Talk:Curvelet

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Computational cost

The computational cost of a curvlet transform is approximately 10-20 times that of an FFT, and has the same $O(n^{2}log(n))$ dependence on the size of the image.

Isn't the cost of an FFT $O(nlog(n))$ ? c.f. Fft --128.8.120.139 (talk) 20:26, 23 March 2010 (UTC) IBB[reply]

That's right. I just added the "dubious" tag to that. —Ben FrantzDale (talk) 23:04, 23 March 2010 (UTC)[reply]

The problem is the definition of n. In Fast Fourier transform N is the product of the dimensions of the array to be processed, while in the estimate on this page it is an nxn array that is being processed. Thus they are the same at least in this 2d case. Holmansf (talk) 13:48, 10 May 2010 (UTC)[reply]