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[lambda.git] / assignment5.mdwn

diff --git a/assignment5.mdwn b/assignment5.mdwn
index 61096c4..f402ec6 100644 (file)
--- a/assignment5.mdwn
+++ b/assignment5.mdwn
@@ -13,9 +13,9 @@ Types and OCaml
                - : int = 1
 
 
                - : 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;;
 
 
                let rec f x = f x;;
 


@@ -35,38 +35,38 @@ Types and OCaml
 
 2.     Throughout this problem, assume that we have
 
 
 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?
 
 
        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
 
 3.     This problem is to begin thinking about controlling order of evaluation.
 The following expression is an attempt to make explicit the


@@ -104,15 +104,15 @@ and that "bool" is any boolean.  Then we can try the following:
 
        However,
 
 
        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
 
 
        terminates, but
 

-               let rec omega x = omega x in
+               let rec blackhole x = blackhole x in

                let b = true 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 |
                match b with true -> y | false -> n;;
 
        does not terminate.  Incidentally, `match bool with true -> yes |


@@ -121,97 +121,132 @@ and that "bool" is any boolean.  Then we can try the following:
        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.
 
        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]]
+       [[hints/assignment 5 hint 1]]

 
 

-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`:
+Booleans, Church numerals, and v3 lists in OCaml
+------------------------------------------------

 
 

-       let bind (x: int option) (f: int -> (int option)) =
-               match x with None -> None | Some n -> f n;;
+(These questions adapted from web materials by Umut Acar. See
+<http://www.mpi-sws.org/~umut/>.)

 
 

+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.)

 
 

-Booleans, Church numbers, and Church lists in OCaml
----------------------------------------------------
-
-(These questions adapted from web materials by Umut Acar. See <http://www.mpi-sws.org/~umut/>.)
+Recall from class System F, or the polymorphic λ-calculus.

 
 

-The idea is to get booleans, Church numbers, "Church" lists, and
-binary trees working in OCaml.
+       types τ ::= c | 'a | τ1 → τ2 | ∀'a. τ
+       expressions e ::= x | λx:τ. e | e1 e2 | Λ'a. e | e [τ]

 
 

-Recall from class System F, or the polymorphic λ-calculus.
+The boolean type, and its two values, may be encoded as follows:

 
 

-       τ ::= α | τ1 → τ2 | ∀α. τ
-       e ::= x | λx:τ. e | e1 e2 | Λα. e | e [τ ]
+       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.
 
 
 (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:
 
 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.
+**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:
 
 
 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:
 
 
 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)
+
+More generally, the polymorphic list type is:
+
+       list := ∀'b. ∀'a. 'a → ('b → 'a → 'a) → 'a

 
 As with nats, recursion is built into the datatype.
 
 We can write functions like map:
 
        map : (σ → τ ) → σ list → τ list
 
 As with nats, recursion is built into the datatype.
 
 We can write functions like map:
 
        map : (σ → τ ) → σ list → τ list

-               = 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.
+<!--
+               = λf :σ → τ. λl:σ list. l [τ list] nil τ (λx:σ. λy:τ list. make_list τ (f x) y
+-->
+
+**Excercise** convert this function to OCaml. We've given you the type; you
+only need to give the term.

 
 

+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. 
+
+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)
+
+
+<!--

 Consider the following simple binary tree type:
 
 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**
 
 **Excercise**

-Write a function `sumLeaves` that computes the sum of all the
-leaves in an int tree.
+Write a function `sum_leaves` that computes the sum of all the leaves in an int
+tree.
+
+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 material on dividing by zero/towards monads from <strike>the end of lecture
+notes for week 6</strike> 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:
+
+       (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' (u: int option) (f: int -> (int option)) =
+               match u with None -> None | Some x -> f x;;

 
 

-Write a function `inOrder` : 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.