week11 tweaks

[lambda.git] / manipulating_trees_with_monads.mdwn

diff --git a/manipulating_trees_with_monads.mdwn b/manipulating_trees_with_monads.mdwn
index de8fc5f..d0897ef 100644 (file)
--- a/manipulating_trees_with_monads.mdwn
+++ b/manipulating_trees_with_monads.mdwn
@@ -46,7 +46,7 @@ Our first task will be to replace each leaf with its double:
 
        let rec treemap (newleaf : 'a -> 'b) (t : 'a tree) : 'b tree =
          match t with
 
        let rec treemap (newleaf : 'a -> 'b) (t : 'a tree) : 'b tree =
          match t with

-           | Leaf x -> Leaf (newleaf x)
+           | Leaf i -> Leaf (newleaf i)

            | Node (l, r) -> Node (treemap newleaf l,
                                   treemap newleaf r);;
 
            | Node (l, r) -> Node (treemap newleaf l,
                                   treemap newleaf r);;
 


@@ -76,7 +76,7 @@ decide to do something else to the leaves without needing to rewrite
 `treemap`.  For instance, we can easily square each leaf instead by
 supplying the appropriate `int -> int` operation in place of `double`:
 
 `treemap`.  For instance, we can easily square each leaf instead by
 supplying the appropriate `int -> int` operation in place of `double`:
 

-       let square x = x * x;;
+       let square i = i * i;;

        treemap square t1;;
        - : int tree =ppp
        Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
        treemap square t1;;
        - : int tree =ppp
        Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))


@@ -92,6 +92,7 @@ a reader monad---is to have the treemap function return a (monadized)
 tree that is ready to accept any `int -> int` function and produce the
 updated tree.
 
 tree that is ready to accept any `int -> int` function and produce the
 updated tree.
 

+\tree (. (. (f 2) (f 3)) (. (f 5) (. (f 7) (f 11))))

 
        \f      .
           _____|____
 
        \f      .
           _____|____


@@ -107,7 +108,7 @@ updated tree.
 That is, we want to transform the ordinary tree `t1` (of type `int
 tree`) into a reader object of type `(int -> int) -> int tree`: something
 that, when you apply it to an `int -> int` function `f` returns an `int
 That is, we want to transform the ordinary tree `t1` (of type `int
 tree`) into a reader object of type `(int -> int) -> int tree`: something
 that, when you apply it to an `int -> int` function `f` returns an `int

-tree` in which each leaf `x` has been replaced with `f x`.
+tree` in which each leaf `i` has been replaced with `f i`.

 
 With previous readers, we always knew which kind of environment to
 expect: either an assignment function (the original calculator
 
 With previous readers, we always knew which kind of environment to
 expect: either an assignment function (the original calculator


@@ -132,7 +133,7 @@ function of type `int -> int` to.
 
        let rec treemonadizer (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =
            match t with
 
        let rec treemonadizer (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =
            match t with

-           | Leaf x -> reader_bind (f x) (fun x' -> reader_unit (Leaf x'))
+           | Leaf i -> reader_bind (f i) (fun i' -> reader_unit (Leaf i'))

            | Node (l, r) -> reader_bind (treemonadizer f l) (fun x ->
                               reader_bind (treemonadizer f r) (fun y ->
                                 reader_unit (Node (x, y))));;
            | Node (l, r) -> reader_bind (treemonadizer f l) (fun x ->
                               reader_bind (treemonadizer f r) (fun y ->
                                 reader_unit (Node (x, y))));;


@@ -164,8 +165,8 @@ For instance, we can use a state monad to count the number of nodes in
 the tree.
 
        type 'a state = int -> 'a * int;;
 the tree.
 
        type 'a state = int -> 'a * int;;

-       let state_unit a = fun i -> (a, i);;
-       let state_bind u f = fun i -> let (a, i') = u i in f a (i' + 1);;
+       let state_unit a = fun s -> (a, s);;
+       let state_bind_and_count u f = fun s -> let (a, s') = u s in f a (s' + 1);;

 
 Gratifyingly, we can use the `treemonadizer` function without any
 modification whatsoever, except for replacing the (parametric) type
 
 Gratifyingly, we can use the `treemonadizer` function without any
 modification whatsoever, except for replacing the (parametric) type


@@ -173,9 +174,9 @@ modification whatsoever, except for replacing the (parametric) type
 
        let rec treemonadizer (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
            match t with
 
        let rec treemonadizer (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
            match t with

-           | Leaf x -> state_bind (f x) (fun x' -> state_unit (Leaf x'))
-           | Node (l, r) -> state_bind (treemonadizer f l) (fun x ->
-                              state_bind (treemonadizer f r) (fun y ->
+           | Leaf i -> state_bind_and_count (f i) (fun i' -> state_unit (Leaf i'))
+           | Node (l, r) -> state_bind_and_count (treemonadizer f l) (fun x ->
+                              state_bind_and_count (treemonadizer f r) (fun y ->

                                 state_unit (Node (x, y))));;
 
 Then we can count the number of nodes in the tree:
                                 state_unit (Node (x, y))));;
 
 Then we can count the number of nodes in the tree:


@@ -210,7 +211,7 @@ But I assume Chris means here, adjust the code so that no corrections of this so
 One more revealing example before getting down to business: replacing
 `state` everywhere in `treemonadizer` with `list` gives us
 
 One more revealing example before getting down to business: replacing
 `state` everywhere in `treemonadizer` with `list` gives us
 

-       # treemonadizer (fun x -> [ [x; square x] ]) t1;;
+       # treemonadizer (fun i -> [ [i; square i] ]) t1;;

        - : int list tree list =
        [Node
          (Node (Leaf [2; 4], Leaf [3; 9]),
        - : int list tree list =
        [Node
          (Node (Leaf [2; 4], Leaf [3; 9]),


@@ -220,16 +221,21 @@ Unlike the previous cases, instead of turning a tree into a function
 from some input to a result, this transformer replaces each `int` with
 a list of `int`'s.
 
 from some input to a result, this transformer replaces each `int` with
 a list of `int`'s.
 

+<!--
+We don't make it clear why the fun has to be int -> int list list, instead of int -> int list
+-->
+
+

 Now for the main point.  What if we wanted to convert a tree to a list
 of leaves?
 
        type ('a, 'r) continuation = ('a -> 'r) -> 'r;;
 Now for the main point.  What if we wanted to convert a tree to a list
 of leaves?
 
        type ('a, 'r) continuation = ('a -> 'r) -> 'r;;

-       let continuation_unit x c = c x;;
-       let continuation_bind u f c = u (fun a -> f a c);;
+       let continuation_unit a = fun k -> k a;;
+       let continuation_bind u f = fun k -> u (fun a -> f a k);;

        
        let rec treemonadizer (f : 'a -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation =
            match t with
        
        let rec treemonadizer (f : 'a -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation =
            match t with

-           | Leaf x -> continuation_bind (f x) (fun x' -> continuation_unit (Leaf x'))
+           | Leaf i -> continuation_bind (f i) (fun i' -> continuation_unit (Leaf i'))

            | Node (l, r) -> continuation_bind (treemonadizer f l) (fun x ->
                               continuation_bind (treemonadizer f r) (fun y ->
                                 continuation_unit (Node (x, y))));;
            | Node (l, r) -> continuation_bind (treemonadizer f l) (fun x ->
                               continuation_bind (treemonadizer f r) (fun y ->
                                 continuation_unit (Node (x, y))));;


@@ -238,7 +244,7 @@ We use the continuation monad described above, and insert the
 `continuation` type in the appropriate place in the `treemonadizer` code.
 We then compute:
 
 `continuation` type in the appropriate place in the `treemonadizer` code.
 We then compute:
 

-       # treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);;
+       # treemonadizer (fun a k -> a :: (k a)) t1 (fun t -> []);;

        - : int list = [2; 3; 5; 7; 11]
 
 We have found a way of collapsing a tree into a list of its leaves.
        - : int list = [2; 3; 5; 7; 11]
 
 We have found a way of collapsing a tree into a list of its leaves.


@@ -249,7 +255,7 @@ note that an interestingly uninteresting thing happens if we use the
 continuation unit as our first argument to `treemonadizer`, and then
 apply the result to the identity function:
 
 continuation unit as our first argument to `treemonadizer`, and then
 apply the result to the identity function:
 

-       # treemonadizer continuation_unit t1 (fun x -> x);;
+       # treemonadizer continuation_unit t1 (fun i -> i);;

        - : int tree =
        Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
 
        - : int tree =
        Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
 


@@ -257,25 +263,25 @@ That is, nothing happens.  But we can begin to substitute more
 interesting functions for the first argument of `treemonadizer`:
 
        (* Simulating the tree reader: distributing a operation over the leaves *)
 interesting functions for the first argument of `treemonadizer`:
 
        (* Simulating the tree reader: distributing a operation over the leaves *)

-       # treemonadizer (fun a c -> c (square a)) t1 (fun x -> x);;
+       # treemonadizer (fun a k -> k (square a)) t1 (fun i -> i);;

        - : int tree =
        Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
 
        (* Simulating the int list tree list *)
        - : int tree =
        Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
 
        (* Simulating the int list tree list *)

-       # treemonadizer (fun a c -> c [a; square a]) t1 (fun x -> x);;
+       # treemonadizer (fun a k -> k [a; square a]) t1 (fun i -> i);;

        - : int list tree =
        Node
         (Node (Leaf [2; 4], Leaf [3; 9]),
          Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
 
        (* Counting leaves *)
        - : int list tree =
        Node
         (Node (Leaf [2; 4], Leaf [3; 9]),
          Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
 
        (* Counting leaves *)

-       # treemonadizer (fun a c -> 1 + c a) t1 (fun x -> 0);;
+       # treemonadizer (fun a k -> 1 + k a) t1 (fun i -> 0);;

        - : int = 5
 
 We could simulate the tree state example too, but it would require
 generalizing the type of the continuation monad to
 
        - : int = 5
 
 We could simulate the tree state example too, but it would require
 generalizing the type of the continuation monad to
 

-       type ('a -> 'b -> 'c) continuation = ('a -> 'b) -> 'c;;
+       type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;;

 
 The binary tree monad
 ---------------------
 
 The binary tree monad
 ---------------------


@@ -284,37 +290,37 @@ Of course, by now you may have realized that we have discovered a new
 monad, the binary tree monad:
 
        type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
 monad, the binary tree monad:
 
        type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;

-       let tree_unit (x: 'a) = Leaf x;;
+       let tree_unit (a: 'a) = Leaf a;;

        let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =
            match u with
        let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =
            match u with

-           | Leaf x -> f x
+           | Leaf a -> f a

            | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));;
 
 For once, let's check the Monad laws.  The left identity law is easy:
 
            | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));;
 
 For once, let's check the Monad laws.  The left identity law is easy:
 

-    Left identity: bind (unit a) f = bind (Leaf a) f = fa
+    Left identity: bind (unit a) f = bind (Leaf a) f = f a

 
 To check the other two laws, we need to make the following
 observation: it is easy to prove based on `tree_bind` by a simple
 induction on the structure of the first argument that the tree
 resulting from `bind u f` is a tree with the same strucure as `u`,
 
 To check the other two laws, we need to make the following
 observation: it is easy to prove based on `tree_bind` by a simple
 induction on the structure of the first argument that the tree
 resulting from `bind u f` is a tree with the same strucure as `u`,

-except that each leaf `a` has been replaced with `fa`:
+except that each leaf `a` has been replaced with `f a`:

 
 

-\tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5)))
+\tree (. (f a1) (. (. (. (f a2) (f a3)) (f a4)) (f a5)))

 
                        .                         .
                      __|__                     __|__
                      |   |                     |   |
 
                        .                         .
                      __|__                     __|__
                      |   |                     |   |

-                     a1  .                    fa1  .
+                     a1  .                   f a1  .

                         _|__                     __|__
                         |  |                     |   |
                         _|__                     __|__
                         |  |                     |   |

-                        .  a5                    .  fa5
+                        .  a5                    .  f a5

           bind         _|__       f   =        __|__
                        |  |                    |   |
           bind         _|__       f   =        __|__
                        |  |                    |   |

-                       .  a4                   .  fa4
+                       .  a4                   .  f a4

                      __|__                   __|___
                      |   |                   |    |
                      __|__                   __|___
                      |   |                   |    |

-                     a2  a3                 fa2  fa3
+                     a2  a3                f a2  f a3

 
 Given this equivalence, the right identity law
 
 
 Given this equivalence, the right identity law
 


@@ -331,26 +337,26 @@ As for the associative law,
 we'll give an example that will show how an inductive proof would
 proceed.  Let `f a = Node (Leaf a, Leaf a)`.  Then
 
 we'll give an example that will show how an inductive proof would
 proceed.  Let `f a = Node (Leaf a, Leaf a)`.  Then
 

-\tree (. (. (. (. (a1)(a2)))))
-\tree (. (. (. (. (a1) (a1)) (. (a1) (a1)))  ))
+\tree (. (. (. (. (a1) (a2)))))
+\tree (. (. (. (. (a1) (a1)) (. (a1) (a1)))))

 
                                                   .
                                               ____|____
                  .               .            |       |
        bind    __|__   f  =    __|_    =      .       .
                |   |           |   |        __|__   __|__
 
                                                   .
                                               ____|____
                  .               .            |       |
        bind    __|__   f  =    __|_    =      .       .
                |   |           |   |        __|__   __|__

-               a1  a2         fa1 fa2       |   |   |   |
+               a1  a2        f a1 f a2      |   |   |   |

                                             a1  a1  a1  a1
 
 Now when we bind this tree to `g`, we get
 
                                             a1  a1  a1  a1
 
 Now when we bind this tree to `g`, we get
 

-                  .
-              ____|____
-              |       |
-              .       .
-            __|__   __|__
-            |   |   |   |
-           ga1 ga1 ga1 ga1
+                   .
+              _____|______
+              |          |
+              .          .
+            __|__      __|__
+            |   |      |   |
+          g a1 g a1  g a1 g a1

 
 At this point, it should be easy to convince yourself that
 using the recipe on the right hand side of the associative law will
 
 At this point, it should be easy to convince yourself that
 using the recipe on the right hand side of the associative law will