On Mon, 23 Nov 2009, John Peterson wrote:
> {lu,cholesky}_solve(rhs, soln);
>
> This seems fairly reasonable on the surface, but there is a gotcha in
> that both routines modify the rhs *as well as* the soln vector
> internally.
That's surprising. Do we need to do this? On first glance:
It looks like the cholesky back substitution never reuses b(i) outside
of (or at later iterations in) the i loop; surely we can replace that
with a single scalar on the stack. The LU back substitution is
trickier, since that's storing a full vector's worth of temporary
data, but do we need to store said temporary data in b? It looks like
we can probably get away with using the unfinished parts of x instead.
---
Roy
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