let rec move_botleft (z : 'a zipper) : 'a zipper =
(* returns z if the targetted node in z has no children *)
- (* else returns move_botleft (zipper which results from moving down and left in z) *)
- YOU SUPPLY THE DEFINITION
+ (* else returns move_botleft (zipper which results from moving down from z to the leftmost child) *)
+ _____
+ (* YOU SUPPLY THE DEFINITION *)
let rec move_right_or_up (z : 'a zipper) : 'a zipper option =
(* if it's possible to move right in z, returns Some (the result of doing so) *)
(* else if it's not possible to move any further up in z, returns None *)
(* else returns move_right_or_up (result of moving up in z) *)
- YOU SUPPLY THE DEFINITION
+ _____
+ (* YOU SUPPLY THE DEFINITION *)
let new_zipper (t : 'a tree) : 'a zipper =
;;
-2. Here's another implementation of the same-fringe function, in Scheme. It's taken from <http://c2.com/cgi/wiki?SameFringeProblem>. It uses thunks to delay the evaluation of code that computes the tail of a list of a tree's fringe. It also involves passing continuations as arguments. Your assignment is to supply comments to the code, to explain what every significant piece is doing.
+2. Here's another implementation of the same-fringe function, in Scheme. It's taken from <http://c2.com/cgi/wiki?SameFringeProblem>. It uses thunks to delay the evaluation of code that computes the tail of a list of a tree's fringe. It also involves passing continuations as arguments. Your assignment is to fill in the blanks in the code, and also to supply comments to the code, to explain what every significant piece is doing.
This code uses Scheme's `cond` construct. That works like this;
is equivalent to:
(if (test1 argument argument)
+ ; then
result1
+ ; else
(if (test2 argument argument)
+ ; then
result2
+ ; else
(if (test3 argument argument)
+ ; then
result3
+ ; else
result4)))
Some other Scheme details:
* `(null? lst)` tests whether `lst` is the empty list
* non-empty lists are implemented as pairs whose second member is a list
* `'()` `'(1)` `'(1 2)` `'(1 2 3)` are all lists
- * `(list)` `(list 1)` `(list 1 2)` `(list 1 2 3)` are the same lists
+ * `(list)` `(list 1)` `(list 1 2)` `(list 1 2 3)` are the same lists as the preceding
* `'(1 2 3)` and `(cons 1 '(2 3))` are both pairs and lists (the same list)
* `(pair? lst)` tests whether `lst` is a pair; if `lst` is a non-empty list, it will also pass this test; if `lst` fails this test, it may be because `lst` is the empty list, or because it's not a list or pair at all
* `(car lst)` extracts the first member of a pair / head of a list
(letrec ([helper (lambda (tree tailk)
(cond
[(pair? tree)
+ (helper (car tree) (lambda () (helper _____ tailk)))]
+ [else (cons tree tailk)]))])
+ (helper tree (lambda () _____))))
+
+ (define (stream-equal? stream1 stream2)
+ (cond
+ [(and (null? stream1) (null? stream2)) _____]
+ [(and (pair? stream1) (pair? stream2))
+ (and (equal? (car stream1) (car stream2))
+ _____)]
+ [else #f]))
+
+ (define (same-fringe? tree1 tree2)
+ (stream-equal? (lazy-flatten tree1) (lazy-flatten tree2)))
+
+ (define tree1 '(((1 . 2) . (3 . 4)) . (5 . 6)))
+ (define tree2 '(1 . (((2 . 3) . (4 . 5)) . 6)))
+
+ (same-fringe? tree1 tree2)
+
+
+<!--
+ (define (lazy-flatten tree)
+ (letrec ([helper (lambda (tree tailk)
+ (cond
+ [(pair? tree)
(helper (car tree) (lambda () (helper (cdr tree) tailk)))]
[else (cons tree tailk)]))])
(helper tree (lambda () (list)))))
(define tree2 '(1 . (((2 . 3) . (4 . 5)) . 6)))
(same-fringe? tree1 tree2)
-
-
+-->
type 'a reader = env -> 'a
-then we can deduce the unit and the bind:
+then the choice of unit and bind is natural:
- let r_unit (x : 'a) : 'a reader = fun (e : env) -> x
+ let r_unit (a : 'a) : 'a reader = fun (e : env) -> a
Since the type of an `'a reader` is `env -> 'a` (by definition),
the type of the `r_unit` function is `'a -> env -> 'a`, which is a
-specific case of the type of the *K* combinator. So it makes sense
+specific case of the type of the *K* combinator. It makes sense
that *K* is the unit for the reader monad.
Since the type of the `bind` operator is required to be
r_bind : ('a reader) -> ('a -> 'b reader) -> ('b reader)
-We can deduce the correct `bind` function as follows:
+We can reason our way to the correct `bind` function as follows. We start by declaring the type:
let r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) =
-We have to open up the `u` box and get out the `'a` object in order to
+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
objects of type `'a`, the way we open a box in this monad is
by applying it to an environment:
- .... f (u e) ...
+ ... f (u e) ...
This subexpression types to `'b reader`, which is good. The only
problem is that we invented an environment `e` that we didn't already have ,
so we have to abstract over that variable to balance the books:
- fun e -> f (u e) ...
+ fun e -> f (u e) ...
This types to `env -> 'b reader`, but we want to end up with `env ->
-'b`. Once again, the easiest way to turn a `'b reader` into a `'b` is to apply it to
-an environment. So we end up as follows:
+'b`. Once again, the easiest way to turn a `'b reader` into a `'b` is to apply it to an environment. So we end up as follows:
- r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) = f (u e) e
+ r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) =
+ f (u e) e
-And we're done.
+And we're done. This gives us a bind function of the right type. We can then check whether, in combination with the unit function we chose, it satisfies the monad laws, and behaves in the way we intend. And it does.
-[This bind is a condensed version of the careful `let a = u e in ...`
+[The bind we cite here is a condensed version of the careful `let a = u e in ...`
constructions we provided in earlier lectures. We use the condensed
version here in order to emphasize similarities of structure across
monads.]
-The **State Monad** is similar. We somehow intuit that we want to use
-the following type constructor:
+The **State Monad** is similar. Once we've decided to use the following type constructor:
type 'a state = store -> ('a, store)
-So our unit is naturally
+Then our unit is naturally:
- let s_unit (x : 'a) : ('a state) = fun (s : store) -> (x, s)
+ let s_unit (a : 'a) : ('a state) = fun (s : store) -> (a, s)
-And we deduce the bind in a way similar to the reasoning given above.
-First, we need to apply `f` to the contents of the `u` box:
+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 =
+ ... f (...) ...
But unlocking the `u` box is a little more complicated. As before, we
need to posit a state `s` that we can apply `u` to. Once we do so,
however, we won't have an `'a`, we'll have a pair whose first element
is an `'a`. So we have to unpack the pair:
- ... let (a, s') = u s in ... (f a) ...
+ ... let (a, s') = u s in ... (f a) ...
Abstracting over the `s` and adjusting the types gives the result:
- let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
- fun (s : store) -> let (a, s') = u s in f a s'
+ let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
+ fun (s : store) -> let (a, s') = u s in f a s'
-The **Option Monad** doesn't follow the same pattern so closely, so we
+The **Option/Maybe Monad** doesn't follow the same pattern so closely, so we
won't pause to explore it here, though conceptually its unit and bind
follow just as naturally from its type constructor.
looks like this:
type 'a list = ['a];;
- l_unit (x : 'a) = [x];;
+ l_unit (a : 'a) = [a];;
l_bind u f = List.concat (List.map f u);;
Recall that `List.map` take a function and a list and returns the
principle types of these functions (rather than deducing what the
types should be):
-<pre>
-# fun f z -> z;;
-- : 'a -> 'b -> 'b = <fun>
-# fun f z -> f 1 z;;
-- : (int -> 'a -> 'b) -> 'a -> 'b = <fun>
-# fun f z -> f 2 (f 1 z);;
-- : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
-# fun f z -> f 3 (f 2 (f 1 z))
-- : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
-</pre>
+ # fun f z -> z;;
+ - : 'a -> 'b -> 'b = <fun>
+ # fun f z -> f 1 z;;
+ - : (int -> 'a -> 'b) -> 'a -> 'b = <fun>
+ # fun f z -> f 2 (f 1 z);;
+ - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
+ # fun f z -> f 3 (f 2 (f 1 z))
+ - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
-Finally, we're getting consistent principle types, so we can stop.
-These types should remind you of the simply-typed lambda calculus
-types for Church numerals (`(o -> o) -> o -> o`) with one extra bit
-thrown in (in this case, an int).
+We can see what the consistent, general principle types are at the end, so we
+can stop. These types should remind you of the simply-typed lambda calculus
+types for Church numerals (`(o -> o) -> o -> o`) with one extra bit thrown in
+(in this case, an int).
So here's our type constructor for our hand-rolled lists:
- type 'a list' = (int -> 'a -> 'a) -> 'a -> 'a
+ type 'b list' = (int -> 'b -> 'b) -> 'b -> 'b
Generalizing to lists that contain any kind of element (not just
ints), we have
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 (x : 'a) : ('a, 'b) list = fun x -> fun f z -> f x z
+ l'_unit (a : 'a) : ('a, 'b) list = fun a -> fun f z -> f 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
l'_bind (u : ('a,'b) list') (f : 'a -> ('c, 'd) list') : ('c, 'd) list' = ...
-Unfortunately, we'll need to spell out the types:
+Unpacking the types gives:
l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
(f : 'a -> ('c -> 'd -> 'd) -> 'd -> 'd)
: ('c -> 'd -> 'd) -> 'd -> 'd = ...
-It's a rookie mistake to quail before complicated types. You should
+But it's a rookie mistake to quail before complicated types. You should
be no more intimiated by complex types than by a linguistic tree with
deeply embedded branches: complex structure created by repeated
application of simple rules.
As usual, we need to unpack the `u` box. Examine the type of `u`.
This time, `u` will only deliver up its contents if we give `u` as an
-argument a function expecting an `'a`. Once that argument is applied
-to an object of type `'a`, we'll have what we need. Thus:
+argument a function expecting an `'a` and a `'b`. `u` will fold that function over its type `'a` members, and that's how we'll get the `'a`s we need. Thus:
- .... u (fun (a : 'a) -> ... (f a) ... ) ...
+ ... u (fun (a : 'a) (b : 'b) -> ... f a ... ) ...
-In order for `u` to have the kind of argument it needs, we have to
-adjust `(f a)` (which has type `('c -> 'd -> 'd) -> 'd -> 'd`) in
-order to deliver something of type `'b -> 'b`. The easiest way is to
-alias `'d` to `'b`, and provide `(f a)` with an argument of type `'c
--> 'b -> 'b`. Thus:
+In order for `u` to have the kind of argument it needs, the `... (f a) ...` has to evaluate to a result of type `'b`. The easiest way to do this is to collapse (or "unify") the types `'b` and `'d`, with the result that `f a` will have type `('c -> 'b -> 'b) -> 'b -> 'b`. Let's postulate an argument `k` of type `('c -> 'b -> 'b)` and supply it to `(f a)`:
- l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
- (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b)
- : ('c -> 'b -> 'b) -> 'b -> 'b =
- .... u (fun (a : 'a) -> f a k) ...
+ ... u (fun (a : 'a) (b : 'b) -> ... f a k ... ) ...
+
+Now we have an argument `b` of type `'b`, so we can supply that to `(f a) k`, getting a result of type `'b`, as we need:
-[Exercise: can you arrive at a fully general bind for this type
-constructor, one that does not collapse `'d`'s with `'b`'s?]
+ ... u (fun (a : 'a) (b : 'b) -> f a k b) ...
-As usual, we have to abstract over `k`, but this time, no further
-adjustments are needed:
+Now, we've used a `k` that we pulled out of nowhere, so we need to abstract over it:
+
+ fun (k : 'c -> 'b -> 'b) -> u (fun (a : 'a) (b : 'b) -> f a k b)
+
+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 we 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 : 'c -> 'b -> 'b) -> u (fun (a : 'a) -> f a k)
+ 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
+
+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?
+
+Suppose we have a list' whose contents are `[1; 2; 4; 8]`---that is, our list' will be `fun f z -> f 1 (f 2 (f 4 (f 8 z)))`. We call that list' `u`. Suppose we also have a function `f` that for each `int` we give it, gives back a list of the divisors of that `int` that are greater than 1. Intuitively, then, binding `u` to `f` should give us:
+
+ concat (map f u) =
+ concat [[]; [2]; [2; 4]; [2; 4; 8]] =
+ [2; 2; 4; 2; 4; 8]
+
+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
+
+do? Well, for each element `a` in `u`, it applies `f` to that `a`, getting one of the lists:
+
+ []
+ [2]
+ [2; 4]
+ [2; 4; 8]
+
+(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
+
+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
+
+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
+
+ fun k z -> List.fold_right k (concat (map f u)) z
+
+would.
+
+For future reference, we might make two eta-reductions to our formula, so that we have instead:
-You should carefully check to make sure that this term is consistent
-with the typing.
+ let l'_bind = fun k -> u (fun a -> f a k);;
-Our theory is that this monad should be capable of exactly
-replicating the behavior of the standard List monad. Let's test:
+Let's make some more tests:
l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
Let's write a general function that will map individuals into their
corresponding generalized quantifier:
- gqize (x : e) = fun (p : e -> t) -> p x
+ gqize (a : e) = fun (p : e -> t) -> p a
This function wraps up an individual in a fancy box. That is to say,
we are in the presence of a monad. The type constructor, the unit and
similar to the List monad just given:
type 'a continuation = ('a -> 'b) -> 'b
- c_unit (x : 'a) = fun (p : 'a -> 'b) -> p x
+ 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 (x : 'a) -> f x k)
+ fun (k : 'a -> 'b) -> u (fun (a : 'a) -> f a k)
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 x = fun f -> f x
- l'_bind u f = fun k -> u (fun x -> f x k)
+ l'_unit a = fun f -> f a
+ l'_bind u f = fun k -> u (fun a -> f a k)
(We performed a sneaky but valid eta reduction in the unit term.)