From: Chris Barker <barker@kappa.linguistics.fas.nyu.edu>
Date: Sat, 2 Oct 2010 19:19:28 +0000 (-0400)
Subject: sigh
X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=commitdiff_plain;h=86a7982ef67a424a823da45ad03a839f64ec56a7

sigh
---

86a7982ef67a424a823da45ad03a839f64ec56a7
diff --cc assignment3.mdwn
index f93afd61,e91eeee5..6f4f3f64
--- a/assignment3.mdwn
+++ b/assignment3.mdwn
@@@ -35,12 -36,11 +36,12 @@@ let isZero = \n. n (\x. false) true i
  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>
  
  
@@@ -57,41 -57,78 +58,84 @@@ greater than 2 (it does't provide enoug
  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.