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;;
142 Church lists in System F
143 ------------------------
145 These questions adapted from web materials written by some dude named Acar.
147 Recall from class System F, or the polymorphic λ-calculus.
149 τ ::= α | τ1 → τ2 | ∀α. τ
150 e ::= x | λx:τ. e | e1 e2 | Λα. e | e [τ ]
151 Despite its simplicity, System F is quite expressive. As discussed in class, it has sufficient expressive power
152 to be able to encode many datatypes found in other programming languages, including products, sums, and
154 For example, recall that bool may be encoded as follows:
155 bool := ∀α. α → α → α
156 true := Λα. λt:α. λf :α. t
157 false := Λα. λt:α. λf :α. f
158 ifτ e then e1 else e2 := e [τ ] e1 e2
159 (where τ indicates the type of e1 and e2)
160 Exercise 1. Show how to encode the following terms. Note that each of these terms, when applied to the
161 appropriate arguments, return a result of type bool.
162 (a) the term not that takes an argument of type bool and computes its negation;
163 (b) the term and that takes two arguments of type bool and computes their conjunction;
164 (c) the term or that takes two arguments of type bool and computes their disjunction.
165 The type nat may be encoded as follows:
166 nat := ∀α. α → (α → α) → α
167 zero := Λα. λz:α. λs:α → α. z
168 succ := λn:nat. Λα. λz:α. λs:α → α. s (n [α] z s)
169 A nat n is defined by what it can do, which is to compute a function iterated n times. In the polymorphic
170 encoding above, the result of that iteration can be any type α, as long as you have a base element z : α and
171 a function s : α → α.
172 Conveniently, this encoding “is” its own elimination form, in a sense:
173 rec(e, e0, x:τ. e1) := e [τ ] e0 (λx:τ. e1)
174 The case analysis is baked into the very definition of the type.
175 Exercise 2. Verify that these encodings (zero, succ , rec) typecheck in System F. Write down the typing
176 derivations for the terms.
179 ══════════════════════════════════════════════════════════════════════════
181 As mentioned in class, System F can express any inductive datatype. Consider the following list type:
184 | Cons of ’a * ’a list
185 We can encode τ lists, lists of elements of type τ as follows:1
186 τ list := ∀α. α → (τ → α → α) → α
187 nilτ := Λα. λn:α. λc:τ → α → α. n
188 consτ := λh:τ. λt:τ list. Λα. λn:α. λc:τ → α → α. c h (t [α] n c)
189 As with nats, The τ list type’s case analyzing elimination form is just application. We can write functions
191 map : (σ → τ ) → σ list → τ list
192 := λf :σ → τ. λl:σ list. l [τ list] nilτ (λx:σ. λy:τ list. consτ (f x) y
193 Exercise 3. Consider the following simple binary tree type:
196 | Node of ’a tree * ’a * ’a tree
197 (a) Give a System F encoding of binary trees, including a definition of the type τ tree and definitions of
198 the constructors leaf : τ tree and node : τ tree → τ → τ tree → τ tree.
199 (b) Write a function height : τ tree → nat. You may assume the above encoding of nat as well as definitions
200 of the functions plus : nat → nat → nat and max : nat → nat → nat.
201 (c) Write a function in-order : τ tree → τ list that computes the in-order traversal of a binary tree. You
202 may assume the above encoding of lists; define any auxiliary functions you need.