From 5c9b178bf3c8376b621fdf5c21c80af1a66ef56b Mon Sep 17 00:00:00 2001
From: Jim Pryor
Date: Wed, 1 Dec 2010 01:35:18 0500
Subject: [PATCH 1/1] listsmonad tweaks
Signedoffby: Jim Pryor

list_monad_as_continuation_monad.mdwn  108 ++++++++++++++++++
1 file changed, 56 insertions(+), 52 deletions()
diff git a/list_monad_as_continuation_monad.mdwn b/list_monad_as_continuation_monad.mdwn
index acc05099..b4158b16 100644
 a/list_monad_as_continuation_monad.mdwn
+++ b/list_monad_as_continuation_monad.mdwn
@@ 19,11 +19,11 @@ version 3 lists and monads).
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
@@ 33,13 +33,13 @@ for the reader monad.
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
@@ 72,15 +72,15 @@ monads.]
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
@@ 102,9 +102,9 @@ follow just as naturally from its type constructor.
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.
@@ 112,16 +112,16 @@ 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
@@ 153,10 +153,10 @@ intriguing to wonder how things would change if we used some other
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
@@ 179,12 +179,12 @@ thrown in, the type of the element a the head of the list
So here's our type constructor for our handrolled 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
@@ 192,19 +192,19 @@ general than an ordinary OCaml list, but we'll see how to map them
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
@@ 237,16 +237,16 @@ Now, we've used a `k` that we pulled out of nowhere, so we need to abstract over
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 etaexpand 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?
@@ 258,28 +258,28 @@ Suppose we have a list' whose contents are `[1; 2; 4; 8]`that is, our list' w
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 ==>
 rightfold + and 0 over [2; 4; 8] = 2+4+8+0 ==>
 rightfold + and 2+4+8+0 over [2; 4] = 2+4+2+4+8+0 ==>
 rightfold + and 2+4+2+4+8+0 over [2] = 2+2+4+2+4+8+0 ==>
 rightfold + and 2+2+4+2+4+8+0 over [] = 2+2+4+2+4+8+0
+ rightfold + and 0 over [2; 4; 8] = ((2+4+8+0) ==>
+ rightfold + and 2+4+8+0 over [2; 4] = 2+4+(2+4+8+0) ==>
+ rightfold + and 2+4+2+4+8+0 over [2] = 2+(2+4+(2+4+8+0)) ==>
+ rightfold + and 2+2+4+2+4+8+0 over [] = 2+(2+4+(2+4+8+0))
which indeed is the result of rightfolding + 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 appropriatelytyped `f`s, as
@@ 289,15 +289,14 @@ would.
For future reference, we might make two etareductions 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)) ~~>
+ 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)) ~~>
Sigh. OCaml won't show us our own list. So we have to choose an `f`
and a `z` that will turn our handcrafted lists into standard OCaml
@@ 318,7 +317,7 @@ We've hinted that Montague's treatment of DPs as generalized
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 individualdenoting 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
@@ 342,7 +341,7 @@ highly similar to the List monad just given:
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.
@@ 352,15 +351,20 @@ it in the List monad. How similar is it to the List monad? Let's
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 etareduced 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 etaexpanded: *)
+ fun k z > u (fun a b > f a k b) z
+ (* an intermediate version, and the fully etareduced: *)
+ 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 etareduced 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 v3list monad and the behavior of the continuations monad are
parallel in a deep sense.
Have we really discovered that lists are secretly continuations? Or

2.11.0