Re: Decorations Challenge
John Pryce wrote:
Nate, P1788
On 20 Oct 2010, at 07:25, Nate Hayes wrote:
John Pryce wrote:
You have been very quiet about my Decorations Challenge. Do you approve
of
Arnold Neumaier's solution? If not, how would you amend it?
... I simply considered your example verbatim without questioning
whether or not the example, as given, had an intentional bug in it or
not.
...
To the extent I am able to read your pseudo-code correctly, it appears
the
algorithm begins by constructing the decorated interval:
xx = ([-1,4],safe)
...
Since your code (taken verbatim) does not reset the decoration of xx (as
Arnold did in his example), the remaining iterations through the loop
will
never reach the msg = "success" break-point.
The problem is stated at a mathematical level, above any particular
decoration scheme. The test for termination is what a mathematician wants
to know:
"Is (yy contained in xx) and (g everywhere defined and continuous on
xx)?"
Thus stated it has a unique correct solution, "success" when count=2 and
xx=[2,1+sqrt(3)] and yy=whatever. It's no business of the mathematical
algorithm to tell the programmer how to use his/her decoration scheme in
order to correctly implement the above test.
The programmer explicitly copies the "possiblyDefined" decoration from the
yy output to the next xx input. Hence, the verbatim mathematical meaning of
the algorithm is that xx is always "possiblyDefined" on all succesive loop
iterations. The output I give is therefore the correct output of the
mathematical algorithm as it was given.
If this was not the intent -- as I've already mentioned -- then it is a bug
and you should have reset the decorations like Arnold did.
It is too much to ask for that the computer should have ESP and know what
are your private wishes and thoughts.
Even so, the important fact to notice is that the decorations did thier job:
even though it is a programming bug the decorations found the bug and
alerted the user to the problem.
The correct solution is essentially as Arnold's code produces:
On 15 Oct 2010, at 12:00, John Pryce wrote:
> AN_Solution1
info =
{
fail = 0
nf = 2
}
x =
{
inf = 2
sup = 2.7321
dec = 0
}
(The other piece of output I quoted there is the _wrong_ answer one gets,
as expected, if one doesn't reset the decoration each iteration.)
The point of the exercise is that, to get the right result using
decorations, one _must_ reset the decoration of xx at each iteration round
the loop, as Arnold did.
I have no problem with that.
As I stated in my orignal response to the challenge, I suspected the program
had a bug and should have been written this way.
You had said more than once in emails to the group (I will seek chapter
and verse if you wish) that such resetting is actually FORBIDDEN in your
scheme. Will you clarify your position on this?
I have NEVER said such a thing.
As we see from a history of the public "red and blue sticky" debates, one of
your favorite passtimes is misrepresenting my point of view. ;-)
So it is sometimes a full-time job for me to keep setting the record
straight...
In fact, if you look through some of our private e-mails from a few months
back you will see I explicitly said just the opposite: I even gave examples
how one can see how decorations can be reset by analyzing the problem in SSA
form.
Nate hayes