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. This problem is to begin thinking about controlling order of evaluation.
72 The following expression is an attempt to make explicit the
73 behavior of `if`-`then`-`else` explored in the previous question.
74 The idea is to define an `if`-`then`-`else` expression using
75 other expression types. So assume that "yes" is any OCaml expression,
76 and "no" is any other OCaml expression (of the same type as "yes"!),
77 and that "bool" is any boolean. Then we can try the following:
78 "if bool then yes else no" should be equivalent to
83 match b with true -> y | false -> n
85 This almost works. For instance,
87 if true then 1 else 2;;
91 let b = true in let y = 1 in let n = 2 in
92 match b with true -> y | false -> n;;
94 also evaluates to 1. Likewise,
96 if false then 1 else 2;;
100 let b = false in let y = 1 in let n = 2 in
101 match b with true -> y | false -> n;;
107 let rec omega x = omega x in
108 if true then omega else omega ();;
112 let rec omega x = omega x in
116 match b with true -> y | false -> n;;
118 does not terminate. Incidentally, `match bool with true -> yes |
119 false -> no;;` works as desired, but your assignment is to solve it
120 without using the magical evaluation order properties of either `if`
121 or of `match`. That is, you must keep the `let` statements, though
122 you're allowed to adjust what `b`, `y`, and `n` get assigned to.
124 [[Hint assignment 5 problem 3]]
129 Read the lecture notes for week 6, then write a
130 function `lift` that generalized the correspondence between + and
131 `add`: that is, `lift` takes any two-place operation on integers
132 and returns a version that takes arguments of type `int option`
133 instead, returning a result of `int option`. In other words,
134 `lift` will have type
136 (int -> int -> int) -> (int option) -> (int option) -> (int option)
138 so that `lift (+) (Some 3) (Some 4)` will evalute to `Some 7`.
139 Don't worry about why you need to put `+` inside of parentheses.
140 You should make use of `bind` in your definition of `lift`:
142 let bind (x: int option) (f: int -> (int option)) =
143 match x with None -> None | Some n -> f n;;
146 Booleans, Church numbers, and Church lists in OCaml
147 ---------------------------------------------------
149 These questions adapted from web materials written by some smart dude named Acar.
150 The idea is to get booleans, Church numbers, "Church" lists, and
151 binary trees working in OCaml.
153 Recall from class System F, or the polymorphic λ-calculus.
155 τ ::= 'α | τ1 → τ2 | ∀'α. τ | c
156 e ::= x | λx:τ. e | e1 e2 | Λ'α. e | e [τ ]
158 Recall that bool may be encoded as follows:
160 bool := ∀α. α → α → α
161 true := Λα. λt:α. λf :α. t
162 false := Λα. λt:α. λf :α. f
164 (where τ indicates the type of e1 and e2)
166 Note that each of the following terms, when applied to the
167 appropriate arguments, return a result of type bool.
169 (a) the term not that takes an argument of type bool and computes its negation;
170 (b) the term and that takes two arguments of type bool and computes their conjunction;
171 (c) the term or that takes two arguments of type bool and computes their disjunction.
173 The type nat (for "natural number") may be encoded as follows:
175 nat := ∀α. α → (α → α) → α
176 zero := Λα. λz:α. λs:α → α. z
177 succ := λn:nat. Λα. λz:α. λs:α → α. s (n [α] z s)
179 A nat n is defined by what it can do, which is to compute a function iterated n times. In the polymorphic
180 encoding above, the result of that iteration can be any type α, as long as you have a base element z : α and
181 a function s : α → α.
183 **Excercise**: get booleans and Church numbers working in OCaml,
184 including OCaml versions of bool, true, false, zero, succ, and pred.
185 It's especially useful to do a version of pred, starting with one
186 of the (untyped) versions available in the lambda library
187 accessible from the main wiki page. The point of the excercise
188 is to do these things on your own, so avoid using the built-in
189 OCaml booleans and list predicates.
191 Consider the following list type:
193 type ’a list = Nil | Cons of ’a * ’a list
195 We can encode τ lists, lists of elements of type τ as follows:
197 τ list := ∀α. α → (τ → α → α) → α
198 nilτ := Λα. λn:α. λc:τ → α → α. n
199 makeListτ := λh:τ. λt:τ list. Λα. λn:α. λc:τ → α → α. c h (t [α] n c)
201 As with nats, recursion is built into the datatype.
203 We can write functions like head, isNil, and map:
205 map : (σ → τ ) → σ list → τ list
207 We've given you the type for map, you only need to give the term.
209 With regard to `head`, think about what value to give back if the
210 argument is the empty list. Ultimately, we might want to make use
211 of our `'a option` technique, but for this assignment, just pick a
212 strategy, no matter how clunky.
214 Please provide both the terms and the types for each item.