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How do i reduce running time in this code

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Tzach Berlinsky
Tzach Berlinsky on 17 Dec 2020
Answered: Swetha Polemoni on 20 Dec 2020
function [f , t] = jacobi2(n)
tic;
if n < 4
error ('n not in the range')
end
n=n^2;
b=zeros(n,1);
b(1,1)=1;
b(n,1)=1;
A = zeros(n,n);
for i=4:(n-3)
for j=4:(n-3)
if i==j
A(i,i)=4;
A(i,i+1)=-1;
A(i,i-1)=-1;
A(i,i+3)=-1;
A(i,i-3)=-1;
A(i+1,i)=-1;
A(i-1,i)=-1;
A(i+3,i)=-1;
A(i-3,i)=-1;
end
end
end
A(1,1)=4;
A(1,2)=-1;
A(3,2)=-1;
A(2,3)=-1;
A(2,1)=-1;
A(2,2)=4;
A(3,3)=4;
A(end,end)=4;
A(end-1,end-1)=4;
A(end-2,end-2)=4;
A(end-3,end-3)=4;
A(end-1,end)=-1;
A(end,end-1)=-1;
A(end-1,end-2)=-1;
A(end-2,end-1)=-1;
epsilon = 1e-3;
f = zeros(n,1);
counter = 0;
flag = 0;
L = tril(A,-1);
D = diag(A);
U = triu(A,1);
B = -1./D.*(L+U);
C = (1./D).*b;
while flag == 0
counter = counter+1;
if counter > 10000
error ('Too many iteration')
end
f_n = (B*f) + C;
if max(abs(f_n-f)/(f_n))<epsilon
flag = 1;
else
f = f_n;
end
end
t=toc;
end

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Answers (1)

Swetha Polemoni
Swetha Polemoni on 20 Dec 2020
Hi Tzach Berlinsky,
In the code you have provided using nested loops is not necessary. Since the body of loops is executed only when i==j, elimination of one loop can be an option. Replace the nested loops with the following.
for i=4:(n-3)
A(i,i)=4;
A(i,i+1)=-1;
A(i,i-1)=-1;
A(i,i+3)=-1;
A(i,i-3)=-1;
A(i+1,i)=-1;
A(i-1,i)=-1;
A(i+3,i)=-1;
A(i-3,i)=-1;
end
Here is the link to best practices that can be followed to improve performance. You may find it helpful.

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