How to convert the for loop to a while loop?

Question

Avinash Gupta 2021-3-30

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/788244-how-to-convert-the-for-loop-to-a-while-loop

编辑： MG 2021-4-1

A = [7 3 2 1; 2 9 4 5; 1 3 13 4; 4 5 8 14 ];        % A is diagonally dominant
b = [1;2;3;4];
x = zeros(4,1);
x_new = zeros(4,1);
% Gauss Seidel using for loop
n = 4;
for iter = 1:25
    for i = 1:n
        num = b(i)-A(i,1:i-1)*x_new(1:i-1) - A(i,i+1:n)*x(i+1:n);
        x_new(i) = num/A(i,i); 
        x = x_new;
    end
    disp(['At iteration = ',num2str(iter), '    x = ', num2str(x_new')]);
end
    
% The for loop works but the while loop doesn't
%% This is what I have done in the while loop.
% Gauss seidel using while loop
n = 4;
error = 0;
iter = 0;
while error >=1e-6
     for i = 1:n
        x_new(i) = (b(i)-A(i,1:i-1)*x_new(1:i-1)- A(i,i+1:n)*x(i+1:n))/A(i,i); 
        x = x_new;
        error = abs(x-x_new);
     end
     iter = iter+1;
     disp(['At iteration = ',num2str(iter), '    x = ', num2str(x_new')]);
end
        
  
   
           
 
     
 

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Answer 1

MG 2021-3-30

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/788244-how-to-convert-the-for-loop-to-a-while-loop#answer_662834

编辑：MG 2021-3-30

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Hi Avinash,

I modified your code into something I believe is what you might want:

A = [7 3 2 1; 2 9 4 5; 1 3 13 4; 4 5 8 14 ];        % A is diagonally dominant
b = [1;2;3;4];
x = zeros(4,1);
x_new = zeros(4,1); %<---- modified
% Gauss seidel using while loop
n = 4;
error = ones(4,1); %<---- modified
iter = 0;
while prod( error >= 1e-6*ones(4,1) ) %<---- modified
     for i = 1:n
        x_new(i) = (b(i)-A(i,1:i-1)*x_new(1:i-1)- A(i,i+1:n)*x(i+1:n))/A(i,i); 
        error(i) = abs(x(i)-x_new(i)); %<---- modified
        x(i)     = x_new(i); %<---- modified to be after error
     end
     iter = iter+1;
     disp(['At iteration = ',num2str(iter), '    x = ', num2str(x_new')]);
end

I then just notice that you have in your code this code sequence

A(i,1:i-1)*x_new(1:i-1)

and I don't know if that does what you expect, beceuse with i =1 then A(1,1:0) *x_new(1:0) is a multiplication of two empty matrices (that probably returns 0).

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Avinash Gupta 2021-3-31

Hi, Michael. Thanks for the prompt response.

1. However, I do not understand this modification: while prod( error >= 1e-6*ones(4,1) )

Why can't we go with: while error >= 1e-5*? Kindly explain the difference

2. A(i,1:i-1)*x_new(1:i-1)-----> This is not so useful for x_new(1) and getting an empty matrix is fine at that step.It is redundant for x_new(1). However, it becomes useful for other x_new(i), i= 2,3,4

3. Also the error should be calculated before the update. Thanks for pointing it out

Ps: I am quite new to MATLAB and I am still navigating my way around

MG 2021-3-31

编辑：MG 2021-4-1

You can use your original 'error >= 1e-6', it should do the same thing. I only did it to more explicitely show that it checks individually each of the elements in your 4 component 'error' array. If any is false, the loop stops. Replace 'prod' with 'any' if you instead want that every error(i) < 1e-6 before the while-loop ends.
Fine, as long as it does what you intend (in practice that is a a multiplication of two empty arrows that here return 0, but I'm not really sure why it returns 0 and not an empty matrix).

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