This code is splitting a list in two pieces by a predicate that take a list and return false in the moment of splitting.
let split pred ys =
let rec split' l r =
match r with
| [] -> []
| x::xs -> if pred (x::l) then x::(split' (x::l) xs) else []
let res = split' [] ys
let last = ys |> Seq.skip (Seq.length res) |> Seq.toList
(res, last)
Do someone knows more optimal and simpler ways to do that in F#?
Well you can make it tail recursive but then you have to reverse the list. You wouldn't want to fold it since it can exit out of the recursive loop at any time. I did a little testing and reversing the list is more than made up for by tail recursion.
// val pred : ('a list -> bool)
let split pred xs =
let rec split' l xs ys =
match xs with
| [] -> [], ys
| x::xs -> if pred (x::l) then (split' (x::l) xs (x::ys)) else x::xs, ys
let last, res = split' [] xs []
(res |> List.rev, last)
A version similar to Brian's that is tail recursive and takes a single value predicate.
// val pred : ('a -> bool)
let split pred xs =
let rec split' xs ys =
match xs with
| [] -> [], ys
| x::xs -> if pred x then (split' xs (x::ys)) else (x::xs), ys
let last, res = split' xs []
(res |> List.rev, last)
This is different from the library function partition in that it stops taking elements as soon as the predicate returns false kind of like Seq.takeWhile.
// library function
let x, y = List.partition (fun x -> x < 5) li
printfn "%A" x // [1; 3; 2; 4]
printfn "%A" y // [5; 7; 6; 8]
let x, y = split (fun x -> x < 5) li
printfn "%A" x // [1; 3]
printfn "%A" y // [5; 7; 2; 4; 6; 8]
Not tail-recursive, but:
let rec Break pred list =
match list with
| [] -> [],[]
| x::xs when pred x ->
let a,b = Break pred xs
x::a, b
| x::xs -> [x], xs
let li = [1; 3; 5; 7; 2; 4; 6; 8]
let a, b = Break (fun x -> x < 5) li
printfn "%A" a // [1; 3; 5]
printfn "%A" b // [7; 2; 4; 6; 8]
// Also note this library function
let x, y = List.partition (fun x -> x < 5) li
printfn "%A" x // [1; 3; 2; 4]
printfn "%A" y // [5; 7; 6; 8]
Here is some foldr way:
let split' pred xs = let f (ls,rs,cond) x = if cond (ls#[x]) then (ls#[x],rs,cond) else (ls,rs#[x],(fun _->false))
let ls,rs,_ = List.fold f ([],[],pred) xs
ls, rs
Related
I have Coord function that transforms an n-dimensional size to the type of coordinates bounded by given size: Coord [2,3] = (Fin 2, Fin 3).
import Data.Fin
import Data.List
Size : Type
Size = List Nat
Coord : Size -> Type
Coord [] = ()
Coord s#(_ :: _) = foldr1 (,) $ map Fin s
I'd like to use show and other functions like (==) with Coord s:
foo : Coord s -> String
foo x = show x
Error: While processing right hand side of foo. Can't find an implementation for Show (Coord s).
22 | foo : Coord s -> String
23 | foo x = show x
^^^^^^
Earlier I tried to implement Show (Coord s), but looks like it's impossible. Here is linked question about it.
You can make your own list like data type:
data Coords : List Nat -> Type where
Nil : Coords []
(::) : Fin x -> Coords xs -> Coords (x :: xs)
toList : Coords xs -> List Nat
toList [] = []
toList (x::xs) = finToNat x :: toList xs
example : Coords [2, 3]
example = [1, 2]
Show (Coords xs) where
show cs = show $ toList cs
You can also try using Data.Vect.Quantifiers.All or Data.List.Quantifiers.All:
import Data.Vect
import Data.Vect.Quantifiers
example : All Fin [1, 2, 3]
example = [0, 1, 2]
-- not sure why this is isn't included with Idris
export
All (Show . p) xs => Show (All p xs) where
show pxs = "[" ++ show' "" pxs ++ "]"
where
show' : String -> All (Show . p) xs' => All p xs' -> String
show' acc #{[]} [] = acc
show' acc #{[_]} [px] = acc ++ show px
show' acc #{_ :: _} (px :: pxs) = show' (acc ++ show px ++ ", ") pxs
string : String
string = show example
I was trying to prove that replicate1 works correctly by showing that all elements of replicate1 n x are x:
all1 : (p : a -> Bool) -> List a -> Bool
all1 p [] = True
all1 p (x :: xs) = p x && all1 p xs
replicate1 : (n: Nat) -> a -> List a
replicate1 Z x = [x]
replicate1 (S k) x = x :: replicate1 k x
all_replicate_is_x : Eq a => {x: a} -> all1 (== x) (replicate1 n x) = True
all_replicate_is_x {n = Z} = ?hole
all_replicate_is_x {n = (S k)} = ?all_replicate_is_x_rhs_2
The base case hole is
Test.hole [P]
`-- a : Type
constraint : Eq a
x : a
-----------------------------------------
Test.hole : x == x && Delay True = True
How to prove this?
I have a function count in idris, defined as :
count : Eq a => a -> Vect n a -> Nat
count x [] = Z
count x (y::ys) = with (x == y)
| True = S (count x ys)
| False = count x ys
And a proof of the maximum value count can return:
countLTELen : Eq a => (x : a) -> (l : Vect n a) -> LTE (count x l) n
countLTELen x [] = lteRefl
countLteLen x (y::ys) with (x == y)
| True = LTESucc (countLTELen x ys)
| False = lteSuccRight (countLTELen x ys)
which is all well and good. I now want to write a function which removes all of an element from a list, removeAll :
removeAll : Eq a => (x : a) -> (l : Vect n a) -> Vect (n - (count x l)) a
removeAll x [] = []
removeAll x (y::ys) with (x == y)
| True = removeAll x ys
| False = x :: removeAll x ys
But this definition gives an error:
|
56 | removeAll : Eq a => (x : a) -> (l : Vect n a) -> Vect (n - (count x l)) a
| ^
When checking type of Proof.removeAll:
When checking argument smaller to function Prelude.Nat.-:
Can't find a value of type
LTE (count a n constraint x l) n
How can I use my proof to inform Idris that this type signature is correct?
Right now, Idris can't find the proof {auto smaller : LTE n m} for (-).
So either you need to be explicit:
removeAll : Eq a => (x : a) -> (l : Vect n a) ->
Vect ((-) {smaller=countLTELen x l} n (count x l) ) a
Or, because smaller is an auto-argument, you can hint the compiler to your proof function. Then this function will be tried when auto-finding a value for LTE (count x l) n.
%hint
countLTELen : Eq a => (x : a) -> (l : Vect n a) -> LTE (count x l) n
When I run the following code I get a syntax error, although as far as I can tell the syntax is correct. This attempts to implement a queue structure, where the function from_list converts a list to a queue with the corresponding values. I wrote str_of_int_q to print the contents of a queue. x and y are supposed to be two nodes, with x at the head and y at the tail.
;; open Assert
type 'a qnode = {v: 'a;
mutable next: 'a qnode option}
type 'a queue = {mutable head: 'a qnode option;
mutable tail: 'a qnode option}
let from_list (l: 'a list) : 'a queue =
let rec loop (l2: 'a list) (qu: 'a queue) =
begin match l2 with
| [] -> qu
| [x] -> let y = {v = x; next = None} in
qu.head <- Some y; qu.tail <- Some y;
qu
| h1::h2::t -> let y = qu.head in
let z = {v = h1; next = y} in
qu.head <- Some z;
qu
end
in loop l {head = None; tail = None}
let str_of_int_q (q: int queue) : string =
let rec loop (r: int qnode option) (s: string) : string =
begin match r with
| None -> s
| Some n -> loop n.next (s ^ (string_of_int n.v))
end
in loop q.head ""
let x = {v = 1; next = None}
let y = {v = 2; next = None}
x.next <- Some y;
let z = {head = Some x; tail = Some y}
;; print_endline (str_of_int_q z)
My error:
line 32, characters 7-9:
Error: Syntax error
Line 32 is the line x.next <- Some y; and characters 7-9 indicate the <-. But I'm storing into a mutable field an object of the appropriate type, so I don't see what's going wrong.
Top-level statements are separated by ;; in OCaml. However, ;; is optional before several keywords, such as let, open, type, etc. This is why you don't need ;; most of the time.
In your case, ;; is needed to disambiguate between let y = {v = 2; next = None} and x.next <- Some y. The latter is an expression and doesn't start with a special keyword, so OCaml doesn't know to insert an implicit ;; here.
See also http://ocaml.org/learn/tutorials/structure_of_ocaml_programs.html#The-disappearance-of.
As explained there, you can either do
let y = {v = 2; next = None}
;; x.next <- Some y
or
let y = {v = 2; next = None}
let () = x.next <- Some y
This latter solution works because by introducing a dummy binding we're starting our statement with let, which disambiguates again.
Note: I've also removed the trailing ; from your code. ; is actually an infix operator that combines two expressions (by throwing the result of the first one away and returning the result of the second one). This is not what you want here.
Following my other question, I tried to implement the actual exercise in Type-Driven Development with Idris for same_cons to prove that, given two equal lists, prepending the same element to each list results in two equal lists.
Example:
prove that 1 :: [1,2,3] == 1 :: [1,2,3]
So I came up with the following code that compiles:
sameS : {xs : List a} -> {ys : List a} -> (x: a) -> xs = ys -> x :: xs = x :: ys
sameS {xs} {ys} x prf = cong prf
same_cons : {xs : List a} -> {ys : List a} -> xs = ys -> x :: xs = x :: ys
same_cons prf = sameS _ prf
I can call it via:
> same_cons {x=5} {xs = [1,2,3]} {ys = [1,2,3]} Refl
Refl : [5, 1, 2, 3] = [5, 1, 2, 3]
Regarding the cong function, my understanding is that it takes a proof, i.e. a = b, but I don't understand its second argument: f a.
> :t cong
cong : (a = b) -> f a = f b
Please explain.
If you have two values u : c and v : c, and a function f : c -> d, then if you know that u = v, it has to follow that f u = f v, following simply from referential transparency.
cong is the proof of the above statement.
In this particular use case, you are setting (via unification) c and d to List a, u to xs, v to ys, and f to (:) x, since you want to prove that xs = ys -> (:) x xs = (:) x ys.