[Haskell-cafe] A quick question about distribution of finite maps.

[Tom Ellis]

And see

https://www.youtube.com/watch?v=4aQlFMvKgdc

for a related talk

On Sun, Jan 10, 2021 at 06:53:31AM -0500, David Feuer wrote:
> You found it! See specifically the documentation for Data.Can.
> 
> On Sun, Jan 10, 2021, 2:54 AM <jack at jackkelly.name> wrote:
> 
> > You may be thinking of https://hackage.haskell.org/package/smash and
> > related packages.
> >
> > -- Jack
> >
> > January 10, 2021 5:19 PM, "David Feuer" <david.feuer at gmail.com> wrote:
> >
> > > I just don't want to take credit for anyone else's work. I have a very
> > vague recollecntion that
> > > Emily Pillmore may have looked at some related categorical stuff, but I
> > could easily be mistaken.
> > >
> > > On Sun, Jan 10, 2021, 2:16 AM Olaf Klinke <olf at aatal-apotheke.de> wrote:
> > >
> > >> On Sat, 2021-01-09 at 18:04 -0500, David Feuer wrote:
> > >>> `These` is not my own invention. It's in the `these` package.
> > >>
> > >> Then please interpret my "due to David Feuer" as "brought into this
> > >> discussion by David Feuer". I wonder whether These has ever been linked
> > >> to categorical products in Kleisli Maybe before. There are
> > >> Data.These.Combinators.justHere and justThere which are the same as my
> > >> fstMaybe and sndMaybe. But combinators resulting in pairKleisli and
> > >> iso1 seem to be missing from the these package.
> > >>
> > >> Olaf
> > >>
> > >>>
> > >>> On Sat, Jan 9, 2021, 5:41 PM Olaf Klinke <olf at aatal-apotheke.de>
> > >>> wrote:
> > >>>
> > >>>> On Sat, 2021-01-09 at 22:26 +0100, MigMit wrote:
> > >>>>> Actually, it's pretty easy to construct a type `P x y`, so that
> > >>>>> Maybe
> > >>>>> (P x y) ~ (Maybe x, Maybe y). It would be
> > >>>>>
> > >>>>> data OneOrBoth x y = Left' x | Right' y | Both x y
> > >>>>>
> > >>>>> The isomorphism is, I think, obvious
> > >>>>>
> > >>>>> iso1 :: Maybe (OneOrBoth x y) -> (Maybe x, Maybe y)
> > >>>>> iso1 Nothing = (Nothing, Nothing)
> > >>>>> iso1 (Just (Left' x)) = (Just x, Nothing)
> > >>>>> iso1 (Just (Right' y)) = (Nothing, Just y)
> > >>>>> iso1 (Just (Both x y)) = (Just x, Just y)
> > >>>>>
> > >>>>> iso2 :: (Maybe x, Maybe y) -> Maybe (OneOrBoth x y)
> > >>>>> iso2 = -- left as an excersize for the reader
> > >>>>>
> > >>>>> And indeed, "OneOrBoth" would be a cartesian product functor in
> > >>>>> the
> > >>>>> category of finite types (and maps).
> > >>>>
> > >>>> I stand corrected. Below is the full code, in case anyone wants to
> > >>>> play
> > >>>> with it.
> > >>>>
> > >>>> import Control.Arrow ((&&&))
> > >>>>
> > >>>> -- due to David Feuer
> > >>>> data OneOrBoth x y = Left' x | Right' y | Both x y
> > >>>>
> > >>>> iso2 :: (Maybe x,Maybe y) -> Maybe (OneOrBoth x y)
> > >>>> iso2 p = case p of
> > >>>> (Nothing,Nothing) -> Nothing
> > >>>> (Just x,Nothing) -> (Just (Left' x))
> > >>>> (Nothing,Just y) -> (Just (Right' y))
> > >>>> (Just x, Just y) -> (Just (Both x y))
> > >>>>
> > >>>> -- (OneOrBoth x) is a functor on Kleisli Maybe
> > >>>> fmapKleisli :: (a -> Maybe b) ->
> > >>>> OneOrBoth x a -> Maybe (OneOrBoth x b)
> > >>>> fmapKleisli k (Left' x) = Just (Left' x)
> > >>>> fmapKleisli k (Right' a) = fmap Right' (k a)
> > >>>> fmapKleisli k (Both x a) = fmap (Both x) (k a)
> > >>>>
> > >>>> -- is OneOrBoth the cartesian product? Seems so:
> > >>>>
> > >>>> pairKleisli :: (a -> Maybe x) ->
> > >>>> (a -> Maybe y) ->
> > >>>> a -> Maybe (OneOrBoth x y)
> > >>>> pairKleisli f g = iso2 . (f &&& g)
> > >>>>
> > >>>> fstMaybe :: OneOrBoth x y -> Maybe x
> > >>>> fstMaybe (Left' x) = Just x
> > >>>> fstMaybe (Both x _) = Just x
> > >>>> fstMaybe (Right' _) = Nothing
> > >>>>
> > >>>> sndMaybe :: OneOrBoth x y -> Maybe y
> > >>>> sndMaybe (Left' _) = Nothing
> > >>>> sndMaybe (Right' y) = Just y
> > >>>> sndMaybe (Both _ y) = Just y
> > >>>>
> > >>>>> But it won't be cartesian closed. If it were, then for any finite
> > >>>>> X
> > >>>>> and Y we should have
> > >>>>>
> > >>>>> Maybe (X^Y) ~
> > >>>>> () -> Maybe (X^Y) ~
> > >>>>> OneOrBoth () Y -> Maybe X ~
> > >>>>> (() -> Maybe X, Y -> Maybe X, ((), Y) -> Maybe X) ~
> > >>>>> (Maybe X, Y -> Maybe X, Y -> Maybe X)
> > >>>>>
> > >>>>> so
> > >>>>>
> > >>>>> X^Y ~ (X, Y -> Maybe X, Y -> Maybe X)
> > >>>>>
> > >>>>> But then
> > >>>>>
> > >>>>> Z -> Maybe (X^Y) ~
> > >>>>> Z -> (Maybe X, Y -> Maybe X, Y -> Maybe X) ~
> > >>>>> (Z -> Maybe X, (Z, Y) -> Maybe X, (Z, Y) -> Maybe X) ~
> > >>>>>
> > >>>>> and
> > >>>>>
> > >>>>> OneOrBoth Z Y -> Maybe X ~
> > >>>>> (Z -> Maybe X, Y -> Maybe X, (Z, Y) -> Maybe X)
> > >>>>>
> > >>>>> We see that those aren't the same, they have a different number
> > >>>>> of
> > >>>>> elements, so, no luck.
> > >>>>
> > >>>> Doesn't this chain of isomorphisms require () to be the terminal
> > >>>> object, or did you take () as synonym for the terminal object in
> > >>>> the
> > >>>> category? For example, there are several functions of the type
> > >>>> Bool -> Maybe ().
> > >>>> So the terminal object in Kleisli Maybe would be Void, because
> > >>>> Maybe
> > >>>> Void ~ (). We'd need to make fields in OneOrBoth strict to have an
> > >>>> isomorphism OneOrBoth Void a ~ a, just as the true categorical
> > >>>> product
> > >>>> in (->) is the strict pair.
> > >>>>
> > >>>> Olaf
> > >>>>
> > >>>>>> On 9 Jan 2021, at 22:01, Olaf Klinke <olf at aatal-apotheke.de>
> > >>>>>> wrote:
> > >>>>>>
> > >>>>>>> Hello!
> > >>>>>>>
> > >>>>>>> Finite maps from `"containers" Data.Map` look like they may
> > >>>>>>> form
> > >>>>>>> a
> > >>>>>>> Cartesian closed category. So it makes sense to ask if the
> > >>>>>>> rule
> > >>>>>>> _α ⇸
> > >>>>>>> (β ⇸ γ) ≡ ⟨α; β⟩ ⇸ γ ≡ ⟨β; α⟩ ⇸ γ ≡ β ⇸ (α ⇸ γ)_ that holds
> > >>>>>>> in
> > >>>>>>> such
> > >>>>>>> categories does hold for finite maps. Note that, a map being
> > >>>>>>> a
> > >>>>>>> functor, this also may be seen as _f (g α) ≡ g (f α)_, which
> > >>>>>>> would
> > >>>>>>> work if maps were `Distributive` [1].
> > >>>>>>>
> > >>>>>>> It looks to me as though the following definition might work:
> > >>>>>>>
> > >>>>>>> distribute = unionsWith union . mapWithKey (fmap .
> > >>>>>>> singleton)
> > >>>>>>>
> > >>>>>>> — And it works on simple examples. _(I checked the law
> > >>>>>>> `distribute ∘
> > >>>>>>> distribute ≡ id` — it appears to be the only law required.)_
> > >>>>>>>
> > >>>>>>> Is this definition correct? Is it already known and defined
> > >>>>>>> elsewhere?
> > >>>>>>>
> > >>>>>>> [1]:
> > >>>>>>>
> > >>>>
> > >>
> > https://hackage.haskell.org/package/distributive-0.6.2.1/docs/Data-Distributive.html#t:Distributive
> > >>>>>> Hi Ignat,
> > >>>>>>
> > >>>>>> TL;DR: No and no.
> > >>>>>>
> > >>>>>> The documentation says that every distributive functor is of
> > >>>>>> the
> > >>>>>> form
> > >>>>>> (->) x for some x, and (Map a) is not like this.
> > >>>>>>
> > >>>>>> If Maps were a category, what is the identity morphism?
> > >>>>>>
> > >>>>>> Let's put the Ord constraint on the keys aside, Tom Smeding has
> > >>>>>> already
> > >>>>>> commented on that. Next, a Map is always finite, hence let's
> > >>>>>> pretend
> > >>>>>> that we are working inside the category of finite types and
> > >>>>>> functions.
> > >>>>>> Then the problems of missing identity and missing Ord go away.
> > >>>>>> Once
> > >>>>>> that all types are finite, we can assume an enumerator. That
> > >>>>>> is,
> > >>>>>> each
> > >>>>>> type x has an operation
> > >>>>>> enumerate :: [x]
> > >>>>>> which we will use to construct the inverse of
> > >>>>>> flip Map.lookup :: Map a b -> a -> Maybe b
> > >>>>>> thereby showing that a Map is nothing but a memoized version of
> > >>>>>> a
> > >>>>>> Kleisli map (a -> Maybe b). Convince yourself that Map
> > >>>>>> concatenation
> > >>>>>> has the same semantics as Kleisli composition (<=<). Given a
> > >>>>>> Kleisli
> > >>>>>> map k between finite types, we build a Map as follows.
> > >>>>>> \k -> Map.fromList (enumerate >>= (\a -> maybe [] (pure.(,) a)
> > >>>>>> (k
> > >>>>>> a)))
> > >>>>>>
> > >>>>>> With that knowledge, we can answer your question by deciding:
> > >>>>>> Is
> > >>>>>> the
> > >>>>>> Kleisli category of the Maybe monad on finite types Cartesian
> > >>>>>> closed?
> > >>>>>> Short answer: It is not even Cartesian.
> > >>>>>> There is an adjunction between the categories (->) and (Kleisli
> > >>>>>> m)
> > >>>>>> for
> > >>>>>> every monad m, where
> > >>>>>> * The left adjoint functor takes
> > >>>>>> types x to x,
> > >>>>>> functions f to return.f
> > >>>>>> * The right adjoint functor takes
> > >>>>>> types x to m x,
> > >>>>>> Kleisli maps f to (=<<) f
> > >>>>>> Right adjoint functors preserve all existing limits, which
> > >>>>>> includes
> > >>>>>> products. Therefore, if (Kleisli m) has binary products, then m
> > >>>>>> must
> > >>>>>> preserve them. So if P x y was the product of x and y in
> > >>>>>> Kleisli m,
> > >>>>>> then m (P x y) would be isomorphic to (m x,m y). This seems not
> > >>>>>> to
> > >>>>>> hold
> > >>>>>> for m = Maybe: I can not imagine a type constructor P where
> > >>>>>> Maybe (P x y) ~ (Maybe x,Maybe y).
> > >>>>>> In particular, P can not be (,). The only sensible Kleisli
> > >>>>>> projections
> > >>>>>> from (x,y) would be fst' = return.fst and snd' = return.snd.
> > >>>>>> Now
> > >>>>>> think
> > >>>>>> of two Kleisli maps f :: Bool -> Maybe x, g :: Bool -> Maybe y.
> > >>>>>> Assume
> > >>>>>> that f True = Just x for some x and g True = Nothing. In order
> > >>>>>> to
> > >>>>>> satisfy g True = (snd' <=< (f&&&g))True, the unique pair arrow
> > >>>>>> (f&&&g)
> > >>>>>> would need to map True to Nothing, but then f True = (fst' <=<
> > >>>>>> (f&&&g))
> > >>>>>> True can not hold. We conclude that (Kleisli Maybe) does not
> > >>>>>> even
> > >>>>>> have
> > >>>>>> categorical products, so asking for Cartesian closure does not
> > >>>>>> make
> > >>>>>> sense.
> > >>>>>>
> > >>>>>> You might ask for a weaker property: For every type x, ((,) x)
> > >>>>>> is a
> > >>>>>> functor on (Kleisli Maybe). Indeed, the following works because
> > >>>>>> ((,) x)
> > >>>>>> is a polynomial functor.
> > >>>>>> fmapKleisli :: Functor m => (a -> m b) -> (x,a) -> m (x,b)
> > >>>>>> fmapKleisli f (x,a) = fmap ((,) x) (f a)
> > >>>>>> Thus you may ask whether this functor has a right adjoint in
> > >>>>>> the
> > >>>>>> Kleisli category of Maybe. This would be a type constructor g
> > >>>>>> with
> > >>>>>> a
> > >>>>>> natural isomorphism
> > >>>>>>
> > >>>>>> (x,a) -> Maybe b ~ a -> Maybe (g b).
> > >>>>>>
> > >>>>>> The first thing that comes to mind is to try
> > >>>>>> g b = x -> Maybe b and indeed djinn can provide two functions
> > >>>>>> going
> > >>>>>> back and forth that have the right type, but they do not
> > >>>>>> establish
> > >>>>>> an
> > >>>>>> isomorphism. I doubt there is such a right adjoint g, but can
> > >>>>>> not
> > >>>>>> prove
> > >>>>>> it at the moment. The idea is that a function (x,a) -> Maybe b
> > >>>>>> may
> > >>>>>> decide for Nothing depending on both x and a, and therefore the
> > >>>>>> image
> > >>>>>> function under the isomorphism must map every a to Just (g b)
> > >>>>>> and
> > >>>>>> delay
> > >>>>>> the Nothing-decision to the g b. But for the reverse
> > >>>>>> isomorphism
> > >>>>>> you
> > >>>>>> can have functions that do not always return Just (g b) and
> > >>>>>> there
> > >>>>>> is no
> > >>>>>> preimage for these.