How can calcuate the lines of code efficiently?

Hello. This code calculates the Laplacian. I am looking for reducing the time running of it. My main aim is improving the performance of loop running time.

Thanks

%%%code
nx=512;
ny=nx;
dx=1;
dy=dx;
format long;
nxny=nx*ny;
r=zeros(1,nx);
r(1:2)=[2,-1];
T=toeplitz(r);
E=speye(nx);
grad=-(kron(T,E)+kron(E,T));%Tensor

%-- for periodic boundaries

for i=1:nx
    ii=(i-1)*nx+1;
    jj=ii+nx-1;
    grad(ii,jj)=1.0;
    grad(jj,ii)=1.0;
    kk=nxny-nx+i;
    grad(i,kk)=1.0;
    grad(kk,i)=1.0;
end%for

grad = grad /(dx*dy);

[Ahmad Nadeem]

hello brother!

I want to contact you. Can you please share your email or any other ID?

Thank You

Best Regards

Ahmad Nadeem

Hongik University, Seoul, South Korea

Answer 1

[Stephan]

Yes, you can if you avoid using a loop and vectorize your code - see here:

tic
%%%code
nx=512;
ny=nx;
dx=1;
dy=dx;
format long;
nxny=nx*ny;
r=zeros(1,nx);
r(1:2)=[2,-1];
T=toeplitz(r);
E=speye(nx);
grad=-(kron(T,E)+kron(E,T));%Tensor
%-- for periodic boundaries

ii=(0:nx-1).*nx+1;
jj = ii+nx-1;
grad(ii,jj)=1.0;
grad(jj,ii)=1.0;
kk=(nxny-nx+1):nxny;
grad(1:nx,kk)=1.0;
grad(kk,1:nx)=1.0;
grad = grad /(dx*dy);
toc

Elapsed time is 0.145899 seconds.

Your code:

tic
%%%code
nx=512;
ny=nx;
dx=1;
dy=dx;
format long;
nxny=nx*ny;
r=zeros(1,nx);
r(1:2)=[2,-1];
T=toeplitz(r);
E=speye(nx);
grad=-(kron(T,E)+kron(E,T));%Tensor
%-- for periodic boundaries
for i=1:nx
    ii=(0:nx-1).*nx+1;
    jj=ii+nx-1;
    grad(ii,jj)=1.0;
    grad(jj,ii)=1.0;
    kk=nxny-nx+i;
    grad(i,kk)=1.0;
    grad(kk,i)=1.0;
end 
toc

Elapsed time is 11.333450 seconds.

Are the results correct? - to check this we call one of the results grad1 and subtract grad1 from grad:

nx=512;
ny=nx;
dx=1;
dy=dx;
format long;
nxny=nx*ny;
r=zeros(1,nx);
r(1:2)=[2,-1];
T=toeplitz(r);
E=speye(nx);
grad=-(kron(T,E)+kron(E,T));%Tensor
%-- for periodic boundaries
for i=1:nx
    ii=(0:nx-1).*nx+1;
    jj=ii+nx-1;
    grad(ii,jj)=1.0;
    grad(jj,ii)=1.0;
    kk=nxny-nx+i;
    grad(i,kk)=1.0;
    grad(kk,i)=1.0;
end 

ii=(0:nx-1).*nx+1;
jj = ii+nx-1;
grad(ii,jj)=1.0;
grad(jj,ii)=1.0;
kk=(nxny-nx+1):nxny;
grad(1:nx,kk)=1.0;
grad(kk,1:nx)=1.0;
grad1 = grad /(dx*dy);

expect_all_zeros = grad1-grad

expect_all_zeros =

   All zero sparse: 262144×262144

Appears to work fine...

[Amir Torabi]

kk in your code first code is an empty array!.

[Stephan]

Sure, for me not. Check it with

whos kk

For me kk is 1x512 sized. Also if it would be empty, why are the results the same like in the code you provided?

[Amir Torabi]

i checked it again. yes your answer is correct. Thanks.

[Ahmad Nadeem]

I didnt understand this code can you please help me to understand this code?