[Numpy-discussion] Numpy and iterative procedures

[Nadav Horesh]

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 at 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 at scipy.org
[mailto:numpy-discussion-bounces at 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 at 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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