X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=assignment5.mdwn;h=f402ec61a70bbe9ebe4e5f7c2f4a2f1ecc853ede;hp=ccf402abaa4a8b890b601146539978d6c2eca0fb;hb=3ed8e56bf33b1c0cfa09b672abf26e9c397ea0f1;hpb=e735d4bc20d0d6147ca24de3ce4bf0843bd2ab17 diff --git a/assignment5.mdwn b/assignment5.mdwn index ccf402ab..f402ec61 100644 --- a/assignment5.mdwn +++ b/assignment5.mdwn @@ -1,110 +1,252 @@ -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. + + [[hints/assignment 5 hint 1]] + +Booleans, Church numerals, and v3 lists in OCaml +------------------------------------------------ + +(These questions adapted from web materials by Umut Acar. See +.) + +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 Î»-calculus. + + types Ï ::= c | 'a | Ï1 â Ï2 | â'a. Ï + expressions e ::= x | Î»x:Ï. e | e1 e2 | Î'a. e | e [Ï] + +The boolean type, and its two values, may be encoded as follows: + + bool := â'a. 'a â 'a â 'a + true := Î'a. Î»t:'a. Î»f :'a. t + false := Î'a. Î»t:'a. Î»f :'a. f + +It's used like this: + + b [Ï] e1 e2 + +where b is a boolean value, and Ï 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 := â'a. 'a â ('a â 'a) â 'a + zero := Î'a. Î»z:'a. Î»s:'a â 'a. z + succ := Î»n:nat. Î'a. Î»z:'a. Î»s:'a â 'a. s (n ['a] z s) + +A nat n is deï¬ned 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 â '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 Ï lists, lists of elements of type Ï as follows: + + Ï list := â'a. 'a â (Ï â 'a â 'a) â 'a + nil Ï := Î'a. Î»n:'a. Î»c:Ï â 'a â 'a. n + make_list Ï := Î»h:Ï. Î»t:Ï list. Î'a. Î»n:'a. Î»c:Ï â 'a â 'a. c h (t ['a] n c) + +More generally, the polymorphic list type is: + + list := â'b. â'a. 'a â ('b â 'a â 'a) â 'a + +As with nats, recursion is built into the datatype. + +We can write functions like map: + + map : (Ï â Ï ) â Ï list â Ï list + + + +**Excercise** convert this function to OCaml. We've given you the type; you +only need to give the term. -This almost works. For instance, +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. - if true then 1 else 2;; +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) -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 = false in let y = 1 in let n = 2 in - match b with true -> y | false -> n;; -both evaluate to 2. +Baby monads +----------- -However, +Read the material on dividing by zero/towards monads from the end of lecture +notes for week 6 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: - let rec omega x = omega x in - if true then omega else omega ();; + (int -> int -> int) -> (int option) -> (int option) -> (int option) -terminates, but +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 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 bind' (u: int option) (f: int -> (int option)) = + match u with None -> None | Some x -> f x;; -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]]