Recall that version 1 style lists are constructed like this:
<pre>
+; 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 nil = makePair true meh in
-let makeList = \h t. makePair false (makePair h t) in
-let mylist = makeList 1 (makeList 2 (makeList 3 nil)) 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
-let succ = \n s z. s (n s z) 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
-length mylist
+eq 3 3
</pre>
+
Then `length mylist` evaluates to 3.
-What does `head (tail (tail mylist))` evaluate to?
+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. (Easy) 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. (Still easy) Now write the same function, but don't use the length function (hint: use `leq` as a model).
+
+5. In assignment 2, we discovered that version 3-type lists (the ones that
+work like Church numerals) made it much easier to define operations
+like map and filter. But now that we have recursion in our toolbox,
+reasonable map and filter functions for version 3 lists are within our
+reach. Give definitions for such a map and a filter.
+
+6. Linguists analyze natural language expressions into trees.
+We'll need trees in future weeks, and tree structures provide good
+opportunities for learning how to write recursive functions.
+Making use of the resources we have at the moment, we can approximate
+trees as follows: instead of words, we'll use Church numerals.
+Then a tree will be a (version 1 type) list in which each element is
+itself a tree. For simplicity, we'll adopt the convention that
+a tree of length 1 must contain a number as its only element.
+Then we have the following representations:
+
+<pre>
+ (a) (b) (c)
+ .
+ /|\ /\ /\
+ / | \ /\ 3 1/\
+ 1 2 3 1 2 2 3
+
+[[1];[2];[3]] [[[1];[2]];[3]] [[1];[[2];[3]]]
+</pre>
+
+Limitations of this scheme include the following: there is no easy way
+to label a constituent (typically a syntactic category, S or NP or VP,
+etc.), and there is no way to represent a tree in which a mother has a
+single daughter.
+
+When processing a tree, you can test for whether the tree contains
+only a numeral (in which case the tree is leaf node) by testing for
+whether the length of the list is less than or equal to 1. This will
+be your base case for your recursive functions that operate on trees.
+
+Write a function that sums the number of leaves in a tree.
+Expected behavior:
+
+<pre>
+
+let t1 = (make-list 1 nil) in
+let t2 = (make-list 2 nil) in
+let t3 = (make-list 3 nil) in
+let t12 = (make-list t1 (make-list t2 nil)) in
+let t23 = (make-list t2 (make-list t3 nil)) in
+let ta = (make-list t1 t23) in
+let tb = (make-list t12 t3) in
+let tc = (make-list t1 (make-list t23 nil)) in
+
+count-leaves t1 ~~> 1
+count-leaves t2 ~~> 2
+count-leaves t3 ~~> 3
+count-leaves t12 ~~> 3
+count-leaves t23 ~~> 5
+count-leaves ta ~~> 6
+count-leaves tb ~~> 6
+count-leaves tc ~~> 6
+<pre>
+
+Write a function that counts the number of leaves.
+
+
+
+
+[The following should be correct, but won't run in my browser:
+
+<pre>
+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)))
+</pre>
+
+It may require more resources than my browser is willing to devote to
+JavaScript.]
+
+; trees
+let t1 = (makeList 1 nil) in
+let t2 = (makeList 2 nil) in
+let t3 = (makeList 3 nil) in
+let t12 = (makeList t1 (makeList t2 nil)) in
+let t23 = (makeList t2 (makeList t3 nil)) in
+let ta = (makeList t1 t23) in
+let tb = (makeList t12 t3) in
+let tc = (makeList t1 (makeList t23 nil)) in