Implementing Applicative (<*>) for StateT - haskell

This question has been asked before, but without a real answer. In fact the accepted answer suggests that it is not possible, despite the fact that
StateT is a Monad, and hence a superset of Applicative. As a result, the standard libraries simply use (<*>) = ap
(as Petr notes) composing applicatives always yields an applicative.
One of the implementations of MaybeT I've read about used
liftA2 (<*>) :: (Applicative f, Applicative f1) => f (f1 (a -> b)) -> f (f1 a) -> f (f1 b)
to implement Applicative but I can't make that work here. My work in progress has tried lots of options around the following:
-- (<*>) :: StateT s f (a -> b) -> State s f a -> State s f b
instance (Applicative f) => Applicative (StateT s f) where
pure a = StateT $ \s -> pure (a, s)
(StateT f) <*> (StateT g) = StateT $ \s -> -- f :: s -> m (a -> b, s), g :: s -> m (a, s)
let
mabs = f s -- mabs :: m (a -> b, s)
mab = fmap fst mabs
ms' = fmap snd mabs
in undefined
I'm wondering what I am missing, and hoping that I will learn something about Applicative in the process.

Tony uses some alternative notation, and Simon's answer is very terse, so here is what I ended up with:
-- (<*>) :: StateT s f (a -> b) -> State s f a -> State s f b
instance (Monad f, Applicative f) => Applicative (StateT s f) where
pure a = StateT $ \s -> pure (a, s)
StateT f <*> StateT a =
StateT $ \s ->
f s >>= \(g, t) -> -- (f s) :: m (a->b, s)
let mapper = \(z, u) -> (g z, u) -- :: (a, s) -> (b, s)
in fmap mapper (a t) -- (a t) :: m (a, s)
I had to declare f also a Monad, but that is OK as it is part of the definition of a Monad transformer, as I understand it.

An implementation (taken from Tony Morris' functional programming course) could be
(<*>) :: (Functor f, Monad f) =>
StateT s f (a -> b)
-> StateT s f a
-> StateT s f b
StateT f <*> StateT a =
StateT (\s -> (\(g, t) -> (\(z, u) -> (g z, u)) <$> a t) =<< f s)

Related

Traversable#mapM & sequence - Necessary?

Typeclassopedia presents Traversable:
class (Functor t, Foldable t) => Traversable t where
traverse :: Applicative f => (a -> f b) -> t a -> f (t b)
sequenceA :: Applicative f => t (f a) -> f (t a)
mapM :: Monad m => (a -> m b) -> t a -> m (t b)
sequence :: Monad m => t (m a) -> m (t a)
A good exercise is to figure out what the default implementations should be: given either traverse or sequenceA, how would you define the other three methods?
I came up with the following code that type-checks:
class (Functor t, Foldable t) => MyTraversable t where
traverse' :: Applicative f => (a -> f b) -> t a -> f (t b)
traverse' = error "..."
sequenceA' :: Applicative f => t (f a) -> f (t a)
sequenceA' f = traverse' id f
mapM :: Monad m => (a -> m b) -> t a -> m (t b)
mapM = traverse'
sequence' :: Monad m => t (m a) -> m (t a)
sequence' = sequenceA'
If my implementations of mapM and sequence' are correct, and, since every Monad is an Applicative:
λ: :i Monad
class Applicative m => Monad (m :: * -> *) where
...
then it's not clear to me why mapM and sequence' are even necessary. Why are they?
P.S. - Credit and my thanks to haoformayor for helping me out with sequenceA.
As mentioned in the comments, there was a time when not every Monad was an Applicative.
Just recently (in base-4.8 which shipped with GHC 7.10) the Applicative was made a superclass of Monad. You might have seen Applicative-Monad-Proposal, AMP.
Currently the Monad of no return, MRP (email-thread) is worked on, which will make
Monad's >> and Applicative's *> the same. Which eventlually would make possible to say that traverse = mapM and sequence = sequenceA. But that is a long process, which will take time.
So to answer your question: for history reasons.

Implementing Applicative's (<*>) for Monad

Applicative's has the (<*>) function:
(<*>) :: (Applicative f) => f (a -> b) -> f a -> f b
Learn You a Haskell shows the following function.
Given:
ap :: (Monad m) => m (a -> b) -> m a -> m b
ap f m = do
g <- f -- '<-' extracts f's (a -> b) from m (a -> b)
m2 <- m -- '<-' extracts a from m a
return (g m2) -- g m2 has type `b` and return makes it a Monad
How could ap be written with bind alone, i.e. >>=?
I'm not sure how to extract the (a -> b) from m (a -> b). Perhaps once I understand how <- works in do notation, I'll understand the answer to my above question.
How could ap be written with bind alone, i.e. >>= ?
This is one sample implementation I can come up with:
ap :: (Monad m) => m (a -> b) -> m a -> m b
ap xs a = xs >>= (\f -> liftM f a)
Of if you don't want to even use liftM then:
ap :: (Monad m) => m (a -> b) -> m a -> m b
ap mf ma = mf >>= (\f -> ma >>= (\a' -> return $ f a'))
Intially these are the types:
mf :: m (a -> b)
ma :: m a
Now, when you apply bind (>>=) operator to mf: mf >>= (\f-> ..., then f has the type of:
f :: (a -> b)
In the next step, ma is also applied with >>=: ma >>= (\a'-> ..., here a' has the type of:
a' :: a
So, now when you apply f a', you get the type b from that because:
f :: (a -> b)
a' :: a
f a' :: b
And you apply return over f a' which will wrap it with the monadic layer and hence the final type you get will be:
return (f a') :: m b
And hence everything typechecks.

Monad vs Applicative in the case of Reader resp. ((->) a)

In Haskell, what does the monad instance of functions give over just applicative? Looking at their implementations, they seem almost identical:
(<*>) f g x = f x (g x)
(>>=) f g x = g (f x) x
Is there anything you can do with >>= that you can't do with just <*>?
They are equivalent in power for the function instance: flip f <*> g == g >>= f. This is not true for most types that are instances of Monad though.
It's a little more clear if we compare <*> and =<< (which is flip (>>=)) specialized to the ((->) r) instance:
(<*>) :: Applicative f => f (a -> b) -> f a -> f b
-- Specialized to ((->) r):
(<*>) :: (r -> a -> b) -> (r -> a) -> r -> b
(=<<) :: Monad m => (a -> m b) -> m a -> m b
-- Specialized to ((->) r):
(=<<) :: (a -> r -> b) -> (r -> a) -> r -> b

How to implement this Traversable use pattern?

When use Data.Traversable I frequently requires some code like
import Control.Applicative (Applicative,(<*>),pure)
import Data.Traversable (Traversable,traverse,sequenceA)
import Control.Monad.State (state,runState)
traverseF :: Traversable t => ((a,s) -> (b,s)) -> (t a, s) -> (t b, s)
traverseF f (t,s) = runState (traverse (state.curry f) t) s
to traverse the structure and build up a new one driven by some state. And I notice the type signature pattern and believe it could be able to generalized as
fmapInner :: (Applicative f,Traversable t) => (f a -> f b) -> f (t a) -> f (t b)
fmapInner f t = ???
But I fail to implement this with just traverse, sequenceA, fmap, <*> and pure. Maybe I need stronger type class constrain? Do I absolutely need a Monad here?
UPDATE
Specifically, I want to know if I can define fmapInner for a f that work for any Traversable t and some laws for intuition applied (I don't know what the laws should be yet), is it imply that the f thing is a Monad? Since, for Monads the implementation is trivial:
--Monad m implies Applicative m but we still
-- have to say it unless we use mapM instead
fmapInner :: (Monad m,Traversable t) => (m a -> m b) -> m (t a) -> m (t b)
fmapInner f t = t >>= Data.Traversable.mapM (\a -> f (return a))
UPDATE
Thanks for the excellent answer. I have found that my traverseF is just
import Data.Traversable (mapAccumL)
traverseF1 :: Traversable t => ((a, b) -> (a, c)) -> (a, t b) -> (a, t c)
traverseF1 =uncurry.mapAccumL.curry
without using Monad.State explicitly and have all pairs flipped. Previously I though it was mapAccumR but it is actually mapAccumL that works like traverseF.
I've now convinced myself that this is impossible. Here's why,
tF ::(Applicative f, Traversable t) => (f a -> f b) -> f (t a) -> f (t b)
So we have this side-effecting computation that returns t a and we want to use this to determine what side effects happen. In other words, the value of type t a will determine what side effects happen when we apply traverse.
However this isn't possible possible with the applicative type class. We can dynamically choose values, but the side effects of out computations are static. To see what I mean,
pure :: a -> f a -- No side effects
(<*>) :: f (a -> b) -> f a -> f b -- The side effects of `f a` can't
-- decide based on `f (a -> b)`.
Now there are two conceivable ways to determine side effects at depending on previous values,
smash :: f (f a) -> f a
Because then we can simply do
smash $ (f :: a -> f a) <$> (fa :: f a) :: f a
Now your function becomes
traverseF f t = smash $ traverse (f . pure) <$> t
Or we can have
bind :: m a -> (a -> m b) -> m b -- and it's obvious how `a -> m b`
-- can choose side effects.
and
traverseF f t = bind t (traverse $ f . pure)
But these are join and >>= respectively and are members of the Monad typeclass. So yes, you need a monad. :(
Also, a nice, pointfree implementation of your function with monad constraints is
traverseM = (=<<) . mapM . (.return)
Edit,
I suppose it's worth noting that
traverseF :: (Applicative f,Traversable t) => (f a -> f b) -> t a -> f (t a)
traverseF = traverse . (.pure)

how to implement mapAccumM?

I want mapM over something that is traversable while passing an accumulator. I came up with:
import Control.Applicative
import Data.Traversable
import Control.Monad.State
mapAccumM :: (Applicative m, Traversable t, MonadState s m)
=> (s -> a -> m (s, b)) -> s -> t a -> m (t b)
mapAccumM f acc0 xs = put acc0 >> traverse g xs
where
g x = do
oldAcc <- get
(newAcc, y) <- f oldAcc x
put newAcc
return y
How can this be done without State monad?
roconnor answered this for me on #haskell
this solves my problem but notice that accumulator is returned in the second element of the tuple instead of the first
mapAccumM :: (Monad m, Functor m, Traversable t) => (a -> b -> m (c, a)) -> a -> t b -> m (t c)
mapAccumM f = flip (evalStateT . (Data.Traversable.traverse (StateT . (flip f))))
or to also return the accumulator:
mapAccumM' :: (Monad m, Functor m, Traversable t) => (a -> b -> m (c, a)) -> a -> t b -> m (t c, a)
mapAccumM' f = flip (runStateT . (Data.Traversable.traverse (StateT . (flip f))))

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