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 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>
Then `length mylist` evaluates to 3.
- 1. Warm-up: 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!,
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
++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, but don't use the length function (hint: use `leq` as a model).
- ##Trees##
-4. (Still easy) Now write the same function, but don't use the length function.
++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
++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,
++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.
++reach. Give definitions for `map` and a `filter` for verson 1 type
++lists.
+
+ #Computing with trees#
- Since we'll be working with linguistic objects, let's approximate
- trees as follows: a tree is a version 1 list
- a Church number is a tree, and
- if A and B are trees, then (make-pair A B) is a tree.
+ 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.
+ 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.