+++ /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 blackhole x = blackhole x;;
-
- All of the following are well-typed.
- Which ones terminate? What are the generalizations?
-
- blackhole;;
-
- blackhole ();;
-
- fun () -> blackhole ();;
-
- (fun () -> blackhole ()) ();;
-
- if true then blackhole else blackhole;;
-
- if false then blackhole else blackhole;;
-
- if true then blackhole else blackhole ();;
-
- if false then blackhole else blackhole ();;
-
- if true then blackhole () else blackhole;;
-
- if false then blackhole () else blackhole;;
-
- 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"!),
-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 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;;
-
-