X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=assignment3.mdwn;h=9039a957b482b1c7e32441f5782a9bee57467bf3;hp=f89f5017a3363236c5e4fc3ba52c45e7ba776818;hb=17c72bcdce271a9975bc4cad438754fa4b70c6fc;hpb=5c7dbf2ff40b4cf646b9fc3bb9290d19e0a8e388 diff --git a/assignment3.mdwn b/assignment3.mdwn index f89f5017..9039a957 100644 --- a/assignment3.mdwn +++ b/assignment3.mdwn @@ -6,7 +6,8 @@ assignment much faster and more secure. *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 +36,11 @@ 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@@ -57,24 +57,86 @@ 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 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 (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 3 lists are within our +reach. Give definitions for such a map and a filter. + +6. 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 (typically 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 trees. + +Write a function that sums the number of leaves in a tree. +Expected behavior: + +
+ +let t1 = (make-list 1 nil) in +let t2 = (make-list 2 nil) in +let t3 = (make-list 3 nil) in +let t12 = (make-list t1 (make-list t2 nil)) in +let t23 = (make-list t2 (make-list t3 nil)) in +let ta = (make-list t1 t23) in +let tb = (make-list t12 t3) in +let tc = (make-list t1 (make-list t23 nil)) in + +count-leaves t1 ~~> 1 +count-leaves t2 ~~> 2 +count-leaves t3 ~~> 3 +count-leaves t12 ~~> 3 +count-leaves t23 ~~> 5 +count-leaves ta ~~> 6 +count-leaves tb ~~> 6 +count-leaves tc ~~> 6 ++ +Write a function that counts the number of leaves. + + - That is, (makeList 1 (makeList 2 (makeList 3 nil))) [The following should be correct, but won't run in my browser: +let factorial = Y (\fac n. isZero n 1 (mult n (fac (predecessor n)))) in -let reverse = Y (\rev l. isNil l nil (isNil (tail l) l @@ -88,3 +150,12 @@ reverse (makeList 1 (makeList 2 (makeList 3 nil))) It may require more resources than my browser is willing to devote to JavaScript.] +; trees +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