From: Jim Pryor <profjim@jimpryor.net>
Date: Sun, 3 Oct 2010 22:11:51 +0000 (-0400)
Subject: continuing assignment4
X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=commitdiff_plain;h=d921bb783f9bbdb90a4b849a2846e4b59a4626a7

continuing assignment4

Signed-off-by: Jim Pryor <profjim@jimpryor.net>
---

diff --git a/assignment4.mdwn b/assignment4.mdwn
index cd89a134..cfe621cd 100644
--- a/assignment4.mdwn
+++ b/assignment4.mdwn
@@ -12,46 +12,73 @@ can use.
 
 #Comparing lists for equality#
 
+
 <OL start=2>
-<LI>blah
+<LI>Suppose you have two lists of integers, `left` and `right`. You want to determine whether those lists are equal: that is, whether they have all the same members in the same order. (Equality for the lists we're working with is *extensional*, or parasitic on the equality of their members, and the list structure. Later in the course we'll see lists which aren't extensional in this way.)
+
+How would you implement such a list comparison?
+
+(See [[hints/Assignment 4 hint 2]] if you need some hints.)
 </OL>
-<!--
-let list_equal =
-    \left right. left
-                ; here's our f
-                (\hd sofar.
-                    ; deconstruct our sofar-pair
-                    sofar (\might_be_equal right_tail.
-                        ; our new sofar
-                        make_pair
-                        (and (and might_be_equal (not (isempty right_tail))) (eq? hd (head right_tail)))
-                        (tail right_tail)
-                    ))
-                ; here's our z
-                ; we pass along the fold a pair
-                ; (might_for_all_i_know_still_be_equal?, tail_of_reversed_right)
-                ; when left is empty, the lists are equal if right is empty
-                (make_pair
-                    true ; for all we know so far, they might still be equal
-                    (reverse right)
-                )
-                ; when fold is finished, check sofar-pair
-                (\might_be_equal right_tail. and might_be_equal (isempty right_tail))
--->
 
-#Mutually-recursive functions#
+
+#Enumerating the fringe of a leaf-labeled tree#
+
+First, read this: [[Implementing trees]]
 
 <OL start=3>
 <LI>blah
 </OL>
 
 
-#Enumerating the fringe of a leaf-labeled tree#
+#Mutually-recursive functions#
 
-[[Implementing trees]]
+<OL start=4>
+<LI>(Challenging.) One way to define the function `even` is to have it hand off part of the work to another function `odd`:
 
+	let even = \x. iszero x
+					; if x == 0 then result is
+					true
+					; else result turns on whether x's pred is odd
+					(odd (pred x))
 
-<OL start=4>
-<LI>blah
+At the same tme, though, it's natural to define `odd` in such a way that it hands off part of the work to `even`:
+
+	let odd = \x. iszero x
+					; if x == 0 then result is
+					false
+					; else result turns on whether x's pred is even
+					(even (pred x))
+
+Such a definition of `even` and `odd` is called **mutually recursive**. If you trace through the evaluation of some sample numerical arguments, you can see that eventually we'll always reach a base step. So the recursion should be perfectly well-grounded:
+
+	even 3
+	~~> iszero 3 true (odd (pred 3))
+	~~> odd 2
+	~~> iszero 2 false (even (pred 2))
+	~~> even 1
+	~~> iszero 1 true (odd (pred 1))
+	~~> odd 0
+	~~> iszero 0 false (even (pred 0))
+	~~> false
+
+But we don't yet know how to implement this kind of recursion in the lambda calculus.
+
+The fixed point operators we've been working with so far worked like this:
+
+	let X = Y T in
+	X <~~> T X
+
+Suppose we had a pair of fixed point operators, `Y1` and `Y2`, that operated on a *pair* of functions `T1` and `T2`, as follows:
+
+	let X1 = Y1 T1 T2 in
+	let X2 = Y2 T1 T2 in
+	X1 <~~> T1 X1 X2 and
+	X2 <~~> T2 X1 X2
+
+If we gave you such a `Y1` and `Y2`, how would you implement the above definitions of `even` and `odd`?
+
+					
+<LI>(More challenging.) Using our derivation of Y from the [[Week2]] notes as a model, construct a pair `Y1` and `Y2` that behave in the way described.
 </OL>
 
diff --git a/hints/assignment_4_hint_2.mdwn b/hints/assignment_4_hint_2.mdwn
new file mode 100644
index 00000000..6bcd7701
--- /dev/null
+++ b/hints/assignment_4_hint_2.mdwn
@@ -0,0 +1,24 @@
+
+<!--
+let list_equal =
+    \left right. left
+                ; here's our f
+                (\hd sofar.
+                    ; deconstruct our sofar-pair
+                    sofar (\might_be_equal right_tail.
+                        ; our new sofar
+                        make_pair
+                        (and (and might_be_equal (not (isempty right_tail))) (eq? hd (head right_tail)))
+                        (tail right_tail)
+                    ))
+                ; here's our z
+                ; we pass along the fold a pair
+                ; (might_for_all_i_know_still_be_equal?, tail_of_reversed_right)
+                ; when left is empty, the lists are equal if right is empty
+                (make_pair
+                    true ; for all we know so far, they might still be equal
+                    (reverse right)
+                )
+                ; when fold is finished, check sofar-pair
+                (\might_be_equal right_tail. and might_be_equal (isempty right_tail))
+-->