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Re: Uniform partition of an interval
From: |
Dirk Laurie |
Subject: |
Re: Uniform partition of an interval |
Date: |
Wed, 31 May 2000 12:07:57 +0200 |
J.C. Gonzalez skryf:
> Dirk Laurie wrote:
> > and people will write things like 'y=1.8:0.05:1.9'. We should agree what
> > that should do. Intuitively one feels that a:h:b with h>0 should be
> > equivalent to:
> > y=[]; x=a;
> > while x<=b, y=[y x]; x += h; end
> > And indeed, if I run the above in Octave on my i686 machine, I get
> > [1.8000 1.8500]. Yet it is unsatisfactory, because with pencil and
> > paper, or on a decimal machine, or on some binary machines, I would have
> > got [1.8000 1.8500 1.9000].
> >
> > One can get round the problem by saying it should be equivalent to:
> > r=(b-a)/h; n=round(r);
> > if h*abs(n-r)>max(a,b)*eps, n=floor(r); end
> > y=a+h*(0:n);
> >
> > But doing so would treat one case of a pervasive problem: the
> > non-intuitiveness of floating-point comparison. A good cure should work
> > in other places too.
> >
> > I think Octave should borrow an idea from the grandfather of interactive
> > matrix languages, namely APL. This language has a built-in variable which
> > in Octave we would call 'comparison_tolerance'. Then we could write:
> >
>
> I agree, but ... shouldn't we use better "linspace" ?
My point is *not*
"what is the best way to make 1.8:0.05:1.9 deliver [1.8 1.85 1.9]"
My point is: Octave should have a technique that allows all floating-point
tests to be tolerant. Then the previous question does not even arise.
Dirk
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