[Numpy-discussion] (no subject)
gzhu@peak...
gzhu@peak...
Fri Feb 16 10:34:50 CST 2007
Hi Nadav,
The code is attached at the end. There is probably still bugs in there
but it does not prevent me from showing the difficulity.
If you look at the inner loop below, you will see that vector v is
updated element by element. The new value of v[i] depends on the new
value of v[i-1] and the old value of v[i+1]. Updating an element
involves the new values of the already updated elements and the old
values of the rest of the elements that we have yet to update. This
makes vectorization difficult.
for i in range(1,N-1):
temp[i]=(1-w)*v[i]+w/D[i]*(q[i]-L[i-1]*v[i-1]-U[i]*v[i+1])
err += (temp[i]-v[i])**2
v[i]=temp[i]
Thanks,
Geoffrey
Complete code here;
def sor(v, L, D, U ,q, tol, w):
'''solve M*v=q. return v.
L, D, U are the sub-diagonal, diagonal, and super-diagonal of the
matrix M.
'''
err=9999999
N=D.shape[0] #number of elements
temp=empty(N)
while err> tol :
err=0
temp[0]=(1-w)*v[0]+w/D[0]*(q[0]-U[0]*v[1])
err += (temp[0]-v[0])**2
v[0]=temp[0]
for i in range(1,N-1):
temp[i]=(1-w)*v[i]+w/D[i]*(q[i]-L[i-1]*v[i-1]-U[i]*v[i+1])
err += (temp[i]-v[i])**2
v[i]=temp[i]
temp[N-1]=(1-w)*v[N-1]+w/D[N-1]*(q[N-1]-L[N-2]*v[N-2])
err += (temp[N-1]-v[N-1])**2
v[N-1]=temp[N-1]
return v
-----Original Message-----
From: numpy-discussion-bounces@scipy.org
[mailto:numpy-discussion-bounces@scipy.org] On Behalf Of Nadav Horesh
Sent: Friday, February 16, 2007 8:52 AM
To: Discussion of Numerical Python
Subject: RE: [Numpy-discussion] Numpy and iterative procedures
At first glance it doesn't look hard to, at least, avoid looping over i,
by replacing [i] by [:-2], [i+1] by [1:-1] and [i+2] by [2:]. But I
might be wrong. Can you submit the piece of code with at least the most
internal loop?
Nadav.
-----Original Message-----
From: numpy-discussion-bounces@scipy.org on behalf of Geoffrey Zhu
Sent: Thu 15-Feb-07 18:32
To: Discussion of Numerical Python
Cc:
Subject: Re: [Numpy-discussion] Numpy and iterative procedures
Thanks Chuck.
I am trying to use Successive Over-relaxation to solve linear equations
defined by M*v=q.
There are several goals:
1. Eventually (in production) I need it to be fast.
2. I am playing with the guts of the algorithm for now, to see how it
works. that means i need some control for now.
3. Even in production, there is a chance i'd like to have the ability to
tinker with the algorithm.
_____
From: numpy-discussion-bounces@scipy.org
[mailto:numpy-discussion-bounces@scipy.org] On Behalf Of Charles R
Harris
Sent: Thursday, February 15, 2007 10:11 AM
To: Discussion of Numerical Python
Subject: Re: [Numpy-discussion] Numpy and iterative procedures
On 2/15/07, Geoffrey Zhu <gzhu@peak6.com> wrote:
Hi,
I am new to numpy. I'd like to know if it is possible to code efficient
iterative procedures with numpy.
Specifically, I have the following problem.
M is an N*N matrix. Q is a N*1 vector. V is an N*1 vector I am trying to
find iteratively from the initial value V_0. The procedure is simply to
calculate
V_{n+1}[i]=3D1/M[I,i]*(q[i]-
(M[i,1]*v_{n+1}[1]+M[I,2]*v_{n+1}[2]+..+M[i,i-1]*v_{n+1}[i-1]) -
(M[I,i+1]*v_{n}[i+1]+M[I,i+2]*v_{n}[i+2]+..+M[I,N]*v_{n}[N]))
I do not see that this is something that can esaily be vectorized, is
it?
I think it would be better if you stated what the actual problem is. Is
it a differential equation, for instance. That way we can determine what
the problem class is and what algorithms are available to solve it.
Chuck
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