X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=assignment5.mdwn;h=5a2a488c4d01befe7a47f023710eea2117bfa641;hp=ccf402abaa4a8b890b601146539978d6c2eca0fb;hb=6ea619babdb3656bf940636bd6823c6a2b8697c1;hpb=96a8c8c9b81fc914ac7ec368fab0ffa4bcf4177a;ds=sidebyside diff --git a/assignment5.mdwn b/assignment5.mdwn index ccf402ab..5a2a488c 100644 --- a/assignment5.mdwn +++ b/assignment5.mdwn @@ -1,110 +1,217 @@ -Types and OCAML +Assignment 5 -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? +Types and OCaml +--------------- - let rec f x = f x;; +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 rec f x = f f;; + # let k (y:'a) (n:'b) = y;; + val k : 'a -> 'b -> 'a = [fun] + # k 1 true;; + - : int = 1 - let rec f x = f x in f f;; - let rec f x = f x in f ();; +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 () = f f;; + let rec f x = f x;; - let rec f () = f ();; + let rec f x = f f;; - let rec f () = f () in f f;; + let rec f x = f x in f f;; - let rec f () = f () in f ();; + let rec f x = f x in f ();; -2. Throughout this problem, assume that we have + let rec f () = f f;; - let rec omega x = omega x;; + let rec f () = f ();; - All of the following are well-typed. - Which ones terminate? What are the generalizations? + let rec f () = f () in f f;; - omega;; + let rec f () = f () in f ();; - omega ();; +2. Throughout this problem, assume that we have - fun () -> omega ();; + let rec blackhole x = blackhole x;; - (fun () -> omega ()) ();; + All of the following are well-typed. + Which ones terminate? What are the generalizations? - if true then omega else omega;; + blackhole;; - if false then omega else omega;; + blackhole ();; - if true then omega else omega ();; + fun () -> blackhole ();; - if false then omega else 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 ();; - let _ = omega in 2;; + if true then blackhole () else blackhole;; - let _ = omega () in 2;; + if false then blackhole () else blackhole;; -3. The following expression is an attempt to make explicit the + 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"!), +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, + + 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]] + +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 by Umut Acar. See .) + +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) -This almost works. For instance, +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. - if true then 1 else 2;; +**Excercise**: get booleans and Church numbers working in OCaml, +including OCaml versions of bool, true, false, zero, succ, add. -evaluates to 1, and +Consider the following list type: - let b = true in let y = 1 in let n = 2 in - match b with true -> y | false -> n;; + type ’a list = Nil | Cons of ’a * ’a list -also evaluates to 1. Likewise, +We can encode τ lists, lists of elements of type τ as follows: - if false then 1 else 2;; + 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) -and +As with nats, recursion is built into the datatype. - let b = false in let y = 1 in let n = 2 in - match b with true -> y | false -> n;; +We can write functions like map: -both evaluate to 2. + 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 -However, +**Excercise** convert this function to OCaml. Also write an `append` function. +Test with simple lists. - let rec omega x = omega x in - if true then omega else omega ();; +Consider the following simple binary tree type: -terminates, but + type ’a tree = Leaf | Node of ’a tree * ’a * ’a tree - 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;; +**Excercise** +Write a function `sumLeaves` that computes the sum of all the +leaves in an int tree. -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. +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. -[[Hint assignment 5 problem 3]]