FLOPS in Complex Array Multiplication
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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
采纳的回答
Roger Stafford
2016-6-5
1 个投票
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.
3 个评论
nauman
2016-6-5
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
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.
更多回答（1 个）
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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