How do I replace "for" loops by matrix multiplication? I am getting confused because of 4 "for" loops. Any suggestions?

Hello, the code is as follows.

value = zeros();
E2 = zeros();
dxdy = 1;
x1=1;y1=1;u1=1;v1=1;

for u = -1:1:1
     for v = -1:1:1

          for x = -1:1:1
            for y = -1:1:1

               value(x1,y1) = input_data(x1,y1)*exp( ( -1i*k*(((u-x)^2) + (v-y)^2)) / (2*0.2))*(dxdy); 
               y1 = y1+1;
            end
             x1 = x1+1;
             y1 = 1;
          end
              S_rows = sum(value,2);           %Adding components of matrix "value"    
              S_complete = sum(S_rows);
              E2(u1,v1) = S_complete;  

            x1 = 1;
            v1 = v1 + 1;
        end
          u1 = u1+1;
          v1 = 1;
   end

      disp(E2)

[madhan ravi]

what's input_data?

[Dhananjay Mishra]

input_data = [1 2 3 ; 4 5 6 ; 7 8 9]

[Torsten]

"input_data" must be 27x3, not 3x3.

[Dhananjay Mishra]

If you see the program the values of u1 and v1 is reset everytime. Hence the input is 3*3. And the output is also 3*3

Answer 1

[Miriam]

dxdy = 1;
[x,y,u,v] = ndgrid(-1:1);
value = input_data.*exp((-1i*k*(((u-x).^2) + (v-y).^2))/(2*0.2))*(dxdy); 
S_rows = sum(value,2);
S_complete = sum(S_rows);
E2 = squeeze(S_complete);
disp(E2)

[Dhananjay Mishra]

Thanks a lot for your answer. However when I use the code it throws an error saying : Any insights? Thank you

Error using  - 
Matrix dimensions must agree.

[Miriam]

What is k in your code? I assumed it was scalar and just used 1.

[Dhananjay Mishra]

k is a scalar quantity. However the input data is 3*3 matrix. Hence now I am getting the error :

Error using  .* 
Matrix dimensions must agree.

[Miriam]

Do you get the same error with the following command?

[1 2].*[3 4;5 6]

This causes no issues with my version of MATLAB.

[Dhananjay Mishra]

Yes, i get the same error using [1 2].*[3 4;5 6].

[Miriam]

Odd there are no issues on my end. Perhaps try this instead:

value = repmat(input_data,1,1,3,3).*exp((-1i*k*(((u-x).^2) + (v-y).^2))/(2*0.2))*(dxdy);

[Dhananjay Mishra]

Now I don't get any error but the answer doesn't matches for both the codes.

[Miriam]

I'm not sure what to tell you then. It works fine for me.