Once again, the lambda evaluator will make working through this
assignment much faster and more secure.
-*Writing recursive functions on version 1 style lists*
+#Writing recursive functions on version 1 style lists#
-Recall that version 1 style lists are constructed like this:
+Recall that version 1 style lists are constructed like this (see
+[[lists and numbers]]):
<pre>
; booleans
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 pred = \n. isZero n 0 (length (tail (n (\p. makeList meh p) nil)))
+in
+let leq = \m n. isZero(n pred m) in
let eq = \m n. and (leq m n)(leq n m) in
-eq 3 3
+eq 2 2 yes no
</pre>
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
+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 (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).
+4. (Still easy) Now write the same function, but don't use the length
+function.
+
+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#
+
+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 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.
- That is, (makeList 1 (makeList 2 (makeList 3 nil)))
+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.
-[The following should be correct, but won't run in my browser:
+1. Write a function that sums the number of leaves in a tree.
-let factorial = Y (\fac n. isZero n 1 (mult n (fac (predecessor n)))) in
+Expected behavior:
<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)))
+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
+
+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
</pre>
-It may require more resources than my browser is willing to devote to
-JavaScript.]
+2. Write a function that counts the number of leaves.