Re: Motion 31 draft text V04.4, extra notes
On 2012-04-11 08:46:42 -0500, Nate Hayes wrote:
> Vincent Lefevre wrote:
> >On 2012-04-05 15:35:18 -0500, Nate Hayes wrote:
> >>Vincent Lefevere wrote:
> >>>I agree. But there will be no "overflown" intervals at Level 2.
> >>>AFAIK, the Level 2 choice for the midpoint on unbounded intervals
> >>>has been done for practical reasons, not because of some notion
> >>>of "overflown" intervals (if this is what you meant).
> >>
> >>Nope. That's not what I meant.
> >>
> >>What I meant is that at Level 1 there is no definition of midpoint for
> >>unbounded intervals. So why include unbounded intervals in the Level 1
> >>model?
> >
> >because the midpoint is not the only function. For instance:
> >
> > 1 / [0,1] = [1,+oo]
> >
> >at Level 1 and Level 2 (+ some decorations).
>
> We don't actually perform arithmetic in a computer at Level 1, and at
> Level 2 with overflow one has
> 1 / [0,1] = [1,+OVR]
This is wrong. At Level 1, 1 / [0,1] is defined as the smallest
(closed) interval that contains { 1 / x | x in [0,1] and x <> 0 },
and this is [1,+oo]. At Level 2, the interval must contain [1,+oo]
entirely. If [1,+OVR] means some interval [1,K] with K finite (but
we don't which one), the result is incorrect, because for any K,
[1,K] does not contain [1,+oo].
> >>Especially when a similar treatment at Level 2 of "overflown"
> >>intervals can provide the same practical benefits? IMO the Level 1
> >>model is then cleaner and simpler.
> >
> >I disagree: without unbounded intervals, you wouldn't have a closed
> >arithmetic
>
> It seems academic to me: we don't perform arithmetic in a computer at
> Level 1, (and the overflown arithmetic is closed at Level 2, btw).
But from the Level 2 results, one can deduce Level 1 results. This
is what matters.
> When restricted to bounded intervals, the Level 1 arithmetic is closed and
> cancellative for addition, subtraction, multiplication and division with 0
> not in the denominator.
This is not the definition of a closed arithmetic. If you can get
an interval as a result, say [0,1], you mustn't remove it from the
possible inputs of an operation.
> This is the oldest interval arithmetic of Ramon Moore, etc. and has
> withstood the test of time already. IMO this is better than the
> current P1788 model with unbounded intervals.
P1788 needs a closed arithmetic.
--
Vincent Lefèvre <vincent@xxxxxxxxxx> - Web: <http://www.vinc17.net/>
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Work: CR INRIA - computer arithmetic / AriC project (LIP, ENS-Lyon)