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-Assignment 4
-------------
-
 #Reversing a list#
 
-How would you define an operation to reverse a list? (Don't peek at the
+<OL>
+<LI>How would you define an operation to reverse a list? (Don't peek at the
 [[lambda_library]]! Try to figure it out on your own.) Choose whichever
 implementation of list you like. Even then, there are various strategies you
 can use.
 
+(See [[hints/Assignment 4 hint 1]] if you need some hints.)
+</OL>
+
 
 #Comparing lists for equality#
 
-<!--
-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
-                    (not (isempty right))
-                    (reverse right)
-                )
-                ; when fold is finished, check sofar-pair
-                (\might_be_equal right_tail. and might_be_equal (isempty right_tail))
--->
+
+<OL start=2>
+<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>
+
 
 #Enumerating the fringe of a leaf-labeled tree#
 
-[[Implementing trees]]
+First, read this: [[Implementing trees]]
+
+<OL start=3>
+<LI>blah
+
+(See [[hints/Assignment 4 hint 3]] if you need some hints.)
+</OL>
+
+
+#Mutually-recursive functions#
+
+<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))
+
+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 [Week3 notes](/week3/#index4h2) as a model, construct a pair `Y1` and `Y2` that behave in the way described.
 
+(See [[hints/Assignment 4 hint 4]] if you need some hints.)
 
+</OL>