Re: FTDIA revisited
Nate & P1788
Sorry to be late replying to this, but I'm visiting Ned Nedialkov at McMaster. He is keeping me very busy on other research topics, which has put me behind. That, and a trip to Toronto yesterday. I saw Ken Jackson at U of T, who sends his best wishes to P1788.
On 2 Aug 2011, at 11:03, Nate Hayes wrote:
> BTW:
>
> I don't object to the idea of FTDIA, but I am rather convinced the "containment order" formulation in Motion 26 is not valid. For example, def is set of all (f,xx) pairs with nonempty xx and ein is set of all (f,xx) pairs with empty xx. But ein is a subset of def??
I think that was an error once, but corrected some time ago, certainly in the current version (I hope), which says
> p_ein(f,xx) : xx is empty (i.e., has some empty component);
> p_bnd(f,xx) : xx is a subset of Domain f, and the restriction of f
> to xx is continuous and bounded;
> p_dac(f,xx) : xx is a subset of Domain f, and the restriction of f
> to xx is continuous;
> p_def(f,xx) : xx is a subset of Domain f;
> p_con(f,xx) : always true;
> p_emp(f,xx) : xx is nonempty and disjoint from Domain f;
> p_ill(f,xx) : xx is nonempty and Domain f is empty.
Note, in Motion 27 Definition 1, you have chosen to make the different p_d predicates of its scheme *logically disjoint* (exclusive), while ours get logically narrower as one goes away from con. So
your "def" is our "def and not dac"
your "con" is our "con and not def and not emp"
and so on.
As you raise technicalities ... I raise a corresponding issue about Motion 27. Theorem 3 is FALSE as regards ndf, and I'm still waiting for your revision that will correct it.
Suppose you do an evaluation as described in Motion 27, with the 4-decoration system.
- If you get the result saf, then certainly Theorem 3's
"the decoration" (which I take to mean the unique d in
{saf,def,con,ndf} such that p_d(f,xx) is true) equals saf.
- If you get def then "the decoration" can be either def or saf.
- But if you get con? Ah, there's the rub of Motion 27.
Theorem 3's "lower bound" implies "the decoration" can be
any of con, def or saf.
NO! It could be ndf, as numerous worked examples have illustrated.
- And if you get ndf? Theorem 3 implies "the decoration" can be
any of ndf, con, def or saf.
NO! again. If you get ndf, then the result is CERTAINLY ndf.
Nate knows this perfectly well, and knows his B&B algorithm
would not work if this were not the case.
So, Motion 26 guys, get your act together and produce a Theorem 3 that is both understandable and true.
John P