Assignment 3
------------
Once again, the lambda evaluator will make working through this
assignment much faster and more secure.
*Writing recursive functions on version 1 style lists*
Recall that version 1 style lists are constructed like this:
; booleans
let true = \x y. x in
let false = \x y. y in
let and = \l r. l (r true false) false in
; version 1 lists
let makePair = \f s g. g f s in
let fst = true in
let snd = false in
let nil = makePair true meh in
let isNil = \x. x fst in
let makeList = \h t. makePair false (makePair h t) in
let head = \l. isNil l err (l snd fst) in
let tail = \l. isNil l err (l snd snd) in
; a list of numbers to experiment on
let mylist = makeList 1 (makeList 2 (makeList 3 nil)) in
; a fixed-point combinator for defining recursive functions
let Y = \f. (\h. f (h h)) (\h. f (h h)) in
; church numerals
let isZero = \n. n (\x. false) true in
let succ = \n s z. s (n s z) in
let mult = \m n s. m (n s) in
let length = Y (\length l. isNil l 0 (succ (length (tail l)))) in
let predecessor = \n. length (tail (n (\p. makeList meh p) nil)) in
let leq = ; (leq m n) will be true iff m is less than or equal to n
Y (\leq m n. isZero m true (isZero n false (leq (predecessor m)(predecessor n)))) in
let eq = \m n. and (leq m n)(leq n m) in
eq 3 3

Then `length mylist` evaluates to 3.
1. What does `head (tail (tail mylist))` evaluate to?
2. Using the `length` function as a model, and using the predecessor
function, write a function that computes factorials. (Recall that n!,
the factorial of n, is n times the factorial of n-1.)
Warning: my browser isn't able to compute factorials of numbers
greater than 2 (it does't provide enough resources for the JavaScript
interpreter; web pages are not supposed to be that computationally
intensive).
3. Write a function `listLenEq` that returns true just in case two lists have the
same length. That is,
listLenEq mylist (makeList meh (makeList meh (makeList meh nil))) ~~> true
listLenEq mylist (makeList meh (makeList meh nil))) ~~> false
4. Now write the same function (true iff two lists have the same
length) but don't use the length function (hint: use `leq` as a model).
That is, (makeList 1 (makeList 2 (makeList 3 nil)))
[The following should be correct, but won't run in my browser:
let factorial = Y (\fac n. isZero n 1 (mult n (fac (predecessor n)))) in
let reverse =
Y (\rev l. isNil l nil
(isNil (tail l) l
(makeList (head (rev (tail l)))
(rev (makeList (head l)
(rev (tail (rev (tail l))))))))) in
reverse (makeList 1 (makeList 2 (makeList 3 nil)))

It may require more resources than my browser is willing to devote to
JavaScript.]