Assignment 3
------------
-Once again, the lambda evaluator will make working through this
-assignment much faster and more secure.
+Erratum corrected 11PM Sun 3 Oct: the following line
-*Writing recursive functions on version 1 style lists*
+ let tb = (make_list t12 (make_list t3 empty)) in
-Recall that version 1 style lists are constructed like this:
+originally read
-<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 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
-</pre>
+ let tb = (make_list t12 t3) in
+
+This has been corrected below, and in the preloaded evaluator for
+working on assignment 3, available here: [[assignment 3 evaluator]].
+
+<hr>
+
+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 (see
+[[lists and numbers]]):
+
+ ; booleans
+ let true = \x y. x in
+ let false = \x y. y in
+ let and = \l r. l (r true false) false in
+
+ let make_pair = \f s g. g f s in
+ let get_fst = true in
+ let get_snd = false in
+ let empty = make_pair true junk in
+ let isempty = \x. x get_fst in
+ let make_list = \h t. make_pair false (make_pair h t) in
+ let head = \l. isempty l err (l get_snd get_fst) in
+ let tail = \l. isempty l err (l get_snd get_snd) in
+
+ ; a list of numbers to experiment on
+ let mylist = make_list 1 (make_list 2 (make_list 3 empty)) in
+
+ ; church numerals
+ let iszero = \n. n (\x. false) true in
+ let succ = \n s z. s (n s z) in
+ let add = \l r. l succ r in
+ let mul = \m n s. m (n s) in
+ let pred = (\shift n. n shift (make\_pair 0 0) get\_snd) (\p. p (\x y. make\_pair (succ x) x)) in
+ let leq = \m n. iszero(n pred m) in
+ let eq = \m n. and (leq m n)(leq n m) in
+
+ ; a fixed-point combinator for defining recursive functions
+ let Y = \f. (\h. f (h h)) (\h. f (h h)) in
+ let length = Y (\length l. isempty l 0 (succ (length (tail l)))) in
+ let fold = Y (\f l g z. isempty l z (g (head l)(f (tail l) g z))) in
+
+ eq 2 2 yes no
Then `length mylist` evaluates to 3.
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).
+ Warning: it takes a long time for my browser to compute factorials larger than 4!
+3. (Easy) Write a function `equal_length` that returns true just in case
+two lists have the same length. That is,
-3. Write a function `listLenEq` that returns true just in case two lists have the
-same length. That is,
+ equal_length mylist (make_list junk (make_list junk (make_list junk empty))) ~~> true
- listLenEq mylist (makeList meh (makeList meh (makeList meh nil))) ~~> true
+ equal_length mylist (make_list junk (make_list junk empty))) ~~> false
- listLenEq mylist (makeList meh (makeList meh nil))) ~~> false
+4. (Still easy) Now write the same function, but don't use the length
+function.
-4. 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 1 lists are within our
+reach. Give definitions for `map` and a `filter` for verson 1 type
+lists.
+#Computing with trees#
-[The following should be correct, but won't run in my browser:
+Linguists analyze natural language expressions into trees.
-let factorial = Y (\fac n. isZero n 1 (mult n (fac (predecessor n)))) in
+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>
-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)))
+ (a) (b) (c)
+ .
+ /|\ /\ /\
+ / | \ /\ 3 1 /\
+ 1 2 3 1 2 2 3
+
+[[1];[2];[3]] [[[1];[2]];[3]] [[1];[[2];[3]]]
</pre>
-It may require more resources than my browser is willing to devote to
-JavaScript.]
+Limitations of this scheme include the following: there is no easy way
+to label a constituent with 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 these
+trees.
+
+<OL start=6>
+<LI>Write a function that sums the values at the leaves in a tree.
+
+Expected behavior:
+
+ let t1 = (make_list 1 empty) in
+ let t2 = (make_list 2 empty) in
+ let t3 = (make_list 3 empty) in
+ let t12 = (make_list t1 (make_list t2 empty)) in
+ let t23 = (make_list t2 (make_list t3 empty)) in
+ let ta = (make_list t1 t23) in
+ let tb = (make_list t12 (make_list t3 empty)) in
+ let tc = (make_list t1 (make_list t23 empty)) in
+
+ sum-leaves t1 ~~> 1
+ sum-leaves t2 ~~> 2
+ sum-leaves t3 ~~> 3
+ sum-leaves t12 ~~> 3
+ sum-leaves t23 ~~> 5
+ sum-leaves ta ~~> 6
+ sum-leaves tb ~~> 6
+ sum-leaves tc ~~> 6
+
+
+<LI>Write a function that counts the number of leaves.
+
+</OL>