Re: A few examples
Dan Zuras wrote:
But ignoring that, if I throw in a few examples of my own
together with your case analysis with xx = {[0,4],D3} we
get (please check my work):
D4 ==> empty & result = empty:
sqrt(xx - 5) + sqrt(xx - 5) = {empty,D4}
{empty,D0} + {empty,D0} = {empty,D4}
By Definition 3, the tracking decoration of the addition should be:
(Empty,D0) + (Empty,D0)
= (Empty+Empty,inf(S(+,Empty,Empty),D0,D0))
= (Empty,inf(D4,D0,D0))
= (Empty,D0) // this does not agree with your answer
D4 ==> empty & result = nonempty:
f(xx) == xx \union {[5,6],doesItMatter?}
==> f(xx) = {[0,6],D4}
By Def. 3, the tracking decoration should be
([0,4],D3) \union ([5,6],??)
= ([0,4] \union [5,6],inf(S(\union,[0,4],[5,6]),D3,??)
= ([0,6],inf(D3,D3,??)) // the answer depends on ??
D3 ==> inDomain&Continuous & result = empty:
xx \intersect (xx + 10) = {empty,D3}
By Def. 3:
([0,4],D3) \intersect ([10,14],D3)
= ([0,4] \intersect [10,14], inf(S(\intersect,[0,4],[10,14]),D3,D3)
= (Empty,inf(D3,D3,D3))
= (Empty,D3) // agrees with your answer
D3 ==> inDomain&Continuous & result = nonempty:
xx \intersect (xx + 1) = {[1,4],D3}
yes
xx + xx = {[0,8],D3}
yes
sqrt(xx - 3) + (xx + 10) = {[10,15],D3}
{[0,1],D1} + {[10,14],D3} = {[10,15],D3}
sqrt(([-3,1],D3)) + ([10,14],D3)
= ([0,1],D1)+([10,14],D3)
= ([0,1]+[10,14],inf(S(+,[0,1],[10,14]),D1,D3))
= ([10,15],inf(D3,D1,D3))
= ([10,15],D1) // does not agree with your result
floor(xx) + (xx + 10) = {[10,18],D3}
{[0,4],D2} + {[10,14],D3} = {[10,18],D3}
floor(xx) + (xx+10)
= ([0,4],D2)+([10,14],D3)
= ([0,4]+[10,14],inf(S(+,[0,4],[10,14]),D2,D3))
= ([10,18],inf(D3,D2,D3))
= ([10,18],D2) // does not agree with your result
...snip...
Do I have all these correct?
No.
. . .
> Well, as I recall only a partial ordering was necessary to
> prove FTDIA. It was:
>
> empty < inDomain&Continuous < inDomain < in&Out
> empty < outDomain < in&Out
>
> Where, as you pointed out, 'ill' need only be placed at
> some level below level 1.
>
> So anything that fits this description would do the job.
Agreed.
IMHO, Dan, it seems you understand pretty well.
Nate
Well, it *appeared* you were understanding, but now I see I was wrong, and
that you've completely missed the point... :-(
Then, in keeping with your assertion that this is a
level 1 only motion, let me suggest an amendment.
...
So the amendment is: Eliminate discussion of implementation
details such as bits & numbering of decorations; rename the
decorations as John once did; & make an explicit statement
about the partial ordering required.
That way this motion gives us everything we need for an
FTDIA at level 1 with as little controversy as possible.
Will you accept this amendment as friendly or should I ask
for a second?
Dan, what you are proposing is to eliminate the whole purpose of the motion:
property tracking. Clearly when you don't follow the Definition 3, you get
lots of wrong answers that will quickly cause most interval algorithms to
fail.
So I will certainly not accept this as a friendly amendment!!
Nate