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Cell contents assignment to a non-cell array object.

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José
José on 29 May 2014
Commented: Udit Gupta on 29 May 2014
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.

  5 Comments

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James Tursa
James Tursa on 29 May 2014
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
José on 29 May 2014
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.

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Answers (2)

Azzi Abdelmalek
Azzi Abdelmalek on 29 May 2014
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

  3 Comments

Mahdi
Mahdi on 29 May 2014
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
José on 29 May 2014
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

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Udit Gupta
Udit Gupta on 29 May 2014
Instead of
x=input('Dame el valor x0 inicial');
x{1}=x;
use
temp=input('Dame el valor x0 inicial');
x{1}=temp;

  3 Comments

José
José on 29 May 2014
it gives me the same error
Cell contents assignment to a non-cell array object.
thanks
Udit Gupta
Udit Gupta on 29 May 2014
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

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