+++ /dev/null
-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. 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 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]]
-
-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;;
-
-
-Booleans, Church numbers, and Church lists in OCaml
----------------------------------------------------
-
-These questions adapted from web materials written by some smart dude named Acar.
-The idea is to get booleans, Church numbers, "Church" lists, and
-binary trees working in OCaml.
-
- Recall from class System F, or the polymorphic λ-calculus.
-
- τ ::= 'α | τ1 → τ2 | ∀'α. τ | c
- e ::= x | λx:τ. e | e1 e2 | Λ'α. e | e [τ ]
-
- 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
-
- (where τ indicates the type of e1 and e2)
-
- Note that each of the following 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 (for "natural number") 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.
-
- **Excercise**: get booleans and Church numbers working in OCaml,
- including OCaml versions of bool, true, false, zero, 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 list predicates.
-
- 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 := ∀α. α → (τ → α → α) → α
- nilτ := Λα. λn:α. λc:τ → α → α. n
- makeListτ := λh:τ. λt:τ list. Λα. λn:α. λc:τ → α → α. c h (t [α] n c)
-
- As with nats, recursion is built into the datatype.
-
- We can write functions like head, isNil, and map:
-
- map : (σ → τ ) → σ list → τ list
-
- We've given you the type for map, you only need to give the term.
-
- With regard to `head`, 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.
-
- Please provide both the terms and the types for each item.