Re: [ublas] Solving linear equations in a loop (Optimization issues)

From: Sourabh (sourabh_at_[hidden])
Date: 2007-03-05 03:35:16

I am sorry for this mail. I did not check it properly.

On Mon, 5 Mar 2007, Sourabh wrote:

> But, when I use inplace_solve of trisolve.hpp, 
> 
> slopes = omega; // omega is a vector
> inplace_solve(lembda, slopes, unit_lower_tag ());
> 
> it does not give me the same result as:
> 
> lembda1 = I - lembda;
> slopes = solve (lembda1, omega, lower_tag);
> 
> 
> 
> On Sat, 3 Mar 2007, Nico Galoppo wrote:
> 
> > Hi Sourabh,
> > 
> > I believe Gunther is trying to say that using unit_lower_tag() makes the 
> > following call solve the system (I+A)Y=B
> > 
> > solve(A, B, unit_lower_tag());
> > 
> > So, for your case, if you fill A with -lembda, you will effectively solve 
> > (I-lembda)Y=B
> > 
> > minuslembda = -lembda;
> > for (;;)
> >    solve(minuslembda, omega, unit_lower_tag());
> > 
> > It will be equivalent to the statement you gave below.
> > 
> > --nico
> > 
> > Sourabh wrote:
> > > Do you mean that I fill lembda with negative values and use this 
> > > statement:
> > > 
> > > solve (lembda, omega, unit_lower_tag ()) ?
> > > 
> > > will it solve the equation treating lemda as (I+lembda)
> > > Will it be equivalent to the following statement ?
> > > 
> > > lembda = I - lembda
> > > solve (lembda, omega, lower_tag ())
> > > 
> > > 
> > 
> > 
> 
> 

-- 
-- Sourabh