<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.)
+<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?
#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`:
+<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
; 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`:
+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
; 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:
+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))
~~> iszero 0 false (even (pred 0))
~~> false
-But we don't yet know how to implement this kind of recursion in the lambda calculus.
+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:
+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`?
+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.
+<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.)