Re: Proposal: 6NF

On Oct 20, 3:12 am, "vc" <boston..._at_hotmail.com> wrote:
> Marshall wrote:
> > On Oct 19, 8:40 pm, "vc" <boston..._at_hotmail.com> wrote:
> > > Marshall wrote:
>
> > > > > "return new ModFour((val+a.val)%4)" is called a bug or cheating
> > > > > because you used a function defined on the entire domain NxN (where N
> > > > > is a subset of the natural numbers implemented by the computer) to
> > > > > generate a result which is undefined for the (2,3) pair.
>
> > > > It's not a bug, since the program produces the correct result, which is
> > > > 1. It's not "cheating" because WTF does cheating even mean in
> > > > this context. What it is, is the absence of closure over the subclass.
> > > > Which is no big deal.
>
> > > So producing an absurd result is no big deal ? You seem to have agreed
> > > that in the {2,3} subset there is no '1', how come your java
> > > implementation manages to extract '1' out of nowhere ?
>
> > Consider the below function:
>
> > Let S4 = {0, 1, 2, 3}
> > Let S23 = {2, 3}
> > f : S23, S23 -> S4
>
> The binary operation over S23 is g:S23:S23 -> S23, your f does not
> qualify. A common example is '+'.
>
>
> > f(2,2) = 0
> > f(2,3) = 1
> > f(3,2) = 1
> > f(3,3) = 2
>
> > Your claim is what, exactly? That this is not a function? That
> > it doesn't exist? That it's cheating? That it's invalidly defined
> > because the result type is not the same as the types of
> > both parameters?
> It's not a binary opeartion that we've been talking about all along.
> Did you read the reference I gave earlier or not ?

So we agree that my f does not meet the definition of binary operation.

Do we also agree that:

1. f is a function
2. f exists
3. f is well-defined ?

Apparently you also think that f is "absurd." Can we take as a definition of "absurd" to be any function that is not of the type g:A,A->A? Because it seems to me that there are many such functions.

> > > > > At thispoint, it may be useful to recall what a function is and
> > > > > apply the memory, if any, to the addMod4 operation.
>
> > > > Indeed. A function is a mapping from one set to another.
> > > > In this case, the domain is pairs of values from {0,1,2,3}
> > > > and the range is that same set. This particular
> > > > function qualifies as a binary function, and also
> > > > exhibits closure over the set {0,1,2,3}. Since {2,3} is a
> > > > subset of {0,1,2,3}, the function is fully defined over
> > > > that subset. However, since that function is not closed
>
> > > How can it be 'fully defined' over {2,3} and 'not closed' at the same
> > > time ? What does (2,3) map to ? If you say to one, then you
> > > exctract '1' out of nowhere again.
>
> > I don't understand this "out of nowhere" business. It's just
> > a plain, regular number one. Nothing could be simpler.
>
> Where is '1' in {2,3} ?

It's not there, but so what? There's no sqrt(2) in there either; does that mean the sqrt function is absurd? Or am I simply not allowed to take the square root of values in {2,3}? What requires me to have *every* function have {2,3} as its range?

You ask "How can it be 'fully defined' over {2,3} and 'not closed' at the same time ?" It appears that you believe that for a function to be fully defined, it has to have its return type the same as its two argument types. Is closure required for a function to be defined? We agree that closure is required to meet the definition of "binary operation" but I'm asking about the more general definition of "function." (I claim that to qualify as a function something has to be well-defined.)

> > > > > > The *only* trouble anywhere in here is if you want to
> > > > > > have methods which are mutator methods; that is,
> > > > > > methods which modify the distinguished object in place.
> > > > > > And yes, in that case, closure is an issue. I have not
> > > > > > seen any math books that give functions the same
> > > > > > semantics as inherited OOPL mutator methods however.
>
> > > > I note you did not respond to this important paragraph. Yes,
> > > > in OOPLs, there is a problem with closure, subtyping,
> > > > and mutator methods. Outside of OOPLs (in math, that
> > > > is) this problem does not apply.
>
> > > We may tackle those issues later after we come to some agreement re.
> > > the trivial stuff like functions and binary operations.
>
> > Okay. What is your definition of a function, then, and how
> > does my f above stack up with regards to your
> > definition?
> Your 'f' has nothing to do with the whole discussion of 'inheriting'
> a binary operation from a type.

Why doesn't it? I already showed an example of working Java code in which a function is defined over pairs of ints mod 4, and that function is *inherited,* in the literal concrete Java Language Spec sense, by a subclass consisting of only the values {2, 3}.

Now, you called that "cheating" but I haven't heard your definition of cheating yet, so I'm unclear as to how to respond. I don't recall the word "cheating" being used in the JLS.

Again, the only problem here is your assertion that sets are inextricably bound to functions defined over them, in some OOPL (but not very mathematical) sense. The algebra books I have read (admitedly few in number) have always made a clear distinction between a set and algebraic structures that include a set with binary operations defined over the set.

I bet there's a relevant wikipedia article. Let's keep it simple:

http://en.wikipedia.org/wiki/Monoid

"A monoid is a set M with binary operation ∗ : M × M → M, obeying the following axioms: [associativity, identity]"

Here, an algebraic structure is defined as a tuple consisting of a set and a binary operation. It's not just the set itself; it can't be. This example is particularly telling to me, because we can immediately see that Z participates in (at least) two monoids: (Z, +) and (Z, *). So if I say "the monoid Z" I'm being appallingly imprecise; I *cannot* assume an implicit relationship between the set and its operations. While the definition of +:Z,Z->Z depends on the definition of Z, it is entirely *distinct* from it; the relation + is a set, and Z is a set, and they are two distinct sets; + is not an element of Z.

> > > > > > Also, I found this equation in a math book on page 3:
>
> > > > > > P ⊆ N ⊆ Z ⊆ Q ⊆ R
>
> > > > > The equation looks like a random collection of characters. What do
> > > > > you suppose it means ?
>
> > > > I suppose your browser doesn't have the necessary font.
>
> > > > In prose, it means:
>
> > > > The set of positive integers is a subset of the set of nonnegative
> > > > integers which is a subset of the integers which is a subset
> > > > of the rationals which is a subset of the reals.
>
> > > That's truly a trivial observation which is not strictly speaking
> > > correct (see the earlier discussion re. real construction).
>
> > Wait, you're saying it's both trivial *and* incorrect? Earlier
> > I recall you suggesting various people consult math books
> > for such questions; I did consult one and it was quite clear
> > on the question: every integer is also a real.
> I guess you need another book because integers as commonly constructed
> are not a subset of reals but rather are isomorphic to some subset of
> reals. In cases when the subtlety is unimportant, one can say that
> the integers are a subset of real, we discussed that before.

You're very hung up on the construction issue; it's not the only way of thinking about this. I agree that we can validly take that viewpoint; I also agree that we can take a type-theoretic approach and reach the same conclusion that 2 and 2.0 are not the same number. Those are valid conclusions when embedded within the appropriate theoretical framework. There also exist (fairly simple) set-theoretic frameworks under which 2 and 2.0 *are* the same number, and that is equally a valid conclusion when embedded within the appropriate theoretical framework.

This isn't Highlander; it's not the case that "there can be only one!"

Marshall Received on Fri Oct 20 2006 - 19:17:07 CEST