X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=manipulating_trees_with_monads.mdwn;h=d0897efd40ef34360d051c96a349accffb6fdca5;hp=2debebe6aa04ad42b1efd92dc50e26e71ab367fb;hb=12ba49b7826c64a85032e1640db29d4c947347f9;hpb=c6ac543d21f9a20d334e3eb9b48120713881bd47 diff --git a/manipulating_trees_with_monads.mdwn b/manipulating_trees_with_monads.mdwn index 2debebe6..d0897efd 100644 --- a/manipulating_trees_with_monads.mdwn +++ b/manipulating_trees_with_monads.mdwn @@ -28,14 +28,14 @@ internal nodes?] We'll be using trees where the nodes are integers, e.g., - let t1 = Node ((Node ((Leaf 2), (Leaf 3))), - (Node ((Leaf 5),(Node ((Leaf 7), - (Leaf 11)))))) + let t1 = Node (Node (Leaf 2, Leaf 3), + Node (Leaf 5, Node (Leaf 7, + Leaf 11))) . ___|___ | | . . - _|__ _|__ + _|_ _|__ | | | | 2 3 5 . _|__ @@ -46,9 +46,9 @@ 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 - | Leaf x -> Leaf (newleaf x) - | Node (l, r) -> Node ((treemap newleaf l), - (treemap newleaf r));; + | Leaf i -> Leaf (newleaf i) + | Node (l, r) -> Node (treemap newleaf l, + treemap newleaf r);; `treemap` takes a function that transforms old leaves into new leaves, and maps that function over all the leaves in the tree, leaving the @@ -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`: - 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))) @@ -89,50 +89,51 @@ behavior of a reader monad. Let's make that explicit. In general, we're on a journey of making our treemap function more and more flexible. So the next step---combining the tree transformer with 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 +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 . - ____|____ - | | - . . - __|__ __|__ - | | | | - f2 f3 f5 . - __|___ - | | - f7 f11 + \f . + _____|____ + | | + . . + __|___ __|___ + | | | | + f 2 f 3 f 5 . + __|___ + | | + f 7 f 11 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 returns an `int -tree` in which each leaf `x` has been replaced with `(f x)`. +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 `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 Jacobson-inspired 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`. +type `int -> int`. - type 'a reader = (int->int) -> 'a;; (* mnemonic: e for environment *) - let reader_unit (x : 'a) : 'a reader = fun _ -> x;; - let reader_bind (u: 'a reader) (f : 'a -> 'c reader) : 'c reader = fun e -> f (u e) e;; + 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`: - let int2int_reader (x : 'a): 'b reader = fun (op : 'a -> 'b) -> op x;; + let int2int_reader : 'a -> 'b reader = fun (a : 'a) -> fun (op : 'a -> 'b) -> op a;; int2int_reader 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 -function of type `int->int` to. +function of type `int -> int` to. 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))));; @@ -142,7 +143,7 @@ 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 -monad through the leaves. +`'b reader` monad through the leaves. # treemonadizer int2int_reader t1 (fun i -> i + i);; - : int tree = @@ -150,7 +151,7 @@ monad through the leaves. 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` function---say, the +int2int_reader t1`) to a different `int -> int` function---say, the squaring function, `fun i -> i * i`---we get an entirely different result: @@ -158,24 +159,24 @@ result: - : 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 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 the tree. type 'a state = int -> 'a * int;; - let state_unit x i = (x, i+.5);; - let state_bind u f i = let (a, i') = u i in f a (i'+.5);; + 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 -`reader` with `state`: +`'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 = 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: @@ -198,10 +199,19 @@ Then we can count the number of nodes in the tree: Notice that we've counted each internal node twice---it's a good exercise to adjust the code to count each node once. + + + 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]), @@ -211,16 +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. + + + 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 - | 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))));; @@ -229,7 +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: - # 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. @@ -240,7 +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: - # 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))) @@ -248,25 +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 *) - # 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 *) - # 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 *) - # 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 - type ('a -> 'b -> 'c) continuation = ('a -> 'b) -> 'c;; + type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;; The binary tree monad --------------------- @@ -275,37 +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);; - 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 - | 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: - 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`, -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 = __|__ | | | | - . a4 . fa4 + . a4 . f a4 __|__ __|___ | | | | - a2 a3 fa2 fa3 + a2 a3 f a2 f a3 Given this equivalence, the right identity law @@ -322,26 +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 -\tree (. (. (. (. (a1)(a2))))) -\tree (. (. (. (. (a1) (a1)) (. (a1) (a1))) )) +\tree (. (. (. (. (a1) (a2))))) +\tree (. (. (. (. (a1) (a1)) (. (a1) (a1))))) . ____|____ . . | | 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 - . - ____|____ - | | - . . - __|__ __|__ - | | | | - 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