Help ,urgent.

Question

Anshuman S 2017-6-14

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/344782-help-urgent

回答： Walter Roberson 2017-6-14

Hi this error is being generated:

In an assignment  A(:) = B, the number of elements in A and B must be the same.
Error in q1_gs_150115 (line 42)
         x(k) = num/At(k,k);

My code :

function [GS] = q1_gs_150115(~)
m=input('variables');
r=input('relaxation_number');
for i=1:m
for j=1:m
A(i,j)=input('element_i,j');
end
end
A=reshape(A,m,m);
for i = 1:m
j = 1:m;
    j(i) = [];
    B = abs(A(i,j));
   c(i) = abs(A(i,i)) - sum(B); % Is the diagonal value greater than the remaining row values combined?
    if c(i) > 0
        fprintf('The matrix is diagonally dominant at row %2i\n\n',i)
    end
end
for i=1:m
s(i,1)=input('constants_i');
end
p=input('number_of_iterations');
for i=1:m
x(i,1)=input('initial_values_i');
end
err = zeros(m,1);
At = [A,s];
for iter = 1:p
     for k = 1:m
         xold = x(k);
         num =  (1-r)*At(k,k)*x(k+1:m)+ r*At(k,end) - r*At(k,1:k-1)*x(1:k-1) - r*At(k,k+1:m)*x(k+1:m);
         x(k) = r*num/At(k,k);
         err(k) = abs(x(k)-xold);
     end
     disp(['Iter ',num2str(iter), '; Error =', num2str(max(err))]); 
end
   disp('The result is:')
disp(x)

2 个评论
显示无隐藏无

Stephen23 2017-6-14

The line shown in the error message does not exist in your code.

Guillaume 2017-6-14

why-your-question-is-not-urgent-or-an-emergency

Doubly ironical, since you haven't posted the relevant information for us to even give you an answer.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2017-6-14

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/344782-help-urgent#answer_270724

在 MATLAB Online 中打开

         num =  (1-r)*At(k,k)*x(k+1:m)+ r*At(k,end) - r*At(k,1:k-1)*x(1:k-1) - r*At(k,k+1:m)*x(k+1:m);

r is a scaar. At(k,k) is a scalar. At(k,end) is a scalar. x(k+1:m) is a column vector. So we have to conclude that (1-r)*At(k,k)*x(k+1:m) is a column vector.

At(k,1:k-1) is a row vector but it is being used with "*" against a column vector, so the result of the "*" is a scalar, so r*At(k,1:k-1)*x(1:k-1) is a scalar.

At(k,k+1:m) is a row vector, but it is being used with "*" against a column vector, so the result of the "*" is a scalar, so r*At(k,k+1:m)*x(k+1:m) is a scalar.

num is therefore column vector minus scalar minus scalar. That is going to give a column vector result.

Then in the next line,

x(k) = r*num/At(k,k);

r is a scalar, At(k,k) is a scalar, and we found from above that num is a column vector. r*num/At(k,k) must therefore be a column vector. But you are trying to store the column vector into a scalar location, x(k)