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#Comparing lists for equality#
+
 blah
+
 Suppose you have two lists of integers, `left` and `right`. You want to
+determine whether those lists are equal: that is, whether they have all the
+same members in the same order. (Equality for the lists we're working with is
+*extensional*, or parasitic on the equality of their members, and the list
+structure. Later in the course we'll see lists which aren't extensional in this
+way.)
+
+How would you implement such a list comparison?
+
+(See [[hints/Assignment 4 hint 2]] if you need some hints.)

#Mutuallyrecursive functions#
+
+#Enumerating the fringe of a leaflabeled tree#
+
+First, read this: [[Implementing trees]]
 blah
+
 Write an implementation of leaflabeled trees. You can do something v3like, or use the Y combinator, as you prefer.
+
+You'll need an operation `make_leaf` that turns a label into a new leaf. You'll
+need an operation `make_node` that takes two subtrees (perhaps leaves, perhaps
+other nodes) and joins them into a new tree. You'll need an operation `isleaf`
+that tells you whether a given tree is a leaf. And an operation `extract_label`
+that tells you what value is associated with a given leaf. And an operation
+`extract_left` that tells you what the left subtree is of a tree that isn't a
+leaf. (Presumably, `extract_right` will work similarly.)
+
+
 The **fringe** of a leaflabeled tree is the list of values at its leaves,
+ordered from left to right. For example, the fringe of this tree:
+
+ .
+ / \
+ . 3
+ / \
+ 1 2
+
+is `[1;2;3]`. And that is also the fringe of this tree:
+
+ .
+ / \
+ 1 .
+ / \
+ 2 3
+
+The two trees are different, but they have the same fringe. We're going to
+return later in the term to the problem of determining when two trees have the
+same fringe. For now, one straightforward way to determine this would be:
+enumerate the fringe of the first tree. That gives you a list. Enumerate the
+fringe of the second tree. That also gives you a list. Then compare the two
+lists to see if they're equal.
+
+Write the fringeenumeration function. It should work on the
+implementation of trees you designed in the previous step.
+
+Then combine this with the list comparison function you wrote for question 2,
+to yield a samefringe detector. (To use your list comparison function, you'll
+have to make sure you only use Church numerals as the labels of your leaves,
+though nothing enforces this selfdiscipline.)
#Enumerating the fringe of a leaflabeled tree#
[[Implementing trees]]
+#Mutuallyrecursive functions#
+
+
+ (Challenging.) One way to define the function `even` is to have it hand off
+part of the work to another function `odd`:
+
+ let even = \x. iszero x
+ ; if x == 0 then result is
+ true
+ ; else result turns on whether x's pred is odd
+ (odd (pred x))
+
+At the same tme, though, it's natural to define `odd` in such a way that it
+hands off part of the work to `even`:
+
+ let odd = \x. iszero x
+ ; if x == 0 then result is
+ false
+ ; else result turns on whether x's pred is even
+ (even (pred x))
+
+Such a definition of `even` and `odd` is called **mutually recursive**. If you
+trace through the evaluation of some sample numerical arguments, you can see
+that eventually we'll always reach a base step. So the recursion should be
+perfectly wellgrounded:
+
+ even 3
+ ~~> iszero 3 true (odd (pred 3))
+ ~~> odd 2
+ ~~> iszero 2 false (even (pred 2))
+ ~~> even 1
+ ~~> iszero 1 true (odd (pred 1))
+ ~~> odd 0
+ ~~> iszero 0 false (even (pred 0))
+ ~~> false
+
+But we don't yet know how to implement this kind of recursion in the lambda
+calculus.
+
+The fixed point operators we've been working with so far worked like this:
+
+ let X = Y T in
+ X <~~> T X
+
+Suppose we had a pair of fixed point operators, `Y1` and `Y2`, that operated on
+a *pair* of functions `T1` and `T2`, as follows:
+
+ let X1 = Y1 T1 T2 in
+ let X2 = Y2 T1 T2 in
+ X1 <~~> T1 X1 X2 and
+ X2 <~~> T2 X1 X2
+
+If we gave you such a `Y1` and `Y2`, how would you implement the above
+definitions of `even` and `odd`?
+
+
+
 (More challenging.) Using our derivation of Y from the [Week3
+notes](/week3/#index4h2) as a model, construct a pair `Y1` and `Y2` that behave
+in the way described.
+(See [[hints/Assignment 4 hint 3]] if you need some hints.)

 blah