I'd like to propose a more systematic approach to answering this question, and also to show examples that do not use any special tricks like the "bottom" values or infinite data types or anything like that.

When do type constructors fail to have type class instances?

In general, there are two reasons why a type constructor could fail to have an instance of a certain type class:

Cannot implement the type signatures of the required methods from the type class. Can implement the type signatures but cannot satisfy the required laws.

Examples of the first kind are easier than those of the second kind because for the first kind, we just need to check whether one can implement a function with a given type signature, while for the second kind, we are required to prove that no implementation could possibly satisfy the laws.

Specific examples

A type constructor that cannot have a functor instance because the type cannot be implemented: data F z a = F (a -> z)

This is a contrafunctor, not a functor, with respect to the type parameter a , because a in a contravariant position. It is impossible to implement a function with type signature (a -> b) -> F z a -> F z b .

A type constructor that is not a lawful functor even though the type signature of fmap can be implemented: data Q a = Q(a -> Int, a) fmap :: (a -> b) -> Q a -> Q b fmap f (Q(g, x)) = Q(\_ -> g x, f x) -- this fails the functor laws!

The curious aspect of this example is that we can implement fmap of the correct type even though F cannot possibly be a functor because it uses a in a contravariant position. So this implementation of fmap shown above is misleading - even though it has the correct type signature (I believe this is the only possible implementation of that type signature), the functor laws are not satisfied. For example, fmap id ≠ id , because let (Q(f,_)) = fmap id (Q(read,"123")) in f "456" is 123 , but let (Q(f,_)) = id (Q(read,"123")) in f "456" is 456 .

In fact, F is only a profunctor, - it is neither a functor nor a contrafunctor.

A lawful functor that is not applicative because the type signature of pure cannot be implemented: take the Writer monad (a, w) and remove the constraint that w should be a monoid. It is then impossible to construct a value of type (a, w) out of a .

A functor that is not applicative because the type signature of <*> cannot be implemented: data F a = Either (Int -> a) (String -> a) .

A functor that is not lawful applicative even though the type class methods can be implemented: data P a = P ((a -> Int) -> Maybe a)

The type constructor P is a functor because it uses a only in covariant positions.

instance Functor P where fmap :: (a -> b) -> P a -> P b fmap fab (P pa) = P (\q -> fmap fab $ pa (q . fab))

The only possible implementation of the type signature of <*> is a function that always returns Nothing :

(<*>) :: P (a -> b) -> P a -> P b (P pfab) <*> (P pa) = \_ -> Nothing -- fails the laws!

But this implementation does not satisfy the identity law for applicative functors.

A functor that is Applicative but not a Monad because the type signature of bind cannot be implemented.

I do not know any such examples!

A functor that is Applicative but not a Monad because laws cannot be satisfied even though the type signature of bind can be implemented.

This example has generated quite a bit of discussion, so it is safe to say that proving this example correct is not easy. But several people have verified this independently by different methods. See Is `data PoE a = Empty | Pair a a` a monad? for additional discussion.

data B a = Maybe (a, a) deriving Functor instance Applicative B where pure x = Just (x, x) b1 <*> b2 = case (b1, b2) of (Just (x1, y1), Just (x2, y2)) -> Just((x1, x2), (y1, y2)) _ -> Nothing

It is somewhat cumbersome to prove that there is no lawful Monad instance. The reason for the non-monadic behavior is that there is no natural way of implementing bind when a function f :: a -> B b could return Nothing or Just for different values of a .

It is perhaps clearer to consider Maybe (a, a, a) , which is also not a monad, and to try implementing join for that. One will find that there is no intuitively reasonable way of implementing join .

join :: Maybe (Maybe (a, a, a), Maybe (a, a, a), Maybe (a, a, a)) -> Maybe (a, a, a) join Nothing = Nothing join Just (Nothing, Just (x1,x2,x3), Just (y1,y2,y3)) = ??? join Just (Just (x1,x2,x3), Nothing, Just (y1,y2,y3)) = ??? -- etc.

In the cases indicated by ??? , it seems clear that we cannot produce Just (z1, z2, z3) in any reasonable and symmetric manner out of six different values of type a . We could certainly choose some arbitrary subset of these six values, -- for instance, always take the first nonempty Maybe - but this would not satisfy the laws of the monad. Returning Nothing will also not satisfy the laws.

A tree-like data structure that is not a monad even though it has associativity for bind - but fails the identity laws.

The usual tree-like monad (or "a tree with functor-shaped branches") is defined as

data Tr f a = Leaf a | Branch (f (Tr f a))

This is a free monad over the functor f . The shape of the data is a tree where each branch point is a "functor-ful" of subtrees. The standard binary tree would be obtained with type f a = (a, a) .

If we modify this data structure by making also the leaves in the shape of the functor f , we obtain what I call a "semimonad" - it has bind that satisfies the naturality and the associativity laws, but its pure method fails one of the identity laws. "Semimonads are semigroups in the category of endofunctors, what's the problem?" This is the type class Bind .

For simplicity, I define the join method instead of bind :

data Trs f a = Leaf (f a) | Branch (f (Trs f a)) join :: Trs f (Trs f a) -> Trs f a join (Leaf ftrs) = Branch ftrs join (Branch ftrstrs) = Branch (fmap @f join ftrstrs)

The branch grafting is standard, but the leaf grafting is non-standard and produces a Branch . This is not a problem for the associativity law but breaks one of the identity laws.

When do polynomial types have monad instances?

Neither of the functors Maybe (a, a) and Maybe (a, a, a) can be given a lawful Monad instance, although they are obviously Applicative .

These functors have no tricks - no Void or bottom anywhere, no tricky laziness/strictness, no infinite structures, and no type class constraints. The Applicative instance is completely standard. The functions return and bind can be implemented for these functors but will not satisfy the laws of the monad. In other words, these functors are not monads because a specific structure is missing (but it is not easy to understand what exactly is missing). As an example, a small change in the functor can make it into a monad: data Maybe a = Nothing | Just a is a monad. Another similar functor data P12 a = Either a (a, a) is also a monad.

Constructions for polynomial monads

In general, here are some constructions that produce lawful Monad s out of polynomial types. In all these constructions, M is a monad:

type M a = Either c (w, a) where w is any monoid type M a = m (Either c (w, a)) where m is any monad and w is any monoid type M a = (m1 a, m2 a) where m1 and m2 are any monads type M a = Either a (m a) where m is any monad

The first construction is WriterT w (Either c) , the second construction is WriterT w (EitherT c m) . The third construction is a component-wise product of monads: pure @M is defined as the component-wise product of pure @m1 and pure @m2 , and join @M is defined by omitting cross-product data (e.g. m1 (m1 a, m2 a) is mapped to m1 (m1 a) by omitting the second part of the tuple):

join :: (m1 (m1 a, m2 a), m2 (m1 a, m2 a)) -> (m1 a, m2 a) join (m1x, m2x) = (join @m1 (fmap fst m1x), join @m2 (fmap snd m2x))

The fourth construction is defined as

data M m a = Either a (m a) instance Monad m => Monad M m where pure x = Left x join :: Either (M m a) (m (M m a)) -> M m a join (Left mma) = mma join (Right me) = Right $ join @m $ fmap @m squash me where squash :: M m a -> m a squash (Left x) = pure @m x squash (Right ma) = ma

I have checked that all four constructions produce lawful monads.

I conjecture that there are no other constructions for polynomial monads. For example, the functor Maybe (Either (a, a) (a, a, a, a)) is not obtained through any of these constructions and so is not monadic. However, Either (a, a) (a, a, a) is monadic because it is isomorphic to the product of three monads a , a , and Maybe a . Also, Either (a,a) (a,a,a,a) is monadic because it is isomorphic to the product of a and Either a (a, a, a) .

The four constructions shown above will allow us to obtain any sum of any number of products of any number of a 's, for example Either (Either (a, a) (a, a, a, a)) (a, a, a, a, a)) and so on. All such type constructors will have (at least one) Monad instance.

It remains to be seen, of course, what use cases might exist for such monads. Another issue is that the Monad instances derived via constructions 1-4 are in general not unique. For example, the type constructor type F a = Either a (a, a) can be given a Monad instance in two ways: by construction 4 using the monad (a, a) , and by construction 3 using the type isomorphism Either a (a, a) = (a, Maybe a) . Again, finding use cases for these implementations is not immediately obvious.