- How would you define an operation to reverse a list? (Don't peek at the [[lambda_library]]! Try to figure it out on your own.) Choose whichever implementation of list you like. Even then, there are various strategies you can use. (See [[hints/Assignment 4 hint 1]] if you need some hints.)

- 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.)

- Write an implementation of leaf-labeled trees. You can do something v3-like, 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 leaf-labeled 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. (You just programmed this above.) Write the fringe-enumeration function. It should work on the implementation of trees you designed in the previous step.

- (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 well-grounded: 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.)