Re: Unbounded intervals
On 2012-04-26 10:46:04 -0500, Nate Hayes wrote:
> Vincent Lefevre wrote:
> >This question is pointless. If there is an unbounded interval in
> >the problem, then you need unbounded intervals to be able to express
> >the problem!
>
> You've missed the point again: there is always some finite limit to
> which we can prove anything useful.
That's wrong. If you don't believe me, try to prove that.
Also there is a difference on the limit of what you can handle in
a given representation and what you can handle on a given machine.
For instance, some people think that it is not possible to work
with 1/3 exactly because it has an infinite number of non-zero
digits, while it is sufficient to switch to a rational arithmetic
(or a FP arithmetic in a radix divisible by 3) to represent it
exactly.
Similarly, there are no problems to work with unbounded intervals
or with some set having an infinite number of elements, as long as
the representation is suitable.
> >>At Level 1, midpoint([1,OVR]) could for the same reasons be
> >>similarly be defined (1+H_f)/2.
> >
> >This would be bad, even if you define it as H_f (a bit like in Level 2),
> >because at Level 1, you have intervals like [1,4*H_f], whose Level 1
> >midpoint is 2*H_f+1/2 > H_f.
> No. H_f is the overflow threshold, just like REALMAX at Level 2. So, for
> example
> 4*[1,OVR]=[4,OVR]
> and midpoint([4,OVR]) would then still be H_f.... just like at Level 2.
There is no overflow threshold at Level 1. See what has already been
voted on.
There would be no point to introduce an overflow threshold at Level 1.
If this is because some implementation limit, then you would need to
introduce some form of precision threshold too... and you would get
Level 2! Level 1 is the mathematical level, without taking into account
limits related to the implementation.
> >This is not a problem. You don't necessarily need a midpoint at
> >Level 1 for this algorithm.
>
> Of course not, but if I choose midpoint the algorithm is undefined at
> Level 1. That's a fact, Vincent.
You are free to choose an algorithm that doesn't mathematically
make sense, but that's entirely your problem. And that's just a
mathematical point of view. Whatever choice will be made in P1788
can't change that.
Probably the end of discussion with you for me.
--
Vincent Lefèvre <vincent@xxxxxxxxxx> - Web: <http://www.vinc17.net/>
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Work: CR INRIA - computer arithmetic / AriC project (LIP, ENS-Lyon)