Cell contents assignment to a non-cell array object.

hi.

i dont know why in my code is keeping me sending this error anybody can help me please my code is:

   A=input('Dame el sistema de ecuaciones expresado en forma matricial');
   b=input('Dame el valor de b');
   k=1;
   x=input('Dame el valor x0 inicial');
   x{1}=x;
   r{k}=(A*x{k})-b;
   p{k}=-(r{k});
   for k=0:10000000000;
       alfa{k}=(-(((r{k}')*r{k})/((p{k}')*(A*p{k}))));
       x{k+1}=x{k}+(alfa{k}*p{k});
       r{k+1}=r{k}+(alfa{k}*A*p{k});
       beta{k+1}=((r{k+1})'*r{k})/((r{k}')*(r{k}));
       p{k+1}=-(r{k+1})+(beta{k+1}*p{k});
       k=k+1;
   if (r{k}<=0)
    break;
   end
   end

and the error is the next

 *Cell contents assignment to a non-cell array object.*
 *Error in gradientesconjugados (line 5)*
 *x{1}=x;*

if anybody can help me i would apreciate so much.

[Mahdi]

Why are you defining the variables in a cell when you want to perform calculations on them? Why not just use matrices and then convert the result into a cell?

[José]

one question

how i do that? o.o

i need to use the result of the last calculation to do the next one

for example

i need r1 to calculate r2

thanks

[James Tursa]

How many iterations do you expect this to take before the break happens? If the iterations go all the way to 10000000000, that will take about 2.7 terabytes of memory just to hold the variable header information (60 bytes each), and probably a gazillion hours of time since you are increasing the size of these cell arrays within a loop.

[José]

i think that i understood you and now i change the code to this because i want to calculate matrices

 A=input('Dame el sistema de ecuaciones expresado en forma matricial');
 b=input('Dame el valor de b');
 k=1;
 x0=input('Dame el valor x0 inicial');
 r(k)=(A*x0)-b;
 p(k)=-(r{k});

but now i get this error

In an assignment A(I) = B, the number of elements in B and I must be the same

if anyone can help me i would apreciate so much.

[José-Luis]

Have you tried the debugger?

Answer 1

[Azzi Abdelmalek]

Try this

clear
A=input('Dame el sistema de ecuaciones expresado en forma matricial');
 b=input('Dame el valor de b');
 k=1;
 xx=input('Dame el valor x0 inicial');
 x{1}=xx;
 r{k}=(A*x{k})-b;
 p{k}=-(r{k});
 for k=0:10000000000;
     alfa{k}=(-(((r{k}')*r{k})/((p{k}')*(A*p{k}))));
     x{k+1}=x{k}+(alfa{k}*p{k});
     r{k+1}=r{k}+(alfa{k}*A*p{k});
     beta{k+1}=((r{k+1})'*r{k})/((r{k}')*(r{k}));
     p{k+1}=-(r{k+1})+(beta{k+1}*p{k});
     k=k+1;
 if (r{k}<=0)
  break;
 end
 end

[Mahdi]

Also, change the for statement to be:

for k=2:10000000000;

If your goal is to populate the elements after the first one you inputted.

Even then, I don't think it will work because you'll get NaN's as outputs.

[José]

a change in the code the xx and the 2 but i get the same error

Cell contents assignment to a non-cell array object.

thanks for your time

[Azzi Abdelmalek]

Have you cleared your variable x?

Answer 2

[Udit Gupta]

Instead of

x=input('Dame el valor x0 inicial');
x{1}=x;

use

temp=input('Dame el valor x0 inicial');
x{1}=temp;

[José]

it gives me the same error

Cell contents assignment to a non-cell array object.

thanks

[Udit Gupta]

In the same lines as before?

[Udit Gupta]

I ran this code without any error. You need to start your loop from k=1 instead of k=0.

A=input('Dame el sistema de ecuaciones expresado en forma matricial');
 b=input('Dame el valor de b');
 k=1;
 temp=input('Dame el valor x0 inicial');
 x{1}=temp;
 r{k}=(A*x{k})-b;
 p{k}=-(r{k});
 for k=1:10000000000;
     alfa{k}=(-(((r{k}')*r{k})/((p{k}')*(A*p{k}))));
     x{k+1}=x{k}+(alfa{k}*p{k});
     r{k+1}=r{k}+(alfa{k}*A*p{k});
     beta{k+1}=((r{k+1})'*r{k})/((r{k}')*(r{k}));
     p{k+1}=-(r{k+1})+(beta{k+1}*p{k});
     k=k+1;
 if (r{k}<=0)
  break;
 end
 end