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? For those that are, give the type of the expression as a whole. For those that are not, why not?
22 let rec f x = f x in f f;;
24 let rec f x = f x in f ();;
30 let rec f () = f () in f f;;
32 let rec f () = f () in f ();;
34 2. Throughout this problem, assume that we have
36 let rec blackhole x = blackhole x;;
38 All of the following are well-typed.
39 Which ones terminate? What are the generalizations?
45 fun () -> blackhole ();;
47 (fun () -> blackhole ()) ();;
49 if true then blackhole else blackhole;;
51 if false then blackhole else blackhole;;
53 if true then blackhole else blackhole ();;
55 if false then blackhole else blackhole ();;
57 if true then blackhole () else blackhole;;
59 if false then blackhole () else blackhole;;
61 if true then blackhole () else blackhole ();;
63 if false then blackhole () else blackhole ();;
65 let _ = blackhole in 2;;
67 let _ = blackhole () in 2;;
69 3. This problem is to begin thinking about controlling order of evaluation.
70 The following expression is an attempt to make explicit the
71 behavior of `if`-`then`-`else` explored in the previous question.
72 The idea is to define an `if`-`then`-`else` expression using
73 other expression types. So assume that "yes" is any OCaml expression,
74 and "no" is any other OCaml expression (of the same type as "yes"!),
75 and that "bool" is any boolean. Then we can try the following:
76 "if bool then yes else no" should be equivalent to
81 match b with true -> y | false -> n
83 This almost works. For instance,
85 if true then 1 else 2;;
89 let b = true in let y = 1 in let n = 2 in
90 match b with true -> y | false -> n;;
92 also evaluates to 1. Likewise,
94 if false then 1 else 2;;
98 let b = false in let y = 1 in let n = 2 in
99 match b with true -> y | false -> n;;
105 let rec blackhole x = blackhole x in
106 if true then blackhole else blackhole ();;
110 let rec blackhole x = blackhole x in
113 let n = blackhole () in
114 match b with true -> y | false -> n;;
116 does not terminate. Incidentally, `match bool with true -> yes |
117 false -> no;;` works as desired, but your assignment is to solve it
118 without using the magical evaluation order properties of either `if`
119 or of `match`. That is, you must keep the `let` statements, though
120 you're allowed to adjust what `b`, `y`, and `n` get assigned to.
122 [[Hint assignment 5 problem 3]]
124 Booleans, Church numerals, and v3 lists in OCaml
125 ------------------------------------------------
127 (These questions adapted from web materials by Umut Acar. See <http://www.mpi-sws.org/~umut/>.)
129 Let's think about the encodings of booleans, numerals and lists in System F, and get datastructures with the same explicit form working in OCaml. (The point... so we won't rely on OCaml's native booleans, integers, or lists.)
131 Recall from class System F, or the polymorphic λ-calculus.
133 types τ ::= c | 'a | τ1 → τ2 | ∀'a. τ
134 expressions e ::= x | λx:τ. e | e1 e2 | Λ'a. e | e [τ]
136 The boolean type, and its two values, may be encoded as follows:
138 bool := ∀'a. 'a → 'a → 'a
139 true := Λ'a. λt:'a. λf :'a. t
140 false := Λ'a. λt:'a. λf :'a. f
146 where b is a boolean value, and τ is the shared type of e1 and e2.
148 **Exercise**. How should we implement the following terms. Note that the result of applying them to the appropriate arguments should also give us a term 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.
155 The type nat (for "natural number") may be encoded as follows:
157 nat := ∀'a. 'a → ('a → 'a) → 'a
158 zero := Λ'a. λz:'a. λs:'a → 'a. z
159 succ := λn:nat. Λ'a. λz:'a. λs:'a → 'a. s (n ['a] z s)
161 A nat n is defined by what it can do, which is to compute a function iterated n
162 times. In the polymorphic encoding above, the result of that iteration can be
163 any type 'a, as long as you have a base element z : 'a and a function s : 'a → 'a.
165 **Excercise**: get booleans and Church numbers working in OCaml,
166 including OCaml versions of bool, true, false, zero, iszero, succ, and **pred**.
168 Consider the following list type:
170 type 'a list = Nil | Cons of 'a * 'a list
172 We can encode τ lists, lists of elements of type τ as follows:
174 τ list := ∀'a. 'a → (τ → 'a → 'a) → 'a
175 nil τ := Λ'a. λn:'a. λc:τ → 'a → 'a. n
176 make_list τ := λh:τ. λt:τ list. Λ'a. λn:'a. λc:τ → 'a → 'a. c h (t ['a] n c)
178 As with nats, recursion is built into the datatype.
180 We can write functions like map:
182 map : (σ → τ ) → σ list → τ list
183 = λf :σ → τ. λl:σ list. l [τ list] nilτ (λx:σ. λy:τ list. consτ (f x) y
185 **Excercise** convert this function to OCaml. Also write an `append` function.
186 Also write a **head** function. Also write nil??? Test with simple lists.
189 Consider the following simple binary tree type:
191 type 'a tree = Leaf | Node of 'a tree * 'a * 'a tree
194 Write a function `sum_leaves` that computes the sum of all the
195 leaves in an int tree.
197 Write a function `in_order` : τ tree → τ list that computes the in-order traversal of a binary tree. You
198 may assume the above encoding of lists; define any auxiliary functions you need.
203 Read the lecture notes for week 6, then write a
204 function `lift'` that generalized the correspondence between + and
205 `add'`: that is, `lift'` takes any two-place operation on integers
206 and returns a version that takes arguments of type `int option`
207 instead, returning a result of `int option`. In other words,
208 `lift'` will have type
210 (int -> int -> int) -> (int option) -> (int option) -> (int option)
212 so that `lift' (+) (Some 3) (Some 4)` will evalute to `Some 7`.
213 Don't worry about why you need to put `+` inside of parentheses.
214 You should make use of `bind'` in your definition of `lift'`:
216 let bind' (x: int option) (f: int -> (int option)) =
217 match x with None -> None | Some n -> f n;;