For instance, take the **Reader Monad**. Once we decide that the type
constructor is
- type 'a reader = env -> 'a
+ type 'a reader = env -> 'a
then the choice of unit and bind is natural:
- let r_unit (a : 'a) : 'a reader = fun (e : env) -> a
+ let r_unit (a : 'a) : 'a reader = fun (e : env) -> a
The reason this is a fairly natural choice is that because the type of
an `'a reader` is `env -> 'a` (by definition), the type of the
Since the type of the `bind` operator is required to be
- r_bind : ('a reader) -> ('a -> 'b reader) -> ('b reader)
+ r_bind : ('a reader) -> ('a -> 'b reader) -> ('b reader)
We can reason our way to the traditional reader `bind` function as
follows. We start by declaring the types determined by the definition
of a bind operation:
- let r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) = ...
+ let r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) = ...
Now we have to open up the `u` box and get out the `'a` object in order to
feed it to `f`. Since `u` is a function from environments to
The **State Monad** is similar. Once we've decided to use the following type constructor:
- type 'a state = store -> ('a, store)
+ type 'a state = store -> ('a, store)
Then our unit is naturally:
- let s_unit (a : 'a) : ('a state) = fun (s : store) -> (a, s)
+ let s_unit (a : 'a) : ('a state) = fun (s : store) -> (a, s)
And we can reason our way to the bind function in a way similar to the reasoning given above. First, we need to apply `f` to the contents of the `u` box:
- let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
+ let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
... f (...) ...
But unlocking the `u` box is a little more complicated. As before, we
Our other familiar monad is the **List Monad**, which we were told
looks like this:
- type 'a list = ['a];;
- l_unit (a : 'a) = [a];;
- l_bind u f = List.concat (List.map f u);;
+ type 'a list = ['a];;
+ l_unit (a : 'a) = [a];;
+ l_bind u f = List.concat (List.map f u);;
Thinking through the list monad will take a little time, but doing so
will provide a connection with continuations.
Recall that `List.map` takes a function and a list and returns the
result to applying the function to the elements of the list:
- List.map (fun i -> [i;i+1]) [1;2] ~~> [[1; 2]; [2; 3]]
+ List.map (fun i -> [i;i+1]) [1;2] ~~> [[1; 2]; [2; 3]]
and List.concat takes a list of lists and erases the embdded list
boundaries:
- List.concat [[1; 2]; [2; 3]] ~~> [1; 2; 2; 3]
+ List.concat [[1; 2]; [2; 3]] ~~> [1; 2; 2; 3]
And sure enough,
- l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
+ l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
Now, why this unit, and why this bind? Well, ideally a unit should
not throw away information, so we can rule out `fun x -> []` as an
strategy for implementating lists). These were the lists that made
lists look like Church numerals with extra bits embdded in them:
- empty list: fun f z -> z
- list with one element: fun f z -> f 1 z
- list with two elements: fun f z -> f 2 (f 1 z)
- list with three elements: fun f z -> f 3 (f 2 (f 1 z))
+ empty list: fun f z -> z
+ list with one element: fun f z -> f 1 z
+ list with two elements: fun f z -> f 2 (f 1 z)
+ list with three elements: fun f z -> f 3 (f 2 (f 1 z))
and so on. To save time, we'll let the OCaml interpreter infer the
principle types of these functions (rather than inferring what the
So here's our type constructor for our hand-rolled lists:
- type 'b list' = (int -> 'b -> 'b) -> 'b -> 'b
+ type 'b list' = (int -> 'b -> 'b) -> 'b -> 'b
Generalizing to lists that contain any kind of element (not just
ints), we have
- type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
+ type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
So an `('a, 'b) list'` is a list containing elements of type `'a`,
where `'b` is the type of some part of the plumbing. This is more
into OCaml lists soon. We don't need to fully grasp the role of the `'b`'s
in order to proceed to build a monad:
- l'_unit (a : 'a) : ('a, 'b) list = fun a -> fun f z -> f a z
+ l'_unit (a : 'a) : ('a, 'b) list = fun a -> fun k z -> k a z
No problem. Arriving at bind is a little more complicated, but
exactly the same principles apply, you just have to be careful and
systematic about it.
- l'_bind (u : ('a,'b) list') (f : 'a -> ('c, 'd) list') : ('c, 'd) list' = ...
+ l'_bind (u : ('a,'b) list') (f : 'a -> ('c, 'd) list') : ('c, 'd) list' = ...
Unpacking the types gives:
- l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
- (f : 'a -> ('c -> 'd -> 'd) -> 'd -> 'd)
- : ('c -> 'd -> 'd) -> 'd -> 'd = ...
+ l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
+ (f : 'a -> ('c -> 'd -> 'd) -> 'd -> 'd)
+ : ('c -> 'd -> 'd) -> 'd -> 'd = ...
Perhaps a bit intimiating.
But it's a rookie mistake to quail before complicated types. You should
This whole expression has type `('c -> 'b -> 'b) -> 'b -> 'b`, which is exactly the type of a `('c, 'b) list'`. So we can hypothesize that our bind is:
- l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
- (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b)
- : ('c -> 'b -> 'b) -> 'b -> 'b =
- fun k -> u (fun a b -> f a k b)
+ l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
+ (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b)
+ : ('c -> 'b -> 'b) -> 'b -> 'b =
+ fun k -> u (fun a b -> f a k b)
That is a function of the right type for our bind, but to check whether it works, we have to verify it (with the unit we chose) against the monad laws, and reason whether it will have the right behavior.
Here's a way to persuade yourself that it will have the right behavior. First, it will be handy to eta-expand our `fun k -> u (fun a b -> f a k b)` to:
- fun k z -> u (fun a b -> f a k b) z
+ fun k z -> u (fun a b -> f a k b) z
Now let's think about what this does. It's a wrapper around `u`. In order to behave as the list which is the result of mapping `f` over each element of `u`, and then joining (`concat`ing) the results, this wrapper would have to accept arguments `k` and `z` and fold them in just the same way that the list which is the result of mapping `f` and then joining the results would fold them. Will it?
Or rather, it should give us a list' version of that, which takes a function `k` and value `z` as arguments, and returns the right fold of `k` and `z` over those elements. What does our formula
- fun k z -> u (fun a b -> f a k b) z
+ fun k z -> u (fun a b -> f a k b) z
do? Well, for each element `a` in `u`, it applies `f` to that `a`, getting one of the lists:
- []
+ [] ; result of applying f to leftmost a
[2]
[2; 4]
- [2; 4; 8]
+ [2; 4; 8] ; result of applying f to rightmost a
(or rather, their list' versions). Then it takes the accumulated result `b` of previous steps in the fold, and it folds `k` and `b` over the list generated by `f a`. The result of doing so is passed on to the next step as the accumulated result so far.
So if, for example, we let `k` be `+` and `z` be `0`, then the computation would proceed:
0 ==>
- right-fold + and 0 over [2; 4; 8] = 2+4+8+0 ==>
- right-fold + and 2+4+8+0 over [2; 4] = 2+4+2+4+8+0 ==>
- right-fold + and 2+4+2+4+8+0 over [2] = 2+2+4+2+4+8+0 ==>
- right-fold + and 2+2+4+2+4+8+0 over [] = 2+2+4+2+4+8+0
+ right-fold + and 0 over [2; 4; 8] = ((2+4+8+0) ==>
+ right-fold + and 2+4+8+0 over [2; 4] = 2+4+(2+4+8+0) ==>
+ right-fold + and 2+4+2+4+8+0 over [2] = 2+(2+4+(2+4+8+0)) ==>
+ right-fold + and 2+2+4+2+4+8+0 over [] = 2+(2+4+(2+4+8+0))
which indeed is the result of right-folding + and 0 over `[2; 2; 4; 2; 4; 8]`. If you trace through how this works, you should be able to persuade yourself that our formula:
- fun k z -> u (fun a b -> f a k b) z
+ fun k z -> u (fun a b -> f a k b) z
will deliver just the same folds, for arbitrary choices of `k` and `z` (with the right types), and arbitrary list's `u` and appropriately-typed `f`s, as
For future reference, we might make two eta-reductions to our formula, so that we have instead:
- let l'_bind = fun k -> u (fun a -> f a k);;
+ let l'_bind = fun k -> u (fun a -> f a k);;
Let's make some more tests:
-
- l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
-
- l'_bind (fun f z -> f 1 (f 2 z))
- (fun i -> fun f z -> f i (f (i+1) z)) ~~> <fun>
+ l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
+
+ l'_bind (fun f z -> f 1 (f 2 z))
+ (fun i -> fun f z -> f i (f (i+1) z)) ~~> <fun>
Sigh. OCaml won't show us our own list. So we have to choose an `f`
and a `z` that will turn our hand-crafted lists into standard OCaml
quantifiers embodies the spirit of continuations (see de Groote 2001,
Barker 2002 for lengthy discussion). Let's see why.
-First, we'll need a type constructor. As you probably know,
+First, we'll need a type constructor. As we've said,
Montague replaced individual-denoting determiner phrases (with type `e`)
with generalized quantifiers (with [extensional] type `(e -> t) -> t`.
In particular, the denotation of a proper name like *John*, which
type 'a continuation = ('a -> 'b) -> 'b
c_unit (a : 'a) = fun (p : 'a -> 'b) -> p a
c_bind (u : ('a -> 'b) -> 'b) (f : 'a -> ('c -> 'd) -> 'd) : ('c -> 'd) -> 'd =
- fun (k : 'a -> 'b) -> u (fun (a : 'a) -> f a k)
+ fun (k : 'a -> 'b) -> u (fun (a : 'a) -> f a k)
Note that `c_unit` is exactly the `gqize` function that Montague used
to lift individuals into the continuation monad.
examine the type constructor and the terms from the list monad derived
above:
- type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
- l'_unit a = fun f -> f a
- l'_bind u f = fun k -> u (fun a -> f a k)
+ type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
+ let l'_unit a = fun k z -> k a z
+
+This can be eta-reduced to:
+
+ let l'_unit a = fun k -> k a
-(We performed a sneaky but valid eta reduction in the unit term.)
+ let l'_bind u f =
+ (* we mentioned three versions of this, the fully eta-expanded: *)
+ fun k z -> u (fun a b -> f a k b) z
+ (* an intermediate version, and the fully eta-reduced: *)
+ fun k -> u (fun a -> f a k)
-The unit and the bind for the Montague continuation monad and the
-homemade List monad are the same terms! In other words, the behavior
-of the List monad and the behavior of the continuations monad are
+Consider the most eta-reduced versions of `l'_unit` and `l'_bind`. They're the same as the unit and bind for the Montague continuation monad! In other words, the behavior of our v3-list monad and the behavior of the continuations monad are
parallel in a deep sense.
Have we really discovered that lists are secretly continuations? Or