Re: V02.1Decorations question

[Arnold Neumaier <Arnold.Neumaier@xxxxxxxxxxxx>]

Dominique Lohez wrote:
> Arnold Neumaier a écrit :
> > The basic question was whether floor([0.5,1.5]) should return the 
> > decoration PossiblyContinuous (my conclusion) or NotContinuous
> > (Nate's conclusion).
>
> IMHO the trouble arises from the fact that the decoration is considered 
> independently   of the value of the      interval to which it is 
> associated.
>
> In the present situation the decoration reflect a property of the 
> history of  the calculations including the application of the  the floor 
> function
>
> Given X = [0.5, 1.5]  and R= floor(X) , we have R= [0,1]
>
> The decoration of R is NotContinuous   whatever the decoration of X may be.
>
> If we consider an expression such R=floor(S) with S=f(X)
>
>      A smart   implementation of  f(X) can produce a result S strictly 
> included between 0 and 1 , we obtain  R=[0,0] .
>      The decoration of  R is  continuous provided  both  X and S bear 
> this decoration
>
>    A less smart implementation of f(X) could produce R =[0,1] or even 
> R=[-1, 1].
> In these  cases,  the decoration of R is necessarily NotContinuous .
> This decoration does not depend on the decorations of X and  S.

No.

We do not know how X was produced. It could have come from a prior
computation yielding overestimation. For example, the arithmetic must 
produce a correct result for fcn([0,1]) given the function defined by
the Matlab program
   R=fcn(z);
   X=0.5+(z-z)^2;
   R=floor(X);
   end;
The function fcn(z) is continuous for z in [0,1].

Your and Nate's recipe produce the decoration NotContinuous, which is
wrong and misleading, while my recipe produces the decoration 
PossiblyContinuous, which is valid though not very informative.


Arnold Neumaier