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Re: strange problem


From: Tomer Altman
Subject: Re: strange problem
Date: Fri, 28 Nov 2003 01:11:14 -0800 (PST)

Laurent, thanks for reminding me of a question I wanted to ask:

If Octave uses LU-decomposition for matrix multiplication, does it
also "cache/memoize" the resulting decomposition, so that further
multiplications are only O(n^2), and not O(n^3) if done from scratch?

Also, are interpreted results "cached/memoized" in general? This could
save time in referentially-transparent which are relatively expensive
to call.

Just curious,

~Tomer



On Nov 28, 2003 at 10:01am, Laurent Jacques wrote:

ljacqu >Date: Fri, 28 Nov 2003 10:01:28 +0100
ljacqu >From: Laurent Jacques <address@hidden>
ljacqu >To: address@hidden
ljacqu >Subject: Re: strange problem
ljacqu >Resent-Date: Fri, 28 Nov 2003 03:01:31 -0600
ljacqu >Resent-From: address@hidden
ljacqu >
ljacqu >On Friday 28 November 2003 00:03, Geraint Paul Bevan wrote:
ljacqu >| The number -2.0817e-17 is -0.000000000000000020817, which is very, 
very
ljacqu >| close to zero. The error is due to the way that numbers are 
represented
ljacqu >| in computers.
ljacqu >
ljacqu >Of course but I think that the question was:
ljacqu >
ljacqu >Why the top-left entry of 
ljacqu >        [1,100;0,1]*[1,0;(-1/100),1] 
ljacqu >is 
ljacqu >        -2.0817e-17
ljacqu >while octave returns the true 0 if you enter
ljacqu >        1 + (100*-1/100)
ljacqu >
ljacqu >Does Octave perform some LU decomposing of matrix to realize matrix 
ljacqu >multiplication ? 
ljacqu >This could explain the difference between the two results.
ljacqu >
ljacqu >Laurent.
ljacqu >
ljacqu >
ljacqu >
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ljacqu >
ljacqu >Octave's home on the web:  http://www.octave.org
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ljacqu >
ljacqu >



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