From: jim <jim@web>
Date: Tue, 10 Feb 2015 17:37:59 +0000 (-0500)
Subject: fill in answers to 25
X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=commitdiff_plain;h=663dd241274261fa15316bc9621eed9b5c71791f;hp=13d1f0c86e0ffb3fdeca38f40640160b579049aa

fill in answers to 25
---

diff --git a/exercises/_assignment2_answers.mdwn b/exercises/_assignment2_answers.mdwn
index e131311c..bb3b724f 100644
--- a/exercises/_assignment2_answers.mdwn
+++ b/exercises/_assignment2_answers.mdwn
@@ -182,8 +182,58 @@ Folds and Lists
 
 25. We mentioned in the Encoding notes that `fold_left (flipped_cons, []) xs` would give us the elements of `xs` but in the reverse order. So that's how we can express `reverse` in terms of `fold_left`. How would you express `reverse` in terms of `fold_right`? As with problem 22, don't use `letrec`!
 
-    See the [[hint|assignment2 hint]].
+    *Here is a boring, inefficient answer*
 
+        let
+          append (ys, zs) = fold_right ((&), zs) ys;  # aka (&&)
+          f (y, prev) = append (prev, [y]);
+          reverse xs = fold_right (f, []) xs
+        in reverse
+
+    or (same basic idea, just written differently):
+
+        let
+          f (y, prev) = fold_right ((&), [y]) prev;
+          reverse xs = fold_right (f, []) xs
+        in reverse
+
+    *Here is an elegant, efficient answer following the [[hint|assignment2 hint]]*
+
+    Suppose the list we want to reverse is `[10, 20, 30]`. Applying `fold_right` to this will begin by computing `f (30, z)` for some `f` and `z` that we specify. If we made the result of that be something like `30 & blah`, or any larger structure that contained something of that form, it's not clear how we could, using just the resources of `fold_right`, reach down into that structure and replace the `blah` with some other element, as we'd evidently need to, since after the next step we should get `30 & (20 & blah)`. What we'd like instead is something like this:
+
+        30 & < >
+
+    Where `< >` isn't some *value* but rather a *hole*. Then with the next step, we want to plug into that hole `20 & < >`, which contains its own hole. Getting:
+
+        30 & (20 & < >)
+
+    And so on. That is the key to the solution. The questions you need to answer, to turn this into something executable, are:
+
+    1.  What is a hole? How can we implement it?
+
+        A hole is a bound variable. `30 & < >` is `lambda x. 30 & x`.
+
+    2.  What should `f` be, so that the result of the second step, namely `f (20, 30 & < >)`, is `30 & (20 & < >)`?
+
+            let
+              f (y, prev) = lambda x. prev (y & x)
+            in ...
+
+    3.  Given that choice of `f`, what should `z` be, so that the result of the first step, namely `f (30, z)` is `30 & < >`?
+
+        The identity function: `f (30, (lambda y. y))` will reduce to `lambda x. (lambda y. y) (30 & x)`, which will reduce to `lambda x. 30 & x`.
+
+    4.  At the end of the `fold_right`, we're going to end with something like `30 & (20 & (10 & < >))`. But what we want is `[30, 20, 10]`. How can we turn what we've gotten into what we want?
+
+        Supply it with `[]` as an argument.
+
+    5.  So now put it all together, and explain how to express `reverse xs` using `fold_right` and primitive syntax like `lambda`, `&`, and `[]`?
+
+            let
+              f (y, prev) = lambda x. prev (y & x);
+              id match lambda y. y;
+              reverse xs = (fold_right (f, id) xs) []
+            in reverse
 
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