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
17 are, give the type of the expression as a whole. For those that are not, why
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 [[hints/assignment 5 hint 1]]
126 Booleans, Church numerals, and v3 lists in OCaml
127 ------------------------------------------------
129 (These questions adapted from web materials by Umut Acar. See
130 <http://www.mpi-sws.org/~umut/>.)
132 Let's think about the encodings of booleans, numerals and lists in System F,
133 and get data-structures with the same form working in OCaml. (Of course, OCaml
134 has *native* versions of these datas-structures: its `true`, `1`, and `[1;2;3]`.
135 But the point of our exercise requires that we ignore those.)
137 Recall from class System F, or the polymorphic λ-calculus.
139 types τ ::= c | 'a | τ1 → τ2 | ∀'a. τ
140 expressions e ::= x | λx:τ. e | e1 e2 | Λ'a. e | e [τ]
142 The boolean type, and its two values, may be encoded as follows:
144 bool := ∀'a. 'a → 'a → 'a
145 true := Λ'a. λt:'a. λf :'a. t
146 false := Λ'a. λt:'a. λf :'a. f
152 where b is a boolean value, and τ is the shared type of e1 and e2.
154 **Exercise**. How should we implement the following terms. Note that the result
155 of applying them to the appropriate arguments should also give us a term of
158 (a) the term not that takes an argument of type bool and computes its negation;
159 (b) the term and that takes two arguments of type bool and computes their conjunction;
160 (c) the term or that takes two arguments of type bool and computes their disjunction.
163 The type nat (for "natural number") may be encoded as follows:
165 nat := ∀'a. 'a → ('a → 'a) → 'a
166 zero := Λ'a. λz:'a. λs:'a → 'a. z
167 succ := λn:nat. Λ'a. λz:'a. λs:'a → 'a. s (n ['a] z s)
169 A nat n is defined by what it can do, which is to compute a function iterated n
170 times. In the polymorphic encoding above, the result of that iteration can be
171 any type 'a, as long as you have a base element z : 'a and a function s : 'a → 'a.
173 **Exercise**: get booleans and Church numbers working in OCaml,
174 including OCaml versions of bool, true, false, zero, iszero, succ, and pred.
175 It's especially useful to do a version of pred, starting with one
176 of the (untyped) versions available in the lambda library
177 accessible from the main wiki page. The point of the excercise
178 is to do these things on your own, so avoid using the built-in
179 OCaml booleans and integers.
181 Consider the following list type:
183 type 'a list = Nil | Cons of 'a * 'a list
185 We can encode τ lists, lists of elements of type τ as follows:
187 τ list := ∀'a. 'a → (τ → 'a → 'a) → 'a
188 nil τ := Λ'a. λn:'a. λc:τ → 'a → 'a. n
189 make_list τ := λh:τ. λt:τ list. Λ'a. λn:'a. λc:τ → 'a → 'a. c h (t ['a] n c)
191 More generally, the polymorphic list type is:
193 list := ∀'b. ∀'a. 'a → ('b → 'a → 'a) → 'a
195 As with nats, recursion is built into the datatype.
197 We can write functions like map:
199 map : (σ → τ ) → σ list → τ list
202 = λf :σ → τ. λl:σ list. l [τ list] nil τ (λx:σ. λy:τ list. make_list τ (f x) y
205 **Excercise** convert this function to OCaml. We've given you the type; you
206 only need to give the term.
208 Also give us the type and definition for a `head` function. Think about what
209 value to give back if the argument is the empty list. Ultimately, we might
210 want to make use of our `'a option` technique, but for this assignment, just
211 pick a strategy, no matter how clunky.
213 Be sure to test your proposals with simple lists. (You'll have to `make_list`
214 the lists yourself; don't expect OCaml to magically translate between its
215 native lists and the ones you buil.d)
219 Consider the following simple binary tree type:
221 type 'a tree = Leaf | Node of 'a tree * 'a * 'a tree
224 Write a function `sum_leaves` that computes the sum of all the leaves in an int
227 Write a function `in_order` : τ tree → τ list that computes the in-order
228 traversal of a binary tree. You may assume the above encoding of lists; define
229 any auxiliary functions you need.
236 Read the material on dividing by zero/towards monads from <strike>the end of lecture
237 notes for week 6</strike> the start of lecture notes for week 7, then write a function `lift'` that generalized the
238 correspondence between + and `add'`: that is, `lift'` takes any two-place
239 operation on integers and returns a version that takes arguments of type `int
240 option` instead, returning a result of `int option`. In other words, `lift'`
243 (int -> int -> int) -> (int option) -> (int option) -> (int option)
245 so that `lift' (+) (Some 3) (Some 4)` will evalute to `Some 7`.
246 Don't worry about why you need to put `+` inside of parentheses.
247 You should make use of `bind'` in your definition of `lift'`:
249 let bind' (u: int option) (f: int -> (int option)) =
250 match u with None -> None | Some x -> f x;;