From cbf0c724914abae08e00292bdcbd3dea1f9460cd Mon Sep 17 00:00:00 2001 From: Chris Date: Mon, 23 Feb 2015 15:34:13 -0500 Subject: [PATCH] defined system f --- topics/_week5_system_F.mdwn | 357 ++++++++++++++++++++++++++++++++++++++++++-- 1 file changed, 344 insertions(+), 13 deletions(-) diff --git a/topics/_week5_system_F.mdwn b/topics/_week5_system_F.mdwn index f0134f92..2c37ae34 100644 --- a/topics/_week5_system_F.mdwn +++ b/topics/_week5_system_F.mdwn @@ -7,24 +7,355 @@ correspondence. In the meantime, note that types respect modus ponens:
```-Expression    Type     Implication
-----------------------------------
-fn            α -> β   α ⊃ β
-arg           α        α
+Expression    Type      Implication
+-----------------------------------
+fn            α -> β    α ⊃ β
+arg           α         α
------        ------    --------
-fn arg        β        β
+(fn arg)      β         β
```
The implication in the right-hand column is modus ponens, of course. -System F is usually attributed to Girard, but was independently -proposed around the same time by Reynolds. It enhances the -simply-typed lambda calculus with quantification over types. In -System F, you can say things like +System F was discovered by Girard (the same guy who invented Linear +Logic), but it was independently proposed around the same time by +Reynolds, who called his version the *polymorphic lambda calculus*. +(Reynolds was also an early player in the development of +continuations.) -`Λ α (\x.x):(α -> α)` +System F enhances the simply-typed lambda calculus with abstraction +over types. In order to state System F, we'll need to adopt the +notational convention that "`x:α`" represents a +expression whose type is `α`. -This says that the identity function maps arguments of type α to -results of type α, for any choice of α. So the Λ is -a universal quantifier over types. +Then System F can be specified as follows (choosing notation that will +match up with usage in O'Caml, whose type system is based on System F): + System F: + types Ï ::= c | 'a | Ï1 -> Ï2 | â'a. Ï + expressions e ::= x | Î»x:Ï. e | e1 e2 | Î'a. e | e [Ï] + +In the definition of the types, "`c`" is a type constant (e.g., `e` or +`t`). "`'a`" is a type variable (the tick mark just indicates that +the variable ranges over types rather than values). "`Ï1 -> Ï2`" is +the type of a function from expressions of type `Ï1` to expressions of +type `Ï2`. And "`â'a. Ï`" is called a universal type, since it +universally quantifies over the type variable `'a`. + +In the definition of the expressions, we have variables "`x`". +Abstracts "`Î»x:Ï. e`" are similar to abstracts in the simply-typed lambda +calculus, except that they have their shrug variable annotated with a +type. Applications "`e1 e2`" are just like in the simply-typed lambda calculus. +In addition to variables, abstracts, and applications, we have two +additional ways of forming expressions: "`Î'a. e`" is a type +abstraction, and "`e [Ï]`" is a type application. The idea is that +`Λ` is a capital `λ`. Just like +the lower-case `λ`, `Λ` binds +variables in its body; unlike `λ`, +`Λ` binds type variables. So in the expression + +`Λ 'a (λ x:'a . x)` + +the `Λ` binds the type variable `'a` that occurs in +the `λ` abstract. This expression is a polymorphic +version of the identity function. It says that this one general +identity function can be adapted for use with expressions of any +type. In order to get it ready to apply to, say, a variable of type +boolean, just do this: + +`(Λ 'a (λ x:'a . x)) [t]` + +The type application (where `t` is a type constant for Boolean truth +values) specifies the value of the type variable `α`, which is +the type of the variable bound in the `λ` expression. Not +surprisingly, the type of this type application is a function from +Booleans to Booleans: + +`((Λ 'a (λ x:'a . x)) [t]): (b -> b)` + +Likewise, if we had instantiated the type variable as an entity (base +type `e`), the resulting identity function would have been a function +of type `e -> e`: + +`((Λ 'a (λ x:'a . x)) [e]): (e -> e)` + +Clearly, for any choice of a type `'a`, the identity function can be +instantiated as a function from expresions of type `'a` to expressions +of type `'a`. In general, then, the type of the unapplied +(polymorphic) identity function is + +```(Λ 'a (λ x:'a . x)): (\forall 'a . 'a -> 'a) + + + + +## + + + +Types in 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 = + +Since `+` is only defined on integers, it has type + + # (+);; + - : int -> int -> int = + +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. + +[Note: There is a limited way you can compare functions, using the +`==` operator instead of the `=` operator. Later when we discuss mutation, +we'll discuss the difference between these two equality operations. +Scheme has a similar pair, which they name `eq?` and `equal?`. In Python, +these are `is` and `==` respectively. It's unfortunate that OCaml uses `==` for the opposite operation that Python and many other languages use it for. In any case, OCaml will accept `(f) == f` even though it doesn't accept +`(f) = f`. However, don't expect it to figure out in general when two functions +are equivalent. (That question is not Turing computable.) + + # (f) == (fun x -> x + 3);; + - : bool = false + +Here OCaml says (correctly) that the two functions don't stand in the `==` relation, which basically means they're not represented in the same chunk of memory. However as the programmer can see, the functions are extensionally equivalent. The meaning of `==` is rather weird.] + + + +Booleans in OCaml, and simple pattern matching +---------------------------------------------- + +Where we would write `true 1 2` in our pure lambda calculus and expect +it to evaluate to `1`, in OCaml boolean types are not functions +(equivalently, they're functions that take zero arguments). Instead, selection is +accomplished 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 with true -> | false -> + +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 and thunks +--------------- + +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 = + +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 = + +Then the only argument you can possibly apply `f` to that is of the +correct type is the unit: + + # f ();; + - : int = 3 + +Now why would that be useful? + +Let's have some fun: 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 = + # f 5;; + - : int = 120 + +We can't define a function that is exactly analogous to our ω. +We could try `let rec omega x = x x;;` what happens? + +[Note: if you want to learn more OCaml, you might come back here someday and try: + + # let id x = x;; + val id : 'a -> 'a = + # let unwrap (`Wrap a) = a;; + val unwrap : [< `Wrap of 'a ] -> 'a = + # let omega ((`Wrap x) as y) = x y;; + val omega : [< `Wrap of [> `Wrap of 'a ] -> 'b as 'a ] -> 'b = + # unwrap (omega (`Wrap id)) == id;; + - : bool = true + # unwrap (omega (`Wrap omega));; + + +But we won't try to explain this now.] + + +Even if we can't (easily) express omega in OCaml, we can do this: + + # let rec blackhole x = blackhole x;; + +By the way, what's the type of this function? + +If you then apply this `blackhole` function to an argument, + + # blackhole 3;; + +the interpreter goes into an infinite loop, and you have to type control-c +to break the loop. + +Oh, one more thing: lambda expressions look like this: + + # (fun x -> x);; + - : 'a -> 'a = + # (fun x -> x) true;; + - : bool = true + +(But `(fun x -> x x)` still won't work.) + +You may also see this: + + # (function x -> x);; + - : 'a -> 'a = + +This works the same as `fun` in simple cases like this, and slightly differently in more complex cases. If you learn more OCaml, you'll read about the difference between them. + +We can try our usual tricks: + + # (fun x -> true) blackhole;; + - : bool = true + +OCaml declined to try to fully reduce the argument before applying the +lambda function. Question: Why is that? Didn't we say that OCaml is a call-by-value/eager language? + +Remember that `blackhole` is a function too, so we can +reverse the order of the arguments: + + # blackhole (fun x -> true);; + +Infinite loop. + +Now consider the following variations in behavior: + + # let test = blackhole blackhole;; + + + # let test () = blackhole blackhole;; + val test : unit -> 'a = + + # test;; + - : unit -> 'a = + + # test ();; + + +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"). + +Question: why do thunks work? We know that `blackhole ()` doesn't terminate, so why do expressions like: + + let f = fun () -> blackhole () + in true + +terminate? + +Bottom type, divergence +----------------------- + +Expressions that don't terminate all belong to the **bottom type**. This is a subtype of every other type. That is, anything of bottom type belongs to every other type as well. More advanced type systems have more examples of subtyping: for example, they might make `int` a subtype of `real`. But the core type system of OCaml doesn't have any general subtyping relations. (Neither does System F.) Just this one: that expressions of the bottom type also belong to every other type. It's as if every type definition in OCaml, even the built in ones, had an implicit extra clause: + + type 'a option = None | Some of 'a;; + type 'a option = None | Some of 'a | bottom;; + +Here are some exercises that may help better understand this. Figure out what is the type of each of the following: + + fun x y -> y;; + + fun x (y:int) -> y;; + + fun x y : int -> y;; + + let rec blackhole x = blackhole x in blackhole;; + + let rec blackhole x = blackhole x in blackhole 1;; + + let rec blackhole x = blackhole x in fun (y:int) -> blackhole y y y;; + + let rec blackhole x = blackhole x in (blackhole 1) + 2;; + + let rec blackhole x = blackhole x in (blackhole 1) || false;; + + let rec blackhole x = blackhole x in 2 :: (blackhole 1);; + +By the way, what's the type of this: + + let rec blackhole (x:'a) : 'a = blackhole x in blackhole + + +Back to thunks: the reason you'd want to control evaluation with +thunks is to manipulate when "effects" happen. In a strongly +normalizing system, like the simply-typed lambda calculus or System F, +there are no "effects." In Scheme and OCaml, on the other hand, we can +write programs that have effects. One sort of effect is printing. +Another sort of effect is mutation, which we'll be looking at soon. +Continuations are yet another sort of effect. None of these are yet on +the table though. The only sort of effect we've got so far is +*divergence* or non-termination. So the only thing thunks are useful +for yet is controlling whether an expression that would diverge if we +tried to fully evaluate it does diverge. As we consider richer +languages, thunks will become more useful. -- 2.11.0 ```