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;;
+
+ 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.
+
+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.
+Also write a **head** function. Also write nil??? 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 `sum_leaves` that computes the sum of all the
+leaves in an int tree.
+
+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.
+
+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;;
- 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