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]=3D(1-w)*v[i]+w/D[i]*(q[i]-L[i-1]*v[i-1]-U[i]*v[i+1]) err +=3D (temp[i]-v[i])**2 v[i]=3Dtemp[i] Thanks, Geoffrey Complete code here; def sor(v, L, D, U ,q, tol, w): '''solve M*v=3Dq. return v. L, D, U are the sub-diagonal, diagonal, and super-diagonal of the matrix M. ''' err=3D9999999 N=3DD.shape[0] #number of elements temp=3Dempty(N) while err> tol : err=3D0 temp[0]=3D(1-w)*v[0]+w/D[0]*(q[0]-U[0]*v[1]) err +=3D (temp[0]-v[0])**2 v[0]=3Dtemp[0] for i in range(1,N-1): temp[i]=3D(1-w)*v[i]+w/D[i]*(q[i]-L[i-1]*v[i-1]-U[i]*v[i+1]) err +=3D (temp[i]-v[i])**2 v[i]=3Dtemp[i] temp[N-1]=3D(1-w)*v[N-1]+w/D[N-1]*(q[N-1]-L[N-2]*v[N-2]) err +=3D (temp[N-1]-v[N-1])**2 v[N-1]=3Dtemp[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=3Dq. 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]=3D3D1/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 _______________________________________________________=0A= =0A= The information in this email or in any file attached=0A= hereto is intended only for the personal and confiden-=0A= tial use of the individual or entity to which it is=0A= addressed and may contain information that is propri-=0A= etary and confidential. If you are not the intended=0A= recipient of this message you are hereby notified that=0A= any review, dissemination, distribution or copying of=0A= this message is strictly prohibited. This communica-=0A= tion is for information purposes only and should not=0A= be regarded as an offer to sell or as a solicitation=0A= of an offer to buy any financial product. Email trans-=0A= mission cannot be guaranteed to be secure or error-=0A= free. P6070214=0A=