RE: Promotion of bare decorations & comparisons
Jurgen, please see comments below.
Nate
> Nate,
> why are yu throwing away all the nice things we have learnt in P1788
?
> see below
>
> Am 12.01.2013 16:10, schrieb Nathan T. Hayes:
> > John Pryce wrote:
> >> On 3 Jan 2013, at 22:08, Nathan T. Hayes wrote:
> >>>> You can construct an equally silly example with that alternative:
> >>>> [1,2] \subseteq floor([0,6])
> >>>> ...
> >>>> = [1,2] \subseteq Empty
> >>>> = false
> >>>>
> >>>> Now we have a false negative.
> >>>
> >>> I don't agree it is a false negative.
> >>>
> >>> Compressing the decorated interval ([0,6],def) into the bare
decoration
> > def
> >>> means the user has explicitly indicated anything less than dac is an
> > error.
> >>>
> >>> So returning false in this example is exactly what the user expects,
> > since
> >>> [1,2] cannot be a subset of any defined and continuous interval range
of
> >>> floor([0,6]).
> >>
> >> I think "Hmm" on this one. What does "any defined and continuous (dac)
> > interval range
> >> of floor([0,6])" mean? The only meaning I can see is "since floor()
isn't
> > dac on the
> >> input [0,6], such a range doesn't exist; and if we insist on treating
this
> > nonexistent
> >> thing as a set, it must be the empty set".
> >
> > That is exactly my view... so why the "Hmmm"?
> >
> >
> >>
> >> I think the above is for compressed arithmetic with threshold dac.
Change
> > the example
> >> slightly:
> >> [0,0] \subseteq floor([0,0.9])
> >> Here, floor() IS dac on [0,0.9], so the result as a decorated interval
> > [0,0]_dac, which
> >> becomes [0,0] as a compressed interval, so we get
> >> [0,0] \subseteq [0,0]
> >> = true.
> >> But now change [0,0.9] to the large interval [0,6] and according to
Nate's
> > scheme
> >> above
> >> [0,0] \subseteq floor([0,6])
> >> ...
> >> = [0,0] \subseteq Empty
> >> = false
> >> I don't think users would expect this: that (A \subseteq f(B)) is true
for
> > some B, but
> >> becomes false when they make B larger.
> >
> > ???
> That is what we call enclosure property
Of course with bare intervals it is.
However, we are in the middle of a discussion on compressed interval
arithmetic, weren't we?
In that case, when the compression threshold specified by user is DAC, as
John says:
"floor() isn't dac on the input [0,6], such a range doesn't exist;
and if we insist on treating this nonexistent thing as a set, it must be the
empty set".
This is my view exactly, as the original example for compressed interval
arithmetic that started this e-mail thread shows:
[0,0] \subseteq floor([0,6])
= [0,0] \subseteq ([0,6],DEF) // Full decorated result
= [0,0] \subseteq DEF // compress ([0,6],DEF) because it
isn't DAC
= [0,0] \subseteq Empty // promote DEF to Empty because no
DAC range exists, just as John says
= false // result of my original example
This whole discussion is getting really bizarre...