Is there a conceptual way to understand where the Fast Fourier Transform is avoiding redundant computation and thus achieving $O(n\log n)$ instead of $O(n^2)$.

Consider a standard example of the FFT to multiply two polynomials faster. Its not obvious to me conceptually why the FFT should yield a faster way to multiply two polynomials.

  • $\begingroup$ you mean $O(n\log n)$ $\endgroup$ Sep 21, 2011 at 0:35

1 Answer 1


Conceptually the FFT takes advantage of a shortcut similar to the distributive law for multiplication. To compute $$(x_1 + x_2)(x_3 + x_4)$$ on could either add first (twice) and then multiply (once), or one could expand $$sx_1x_3 + x_1x_4 + x_2x_3 + x_2x_4$$ and multiply (four times) and then add (three times). This idea has been spelled out in the paper The Generalized Distributive Law.

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    $\begingroup$ @R Hahn your link seems to be broken $\endgroup$ Sep 21, 2011 at 3:28
  • $\begingroup$ It had a forward slash at the end that it shouldn't have. Should be fixed now. $\endgroup$
    – R Hahn
    Sep 21, 2011 at 5:35

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