-Types and OCAML
+Assignment 5
-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?
+Types and OCaml
+---------------
- let rec f x = f x;;
+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 rec f x = f f;;
+ # let k (y:'a) (n:'b) = y;;
+ val k : 'a -> 'b -> 'a = [fun]
+ # k 1 true;;
+ - : int = 1
- let rec f x = f x in f f;;
- let rec f x = f x in f ();;
+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 () = f f;;
+ let rec f x = f x;;
- let rec f () = f ();;
+ let rec f x = f f;;
- let rec f () = f () in f f;;
+ let rec f x = f x in f f;;
- let rec f () = f () in f ();;
+ let rec f x = f x in f ();;
-2. Throughout this problem, assume that we have
+ let rec f () = f f;;
- let rec omega x = omega x;;
+ let rec f () = f ();;
- All of the following are well-typed.
- Which ones terminate? What are the generalizations?
+ let rec f () = f () in f f;;
- omega;;
+ let rec f () = f () in f ();;
- omega ();;
+2. Throughout this problem, assume that we have
- fun () -> omega ();;
+ let rec blackhole x = blackhole x;;
- (fun () -> omega ()) ();;
+ All of the following are well-typed.
+ Which ones terminate? What are the generalizations?
- if true then omega else omega;;
+ blackhole;;
- if false then omega else omega;;
+ blackhole ();;
- if true then omega else omega ();;
+ fun () -> blackhole ();;
- if false then omega else omega ();;
+ (fun () -> blackhole ()) ();;
- if true then omega () else omega;;
+ if true then blackhole else blackhole;;
- if false then omega () else omega;;
+ if false then blackhole else blackhole;;
- if true then omega () else omega ();;
+ if true then blackhole else blackhole ();;
- if false then omega () else omega ();;
+ if false then blackhole else blackhole ();;
- let _ = omega in 2;;
+ if true then blackhole () else blackhole;;
- let _ = omega () in 2;;
+ if false then blackhole () else blackhole;;
-3. The following expression is an attempt to make explicit the
+ if true then blackhole () else blackhole ();;
+
+ if false then blackhole () else blackhole ();;
+
+ let _ = blackhole in 2;;
+
+ let _ = blackhole () 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"!),
+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
+ 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.
+
+ [[Hint assignment 5 problem 3]]
+
+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 datastructures with the same explicit form working in OCaml. (The point... so we won't rely on OCaml's native booleans, integers, or lists.)
+
+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.
+
+**Excercise**: get booleans and Church numbers working in OCaml,
+including OCaml versions of bool, true, false, zero, iszero, succ, and **pred**.
+
+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)
+
+As with nats, recursion is built into the datatype.
-This almost works. For instance,
+We can write functions like map:
- if true then 1 else 2;;
+ 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
-evaluates to 1, and
+**Excercise** convert this function to OCaml. Also write an `append` function.
+Also write a **head** function. Also write nil??? Test with simple lists.
- let b = true in let y = 1 in let n = 2 in
- match b with true -> 1 | false -> 2;;
-also evaluates to 1. Likewise,
+Consider the following simple binary tree type:
- if false then 1 else 2;;
+ type 'a tree = Leaf | Node of 'a tree * 'a * 'a tree
-and
+**Excercise**
+Write a function `sum_leaves` that computes the sum of all the
+leaves in an int tree.
- let b = false in let y = 1 in let n = 2 in
- match b with true -> y | false -> n;;
+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.
-both evaluate to 2.
+Baby monads
+-----------
-However,
+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
- let rec omega x = omega x in
- if true then omega else omega ();;
+ (int -> int -> int) -> (int option) -> (int option) -> (int option)
-terminates, but
+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 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;;
+ let bind' (x: int option) (f: int -> (int option)) =
+ match x with None -> None | Some n -> f 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]]