6 0. Recall that the S combinator is given by \x y z. x z (y z).
7 Give two different typings for this function in OCAML.
8 To get you started, here's one typing for K:
10 # let k (y:'a) (n:'b) = y;;
11 val k : 'a -> 'b -> 'a = <fun>
16 1. Which of the following expressions is well-typed in OCAML?
17 For those that are, give the type of the expression as a whole.
18 For those that are not, why not?
24 let rec f x = f x in f f;;
26 let rec f x = f x in f ();;
32 let rec f () = f () in f f;;
34 let rec f () = f () in f ();;
36 2. Throughout this problem, assume that we have
38 let rec omega x = omega x;;
40 All of the following are well-typed.
41 Which ones terminate? What are the generalizations?
49 (fun () -> omega ()) ();;
51 if true then omega else omega;;
53 if false then omega else omega;;
55 if true then omega else omega ();;
57 if false then omega else omega ();;
59 if true then omega () else omega;;
61 if false then omega () else omega;;
63 if true then omega () else omega ();;
65 if false then omega () else omega ();;
69 let _ = omega () in 2;;
71 3. The following expression is an attempt to make explicit the
72 behavior of `if`-`then`-`else` explored in the previous question.
73 The idea is to define an `if`-`then`-`else` expression using
74 other expression types. So assume that "yes" is any OCAML expression,
75 and "no" is any other OCAML expression (of the same type as "yes"!),
76 and that "bool" is any boolean. Then we can try the following:
77 "if bool then yes else no" should be equivalent to
82 match b with true -> y | false -> n
84 This almost works. For instance,
86 if true then 1 else 2;;
90 let b = true in let y = 1 in let n = 2 in
91 match b with true -> y | false -> n;;
93 also evaluates to 1. Likewise,
95 if false then 1 else 2;;
99 let b = false in let y = 1 in let n = 2 in
100 match b with true -> y | false -> n;;
106 let rec omega x = omega x in
107 if true then omega else omega ();;
111 let rec omega x = omega x in
115 match b with true -> y | false -> n;;
117 does not terminate. Incidentally, `match bool with true -> yes |
118 false -> no;;` works as desired, but your assignment is to solve it
119 without using the magical evaluation order properties of either `if`
120 or of `match`. That is, you must keep the `let` statements, though
121 you're allowed to adjust what `b`, `y`, and `n` get assigned to.
123 [[Hint assignment 5 problem 3]]
125 4. Baby monads. Read the lecture notes for week 6, then write a
126 function `lift` that generalized the correspondence between + and
127 `add`: that is, `lift` takes any two-place operation on integers
128 and returns a version that takes arguments of type `int option`
129 instead, returning a result of `int option`. In other words,
130 `lift` will have type
132 (int -> int -> int) -> (int option) -> (int option) -> (int option)
134 so that `lift (+) (Some 3) (Some 4)` will evalute to `Some 7`.
135 Don't worry about why you need to put `+` inside of parentheses.
136 You should make use of `bind` in your definition of `lift`:
138 let bind (x: int option) (f: int -> (int option)) =
139 match x with None -> None | Some n -> f n;;