Assignment 5
-Types and OCAML
+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:
+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
+ # 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?
+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 x;;
- let rec f x = f f;;
+ let rec f x = f f;;
- let rec f x = f x in f f;;
+ let rec f x = f x in f f;;
- let rec f x = f x in f ();;
+ let rec f x = f x in f ();;
- let rec f () = f f;;
+ let rec f () = f f;;
- let rec f () = f ();;
+ let rec f () = f ();;
- let rec f () = f () in f f;;
+ let rec f () = f () in f f;;
- let rec f () = f () in f ();;
+ let rec f () = f () in f ();;
-2. Throughout this problem, assume that we have
+2. Throughout this problem, assume that we have
- let rec omega x = omega x;;
+ let rec blackhole x = blackhole x;;
- All of the following are well-typed.
- Which ones terminate? What are the generalizations?
+ All of the following are well-typed.
+ Which ones terminate? What are the generalizations?
- omega;;
+ blackhole;;
- omega ();;
+ blackhole ();;
- fun () -> omega ();;
+ fun () -> blackhole ();;
- (fun () -> 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 ();;
- 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;;
+ let _ = blackhole in 2;;
- let _ = omega () in 2;;
+ let _ = blackhole () in 2;;
-3. The following expression is an attempt to make explicit the
+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,
+ This almost works. For instance,
- if true then 1 else 2;;
+ if true then 1 else 2;;
-evaluates to 1, and
+ evaluates to 1, and
- let b = true in let y = 1 in let n = 2 in
- match b with true -> y | false -> n;;
+ let b = true in let y = 1 in let n = 2 in
+ match b with true -> y | false -> n;;
-also evaluates to 1. Likewise,
+ also evaluates to 1. Likewise,
- if false then 1 else 2;;
+ if false then 1 else 2;;
-and
+ and
- let b = false in let y = 1 in let n = 2 in
- match b with true -> y | false -> n;;
+ let b = false in let y = 1 in let n = 2 in
+ match b with true -> y | false -> n;;
-both evaluate to 2.
+ both evaluate to 2.
-However,
+ However,
- let rec omega x = omega x in
- if true then omega else omega ();;
+ let rec blackhole x = blackhole x in
+ if true then blackhole else blackhole ();;
-terminates, but
+ 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;;
+ 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.
+ 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]]
+ [[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
+Baby monads
+-----------
- (int -> int -> int) -> (int option) -> (int option) -> (int option)
+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
- 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`:
+ (int -> int -> int) -> (int option) -> (int option) -> (int option)
- let bind (x: int option) (f: int -> (int option)) =
- match x with None -> None | Some n -> f n;;
+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 by Umut Acar. See <http://www.mpi-sws.org/~umut/>.)
+
+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
+ e ::= x | λx:τ. e | e1 e2 | Λα. 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, add.
+
+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 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
+
+**Excercise** convert this function to OCaml. Also write an `append` function.
+Test with simple lists.
+
+Consider the following simple binary tree type:
+
+ type ’a tree = Leaf | Node of ’a tree * ’a * ’a tree
+
+**Excercise**
+Write a function `sumLeaves` that computes the sum of all the
+leaves in an int tree.
+
+Write a function `inOrder` : 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.