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 blackhole x = blackhole x;;
40 All of the following are well-typed.
41 Which ones terminate? What are the generalizations?
47 fun () -> blackhole ();;
49 (fun () -> blackhole ()) ();;
51 if true then blackhole else blackhole;;
53 if false then blackhole else blackhole;;
55 if true then blackhole else blackhole ();;
57 if false then blackhole else blackhole ();;
59 if true then blackhole () else blackhole;;
61 if false then blackhole () else blackhole;;
63 if true then blackhole () else blackhole ();;
65 if false then blackhole () else blackhole ();;
67 let _ = blackhole in 2;;
69 let _ = blackhole () 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 blackhole x = blackhole x in
108 if true then blackhole else blackhole ();;
112 let rec blackhole x = blackhole x in
115 let n = blackhole () 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]]
126 Booleans, Church numbers, and Church lists in OCaml
127 ---------------------------------------------------
129 (These questions adapted from web materials by Umut Acar. See <http://www.mpi-sws.org/~umut/>.)
131 The idea is to get booleans, Church numbers, v3 lists, and
132 binary trees working in OCaml.
134 Recall from class System F, or the polymorphic λ-calculus.
136 τ ::= α | τ1 → τ2 | ∀α. τ
137 e ::= x | λx:τ. e | e1 e2 | Λα. e | e [τ ]
139 Recall that bool may be encoded as follows:
141 bool := ∀α. α → α → α
142 true := Λα. λt:α. λf :α. t
143 false := Λα. λt:α. λf :α. f
145 (where τ indicates the type of e1 and e2)
147 Note that each of the following terms, when applied to the
148 appropriate arguments, return a result of type bool.
150 (a) the term not that takes an argument of type bool and computes its negation;
151 (b) the term and that takes two arguments of type bool and computes their conjunction;
152 (c) the term or that takes two arguments of type bool and computes their disjunction.
154 The type nat (for "natural number") may be encoded as follows:
156 nat := ∀α. α → (α → α) → α
157 zero := Λα. λz:α. λs:α → α. z
158 succ := λn:nat. Λα. λz:α. λs:α → α. s (n [α] z s)
160 A nat n is defined by what it can do, which is to compute a function iterated n times. In the polymorphic
161 encoding above, the result of that iteration can be any type α, as long as you have a base element z : α and
162 a function s : α → α.
164 **Excercise**: get booleans and Church numbers working in OCaml,
165 including OCaml versions of bool, true, false, zero, succ, add.
167 Consider the following list type:
169 type ’a list = Nil | Cons of ’a * ’a list
171 We can encode τ lists, lists of elements of type τ as follows:
173 τ list := ∀α. α → (τ → α → α) → α
174 nilτ := Λα. λn:α. λc:τ → α → α. n
175 makeListτ := λh:τ. λt:τ list. Λα. λn:α. λc:τ → α → α. c h (t [α] n c)
177 As with nats, recursion is built into the datatype.
179 We can write functions like map:
181 map : (σ → τ ) → σ list → τ list
182 = λf :σ → τ. λl:σ list. l [τ list] nilτ (λx:σ. λy:τ list. consτ (f x) y
184 **Excercise** convert this function to OCaml. Also write an `append` function.
185 Test with simple lists.
187 Consider the following simple binary tree type:
189 type ’a tree = Leaf | Node of ’a tree * ’a * ’a tree
192 Write a function `sumLeaves` that computes the sum of all the
193 leaves in an int tree.
195 Write a function `inOrder` : τ tree → τ list that computes the in-order traversal of a binary tree. You
196 may assume the above encoding of lists; define any auxiliary functions you need.
201 Read the lecture notes for week 6, then write a
202 function `lift'` that generalized the correspondence between + and
203 `add'`: that is, `lift'` takes any two-place operation on integers
204 and returns a version that takes arguments of type `int option`
205 instead, returning a result of `int option`. In other words,
206 `lift'` will have type
208 (int -> int -> int) -> (int option) -> (int option) -> (int option)
210 so that `lift' (+) (Some 3) (Some 4)` will evalute to `Some 7`.
211 Don't worry about why you need to put `+` inside of parentheses.
212 You should make use of `bind'` in your definition of `lift'`:
214 let bind' (x: int option) (f: int -> (int option)) =
215 match x with None -> None | Some n -> f n;;