% Super Relaxation Iteration (Sorting Method)
function [x k flag]=SOR(A，b，eps，w，maxl)
%A is the coefficient matrix of the equation.
%b is the right vector of the equation.
Accuracy requires %eps.
%maxl is the maximum number of iterations and the default value is 100.
%w is the super relaxation factor, and the default value is 1.
%x is the solution of the system of equations
%k is the number of iterations
%flag is an indicator variable.
Initial value of% allocation:
If narrow & lt5maxl =100; end
If narrow & lt4w =1; end
If narrow & LT3EPS =1e-5; end
N = length (b); k = 0；
X = zero (n,1); Y = zero (n,1); flag = ' OK
% start iteration:
Although it is true.
y = x； % y and z are both auxiliary variables.
For i= 1:n
z = b(I)；
For j= 1:n
If j~=i
z=z-A(i，j)* x(j)；
end
end
If abs(A(i, I)) <; 1e- 10|k==maxl
Flag= "failed"; Return;
end
z=z/A(i，I)； x(I)=( 1-w)* x(I)+w * z；
end
if norm(y-x，INF)& lt； storage battery
Break;
end
k = k+ 1；
end
%Jacobi iterative method for solving linear equations
function [x k flag]=Jacobi(A，b，eps，maxl)
%A is the coefficient matrix of the equation.
%b is the right vector of the equation.
Accuracy requires %eps.
%maxl is the maximum number of iterations and the default value is 100.
%x is the solution of the system of equations
%k is the number of iterations
%flag is an indicator variable.
Initial value of% allocation:
If narrow & lt4maxl =100; end
If narrow & LT3EPS =1e-5; end
N = length (b); k = 0；
X = zero (n,1); Y = zero (n,1); flag = ' OK
% start iteration:
And 1
For i= 1:n
y(I)= b(I)；
For j= 1:n
If j~=i
y(i)=y(i)-A(i，j)* x(j)；
end
end
If abs(A(i, I)) <; 1e- 10|k==maxl
Flag= "failed"; Return;
end
y(i)=y(i)/A(i，I)；
end
if norm(y-x，INF)& lt； storage battery
Break;
end
x = y； k = k+ 1；
end
I am sure I can pass a little debugging. It's all tested. Remember to grade me. Thank you ~ ~ ~