Re: Decorations Challenge
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 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.
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?
Regards
John Pryce