By: Robert Myers (rbmyersusa.delete@this.gmail.com), October 20, 2012 9:35 am
Room: Moderated Discussions
Emil Briggs (me.delete@this.nowherespam.com) on October 20, 2012 8:34 am wrote:
> >
> > I not sure what point
> > you're trying to make here. DFT is
> based on first principles and has a very good
> > track record but it's
> computationally very expensive. You can do a lot of useful
> > work with it on
> systems that range up to a few thousand atoms but with O(N^3)
> > scaling you
> can easily use up any feasible increase in computer power without
> >
> extending the size range you can model very far. Hence the efforts to improve
>
> > the algorithmic scaling. It's not a solved problem yet but it's not
> perpetual
> > motion.
> >
>
No, you don't believe in perpetual motion. You're apparently doing what I suggested was possible: you are using a computationally clumsy algorithm to match the work done by the processors to the wimpy bandwidth available. I wasn't quite certain about what N you were referring to in the N^3 scaling. If you really are referring to the number of atoms (which itself goes as M^3, where M is the number of atoms in a line on a cubic lattice), then you are (and you do refer to it as a DFT and not as an FFT) forgoing the advantages of divide and conquer. Ummm...
Robert.
> >
> > I not sure what point
> > you're trying to make here. DFT is
> based on first principles and has a very good
> > track record but it's
> computationally very expensive. You can do a lot of useful
> > work with it on
> systems that range up to a few thousand atoms but with O(N^3)
> > scaling you
> can easily use up any feasible increase in computer power without
> >
> extending the size range you can model very far. Hence the efforts to improve
>
> > the algorithmic scaling. It's not a solved problem yet but it's not
> perpetual
> > motion.
> >
>
No, you don't believe in perpetual motion. You're apparently doing what I suggested was possible: you are using a computationally clumsy algorithm to match the work done by the processors to the wimpy bandwidth available. I wasn't quite certain about what N you were referring to in the N^3 scaling. If you really are referring to the number of atoms (which itself goes as M^3, where M is the number of atoms in a line on a cubic lattice), then you are (and you do refer to it as a DFT and not as an FFT) forgoing the advantages of divide and conquer. Ummm...
Robert.
