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diff git a/assignment3.mdwn b/assignment3.mdwn
index 4e3c4f33..9f64d808 100644
 a/assignment3.mdwn
+++ b/assignment3.mdwn
@@ 57,8 +57,7 @@ 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
@@ 66,14 +65,75 @@ same length. That is,
listLenEq mylist (makeList meh (makeList meh nil))) ~~> false
4. 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 (hint: use `leq` as a model).
+
+5. In assignment 2, we discovered that version 3type 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 = (makelist 1 nil)
+let t2 = (makelist 2 nil)
+let t3 = (makelist 3 nil)
+let t12 = (makelist t1 (makelist t2 nil))
+let t23 = (makelist t2 (makelist t3 nil))
+let ta = (makelist t1 t23)
+let tb = (makelist t12 t3)
+let tc = (makelist t1 (makelist t23 nil))
+
+countleaves t1 ~~> 1
+countleaves t2 ~~> 2
+countleaves t3 ~~> 3
+countleaves t12 ~~> 3
+countleaves t23 ~~> 5
+countleaves ta ~~> 6
+countleaves tb ~~> 6
+countleaves tc ~~> 6
+
+Write a function that counts the number of leaves.
+
+
[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