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Re: Comments on Motion 27-A "Decorated Intervals"



John Pryce wrote:
Jürgen, Marco, Vincent, P1788

On 15 Jul 2011, at 23:32, Vincent Lefevre wrote:
About Remark 5 (and the whole motion), what do the decorations really
mean in practice? For instance, consider f(x) = tan(floor(x)*pi+pi/2),
implemented as the composition of the various basic functions, and its
interval extension. At Level 1, it is ndf for X = [1,1.5], but not at
Level 2, where it is con (because of rounding). With X = [1,3], it is
con at both Level 1 and Level 2, though f(x) is nowhere defined. So,
what confidence con brings here?

The decoration con tells you that that the level 2 computation of the
function f(x) may have a singularity on the interval x.

I meant: what confidence con brings compared to ndf?

Since
 * one can obtain con for a function that is nowhere defined, and
 * one can obtain ndf for a function that is not nowhere defined
   (due to the tracking rules),
it seems that there isn't a difference between ndf and con.

From my reading of the text, I agree. It's not the _mechanism_ of
propagating decorations that is at fault -- that (Def 7, Def 8) is the one
I think we are all agreed on.

What's wrong is Theorem 3. If that is all one can trust to be true, it
seems to me the current draft can't handle Nate's branch and bound
application. Suppose one computes ndf. According to Theorem 3, con might
be true, or even saf!

???

If one computes ndf *by Definition 8* (which is what Theorem 3 says), then f
is certainly not defined on xx.



But Nate's B&B relies on knowing that if you compute ndf then f is
definitely nowhere defined on xx.

That's what motion 27 does.

I think you missed the point Vincent was making: sometimes in pathological
cases the decoration con might be returned when f is actually undefined over
xx. But that's not a failure (even FTDIA says so).

ndf is never returned when f is at least somehere defined on xx.

So Theorem 3 is fine.

Nate