For the map method, why isn't it like the 2nd line instead of the 1st? It takes 2 parameters, a function, and something else, so why does it have 2 of the arrows?
Filterable f => (a → Boolean) → f a → f a
Filterable f => (a → Boolean) , f a → f a
This is currying. Briefly, you can replace a function of multiple arguments:
const f = (x, y) => x + y;
f(1, 2) // 3
With a series of functions of one argument:
const f = x => y => x + y;
f(1)(2) // 3
One advantage being that it’s easier to partially apply a curried function:
const add1 = f(1);
add1(2) // 3
This is used extensively in Haskell, which seems to be a source of inspiration for ramda.js, and in turn the lambda calculus on which Haskell is based.
Note that the function arrow type is right-associative, so these are equivalent:
Filterable f => (a → Boolean) → f a → f a
Filterable f => (a → Boolean) → (f a → f a)
Seen this way, a function of this type converts a predicate on as into a transformation on fs of as.
There’s also a relation to simple algebra: in type theory, a tuple (a, b) corresponds to a product (a × b), and a function arrow a → b corresponds to exponentiation (ba). You can convert between a → b → c and (a, b) → c for the same reason you can convert between (cb)a and cb × a.
Related
Idris 2 doesn't have Control.ST, only Control.Monad.ST which is a completely different beast (it is basically the same as Haskell's Control.Monad.ST, i.e. mutable references behind a safe, pure interface). It seems that Control.App is roughly-vaguely what is supposed to replace it. However, the Control.App documentation isn't written with Control.ST in mind, and I can't figure out what the migration path is supposed to be.
For example, in this Js frontend library, we have the following Idris 1 API:
public export
interface Dom (m : Type -> Type) where
DomRef : (a:Type) -> (f : a -> Type) -> (g : a -> Type) -> a -> Type
initBody : List (DomOption a f g) -> ((x:a) -> f x -> Html (g x)) -> (z:a) -> f z -> ST m Var [add (DomRef a f g z)]
clearDom : (dom : Var) -> ST m () [remove dom (DomRef a f g z)]
domPut : (dom : Var) -> {x:a} -> f x -> ST m () [dom ::: (DomRef a f g x)]
It is not at all clear what the App version of this interface would be. Is it supposed to simply use App {l} e () everywhere, and the relationship between e and DomRef {e} would be tracked somehow differently? Is initBody / clearDom supposed to use the with pattern, i.e. something like withDom : (App {l} e ()) -> App {l} e ()? If domPut : {x:a} -> f x -> App {l} e (), how is f's type connected to DomRef {e}?
I was reading Idris tutorial. And I can't understand the following code.
disjoint : (n : Nat) -> Z = S n -> Void
disjoint n p = replace {P = disjointTy} p ()
where
disjointTy : Nat -> Type
disjointTy Z = ()
disjointTy (S k) = Void
So far, what I figure out is ...
Void is the empty type which is used to prove something is impossible.
replace : (x = y) -> P x -> P y
replace uses an equality proof to transform a predicate.
My questions are:
which one is an equality proof? (Z = S n)?
which one is a predicate? the disjointTy function?
What's the purpose of disjointTy? Does disjointTy Z = () means Z is in one Type "land" () and (S k) is in another land Void?
In what way can an Void output represent contradiction?
Ps. What I know about proving is "all things are no matched then it is false." or "find one thing that is contradictory"...
which one is an equality proof? (Z = S n)?
The p parameter is the equality proof here. p has type Z = S n.
which one is a predicate? the disjointTy function?
Yes, you are right.
What's the purpose of disjointTy?
Let me repeat the definition of disjointTy here:
disjointTy : Nat -> Type
disjointTy Z = ()
disjointTy (S k) = Void
The purpose of disjointTy is to be that predicate replace function needs. This consideration determines the type of disjointTy, viz. [domain] -> Type. Since we have equality between naturals numbers, [domain] is Nat.
To understand how the body has been constructed we need to take a look at replace one more time:
replace : (x = y) -> P x -> P y
Recall that we have p of Z = S n, so x from the above type is Z and y is S n. To call replace we need to construct a term of type P x, i.e. P Z in our case. This means the type P Z returns must be easily constructible, e.g. the unit type is the ideal candidate for this. We have justified disjointTy Z = () clause of the definition of disjointTy. Of course it's not the only option, we could have used any other inhabited (non-empty) type, like Bool or Nat, etc.
The return value in the second clause of disjointTy is obvious now -- we want replace to return a value of Void type, so P (S n) has to be Void.
Next, we use disjointTy like so:
replace {P = disjointTy} p ()
^ ^ ^
| | |
| | the value of `()` type
| |
| proof term of Z = S n
|
we are saying "this is the predicate"
As a bonus, here is an alternative proof:
disjoint : (n : Nat) -> Z = S n -> Void
disjoint n p = replace {P = disjointTy} p False
where
disjointTy : Nat -> Type
disjointTy Z = Bool
disjointTy (S k) = Void
I have used False, but could have used True -- it doesn't matter. What matters is our ability to construct a term of type disjointTy Z.
In what way can an Void output represent contradiction?
Void is defined like so:
data Void : Type where
It has no constructors! There is no way to create a term of this type whatsoever (under some conditions: like Idris' implementation is correct and the underlying logic of Idris is sane, etc.). So if some function claims it can return a term of type Void there must be something fishy going on. Our function says: if you give me a proof of Z = S n, I will return a term of the empty type. This means Z = S n cannot be constructed in the first place because it leads to a contradiction.
Yes, p : x = y is an equality proof. So p is a equality proof and Z = S k is a equality type.
Also yes, usually any P : a -> Type is called predicate, like IsSucc : Nat -> Type. In boolean logic, a predicate would map Nat to true or false. Here, a predicate holds, if we can construct a proof for it. And it is true, if we can construct it (prf : ItIsSucc 4). And it is false, if we cannot construct it (there is no member of ItIsSucc Z).
At the end, we want Void. Read the replace call as Z = S k -> disjointTy Z -> disjointTy (S k), that is Z = S K -> () -> Void. So replace needs two arguments: the proof p : Z = S k and the unit () : (), and voilà, we have a void. By the way, instead of () you could use any type that you can construct, e.g. disjointTy Z = Nat and then use Z instead of ().
In dependent type theory we construct proofs like prf : IsSucc 4. We would say, we have a proof prf that IsSucc 4 is true. prf is also called a witness for IsSucc 4. But with this alone we could only proove things to be true. This is the definiton for Void:
data Void : Type where
There is no constructor. So we cannot construct a witness that Void holds. If you somehow ended up with a prf : Void, something is wrong and you have a contradiction.
I am trying to create module/interface (i dont exactly know how its called, i am new to the language) for basic operations on BST in OCaml. My goal is to have an implementation that lets me doing something like this:
T.create();;
T.push(2);;
T.push(3);;
T.push(5);;
in order to get a bst tree consisting of 2,3,5.
But at the moment to achieve this i have to write something like this:
let teeBst = T.push(2)(T.push(3)(T.push(5)(T.create())));;
So when I am checking/using my code I have to do it like this:
let tee2 = T.push(2)(T.push(3)(T.push(5)(T.create())));;
T.postorder(tee2);;
The output is fine:
# val tee2 : T.bt = <abstr>
# - : int list = [2; 3; 5]
But, as I said before, I would like to achieve this doing as below:
T.push(2);;
T.push(3);;
T.push(5);;
T.postorder();;
(I realise this requires some changes to my postorder function but the one I am currently using is a temporary one so I can check the tree I have atm )
Below is my implementation. If you see the solution, please let me know ;)
module type Tree =
sig
type bt
val create: unit -> bt
val push: int -> bt -> bt
val find: int -> bt -> bool
val preorder: bt -> int list
val postorder: bt -> int list
val inorder: bt -> int list
end;;
module T : Tree =
struct
type bt = E | B of bt * int * bt
let create () = E
let rec push x = function
| E -> B(E, x, E)
| B (l, y, r) when x<y -> B(push x l, y, r)
| B (l, y, r) when x>y -> B(l, y, push x r)
| xs -> xs;;
let rec find x = function
| E -> false
| B(l, y,_) when x< y -> find x l
| B(_,y,r) when x>y -> find x r
| _ -> true;;
let rec preorder = function
| B(l,v,r) -> v::(preorder r) # (preorder l)
| E -> [];;
let rec inorder = function
| B(l,v,r) ->(inorder r) # v::(inorder l)
| E -> []
let rec postorder = function
| B(l,v,r) -> (postorder r) # (postorder l) # [v]
| E -> []
end;;
It seems like you want modules to be classes, but I'd advise you to consider more idiomatic solutions. Have you considered using the pipe operator?
T.create()
|> T.push(2)
|> T.push(3)
|> T.push(5)
|> T.postorder;;
Or with local open (which makes more sense if you have a module with a longer name than just T of course) you can even do
T.(
create()
|> push(2)
|> push(3)
|> push(5)
|> postorder
);
What you're asking for would require introducing global mutable state, which isn't just "some changes" but an entirely different paradigm. And one that is generally frowned upon because it makes your code unpredictable and hard to debug since it relies on state that might change at any moment from anywhere.
Another possibility is to actually use classes, since OCaml has those too. Then you'd still have mutable state, but it would at least be contained.
What is the syntax to constrain a function argument in an interface which takes a function? I tried:
interface Num a => Color (f : a -> Type) where
defs...
But it says the Name a is not bound in interface...
Your interface actually has two parameters: a and f. But f should be enough to pick an implementation:
interface Num a => Color (a : Type) (f : a -> Type) | f where
f here is called a determining parameter.
Here's a nonsensical full example:
import Data.Fin
interface Num a => Color (a : Type) (f : a -> Type) | f where
foo : (x : a) -> f (1 + x)
Color Nat Fin where
foo _ = FZ
x : Fin 6
x = foo {f = Fin} 5
I'm trying to translate the Haskell core library's Arrows into F# (I think it's a good exercise to understanding Arrows and F# better, and I might be able to use them in a project I'm working on.) However, a direct translation isn't possible due to the difference in paradigms. Haskell uses type-classes to express this stuff, but I'm not sure what F# constructs best map the functionality of type-classes with the idioms of F#. I have a few thoughts, but figured it best to bring it up here and see what was considered to be the closest in functionality.
For the tl;dr crowd: How do I translate type-classes (a Haskell idiom) into F# idiomatic code?
For those accepting of my long explanation:
This code from the Haskell standard lib is an example of what I'm trying to translate:
class Category cat where
id :: cat a a
comp :: cat a b -> cat b c -> cat a c
class Category a => Arrow a where
arr :: (b -> c) -> a b c
first :: a b c -> a (b,d) (c,d)
instance Category (->) where
id f = f
instance Arrow (->) where
arr f = f
first f = f *** id
Attempt 1: Modules, Simple Types, Let Bindings
My first shot at this was to simply map things over directly using Modules for organization, like:
type Arrow<'a,'b> = Arrow of ('a -> 'b)
let arr f = Arrow f
let first f = //some code that does the first op
That works, but it loses out on polymorphism, since I don't implement Categories and can't easily implement more specialized Arrows.
Attempt 1a: Refining using Signatures and types
One way to correct some issues with Attempt 1 is to use a .fsi file to define the methods (so the types enforce easier) and to use some simple type tweaks to specialize.
type ListArrow<'a,'b> = Arrow<['a],['b]>
//or
type ListArrow<'a,'b> = LA of Arrow<['a],['b]>
But the fsi file can't be reused (to enforce the types of the let bound functions) for other implementations, and the type renaming/encapsulating stuff is tricky.
Attempt 2: Object models and interfaces
Rationalizing that F# is built to be OO also, maybe a type hierarchy is the right way to do this.
type IArrow<'a,'b> =
abstract member comp : IArrow<'b,'c> -> IArrow<'a,'c>
type Arrow<'a,'b>(func:'a->'b) =
interface IArrow<'a,'b> with
member this.comp = //fun code involving "Arrow (fun x-> workOn x) :> IArrow"
Aside from how much of a pain it can be to get what should be static methods (like comp and other operators) to act like instance methods, there's also the need to explicitly upcast the results. I'm also not sure that this methodology is still capturing the full expressiveness of type-class polymorphism. It also makes it hard to use things that MUST be static methods.
Attempt 2a: Refining using type extensions
So one more potential refinement is to declare the interfaces as bare as possible, then use extension methods to add functionality to all implementing types.
type IArrow<'a,'b> with
static member (&&&) f = //code to do the fanout operation
Ah, but this locks me into using one method for all types of IArrow. If I wanted a slightly different (&&&) for ListArrows, what can I do? I haven't tried this method yet, but I would guess I can shadow the (&&&), or at least provide a more specialized version, but I feel like I can't enforce the use of the correct variant.
Help me
So what am I supposed to do here? I feel like OO should be powerful enough to replace type-classes, but I can't seem to figure out how to make that happen in F#. Were any of my attempts close? Are any of them "as good as it gets" and that'll have to be good enough?
My brief answer is:
OO is not powerful enough to replace type classes.
The most straightforward translation is to pass a dictionary of operations, as in one typical typeclass implementation. That is if typeclass Foo defines three methods, then define a class/record type named Foo, and then change functions of
Foo a => yadda -> yadda -> yadda
to functions like
Foo -> yadda -> yadda -> yadda
and at each call site you know the concrete 'instance' to pass based on the type at the call-site.
Here's a short example of what I mean:
// typeclass
type Showable<'a> = { show : 'a -> unit; showPretty : 'a -> unit } //'
// instances
let IntShowable =
{ show = printfn "%d"; showPretty = (fun i -> printfn "pretty %d" i) }
let StringShowable =
{ show = printfn "%s"; showPretty = (fun s -> printfn "<<%s>>" s) }
// function using typeclass constraint
// Showable a => [a] -> ()
let ShowAllPretty (s:Showable<'a>) l = //'
l |> List.iter s.showPretty
// callsites
ShowAllPretty IntShowable [1;2;3]
ShowAllPretty StringShowable ["foo";"bar"]
See also
https://web.archive.org/web/20081017141728/http://blog.matthewdoig.com/?p=112
Here's the approach I use to simulate Typeclasses (from http://code.google.com/p/fsharp-typeclasses/ ).
In your case, for Arrows could be something like this:
let inline i2 (a:^a,b:^b ) =
((^a or ^b ) : (static member instance: ^a* ^b -> _) (a,b ))
let inline i3 (a:^a,b:^b,c:^c) =
((^a or ^b or ^c) : (static member instance: ^a* ^b* ^c -> _) (a,b,c))
type T = T with
static member inline instance (a:'a ) =
fun x -> i2(a , Unchecked.defaultof<'r>) x :'r
static member inline instance (a:'a, b:'b) =
fun x -> i3(a, b, Unchecked.defaultof<'r>) x :'r
type Return = Return with
static member instance (_Monad:Return, _:option<'a>) = fun x -> Some x
static member instance (_Monad:Return, _:list<'a> ) = fun x -> [x]
static member instance (_Monad:Return, _: 'r -> 'a ) = fun x _ -> x
let inline return' x = T.instance Return x
type Bind = Bind with
static member instance (_Monad:Bind, x:option<_>, _:option<'b>) = fun f ->
Option.bind f x
static member instance (_Monad:Bind, x:list<_> , _:list<'b> ) = fun f ->
List.collect f x
static member instance (_Monad:Bind, f:'r->'a, _:'r->'b) = fun k r -> k (f r) r
let inline (>>=) x (f:_->'R) : 'R = T.instance (Bind, x) f
let inline (>=>) f g x = f x >>= g
type Kleisli<'a, 'm> = Kleisli of ('a -> 'm)
let runKleisli (Kleisli f) = f
type Id = Id with
static member instance (_Category:Id, _: 'r -> 'r ) = fun () -> id
static member inline instance (_Category:Id, _:Kleisli<'a,'b>) = fun () ->
Kleisli return'
let inline id'() = T.instance Id ()
type Comp = Comp with
static member instance (_Category:Comp, f, _) = (<<) f
static member inline instance (_Category:Comp, Kleisli f, _) =
fun (Kleisli g) -> Kleisli (g >=> f)
let inline (<<<) f g = T.instance (Comp, f) g
let inline (>>>) g f = T.instance (Comp, f) g
type Arr = Arr with
static member instance (_Arrow:Arr, _: _ -> _) = fun (f:_->_) -> f
static member inline instance (_Arrow:Arr, _:Kleisli<_,_>) =
fun f -> Kleisli (return' <<< f)
let inline arr f = T.instance Arr f
type First = First with
static member instance (_Arrow:First, f, _: 'a -> 'b) =
fun () (x,y) -> (f x, y)
static member inline instance (_Arrow:First, Kleisli f, _:Kleisli<_,_>) =
fun () -> Kleisli (fun (b,d) -> f b >>= fun c -> return' (c,d))
let inline first f = T.instance (First, f) ()
let inline second f = let swap (x,y) = (y,x) in arr swap >>> first f >>> arr swap
let inline ( *** ) f g = first f >>> second g
let inline ( &&& ) f g = arr (fun b -> (b,b)) >>> f *** g
Usage:
> let f = Kleisli (fun y -> [y;y*2;y*3]) <<< Kleisli ( fun x -> [ x + 3 ; x * 2 ] ) ;;
val f : Kleisli<int,int list> = Kleisli <fun:f#4-14>
> runKleisli f <| 5 ;;
val it : int list = [8; 16; 24; 10; 20; 30]
> (arr (fun y -> [y;y*2;y*3])) 3 ;;
val it : int list = [3; 6; 9]
> let (x:option<_>) = runKleisli (arr (fun y -> [y;y*2;y*3])) 2 ;;
val x : int list option = Some [2; 4; 6]
> ( (*) 100) *** ((+) 9) <| (5,10) ;;
val it : int * int = (500, 19)
> ( (*) 100) &&& ((+) 9) <| 5 ;;
val it : int * int = (500, 14)
> let x:List<_> = (runKleisli (id'())) 5 ;;
val x : List<int> = [5]
Note: use id'() instead of id
Update: you need F# 3.0 to compile this code, otherwise here's the F# 2.0 version.
And here's a detailed explanation of this technique which is type-safe, extensible and as you can see works even with some Higher Kind Typeclasses.