X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=assignment3.mdwn;h=6f4f3f64f2bdda0ca0a12729271ef413e3a8eeee;hp=013bc3c5052cd781aad78bb5a82656d5e47505c3;hb=e339942e9ebf6bd1650c1640e32298a75318c55b;hpb=f2662f133c25062dc28261f655bb9daf9c2bbb9c diff --git a/assignment3.mdwn b/assignment3.mdwn index 013bc3c5..6f4f3f64 100644 --- a/assignment3.mdwn +++ b/assignment3.mdwn @@ -4,9 +4,10 @@ Assignment 3 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]]):
 ; booleans
@@ -35,12 +36,12 @@ 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 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

@@ -57,35 +58,84 @@ greater than 2 (it does't provide enough resources for the JavaScript 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: + +
+   (a)           (b)             (c)
+    .
+   /|\            /\              /\
+  / | \          /\ 3            1 /\
+  1 2  3        1  2               2 3
+
+[[1];[2];[3]]  [[[1];[2]];[3]]   [[1];[[2];[3]]]
+
+ +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:
-let reverse =
-  Y (\rev l. isNil l nil
-                   (isNil (tail l) l
-                          (makeList (head (rev (tail 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

-It may require more resources than my browser is willing to devote to -JavaScript.] +2. Write a function that counts the number of leaves.