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 numerals, and v3 lists in OCaml
127 ------------------------------------------------
129 (These questions adapted from web materials by Umut Acar. See <http://www.mpi-sws.org/~umut/>.)
131 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.)
133 Recall from class System F, or the polymorphic λ-calculus.
135 types τ ::= c | 'a | τ1 → τ2 | ∀'a. τ
136 expressions e ::= x | λx:τ. e | e1 e2 | Λ'a. e | e [τ]
138 The boolean type, and its two values, may be encoded as follows:
140 bool := ∀'a. 'a → 'a → 'a
141 true := Λ'a. λt:'a. λf :'a. t
142 false := Λ'a. λt:'a. λf :'a. f
148 where b is a boolean value, and τ is the shared type of e1 and e2.
150 **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.
152 (a) the term not that takes an argument of type bool and computes its negation;
153 (b) the term and that takes two arguments of type bool and computes their conjunction;
154 (c) the term or that takes two arguments of type bool and computes their disjunction.
157 The type nat (for "natural number") may be encoded as follows:
159 nat := ∀'a. 'a → ('a → 'a) → 'a
160 zero := Λ'a. λz:'a. λs:'a → 'a. z
161 succ := λn:nat. Λ'a. λz:'a. λs:'a → 'a. s (n ['a] z s)
163 A nat n is defined by what it can do, which is to compute a function iterated n
164 times. In the polymorphic encoding above, the result of that iteration can be
165 any type 'a, as long as you have a base element z : 'a and a function s : 'a → 'a.
167 **Excercise**: get booleans and Church numbers working in OCaml,
168 including OCaml versions of bool, true, false, zero, succ, add.
170 Consider the following list type:
172 type 'a list = Nil | Cons of 'a * 'a list
174 We can encode τ lists, lists of elements of type τ as follows:
176 τ list := ∀'a. 'a → (τ → 'a → 'a) → 'a
177 nilτ := Λ'a. λn:'a. λc:τ → 'a → 'a. n
178 makeListτ := λh:τ. λt:τ list. Λ'a. λn:'a. λc:τ → 'a → 'a. c h (t ['a] n c)
180 As with nats, recursion is built into the datatype.
182 We can write functions like map:
184 map : (σ → τ ) → σ list → τ list
185 = λf :σ → τ. λl:σ list. l [τ list] nilτ (λx:σ. λy:τ list. consτ (f x) y
187 **Excercise** convert this function to OCaml. Also write an `append` function.
188 Test with simple lists.
190 Consider the following simple binary tree type:
192 type 'a tree = Leaf | Node of 'a tree * 'a * 'a tree
195 Write a function `sumLeaves` that computes the sum of all the
196 leaves in an int tree.
198 Write a function `inOrder` : τ tree → τ list that computes the in-order traversal of a binary tree. You
199 may assume the above encoding of lists; define any auxiliary functions you need.
204 Read the lecture notes for week 6, then write a
205 function `lift'` that generalized the correspondence between + and
206 `add'`: that is, `lift'` takes any two-place operation on integers
207 and returns a version that takes arguments of type `int option`
208 instead, returning a result of `int option`. In other words,
209 `lift'` will have type
211 (int -> int -> int) -> (int option) -> (int option) -> (int option)
213 so that `lift' (+) (Some 3) (Some 4)` will evalute to `Some 7`.
214 Don't worry about why you need to put `+` inside of parentheses.
215 You should make use of `bind'` in your definition of `lift'`:
217 let bind' (x: int option) (f: int -> (int option)) =
218 match x with None -> None | Some n -> f n;;