I can not find the error in this Gauss Seidal equation.

1 view (last 30 days)
%
function x1 = GaussSeidal(a,b,x0,tol)
N=100000;
k=1;
x=x0;
while(k<N)
n = length(b);
for j = 1 : n
x(j)=(b(j)-a(j,1:j-1).*x(1:j-1)-a(j,j+1:n)*x0(j+1:n))/a(j,j);
end
x1=x';
error=x1-x0;
if norm(error)< tol
disp('succefful');
break
end
k=k+1;
x0=x1;
end
disp(' The number of iteration is');disp(k)
tic;
gs_time=toc;
disp(gs_time);
end
end
%%Example
a=[5 -2 3;-3 9 1;2 -1 -7];b=[-1 2 3]';x0=[0 0 0]';tol=0.0001;

Answers (1)

Walter Roberson
Walter Roberson on 26 Feb 2018
You have one too many end statements in your function.
In the line
x(j)=(b(j)-a(j,1:j-1)*x(1:j-1)-a(j,j+1:n)*x0(j+1:n))/a(j,j);
when j = 1, a(j,j+1:n) is a 1 x 2 vector, and x0(j+1:n) is a 1 x 2 vector. You have the * operator between two 1 x 2 vectors. The * operator is algebraic matrix multiplication, which requires that for an M x N array "*" a P x Q array, that N and P are the same -- M x N * N x Q giving an M x Q result. 1 x 2 * 1 * 2 does not follow that rule and so is an error. Possibly you want the .* operator instead of * but possibly not.
You also have the problem in that line a(j,1:j-1) is empty, which would make the result of a(j,1:j-1)*x(1:j-1) empty as well. An empty array minus something is empty, so you are down to (b(j) - empty)/a(j,j) but a scalar minus empty is empty too, so you are now down to empty / a(j,j) which will be empty as well, leaving you with x(1) = empty . Which is a problem: you cannot store a 0 x 0 array into a slot that expects 1 value.
  5 Comments
Walter Roberson
Walter Roberson on 26 Feb 2018
It does not matter that a*x0 produces a value. The a*x part is producing empty, and empty minus a value is either empty or an error.
Looks like I need to be more explicit for you:
Your problem is that your code needs to calculate based upon the content of the previous columns, which is a problem when you start your calculation from column 1 because there are no previous columns. So do not start from column 1.
Dereje
Dereje on 27 Feb 2018
Exactly this is my problem from the beginning. I tried separately calculate x(1), to use for the next iteration. I will try in different ways. Thanks for the detailed explanation and of course for your fast response!

Sign in to comment.

Categories

Find more on Numerical Integration and Differential Equations in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!