Re: [Haskell-cafe] 'Associative' order of calling

29 Oct 2015

      On Wed, Oct 28, 2015 at 7:50 PM, Janis Voigtländer <
janis.voigtlaender@gmail.com> wrote:
...
Your message contained lots of category-theory-speak that wasn’t called
for.
Wait, what? You're alluding to my mention of F-algebras?

Now I know that your discussion with Matteo that cited two papers is
pertinent and also kinda off on the side. As we both know, those papers go
far deeper into CT than F-algebras, which is like, baby stuff in comparison.

Baby stuff that they are, F-algebras are nice because it's a theory of data
where the folds fall out for free. It's a rock solid theory in a sea of
nebulous generalizations about fold. That's why I mentioned them.

Question: What is a fold of type “(a -> a -> a) -> [a] -> a” that promises
...
to call its first parameter “associatively”?
Answer: It is a fold equivalent to a fold1 (left or right, it doesn’t
matter).
No, it seems you haven’t understood the answer. At least, that was not the
answer.
Well, look. I thought otherwise, but I may be wrong. I frequently am. And
if I am wrong, I want to be put right. Let's examine the facts.

In an earlier response to Tom, you wrote:

'So, for the special case of non-empty lists, how about expressing the
desired property as follows: "The function h must satisfy: for all
associative f and all lists xs, it holds that h f xs = foldl1 f xs."'

I thought impeccable the logic you used to arrive at this.

Over two eta-contractions, I get h = foldl.

The folds are monotyped, the lists finite, and so foldl = foldr.

Putting it all together, finally:

The property of the fold in question is that it is equivalent to a fold
(left or right, it doesn't matter which).
...

2015-10-28 13:41 GMT+01:00 Kim-Ee Yeoh :
...
On Sun, Oct 25, 2015 at 1:17 PM, Janis Voigtländer <
janis.voigtlaender@gmail.com> wrote:
...
Kim-Ee, I think you are overcomplicating things
No.
Matteo cited a paper on traversables and you added a rejoinder. So it
wasn't me who introduced data abstraction.
Also, OP explicitly asked about a fold "on a set of data". Only later,
when requested for an example, did he give lists.
In fact, the discussion of "arbitrary data structures" (in this thread in
...
general) distracts a lot from the properties under consideration, and from
even first finding out whether they are properties of foldLike or of f in
foldLike f.
So, let's specialize. Let's consider only the case of non-empty lists.
And look at what we have: A definitive answer to OP's question:
Question: What is a fold of type "(a -> a -> a) -> [a] -> a" that
promises to call its first parameter "associatively"?
Answer: It is a fold equivalent to a fold1 (left or right, it doesn't
matter).
That's nice.
But is that all we can come up with? Does it really do justice to the
original question? Viz.
"Is there a name for a fold that promises to call a function such that
only an associative function will always return the same result. Or in
other words, it has the property that it promises to call a function
"associatively" on a set of data?"
The challenge becomes:
* generalize the result to fearsomely complicated "arbitrary data
structures"
* answer the original question in a holistic spirit
...
Then, candidate functions for "foldlike functions" will be functions of
this type:
foldLike : (a -> a -> a) -> [a] -> a
Here are some candidates (I give only an equation for four element lists
in each case, but I assume everyone has enough imagination to see how these
could be extended to other lists in an intended way):
foldLike1 f [a,b,c,d] = [] -- throws away stuff
foldLike2 f [a,b,c,d] = f a (f b (f c d)) -- we have all ancountered
this one
foldLike3 f [a,b,c,d] = f (f (f a b) c) d -- also looks familiar
foldLike4 f [a,b,c,d] = f (f a b) (f c d) -- that's a "new" one, but
looks good
foldLike5 f [a,b,c,d] = f a a -- probably not a very popular one
foldLike6 f [a,b,c,d] = f (f c a) (f b d) -- a reasonable one, for
example there's a Traversable instance that leads to this one; but still,
it's not one that Charles would like, I think
So now we can ask which of these satisfy Charles's 1., 2., 3. points.
Can't we?
There was:
...
1. Promises to call f on all data (does not have any guarantees on
order)
This is satisfied by foldLike2, foldLike3, foldLike4, and foldLike6, but
not by the others.
...
2. Promises to call f on all data in order (like a left fold)
This is satisfied by foldLike3, but not by the others.
...
3. Promises to call f "associatively" (perhaps can be formalized as an
in order break down of the data into tree structures)
This is satisfied by foldLike2, foldLike3, and foldLike4, but not by the
others.
Since I am able to tell, for a given foldLike candidate, whether or not
it satisfies 3. (for example, I could specifically see that foldLike6 does
not satisfy 3., while it does satisfy 1.), it cannot be maintained that 3.
has no meaning.
Note: Nothing in the above makes any assumptions about f. Whether or not
f is an associative function is irrelevant for what is asked here.
2015-10-25 4:19 GMT+01:00 Kim-Ee Yeoh :
...
On Sun, Oct 25, 2015 at 1:12 AM, Janis Voigtländer <
janis.voigtlaender@gmail.com> wrote:
...
It has already been established in this thread what Charles meant by 3.
He meant that a fold-function that has the property he is after would
guarantee that it:
a) takes all the content elements from a data structure, say x1,...,xn,
b) builds an application tree with the to-be-folded, binary operation
f in the internal nodes of a binary tree, whose leafs, read from left to
right, form exactly the sequence x1,...,xn,
c) evaluates that application tree.
Isn't this what Charles meant by his 2nd property:
...
2. Promises to call f on all data in order (like a left fold)
What about his 3rd?
Do you agree that what I describe above is a property of a given
...
fold-like function, not of the f handed to that fold-like function?
Before discussing a property of X, isnt it worth asking what X even
means?
The folds whose meanings are crystal clear are the arrows out of
initial objects in the category of F-algebras.
They are crystal clear because they couple as one with the data
definition spelled out in full.
In the quest for useful generalizations of catamorphisms, that coupling
with the data definition continues to be assumed.
Observe, for instance:
...
a) takes all the content elements from a data structure, say
x1,...,xn,
Does a foliar data structure have a canonical flattening out of its
leaves? Are there symmetric canonicalizations? How is one selected over the
others?
Is the meaning of "all" referentially transparent? That turns out to be
a subtle point, as this convo shows:
http://haskell.1045720.n5.nabble.com/A-Proposed-Law-for-Foldable-tp5765339.h...
With the theory of F-algebras, we started with precise notions of data
and folds came for free.
But data can be overspecified. And also, the folds among swathes of
data suggest useful generalizations.
So now, a raft of proto-precise and necessarily psychological notions
of Foldable waded in, and since then it's been fun playing sorting games
with shape blocks and holes to squeeze them into.
Fun is good. It's a stage in the journey to knowledge.
And do you agree that what I describe above is a property that is
...
weaker than (and so, in particular different from) for example the property
"this fold-like function is foldl or foldr".
...
2015-10-24 19:55 GMT+02:00 Kim-Ee Yeoh :
...
On Sun, Oct 25, 2015 at 12:42 AM, Matteo Acerbi <
matteo.acerbi@gmail.com> wrote:
> For what concerns question 3, I didn't understand the idea of
> calling a function "associatively".
>
This. Associativity is a property of binary operators. It's not a
property of the catamorphism 'calling' on a given binary operator.
Also, when Charles writes: "Then it goes on to use f in "thisFold f
[0,1,2]" like "f (1 (f 0 2))""
commutativity appears to raise its head.
-- Kim-Ee