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Re: Motion 31 draft text V04.4, extra notes

Vincent Lefevre wrote:

On 2012-04-11 10:03:43 -0500, Nate Hayes wrote:

Vincent Lefevre wrote:
>On 2012-04-11 08:46:42 -0500, Nate Hayes wrote:

[...]

>>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].

In my recent e-mails I've said that the interval [1,+OVR] should be
considered as a family of intervals: the number of elements in the
family is inifinite, but each element is closed and bounded.


So, basically, you get something that is equivalent to an unbounded
interval, but with a more complex definition!


Exactly. The reasons for doing it this way are:
   -- unbounded intervals are then completely unnecessary at any level
   -- the absence of unbounded intervals at Level 1 means A = B
when A + X = B + X
   -- valid ranges for things like 1/[0,1]=[1,+OVR] can still be computed
at Level 2 (which is where for a computational standard like P1788 it
matters in the real world).




>>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.

At Level 2 its not.


Do you mean that you have an operation at Level 2 with no
corresponding operation at Level 1? This doesn't make sense.

Sure it does.
It all depends on the underlying axioms and definitions.




>>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.

The practice of interval arithmetic has thrived for over 50 years based
on
Moore's Level 1 model. Where is the evidence to support your claim to the
contrary?

BTW, where is your answer to my original question about a concrete
example?


I've given an example in another mail.

Where?




All your points seem really academic to me.


No, really, I need to compute ranges of functions (that can be defined
by arbitrary expressions), and I need a correct answer.

So do I. I can compute them at Level 2 with 1/[0,1]=[1,+OVR].
I've never written a computer program that operates at Level 1. That's
nonsense.




The arithmetic can be closed at Level 2, anyhow, with overflown
intervals.


My point is that it must be closed at Level 1 too.

Why? I haven't seen you give any compelling reason for this.




For me, it is more important that the Level 1 arithmetic is cancellative,


It is cancellative on all the instances allowed by your model.
So, you do not miss anything.

So, it sounds to me you agree then that a Level 1 arithmetic that allows
unbounded intervals is not cancellative.




and the current P1788 model with unbounded intervals does not have this
property.


Whatever model you choose, there are properties that will not be
satisfied.

At Level 1, I see it is a choice betwee closure and cancellation.
IMO and experience, cancellation is the much more important property.




Unbounded intervals are unnecessary.


for you.


I'm still waiting for you or anyone else to show why they are *necessary*.
So far I don't seen you've accomplished that.

Nate