Thanks for clarifying this - it seems that timesSquared job is more efficient than I thought.
Regarding the Ritz transformation - it depends on the sparsity of A and the number of iterations whether the final product for computing U should be done using your technique of U ~= A V D^-1 or not. On Mon, Dec 12, 2011 at 8:08 PM, Jake Mannix <[email protected]> wrote: > For reference, look in DistributedRowMatrix#timesSquared(Vector), which is > contained on lines 227-247. JobClient.runJob() is called only one time, > running > the TimesSquaredJob (a single map-reduce job).' > > -jake > > On Mon, Dec 12, 2011 at 9:58 AM, Danny Bickson <[email protected] > >wrote: > > > K passes over the data - where in each pass you multiply once by A and > once > > by A' I call 2K passes over the data. > > > > On Mon, Dec 12, 2011 at 7:48 PM, Jake Mannix <[email protected]> > > wrote: > > > > > On Mon, Dec 12, 2011 at 9:10 AM, Danny Bickson < > [email protected] > > > >wrote: > > > > > > > I meant to write: twice in case of a rectangular matrix. > > > > By the way, if you want to have the two sides matrices [U,D,V]=svd(A) > > > > You will need to run Lanczos twice: once with A and another time with > > A'. > > > > So run time should be doubled. > > > > > > > > > > Neither of these statements are actually correct: for a rectangular > > matrix, > > > if you want the > > > top K singular vectors and values, you will make K passes over the data > > > (not 2K) each > > > one being an operation of (A'A)*v without ever computing A'A itself. > > This > > > operation > > > "timesSquared(Vector)", for a matrix with row i having d_i nonzero > > entries, > > > will scale like > > > sum_i(d_i^2), but still only one pass over the data. > > > > > > Also, once you have run Lanczos over A, and gotten the matrix V out, > you > > > can recover > > > U in O(1) map-reduce operations, by use of the identity: U = A * V * > D^-1 > > > > > > -jake > > > > > > > > > > > > > > On Mon, Dec 12, 2011 at 7:08 PM, Danny Bickson < > > [email protected] > > > > >wrote: > > > > > > > > > In each Lanczos iteration you multiply by the matrix A (in case of > a > > > > > square matrix) > > > > > or twice, by the matrix A' and A. Multiplication is linear in the > > > number > > > > > of non zero edges. > > > > > See http://en.wikipedia.org/wiki/Lanczos_algorithm > > > > > Finally a decomposition of a tridiagonal matrix T for extracting > the > > > > > eigenvalues. > > > > > I think it is also linear in the number of iterations (since the > size > > > of > > > > T > > > > > is number of iterations+1). Note that this code is not distributed > > > since > > > > it > > > > > can be efficiently done on a single node. > > > > > The last step is the Ritz transformation - a product of the > > > intermediate > > > > > vectors v > > > > > with the eigenvectors. This step may be heavy since those matrices > > are > > > > > typically dense. > > > > > > > > > > Best, > > > > > > > > > > DB > > > > > > > > > > > > > > > 2011/12/12 Fernando Fernández < > [email protected] > > > > > > > > > > > > >> Hi all, > > > > >> > > > > >> This is a question for everybody, though it may be better answered > > by > > > > Jake > > > > >> Mannix. Do you guys know what is the complexity of the algorithm > > > > >> implemented in mahout for Lancos SVD? Linear, quadratic, etc.. > > > > >> > > > > >> > > > > >> Thanks in advance!! > > > > >> Fernando. > > > > >> > > > > > > > > > > > > > > > > > > > >
