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What applications require polynomial interpolation with such high order that an FFT is worth it? Why not use sinc interpolation if you're computing an FFT anywa
by duped 4y ago
What applications require polynomial interpolation with such high order that an FFT is worth it? Why not use sinc interpolation if you're computing an FFT anyway?
sidenote: polynomial multiplication != interpolation
- ninepoints 4y agoFFT is almost never not worth it. Even at small scales, you should be using the FFT (and you are a daily user in audio/video/image contexts).
- duped 4y agoWe have very different experiences then. "Small" scale to me is n < 64, which is right on the threshold of whether it's faster to run an n^2 algorithm that could be nlogn with an FFT (for 1-dimensional vectors) - depending on the algorithm, implementation, and hardware of course. I've benchmarked FFT/non-FFT based approaches for almost everything in that scale for the last ten years and gotten quite different results.
- enriquto 4y agoAn interesting case is two-dimensional interpolation. For example, for precise image resampling. In that case, due to the separability, the FFT is still O(nlog(n)), thus just as fast and much more precise than space-domain computations with 2D splines. Try to rotate 10 degrees iteratively an image 36 times with splines and with fft-based methods. Even 7 tap splines (which is a huge order) will utterly destroy the image contents; but it is just peanuts to the fft.
- deleted 4y ago[deleted]
- steppi 4y agoThere's an aside in the linked section of CLRS that mentions that FFT based polynomial multiplication can be used for fast evaluation of Lagrange interpolation. I'm not aware of applications though, Lagrange interpolation isn't numerically stable; there are better interpolation methods for large datasets. It's more of a mathematical curiosity.
- rerdavies 4y agoNot sure why you would use FFTs to evaluate Lagrange interpolators, when Lagrange interpolators have a very pretty O(N) implementation.
- steppi 4y agoExample please?
- rerdavies 4y ago:-( Walking it back a bit. For common cases, O[N], but O[N^2] for initialization. https://news.ycombinator.com/edit?id=34720462 https://news.ycombinator.com/edit?id=34720462
- mjcohen 4y agoThere are n terms with n products so O(n^2).
- foooobaba 4y agoThe FFT is not used for that. For multiplying two polynomials C(x) = A(x) B(x) then FFT is used to evaluate the polynomials (evaluated at complex roots of unity) to FIND the Lagrange interpolating polynomials for A(x), and B(x) in O(nlogn) time. Then the Lagrange interpolation is evaluated in O(n) time compute the y values for y= A(x)B(x), then FFT is used again to compute the coefficients of the final C(x). This is discussed on pages 825-827 of CLRS and the diagram on page 828 shows this in a nice diagram (the pointwise multiplication refers to evaluating the Lagrange interpolating polynomials).
- photon12 4y agoZero-knowledge proof generation, for one. Edit, see for example: https://www.zprize.io/prizes/accelerating-ntt-operations-on-an-fpga https://www.zprize.io/prizes/accelerating-ntt-operations-on-...
- steppi 4y agoThat’s pretty neat. I didn’t even think of finite fields when I made my other comment. Another example where mathematical curiosities prove useful in cryptography.