[Numpy-discussion] curious FFFT timing
Eric Hagemann
ehagemann at home.com
Thu May 3 21:11:26 EDT 2001
Scott,
Right you are. My brain fried -- I forgot about the other "small prime"
stuff
Inverse FFT ( 4096) Number of seconds 5.508000 Time per FFT
0.005508
Forward FFT ( 4096) Number of seconds 3.225000 Time per FFT
0.003225
Inverse FFT ( 4097) Number of seconds 29.723000 Time per FFT
0.029723
Forward FFT ( 4097) Number of seconds 26.788000 Time per FFT
0.026788
Cheers
Eric
----- Original Message -----
From: "Scott Ransom" <ransom at cfa.harvard.edu>
To: "Eric Hagemann" <ehagemann at home.com>
Cc: <Numpy-discussion at lists.sourceforge.net>
Sent: Tuesday, April 03, 2001 9:05 PM
Subject: Re: [Numpy-discussion] curious FFFT timing
> Hi Eric,
>
> > Any one want to speculate on the timing difference for the 4095 vice
> > 4096 long transforms ?
> > Since 4095 is not a power of two I would have expected a greater time
> > difference (DFT vice FFT)
>
> You are correct in that 4095 is not a power of two, but is is the
> product of only small prime factors:
> 3 * 3 * 5 * 7 * 13 = 4095
>
> Since the FFT code implements a N*log_2(N) algorithm for numbers that
> contain only small prime factors, the difference in times is rather
> small. FFTs that have lengths that are powers-of-two tend to be more
> efficient in general (since the decimation routines are cleaner for this
> case).
>
> If you test with 4097 (17 * 241) it will be quite a bit slower I'd
> guess...
>
> Scott
>
>
> --
> Scott M. Ransom Address: Harvard-Smithsonian CfA
> Phone: (617) 495-4142 60 Garden St. MS 10
> email: ransom at cfa.harvard.edu Cambridge, MA 02138
> GPG Fingerprint: 06A9 9553 78BE 16DB 407B FFCA 9BFA B6FF FFD3 2989
>
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