1
$\begingroup$

Consider $N$ particles with massa $m$ spaced out on a $\Omega = [0,1]^2$. The force acting on each particle reads $$F_i = \sum_j C\frac{m_i m_j}{r_{ij}^2}$$ for some constant $C$. The problem here is that we would need to do $O(N^2)$ calculations to compute it fully (since there are $N$ forces to calculate each containing a sum over $N-1$ elements).

Therefore I try to use some handy approximation that will reduce the computational complexity. We subdivide the particles into a quadtree data structure in order to use Barnes - Hut algorithm (alternative) on the problem. All the sources I see state the complexity to be $O(N \log N)$ but I am unable to find any kind of proof for this statement. Also it seems a wrong statement in general - BH has a free parameter $\theta$ which should be included in the analysis because if $\theta = 0$ the complexity is $O(N^2)$.

I'd like to have an (derivation of) expression (or at least a sketch of a derivation) for the complexity as a function of $N$ and $\theta$. Also some bound on error of the approximation as a function of these two variables would be nice to have for case $m_i = 1$ for all $i$ and for case $m_i = \pm 1$ distributed evenly.

$\endgroup$
  • $\begingroup$ I have not carried out a proper analysis, but intuitively the asymptotic complexity should be the same for all $\theta > 0$. The choice of $\theta$ just affects the constant factor and determines how large the system needs to be for the asymptotic scaling to become relevant. $\endgroup$ – Ilmari Karonen Mar 19 '18 at 23:34
  • $\begingroup$ I agree with this intuition. Problem is, I have only a very abstract idea on how such analyse should be carried out. The actual analysis would help me lots with the conceptual understanding of handling such type problems. And also it would solve this very problem - which is also very nice :) $\endgroup$ – Piotr Benedysiuk Mar 19 '18 at 23:40
2
+150
$\begingroup$

Asymptotics often ignore parameters - it's wrong to assume an asymptotic is uniform in all parameters by default.

For masses with consistent signs the multiplicative error is quadratic in $\theta.$ So if the estimate for $F_i$ is $F_i'$ then $F_i'/F_i$ will lie in $[1-c\theta^2,1+c\theta^2],$ for some absolute constant $c$ (and $0<\theta<1/2$ say). By scaling it suffices to look at the error of estimating the sum of $f(x)=1/|x|^2$ for particles with center of mass $(1,0,0)$ and within distance $O(\theta).$ By Taylor series each value is $f(1,0,0)+(x-(1,0,0))\cdot (\nabla f)(1,0,0)+O(|x-(1,0,0)|^2).$ The linear term cancels out because we're using the center of mass (monopole approximation). So the multiplicative error is bounded by $O(\theta^2).$ (I suspect there's a corresponding lower bound given by a uniform distribution.)

For mixed mass signs my instinct is to say there's isn't a sensible a priori error bound - if the forces cancel, the error could be much larger than the actual value.

The $O(N\log N)$ in the Barnes-Hut paper comes from considering particles in a grid (at least as I read it - they just say uniformly distributed). Suppose the universe size is $S$ and any pair of particles are separated by at least $s.$ Consider running Barnes-Hut to calculate the force at a particle at position $x.$ Points at distance $r>0$ from $x$ will be put in a cube of radius at least $\tfrac{1}{100}\theta r$ (being sloppy about constant). So we can upper bound the number of cubes by integrating $(\tfrac{1}{100}\theta r)^{-3}$ over points at distance between $s$ and $S$ from $x.$ Integrating using spherical co-ordinates gives a bound of $O(\theta^{-3})\int^S_s 1/r=O(\theta^{-3}\log(S/s)).$ For a grid we can put $S=1$ and $s=1/N^{1/3}$ for example, giving a bound of $O(\theta^{-3}\log N)$ cubes, for each of the $N$ iterations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.