FLOPS in Complex Array Multiplication

Hi all

I have two complex number arrays of size N. If i do element by element multiplication of these two arrays, how can i calculate total no of "FLOPs" for this array multiplication?

Kindly help me. Thanks

Answer 1

[Roger Stafford]

I count six floating point operations per complex multiplication. If your matrix is N-by-N, that would be a total of 6*N^2 flops for an element-by-element multiplication.

[nauman]

Hi Roger Thanks. It means if i multiply two N size complex arrays, "element by element", total no of flops would be 6N. Am i right?

[Roger Stafford]

It depends on what you mean by "N size". If you mean N-by-1 or 1-by-N, then 6*N is correct. The fundamental fact is that a multiplication of one complex number by another complex number takes six flops, consisting of four floating point multiplications and two floating point additions.

[Walter Roberson]

If the arrays have N elements then Yes.

a .* b = complex( real(a).*real(b) - imag(a).*imag(b), real(a).*imag(b) + imag(a).*real(b) )

2 multiplications and one addition (or subtraction) on each side, for a total of 6 operations.

Answer 2

[Adam]

doc timeit

Put the multiplication in a function, use timeit and divide the number of floating point operations (easy to calculate) by the time you get.

[nauman]

Hi Adam Thanks for help. Actually i need to theoretically calculate FLOPs for N size complex arrays multiplication