Types and OCAML 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. 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]]