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Re: Example...

Dan Zuras wrote:

From: "Nate Hayes" <nh@xxxxxxxxxxxxxxxxx>
To: "Dan Zuras Intervals" <intervals08@xxxxxxxxxxxxxx>
Subject: Example...
Date: Wed, 11 May 2011 14:40:27 -0500

> I'm not sure if this suggests that the decoration should
> be the best (safest) of the input decorations as Nate &
> John have discussed.  But it is a reasonable suggestion.
>
> I would certainly agree in the case of union (something
> that seeks the truth about EITHER of its operands) but
> I'm a bit more iffy about intersection.
>
> Let me ask of the assembled lurkers if there is some
> optimization problem that seeks to find a point (or
> interval) where two distinct things are true.  Something
> for which the natural interval extension might be
> thing1 \intersect thing2.  Something more revealing
> than the Newton's example.
>
> Can anyone present us with such a problem to serve as
> a guide in answering the decoration question?
>

Hi Dan,
FWIW, here is an example about the intersection operation, posted originally 
to P1788 on 1/7/2011.
Union may need a little more thought, but it seems clear to me that for min, 
max and intersection the worst decoration is always required.

Nate



An excellent example Nate.
And your figure 4 suggests we need to be conservative
on our assignment of decorations for intersection to
prevent false positives.
I'm not sure what needs to be done to prevent false
negatives.
To eliminate the false negatives requires that an empty result can sometimes have a "good" decoration, depending on how it was produced. For example, 
   ([1,2],D3) \intersect ([3,4],D3)
should be
   (Empty,D3).
Note the result is empty, but the decoration is "everywhere defined" because both operands were also "everywhere defined". In this regard, min, max and intersection behave much like normal arithmetic operations. That is how Fig. 3 was computed.
This in contrast to v3.01 which requires an empty result must always bear the decoration "emp" (or worse), regardless of the decorations on the operands, e.g., 
   ([1,2],def) \intersect ([3,4],def) = (Empty,emp).
That is how Fig. 4 was computed.

Nate