XGitUrl: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=manipulating_trees_with_monads.mdwn;h=61d296405e87a62bc2cfa2e071ba8f4353fdd756;hp=69b5131e614359dd6f2e6b12a7c1499f4cf7fb54;hb=9efbe94f74c2ea61522fcdb3e3d012fde6034fcd;hpb=5f8ea7e4c56ab00888ca59c5ba0fb6e3d5d5a65c
diff git a/manipulating_trees_with_monads.mdwn b/manipulating_trees_with_monads.mdwn
index 69b5131e..61d29640 100644
 a/manipulating_trees_with_monads.mdwn
+++ b/manipulating_trees_with_monads.mdwn
@@ 44,18 +44,18 @@ We'll be using trees where the nodes are integers, e.g.,
Our first task will be to replace each leaf with its double:
 let rec treemap (newleaf : 'a > 'b) (t : 'a tree) : 'b tree =
+ let rec tree_map (leaf_modifier : 'a > 'b) (t : 'a tree) : 'b tree =
match t with
  Leaf i > Leaf (newleaf i)
  Node (l, r) > Node (treemap newleaf l,
 treemap newleaf r);;
+  Leaf i > Leaf (leaf_modifier i)
+  Node (l, r) > Node (tree_map leaf_modifier l,
+ tree_map leaf_modifier r);;
`treemap` takes a function that transforms old leaves into new leaves,
+`tree_map` takes a function that transforms old leaves into new leaves,
and maps that function over all the leaves in the tree, leaving the
structure of the tree unchanged. For instance:
let double i = i + i;;
 treemap double t1;;
+ tree_map double t1;;
 : int tree =
Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
@@ 70,28 +70,29 @@ structure of the tree unchanged. For instance:
 
14 22
We could have built the doubling operation right into the `treemap`
code. However, because what to do to each leaf is a parameter, we can
+We could have built the doubling operation right into the `tree_map`
+code. However, because we've left what to do to each leaf as a parameter, we can
decide to do something else to the leaves without needing to rewrite
`treemap`. For instance, we can easily square each leaf instead by
+`tree_map`. For instance, we can easily square each leaf instead by
supplying the appropriate `int > int` operation in place of `double`:
let square i = i * i;;
 treemap square t1;;
+ tree_map square t1;;
 : int tree =ppp
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
Note that what `treemap` does is take some global, contextual
+Note that what `tree_map` does is take some unchanging contextual
informationwhat to do to each leafand supplies that information
to each subpart of the computation. In other words, `treemap` has the
+to each subpart of the computation. In other words, `tree_map` has the
behavior of a reader monad. Let's make that explicit.
In general, we're on a journey of making our treemap function more and
+In general, we're on a journey of making our `tree_map` function more and
more flexible. So the next stepcombining the tree transformer with
a reader monadis to have the treemap function return a (monadized)
+a reader monadis to have the `tree_map` function return a (monadized)
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 .
_________
@@ 112,77 +113,106 @@ 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
simulation), a world (the intensionality monad), an integer (the
Jacobsoninspired link monad), etc. In this situation, it will be
enough for now to expect that our reader will expect a function of
type `int > int`.
+Jacobsoninspired link monad), etc. In the present case, we expect that our "environment" will be some function of type `int > int`. "Looking up" some `int` in the environment will return us the `int` that comes out the other side of that function.
type 'a reader = (int > int) > 'a;; (* mnemonic: e for environment *)
let reader_unit (a : 'a) : 'a reader = fun _ > a;;
let reader_bind (u: 'a reader) (f : 'a > 'b reader) : 'b reader = fun e > f (u e) e;;
It's easy to figure out how to turn an `int` into an `int reader`:
+It would be a simple matter to turn an *integer* into an `int reader`:
 let int2int_reader : 'a > 'b reader = fun (a : 'a) > fun (op : 'a > 'b) > op a;;
 int2int_reader 2 (fun i > i + i);;
+ let int_readerize : int > int reader = fun (a : int) > fun (modifier : int > int) > modifier a;;
+ int_readerize 2 (fun i > i + i);;
 : int = 4
But what do we do when the integers are scattered over the leaves of a
tree? A binary tree is not the kind of thing that we can apply a
+But how do we do the analagous transformation when our `int`s are scattered over the leaves of a tree? How do we turn an `int tree` into a reader?
+A tree is not the kind of thing that we can apply a
function of type `int > int` to.
 let rec treemonadizer (f : 'a > 'b reader) (t : 'a tree) : 'b tree reader =
+But we can do this:
+
+ let rec tree_monadize (f : 'a > 'b reader) (t : 'a tree) : 'b tree reader =
match t with
 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 >
+  Node (l, r) > reader_bind (tree_monadize f l) (fun x >
+ reader_bind (tree_monadize f r) (fun y >
reader_unit (Node (x, y))));;
This function says: give me a function `f` that knows how to turn
something of type `'a` into an `'b reader`, and I'll show you how to
turn an `'a tree` into an `'a tree reader`. In more fanciful terms,
the `treemonadizer` function builds plumbing that connects all of the
leaves of a tree into one connected monadic network; it threads the
`'b reader` monad through the leaves.
+something of type `'a` into an `'b reader`this is a function of the same type that you could bind an `'a reader` toand I'll show you how to
+turn an `'a tree` into an `'b tree reader`. That is, if you show me how to do this:
+
+ 
+ 1 >  1 
+ 
+
+then I'll give you back the ability to do this:
+
+ ____________
+ .  . 
+ _____ >  _____ 
+      
+ 1 2  1 2 
+ 
+
+And how will that boxed tree behave? Whatever actions you perform on it will be transmitted down to corresponding operations on its leaves. For instance, our `int reader` expects an `int > int` environment. If supplying environment `e` to our `int reader` doubles the contained `int`:
 # treemonadizer int2int_reader t1 (fun i > i + i);;
+ 
+ 1 >  1  applied to e ~~> 2
+ 
+
+Then we can expect that supplying it to our `int tree reader` will double all the leaves:
+
+ ____________
+ .  .  .
+ _____ >  _____  applied to e ~~> _____
+        
+ 1 2  1 2  2 4
+ 
+
+In more fanciful terms, the `tree_monadize` function builds plumbing that connects all of the leaves of a tree into one connected monadic network; it threads the
+`'b reader` monad through the original tree's leaves.
+
+ # tree_monadize int_readerize t1 double;;
 : int tree =
Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
Here, our environment is the doubling function (`fun i > i + i`). If
we apply the very same `int tree reader` (namely, `treemonadizer
int2int_reader t1`) to a different `int > int` functionsay, the
+we apply the very same `int tree reader` (namely, `tree_monadize
+int_readerize t1`) to a different `int > int` functionsay, the
squaring function, `fun i > i * i`we get an entirely different
result:
 # treemonadizer int2int_reader t1 (fun i > i * i);;
+ # tree_monadize int_readerize t1 square;;
 : int tree =
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
Now that we have a tree transformer that accepts a reader monad as a
+Now that we have a tree transformer that accepts a *reader* monad as a
parameter, we can see what it would take to swap in a different monad.
For instance, we can use a state monad to count the number of nodes in
+
+For instance, we can use a state monad to count the number of leaves in
the tree.
type 'a state = int > 'a * int;;
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);;
+ let state_bind u f = fun s > let (a, s') = u s in f a s';;
Gratifyingly, we can use the `treemonadizer` function without any
+Gratifyingly, we can use the `tree_monadize` function without any
modification whatsoever, except for replacing the (parametric) type
`'b reader` with `'b state`, and substituting in the appropriate unit and bind:
 let rec treemonadizer (f : 'a > 'b state) (t : 'a tree) : 'b tree state =
+ let rec tree_monadize (f : 'a > 'b state) (t : 'a tree) : 'b tree state =
match t with
  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 >
+  Leaf i > state_bind (f i) (fun i' > state_unit (Leaf i'))
+  Node (l, r) > state_bind (tree_monadize f l) (fun x >
+ state_bind (tree_monadize f r) (fun y >
state_unit (Node (x, y))));;
Then we can count the number of nodes in the tree:
+Then we can count the number of leaves in the tree:
 # treemonadizer state_unit t1 0;;
+ # tree_monadize (fun a > fun s > (a, s+1)) t1 0;;
 : int tree * int =
 (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13)
+ (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 5)
.
______
@@ 195,22 +225,12 @@ Then we can count the number of nodes in the tree:
 
7 11
Notice that we've counted each internal node twiceit's a good
exercise to adjust the code to count each node once.



+Why does this work? Because the operation `fun a > fun s > (a, s+1)` takes an `int` and wraps it in an `int state` monadic box that increments the state. When we give that same operations to our `tree_monadize` function, it then wraps an `int tree` in a box, one that does the same stateincrementing for each of its leaves.
One more revealing example before getting down to business: replacing
`state` everywhere in `treemonadizer` with `list` gives us
+`state` everywhere in `tree_monadize` with `list` gives us
 # treemonadizer (fun i > [ [i; square i] ]) t1;;
+ # tree_monadize (fun i > [ [i; square i] ]) t1;;
 : int list tree list =
[Node
(Node (Leaf [2; 4], Leaf [3; 9]),
@@ 218,10 +238,11 @@ One more revealing example before getting down to business: replacing
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.
+a list of `int`'s. We might also have done this with a Reader Monad, though then our environments would need to be of type `int > int list`. Experiment with what happens if you supply the `tree_monadize` based on the List Monad an operation like `fun > [ i; [2*i; 3*i] ]`. Use small trees for your experiment.
+
@@ 232,49 +253,50 @@ of leaves?
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 =
+ let rec tree_monadize (f : 'a > ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation =
match t with
 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 >
+  Node (l, r) > continuation_bind (tree_monadize f l) (fun x >
+ continuation_bind (tree_monadize f r) (fun y >
continuation_unit (Node (x, y))));;
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 `tree_monadize` code. Then if we give the `tree_monadize` function an operation that converts `int`s into continuations expecting `'b` arguments, it will give us back a way to turn `int tree`s into continuations that expect `'b tree` arguments. The effect of giving the continuation such an argument will be to distribute across the `'b tree`'s leaves effects that parallel the effects that the `'b`expecting continuations would have on their `'b`s.
+
+So for example, we compute:
 # treemonadizer (fun a k > a :: (k a)) t1 (fun t > []);;
+ # tree_monadize (fun a > fun 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.
+We have found a way of collapsing a tree into a list of its leaves. Can you trace how this is working?
The continuation monad is amazingly flexible; we can use it to
simulate some of the computations performed above. To see how, first
note that an interestingly uninteresting thing happens if we use the
continuation unit as our first argument to `treemonadizer`, and then
+note that an interestingly uninteresting thing happens if we use
+`continuation_unit` as our first argument to `tree_monadize`, and then
apply the result to the identity function:
 # treemonadizer continuation_unit t1 (fun i > i);;
+ # tree_monadize continuation_unit t1 (fun i > i);;
 : int tree =
Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
That is, nothing happens. But we can begin to substitute more
interesting functions for the first argument of `treemonadizer`:
+interesting functions for the first argument of `tree_monadize`:
(* Simulating the tree reader: distributing a operation over the leaves *)
 # treemonadizer (fun a k > k (square a)) t1 (fun i > i);;
+ # tree_monadize (fun a > fun 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 *)
 # treemonadizer (fun a k > k [a; square a]) t1 (fun i > i);;
+ # tree_monadize (fun a > fun 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 *)
 # treemonadizer (fun a k > 1 + k a) t1 (fun i > 0);;
+ # tree_monadize (fun a > fun k > 1 + k a) t1 (fun i > 0);;
 : int = 5
We could simulate the tree state example too, but it would require
@@ 282,6 +304,9 @@ generalizing the type of the continuation monad to
type ('a, 'b, 'c) continuation = ('a > 'b) > 'c;;
+If you want to see how to parameterize the definition of the `tree_monadize` function, so that you don't have to keep rewriting it for each new monad, see [this code](/code/tree_monadize.ml).
+
+
The binary tree monad

@@ 305,7 +330,7 @@ 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 `f a`:
\tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5)))
+\tree (. (f a1) (. (. (. (f a2) (f a3)) (f a4)) (f a5)))
. .
____ ____
@@ 331,13 +356,13 @@ falls out once we realize that
As for the associative law,
 Associativity: bind (bind u f) g = bind u (\a. bind (fa) g)
+ Associativity: bind (bind u f) g = bind u (\a. bind (f a) g)
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)))))
.
________