In Idris I am trying to write a type-safe(r) version of a splitEvery function for Vect, that on Lists looks like this:
splitEvery : Nat -> List a -> List (List a)
splitEvery _ [] = []
splitEvery n xs = (take n xs) ++ splitEvery n (drop n xs)
So far I've got:
splitEvery : {k : Nat} -> (n : Nat) -> Vect (k * n) a -> Vect k (Vect n a)
splitEvery {k = Z} _ [] = []
splitEvery {k = (S j)} n xs = (take n xs) :: splitEvery {k = j} n (drop n xs)
which will type-check and load in the REPL alright, but when I try to invoke it, it will fail:
*Main> splitEvery 2 (fromList [1..10])
(input):1:1-31:When checking an application of function Main.splitEvery:
Type mismatch between
Vect (length (enumFromTo 1 10))
Integer (Type of fromList (enumFromTo 1 10))
and
Vect (k * 2) Integer (Expected type)
Specifically:
Type mismatch between
length (enumFromTo 1 10)
and
mult k 2
So obviously Idris isn't seeing that a valid choice of k here would be 5. A way to make it work would be to explicitly give it the implicit parameter k, but that's ugly:
*Main> splitEvery {k = 5} 2 (fromList [1..10])
[[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]] : Vect 5 (Vect 2 Integer)
So my question is: is there a way to make this work that isn't ugly, or is more idiomatic than what I've produced so far?
Idris is only failing to infer the value of k in the REPL for some reason. If you amend your code with
result: Vect 5 (Vect 2 Nat)
result = splitEvery 2 (fromList [1..10])
everything typechecks - as the type is already known and k can be infered. You can achieve the same on the REPL by providing a type hint using the:
the (Vect 5 (Vect 2 Nat)) (splitEvery 2 (fromList [1..10])) works as well.
Related
An example value: [[1, 2], [1]]
Here, I would know that there are 2 lists, and that the first list has length 2, while the second one has length 1. Ideally, this function would compute those types:
func : (n ** Vect n Nat) -> Type
But I don't know how to write it. I'm pretty sure it's something to do with dependent pairs, but I'm not sure how to write it.
To clarify, I know it'd be possible to simply use (n ** Vect n (p ** Vect p Double)) as the type of the example value. However, n only constrains the number of lists, not the number of their elements, because inside the list, p could be anything. I would most likely need something where the first element of the dependent pair is a vector of lengths, not just the number of lists. So something like (Vect n Nat ** Vect n (Vect m Double))--where each m is the corresponding element of the first vector.
You could define a new vector type which contains possibly differently indexed elements of an indexed type at each position:
import Prelude
import Data.Vect
-- heterogeneously indexed vector
data IVect : Vect n ix -> (ix -> Type) -> Type where
Nil : IVect Nil b
(::) : b i -> IVect is b -> IVect (i :: is) b
-- of which a special case is a vector of vectors
VVect : Vect n Nat -> Type -> Type
VVect is a = IVect is (flip Vect a)
test1 : VVect [2, 2, 2] Nat
test1 = [[1, 2], [3, 4], [5, 6]]
test2 : VVect [0, 1, 2] Bool
test2 = [[], [True], [False, True]]
Alternatively, you can define VVect using dependent pairs and map, but this is more cumbersome to use:
VVect' : Vect n Nat -> Type -> Type
VVect' {n = n} is a = (xs : Vect n (m ** Vect m a) ** (map fst xs = is))
test3 : VVect' [0, 1, 2] Bool
test3 = ([(_ ** []), (_ ** [True]), (_ ** [False, False])] ** Refl)
You have some choice though whether to use lists or vectors. With lists as the inner container, values look more compact:
VVect'' : Vect n Nat -> Type -> Type
VVect'' {n = n} is a = (xs : Vect n (List a) ** (map length xs = is))
test4 : VVect'' [0, 1, 2] Bool
test4 = ([[], [True], [False, True]] ** Refl)
I defined n-dimensional vectors in Idris as follows:
import Data.Vect
NDVect : (Num t) => (rank : Nat) -> (shape : Vect rank Nat) -> (t : Type) -> Type
NDVect Z [] t = t
NDVect (S n) (x::xs) t = Vect x (NDVect n xs t)
Then I defined the following function which maps a function f to every entry in the tensor.
iterateT : (f : t -> t') -> (v : NDVect r s t) -> NDVect r s t'
iterateT {r = Z} {s = []} f v = f v
iterateT {r = S n} {s = x::xs} f v = map (iterateT f) v
But when I try to call iteratorT in the following function:
scale : Num t => (c : t) -> (v : NDVect rank shape t) -> NDVect rank shape t
scale c v = iterateT (*c) v
I get the following error message saying there is a type mismatched, which seems pretty fine to me
When checking right hand side of scale with expected type
NDVect rank shape t
When checking argument v to function Main.iterateT:
Type mismatch between
NDVect rank shape t (Type of v)
and
NDVect r s t (Expected type)
Specifically:
Type mismatch between
NDVect rank shape t
and
NDVect r s t
Specifically:
Type mismatch between
NDVect rank shape t
and
NDVect r s t
I have also been wondering how to express n-dimensional vectors (i.e. tensors) in Idris. I had a play with the type definition in the question, but encountered various issues, so I expressed the NDVect function as a data type:
data NDVect : (rank : Nat) -> (shape : Vect rank Nat) -> Type -> Type where
NDVZ : (value : t) -> NDVect Z [] t
NDV : (values : Vect n (NDVect r s t)) -> NDVect (S r) (n::s) t
And implemented map as follows:
nmap : (t -> u) -> (NDVect r s t) -> NDVect r s u
nmap f (NDVZ value) = NDVZ (f value)
nmap f (NDV values) = NDV (map (nmap f) values)
The following now works:
*Main> NDVZ 5
NDVZ 5 : NDVect 0 [] Integer
*Main> nmap (+4) (NDVZ 5)
NDVZ 9 : NDVect 0 [] Integer
*Main> NDV [NDVZ 1, NDVZ 2, NDVZ 3]
NDV [NDVZ 1, NDVZ 2, NDVZ 3] : NDVect 1 [3] Integer
*Main> nmap (+4) (NDV [NDVZ 1, NDVZ 2, NDVZ 3])
NDV [NDVZ 5, NDVZ 6, NDVZ 7] : NDVect 1 [3] Integer
Unfortunately, having all the type constructors makes things a bit ugly. I'd love to know if there's a cleaner way to solve this.
Edit:
Here's a slightly shorter type signature that doesn't explicitly encode the tensor rank in the type:
data NDVect : (shape : List Nat) -> Type -> Type where
NDVZ : (value : t) -> NDVect [] t
NDV : (values : Vect n (NDVect s t)) -> NDVect (n::s) t
nmap : (t -> u) -> (NDVect s t) -> NDVect s u
nmap f (NDVZ value) = NDVZ (f value)
nmap f (NDV values) = NDV (map (nmap f) values)
Given:
*section3> :module Data.Vect
*section3> :let e = the (Vect 0 Int) []
*section3> :let xs = the (Vect _ _) [1,2]
*section3> decEq xs e
(input):1:7:When checking argument x2 to function Decidable.Equality.decEq:
Type mismatch between
Vect 0 Int (Type of e)
and
Vect 2 Integer (Expected type)
Specifically:
Type mismatch between
0
and
2
Why must the Nat arguments equal each other for DecEq?
Note - posted in https://groups.google.com/forum/#!topic/idris-lang/qgtImCLka3I originally
decEq is for homogenous propositional equality:
||| Decision procedures for propositional equality
interface DecEq t where
||| Decide whether two elements of `t` are propositionally equal
total decEq : (x1 : t) -> (x2 : t) -> Dec (x1 = x2)
As you can see, x1 and x2 are both of type t. In your case, you have x1 : Vect 2 Integer and x2 : Vect 0 Int. These are two different types.
You can write your own heterogenous equality decider for Vectors of the same element type by first checking their lengths, then delegating to the homogenous version:
import Data.Vect
vectLength : {xs : Vect n a} -> {ys : Vect m a} -> xs = ys -> n = m
vectLength {n = n} {m = n} Refl = Refl
decEqVect : (DecEq a) => (xs : Vect n a) -> (ys : Vect m a) -> Dec (xs = ys)
decEqVect {n = n} {m = m} xs ys with (decEq n m)
decEqVect xs ys | Yes Refl = decEq xs ys
decEqVect xs ys | No notEq = No (notEq . vectLength)
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.
I'm trying to write in Idris a function that create a Vect by passing the size of the Vect and a function taking the index in parameter.
So far, I've this :
import Data.Fin
import Data.Vect
generate: (n:Nat) -> (Nat -> a) ->Vect n a
generate n f = generate' 0 n f where
generate': (idx:Nat) -> (n:Nat) -> (Nat -> a) -> Vect n a
generate' idx Z f = []
generate' idx (S k) f = (f idx) :: generate' (idx + 1) k f
But I would like to ensure that the function passed in parameter is only taking index lesser than the size of the Vect.
I tried that :
generate: (n:Nat) -> (Fin n -> a) ->Vect n a
generate n f = generate' 0 n f where
generate': (idx:Fin n) -> (n:Nat) -> (Fin n -> a) -> Vect n a
generate' idx Z f = []
generate' idx (S k) f = (f idx) :: generate' (idx + 1) k f
But it doesn't compile with the error
Can't convert
Fin n
with
Fin (S k)
My question is : is what I want to do possible and how ?
The key idea is that the first element of the vector is f 0, and for the tail, if you have k : Fin n, then FS k : Fin (S n) is a "shift" of the finite number that increments its value and its type at the same time.
Using this observation, we can rewrite generate as
generate : {n : Nat} -> (f : Fin n -> a) -> Vect n a
generate {n = Z} f = []
generate {n = S _} f = f 0 :: generate (f . FS)
Another possibility is to do what #dfeuer suggested and generate a vector of Fins, then map f over it:
fins : (n : Nat) -> Vect n (Fin n)
fins Z = []
fins (S n) = FZ :: map FS (fins n)
generate' : {n : Nat} -> (f : Fin n -> a) -> Vect n a
generate' f = map f $ fins _
Proving generate f = generate' f is left as en exercise to the reader.
Cactus's answer appears to be about the best way to get what you asked for, but if you want something that can be used at runtime, it will be quite inefficient. The essential reason for this is that to weaken a Fin n to a Fin n+m requires that you completely deconstruct it to change the type of its FZ, and then build it back up again. So there can be no sharing at all between the Fin values produced for each vector element. An alternative is to combine a Nat with a proof that it is below a given bound, which leads to the possibility of erasure:
data NFin : Nat -> Type where
MkNFin : (m : Nat) -> .(LT m n) -> NFin n
lteSuccLeft : LTE (S n) m -> LTE n m
lteSuccLeft {n = Z} prf = LTEZero
lteSuccLeft {n = (S k)} {m = Z} prf = absurd (succNotLTEzero prf)
lteSuccLeft {n = (S k)} {m = (S j)} (LTESucc prf) = LTESucc (lteSuccLeft prf)
countDown' : (m : Nat) -> .(m `LTE` n) -> Vect m (NFin n)
countDown' Z mLTEn = []
countDown' (S k) mLTEn = MkNFin k mLTEn :: countDown' k (lteSuccLeft mLTEn)
countDown : (n : Nat) -> Vect n (NFin n)
countDown n = countDown' n lteRefl
countUp : (n : Nat) -> Vect n (NFin n)
countUp n = reverse $ countDown n
generate : (n : Nat) -> (NFin n -> a) -> Vect n a
generate n f = map f (countUp n)
As in the Fin approach, the function you pass to generate does not need to work on all naturals; it only needs to handle ones less than n.
I used the NFin type to explicitly indicate that I want the LT m n proof to be erased in all cases. If I didn't want/need that, I could just use (m ** LT m n) instead.