Yes, you can if you avoid using a loop and vectorize your code - see here:
tic
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));
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
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));
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));
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...
Best Answer