Re: Noncontinuous examples
John,
Sorry for typos.
Set-based flavour defines that bare evaluation of "case(c,g,h)" must return the rusult that
is sometimes wider than natural interval extension.
It seems to me that we nead also to change decoration propagation rules for this operation.
Let c=[1,2]_com, g=[]_emp, h=[3,4]_com.
Natural bare interval extension of case(c,g,h) is [].
Natural decoration propagation is min(com,com,emp,com)=emp.
So natural decorated interval extension is []_emp.
However, set-based flaveor demands to return bare interval [3,4].
The combination [3,4]_emp is allowed, but it doesn't sound good.
The intention of modified definition of bare result was to evaluate g and h arguments by name instead of by value.
If we want to obtain decoration emp in this example, we still need evaluation by value.
I suggest to accept that arguments of all arithmetic operations are evaluated by value,
and not to consider case(c,g,h) as arithmetic operation.
I see two options:
A) case(c,g,h) returns bare interval as well as intersection and convexHull;
B) case(c,g,h) still returns decorated interval with special propagation rules.
The computation of interval part if case(c,g,h) is already defined in set-based flavour.
I sugget the following rules for decoration propagation:
if (c in empty) {
return dc;
} else if (inf(c) >= 0) {
return min((c and h are bounded nonempty && inf(c) > 0 ? COM: DAC), dc, dh);
} else if (sup(c) < 0) {
return min((c and g are bounded nonempty ? COM: DAC), dc, dg);
} else {
return min(DEF, dc, dg, dh);
}
This rules support evaluation by name of arguments g and h.
By the way, the option (B) restricts common instances of case(c,g,h) to
i) (c,g,h; h) where c,g,h - are common and inf(c) > 0
ii) (c,g,h; g) where c,g,h - are common and sup(c) < 0
-Dima
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Отправленные: Пятница, 28 Декабрь 2012 г 22:12:35 GMT +04:00 Абу-Даби, Маскат
Тема: Re: Noncontinuous examples
Dmitry
On 28 Dec 2012, at 13:21, Dmitry Nadezhin wrote:
> I want to confirm if I understand "common evaluation" correctly.
> So I describe two examples. Please say me if I misunderstand.
>
> The definition in 5.2 says:
> "The common operation instances are those (x1 , x2 , . . . , xk ; y) such that the point
> function phi is everywhere defined and continuous on the closed, bounded, nonempty box
> (x1 , x2 , . . . , xk ), and y equals the range of phi over this box."
> It says "phi is continous" not "restriction of phi is continous".
> I guess that this is reserved for possible cset flavour.
You make a very good point. I thought about this distinction when first designing the flavor concept but forgot about it when writing the text. Yes, your interpretation is the one I support, mainly to allow a cset flavor as you say. To make it clearer I suggest
> phi is everywhere defined and continuous on
should change to
> phi is defined and continuous at each point of
> I try to interpret this definition for floor(x) and case(c,g,h).
<<Typo: you wrote "float">>
> 1) The only common evaluations of floor(x) are
> ([a,b] ; [k,k] ) where k < a <= b < k + 1 , k \in Z .
> Pair ([0,0]; [0,0]), is not common evaluation because floor(x) is discontinous at 0 .
Yes
> 2) The only common evaluations of case(c,g,h) are
> i) ([a,b],[c,d],[e,f]; [c,d] ) where 0 < a <= b, c <= d, e <= f
> ii) ([a,b],[c,d],[e,f]; [e,f]) where a <= b < 0, c <= d, e <= f
Shouldn't the result be [e,f] in (i) and [c,d] in (ii), instead of what you have? Because case(c,g,h) is g if c<0 and h otherwise.
> iii) ([a,b],[c,c],[c,c]; [c,c]) where a <= b
> Pair ([0,1],[2,2],[2,3]; [2,2]) is not common evaluation because case(c,g,h) is discontinous at (0,2,3) .
([0,1],[2,2],[2,3]; [2,3]) I think, see above.
But yes, it seems the points (c,g,h) where case() is continuous are precisely the set
{c \ne 0; g,h arbitrary} union {c = 0; g = h}.
And the boxes contained in the set are just those specified by your (i), (ii), (iii) -- corrected as above.
> Is definition of common evaluation for case(c,g,h) is different from other point functions ?
The standard definition, applied to this function, does give a rather strange result. What do people think? Is there a justification for an alternative definition?
John Pryce