MarLinn monkleyon at gmail.com
Wed May 3 11:56:52 UTC 2017

```It's a nice challenge to come up with datatypes that could be
Traversable but not Foldable.

First of all, you can think of Traversable as an extension of Functor.
Instead of only mapping over the structure with a pure function, you can
now also map over it with a "functorial function" (a -> f b) (I don't
know of a better name, so I'm inventing one.) What that means is to have
a Traversable structure, you need to have the ability to touch every
element. If you can touch every element, why would you not be able to
feed them all into a folding function?

Only a few possible anwers come to mind for me:

1. There might be no obvious ordering. But in practice, you just invent
one.
2. The structure might be infinite, so folding would never succeed. But
laziness ftw.
3. Generalizing the previous answer, the structure might be computed
ad-hoc and contain fixed points. That alone might not suffice, but
it sounds like a more interesting starting point.
4. Generalizing a bit differently, the structure might be computed
ad-hoc… because it's IO. IO surely can't be Foldable. But could it
be Traversable? Well… no actually, because order is important.

So can you come up with a structure that is computed ad-hoc, that
contains fixed points, and where ordering is unimportant? Good luck.

Cheers.

On 2017-05-03 11:56, Jonathon Delgado wrote:
> OK, I understand why Traversable is useful here - thank you Chris and Dmitry!
>
> The next question is why Traversable requires Foldable. I looked at the source, and couldn't see where Foldable is being used, other than as a constraint on Traversable. To put the question differently, what would fail to compile if this constraint was removed?
>
>
>
> From: Dmitry Olshansky <olshanskydr at gmail.com>
> Sent: 03 May 2017 09:53
> Subject: Re: [Haskell-cafe] Foldable for (,)
>
>
>
>
>
> With fmap you can only change all values in some "container".
>
>   With Foldable you can "fold" it, i.e. calculate some "scalar" result.
>
>   With Traversable you can "change order of two containers":
>> sequenceA [[1,2,3],[4,5]]
> [[1,4],[1,5],[2,4],[2,5],[3,4],[3,5]]
>> sequenceA ("test",[2,3,4])
> [("test",2),("test",3),("test",4)]
>> sequenceA ("test",([1,2,3],[4,5,6]))
> ([1,2,3],("test",[4,5,6]))
>
>
>
>
>
> 2017-05-03 12:12 GMT+03:00 Jonathon Delgado  <voldermort at hotmail.com>:
>   Why do you want to traverse a tuple instead of fmap? i.e. what can you do with Foldable/Traversable for (,) that you can't do with Functor?
>
> My background, as you can probably guess, is beginner.
>
>
> Sent: 03 May 2017 08:51
> To: Tony Morris
> Subject: Re: [Haskell-cafe] Foldable for (,)
>
>
>
>
> Replying to myself, I suppose one good answer is that whether or not you care about Foldable instances for tuples, you might care about Traversable instances, and those require Foldable as a superclass.
>
>
> For example, one possible specialization of `traverse` is:
>
>
>      traverse :: (a -> IO b) -> (SideValue, a) -> IO (SideValue, b)
>
>
> Jonathon, I don't know how much background you're coming from, so I'd be happy to explain that in more detail if you need it.
>
>
> On Wed, May 3, 2017 at 1:44 AM, Chris Smith  <cdsmith at gmail.com> wrote:
>
> I'm also interested in Jonathon's question, so let me try to bring things back to the question.  Everyone agrees that there's only one reasonable way to define this instance if it exists.  But the question is: why is it defined at all?
>
>
> That's an easy question to answer for Functor, Applicative, and Monad.  But I am having trouble giving a simple or accessible answer for Foldable.  Do you know one?
>
>
>
>
> On Wed, May 3, 2017 at 1:32 AM, Tony Morris  <tonymorris at gmail.com> wrote:
>   It's Foldable for ((,) a).
>
> It is not Foldable for any of these things:
>
> * (,)
> * tuples
> * pairs
>
> In fact, to talk about a Foldable for (,) or "tuples" is itself a kind
> error. There is no good English name for the type constructor ((,) a)
> which I suspect, along with being unfamiliar with utilising the
> practical purpose of types (and types of types) is the root cause of all
> the confusion in this discussion.
>
> Ask yourself what the length of this value is:
>
> [[1,2,3], [4,5,6]]
>
>
> [(1, 'a'), (undefined, 77)]
>
> Is it 4? No, obviously not, which we can determine by:
>
> :kind Foldable :: (* -> *) -> Constraint
> :kind [] :: * -> *
>
> Therefore, there is no possible way that the Foldable instance for []
> can inspect the elements (and determine that they are pairs in this
> case). By this method, we conclude that the length of the value is 2. It
> cannot be anything else, some assumptions about length itself put aside.
>
> By this ubiquitous and very practical method of reasoning, the length of
> any ((,) a) is not only one, but very obviously so.
>
>
>
> On 03/05/17 17:21, Jonathon Delgado wrote:
>> I sent the following post to the Beginners list a couple of weeks ago (which failed to furnish an actual concrete example that answered the question). Upon request I'm reposting it to Café:
>>
>> I've seen many threads, including the one going on now, about why we need to have:
>>
>> length (2,3) = 1
>> product (2,3) = 3
>> sum (2,3) = 3
>> or (True,False) = False
>>
>> but the justifications all go over my head. Is there a beginner-friendly explanation for why such seemingly unintuitive operations should be allowed by default?
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