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Let $(x_1 \ldots ,x_n) \in \mathbb{R}^n$ and $f_i = \Pi_{j=1, j \neq i }^n ( x_i - x_j )$

I'm trying to evaluate $(f_1, \ldots, f_n)$. A trivial algorithm runs in $\mathcal{O}(n^2)$ but given the very specific form of the problem, there's got to be something faster. Maybe I've overlooked something simple, maybe a fourier transform is in order... What are your thoughts?

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I will be surprised if you get subquadratic for the general case. If the x_I are in arithmetic progression, there is some chance for speedup, but deciding how to take advantage of that will also take time. Gerhard "Ask Me About System Design" Paseman, 2011.11.30 – Gerhard Paseman Nov 30 '11 at 17:40
up vote 2 down vote accepted

The following may be of some help, if you haven't thought of it already. Let $V$ be the Vandermonde matrix with $(i,j)$th entry $x^{i-1}_{j}$, $i,j=1,\ldots,n$. Its inverse $W$ has $(i,1)$th entry $$ (-1)^{n} \frac{x_1 \ldots x_{i-1} x_{i+1} … x_n}{f_i}. $$ Hence, to find $f_1,\ldots,f_n$ we need to solve $V \alpha = e_1$, where $e_1=(1,0,\ldots,0)$. Thus, the question is, how fast can one solve a system of equations with a Vandermonde matrix? There's discussion on this topic in Nick Higham's book "Accuracy and stability of numerical algorithms" (see chpt 22). It appears that there are algorithms that are $O(n (log n)^2)$. However, it's noted that these may not be numerically stable or practical.

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nice, much cleaner than my solution – Federico Poloni Dec 1 '11 at 22:35

I heard there's a method called non-equispaced FFT (though I know very little about it), which should be able to compute the coefficients of the polynomial $p(x)=\prod(x-x_j)$ from the $x_j$ in quasi-linear time.

Using it, we can get a divide-and-conquer algorithm as follows:

  1. Divide your points in two sets $A$ and $B$ of more or less equal magnitude
  2. Compute with non-equispaced FFT $g_{A,i}=\prod_{j\in A} (x_i-x_k)$ for all $i\in B$ and $g_{B,i}$ for all $i\in A$.
  3. solve the two smaller sub-problems for $A$ and $B$.
  4. compute the $f_i$ by multiplying the results in points 2 and 3.

I am not sure about the practicality of this nFFT though.

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I'm not sure how that solves the problem. Say I find coefficients for $P(x) = \Pi(x-x_j)$. I can deduce the coefficient for $P'(x)$. Then what? I still need to evaluate $P'(x)$ over $N$ points. How do you perform step 2 in subquadratic time? Something seems wrong. – Arthur B Dec 1 '11 at 0:39
That evaluation is an inverse non-equispaced FFT. This is analogue to similar evaluation/interpolation tricks for the traditional FFT, e.g.… – Federico Poloni Dec 1 '11 at 8:14

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