[Haskell-cafe] Foldable/Traversable and Applicative/Monoid?
David Feuer
david.feuer at gmail.com
Sun Feb 7 04:26:00 UTC 2016
It's not terribly unusual. Functor can be a superclass of Applicative
because
fmap f xs = pure f <*> xs
Applicative can be a superclass of Monad because
pure = return
(<*>) = ap
Distributive can be a superclass of Representable because
distribute wf = tabulate (\k -> fmap (`index` k) wf)
Obviously, it often *doesn't* work like this. The class structure may be
arranged as it is because the subclass conceptually or practically
represents a refinement of the superclass. But when the methods of a given
class can be implemented using the methods of another, that suggests that
it *might* make sense for it to be a superclass.
On Feb 6, 2016 10:10 AM, "David Banas" <capn.freako at gmail.com> wrote:
> Hi David,
>
> Thanks for your reply!
>
> That’s really interesting; I never would have thought to try and implement
> super-class member functions, in terms of sub-class member functions.
> I was trying to go the other way: implement sequenceA, in terms of
> foldMap, which seemed to require a completely generic way of turning an
> Applicative (guaranteed by the type signature of sequenceA) into a Monoid
> (required by foldMap). I came up with this:
>
>
> {-# LANGUAGE Rank2Types
> FlexibleContexts
> UndecidableInstances
> AllowAmbiguousTypes
> #-}
>
> newtype MonApp = MonApp {getApp :: (Applicative f, Monoid a) => f a}
>
> instance Monoid MonApp where
> mempty = MonApp $ pure mempty
> mappend ma1 ma2 = MonApp $ mappend <$> (getApp ma1) <*> (getApp ma2)
>
> instance (Monoid a) => Monoid (Tree a) where
> mempty = Empty
> mappend Empty t = t
> mappend t Empty = t
> mappend (Leaf x) (Leaf y) = Leaf (x `mappend` y)
> mappend (Leaf x) (Node t1 y t2) = Node t1 (x `mappend` y) t2
> mappend (Node t1 y t2) (Leaf x) = Node t1 (y `mappend` x) t2
> mappend (Node t1 x t2) (Node t3 y t4) = Node (t1 `mappend` t3) (x
> `mappend` y) (t2 `mappend` t4)
>
> instance Monoid (Tree a) => Traversable Tree where
> sequenceA = getApp . foldMap (MonApp . (fmap Leaf))
>
>
> to which the compiler responded:
>
>
> Couldn't match type ‘f (Tree a1)’ with ‘forall (f1 :: * -> *) a2.
> (Applicative f1, Monoid a2) => f1 a2’
> Expected type: f (Tree a1) -> interactive:IHaskell161.MonApp
> Actual type: (forall (f :: * -> *) a. (Applicative f, Monoid a) => f a) ->
> interactive:IHaskell161.MonApp
> Relevant bindings include sequenceA :: Tree (f a1) -> f (Tree a1) (bound
> at :14:3)
> In the first argument of ‘(.)’, namely ‘IHaskell161.MonApp’
> In the first argument of ‘foldMap’, namely
> ‘(interactive:IHaskell161.MonApp . (fmap Leaf))’
>
>
> -db
>
>
> On Feb 5, 2016, at 11:20 AM, David Feuer <david.feuer at gmail.com> wrote:
>
> It's not so much that it's *necessary* as that it's *possible*. The
> existence of two functions in Data.Traversable explains both of the
> superclasses of Traversable:
>
> fmapDefault :: Traversable t => (a -> b) -> t a -> t b
>
> foldMapDefault :: (Traversable t, Monoid m) => (a -> m) -> t a -> m
>
> Each of these is written using only traverse, and they can be used to
> define fmap and foldMap for types when you've written traverse.
>
> Hint: Consider traversing using the following applicative functors:
>
> newtype Const a b = Const a
> instance Monoid a => Applicative (Const a)
>
> newtype Identity a = Identity a
> instance Applicative Identity
> On Feb 5, 2016 1:45 PM, "David Banas" <capn.freako at gmail.com> wrote:
>
>> Hi all,
>>
>> I don't understand why Foldable is a necessary super-class of
>> Traversable, and I suspect that the Applicative/Monoid duality, which I've
>> just begun discovering in the literature, has something to do with why that
>> is so.
>>
>> Can anyone give me a hint, without giving me the answer?
>>
>> Thanks!
>> -db
>>
>>
>> _______________________________________________
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>> Haskell-Cafe at haskell.org
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
>>
>>
>
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