X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=assignment5.mdwn;h=f402ec61a70bbe9ebe4e5f7c2f4a2f1ecc853ede;hp=cc714e90766536fcceb0927d297d276f923856c0;hb=HEAD;hpb=90294e766ccb45391a5d5e9909a0720ed92cca60 diff --git a/assignment5.mdwn b/assignment5.mdwn deleted file mode 100644 index cc714e90..00000000 --- a/assignment5.mdwn +++ /dev/null @@ -1,214 +0,0 @@ -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 omega x = omega x;; - - All of the following are well-typed. - Which ones terminate? What are the generalizations? - - omega;; - - omega ();; - - fun () -> omega ();; - - (fun () -> omega ()) ();; - - if true then omega else omega;; - - if false then omega else omega;; - - if true then omega else omega ();; - - if false then omega else omega ();; - - if true then omega () else omega;; - - if false then omega () else omega;; - - if true then omega () else omega ();; - - if false then omega () else omega ();; - - let _ = omega in 2;; - - let _ = omega () 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 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]] - -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;; - - -Booleans, Church numbers, and Church lists in OCaml ---------------------------------------------------- - -These questions adapted from web materials written by some smart dude named Acar. -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 | c - e ::= x | λx:τ. e | e1 e2 | Λ'α. 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, 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 list predicates. - - 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 head, isNil, and map: - - map : (σ → τ ) → σ list → τ list - - We've given you the type for map, you only need to give the term. - - With regard to `head`, 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. - - Please provide both the terms and the types for each item.