Assignment 5

## Types and OCaml

1. 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

2. 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 ();;

3. Throughout this problem, assume that we have

let rec blackhole x = blackhole x;;


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

blackhole;;

blackhole ();;

fun () -> blackhole ();;

(fun () -> blackhole ()) ();;

if true then blackhole else blackhole;;

if false then blackhole else blackhole;;

if true then blackhole else blackhole ();;

if false then blackhole else blackhole ();;

if true then blackhole () else blackhole;;

if false then blackhole () else blackhole;;

if true then blackhole () else blackhole ();;

if false then blackhole () else blackhole ();;

let _ = blackhole in 2;;

let _ = blackhole () in 2;;

4. This problem is to begin thinking about controlling order of evaluation. 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 blackhole x = blackhole x in
if true then blackhole else blackhole ();;


terminates, but

let rec blackhole x = blackhole x in
let b = true in
let y = blackhole in
let n = blackhole () 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.

assignment 5 hint 1

## Booleans, Church numerals, and v3 lists in OCaml

(These questions adapted from web materials by Umut Acar. See http://www.mpi-sws.org/~umut/.)

Let's think about the encodings of booleans, numerals and lists in System F, and get data-structures with the same form working in OCaml. (Of course, OCaml has native versions of these datas-structures: its true, 1, and [1;2;3]. But the point of our exercise requires that we ignore those.)

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

types τ ::= c | 'a | τ1 → τ2 | ∀'a. τ
expressions e ::= x | λx:τ. e | e1 e2 | Λ'a. e | e [τ]


The boolean type, and its two values, may be encoded as follows:

bool := ∀'a. 'a → 'a → 'a
true := Λ'a. λt:'a. λf :'a. t
false := Λ'a. λt:'a. λf :'a. f


It's used like this:

b [τ] e1 e2


where b is a boolean value, and τ is the shared type of e1 and e2.

Exercise. How should we implement the following terms. Note that the result of applying them to the appropriate arguments should also give us a term 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 (for "natural number") may be encoded as follows:

nat := ∀'a. 'a → ('a → 'a) → 'a
zero := Λ'a. λz:'a. λs:'a → 'a. z
succ := λn:nat. Λ'a. λz:'a. λs:'a → 'a. s (n ['a] 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 'a, as long as you have a base element z : 'a and a function s : 'a → 'a.

Exercise: get booleans and Church numbers working in OCaml, including OCaml versions of bool, true, false, zero, iszero, succ, and pred. It's especially useful to do a version of pred, starting with one of the (untyped) versions available in the lambda library accessible from the main wiki page. The point of the excercise is to do these things on your own, so avoid using the built-in OCaml booleans and integers.

Consider the following list type:

type 'a list = Nil | Cons of 'a * 'a list


We can encode τ lists, lists of elements of type τ as follows:

τ list := ∀'a. 'a → (τ → 'a → 'a) → 'a
nil τ := Λ'a. λn:'a. λc:τ → 'a → 'a. n
make_list τ := λh:τ. λt:τ list. Λ'a. λn:'a. λc:τ → 'a → 'a. c h (t ['a] n c)


More generally, the polymorphic list type is:

list := ∀'b. ∀'a. 'a → ('b → 'a → 'a) → 'a


As with nats, recursion is built into the datatype.

We can write functions like map:

map : (σ → τ ) → σ list → τ list


Excercise convert this function to OCaml. We've given you the type; you only need to give the term.

Also give us the type and definition for a head function. Think about what value to give back if the argument is the empty list. Ultimately, we might want to make use of our 'a option technique, but for this assignment, just pick a strategy, no matter how clunky.

Be sure to test your proposals with simple lists. (You'll have to make_list the lists yourself; don't expect OCaml to magically translate between its native lists and the ones you buil.d)

Read the material on dividing by zero/towards monads from the end of lecture notes for week 6 the start of lecture notes for week 7, 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' (u: int option) (f: int -> (int option)) =