Accelerate Matlab calculation for multi-dimension matrix

I want to solve a problem in space (3-dimensional:x y and z coordinate) and time(t). In general, my problem is like the code below:

A=rand(220);

x=0:1:50;

y=0:1:50;

z=0:1:50;

t=0:1:50;

B=zeros(length(x),length(y),length(z),length(t));

for ix=1:length(x)

for iy=1:length(y)

for iz=1:length(z)

for it=1:length(t)

B(ix,iy,iz,it)=sum(sum(sum(sum((A.*A*x(ix))+((A.^3)*y(iy))+...
    ((A+2*A)*z(iz))+((A.^4)*t(it))))));

end

end

end

end

However, the looping proccess takes much time. I am afraid i can not use repmat to decrease the computational time since it will need large memory to store the variable. Do you have any suggestion to make the simulation faster?

Thank you very much.

Michael

Answer 1

[the cyclist]

A 50x50x50x50 array of doubles is only about 48 MB of storage. I see no memory problem in vectorizing the whole calculation.

[Michael Lubis]

Yeah, you are right. However, i want to multiply matrix with size 220x220 (the matrix A in my command) for each space and time entity. If i use repmat for the vectorization, then i have to create a matrix with dimension 50x50x50x50x220x220. Then i need a lot of memory.

Furthermore, in the real problem, i want to do 3 hours simulation with time interval 0.5 second (t=0:0.5:3*3600) which actually will require more memory. There is also a posibility for me to increase the range for space with smaller interval (for example x=-50:0.5:50). Therefore, is there another way to do all of this calculation instead of using looping?

Answer 2

[Roger Stafford]

As you have written your code, A is a 2D array and yet you have taken four successive sums, which makes no sense. Either you meant A to be 4D or you meant to have only two successive sums. Either way your code is enormously inefficient. The summation of A.*A, A.^3, A+2*A, and A.^4 can be done just once for each of these four rather than 48,400 separate times each.

I will assume that A is to be 2D and you meant only two sums.

   A=rand(220);
   x=0:1:50;
   y=0:1:50;
   z=0:1:50;
   t=0:1:50;
   A1 = sum(sum(A.^2))*x;
   A2 = sum(sum(A.^3))*y;
   A3 = sum(sum(3*A))*z;
   A4 = sum(sum(A.^4))*t;
   B = zeros(51,51,51,51);
   for ix = 1:51, for iy = 1:51, for iz = 1:51, for it = 1:51
     B(ix,iy,iz,it) = A1(ix)+A2(iy)+A3(iz)+A4(it);
   end, end, end, end

This should be much faster than your original code (unless Mathworks' compilers are awfully smart.)

[Michael Lubis]

Yeah you are right. It should be only two sums. And exactly, your scheme calculate faster, eventhough still take takes sometimes for it to finish. Thank you very much for your answer.