How can I stop My Gauss-Seidel loop when the if condition is met.
    3 vues (au cours des 30 derniers jours)
  
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D = diag(diag(A));
L = tril(-A,-1);
U = triu(-A,1);
%transition matrix and constant vector used for iterations
Tg = inv(D-L)*U; 
cg = inv(D-L)*b;
tol = 1e-05;
k = 1;
x = zeros(n,1);      %starting vector
N=100;
while k <= N
   x(:,k+1) = Tg*(x(:,k)) + cg;
   if ((norm(x(:,k+1)-x(:,k),'inf'))/(norm(x(:,k+1),'inf'))) < tol
      break;
   end;
   k = k+1;
end;
disp(['Iteration ' num2str(k)]); 
plot(x,'-o');
  end
Hello, I tried to stop my iteration when ever my condition is met: infinity norm of (x(k+1) - x(k)) / infinity norm of(x(k)) < tol. Somehow, my loop will keep working until it reaches the end of my designated N. Please help me fix my while loop. Thank you.
2 commentaires
  Darshan Ramakant Bhat
      
 le 6 Mar 2017
				Can you please put the complete code. Lots of values are missing, cannot run this script in the MATLAB editor. Also put sample test matrix 'A' in which you are getting the error. Without the data it is hard to debug.
Regards
Darshan
  Bruno Luong
      
      
 le 30 Sep 2018
				
      Modifié(e) : Bruno Luong
      
      
 le 30 Sep 2018
  
			May be it's a slow convergence and you need more iterations.
My suggestion: display
norm(x(:,k+1)-x(:,k),'inf')
and
norm(x(:,k+1),'inf')
along the iterations and see how they evolve.
Réponses (1)
  Jan
      
      
 le 6 Mar 2017
        The INF-norm is not sufficient here:
if ((norm(x(:,k+1)-x(:,k),'inf'))/(norm(x(:,k+1),'inf'))) < tol
What about:
if max(abs(x(:,k+1) - x(:,k)) ./ abs(x(:,k+1))) < tol
2 commentaires
  Leah Youngquist
 le 29 Sep 2018
				Can I ask why the infinity norm is 'insufficient' here? Your solution was exactly what I needed for my problem, but I'm not sure I understand why! ((sorry if this kind of thing is against the Matlab forum rules, I'm a fresh coding baby ha-ha!))
  Bruno Luong
      
      
 le 30 Sep 2018
				
      Modifié(e) : Bruno Luong
      
      
 le 30 Sep 2018
  
			Agree: Jan's expression compute exactly the same than Nguyen's expression.
Voir également
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