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-Types and OCAML
+Assignment 5
-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?
+Types and OCaml
+---------------
- let rec f x = f x;;
+0. Recall that the S combinator is given by \x y z. x z (y z).
+ Give two different typings for this function in OCaml.
+ To get you started, here's one typing for K:
- let rec f x = f f;;
+ # let k (y:'a) (n:'b) = y;;
+ val k : 'a -> 'b -> 'a = [fun]
+ # k 1 true;;
+ - : int = 1
- let rec f x = f x in f f;;
- let rec f x = f x in f ();;
+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 () = f f;;
+ let rec f x = f x;;
- let rec f () = f ();;
+ let rec f x = f f;;
- let rec f () = f () in f f;;
+ let rec f x = f x in f f;;
- let rec f () = f () in f ();;
+ let rec f x = f x in f ();;
-2. Throughout this problem, assume that we have
+ let rec f () = f f;;
- let rec omega x = omega x;;
+ let rec f () = f ();;
- All of the following are well-typed.
- Which ones terminate? What are the generalizations?
+ let rec f () = f () in f f;;
- omega;;
+ let rec f () = f () in f ();;
- omega ();;
+2. Throughout this problem, assume that we have
- fun () -> omega ();;
+ let rec blackhole x = blackhole x;;
- (fun () -> omega ()) ();;
+ All of the following are well-typed.
+ Which ones terminate? What are the generalizations?
- if true then omega else omega;;
+ blackhole;;
- if false then omega else omega;;
+ blackhole ();;
- if true then omega else omega ();;
+ fun () -> blackhole ();;
- if false then omega else 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 ();;
- let _ = omega in 2;;
+ if true then blackhole () else blackhole;;
- let _ = omega () in 2;;
+ if false then blackhole () else blackhole;;
-3. The following expression is an attempt to make explicit the
+ if true then blackhole () else blackhole ();;
+
+ if false then blackhole () else blackhole ();;
+
+ let _ = blackhole 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
behavior of `if`-`then`-`else` explored in the previous question.
-The idea is to define an `if`-`then`-`else` expression using
-other expression types. So assume that "yes" is any OCAML expression,
-and "no" is any other OCAML expression (of the same type as "yes"!),
+The idea is to define an `if`-`then`-`else` expression using
+other expression types. So assume that "yes" is any OCaml expression,
+and "no" is any other OCaml expression (of the same type as "yes"!),
and that "bool" is any boolean. Then we can try the following:
"if bool then yes else no" should be equivalent to
- let b = bool in
- let y = yes in
- let n = no in
- match b with true -> y | false -> n
+ let b = bool in
+ let y = yes in
+ let n = no in
+ match b with true -> y | false -> n
+
+ This almost works. For instance,
+
+ if true then 1 else 2;;
+
+ evaluates to 1, and
+
+ let b = true in let y = 1 in let n = 2 in
+ match b with true -> y | false -> n;;
+
+ also evaluates to 1. Likewise,
+
+ if false then 1 else 2;;
+
+ and
+
+ let b = false in let y = 1 in let n = 2 in
+ match b with true -> y | false -> n;;
+
+ both evaluate to 2.
+
+ However,
+
+ let rec blackhole x = blackhole x in
+ if true then blackhole else blackhole ();;
+
+ terminates, but
+
+ let rec blackhole x = blackhole x in
+ let b = true 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 |
+ false -> no;;` works as desired, but your assignment is to solve it
+ without using the magical evaluation order properties of either `if`
+ or of `match`. That is, you must keep the `let` statements, though
+ you're allowed to adjust what `b`, `y`, and `n` get assigned to.
+
+ [[Hint assignment 5 problem 3]]
+
+Booleans, Church numerals, and v3 lists in OCaml
+------------------------------------------------
+
+(These questions adapted from web materials by Umut Acar. See
+.)
+
+Let's think about the encodings of booleans, numerals and lists in System F,
+and get data-structures with the same form working in OCaml. (Of course, OCaml
+has *native* versions of these datas-structures: its `true`, `1`, and `[1;2;3]`.
+But the point of our exercise requires that we ignore those.)
+
+Recall from class System F, or the polymorphic Î»-calculus.
+
+ types Ï ::= c | 'a | Ï1 â Ï2 | â'a. Ï
+ expressions e ::= x | Î»x:Ï. e | e1 e2 | Î'a. e | e [Ï]
+
+The boolean type, and its two values, may be encoded as follows:
+
+ bool := â'a. 'a â 'a â 'a
+ true := Î'a. Î»t:'a. Î»f :'a. t
+ false := Î'a. Î»t:'a. Î»f :'a. f
+
+It's used like this:
+
+ b [Ï] e1 e2
+
+where b is a boolean value, and Ï is the shared type of e1 and e2.
+
+**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 := â'a. 'a â ('a â 'a) â 'a
+ zero := Î'a. Î»z:'a. Î»s:'a â 'a. z
+ succ := Î»n:nat. Î'a. Î»z:'a. Î»s:'a â 'a. s (n ['a] z s)
+
+A nat n is deï¬ned 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 â 'a.
+
+**Exercise**: get booleans and Church numbers working in OCaml,
+including OCaml versions of bool, true, false, zero, iszero, succ, and pred.
+It's especially useful to do a version of pred, starting with one
+of the (untyped) versions available in the lambda library
+accessible from the main wiki page. The point of the excercise
+is to do these things on your own, so avoid using the built-in
+OCaml booleans and integers.
+
+Consider the following list type:
+
+ type 'a list = Nil | Cons of 'a * 'a list
+
+We can encode Ï lists, lists of elements of type Ï as follows:
+
+ Ï list := â'a. 'a â (Ï â 'a â 'a) â 'a
+ nil Ï := Î'a. Î»n:'a. Î»c:Ï â 'a â 'a. n
+ make_list Ï := Î»h:Ï. Î»t:Ï list. Î'a. Î»n:'a. Î»c:Ï â 'a â 'a. c h (t ['a] n c)
+
+More generally, the polymorphic list type is:
+
+ list := â'b. â'a. 'a â ('b â 'a â 'a) â 'a
+
+As with nats, recursion is built into the datatype.
+
+We can write functions like map:
+
+ map : (Ï â Ï ) â Ï list â Ï list
+
+
+
+**Excercise** convert this function to OCaml. We've given you the type; you
+only need to give the term.
-This almost works. For instance,
+Also give us the type and definition for a `head` function. Think about what
+value to give back if the argument is the empty list. Ultimately, we might
+want to make use of our `'a option` technique, but for this assignment, just
+pick a strategy, no matter how clunky.
- if true then 1 else 2;;
+Be sure to test your proposals with simple lists. (You'll have to `make_list`
+the lists yourself; don't expect OCaml to magically translate between its
+native lists and the ones you buil.d)
-evaluates to 1, and
- let b = true in let y = 1 in let n = 2 in
- match b with true -> y | false -> n;;
+
- let b = false in let y = 1 in let n = 2 in
- match b with true -> y | false -> n;;
-both evaluate to 2.
+Baby monads
+-----------
-However,
+Read the material on dividing by zero/towards monads from ~~the end of lecture
+notes for week 6~~ the start of lecture notes for week 7, 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:
- let rec omega x = omega x in
- if true then omega else omega ();;
+ (int -> int -> int) -> (int option) -> (int option) -> (int option)
-terminates, but
+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 rec omega x = omega x in
- let b = true in
- let y = omega in
- let n = omega () in
- match b with true -> y | false -> n;;
+ let bind' (u: int option) (f: int -> (int option)) =
+ match u with None -> None | Some x -> f x;;
-does not terminate. Incidentally, `match bool with true -> yes |
-false -> no;;` works as desired, but your assignment is to solve it
-without using the magical evaluation order properties of either `if`
-or of `match`. That is, you must keep the `let` statements, though
-you're allowed to adjust what `b`, `y`, and `n` get assigned to.
-[[Hint assignment 5 problem 3]]