Can any one explain me how to apply vectorized operation on the outer for loop of the given program?

a=[1+1i 2+2i 3+3i 4+4i 5+5i 6+6i 7+7i 8+8i 9+9i 1+1i 2+2i 3+3i 4+4i 5+5i 6+6i 7+7i 8+8i 9+9i];
for m=1:6
    m=m-1;
    k=1:1:4;
    out_1(m+1)=(sum((a(k+m)).*(conj(a(k+12+m)))))/12;
end

[Alexandra Harkai]

Does this code not give you an indexing error for a(k+12+m) after the 3rd loop? (k+12+3 = 1:4+12+3=16:19 and a has only 18 elements)

[Tamil selvan]

Hello Alexandra Harkai,

Sorry. Just now I seen it is as 12 instead of 8.

Answer 1

[Guillaume]

As Alexandra pointed out, your example is indexing past the end of the array.

Assuming a sufficiently large array, the following would work:

%input: %a, a vector with enough elements
%e.g:
a = repmat((1:9).*(1+1i), 1, 3)
m = 1:6;      %row vector
k = (0:3).';  %column vector
assert(numel(a) >= max(m) + max(k) + 12, 'Not enough elements in a');
%in R2016b:
out = sum(a(m+k) .* conj(a(m+k+12))) / 12;
%prior to R2016b:
%indices = bsxfun(@plus, m, k);
%out = sum(a(indices) .* conj(a(indices+12))) / 12;

[Tamil selvan]

Hello Guillaume,

Thanks for your code its working fine. When I am checking 100000 samples input, the execution time for the above code it takes 4times more than the original code had taken. Can you tell me any other execution time optimization for the above code?