Moreover, H is bigger than the original data due to partial fill-in and the computation is almost always going to be I/O bound anyway. Thus, FLOPS will be as Jake says, wall clock will probably probably scale in the ratio of the sizes of H to the original data.
On Wed, Feb 24, 2010 at 10:17 PM, Jake Mannix <[email protected]> wrote: > > If instead you computed kernelized Lanczos all in one fell > swoop, you'll be making k passes over the data, and computing > the kernel each time, yes, but the scaling is just that of Lanczos: > > O(k * numUsers * kernelCost(userSparsity) ) > > (as you mention, I forgot the kernel cost here, yes) > > This is tremendously better than computing the full kernel matrix, > by a factor of k / numUsers = 5000 or so. Unless I'm missing > a factor somewhere. -- Ted Dunning, CTO DeepDyve
