#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
+
+(See [[hints/Assignment 4 hint 3]] if you need some hints.)
</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))
+
+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.
+
+(See [[hints/Assignment 4 hint 4]] if you need some hints.)
-<OL start=4>
-<LI>blah
</OL>