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]]
128 Read the lecture notes for week 6, then write a
129 function `lift` that generalized the correspondence between + and
130 `add`: that is, `lift` takes any two-place operation on integers
131 and returns a version that takes arguments of type `int option`
132 instead, returning a result of `int option`. In other words,
133 `lift` will have type
135 (int -> int -> int) -> (int option) -> (int option) -> (int option)
137 so that `lift (+) (Some 3) (Some 4)` will evalute to `Some 7`.
138 Don't worry about why you need to put `+` inside of parentheses.
139 You should make use of `bind` in your definition of `lift`:
141 let bind (x: int option) (f: int -> (int option)) =
142 match x with None -> None | Some n -> f n;;
145 Booleans, Church numbers, and Church lists in System F
146 ------------------------------------------------------
148 These questions adapted from web materials written by some smart dude named Acar.
150 Recall from class System F, or the polymorphic λ-calculus.
152 τ ::= α | τ1 → τ2 | ∀α. τ
153 e ::= x | λx:τ. e | e1 e2 | Λα. e | e [τ ]
155 Recall that bool may be encoded as follows:
157 bool := ∀α. α → α → α
158 true := Λα. λt:α. λf :α. t
159 false := Λα. λt:α. λf :α. f
160 ifτ e then e1 else e2 := e [τ ] e1 e2
162 (where τ indicates the type of e1 and e2)
164 Exercise 1. Show how to encode the following terms. Note that each of these terms, when applied to the
165 appropriate arguments, return a result of type bool.
167 (a) the term not that takes an argument of type bool and computes its negation;
168 (b) the term and that takes two arguments of type bool and computes their conjunction;
169 (c) the term or that takes two arguments of type bool and computes their disjunction.
171 The type nat (for "natural number") may be encoded as follows:
173 nat := ∀α. α → (α → α) → α
174 zero := Λα. λz:α. λs:α → α. z
175 succ := λn:nat. Λα. λz:α. λs:α → α. s (n [α] z s)
177 A nat n is defined by what it can do, which is to compute a function iterated n times. In the polymorphic
178 encoding above, the result of that iteration can be any type α, as long as you have a base element z : α and
179 a function s : α → α.
181 Exercise 2. Verify that these encodings (zero, succ , rec) typecheck in System F.
182 (Draw a type tree for each term.)
184 Consider the following list type:
186 datatype ’a list = Nil | Cons of ’a * ’a list
188 We can encode τ lists, lists of elements of type τ as follows:
190 τ list := ∀α. α → (τ → α → α) → α
191 nilτ := Λα. λn:α. λc:τ → α → α. n
192 consτ := λh:τ. λt:τ list. Λα. λn:α. λc:τ → α → α. c h (t [α] n c)
194 As with nats, The τ list type’s case analyzing elimination form is just application.
196 We can write functions like map:
198 map : (σ → τ ) → σ list → τ list
199 := λf :σ → τ. λl:σ list. l [τ list] nilτ (λx:σ. λy:τ list. consτ (f x) y
201 Exercise 3. Consider the following simple binary tree type:
203 datatype ’a tree = Leaf | Node of ’a tree * ’a * ’a tree
205 (a) Give a System F encoding of binary trees, including a definition of the type τ tree and definitions of
206 the constructors leaf : τ tree and node : τ tree → τ → τ tree → τ tree.
208 (b) Write a function height : τ tree → nat. You may assume the above encoding of nat as well as definitions
209 of the functions plus : nat → nat → nat and max : nat → nat → nat.
211 (c) Write a function in-order : τ tree → τ list that computes the in-order traversal of a binary tree. You
212 may assume the above encoding of lists; define any auxiliary functions you need.