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.
+
+ [[hints/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
+
+<!--
+ = λf :σ → τ. λl:σ list. l [τ list] nil τ (λx:σ. λy:τ list. make_list τ (f x) y
+-->
+
+**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)
+
+
+<!--
+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 material on dividing by zero/towards monads from <strike>the end of lecture
+notes for week 6</strike> 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)) =
+ match u with None -> None | Some x -> f x;;
- 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