Monad theory and Haskell

Question

Most tutorials seem to give a lot of examples of monads (IO, state, list and so on) and then expect the reader to be able to abstract the overall principle and then they mention category theory. I don't tend to learn very well by trying generalise from examples and I would like to understand from a theoretical point of view why this pattern is so important.

Judging from this thread: Can anyone explain Monads? this is a common problem, and I've tried looking at most of the tutorials suggested (except the Brian Beck videos which won't play on my linux machine):

Does anyone know of a tutorial that starts from category theory and explains IO, state, list monads in those terms? the following is my unsuccessful attempt to do so:

As I understand it a monad consists of a triple: an endo-functor and two natural transformations.

The functor is usually shown with the type: (a -> b) -> (m a -> m b) I included the second bracket just to emphasise the symmetry.

But, this is an endofunctor, so shouldn't the domain and codomain be the same like this?:

(a -> b) -> (a -> b)

I think the answer is that the domain and codomain both have a type of:

(a -> b) | (m a -> m b) | (m m a -> m m b) and so on ...

But I'm not really sure if that works or fits in with the definition of the functor given?

When we move on to the natural transformation it gets even worse. If I understand correctly a natural transformation is a second order functor (with certain rules) that is a functor from one functor to another one. So since we have defined the functor above the general type of the natural transformations would be: ((a -> b) -> (m a -> m b)) -> ((a -> b) -> (m a -> m b))

But the actual natural transformations we are using have type:

a -> m a

m a -> (a ->m b) -> m b

Are these subsets of the general form above? and why are they natural transformations?

Martin

Answer 1

A quick disclaimer: I'm a little shaky on category theory in general, while I get the impression you have at least some familiarity with it. Hopefully I won't make too much of a hash of this...

Does anyone know of a tutorial that starts from category theory and explains IO, state, list monads in those terms?

First of all, ignore IO for now, it's full of dark magic. It works as a model of imperative computations for the same reasons that State works for modelling stateful computations, but unlike the latter IO is a black box with no way to deduce the monadic structure from the outside.

The functor is usually shown with the type: (a -> b) -> (m a -> m b) I included the second bracket just to emphasise the symmetry.

But, this is an endofunctor, so shouldn't the domain and codomain be the same like this?:

I suspect you are misinterpreting how type variables in Haskell relate to the category theory concepts.

First of all, yes, that specifies an endofunctor, on the category of Haskell types. A type variable such as a is not anything in this category, however; it's a variable that is (implicitly) universally quantified over all objects in the category. Thus, the type (a -> b) -> (a -> b) describes only endofunctors that map every object to itself.

Type constructors describe an injective function on objects, where the elements of the constructor's codomain cannot be described by any means except as an application of the type constructor. Even if two type constructors produce isomorphic results, the resulting types remain distinct. Note that type constructors are not, in the general case, functors.

The type variable m in the functor signature, then, represents a one-argument type constructor. Out of context this would normally be read as universal quantification, but that's incorrect in this case since no such function can exist. Rather, the type class definition binds m and allows the definition of such functions for specific type constructors.

The resulting function, then, says that for any type constructor m which has fmap defined, for any two objects a and b and a morphism between them, we can find a morphism between the types given by applying m to a and b.

Note that while the above does, of course, define an endofunctor on Hask, it is not even remotely general enough to describe all such endofunctors.

But the actual natural transformations we are using have type:

a -> m a

m a -> (a ->m b) -> m b

Are these subsets of the general form above? and why are they natural transformations?

Well, no, they aren't. A natural transformation is roughly a function (not a functor) between functors. The two natural transformations that specify a monad M look like I -> M where I is the identity functor, and M ∘ M -> M where ∘ is functor composition. In Haskell, we have no good way of working directly with either a true identity functor or with functor composition. Instead, we discard the identity functor to get just (Functor m) => a -> m a for the first, and write out nested type constructor application as (Functor m) => m (m a) -> m a for the second.

The first of these is obviously return; the second is a function called join, which is not part of the type class. However, join can be written in terms of (>>=), and the latter is more often useful in day-to-day programming.

---

As far as specific monads go, if you want a more mathematical description, here's a quick sketch of an example:

For some fixed type S, consider two functors F and G where F(x) = (S, x) and G(x) = S -> x (It should hopefully be obvious that these are indeed valid functors).

These functors are also adjoints; consider natural transformations unit :: x -> G (F x) and counit :: F (G x) -> x. Expanding the definitions gives us unit :: x -> (S -> (S, x)) and counit :: (S, S -> x) -> x. The types suggest uncurried function application and tuple construction; feel free to verify that those work as expected.

An adjunction gives rise to a monad by composition of the functors, so taking G ∘ F and expanding the definition, we get G (F x) = S -> (S, x), which is the definition of the State monad. The unit for the adjunction is of course return; and you should be able to use counit to define join.

Answer 2

This page does exactly that. I think your main confusion is that the class doesn't really make the Type a functor, but it defines a functor from the category of Haskell types into the category of that type.

Following the notation of the link, assuming F is a Haskell Functor, it means that there is a functor from the category of Hask to the category of F.

Answer 3

Roughly speaking, Haskell does its category theory all in just one category, whose objects are Haskell types and whose arrows are functions between these types. It's definitely not a general-purpose language for modelling category theory.

A (mathematical) functor is an operation turning things in one category into things in another, possibly entirely different, category. An endofunctor is then a functor which happens to have the same source and target categories. In Haskell, a functor is an operation turning things in the category of Haskell types into other things also in the category of Haskell types, so it is always an endofunctor.

[If you're following the mathematical literature, technically, the operation '(a->b)->(m a -> m b)' is just the arrow part of the endofunctor m, and 'm' is the object part]

When Haskellers talk about working 'in a monad' they really mean working in the Kleisli category of the monad. The Kleisli category of a monad is a thoroughly confusing beast at first, and normally needs at least two colours of ink to give a good explanation, so take the following attempt for what it is and check out some references (unfortunately Wikipedia is useless here for all but the straight definitions).

Suppose you have a monad 'm' on the category C of Haskell types. Its Kleisli category Kl(m) has the same objects as C, namely Haskell types, but an arrow a ~(f)~> b in Kl(m) is an arrow a -(f)-> mb in C. (I've used a squiggly line in my Kleisli arrow to distinguish the two). To reiterate: the objects and arrows of the Kl(C) are also objects and arrows of C but the arrows point to different objects in Kl(C) than in C. If this doesn't strike you as odd, read it again more carefully!

Concretely, consider the Maybe monad. Its Kleisli category is just the collection of Haskell types, and its arrows a ~(f)~> b are functions a -(f)-> Maybe b. Or consider the (State s) monad whose arrows a ~(f)~> b are functions a -(f)-> (State s b) == a -(f)-> (s->(s,b)). In any case, you're always writing a squiggly arrow as a shorthand for doing something to the type of the codomain of your functions.

[Note that State is not a monad, because the kind of State is * -> * -> *, so you need to supply one of the type parameters to turn it into a mathematical monad.]

So far so good, hopefully, but suppose you want to compose arrows a ~(f)~> b and b ~(g)~> c. These are really Haskell functions a -(f)-> mb and b -(g)-> mc which you cannot compose because the types don't match. The mathematical solution is to use the 'multiplication' natural transformation u:mm->m of the monad as follows: a ~(f)~> b ~(g)~> c == a -(f)-> mb -(mg)-> mmc -(u_c)-> mc to get an arrow a->mc which is a Kleisli arrow a ~(f;g)~> c as required.

Perhaps a concrete example helps here. In the Maybe monad, you cannot compose functions f : a -> Maybe b and g : b -> Maybe c directly, but by lifting g to

Maybe_g :: Maybe b -> Maybe (Maybe c)
Maybe_g Nothing = Nothing
Maybe_g (Just a) = Just (g a)

and using the 'obvious'

u :: Maybe (Maybe c) -> Maybe c
u Nothing = Nothing
u (Just Nothing) = Nothing
u (Just (Just c)) = Just c

you can form the composition u . Maybe_g . f which is the function a -> Maybe c that you wanted.

In the (State s) monad, it's similar but messier: Given two monadic functions a ~(f)~> b and b ~(g)~> c which are really a -(f)-> (s->(s,b)) and b -(g)-> (s->(s,c)) under the hood, you compose them by lifting g into

State_s_g :: (s->(s,b)) -> (s->(s,(s->(s,c))))
State_s_g p s1 = let (s2, b) = p s1 in (s2, g b)

then you apply the 'multiplication' natural transformation u, which is

u :: (s->(s,(s->(s,c)))) -> (s->(s,c))
u p1 s1 = let (s2, p2) = p1 s1 in p2 s2

which (sort of) plugs the final state of f into the initial state of g.

In Haskell, this turns out to be a bit of an unnatural way to work so instead there's the (>>=) function which basically does the same thing as u but in a way that makes it easier to implement and use. This is important: (>>=) is not the natural transformation 'u'. You can define each in terms of the other, so they're equivalent, but they're not the same thing. The Haskell version of 'u' is written join.

The other thing missing from this definition of Kleisli categories is the identity on each object: a ~(1_a)~> a which is really a -(n_a)-> ma where n is the 'unit' natural transformation. This is written return in Haskell, and doesn't seem to cause as much confusion.

I learned category theory before I came to Haskell, and I too have had difficulty with the mismatch between what mathematicians call a monad and what they look like in Haskell. It's no easier from the other direction!

Answer 4

Not sure I understand what was the question but yes, you are right, monad in Haskell is defined as a triple:

m :: * -> * -- this is endofunctor from haskell types to haskell types!
return :: a -> m a
(>>=) :: m a -> (a -> m b) -> m b

but common definition from category theory is another triple:

m :: * -> *
return :: a -> m a
join ::  m (m a) -> m a

It is slightly confusing but it's not so hard to show that these two definitions are equal. To do that we need to define join in terms of (>>=) (and vice versa). First step:

join :: m (m a) -> m a
join x = ?

This gives us x :: m (m a).
All we can do with something that have type m _ is to aply (>>=) to it:

(x >>=) :: (m a -> m b) -> m b

Now we need something as a second argument for (>>=), and also, from the type of join we have constraint (x >>= y) :: ma.
So y here will have type y :: ma -> ma and id :: a -> a fits it very well:

join x = x >>= id

The other way

(>>=) :: ma -> (a -> mb) -> m b
(>>=) x y = ?

Where x :: m a and y :: a -> m b. To get m b from x and y we need something of type a.
Unfortunately, we can't extract a from m a. But we can substitute it for something else (remember, monad is a functor also):

fmap :: (a -> b) -> m a -> m b
fmap y x :: m (m b)

And it's perfectly fits as argument for join: (>>=) x y = join (fmap y x).

Answer 5

The best way to look at monads and computational effects is to start with where Haskell got the notion of monads for computational effects from, and then look at Haskell after you understand that. See this paper in particular: Notions of Computation and Monads, by E. Moggi.

See also Moggi's earlier paper which shows how monads work for the lambda calculus alone: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.26.2787

The fact that monads capture substitution, among other things (http://blog.sigfpe.com/2009/12/where-do-monads-come-from.html), and substitution is key to the lambda calculus, should give a good clue as to why they have so much expressive power.

Answer 6

While monads originally came from category theory, this doesn't mean that category theory is the only abstract context in which you can view them. A different viewpoint is given by operational semantics. For an introduction, have a look at my Operational Monad Tutorial.

Answer 7

One way to look at IO is to consider it as a strange kind of state monad. Remember that the state monad looks like:

data State s a = State (s -> (s, a))

where the "s" argument is the data type you want to thread through the computation. Also, this version of "State" doesn't have "get" and "put" actions and we don't export the constructor.

Now imagine a type

data RealWorld = RealWorld ......

This has no real definition, but notionally a value of type RealWorld holds the state of the entire universe outside the computer. Of course we can never have a value of type RealWorld, but you can imagine something like:

getChar :: RealWorld -> (RealWorld, Char)

In other words the "getChar" function takes a state of the universe before the keyboard button has been pressed, and returns the key pressed plus the state of the universe after the key has been pressed. Of course the problem is that the previous state of the world is still available to be referenced, which can't happen in reality.

But now we write this:

type IO = State RealWorld

getChar :: IO Char

Notionally, all we have done is wrap the previous version of "getChar" as a state action. But by doing this we can no longer access the "RealWorld" values because they are wrapped up inside the State monad.

So when a Haskell program wants to change a lightbulb it takes hold of the bulb and applies a "rotate" function to the RealWorld value inside IO.

Answer 8

For me, so far, the explanation that comes closest to tie together monads in category theory and monads in Haskell is that monads are a monid whose objects have the type a->m b. I can see that these objects are very close to an endofunctor and so the composition of such functions are related to an imperative sequence of program statements. Also functions which return IO functions are valid in pure functional code until the inner function is called from outside.

This id element is 'a -> m a' which fits in very well but the multiplication element is function composition which should be:

(>=>) :: Monad m => (a -> m b) -> (b -> m c) -> (a -> m c)

This is not quite function composition, but close enough (I think to get true function composition we need a complementary function which turns m b back into a, then we get function composition if we apply these in pairs?), I'm not quite sure how to get from that to this:

(>>=) :: Monad m => m a -> (a -> m b) -> m b

I've got a feeling I may have seen an explanation of this in all the stuff that I read, without understanding its significance the first time through, so I will do some re-reading to try to (re)find an explanation of this.

The other thing I would like to do is link together all the different category theory explanations: endofunctor+2 natural transformations, Kleisli category, a monoid whose objects are monoids and so on. For me the thing that seems to link all these explanations is that they are two level. That is, normally we treat category objects as black-boxes where we imply their properties from their outside interactions, but here there seems to be a need to go one level inside the objects to see what’s going on? We can explain monads without this but only if we accept apparently arbitrary constructions.

Martin

Answer 9

See this question: is chaining operations the only thing that the monad class solves?

In it, I explain my vision that we must differentiate between the Monad class and individual types that solve individual problems. The Monad class, by itself, only solve the important problem of "chaining operations with choice" and mades this solution available to types being instance of it (by means of "inheritance").

On the other hand, if a given type that solves a given problem faces the problem of "chaining operations with choice" then, it should be made an instance (inherit) of the Monad class.

The fact is that problems not get solved merely by being a Monad. It would be like saying that "wheels" solve many problems, but actually "wheels" only solve a problem, and things with wheels solve many different problems.