How do i reduce running time in this code

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Tzach Berlinsky
Tzach Berlinsky 2020-12-17
回答： Swetha Polemoni 2020-12-20
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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回答（1 个）

Swetha Polemoni
Swetha Polemoni 2020-12-20
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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