week 6 start

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+Types and OCAML
+
+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 f;;
+
+    let rec f x = f x in f f;;
+
+    let rec f x = f x in f ();;
+
+    let rec f () = f f;;
+
+    let rec f () = f ();;
+
+    let rec f () = f () in f f;;
+
+    let rec f () = f () in f ();;
+
+2. Throughout this problem, assume that we have 
+
+    let rec omega x = omega x;;
+
+   All of the following are well-typed.
+   Which ones terminate?  What are the generalizations?
+
+    omega;;
+
+    omega ();;
+
+    fun () -> omega ();;
+
+    (fun () -> omega ()) ();;
+
+    if true then omega else omega;;
+
+    if false then omega else omega;;
+
+    if true then omega else omega ();;
+
+    if false then omega else omega ();;
+
+    if true then omega () else omega;;
+
+    if false then omega () else omega;;
+
+    if true then omega () else omega ();;
+
+    if false then omega () else omega ();;
+
+    let _ = omega in 2;;
+
+    let _ = omega () in 2;;
+
+3. 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"!),
+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
+
+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 -> 1 | false -> 2;;
+
+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 omega x = omega x in 
+    if true then omega else omega ();;
+
+terminates, but 
+
+    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;;
+
+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]]

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+[[!toc]]
+
+Types, OCAML
+------------
+
+OCAML has type inference: the system can often infer what the type of
+an expression must be, based on the type of other known expressions.
+
+For instance, if we type 
+
+    # let f x = x + 3;;
+
+The system replies with 
+
+    val f : int -> int = <fun>
+
+Since `+` is only defined on integers, it has type
+
+     # (+);;
+     - : int -> int -> int = <fun>
+
+The parentheses are there to turn off the trick that allows the two
+arguments of `+` to surround it in infix (for linguists, SOV) argument
+order. That is,
+
+    # 3 + 4 = (+) 3 4;;
+    - : bool = true
+
+In general, tuples with one element are identical to their one
+element:
+
+    # (3) = 3;;
+    - : bool = true
+
+though OCAML, like many systems, refuses to try to prove whether two
+functional objects may be identical:
+
+    # (f) = f;;
+    Exception: Invalid_argument "equal: functional value".
+
+Oh well.
+
+
+Booleans in OCAML, and simple pattern matching
+----------------------------------------------
+
+Where we would write `true 1 2` and expect it to evaluate to `1`, in
+OCAML boolean types are not functions (equivalently, are functions
+that take zero arguments).  Choices are made as follows:
+
+    # if true then 1 else 2;;
+    - : int = 1
+
+The types of the `then` clause and of the `else` clause must be the
+same.
+
+The `if` construction can be re-expressed by means of the following
+pattern-matching expression:
+
+    match <bool expression> with true -> <expression1> | false -> <expression2>
+
+That is,
+
+    # match true with true -> 1 | false -> 2;;
+    - : int = 1
+
+Compare with 
+
+    # match 3 with 1 -> 1 | 2 -> 4 | 3 -> 9;;
+    - : int = 9
+
+Unit
+----
+
+All functions in OCAML take exactly one argument.  Even this one:
+
+    # let f x y = x + y;;
+    # f 2 3;;
+    - : int = 5
+
+Here's how to tell that `f` has been curry'd:
+
+    # f 2;;
+    - : int -> int = <fun>
+
+After we've given our `f` one argument, it returns a function that is
+still waiting for another argument.
+
+There is a special type in OCAML called `unit`.  There is exactly one
+object in this type, written `()`.  So
+
+    # ();;
+    - : unit = ()
+
+Just as you can define functions that take constants for arguments
+
+    # let f 2 = 3;;
+    # f 2;;
+    - : int = 3;;
+
+you can also define functions that take the unit as its argument, thus
+
+    # let f () = 3;;
+    val f : unit -> int = <fun>
+
+Then the only argument you can possibly apply `f` to that is of the
+correct type is the unit:
+
+    # f ();;
+    - : int = 3
+
+Let's have some fn: think of `rec` as our `Y` combinator.  Then
+
+    # let rec f n = if (0 = n) then 1 else (n * (f (n - 1)));; 
+    val f : int -> int = <fun>
+    # f 5;;
+    - : int = 120
+
+We can't define a function that is exactly analogous to our &omega;.
+We could try `let rec omega x = x x;;` what happens?  However, we can
+do this:
+
+    # let rec omega x = omega x;;
+
+By the way, what's the type of this function?
+If you then apply this omega to an argument,
+
+    # omega 3;;
+
+the interpreter goes into an infinite loop, and you have to control-C
+to break the loop.
+
+Oh, one more thing: lambda expressions look like this:
+
+    # (fun x -> x);;
+    - : 'a -> 'a = <fun>
+    # (fun x -> x) true;;
+    - : book = true
+
+(But `(fun x -> x x)` still won't work.)
+
+So we can try our usual tricks:
+
+    # (fun x -> true) omega;;
+    - : bool = true
+
+OCAML declined to try to evaluate the argument before applying the
+functor.  But remember that `omega` is a function too, so we can
+reverse the order of the arguments:
+
+    # omega (fun x -> true);;
+
+Infinite loop.
+
+Now consider the following differences:
+
+    # let test = omega omega;;
+    [Infinite loop, need to control c out]
+
+    # let test () = omega omega;;
+    val test : unit -> 'a = <fun>
+
+    # test;;
+    - : unit -> 'a = <fun>
+
+    # test ();;
+    [Infinite loop, need to control c out]
+
+We can use functions that take arguments of type unit to control
+execution.  In Scheme parlance, functions on the unit type are called
+*thunks* (which I've always assumed was a blend of "think" and "chunk").
+