Is there a monad that doesn't have a corresponding monad transformer (except IO)?

问题

So far, every monad (that can be represented as a data type) that I have encountered had a corresponding monad transformer, or could have one. Is there such a monad that can't have one? Or do all monads have a corresponding transformer?

By a transformer t corresponding to monad m I mean that t Identity is isomorphic to m. And of course that it satisfies the monad transformer laws and that t n is a monad for any monad n.

I'd like to see either a proof (ideally a constructive one) that every monad has one, or an example of a particular monad that doesn't have one (with a proof). I'm interested in both more Haskell-oriented answers, as well as (category) theoretical ones.

As a follow-up question, is there a monad m that has two distinct transformers t1 and t2? That is, t1 Identity is isomorphic to t2 Identity and to m, but there is a monad n such that t1 n is not isomorphic to t2 n.

(IO and ST have a special semantics so I don't take them into account here and let's disregard them completely. Let's focus only on "pure" monads that can be constructed using data types.)

回答1:

I'm with @Rhymoid on this one, I believe all Monads have two (!!) transformers. My construction is a bit different, and far less complete. I'd like to be able to take this sketch into a proof, but I think I'm either missing the skills/intuition and/or it may be quite involved.

Due to Kleisli, every monad (m) can be decomposed into two functors F_k and G_k such that F_k is left adjoint to G_k and that m is isomorphic to G_k * F_k (here * is functor composition). Also, because of the adjunction, F_k * G_k forms a comonad.

I claim that t_mk defined such that t_mk n = G_k * n * F_k is a monad transformer. Clearly, t_mk Id = G_k * Id * F_k = G_k * F_k = m. Defining return for this functor is not difficult since F_k is a "pointed" functor, and defining join should be possible since extract from the comonad F_k * G_k can be used to reduce values of type (t_mk n * t_mk n) a = (G_k * n * F_k * G_k * n * F_k) a to values of type G_k * n * n * F_k, which is then further reduces via join from n.

We do have to be a bit careful since F_k and G_k are not endofunctors on Hask. So, they are not instances of the standard Functor typeclass, and also are not directly composable with n as shown above. Instead we have to "project" n into the Kleisli category before composition, but I believe return from m provides that "projection".

I believe you can also do this with the Eilenberg-Moore monad decomposition, giving m = G_em * F_em, tm_em n = G_em * n * F_em, and similar constructions for lift, return, and join with a similar dependency on extract from the comonad F_em * G_em.

回答2:

Here's a hand-wavy I'm-not-quite-sure answer.

Monads can be thought of as the interface of imperative languages. return is how you inject a pure value into the language, and >>= is how you splice pieces of the language together. The Monad laws ensure that "refactoring" pieces of the language works the way you would expect. Any additional actions provided by a monad can be thought of as its "operations."

Monad Transformers are one way to approach the "extensible effects" problem. If we have a Monad Transformer t which transforms a Monad m, then we could say that the language m is being extended with additional operations available via t. The Identity monad is the language with no effects/operations, so applying t to Identity will just get you a language with only the operations provided by t.

So if we think of Monads in terms of the "inject, splice, and other operations" model, then we can just reformulate them using the Free Monad Transformer. Even the IO monad could be turned into a transformer this way. The only catch is that you probably want some way to peel that layer off the transformer stack at some point, and the only sensible way to do it is if you have IO at the bottom of the stack so that you can just perform the operations there.

回答3:

Update: Statements 1 and 2 below are most likely incorrect; I think I found monad transformers for those cases. I have not yet finished the required calculations but they look promising. Statements 3 and 4 still stand. I added Statement 5 to show an example of a monad with two distinct transformers.

The transformer for Either a (z -> a) is (most likely) n (Either a (z -> m a), where m is an arbitrary foreign monad. The transformer for (a -> n p) -> n a is (most likely) (a -> t m p) -> t m a where t m is the transformer for the monad n.

1. I think I found at least one counterexample: a simple and explicit monad that has no simple and explicit monad transformer.

The monad type constructor L for this counterexample is defined by

  type L z a  = Either a (z -> a)

The intent of this monad is to embellish the ordinary reader monad z -> a with an explicit pure value (Left x). The ordinary reader monad's pure value is a constant function pure x = _ -> x. However, if we are given a value of type z -> a, we will not be able to determine whether this value is a constant function. With L z a, the pure value is represented explicitly as Left x. Users can now pattern-match on L z a and determine whether a given monadic value is pure or has an effect. Other than that, the monad L z does exactly the same thing as the reader monad.

The monad instance:

  instance Monad (L z) where
     return x = Left x
     (Left x) >>= f = f x
     (Right q) >>= f = Right(join merged) where
        join :: (z -> z -> r) -> z -> r
        join f x = f x x -- the standard `join` for Reader monad
        merged :: z -> z -> r
        merged = merge . f . q -- `f . q` is the `fmap` of the Reader monad
        merge :: Either a (z -> a) -> z -> a 
        merge (Left x) _ = x
        merge (Right p) z = p z

This monad L z is a specific case of a more general construction, (Monad m) => Monad (L m) where L m a = Either a (m a). This construction embellishes a given monad m by adding an explicit pure value (Left x), so that users can now pattern-match on L m to decide whether the value is pure. In all other ways, L m represents the same computational effect as the monad m.

The monad instance for L m is almost the same as for the example above, except the join and fmap of the monad m need to be used, and the helper function merge is defined by

    merge :: Either a (m a) -> m a
    merge (Left x) = return @m x
    merge (Right p) = p

I checked that the laws of the monad hold for L m with an arbitrary monad m.

So, I think L m has no monad transformer, either for general m or even for a simple monad m = Reader. It suffices to consider L z as defined above; even that simple monad does not seem to have a transformer.

The (heuristic) reason for the non-existence of a monad transformer is that this monad has a Reader inside an Either. The Either monad transformer needs its base monad to be composed inside the foreign monad, EitherT e m a = m (Either e a), because the monad transformer operates using the traversal. It appears that any monad that contains an Either in its data type will need a traversal for the monad transformer to work, and so there must be an "inside" composition in the transformer. However, the Reader monad transformer needs its base monad to be composed outside the foreign monad, ReaderT r m a = r -> m a. The monad L is a composition of a data type that demands a composed-inside transformer and a monad that demands a composed-outside transformer, and the second monad is inside the first one, which is impossible to reconcile. No matter how we try to define an L-transformer LT, it seems that we cannot satisfy the laws of the monad transformers.

One possibility of defining a type constructor LT would be LT z m a = Either a (z -> m a). The result is a lawful monad, but the morphism m a -> LT z m a does not preserve m's return because return x is mapped into a Right (\z -> return x), which is not the L's return (always a Left x).

Another possibility is LT z m a = z -> Either a (m a). The result is a monad, but again m's return is mapped into \_ -> Right (...) instead of the Left (...) as required for the monad z -> Either a (m a).

Yet another possibility of combining the available type constructors is LT z m a = Either a (m (z -> a) ), but this is not a monad for arbitrary monad m.

I am not sure how to prove rigorously that L has no monad transformer, but no combination of the type constructors Either, ->, and m seems to work correctly.

So, the monad L z and generally the monads of the form L m seem to have no simple and easy to use monad transformer that would be an explicit type constructor (a combination of Either, -> and m).

2. Another example of a monad that does not seem to have an explicit monad transformer:

type S a = (a -> Bool) -> Maybe a

This monad appeared in the context of "search monads" here. The paper by Jules Hedges also mentions the search monad, and more generally, "selection" monads of the form

 type Sq n q a = (a -> n q) -> n a

for a given monad n and a fixed type q. The search monad above is a particular case of the selection monad with n a = Maybe a and q = (). However, the paper by Hedges (in my view incorrectly) claims that Sq is a monad transformer for the monad (a -> q) -> a.

My opinion is that the monad (a -> q) -> a has the monad transformer (m a -> q) -> m a of the "composed outside" type. This is related to the property of "rigidity" explored in the question Is this property of a functor stronger than a monad? Namely, (a -> q) -> a is a rigid monad, and all rigid monads have monad transformers of the "composed-outside" type.

However, (a -> n q) -> n a is not rigid unless the monad n is rigid. Since not all monads are rigid (e.g. Maybe and Cont are not rigid), the monad (a -> n q) -> n a will not have a monad transformer of the "composed-outside" type, (m a -> n q) -> n (m a). Neither will it have a "composed-inside" transformer, m((a -> n q) -> n a) - this is not a monad for arbitrary monad m; take m = Maybe for a counterexample. The type (a -> m (n q)) -> m (n a) is similarly not a monad for arbitrary monads m and n. The type m(a -> n q) -> n a is a monad for any m but does not admit a lifting m a -> m (a -> n q) -> n a because we cannot compute a value of n a given only some values wrapped into an arbitrary monad m.

Both S and Sq are lawful monads (I checked it manually) but neither of them seems to have a lawful monad transformer.

Here is a heuristic argument for the non-existence of the monad transformer. If there were a data type definition for the monad transformer of (a -> n q) -> n a that works for all monads n, that data type definition would have yielded the "composed-outside" transformer for rigid n and some other transformer for non-rigid n. But this kind of choice is impossible for a type expression that uses n naturally and parametrically (i.e. as an opaque type constructor with a monad instance).

3. Generally, transformed monads don't themselves automatically possess a monad transformer. That is, once we take some foreign monad m and apply some monad transformer t to it, we obtain a new monad t m, and this monad doesn't have a transformer: given a new foreign monad n, we don't know how to transform n with the monad t m. If we know the transformer mt for the monad m, we can first transform n with mt and then transform the result with t. But if we don't have a transformer for the monad m, we are stuck: there is no construction that creates a transformer for the monad t m out of the knowledge of t alone and works for arbitrary foreign monads m.

4. @JamesCandy's answer suggests that for any monad (including IO?!), one can write a (general but complicated) type expression that represents the corresponding monad transformer. Namely, you first need to Church-encode your monad type, which makes the type look like a continuation monad, and then define its monad transformer as if for the continuation monad. But I think this is incorrect - it does not give a recipe for producing a monad transformer in general.

Taking the Church encoding of a type a means writing down the type

 type ca = forall r. (a -> r) -> r

This type ca is completely isomorphic to a by Yoneda's lemma. So far we have achieved nothing other than made the type a lot more complicated by introducing a quantified type parameter forall r.

Now let's Church-encode a base monad L:

 type CL a = forall r. (L a -> r) -> r

Again, we have achieved nothing so far, since CL a is fully equivalent to L a.

Now pretend for a second that CL a a continuation monad (which it isn't!), and write the monad transformer as if it were a continuation monad transformer, by replacing the result type r through m r:

 type TCL m a = forall r. (L a -> m r) -> m r

This is claimed to be the "Church-encoded monad transformer" for L. But this seems to be incorrect. We need to check the properties:

• TCL m is a lawful monad for any foreign monad m and for any base monad L
• m a -> TCL m a is a lawful monadic morphism

The second property holds, but I believe the first property fails, - in other words, TCL m is not a monad for an arbitrary monad m. Perhaps some monads m admit this but others do not. I was not able to find a general monad instance for TCL m corresponding to an arbitrary base monad L.

Another way to argue that TCL m is not in general a monad is to note that forall r. (a -> m r) -> m r is indeed a monad for any type constructor m. Denote this monad by CM. Now, TCL m a = CM (L a). If TCL m were a monad, it would imply that CM can be composed with any monad L and yields a lawful monad CM (L a). However, it is highly unlikely that a nontrivial monad CM (in particular, one that is not equivalent to Reader) will compose with all monads L. Monads usually do not compose without stringent further constraints.

A specific example where this does not work is for reader monads. Consider L a = r -> a and m a = s -> a where r and s are some fixed types. Now, we would like to consider the "Church-encoded monad transformer" forall t. (L a -> m t) -> m t. We can simplify this type expression using the Yoneda lemma,

 forall t. (x -> t) -> Q t  = Q x

(for any functor Q) and obtain

 forall t. (L a -> s -> t) -> s -> t
 = forall t. ((L a, s) -> t) -> s -> t
 = s -> (L a, s)
 = s -> (r -> a, s)

So this is the type expression for TCL m a in this case. If TCL were a monad transformer then P a = s -> (r -> a, s) would be a monad. But one can check explicitly that this P is actually not a monad (one cannot implement return and bind that satisfy the laws).

Even if this worked (i.e. assuming that I made a mistake in claiming that TCL m is in general not a monad), this construction has certain disadvantages:

• It is not functorial (i.e. not covariant) with respect to the foreign monad m, so we cannot do things like interpret a transformed free monad into another monad, or merge two monad transformers as explained here Is there a principled way to compose two monad transformers if they are of different type, but their underlying monad is of the same type?
• The presence of a forall r makes the type quite complicated to reason about and may lead to performance degradation (see the "Church encoding considered harmful" paper) and stack overflows (since Church encoding is usually not stack-safe)
• The Church-encoded monad transformer for an identity base monad (L = Id) does not yield the unmodified foreign monad: T m a = forall r. (a -> m r) -> m r and this is not the same as m a. In fact it's quite difficult to figure out what that monad is, given a monad m.

As an example showing why forall r makes reasoning complicated, consider the foreign monad m a = Maybe a and try to understand what the type forall r. (a -> Maybe r) -> Maybe r actually means. I was not able to simplify this type or to find a good explanation about what this type does, i.e. what kind of "effect" it represents (since it's a monad, it must represent some kind of "effect") and how one would use such a type.

• The Church-encoded monad transformer is not equivalent to the standard well-known monad transformers such as ReaderT, WriterT, EitherT, StateT and so on.

It is not clear how many other monad transformers exist and in what cases one would use one or another transformer.

5. One of the questions in the post is to find an explicit example of a monad m that has two transformers t1 and t2 such that for some foreign monad n, the monads t1 n and t2 n are not equivalent.

I believe that the Search monad provides such an example.

 type Search a = (a -> p) -> a

where p is a fixed type.

The transformers are

 type SearchT1 n a = (a -> n p) -> n a
 type SearchT2 n a = (n a -> p) -> n a

I checked that both SearchT1 n and SearchT2 n are lawful monads for any monad n. We have liftings n a -> SearchT1 n a and n a -> SearchT2 n a that work by returning constant functions (just return n a as given, ignoring the argument). We have SearchT1 Identity and SearchT2 Identity obviously equivalent to Search.

The big difference between SearchT1 and SearchT2 is that SearchT1 is not functorial in n, while SearchT2 is. This may have implications for "running" ("interpreting") the transformed monad, since normally we would like to be able to lift an interpreter n a -> n' a into a "runner" SearchT n a -> SearchT n' a. This is possibly only with SearchT2.

A similar deficiency is present in the standard monad transformers for the continuation monad and the codensity monad: they are not functorial in the foreign monad.

回答4:

My solution exploits the logical structure of Haskell terms etc.

I looked at right Kan extensions as possible representations of the monad transformer. As everyone knows, right Kan extensions are limits, so it makes sense that they should serve as universal encoding of any object of interest. For monadic functors F and M, I looked at the right Kan extension of MF along F.

First I proved a lemma, "rolling lemma:" a procomposed functor to the Right kan extension can be rolled inside it, giving the map F(Ran G H) -> Ran G(FH) for any functors F, G, and H.

Using this lemma, I computed a monadic join for the right Kan extension Ran F (MF), requiring the distributive law FM -> MF. It is as follows:

Ran F(MF) . Ran F(MF) [rolling lemma] =>
  Ran F(Ran F(MF)MF) [insert eta] =>
  Ran F(Ran F(MF)FMF) [gran] =>
  Ran F(MFMF) [apply distributive law] =>
  Ran F(MMFF) [join Ms and Fs] =>
  Ran F(MF).

What seems to be interesting about this construction is that it admits of lifts of both functors F and M as follows:

(1) F [lift into codensity monad] =>
  Ran F F [procompose with eta] =>
  Ran F(MF).

(2) M [Yoneda lemma specialized upon F-] =>
  Ran F(MF).

I also investigated the right Kan extension Ran F(FM). It seems to be a little better behaved achieving monadicity without appeal to the distributive law, but much pickier in what functors it lifts. I determined that it will lift monadic functors under the following conditions:

1) F is monadic.

2) F |- U, in which case it admits the lift F ~> Ran U(UM). This can be used in the context of a state monad to "set" the state.

3) M under certain conditions, for instance when M admits a distributive law.

来源：https://stackoverflow.com/questions/24515876/is-there-a-monad-that-doesnt-have-a-corresponding-monad-transformer-except-io