HiI saw some distributed matrix functions included in Samsara now. Wondering if
we have a plan to support matrix inversion ?BTW, am I correct that it is
distributed memory based, not out-of-core ? thanks, canal
I doubt seriously that Samsara will support matrix inversion per se. The
problem is
a) it densifies sparse matrices
b) it is much more costly than solving a linear system
Samsara is roughly memory based, but different back-ends will try to spill
to disk if necessary. It is likely that the resul
oh, it is so unfortunate that the first step of my project requires the
inversion of a very large matrix. will have to revert back to scalapack or MR
based solutions I guess.
thanks, canal
On Saturday, October 3, 2015 11:31 PM, Ted Dunning
wrote:
I doubt seriously that Samsara wi
Can you explain why you feel you must invert a very large matrix. This
can be a Bad Idea.
On 10/03/2015 08:09 PM, go canal wrote:
> oh, it is so unfortunate that the first step of my project requires
> the inversion of a very large matrix. will have to revert back to
> scalapack or MR based solut
If there's a need for inversion that's good info; would love to know the
purpose to get a sense of how people want to use the product.
On Saturday, October 3, 2015, Allen McIntosh
wrote:
> Can you explain why you feel you must invert a very large matrix. This
> can be a Bad Idea.
>
> On 10/03/2
It is almost certain that starting with an inversion is a serious error.
Are you sure you don't want a matrix solver instead?
Sent from my iPhone
> On Oct 3, 2015, at 20:09, go canal wrote:
>
> oh, it is so unfortunate that the first step of my project requires the
> inversion of a very lar
This should be done with a matrix solver indeed!!!
On Oct 4, 2015 11:53 AM, "Ted Dunning" wrote:
>
>
> It is almost certain that starting with an inversion is a serious error.
>
> Are you sure you don't want a matrix solver instead?
>
> Sent from my iPhone
>
> > On Oct 3, 2015, at 20:09, go can
Thank you all, the solver is something like this, am I correct:
Matrix m =
Matrix inverse = new QRDecomposition(m).solve(new DiagonalMatrix(1,
m.rowSize()));
The problem I have is that the matrix is too big, I need distributed, or
out-of-core solution.
thanks, canal
On Monday, Oct
1) Is m sparse?
2) Once you have computed "inverse", what are you going to do with it?
On 10/04/2015 10:31 PM, go canal wrote:
> Thank you all, the solver is something like this, am I correct:
> Matrix m =
> Matrix inverse = new QRDecomposition(m).solve(new DiagonalMatrix(1,
> m.rowSize()));
both sparse and dense.
there are a few steps for the whole calculation:. inverse of M. get degree of
freedom. multiplication. addition
in fact i need to support both double and complex double for either distributed
memory based or out-of-core.
I found one MR based solution for large matrix inver
This sort of thing would definitely compute the inverse.
And it is definitely to be avoided.
How about you give some specifics so I can say what should be done?
On Sun, Oct 4, 2015 at 7:31 PM, go canal wrote:
> Thank you all, the solver is something like this, am I correct:
> Matrix m =
On Sun, Oct 4, 2015 at 10:32 PM, go canal wrote:
> in fact i need to support both double and complex double for either
> distributed memory based or out-of-core.
Ahh...
Well Mahout doesn't support complex anything. So this isn't going to help
you.
many thanks, here is the details:
given square matrix Z, . get Z inverse Zi. get degree of freedome of Zi : U.
finally, calculate final which is: A + B * U * C
thanks, canal
On Monday, October 5, 2015 2:00 PM, Ted Dunning
wrote:
This sort of thing would definitely compute the i
I will be more than interested to extend to complex double, when the solver is
ready for double data type. thanks, canal
On Monday, October 5, 2015 2:02 PM, Ted Dunning
wrote:
On Sun, Oct 4, 2015 at 10:32 PM, go canal wrote:
> in fact i need to support both double and complex do
That isn't enough detail.
How do you mean to compute degrees of freedom? WHy do you need the inverse
to do this?
Where did you get this algorithm?
Is this even appropriate at large scale?
Is this a stable computation?
On Sun, Oct 4, 2015 at 11:18 PM, go canal wrote:
> I will be more than
Unfortunately I do not know much details of these. The steps of these
calculation is passed to me from a research team. I am helping them with coding
part only. I myself is not good at math :-(
btw, I think Mahout supports out-of-core SVD, am I correct ? If so, I can get
inverse of matrix from
Yes. You can get the inverse from an SVD or emulate its effect.
Can you share the actual mathematical specification for your problem?
If you can't, then there is little we can do to help.
On Wed, Oct 7, 2015 at 11:35 PM, go canal wrote:
> Unfortunately I do not know much details of these. Th
Yeah, nice trick Ted; here's a how-to for the list:
http://www.cse.unr.edu/~bebis/CS791E/Notes/SVD.pdf
On Thu, Oct 8, 2015 at 2:31 PM, Ted Dunning wrote:
> Yes. You can get the inverse from an SVD or emulate its effect.
>
> Can you share the actual mathematical specification for your problem?
>
Mahout translation (approximation, since ssvd is reduced-rank, not the true
thing):
val (drmU, drmV, s) = dssvd(drmA, k = 100)
val drmInvA = drmV %*% diagv(1 /=: s) %*% drmU.t
Still, technically, it is a right inverse as in reality m is rarely the
same as n. Also, k must be k<= drmA.nrow min drmA
Go, if you'd like to put your research team in touch with the list we may
be able to help work through a good approach; let us know.
On Thu, Oct 8, 2015 at 3:58 PM, Dmitriy Lyubimov wrote:
> Mahout translation (approximation, since ssvd is reduced-rank, not the true
> thing):
>
> val (drmU, drmV
or pseudoinverse really, i guess
On Thu, Oct 8, 2015 at 3:58 PM, Dmitriy Lyubimov wrote:
> Mahout translation (approximation, since ssvd is reduced-rank, not the
> true thing):
>
> val (drmU, drmV, s) = dssvd(drmA, k = 100)
> val drmInvA = drmV %*% diagv(1 /=: s) %*% drmU.t
>
> Still, technicall
Totally an approximation; depends on why people are asking for the inverse
and whether it'd do.
On Thu, Oct 8, 2015 at 4:20 PM, Dmitriy Lyubimov wrote:
> or pseudoinverse really, i guess
>
> On Thu, Oct 8, 2015 at 3:58 PM, Dmitriy Lyubimov
> wrote:
>
> > Mahout translation (approximation, since
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