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FLOPS in Complex Array Multiplication

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nauman
nauman 2016-6-4
评论： Walter Roberson 2016-6-5
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

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Roger Stafford
Roger Stafford 2016-6-5
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.
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Roger Stafford
Roger Stafford 2016-6-5
编辑：Roger Stafford 2016-6-5
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
Walter Roberson 2016-6-5
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.

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Adam
Adam 2016-6-4
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.
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nauman
nauman 2016-6-5
Hi Adam Thanks for help. Actually i need to theoretically calculate FLOPs for N size complex arrays multiplication

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