X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=assignment5.mdwn;h=02c0ac4dd8bcd45d6e5d470eaa64dcdcb24a54f6;hp=4a4e06d2a36f5b6ad49072453ffc9b0533b11ff5;hb=0d85c76d0d37b32bf99483b86828a7d2829db44e;hpb=ac9dca9445dd794a3cbda9cfa80d48cb521a96d7 diff --git a/assignment5.mdwn b/assignment5.mdwn index 4a4e06d2..02c0ac4d 100644 --- a/assignment5.mdwn +++ b/assignment5.mdwn @@ -1,215 +1,217 @@ 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. This problem is to begin thinking about controlling order of evaluation. +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;; + if true then 1 else 2;; -evaluates to 1, and + evaluates to 1, and - let b = true in let y = 1 in let n = 2 in - match b with true -> y | false -> n;; + let b = true in let y = 1 in let n = 2 in + match b with true -> y | false -> n;; -also evaluates to 1. Likewise, + also evaluates to 1. Likewise, - if false then 1 else 2;; + if false then 1 else 2;; -and + and - let b = false in let y = 1 in let n = 2 in - match b with true -> y | false -> n;; + let b = false in let y = 1 in let n = 2 in + match b with true -> y | false -> n;; -both evaluate to 2. + both evaluate to 2. -However, + However, - let rec omega x = omega x in - if true then omega else omega ();; + let rec blackhole x = blackhole x in + if true then blackhole else blackhole ();; -terminates, but + 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;; + 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. + 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]] + [[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 +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) + (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`: +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;; + 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 +Booleans, Church numbers, and Church lists in OCaml --------------------------------------------------- -These questions adapted from web materials written by some smart dude named Acar. +(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. +binary trees working in OCaml. + +Recall from class System F, or the polymorphic λ-calculus. - 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 ] - 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: - 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 - 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) - (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. - 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. - (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: - 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) - 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. - 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, add. - **Excercise**: get booleans and Church numbers working in OCAML, - including OCAML versions of bool, true, false, zero, succ, add. +Consider the following list type: - Consider the following list type: + type ’a list = Nil | Cons of ’a * ’a list - type ’a list = Nil | Cons of ’a * ’a list +We can encode τ lists, lists of elements of type τ as follows: - 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) - 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. - As with nats, recursion is built into the datatype. +We can write functions like map: - 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 - 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. +Test with simple lists. - **Excercise** convert this function to OCAML. Also write an `append` function. - Test with simple lists. +Consider the following simple binary tree type: - Consider the following simple binary tree type: + type ’a tree = Leaf | Node of ’a tree * ’a * ’a tree - type ’a tree = Leaf | Node of ’a tree * ’a * ’a tree +**Excercise** +Write a function `sumLeaves` that computes the sum of all the +leaves in an int tree. - **Excercise** - Write a function `sumLeaves` that computes the sum of all the - leaves in an int tree. +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. - 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.