haskellfunctional-programmingmonadsarrow-abstraction

Can someone explain to me why the app function of ArrowApply makes them as powerful as monads?


So I'll break my question into 4 parts, but first some background:

I feel relatively comfortable with Monads, but not very comfortable with Arrows. I suppose the main problem I have with them is, I don't see what they are useful for. Whether formally correct or not, I understand Monads to be a tool that allows us to introduce side effects from computation. As they generalize program fragments from pure values to values boxed with other actions. From my shotgun "read all the papers" approach to learning about arrows, I've come across two conflicting viewpoints:

A. Arrows are more powerful than Monads/ are generalizations of Monads. The haskell wiki starts off with "They can do everything monads can do, and more. They are roughly comparable to monads with a static component."

B. Arrows are a subset of Monads With ArrowApply we can define a monad

  1. Is there any truth to viewpoint A?
  2. What kind of functionality do arrows not have, I've read that the difference has to do with composition, so what does the >>> operator allow us to do that >>= doesn't?
  3. What does app exactly do? it's type doesn't even have an (->)
  4. Why would we ever want to use applicative arrows over monads?

Solution

  • Multiply-interpretable statements alert:

    "A is more powerful than B" ... "C is a generalisation of D" ... "E can do everything F can do, and more" ... "G is a subset of H" ...

    First we need to get a grip on what we mean by powerful etc. Suppose we had a class GripHandle for things which had a grip handle, and another class Screwdriver for screwdrivers. Which is more powerful?

    OK, that's a silly argument, but it raises a good point about how the phrase "You can do more with a _____" is rather ambiguous.

    If you stick to the interface alone, a screwdriver is more useful than a handle, but all things with griphandles are together more useful than all screwdrivers if you use more functions than just the interface.

    How hierarchy works

    A is more general than B
    = A's interface-only capabilities are a subset of B's = you can do more with a B instance (alone)
    = The class of all Bs is a subset of the class of all As = there are more As than Bs = you can do more with the A class

    more general
    = more possible instances
    = able to be more widely used
    = can do extra things behind the scenes
    = fewer capabilities are specified in the interface
    = can do fewer things via the interface

    What's the hierarchy between arrows and monads?

    These two statements are proved in full detail in the paper Petr Pudlák linked to in his comment: Idioms are oblivious, arrows are meticulous, monads are promiscuous.

    The assertions in A and B

    Your four questions

    1. Is there any truth to viewpoint A?
      Some. See above. There's some truth in B too. Both are misleading in some way or other.
    2. What kind of functionality do arrows not have, I've read that the difference has to do with composition, so what does the >>> operator allow us to do that >>= doesn't?
      In fact >>= allows you do to more than >>> (more interface-supplied capability). It allows you to context-switch. This is because Monad m => a -> m b is a function, so you can execute arbitrary pure code on the input a before deciding which monadic thing to run, whereas Arrow m => m a b isn't a function, and you've decided which arrow thing is going to run before you examined the input a.

      monadSwitch :: Monad m => m a -> m a -> (Bool -> m a)
      monadSwitch computation1 computation2 test 
            = if test then computation1 else computation2
      

      It's not possible to simulate this using Arrow without using app from ArrowApply

    3. What does app exactly do? it's type doesn't even have an (->)
      It lets you use the output of an arrow as an arrow. Let's look at the type.

      app :: ArrowApply m => m (m b c, b) c
      

      I prefer to use m to a because m feels more like a computation and a feels like a value. Some people like to use a type operator (infix type constructor), so you get

      app :: ArrowApply (~>) => (b ~> c, b) ~> c
      

      We think of b ~> c as an arrow, and we think of an arrow as a thing which takes bs, does something and gives cs. So this means app is an arrow that takes an arrow and a value, and can produce the value that the first arrow would have produced on that input.

      It doesn't have -> in the type signature because when programming with arrows, we can turn any function into an arrow using arr :: Arrow (~>) => (b -> c) -> b ~> c, but you can't turn every arrow into a function, thus (b ~> c, b) ~> c is usable where (b ~> c, b) -> c or (b -> c, b) ~> c would not be.

      We can easily make an arrow that produces an arrow or even multiple arrows, even without ArrowApply, just by doing produceArrow :: Arrow (~>) => (b ~> c) -> (any ~> (b ~> c)) defined with produceArrow a = arr (const a). The difficulty is in making that arrow do any arrow work - how to you get an arrow that you produced to be the next arrow? You can't pop it in as the next computation using >>> like you can do with a monadic function Monad m => a -> m b (just do id :: m a -> m a!), because, crucially, arrows aren't functions, but using app, we can make the next arrow do whatever the arrow produced by the previous arrow would have done.

      Thus ArrowApply gives you the runtime-generated computation runnability that you have from Monad.

    4. Why would we ever want to use applicative arrows over monads?
      Er, do you mean Arrows or Applicative Functors? Applicative Functors are great. They're more general than either Monad or Arrow (see the paper) so have less interface-specified functionality, but are more widely applicable (get it? applicable/applicative chortle chortle lol rofl category theory humor hahahaha).

      Applicative Functors have a lovely syntax that looks very like pure function application. f <$> ma <*> mb <*> mc runs ma then mb then mc and applies the pure function f to the three results. For example. (+) <$> readLn <*> readLn reads two integers from the user and adds them.

      You can use Applicative to get the generality, and you can use Monads to get the interface-functionality, so you could argue that theoretically we don't need them, but some people like the notation for arrows because it's like do notation, and you can indeed use Arrow to implement parsers that have a static component, thus apply compile-time optimisations. I believe you can do that with Applicative, but it was done with Arrow first.

      A note about Applicative being "less powerful":
      The paper points out that Applicative is more general than Monad, but you could make Applicative functors have the same abilities by providing a function run :: Applicative f => f (f b) -> f b that lets you run a produced computation, or use :: Applicative f => f (a -> f b) -> f a -> f b that allows you to promote a produced computation to a computation. If we define join = run and unit = (<$>) we get the two functions that make one theoretical basis for Monads, and if we define (>>=) = flip (use.pure) and return = unit we get the other one that's used in Haskell. There isn't an ApplicativeRun class, simply because if you can make that, you can make a monad, and the type signatures are almost identical. The only reason we have ArrowApply instead of reusing Monad is that the types aren't identical; ~> is abstracted (generalised) into the interface in ArrowApply but function application -> is used directly in Monad. This distinction is what makes programming with Arrows feel different in many ways to programming in monads, despite the equivalence of ArrowApply and Monad.

    5. < cough > Why would we ever want to use Arrows/ArrowApply over Monad?
      OK, I admit I knew that's what you meant, but wanted to talk about Applicative functors and got so carried away I forgot to answer!

      Capability reasons: Yes, you would want to use Arrow over Monad if you had something that can't be made into a monad. The motivating example that brought us Arrows in the first place was parsers - you can use Arrow to write a parser library that does static analysis in the combinators, thus making more efficient parsers. The previous Monadic parsers can't do this because they represent a parser as a function, which can do arbitrary things to the input without recording them statically, so you can't analyse them at compile time/combine time.

      Syntactic reasons: No, I personally wouldn't want to use Arrow based parsers, because I dislike the arrow proc/do notation - I find it even worse than the monadic notation. My preferred notation for parsers is Applicative, and you might be able to write an Applicative parser library that does the efficient static analysis that the Arrow one does, although I freely admit that the parser libraries I commonly use don't, possibly because they want to supply a Monadic interface.

      • Monad:

            parseTerm = do
                 x <- parseSubterm
                 o <- parseOperator
                 y <- parseSubterm
                 return $ Term x o y
        
      • Arrow:

            parseTerm = proc _ -> do
                 x <- parseSubterm -< ()
                 o <- parseOperator -< ()
                 y <- parseSubterm -< ()
                 returnA -< Term x o y
        
      • Applicative:

            parseTerm = Term <$> parseSubterm <*> parseOperator <*> parseSubterm
        

        That just looks like function application using $ a few times. Mmmmm. Neat. Clear. Low syntax. Reminds me why I prefer Haskell to any imperative programming language.

    Why does app in ArrowApply make a Monad?

    There's a Monad instance in the ArrowApply section of the Control.Arrow module, and I'll edit in (~>) instead of a for my clarity of thought. (I've left Functor in because it's silly to define Monad without Functor anyway - you should define fmap f xs = xs >>= return . f.):

    newtype ArrowMonad (~>) b = ArrowMonad (() ~> b)
    
    instance Arrow (~>) => Functor (ArrowMonad (~>)) where
        fmap f (ArrowMonad m) = ArrowMonad $ m >>> arr f
    
    instance ArrowApply (~>) => Monad (ArrowMonad (~>)) where
        return x = ArrowMonad (arr (\_ -> x))
        ArrowMonad m >>= f = ArrowMonad $
            m >>> arr (\x -> let ArrowMonad h = f x in (h, ())) >>> app
    

    What does that do? Well, first, ArrowMonad is a newtype instead of a type synonym just so we can make the instance without all sorts of nasty type system problems, but lets ignore that to go for conceptual clarity over compilability by substituting in as if it were type ArrowMonad (~>) b = () ~> b

    instance Arrow (~>) => Functor (() ~>) where
        fmap f m = m >>> arr f
    

    (using an uncompilable type operator section (()~>) as a type constructor)

    instance ArrowApply (~>) => Monad (() ~>) where
     -- return :: b -> (() ~> b)
        return x = arr (\_ -> x)
     -- (>>=) ::   ()~>a   ->    (a  ->  ()~>b )   ->   ()~>b 
        m >>= f = 
            m >>> arr (\x ->  (f x, ()) ) >>> app
    

    OK, that's a bit clearer what's going on. Notice first that the correspondence between arrows and monads is between Monad m => b -> m c and Arrow (~>) => b ~> c, but the monad class doesn't involve the b in the declaration. That's why we need to supply the dummy value () in () ~> b to get things started on zero input and replicate something of type m b.