id(k); B3(i1)=coef(k);
    end    
end
for i=ni+1:n  %indentity matrix for boundary points
    i1=i1+1;
    B1(i1)=i; B2(i1)=i;  B3(i1)=1;  
end
localmpsmatrix = sparse(B1(1:i1),B2(1:i1),B3(1:i1),n,n);
return
%%DistanceMatrix: r^2
function DM = DistanceMatrix(dsites, ctrs)
[M,s] = size(dsites); N = length(ctrs); 
DM = zeros(M,N);
for i=1:s
    [dr,cc]=ndgrid(dsites(:,i),ctrs(:,i));
    DM = DM +(dr-cc).^2;
end
return

[John D'Errico]

If you have a question about a specific line of code, then ask it. Otherwise, you need to read the documentation for functions where you have questions. This is far too large a question to have an answer otherwise.

[Walter Roberson]

No, we are not going to do that. You have indicated that you are second years masters student, but you did not mention your area of study or your undergrad concentration, or your previous programming experience, so we would have to assume that your Masters is in the morphology of written ancient Etruscan, and that we would have two write two or three books to provide the basis to explain the code.

You need to ask specific questions to expect an answer.

[Omar Idriss]

OK thank you so much I understood! I will start with x = [intnode (:, 1); PTDB (:, 1)] '; 1-the role of "bdpt" ???? 2-for itnode we will have (n-2) index but for bdpt it's just 2 indices ?? why the scribe of x is true and thank you infinitely someone explain to me the meaning of x and thank you so much

[Rena Berman]

(Answers Dev) Restored edit

[Stephen23]

Original question by Omar Idriss on 17th April 2018:

https://web.archive.org/web/20210422123359/https://webcache.googleusercontent.com/search?q=cache%3AL3FI33M670gJ%3Ahttps%3A%2F%2Fwww.mathworks.com%2Fmatlabcentral%2Fanswers%2F395369-hello-everybody-i-am-a-student-master-second-year-i-have-a-problem-i-have-no-knowledge-about-matlab

hello everybody i am a student master second year i have a problem i have no knowledge about Matlab anyone can help me explain this code and thanks in advance

function kansa1_MQ_1D% 
clear ALL;
% Local MPS method; Bi-Harmonic equation
% ni-----# of interior points
% nb-----# of boundary points
% c------shape parameter
% scale with maximum radius
%generate evenly distribute grid points on a unit square
tic
ns=3;  nn=30; c=50;
n=nn+1 ; nb=2; ni=n-nb; bb=1;
x1=0:bb/nn:bb;
intnode=zeros(ni,1);
j=0;k=0; %generate the interior and boundary nodes evenly
for i=1:n
    if(x1(i)==0 || x1(i)==bb)
       j=j+1;
       bdpt(j,1)=x1(i);
    else 
        k=k+1;
        intnode(k,1)=x1(i);
    end
end
x=[intnode(:,1); bdpt(:,1)]';
B=localmpsmatrix(x,ni,nb,n,ns,c);%produce sparse matrix
spy(B)
% Exemple 1
% f(1:ni)=exp(x(1:ni)); % for excat solution exp(x)
% f(ni+1:n)=exp(x(ni+1:n));% for excat solution exp(x)
%Exemple 2
% f(1:ni)=- sin(x(1:ni)); % for excat solution sin(x)
% f(ni+1:n)=sin(x(ni+1:n));% for excat solution sin(x)
% 
%Exemple 3
% f(1:ni)=- sin(x(1:ni))+exp(x(1:ni)); % for excat solution sin(x)
% f(ni+1:n)=sin(x(ni+1:n))+exp(x(ni+1:n));% for excat solution sin(x)
%Exemple 4
%f(1:ni)=3.*x(1:ni).^2-x(1:ni); % for excat solution x(x-1)
%f(ni+1:n)=0;% for excat solution x(x-1)
f(1:ni)=(x(1:ni).^2/2+x(1:ni)).*exp(x(1:ni))-x(1:ni).^2/2.*sin(x(1:ni))+x(1:ni).*cos(x(1:ni)); % for excat solution x(x-1)
f(ni+1:n)=sin(x(ni+1:n))+exp(x(ni+1:n));% for excat solution x(x-1)
alpha=B\f';
%u=exp(x)'; %exact solution
%u=sin(x)'; %exact solution
%u=(sin(x)+exp(x))'; %exact solution
%u=(x.*(x-1))';
u=(exp(x(1:n))+sin(x(1:n)))';
error1=max(abs((alpha(1:n)-u(1:n)))); %absolute maximu error
rms1=norm((alpha(1:n)-u(1:n)),2)/sqrt(n); % RMS error
toc
fprintf('n = %6d, nb = %4d, ns= %2d, c =%4d, \n',n,nb,ns,c);
fprintf('max error1: %e\nRMS1 error: %e\n', error1,rms1);
%hold on
%semilogy(param,error1,'b'),xlabel 'shape parameter' , ylabel 'error'
return
%%Produce the local sparse matrix
function localmpsmatrix=localmpsmatrix(x,ni,nb,n,ns,c)
t=x'; i1=0;
tree=kdtree_build(t); %search
[id, D]=kdtree_k_nearest_neighbors(tree,t(1,:),ns); 
ctrs=x(id(1:ns))'; %center of locale domain
DM= DistanceMatrix(ctrs, ctrs); %Distance matrix  
cc=D(1)*c;  
c2=cc*cc;
% r2 = D.*D; 
% mq=sqrt(DM+c2);
% A=mq;
% V1=sqrt(r2);
% V=c2./(r2 + c2).^(3/2);
% 
% coef=A\V; 
B1=zeros(1,ns*ni+nb); B2=B1; B3=B1;
for i=1:ni   
   % [id]=kdtree_k_nearest_neighbors(tree,t(i,:),ns); 
   [id, D]=kdtree_k_nearest_neighbors(tree,t(i,:),ns); 
ctrs=x(id(1:ns))'; %center of locale domain
DM= DistanceMatrix(ctrs, ctrs); %Distance matrix  
% cc=D(1)*c;  
% c2=cc*cc;
r2 = D.*D; 
mq=sqrt(DM+c2);
A=mq;
%%for solving -div(K*grad(u))=f
V1=(t(i,:)-ctrs(:))./(r2 + c2).^(1/2);
V2=c2./(r2 + c2).^(3/2);
V=t(i,:).^2*V2/2.+t(i,:).*V1;
coef=A\V; 
    for k=1:ns %sparse matrix storage for interior points
       i1=i1+1;
       B1(i1)=i; B2(i1)=id(k); B3(i1)=coef(k);
    end    
end
for i=ni+1:n  %indentity matrix for boundary points
    i1=i1+1;
    B1(i1)=i; B2(i1)=i;  B3(i1)=1;  
end
localmpsmatrix = sparse(B1(1:i1),B2(1:i1),B3(1:i1),n,n);
return
%%DistanceMatrix: r^2
function DM = DistanceMatrix(dsites, ctrs)
[M,s] = size(dsites); N = length(ctrs); 
DM = zeros(M,N);
for i=1:s
    [dr,cc]=ndgrid(dsites(:,i),ctrs(:,i));
    DM = DM +(dr-cc).^2;
end
return

[Rena Berman]

(Answers Dev) Restored edit

[Rena Berman]

(Answers Dev) Restored edit

Answer 1

[Walter Roberson]

bdpt is a temporary variable used to construct a list of border points. intnode is a temporary variable used to construct a list of interior nodes. The complete list of nodes is constructed as the interior nodes followed by the border nodes.

In a 1D situation, there are exactly two border nodes, one at the beginning and one at the end. The other n-2 nodes are interior nodes.

Answer 2

[Omar Idriss]

Ok thank you very much Walter Roberson, I have another question please I did not understand this writing B\f'; B a matrix and f a vector ? -div(K*grad(u))=f, I did not understand how we resolved this,,,,..??? CORDIALLY

[Walter Roberson]

Sorry, I am not familiar with the computation being done.

B\f is like inv(B)*f for the case where B is invertable and f is a column vector -- that is, one of its main uses is to solve simultaneous equations. It is, though, not restricted to that situation: it can do a least-squared fit for over-determined equations as well.