Re: [stack] Combining Combinators

On 6/19/07, Manfred Von Thun <[email protected]> wrote:
> On 19/6/07 1:05 AM, "William Tanksley, Jr" <[email protected]> wrote:
>
> > Manfred Von Thun <[email protected]> wrote:
>
> > At level zero there are numbers, chars, truth values, lists...
> >> At level one there are functions successor, +, and ...
> >> At level two there are combinators, functions which take functions
> >> as arguments (or as values): map, fold, and many more.
> >> I was looking for something at level three: functions which
> >> take combinators as arguments, but which are distinct from
> >> combinators themselves.
> >
> > How would that distinction be achieved? You hint below that
> > third-order entities would NOT take a function as a parameter; but I
> > must point out that every combinator IS a function.
>
> Yes, every combinator is a function. My wording was too cryptic
> perhaps. I'll try again:
>
> At level one there are functions which take zero level entities
> as arguments.

In Cat this would be the set of functions with types matching: ('A 'b -> 'C)

> At level two there are combinators which take
> level one entities as arguments but cannot take level zero
> entities as arguments. Examples: dip, nullary, map, filter, fold.

These have types that match the pattern: ('A ('B -> 'C) -> 'D)

> At level three there would be functions which take level two
> entities as arguments but cannot take level one entities as
> arguments.

These would have types: ('A ('B ('C -> 'D) -> 'E) -> 'F)

In Cat, unlike Joy, level two combinators are never also level one
combinators.

> If there are such things at all, let foo be one of them.
> Then the first below would be meaningful, but the second would not:
>
> 2 3 [+] [nullary] foo
> 2 3 [+] foo
>
> >> I never found any that are distinct
> >> from the existing bunch of Joy combinators.
> >
> > Those are complete, right?
>
> It may or may not have to do with any kind of completeness, I do
> not know. The existing bunch seems to be pretty useful. But I was
> not looking for new level two entities, I was looking for genuinely
> level three entities which are not also level two.
>
> >> a strange new world to me. But it has not led to any combinator
> >> N (for Œnew¹) such that [C] N is meaningful for a combinator [C]
> >> but [Q] N is not meaningful for a non-combinator Q. This
> >> is what I meant by a higher order combinator. But I do not have
> >> a proof that there cannot be such a beast.
> >
> > It seems to me that you want a function which takes combinators as
> > arguments, but does not take functions (right?). But because all
> > combinators _are_ functions, this can't be possible.
>
> If there are the kind of thing I was looking for, this would open
> up enormous possibilities and complications, not necessarily
> pleasant ones. So I did not really want them. I was pleased (and
> often surprised) when I found ways in which existing combinators
> C, D can be combined as in [C] D. So nothing third level. But
> it is stillan open question as far as I can see.

Is:

DEFINE f = [[dip] cons] dip.

A level-3 combinator in Joy? In general I think the secret to creating
a level-3 combinator would be to apply a level-2 combinator to the
result of calling a level-2 combinator.

If you are concerned about eliminating any possibility of level-3
combinators in Joy, can't you rewrite the definitions of the level-2
primitives so that they accept combinators or non-combinators? I don't
see the reason that "dip" can't accept straight values like "5" on the
top of the stack, if "i" can accept them. A definition of "dip" as
"swap unit cons i" should be equivalent shouldn't it? In which case
you wouldn't actually be able to say "dip" is a level-2 combinator. Or
am I missing something here?

I believe it is either good to go all the way one direction or the
other: either non-combinator values can be "evaluated" or they can't.
As far as I can see (and I could be mistaken) Joy is somewhat
ambiguous about what exactly can and can't be evaluated.

Hopefully I am not sticking my foot in my mouth here.

Cheers,
Christopher


 
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