- : int = 1
-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?
+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?
let rec f x = f x;;
2. Throughout this problem, assume that we have
- let rec omega x = omega x;;
+ let rec blackhole x = blackhole x;;
All of the following are well-typed.
Which ones terminate? What are the generalizations?
- omega;;
+ blackhole;;
- omega ();;
+ blackhole ();;
- fun () -> omega ();;
+ fun () -> blackhole ();;
- (fun () -> omega ()) ();;
+ (fun () -> blackhole ()) ();;
- if true then omega else omega;;
+ if true then blackhole else blackhole;;
- if false then omega else omega;;
+ if false then blackhole else blackhole;;
- if true then omega else omega ();;
+ if true then blackhole else blackhole ();;
- if false then omega else omega ();;
+ if false then blackhole else blackhole ();;
- if true then omega () else omega;;
+ if true then blackhole () else blackhole;;
- if false then omega () else omega;;
+ if false then blackhole () else blackhole;;
- if true then omega () else omega ();;
+ if true then blackhole () else blackhole ();;
- if false then omega () else omega ();;
+ if false then blackhole () else blackhole ();;
- let _ = omega in 2;;
+ let _ = blackhole in 2;;
- let _ = omega () in 2;;
+ let _ = blackhole () in 2;;
3. This problem is to begin thinking about controlling order of evaluation.
The following expression is an attempt to make explicit the
However,
- let rec omega x = omega x in
- if true then omega else omega ();;
+ let rec blackhole x = blackhole x in
+ if true then blackhole else blackhole ();;
terminates, but
- let rec omega x = omega x in
+ let rec blackhole x = blackhole x in
let b = true in
- let y = omega in
- let n = omega () in
+ let y = blackhole in
+ let n = blackhole () in
match b with true -> y | false -> n;;
does not terminate. Incidentally, `match bool with true -> yes |
[[Hint assignment 5 problem 3]]
-Baby monads
------------
-
-Read the lecture notes for week 6, then write a
-function `lift` that generalized the correspondence between + and
-`add`: that is, `lift` takes any two-place operation on integers
-and returns a version that takes arguments of type `int option`
-instead, returning a result of `int option`. In other words,
-`lift` will have type
-
- (int -> int -> int) -> (int option) -> (int option) -> (int option)
-
-so that `lift (+) (Some 3) (Some 4)` will evalute to `Some 7`.
-Don't worry about why you need to put `+` inside of parentheses.
-You should make use of `bind` in your definition of `lift`:
-
- let bind (x: int option) (f: int -> (int option)) =
- match x with None -> None | Some n -> f n;;
-
-
-Booleans, Church numbers, and Church lists in OCaml
----------------------------------------------------
+Booleans, Church numerals, and v3 lists in OCaml
+------------------------------------------------
(These questions adapted from web materials by Umut Acar. See <http://www.mpi-sws.org/~umut/>.)
-The idea is to get booleans, Church numbers, "Church" lists, and
-binary trees working in OCaml.
+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.)
Recall from class System F, or the polymorphic λ-calculus.
- τ ::= α | τ1 → τ2 | ∀α. τ
- e ::= x | λx:τ. e | e1 e2 | Λα. e | e [τ ]
+ types τ ::= c | 'a | τ1 → τ2 | ∀'a. τ
+ expressions e ::= x | λx:τ. e | e1 e2 | Λ'a. e | e [τ]
+
+The boolean type, and its two values, may be encoded as follows:
+
+ bool := ∀'a. 'a → 'a → 'a
+ true := Λ'a. λt:'a. λf :'a. t
+ false := Λ'a. λt:'a. λf :'a. f
-Recall that bool may be encoded as follows:
+It's used like this:
- bool := ∀α. α → α → α
- true := Λα. λt:α. λf :α. t
- false := Λα. λt:α. λf :α. f
+ b [τ] e1 e2
-(where τ indicates the type of e1 and e2)
+where b is a boolean value, and τ is the shared type of e1 and e2.
-Note that each of the following terms, when applied to the
-appropriate arguments, return a result of type bool.
+**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.
(a) the term not that takes an argument of type bool and computes its negation;
(b) the term and that takes two arguments of type bool and computes their conjunction;
(c) the term or that takes two arguments of type bool and computes their disjunction.
+
The type nat (for "natural number") may be encoded as follows:
- nat := ∀α. α → (α → α) → α
- zero := Λα. λz:α. λs:α → α. z
- succ := λn:nat. Λα. λz:α. λs:α → α. s (n [α] z s)
+ nat := ∀'a. 'a → ('a → 'a) → 'a
+ zero := Λ'a. λz:'a. λs:'a → 'a. z
+ succ := λn:nat. Λ'a. λz:'a. λs:'a → 'a. s (n ['a] z s)
-A nat n is defined by what it can do, which is to compute a function iterated n times. In the polymorphic
-encoding above, the result of that iteration can be any type α, as long as you have a base element z : α and
-a function s : α → α.
+A nat n is defined by what it can do, which is to compute a function iterated n
+times. In the polymorphic encoding above, the result of that iteration can be
+any type 'a, as long as you have a base element z : 'a and a function s : 'a → 'a.
**Excercise**: get booleans and Church numbers working in OCaml,
-including OCaml versions of bool, true, false, zero, succ, add.
+including OCaml versions of bool, true, false, zero, iszero, succ, and **pred**.
Consider the following list type:
- type ’a list = Nil | Cons of ’a * ’a list
+ type 'a list = Nil | Cons of 'a * 'a list
We can encode τ lists, lists of elements of type τ as follows:
- τ list := ∀α. α → (τ → α → α) → α
- nilτ := Λα. λn:α. λc:τ → α → α. n
- makeListτ := λh:τ. λt:τ list. Λα. λn:α. λc:τ → α → α. c h (t [α] n c)
+ τ list := ∀'a. 'a → (τ → 'a → 'a) → 'a
+ nil τ := Λ'a. λn:'a. λc:τ → 'a → 'a. n
+ make_list τ := λh:τ. λt:τ list. Λ'a. λn:'a. λc:τ → 'a → 'a. c h (t ['a] n c)
As with nats, recursion is built into the datatype.
= λf :σ → τ. λl:σ list. l [τ list] nilτ (λx:σ. λy:τ list. consτ (f x) y
**Excercise** convert this function to OCaml. Also write an `append` function.
-Test with simple lists.
+Also write a **head** function. Also write nil??? Test with simple lists.
+
Consider the following simple binary tree type:
- type ’a tree = Leaf | Node of ’a tree * ’a * ’a tree
+ type 'a tree = Leaf | Node of 'a tree * 'a * 'a tree
**Excercise**
-Write a function `sumLeaves` that computes the sum of all the
+Write a function `sum_leaves` that computes the sum of all the
leaves in an int tree.
-Write a function `inOrder` : τ tree → τ list that computes the in-order traversal of a binary tree. You
+Write a function `in_order` : τ tree → τ list that computes the in-order traversal of a binary tree. You
may assume the above encoding of lists; define any auxiliary functions you need.
+Baby monads
+-----------
+
+Read the lecture notes for week 6, then write a
+function `lift'` that generalized the correspondence between + and
+`add'`: that is, `lift'` takes any two-place operation on integers
+and returns a version that takes arguments of type `int option`
+instead, returning a result of `int option`. In other words,
+`lift'` will have type
+
+ (int -> int -> int) -> (int option) -> (int option) -> (int option)
+
+so that `lift' (+) (Some 3) (Some 4)` will evalute to `Some 7`.
+Don't worry about why you need to put `+` inside of parentheses.
+You should make use of `bind'` in your definition of `lift'`:
+
+ let bind' (x: int option) (f: int -> (int option)) =
+ match x with None -> None | Some n -> f n;;
+
+