Re: A few examples
> From: "Nate Hayes" <nh@xxxxxxxxxxxxxxxxx>
> To: "Dan Zuras Intervals" <intervals08@xxxxxxxxxxxxxx>
> Subject: Re: A few examples
> Date: Mon, 30 May 2011 13:20:10 -0500
>
> Dan Zuras wrote:
> >> From: "Nate Hayes" <nh@xxxxxxxxxxxxxxxxx>
> >> To: "P-1788" <stds-1788@xxxxxxxxxxxxxxxxx>
> >> Subject: A few examples
> >> Date: Mon, 30 May 2011 10:06:21 -0500
> >>
> >> Dan Zuras wrote:
> >> > I would still like to see the table in section 3 filled
> >> > out with examples. As it is, there is much that is
> >> > non-obvious if not incorrect.
> >>
> >> . . .
> >
> > All good examples.
> > And for the remaining 5 cases?
> > Would decrementing '10' to, say, '1' or '0.5' mostly do
> > the job?
>
> Yup.
OK, I like things out in the open where we all can judge
them. So let me try:
D4 ==> empty
D3 ==> inDomain&Continuous
D2 ==> inDomain
D1 ==> inDomain&OutDomain
D0 ==> OutDomain
Off the top of my head I think we have to swap the last
two to maintain the correct partial order.
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}
(Because all of the add's operands were empty.)
D4 ==> empty & result = nonempty:
f(xx) == xx \union {[5,6],doesItMatter?}
==> f(xx) = {[0,6],D4}
(Because all of F()'s operands were empty.)
D3 ==> inDomain&Continuous & result = empty:
xx \intersect (xx + 10) = {empty,D3}
(Because both operands were D3?)
D3 ==> inDomain&Continuous & result = nonempty:
xx \intersect (xx + 1) = {[1,4],D3}
xx + xx = {[0,8],D3}
sqrt(xx - 3) + (xx + 10) = {[10,15],D3}
{[0,1],D1} + {[10,14],D3} = {[10,15],D3}
floor(xx) + (xx + 10) = {[10,18],D3}
{[0,4],D2} + {[10,14],D3} = {[10,18],D3}
(Except for the first, all of these examples
are D3 not because one of their operands is
D3 but because both of their operands are
contained within the defined & continuous
domain of add.)
D2 ==> inDomain & result = empty:
floor(xx) \intersect (xx + 10) = {empty,D2}
(Because the floor was D2?)
D2 ==> inDomain & result = nonempty:
floor(xx) + xx = {[0,8],D2}
(Because the floor was D2?)
D1 ==> inDomain&OutDomain & result = empty:
sqrt(xx - 3) \intersect (xx + 10) = {empty,D1}
{[0,1],D1} \intersect {[10,14],D3} = {empty,D1}
(Because the sqrt was D1.)
D1 ==> inDomain&OutDomain & result = nonempty:
sqrt(xx - 3) = {[0,1],D1}
(Because the sqrt was D1.)
D0 ==> OutDomain & result = empty:
sqrt(xx - 10) \intersect xx = {empty,D0}
(Because the sqrt was D0.)
D0 ==> OutDomain & result = nonempty:
sqrt(xx - 10) \union xx = {[0,4],D0}
(Because the sqrt was D0.)
Do I have all these correct?
(On rereading this I think the last example is wrong. But
I left it in, in the hopes that someone else can provide a
correct example. There may not be one.)
Does anyone else have simpler or more illuminating examples?
I tried to include examples without set theoretic operations
where I could think of them. But I couldn't think of many.
>
> . . .
>
> > 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
Then, in keeping with your assertion that this is a
level 1 only motion, let me suggest an amendment.
The use of the bits D(f,X)+, D(f,X)-, & C(f,X) amount
to a style of implementation that belongs at level 3
or 4.
Similarly, naming the decorations D0 through D4 also
implies a particular implementation that has no place
in a level 1 discussion.
John had 3 letter abbreviations for these decorations.
D4 = emp = empty
D3 = con = inDomain&Continuous
D2 = dom = inDomain
D1 = bot = inDomain&OutDomain (both?)
D0 = out = outDomain
(I may have them wrong. John, please edit to taste.)
Let me suggest that we use non-pejorative names only
& leave the implementation details until later.
Further, only the partial ordering is necessary for a
proof of FTDIA. Namely:
emp < con < dom < bot
emp < out < bot
Let us so state in this motion.
As you mentioned before, ill & these other things can be
discussed at a lower level & a later date.
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