CUDA fft with prime number lengths

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Rodrigo
Rodrigo on 7 Nov 2015
Answered: Edric Ellis on 9 Nov 2015
Hi, I just ran into a curious problem with R2015a'a GPU fftn function. For some reason fftn and ifftn refuse to transform 512x512x277 gpuArrays. Both a=fftn(b) and a=ifftn(b), where b is 512x512x277, return a=b as answers. The CPU version of the routine does what it's supposed to.
In hindsight, 277 is a rather large prime number, so it's not unreasonable for cuFFT to choke, but why doesn't it produce an error message or non-sense output?
Thanks, Rodrigo

Answers (1)

Edric Ellis
Edric Ellis on 9 Nov 2015
Thankyou for reporting this problem. In R2015b, this transform actually fails with an error, so I suspect that in R2015a the transform fails in a similar manner but without reporting an error, and so we end up returning the transform input.

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