From 842e591b50617da16e6bf1727322a0e5a755b609 Mon Sep 17 00:00:00 2001 From: Jim Pryor Date: Wed, 1 Dec 2010 01:56:03 -0500 Subject: [PATCH 1/1] lists-monad tweaks Signed-off-by: Jim Pryor --- list_monad_as_continuation_monad.mdwn | 22 +++++++++++----------- 1 file changed, 11 insertions(+), 11 deletions(-) diff --git a/list_monad_as_continuation_monad.mdwn b/list_monad_as_continuation_monad.mdwn index d2136dbe..970a587d 100644 --- a/list_monad_as_continuation_monad.mdwn +++ b/list_monad_as_continuation_monad.mdwn @@ -218,15 +218,15 @@ paragraph much easier to follow.] 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` an argument that is 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: +fold that function over its type `'a` members, and that's where we can get at the `'a`s we need. Thus: ... u (fun (a : 'a) (b : 'b) -> ... f a ... ) ... -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)`: +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)`: ... 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: +Now the function we're supplying to `u` also receives an argument `b` of type `'b`, so we can supply that to `f a k`, getting a result of type `'b`, as we need: ... u (fun (a : 'a) (b : 'b) -> f a k b) ... @@ -251,8 +251,8 @@ Now let's think about what this does. It's a wrapper around `u`. In order to beh 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]] = + List.concat (List.map f u) = + List.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 @@ -271,18 +271,18 @@ do? Well, for each element `a` in `u`, it applies `f` to that `a`, getting one o 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)) + 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 +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 + fun k z -> List.fold_right k (List.concat (List.map f u)) z would. -- 2.11.0