Can the following polymorphic functions
let id x = x;;
let compose f g x = f (g x);;
let rec fix f = f (fix f);; (*laziness aside*)
be written for types/type constructors or modules/functors? I tried
type 'x id = Id of 'x;;
type 'f 'g 'x compose = Compose of ('f ('g 'x));;
type 'f fix = Fix of ('f (Fix 'f));;
for types but it doesn't work.
Here's a Haskell version for types:
data Id x = Id x
data Compose f g x = Compose (f (g x))
data Fix f = Fix (f (Fix f))
-- examples:
l = Compose [Just 'a'] :: Compose [] Maybe Char
type Natural = Fix Maybe -- natural numbers are fixpoint of Maybe
n = Fix (Just (Fix (Just (Fix Nothing)))) :: Natural -- n is 2
-- up to isomorphism composition of identity and f is f:
iso :: Compose Id f x -> f x
iso (Compose (Id a)) = a
Haskell allows type variables of higher kind. ML dialects, including Caml, allow type variables of kind "*" only. Translated into plain English,
In Haskell, a type variable g can correspond to a "type constructor" like Maybe or IO or lists. So the g x in your Haskell example would be OK (jargon: "well-kinded") if for example g is Maybe and x is Integer.
In ML, a type variable 'g can correspond only to a "ground type" like int or string, never to a type constructor like option or list. It is therefore never correct to try to apply a type variable to another type.
As far as I'm aware, there's no deep reason for this limitation in ML. The most likely explanation is historical contingency. When Milner originally came up with his ideas about polymorphism, he worked with very simple type variables standing only for monotypes of kind *. Early versions of Haskell did the same, and then at some point Mark Jones discovered that inferring the kinds of type variables is actually quite easy. Haskell was quickly revised to allow type variables of higher kind, but ML has never caught up.
The people at INRIA have made a lot of other changes to ML, and I'm a bit surprised they've never made this one. When I'm programming in ML, I might enjoy having higher-kinded type variables. But they aren't there, and I don't know any way to encode the kind of examples you are talking about except by using functors.
You can do something similar in OCaml, using modules in place of types, and functors (higher-order modules) in place of higher-order types. But it looks much uglier and it doesn't have type-inference ability, so you have to manually specify a lot of stuff.
module type Type = sig
type t
end
module Char = struct
type t = char
end
module List (X:Type) = struct
type t = X.t list
end
module Maybe (X:Type) = struct
type t = X.t option
end
(* In the following, I decided to omit the redundant
single constructors "Id of ...", "Compose of ...", since
they don't help in OCaml since we can't use inference *)
module Id (X:Type) = X
module Compose
(F:functor(Z:Type)->Type)
(G:functor(Y:Type)->Type)
(X:Type) = F(G(X))
let l : Compose(List)(Maybe)(Char).t = [Some 'a']
module Example2 (F:functor(Y:Type)->Type) (X:Type) = struct
(* unlike types, "free" module variables are not allowed,
so we have to put it inside another functor in order
to scope F and X *)
let iso (a:Compose(Id)(F)(X).t) : F(X).t = a
end
Well... I'm not an expert of higher-order-types nor Haskell programming.
But this seems to be ok for F# (which is OCaml), could you work with these:
type 'x id = Id of 'x;;
type 'f fix = Fix of ('f fix -> 'f);;
type ('f,'g,'x) compose = Compose of ('f ->'g -> 'x);;
The last one I wrapped to tuple as I didn't come up with anything better...
You can do it but you need to make a bit of a trick:
newtype Fix f = In{out:: f (Fix f)}
You can define Cata afterwards:
Cata :: (Functor f) => (f a -> a) -> Fix f -> a
Cata f = f.(fmap (cata f)).out
That will define a generic catamorphism for all functors, which you can use to build your own stuff. Example:
data ListFix a b = Nil | Cons a b
data List a = Fix (ListFix a)
instance functor (ListFix a) where
fmap f Nil = Nil
fmap f (Cons a lst) = Cons a (f lst)
Related
Apologies if this is considered a dumb question, but how can I make an Isabelle theory recognise ML code? For example, let's say I have an Isabelle file that looks like
ML ‹
type vec = real * real;
fun addvec ((a,b):vec) ((c,d):vec) : vec = (a+b,c+d);
›
lemma "addvec (1,2) (3,4) = (4,6)"
Unfortunately addvec isn't recognised in the lemma. How can I make the function recognised? I've read through How to get ML values from HOL? as well as the Isabelle Cookbook. The former uses a local_setup to assign the constants to Isabelle constants (as far as I can see) using a function called mk_const
fun mk_const c t =
let
val b = Binding.name c
val defb = Binding.name (c ^ "_def")
in (((b, NoSyn), ((defb, []), t)) |> Local_Theory.define) #> snd end
What do the functions Binding.name and Local_Theory.define do, and what is the local_theory type?
Thanks in advance!
The idea is not to define a function written in ML, it is to define in ML a function that you can use in Isabelle.
ML ‹
fun mk_const c t =
let
val b = Binding.name c
val defb = Binding.name (c ^ "_def")
in (((b, NoSyn), ((defb, []), t)) |> Local_Theory.define) #> snd end
›
record point =
coord_x::int
For example, let us define a function that just calls coord_x:
ML ‹
val f = Abs ("x", #{typ "point"}, Const( \<^const_name>‹coord_x›, #{typ "point"} --> #{typ int}) $ Bound 0)
›
Now we have written the definition, we can bind it a name:
local_setup‹mk_const "c" f›
thm c_def
(*c ≡ coord_x*)
local_setup is a keyword to change the theory (so add constants, change the context and so on).
Now obviously you most likely do not want hard coded constants like coord_x.
Some general comments here: I have never used records and I have written a lot of Isabelle. They can be useful (because they are extensible and so on), but they are weird /because they are extensible/. So if you can work on a datatype, do so. It is nicer, it is a proper type (so locales/instances/... just work).
In the image below there is a quick explanation, why pure functions appear to have only one possible implementation.I don't really get the idea because (++) : ('a -> 'b) -> ('a -> 'b) -> 'a -> 'b for example can clearly be implemented by let (++) (f: ('a -> 'b)) (g: ('a -> 'b)) x = f x orlet (++) (f: ('a -> 'b)) (g: ('a -> 'b)) x = g x
Is that image just wrong or do I miss something here?
You are right. The attached image is incorrect even without type annotations.
At first, it's important to consider what kind of "equality" on implementations is assumed here. Let's consider the following examples.
Is (##) equal to (##+)?
let ( ## ) f x = f x
let ( ##+ ) f x =
let _ = 42 in
f x
Is (|>) equal to (|>+)?
let ( |> ) x f = f x
let ( |>+ ) x f = f ## x
Is (%) equal to (%+)?
let ( % ) f g x = f (g x)
let ( %+ ) p q r = p (q r)
If (##) is not equal to (##+), then we can construct the 5th implementation of a function bool -> bool, such as (fun x -> let _ = 42 in true).
Therefore, the author of the image would have wanted to distinguish functions not by its implementation (or codes), but by some other element such as its behavior (like duck test or the equality on mathematical functions).
Still, the image is incorrect. The image claims "for pure functions that don't have any concrete type in the signature, there is only one possible implementation", but no. For example, there is no pure function 'a -> 'b. This can be shown through the Curry–Howard correspondence.
The image is wrong if you consider the counterexample you just gave. I think the author of the image didn’t consider the possibility of type annotation.
In fact if
there are no type annotations and
all the arguments are polymorphic and or function over polymorphic types,
you don’t consider the existence of polymorphic operators such as = or <>
(otherwise it is wrong since <> and = have the same type and different implementation),
then there is only one pure implementation of your function signature.
(you can probably prove that by saying the only things you can use to define that function are :
pure functions of the same type, that can be inlined, so you can ignore that
match-patterns and let, for which the image’s argument is true
cartesian product (let f a b = a, b)
function composition
infinite recursion
and maybe other things I forget, but you can make an exhaustive list
and that the combination of these used can be guessed from the output and input types.
)
I need a type of tree and a map on those, so I do this:
type 'a grouping =
G of ('a * 'a grouping) list
with
member g.map f =
let (G gs) = g
gs |> List.map (fun (s, g) -> f s, g.map f) |> G
But this makes me wonder:
The map member is boilerplate. In Haskell, GHC would implement fmap for me (... deriving (Functor)). I know F# doesn't have typeclasses, but is there some other way I can avoid writing map myself in F#?
Can I somehow avoid the line let (G gs) = g?
Is this whole construction somehow non-idiomatic? It looks weird to me, but maybe that's just because putting members on sum types is new to me.
I don't think there is a way to derive automatically map, however there's a way to emulate type classes in F#, your code can be written like this:
#r #"FsControl.Core.dll"
#r #"FSharpPlus.dll"
open FSharpPlus
open FsControl.Core.TypeMethods
type 'a grouping =
G of ('a * 'a grouping) list
with
// Add an instance for Functor
static member instance (_:Functor.Map, G gs, _) = fun (f:'b->'c) ->
map (fun (s, g) -> f s, map f g) gs |> G
// TEST
let a = G [(1, G [2, G[]] )]
let b = map ((+) 10) a // G [(11, G [12, G[]] )]
Note that map is really overloaded, the first application you see calls the instance for List<'a> and the second one the instance for grouping<'a>. So it behaves like fmap in Haskell.
Also note this way you can decompose G gs without creating the let (G gs) = g
Now regarding what is idiomatic I think many people would agree your solution is more F# idiomatic, but to me new idioms should also be developed in order to get more features and overcome current language limitations, that's why I consider using a library which define clear conventions also idiomatic.
Anyway I agree with #kvb in that it's slightly more idiomatic to define map into a module, in F#+ that convention is also used, so you have the generic map and the specific ModuleX.map
Is it possible to write recursive anonymous functions in SML? I know I could just use the fun syntax, but I'm curious.
I have written, as an example of what I want:
val fact =
fn n => case n of
0 => 1
| x => x * fact (n - 1)
The anonymous function aren't really anonymous anymore when you bind it to a
variable. And since val rec is just the derived form of fun with no
difference other than appearance, you could just as well have written it using
the fun syntax. Also you can do pattern matching in fn expressions as well
as in case, as cases are derived from fn.
So in all its simpleness you could have written your function as
val rec fact = fn 0 => 1
| x => x * fact (x - 1)
but this is the exact same as the below more readable (in my oppinion)
fun fact 0 = 1
| fact x = x * fact (x - 1)
As far as I think, there is only one reason to use write your code using the
long val rec, and that is because you can easier annotate your code with
comments and forced types. For examples if you have seen Haskell code before and
like the way they type annotate their functions, you could write it something
like this
val rec fact : int -> int =
fn 0 => 1
| x => x * fact (x - 1)
As templatetypedef mentioned, it is possible to do it using a fixed-point
combinator. Such a combinator might look like
fun Y f =
let
exception BlackHole
val r = ref (fn _ => raise BlackHole)
fun a x = !r x
fun ta f = (r := f ; f)
in
ta (f a)
end
And you could then calculate fact 5 with the below code, which uses anonymous
functions to express the faculty function and then binds the result of the
computation to res.
val res =
Y (fn fact =>
fn 0 => 1
| n => n * fact (n - 1)
)
5
The fixed-point code and example computation are courtesy of Morten Brøns-Pedersen.
Updated response to George Kangas' answer:
In languages I know, a recursive function will always get bound to a
name. The convenient and conventional way is provided by keywords like
"define", or "let", or "letrec",...
Trivially true by definition. If the function (recursive or not) wasn't bound to a name it would be anonymous.
The unconventional, more anonymous looking, way is by lambda binding.
I don't see what unconventional there is about anonymous functions, they are used all the time in SML, infact in any functional language. Its even starting to show up in more and more imperative languages as well.
Jesper Reenberg's answer shows lambda binding; the "anonymous"
function gets bound to the names "f" and "fact" by lambdas (called
"fn" in SML).
The anonymous function is in fact anonymous (not "anonymous" -- no quotes), and yes of course it will get bound in the scope of what ever function it is passed onto as an argument. In any other cases the language would be totally useless. The exact same thing happens when calling map (fn x => x) [.....], in this case the anonymous identity function, is still in fact anonymous.
The "normal" definition of an anonymous function (at least according to wikipedia), saying that it must not be bound to an identifier, is a bit weak and ought to include the implicit statement "in the current environment".
This is in fact true for my example, as seen by running it in mlton with the -show-basis argument on an file containing only fun Y ... and the val res ..
val Y: (('a -> 'b) -> 'a -> 'b) -> 'a -> 'b
val res: int32
From this it is seen that none of the anonymous functions are bound in the environment.
A shorter "lambdanonymous" alternative, which requires OCaml launched
by "ocaml -rectypes":
(fun f n -> f f n)
(fun f n -> if n = 0 then 1 else n * (f f (n - 1))
7;; Which produces 7! = 5040.
It seems that you have completely misunderstood the idea of the original question:
Is it possible to write recursive anonymous functions in SML?
And the simple answer is yes. The complex answer is (among others?) an example of this done using a fix point combinator, not a "lambdanonymous" (what ever that is supposed to mean) example done in another language using features not even remotely possible in SML.
All you have to do is put rec after val, as in
val rec fact =
fn n => case n of
0 => 1
| x => x * fact (n - 1)
Wikipedia describes this near the top of the first section.
let fun fact 0 = 1
| fact x = x * fact (x - 1)
in
fact
end
This is a recursive anonymous function. The name 'fact' is only used internally.
Some languages (such as Coq) use 'fix' as the primitive for recursive functions, while some languages (such as SML) use recursive-let as the primitive. These two primitives can encode each other:
fix f => e
:= let rec f = e in f end
let rec f = e ... in ... end
:= let f = fix f => e ... in ... end
In languages I know, a recursive function will always get bound to a name. The convenient and conventional way is provided by keywords like "define", or "let", or "letrec",...
The unconventional, more anonymous looking, way is by lambda binding. Jesper Reenberg's answer shows lambda binding; the "anonymous" function gets bound to the names "f" and "fact" by lambdas (called "fn" in SML).
A shorter "lambdanonymous" alternative, which requires OCaml launched by "ocaml -rectypes":
(fun f n -> f f n)
(fun f n -> if n = 0 then 1 else n * (f f (n - 1))
7;;
Which produces 7! = 5040.
This isn't a homework question, by the way. It got brought up in class but my teacher couldn't think of any. Thanks.
How do you define the identity functions ? If you're only considering the syntax, there are different identity functions, which all have the correct type:
let f x = x
let f2 x = (fun y -> y) x
let f3 x = (fun y -> y) (fun y -> y) x
let f4 x = (fun y -> (fun y -> y) y) x
let f5 x = (fun y z -> z) x x
let f6 x = if false then x else x
There are even weirder functions:
let f7 x = if Random.bool() then x else x
let f8 x = if Sys.argv < 5 then x else x
If you restrict yourself to a pure subset of OCaml (which rules out f7 and f8), all the functions you can build verify an observational equation that ensures, in a sense, that what they compute is the identity : for all value f : 'a -> 'a, we have that f x = x
This equation does not depend on the specific function, it is uniquely determined by the type. There are several theorems (framed in different contexts) that formalize the informal idea that "a polymorphic function can't change a parameter of polymorphic type, only pass it around". See for example the paper of Philip Wadler, Theorems for free!.
The nice thing with those theorems is that they don't only apply to the 'a -> 'a case, which is not so interesting. You can get a theorem out of the ('a -> 'a -> bool) -> 'a list -> 'a list type of a sorting function, which says that its application commutes with the mapping of a monotonous function.
More formally, if you have any function s with such a type, then for all types u, v, functions cmp_u : u -> u -> bool, cmp_v : v -> v -> bool, f : u -> v, and list li : u list, and if cmp_u u u' implies cmp_v (f u) (f u') (f is monotonous), you have :
map f (s cmp_u li) = s cmp_v (map f li)
This is indeed true when s is exactly a sorting function, but I find it impressive to be able to prove that it is true of any function s with the same type.
Once you allow non-termination, either by diverging (looping indefinitely, as with the let rec f x = f x function given above), or by raising exceptions, of course you can have anything : you can build a function of type 'a -> 'b, and types don't mean anything anymore. Using Obj.magic : 'a -> 'b has the same effect.
There are saner ways to lose the equivalence to identity : you could work inside a non-empty environment, with predefined values accessible from the function. Consider for example the following function :
let counter = ref 0
let f x = incr counter; x
You still that the property that for all x, f x = x : if you only consider the return value, your function still behaves as the identity. But once you consider side-effects, you're not equivalent to the (side-effect-free) identity anymore : if I know counter, I can write a separating function that returns true when given this function f, and would return false for pure identity functions.
let separate g =
let before = !counter in
g ();
!counter = before + 1
If counter is hidden (for example by a module signature, or simply let f = let counter = ... in fun x -> ...), and no other function can observe it, then we again can't distinguish f and the pure identity functions. So the story is much more subtle in presence of local state.
let rec f x = f (f x)
This function never terminates, but it does have type 'a -> 'a.
If we only allow total functions, the question becomes more interesting. Without using evil tricks, it's not possible to write a total function of type 'a -> 'a, but evil tricks are fun so:
let f (x:'a):'a = Obj.magic 42
Obj.magic is an evil abomination of type 'a -> 'b which allows all kinds of shenanigans to circumvent the type system.
On second thought that one isn't total either because it will crash when used with boxed types.
So the real answer is: the identity function is the only total function of type 'a -> 'a.
Throwing an exception can also give you an 'a -> 'a type:
# let f (x:'a) : 'a = raise (Failure "aaa");;
val f : 'a -> 'a = <fun>
If you restrict yourself to a "reasonable" strongly normalizing typed λ-calculus, there is a single function of type ∀α α→α, which is the identity function. You can prove it by examining the possible normal forms of a term of this type.
Philip Wadler's 1989 article "Theorems for Free" explains how functions having polymorphic types necessarily satisfy certain theorems (e.g. a map-like function commutes with composition).
There are however some nonintuitive issues when one deals with much polymorphism. For instance, there is a standard trick for encoding inductive types and recursion with impredicative polymorphism, by representing an inductive object (e.g. a list) using its recursor function. In some cases, there are terms belonging to the type of the recursor function that are not recursor functions; there is an example in §4.3.1 of Christine Paulin's PhD thesis.