; 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@@ -57,23 +58,31 @@ greater than 2 (it does't provide enough resources for the JavaScript interpreter; web pages are not supposed to be that computationally intensive). -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. (Still easy) Now write the same function, 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 +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. +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. -6. Linguists analyze natural language expressions into trees. +#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 @@ -87,63 +96,46 @@ Then we have the following representations: (a) (b) (c) . /|\ /\ /\ - / | \ /\ 3 1/\ + / | \ /\ 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, +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 trees. - -Write a function that sums the number of leaves in a tree. -Expected behavior: - -let t1 = (make-list 1 nil) -let t2 = (make-list 2 nil) -let t3 = (make-list 3 nil) -let t12 = (make-list t1 (make-list t2 nil)) -let t23 = (make-list t2 (make-list t3 nil)) -let ta = (make-list t1 t23) -let tb = (make-list t12 t3) -let tc = (make-list t1 (make-list t23 nil)) - -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. +be your base case for your recursive functions that operate on these +trees. +1. Write a function that sums the number of leaves in a tree. - - -[The following should be correct, but won't run in my browser: +Expected behavior:

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