Assignment 5

Types and OCAML
---------------

0. Recall that the S combinator is given by \x y z. x z (y z).
   Give two different typings for this function in OCAML.
   To get you started, here's one typing for K:

    # let k (y:'a) (n:'b) = y;;
    val k : 'a -> 'b -> 'a = [fun]
    # k 1 true;;
    - : int = 1


1. Which of the following expressions is well-typed in OCAML?  
   For those that are, give the type of the expression as a whole.
   For those that are not, why not?

    let rec f x = f x;;

    let rec f x = f f;;

    let rec f x = f x in f f;;

    let rec f x = f x in f ();;

    let rec f () = f f;;

    let rec f () = f ();;

    let rec f () = f () in f f;;

    let rec f () = f () in f ();;

2. Throughout this problem, assume that we have 

    let rec omega x = omega x;;

   All of the following are well-typed.
   Which ones terminate?  What are the generalizations?

    omega;;

    omega ();;

    fun () -> omega ();;

    (fun () -> omega ()) ();;

    if true then omega else omega;;

    if false then omega else omega;;

    if true then omega else omega ();;

    if false then omega else omega ();;

    if true then omega () else omega;;

    if false then omega () else omega;;

    if true then omega () else omega ();;

    if false then omega () else omega ();;

    let _ = omega in 2;;

    let _ = omega () in 2;;

3. The following expression is an attempt to make explicit the
behavior of `if`-`then`-`else` explored in the previous question.
The idea is to define an `if`-`then`-`else` expression using 
other expression types.  So assume that "yes" is any OCAML expression,
and "no" is any other OCAML expression (of the same type as "yes"!),
and that "bool" is any boolean.  Then we can try the following:
"if bool then yes else no" should be equivalent to

    let b = bool in
    let y = yes in 
    let n = no in 
    match b with true -> y | false -> n

This almost works.  For instance, 

    if true then 1 else 2;;

evaluates to 1, and 

    let b = true in let y = 1 in let n = 2 in 
    match b with true -> y | false -> n;;

also evaluates to 1.  Likewise,

    if false then 1 else 2;;

and

    let b = false in let y = 1 in let n = 2 in 
    match b with true -> y | false -> n;;

both evaluate to 2.

However,

    let rec omega x = omega x in 
    if true then omega else omega ();;

terminates, but 

    let rec omega x = omega x in 
    let b = true in
    let y = omega in 
    let n = omega () in 
    match b with true -> y | false -> n;;

does not terminate.  Incidentally, `match bool with true -> yes |
false -> no;;` works as desired, but your assignment is to solve it
without using the magical evaluation order properties of either `if`
or of `match`.  That is, you must keep the `let` statements, though
you're allowed to adjust what `b`, `y`, and `n` get assigned to.

[[Hint assignment 5 problem 3]]

4. Baby monads.  Read the lecture notes for week 6, then write a
   function `lift` that generalized the correspondence between + and
   `add`: that is, `lift` takes any two-place operation on integers
   and returns a version that takes arguments of type `int option`
   instead, returning a result of `int option`.  In other words,
   `lift` will have type

     (int -> int -> int) -> (int option) -> (int option) -> (int option)

   so that `lift (+) (Some 3) (Some 4)` will evalute to `Some 7`.  
   Don't worry about why you need to put `+` inside of parentheses.
   You should make use of `bind` in your definition of `lift`:

    let bind (x: int option) (f: int -> (int option)) = 
      match x with None -> None | Some n -> f n;;


Church lists in System F
------------------------

These questions adapted from web materials written by some dude named Acar.

   Recall from class System F, or the polymorphic λ-calculus.

   τ ::= α | τ1 → τ2 | ∀α. τ
   e ::= x | λx:τ. e | e1 e2 | Λα. e | e [τ ]
   Despite its simplicity, System F is quite expressive. As discussed in class, it has sufficient expressive power
   to be able to encode many datatypes found in other programming languages, including products, sums, and
   inductive datatypes.
   For example, recall that bool may be encoded as follows:
   bool := ∀α. α → α → α
   true := Λα. λt:α. λf :α. t
   false := Λα. λt:α. λf :α. f
   ifτ e then e1 else e2 := e [τ ] e1 e2
   (where τ indicates the type of e1 and e2)
   Exercise 1. Show how to encode the following terms. Note that each of these terms, when applied to the
   appropriate arguments, return a result of type bool.
   (a) the term not that takes an argument of type bool and computes its negation;
   (b) the term and that takes two arguments of type bool and computes their conjunction;
   (c) the term or that takes two arguments of type bool and computes their disjunction.
   The type nat may be encoded as follows:
   nat := ∀α. α → (α → α) → α
   zero := Λα. λz:α. λs:α → α. z
   succ := λn:nat. Λα. λz:α. λs:α → α. s (n [α] z s)
   A nat n is defined by what it can do, which is to compute a function iterated n times. In the polymorphic
   encoding above, the result of that iteration can be any type α, as long as you have a base element z : α and
   a function s : α → α.
   Conveniently, this encoding “is” its own elimination form, in a sense:
   rec(e, e0, x:τ. e1) := e [τ ] e0 (λx:τ. e1)
   The case analysis is baked into the very definition of the type.
   Exercise 2. Verify that these encodings (zero, succ , rec) typecheck in System F. Write down the typing
   derivations for the terms.
   1

   ══════════════════════════════════════════════════════════════════════════

   As mentioned in class, System F can express any inductive datatype. Consider the following list type:
   datatype ’a list =
   Nil
   | Cons of ’a * ’a list
   We can encode τ lists, lists of elements of type τ as follows:1
   τ list := ∀α. α → (τ → α → α) → α
   nilτ := Λα. λn:α. λc:τ → α → α. n
   consτ := λh:τ. λt:τ list. Λα. λn:α. λc:τ → α → α. c h (t [α] n c)
   As with nats, The τ list type’s case analyzing elimination form is just application. We can write functions
   like map:
   map : (σ → τ ) → σ list → τ list
   := λf :σ → τ. λl:σ list. l [τ list] nilτ (λx:σ. λy:τ list. consτ (f x) y
   Exercise 3. Consider the following simple binary tree type:
   datatype ’a tree =
   Leaf
   | Node of ’a tree * ’a * ’a tree
   (a) Give a System F encoding of binary trees, including a definition of the type τ tree and definitions of
   the constructors leaf : τ tree and node : τ tree → τ → τ tree → τ tree.
   (b) Write a function height : τ tree → nat. You may assume the above encoding of nat as well as definitions
   of the functions plus : nat → nat → nat and max : nat → nat → nat.
   (c) Write a function in-order : τ tree → τ list that computes the in-order traversal of a binary tree. You
   may assume the above encoding of lists; define any auxiliary functions you need.

-- 
Jim Pryor
jim@jimpryor.net