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diff --git a/assignment3.mdwn b/assignment3.mdwn
index 71960dc8..f89f5017 100644
--- a/assignment3.mdwn
+++ b/assignment3.mdwn
@@ -9,24 +9,82 @@ assignment much faster and more secure.
 Recall that version 1 style lists are constructed like this:
 
 <pre>
+; 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 nil = makePair true meh in
-let makeList = \h t. makePair false (makePair h t) in
-let mylist = makeList 1 (makeList 2 (makeList 3 nil)) 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
-let succ = \n s z. s (n s z) 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 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 eq = \m n. and (leq m n)(leq n m) in
 
-length mylist
+eq 3 3
 </pre>
 
+
 Then `length mylist` evaluates to 3.
 
-What does `head (tail (tail mylist))` evaluate to?
+1. What does `head (tail (tail mylist))` evaluate to?
+
+2. Using the `length` function as a model, and using the predecessor
+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).
+
+
+3. 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).
+
+   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
+
+<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)))
+</pre>
+
+It may require more resources than my browser is willing to devote to
+JavaScript.]
+