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@cfa.harvard.edu> To: "Eric Hagemann" <ehagemann@home.com> Cc: <Numpy-discussion@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@cfa.harvard.edu Cambridge, MA 02138 GPG Fingerprint: 06A9 9553 78BE 16DB 407B FFCA 9BFA B6FF FFD3 2989
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