+ Once again, the lambda evaluator will make working through this assignment much faster and more secure. @@ -9,39 +22,38 @@ assignment much faster and more secure. Recall that version 1 style lists are constructed like this (see [[lists and numbers]]): -

-; booleans -let true = \x y. x in -let false = \x y. y in -let and = \l r. l (r true false) false in - -; version 1 lists -let makePair = \f s g. g f s in -let fst = true in -let snd = false in -let nil = makePair true meh in -let isNil = \x. x fst in -let makeList = \h t. makePair false (makePair h t) in -let head = \l. isNil l err (l snd fst) in -let tail = \l. isNil l err (l snd snd) in - -; a list of numbers to experiment on -let mylist = makeList 1 (makeList 2 (makeList 3 nil)) in - -; a fixed-point combinator for defining recursive functions -let Y = \f. (\h. f (h h)) (\h. f (h h)) in - -; church numerals -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 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 2 2 yes no -+ ; booleans + let true = \x y. x in + let false = \x y. y in + let and = \l r. l (r true false) false in + + let make_pair = \f s g. g f s in + let get_fst = true in + let get_snd = false in + let empty = make_pair true junk in + let isempty = \x. x get_fst in + let make_list = \h t. make_pair false (make_pair h t) in + let head = \l. isempty l err (l get_snd get_fst) in + let tail = \l. isempty l err (l get_snd get_snd) in + + ; a list of numbers to experiment on + let mylist = make_list 1 (make_list 2 (make_list 3 empty)) in + + ; church numerals + let iszero = \n. n (\x. false) true in + let succ = \n s z. s (n s z) in + let add = \l r. l succ r in + let mul = \m n s. m (n s) in + let pred = (\shift n. n shift (make\_pair 0 0) get\_snd) (\p. p (\x y. make\_pair (succ x) x)) in + let leq = \m n. iszero(n pred m) in + let eq = \m n. and (leq m n)(leq n m) in + + ; a fixed-point combinator for defining recursive functions + let Y = \f. (\h. f (h h)) (\h. f (h h)) in + let length = Y (\length l. isempty l 0 (succ (length (tail l)))) in + let fold = Y (\f l g z. isempty l z (g (head l)(f (tail l) g z))) in + + eq 2 2 yes no Then `length mylist` evaluates to 3. @@ -52,44 +64,47 @@ Then `length mylist` evaluates to 3. function, write a function that computes factorials. (Recall that n!, the factorial of n, is n times the factorial of n-1.) -Warning: my browser isn't able to compute factorials of numbers -greater than 2 (it does't provide enough resources for the JavaScript -interpreter; web pages are not supposed to be that computationally -intensive). + Warning: it takes a long time for my browser to compute factorials larger than 4! -3. (Easy) Write a function `listLenEq` that returns true just in case two lists have the -same length. That is, +3. (Easy) Write a function `equal_length` that returns true just in case +two lists have the same length. That is, - listLenEq mylist (makeList meh (makeList meh (makeList meh nil))) ~~> true + equal_length mylist (make_list junk (make_list junk (make_list junk empty))) ~~> true - listLenEq mylist (makeList meh (makeList meh nil))) ~~> false + equal_length mylist (make_list junk (make_list junk empty))) ~~> false -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, -reasonable map and filter functions for version 3 lists are within our -reach. Give definitions for `map` and a `filter` for verson 1 type lists. +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. +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. +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) + (a) (b) (c) . /|\ /\ /\ - / | \ /\ 3 1/\ + / | \ /\ 3 1 /\ 1 2 3 1 2 2 3 [[1];[2];[3]] [[[1];[2]];[3]] [[1];[[2];[3]]] @@ -106,29 +121,31 @@ 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. -1. Write a function that sums the number of leaves in a tree. ++

- Write a function that sums the values at the 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 -+ let t1 = (make_list 1 empty) in + let t2 = (make_list 2 empty) in + let t3 = (make_list 3 empty) in + let t12 = (make_list t1 (make_list t2 empty)) in + let t23 = (make_list t2 (make_list t3 empty)) in + let ta = (make_list t1 t23) in + let tb = (make_list t12 (make_list t3 empty)) in + let tc = (make_list t1 (make_list t23 empty)) 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 + + +- Write a function that counts the number of leaves. -2. Write a function that counts the number of leaves. +