Re: Ar we succeeding?
Of course, John.
But your choices are entirely arbitrary, too. And they have the serious
disadvantage of failing if the user makes the wrong arbitrary choice.
Nate
----- Original Message -----
From: "John Pryce" <j.d.pryce@xxxxxxxxxxxx>
To: "Nate Hayes" <nh@xxxxxxxxxxxxxxxxx>; "stds-1788"
<stds-1788@xxxxxxxxxxxxxxxxx>
Cc: "Vladik Kreinovich" <vladik@xxxxxxxx>
Sent: Tuesday, May 24, 2011 5:50 AM
Subject: Re: Ar we succeeding?
Nate and all
On 23 May 2011, at 15:06, Nate Hayes wrote:
Hi Vladik,
Here is an example I've posted a couple times. The result of Fig. 4 was
computed with definitions of decorations according to v3.01; and the
result of Fig. 3 was computed with the repaired definitions in the new
motion.
Nate
----- Original Message ----- From: "Kreinovich, Vladik" <vladik@xxxxxxxx>
To: "Nate Hayes" <nh@xxxxxxxxxxxxxxxxx>; "Ralph Baker Kearfott"
<rbk5287@xxxxxxxxxxxxx>
Cc: "John Pryce" <j.d.pryce@xxxxxxxxxxxx>
Sent: Sunday, May 22, 2011 7:27 PM
Subject: RE: Ar we succeeding?
At least to me, this is a completely new information. I am not sure what
kind of failures you got and why, and I did not realize that what you
propose is new definitions. I think before we start understanding nthe
solution we need to understand the problem. Can you clarify the problem?
At last I'm getting round to respond to this example. In your file
<intersection.pdf> giving the example, you have two real-variable functions
f(x), g(x) of vector x=(x0,x1) and define
S_f = {x | f(x) is defined}, S_g = {x | g(x) is defined}
and
S = (S_f intersect S_g) = {x | both are defined}.
Now you use the interval function
h(X) = f(X) intersect g(X)
in your Branch & Bound algorithm to determine S.
But this is totally arbitrary. h(X) has no intrinsic meaning. For a small
box X it will "almost always" be empty, and if you imagine a small X
"gradually increasing", it becomes nonempty at some point dependent on the
vagaries of f and g. Your algorithm succeeds using h, *only* because you
have chosen a particular way to decorate the intersection operation.
Both your scheme and the Neumaier-Pryce scheme work if instead you define
h(x) = f(x)+g(x) (1)
and use its interval extension in the B&B.
I used + in (1) because it is defined on all of R^2. Using - or * would do
equally well.
Cheers
John