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I need to calculate gflops of cufft forward and inverse algorithm .

I have : 

double gflops = 1.0e-9 * dNumOps/gpuTime;

but I don't know how set dNumOps , I need to know total operations of cufft forward and inverse.

I remember that fft usually have a N*logr(N) operations where r is the number of dataset division in fft, but for cufft ?

tulkas85
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  • All FFTs use O(n*log(n)), that is the minimum complexity possible. If the complexity of the implementation was O(n^2) it would make it a **DFT** not an FFT. The question is what the constants are. What do you need the FLOPs for? – Danny Varod Jul 09 '12 at 12:53
  • I need something like this : http://www.sharcnet.ca/~merz/CUDA_benchFFT/p2.ip.1d.png I make a forward and inverse transform of N elements, I have time elapsed, but I don't know how many ops these functions performs... I use cufftExecZ2Z. – tulkas85 Jul 09 '12 at 16:34
  • a simple benchmark with cufft and serial fft – tulkas85 Jul 09 '12 at 21:36
  • Perhaps better benchmarks would be 1. FLOPS in general per GPU, which you can get from the vendor's site or measure with your own kernels instead of built in functions, 2. Time measured. – Danny Varod Jul 10 '12 at 08:41

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